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\begin{document}
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\title{Quadratic Forms}


\author{Marco Schlichting\\
Notes by Florian Bouyer}

\maketitle
\selectlanguage{british}%
\begin{center}
Copyright (C) Bouyer 2012. 
\par\end{center}

\begin{center}
Permission is granted to copy, distribute and/or modify this document 
\par\end{center}

\begin{center}
under the terms of the GNU Free Documentation License, Version 1.3 
\par\end{center}

\begin{center}
or any later version published by the Free Software Foundation; 
\par\end{center}

\begin{center}
with no Invariant Sections, no Front-Cover Texts, and no Back-Cover
Texts. 
\par\end{center}

\begin{center}
A copy of the license can be found at \href{http://www.gnu.org/licenses/fdl.html}{http://www.gnu.org/licenses/fdl.html}
\par\end{center}

\selectlanguage{english}%
\tableofcontents{}

\newpage{}

In this course every ring is commutative with unit. Every module is
a left module.
\begin{defn}
Let $R$ be a (commutative) ring and $M$ a (left) $R$-module. Then
a \emph{bilinear form on $M$} is a map $\beta:M\times M\rightarrow R$
which is $R$-linear in both variables. i.e. $\beta(ax+by,z)=a\beta(x,z)+b\beta(y,z)$
and $\beta(x,by+cz)=b\beta(x,y)+c\beta(x,z)$ $\forall x,y,z\in M,a,b,c\in R$\end{defn}
\begin{example*}
Standard Euclidean scalar product on $\mathbb{R}^{n}$. $\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}$
is a bilinear form on the $\mathbb{R}$-module $\mathbb{R}^{n}$\end{example*}
\begin{defn}
A bilinear form $\beta:M\times M\rightarrow R$ is called \emph{symmetric}
if $\beta(x,y)=\beta(y,x)$ $\forall x,y\in M$. It is called \emph{skewed
symmetric} if $\beta(x,y)=-\beta(y,x)\,\forall x,y\in M$. It is called
\emph{symplectic} if $\beta(x,x)=0\,\forall x\in M$\end{defn}
\begin{example*}
Standard scalar product on $\mathbb{R}^{n}$ is symmetric.\\
On $\mathbb{R}^{2}$ the bilinear form $(x_{1},y_{1})\times(x_{2},y_{2})\mapsto x_{1}y_{2}-x_{2}y_{1}$
is symplectic and skew-symmetric\end{example*}
\begin{rem*}
Any symplectic bilinear form is also skew-symmetric because $\beta$
symplectic $\Rightarrow0=\beta(x+y,x+y)=\beta(x,x)+\beta(x,y)+\beta(y,x)+\beta(y,y)=\beta(x,y)+\beta(y,x)$
hence it is skew-symmetric.

If $2\in R$ is a non-zero divisor (i.e. $2a=0\Rightarrow a=0\,\forall a\in R$)
then any skew-symmetric form is also symplectic because $\beta$ skew-symmetric
$\Rightarrow\beta(x,x)=-\beta(x,x)\Rightarrow2\beta(x,x)=0$ and $2$
a non-zero divisors we can divide by $2$ hence $\beta(x,x)=0\,\forall x\in M\Rightarrow\beta$
symplectic\end{rem*}
\begin{example*}
For $R=\mathbb{F}_{2}$, the form $\mathbb{F}_{2}\times\mathbb{F}_{2}\rightarrow\mathbb{F}_{2}$
defined by $x,y\mapsto xy$ is skewed-symmetric but not symplectic
because $\beta(1,1)=1\neq0\in\mathbb{F}_{2}$\end{example*}
\begin{defn}
A bilinear form $\beta:M\times M\rightarrow R$ is called \emph{regular
or non-degenerate or non-singular} if 
\begin{enumerate}
\item $\forall f:M\rightarrow R$ a $R$-linear map $\exists x_{0},y_{0}\in M$
such that $f(x)=\beta(x_{0},x)$ and $f(x)=\beta(x,y_{0})$
\item $\beta(x,y)=0\,\forall x\in M\Rightarrow y=0$, similarly $\beta(x,y)=0\,\forall y\in M\Rightarrow x=0$
\end{enumerate}
\end{defn}
\begin{rem*}
$r,l:M\rightarrow M^{v}=\mbox{Hom}_{R}(M,R)=\{f:M\mapsto R:f\mbox{ is }R\mbox{-linear}\}$
defined by $x\mapsto r(x)=\beta(x,-):M\rightarrow R:t\mapsto\beta(x,t)$
and $y\mapsto l(y)=\beta(y,-):M\rightarrow R:t\mapsto\beta(t,y)$.
Then
\begin{enumerate}
\item says $r,l$ are surjective
\item says $r,l$ are injective
\end{enumerate}
In particular, if $M=V$ a finite dimensional vector space over a
field $R=F$ then $2.\Rightarrow1.$ (as $\dim V=\dim V^{v}$ and
thus $1.$ injectivitiy$\Rightarrow$ 2. surjectivity)\end{rem*}
\begin{defn}
Let $(M,\beta),(M',\beta')$ be bilinear forms. An isometry from $M$
to $M'$ is an $R$-linear isomorphism $f:M\rightarrow M'$ such that
$\beta(x,y)=\beta'(f(x),f(y))\,\forall x,y\in M$

Two bilinear forms $(M,\beta),(M',\beta')$ are isometric if there
exists an isometry between them\end{defn}
\begin{xca*}
Check that isometry is an equivalence relation

Check if $(M,\beta)$ and $(M',\beta')$ are isometric then $(M,\beta)$
is symmetric (skew-symmetric, symplectic, regular) if and only if
$(M',\beta')$ is.\end{xca*}
\begin{defn}
Let $M$ be an $R$-module. A \emph{quadratic form} on $M$ is a function
$q:M\rightarrow R$ such that
\begin{enumerate}
\item $q(ax)=a^{2}q(x)\,\forall a\in R,x\in M$
\item The form $\beta_{q}:M\times M\rightarrow R$ defined by $\beta_{q}(x,y)=q(x+y)-q(x)-q(y)$
is bilinear
\end{enumerate}
$\beta_{q}$ is called the associated symmetric bilinear form.

The quadratic form $q:M\rightarrow R$ is called regular if $\beta_{q}$
is regular.

Let $(M,q),(M',q')$ be two quadratic forms modules. An isometry from
$M$ to $M'$ is an $R$-linear isomorphism $f:M\rightarrow M'$ such
that $q(x)=q'(f(X))\,\forall x\in M$.\end{defn}
\begin{rem*}
If $\beta$ is a (symmetric) bilinear form then $q_{\beta}(x)=\beta(x,x)$
is a quadratic form because $q_{\beta}(ax)=\beta(ax,ax)=a^{2}\beta(x,x)=a^{2}q_{\beta}(x)$
and $\beta_{q_{\beta}}(x,y)=q_{\beta}(x+y)-q_{\beta}(x)-q_{\beta}(y)=\beta(x+y,x+y)-\beta(x,x)-\beta(y,y)=\beta(x,y)+\beta(y,x)$
is bilinear.

If $\beta$ is symmetric then $\beta_{q_{\beta}}=2\beta$\end{rem*}
\begin{cor}
If $\frac{1}{2}\in R$ (i.e. $2\in R$ is a unit) and $M$ is an $R$-module
then $\{\mbox{quadratic forms on }M\}\rightarrow\{\mbox{symetric bilinear forms on }M\}$
by $q\mapsto\beta_{q}$ is a bijection with inverse $\{\mbox{symetric bilinear forms on }M\}\rightarrow\{\mbox{quadratic forms on }M\}$
defined by $\beta\mapsto\frac{1}{2}q_{\beta}$ \end{cor}
\begin{proof}
Exercise\end{proof}
\begin{rem*}
If $\frac{1}{2}\in R$ then the theory of quadratic forms is the same
as the theory of symmetric bilinear forms.

But if $\frac{1}{2}\notin R$ then the two theories may differ:
\begin{example*}
The symmetric bilinear form $\mathbb{Z\times Z\rightarrow Z}$ defined
by $x,y\mapsto xy$ does not come from a quadratic forms on $\mathbb{Z}$
because if $q:\mathbb{Z\rightarrow Z}$ is a quadratic form then $q(a)=q(a\cdot1)=a^{2}q(1)$
and $\beta_{q}(x,y)=q(x+y)-q(x)-q(y)=(x+y)^{2}q(1)-x^{2}q(1)-y^{2}q(1)=2xyq(1)\neq xy$
\end{example*}
\end{rem*}
Objective of this course: Understand classification of quadratic forms
(or symmetric bilinear forms) up to isometry:\\
How many quadratic forms exists (up to isometry)?\\
Given two quadratic forms how can I decide when they are isometric

A few applications of quadratic forms:
\begin{itemize}
\item Algebra (quaternion algebras)
\item Manifold theory (as products pairing)
\item Number theory
\item Lattice theory (sphere packing)
\end{itemize}

\section{Quadratic forms and homogeneous polynomial of degree 2}
\begin{defn}
A polynomial $f=\sum a_{i_{1},...,i_{n}}x_{1}^{i_{1}}...x_{n}^{i_{n}}\in R[x_{1},...,x_{n}]$
is called \emph{homogenous} of degree $m$ if all occurring monomials
$x_{1}^{i_{1}}...x_{n}^{i_{n}}$ ($a_{i_{1},...i_{n}}\neq0$) has
degree $i_{1}+...+i_{n}=m$\end{defn}
\begin{example*}
$x^{3}+x^{2}y+z^{3}$ is homogeneous of degree 3

$x^{3}+x^{2}+zy^{2}$ is not homogeneous.
\end{example*}
Every homogeneous degree 2 polynomial $f\in R[x_{1},...,x_{n}]$ has
the form $f=\sum_{i=1}^{n}a_{i}x_{i}^{2}+\sum_{i<j}b_{ij}x_{i}x_{j}$.
To every polynomial $f\in R[x_{1},...,x_{n}]$ one can associate a
(polynomial) function $\bar{f}:R^{n}\rightarrow R$ by $(r_{1},...,r_{n})\mapsto f(r_{1},...,r_{n})$ 
\begin{rem*}
In general $R[x_{1},...,x_{n}]\to\mbox{Funtions }(R^{n},R)$ which
maps to $f\mapsto\bar{f}$ is not injective. (find examples!)

But homogenous polynomials in $n$-variables of degree $m$ $\to$
Functions $(R^{n},R)$ is injective\end{rem*}
\begin{claim*}
If $f\in R[x_{1},...,x_{n}]$ is homogeneous of degree 2 then $\bar{f}:R^{n}\to R$
is a quadratic from\end{claim*}
\begin{proof}
Note: If $q_{1},q_{2}$ are quadratic forms on $M$ then $q_{1}+q_{2}$
and $aq_{1}$ are all quadratic forms $\forall a\in R$. Thus can
assume $f=x_{i}x_{j}$. Then $\bar{f}(r_{1},...,r_{n})=r_{i}r_{j}$
so
\begin{enumerate}
\item $\bar{f}(ar)=a^{2}\bar{f}(r)\,\forall a\in R_{1},r\in R^{n}$
\item $\bar{f}(r+s)-\bar{f}(r)-\bar{f}(s)=(r_{i}+s_{i})(r_{j}+s_{j})-r_{i}r_{j}-s_{i}s_{j}=r_{i}s_{j}+s_{i}r_{j}$
is bilinear in $r,s\in R^{n}$
\end{enumerate}
\end{proof}
\begin{lem}
For any (commutative!) ring $R$, the map 
\[
\left\{ \mbox{\ensuremath{\overset{\mbox{homoegenous polynomials of degree}}{2\mbox{ in }n\mbox{ variables}}} }\right\} \to\{\mbox{quadratic forms }R^{n}\}
\]
 defined by $f\mapsto\bar{f}$ is bijective\end{lem}
\begin{proof}
Exercise
\end{proof}
If $\underset{e_{j}\mapsto\sum_{i=1}^{n}a_{ij}e_{i}}{R^{n}\to R^{n}}$
is an $R$-linear map given by a matrix $A=(a_{ij})\in M_{n}(R)$
we can define a ring homomorphism $A_{*}:R[x_{1},...,x_{n}]\to R[x_{1},...,x_{n}]$
by $x_{j}\mapsto A_{*}(x_{j})=\sum_{j=1}^{n}a_{ij}x_{j}$ which sends
homogenous polynomials of degree $m$ to homogenous polynomials of
the degree $m$.
\begin{defn}
Two homogenous polynomials of degree 2 $f,g\in R[x_{1},...,x_{n}]$
are (linearly) equivalent if $\exists A\in M_{n}(R)$ invertible with
$A_{*}(f)=g$\end{defn}
\begin{lem}
The map 
\[
\left\{ \mbox{\ensuremath{\overset{\mbox{homoegenous polynomials of degree}}{2\mbox{ in }n\mbox{ variables}}} }\right\} /\mbox{linear equivalence}\to\{\mbox{quadratic forms on }R^{n}\}/\mbox{isometry}
\]
 is bijective\end{lem}
\begin{proof}
Exercise
\end{proof}

\subsection{Free bilinear form modules}
\begin{defn}
A bilinear $R$-module $(M,\beta)$ is called \emph{free of rank $n$}
($n\in\mathbb{N})$ if $M\cong R^{n}$

If $(M,\beta)$ is free of rank $n$ then $M$ has a basis $e_{1},...,e_{n}$
and we can defined an associated bilinear form matrix $B=\left(\beta(e_{i},e_{j})\right)$.
Note that $B=\left(\beta(e_{i},e_{j})\right)$ determines $\beta$
since if $x,y\in M$ have coordinates (with respect to $e_{1},...,e_{n})$
$x_{1},...,x_{n},y_{1},...,y_{n}\in R$ i.e. $x=\sum x_{i}e_{i},y=\sum y_{i}e_{i}$
then $\beta(x,y)=\beta(\sum x_{i}e_{i},\sum y_{j}e_{j})=\sum x_{i}\beta(e_{i},e_{j})y_{j}=(x_{1},...,x_{n})B(y_{1},...,y_{n})^{T}$\end{defn}
\begin{example*}
Standard scalar product on $\rr^{n}$ has bilinear form matrix with
respect to standard basis $B=\begin{pmatrix}1 &  & 0\\
 & \ddots\\
0 &  & 1
\end{pmatrix}$\end{example*}
\begin{lem}
Let $(M,\beta)$ be a free bilinear form module of rank $n$ with
basis $e_{1},\dots,e_{n}$, then $\beta$ is non-degenerate $\iff$
the associated bilinear form matrix $B=(\beta(e_{i},e_{j}))\in M_{n}(R)$
is invertible\end{lem}
\begin{proof}
Recall that $\beta$ is non degenerate if and only if $r,l:M\to\mbox{Hom}_{R}(M,R)$
defined by 
\begin{eqnarray*}
x & \mapsto & r(x)=\beta(x,-),\, r(x)(y)=\beta(x,y)\\
 & \mapsto & l(x)=\beta(-,x),\, l(x)(y)=\beta(y,x)
\end{eqnarray*}
are bijective. $M$ has basis $e_{1},\dots,e_{n}$. Then $\mbox{Hom}_{R}(M,R)$
has basis $e_{1}^{\#},...,e_{n}^{\#}$ where $e_{i}^{\#}e_{j}=\begin{cases}
1 & i=j\\
0 & i\neq j
\end{cases}$ i.e. $e_{i}^{\#}\left(\sum x_{j}e_{j}\right)=x_{i}$. Let $(r_{ij})$
receptively $(l_{ij})$ be the $n\times n$ of $r,l$ with respect
to basis $e_{1},...,e_{n}$of $M$ and $e_{1}^{\#},...,e_{n}^{\#}$of
$\mbox{Hom}_{R}(M,R)$ i.e. $\sum_{k=1}^{n}r_{kj}e_{k}^{\#}=r(e_{j})$
and $\sum_{k=1}^{n}l_{kj}e_{k}^{\#}=l(e_{j})$ so $\beta(e_{j},e_{i})r(e_{j})(e_{i})=\sum_{k=1}^{n}r_{kj}e_{k}^{\#}(e_{i})=r_{ij}\,\forall i,j\Rightarrow(r_{ij})=B^{T}=$
transpose of $B$. Similarly for $(e_{ij})=B$. So, $\beta$ non degenerated
$\iff r,l$ are $R$-linear isomorphism $\iff(r_{ij})\mbox{ and }(l_{ij})$
are invertible $\iff B^{T},B$ are invertible $\iff B$ is invertible\end{proof}
\begin{lem}
Let $(M,\beta),(M',\beta')$ be two free bilinear form modules over
$R$ of rank $n$ with basis $e_{1},...,e_{n}$ for $M$ and $e_{1}',...,e_{n}'$
for $M'$ then $(M,\beta)$ and $(M',\beta')$ are isometric $\iff$
associated bilinear form matrices $B=(\beta(e_{i},e_{j}))$ and $B'=(\beta'(e_{i}',e_{j}'))$
are congruent. i.e.~$\exists A\in M_{n}(R)$ invertible such that
$B=A^{T}B'A$\end{lem}
\begin{proof}
$"\Rightarrow"$: Let $f:M\to M'$ be an isometry. Let $(f_{ij})$
be the associated matrix with respect to the basis $e_{1},...,e_{n}$
and $e_{1}',..,e_{n}'$ i.e.~$f(e_{j})=\sum_{k=1}^{n}f_{kj}e_{k}'$.
Then $ $$f$ isometry $\Rightarrow f$ isomorphism $\Rightarrow(f_{ij})$
invertible and $\beta(e_{i},e_{j})=\beta'(f(e_{i}),f(e_{j}))=\beta'\left(\sum_{k=1}^{n}f_{ki}e_{k}',\sum_{l=1}^{n}f_{lj}e_{l}'\right)=\sum f_{ki}\beta'(e_{k}',e_{l}')f_{lj}=(A^{T}B'A)_{ij}$.
Hence $B=A^{T}B'A$

$"\Leftarrow"$: $A=(f_{ij})\in M_{n}(R)$ defines an isomorphism
$f:M\to M'$ by $f(e_{j})=\sum f_{kj}e_{k}'$ such that $\beta(e_{i},e_{j})=\beta'(f(e_{i}),f(e_{j}))$
(Calculation as above). Hence $f:M\to M'$ is an isometry\end{proof}
\begin{defn}
Let $B\in M_{n}(R)$ we let $\langle B\rangle$ stand for the bilinear
form module $(R^{n},\beta)$, $\beta:R^{n}\times R^{n}\to R$ defined
by
\[
\underset{\begin{pmatrix}x_{1}\\
\\
x_{n}
\end{pmatrix}}{x},\underset{\begin{pmatrix}y_{1}\\
\\
y_{n}
\end{pmatrix}}{y}\mapsto\beta(x,y)=x^{T}By
\]
This is a free bilinear form module with basis $e_{1},\dots,e_{n}$
with $e_{i}$ has an $1$ in the $i$-th position and associated bilinear
form matrix $B$. Note $\beta(e_{i},e_{j})=e_{i}^{T}Be_{j}=B_{ij}$\end{defn}
\begin{rem*}
$\langle B\rangle\cong\langle B'\rangle$ for $B,B'\in M_{n}(R)\iff\exists A\in M_{n}(R)$
invertible such that $B'=A^{T}BA$\end{rem*}
\begin{defn}
The \emph{determinant} of a non-degenerate free bilinear form module
$M=(M,\beta)$ with basis $e_{1},\dots,e_{n}$ is the determinant
$\det M=\det(\beta(e_{i},e_{j}))_{i,j=1,\dots,n}\in R^{*}/R^{2*}$.
Here $R^{2*}\subset R^{*}$ is the set of units which are squares.
Note that $\det M\in R^{*}/R^{2*}$ does not depend on the choice
of basis because if $e_{1}',\dots,e_{n}'$ is another basis, then
$(\beta(e_{i}',e_{j}'))=A^{T}(\beta(e_{i},e_{j})A\Rightarrow\det(\beta(e_{1}',e_{j}')=(\det(A))^{2}\det(\beta(e_{i},e_{j}))$\end{defn}
\begin{example*}
Recall $\langle a\rangle$ is $R\times R\to R$ defined by $x,y\mapsto axy$. 
\begin{itemize}
\item If $\langle1\rangle=\langle2\rangle\Rightarrow\underset{=1}{\det\langle1\rangle}=\underset{=2}{\det\langle2\rangle}\in R^{*}/R^{2*}\Rightarrow2$
is a square $\Rightarrow\langle1\rangle\ncong\langle2\rangle$ over
$\qq$
\item If $\left\langle \begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}\right\rangle \cong\left\langle \begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}\right\rangle \Rightarrow1=-1\in R^{*}/R^{2*}\Rightarrow(-1)$ is a square $\Rightarrow\left\langle \begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}\right\rangle \ncong\left\langle \begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}\right\rangle $ over $\zz,\qq,\rr$ but $\left\langle \begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}\right\rangle \cong\left\langle \begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}\right\rangle $ over $\cc$ (see below)
\end{itemize}
\end{example*}
\begin{rem*}
if $(M,\beta)$ is a free of rank $n$ with basis $e_{1},\dots,e_{n}$
and bilinear form matrix $B$ then 
\begin{itemize}
\item $(M,\beta)$ is non-degenerate $\iff\det B\in R^{*}$. 
\item $(M,\beta)$ is symmetric $\iff B=B^{T}$
\item $(M,\beta)$ is skew-symmetric $\iff B^{T}=-B$
\item $(M,\beta)$ is symplectic $\iff B^{T}=-B$ and all diagonal entries
of $B$ are 0
\end{itemize}
\end{rem*}
\begin{proof}
Exercise.\end{proof}
\begin{lem}
\label{lem:rk_welldfn}Let $R$ be (commutative!) ring and $f:R^{n}\to R^{l}$
is a surjective $R$-module homomorphism. Then $n\geq l$. In particular,
$R^{n}\cong R^{l}\Rightarrow n=l$\end{lem}
\begin{proof}
Let $m\subset R$ be a maximal ideal. Then $R/m=k$ is a field. Reducing
$f\mod m$ yields a surjective map $f:(R/m)^{n}\to(R/m)^{l}$ of finite
dimensional $k$-vector space $\Rightarrow n=\dim_{k}(R/m)^{n}\geq\dim_{k}(R/m)^{l}=l$\end{proof}
\begin{rem*}
Lemma does not hold for non-commutative ring in general. For example
let $V$ be a $k$-vector space of $\infty$ dimension and $A=\mbox{End}_{k}(V)$
then $A\oplus A\cong A$ as $A$ module
\end{rem*}

\section{Orthogonal sum}
\begin{defn}
Let $(M,\beta)$ and $(M',\beta')$ be two bilinear form modules.
Their \emph{orthogonal sum} $(M,\beta)\perp(M',\beta')$ has underlying
module $M\oplus M'$ and bilinear form $(M\oplus M')\times(M\oplus M')\to R$
defined by $(x,u),(y,v)\mapsto\beta(x,y)+\beta'(u,v)$.\end{defn}
\begin{rem*}
{}
\begin{itemize}
\item $B\in M_{n}(R),B'\in M_{m}(R)$ then $\left\langle B\right\rangle \perp\left\langle B'\right\rangle =\left\langle \begin{pmatrix}B & 0\\
0 & B'
\end{pmatrix}\right\rangle $
\item If $(M,\beta)$ and $(M',\beta')$ are regular (symmetric,skew-symmetric,symplectic)
then so is $(M,\beta)\perp(M',\beta')$
\end{itemize}
\end{rem*}
\begin{defn}
Let $(M,\beta)$ be a symmetric or skew-symmetric bilinear form (so
$\beta(x,y)=0\Rightarrow\beta(y,x)=0$). Let $N\subset M$ be a sub-module.
The \emph{orthogonal complement} of $N$ (in $M$) is the sub-module
$N^{\perp}=\{x\in M|\beta(x,y)=0\,\forall y\in N\}$\end{defn}
\begin{lem}
Let $(M,\beta)$ be symmetric or skew-symmetric and $N\subset M$
a sub-module such that $(N,\beta_{N})$ is non degenerate. Then $M=N\perp N^{\perp}$\end{lem}
\begin{proof}
Have to check that $N\cap N^{\perp}=0$. If $x\in N\cap N^{\perp}$
then $\beta(x,y)=0\,\forall y\in N$ (as $x\in N^{\perp}$). But since
$x\in N$ and $\beta_{N}$ is non degenerate we have $x=0$. So $M\supset N+N^{\perp}=N\oplus N^{\perp}$. 

We next need to check $N+N^{\perp}=M$. Let $x\in M$ then $\beta(x,-)|_{N}\in\Hom_{R}(N,R)\underset{\beta_{N}\mathrm{\, non-degenrate}}{\Rightarrow}\exists x_{0}\in N:\,\beta(x,y)=\beta(x_{0},y)\,\forall y\in N$
then $\beta(x-x_{0},y)=0\,\forall y\in N\Rightarrow x-x_{0}\in N^{\perp}\Rightarrow x=\underset{\in N}{\underbrace{x_{0}}}+\underset{\in N^{\perp}}{\underbrace{x-x_{0}}}$.
Thus $N+N^{\perp}=M\Rightarrow N\oplus N^{\perp}=M$

Lastly we need to check that $(N,\beta_{N})\perp(N^{\perp},\beta_{N}^{\perp})=(M,\beta)$.
If $x,y\in N,u,v\in N^{\perp}$ then $\beta(x+u,y+v)=\beta(x,y)+\underset{=0}{\underbrace{\beta(x,v)}}+\underset{=0}{\underbrace{\beta(v,y)}}+\beta(u,v)=\beta_{N}(x,y)+\beta_{N}^{\perp}(u,v)$\end{proof}
\begin{cor}
If $(M,\beta)$ is a finitely generated symmetric bilinear form module.
Then $M=\left\langle u_{1}\right\rangle \perp\left\langle u_{2}\right\rangle \perp\dots\perp\left\langle u_{k}\right\rangle \perp N$
where $u_{i}\in R^{*}$ and $\beta(x,x)\in R\setminus R^{*}\,\forall x\in N$\end{cor}
\begin{proof}
Set $M_{0}=M$ and if $\beta(x,x)=\in R\setminus R^{*}\,\forall x\in M_{0}=M$
then take $N=M_{0}=M$ and we are done. So assume $\exists x\in M_{0}:\beta(x,x)\in R^{*}$
then $ax\neq0\in M\,\forall a\in R\setminus\{0\}$ (if $ax=0\Rightarrow\beta(ax,x)=a\beta(x,x)\Rightarrow a=0$).
So $Rx\subset M$ is a free module of rank 1 with basis $x$. $Rx$
has bilinear form matrix $(\beta(x,x))\in R^{*}$ invertible. So,
$\beta|_{Rx}$ is non degenerate. $\Rightarrow M=\underset{\left\langle u_{1}\right\rangle }{\underbrace{Rx}}\perp\underset{=:M_{1}}{\underbrace{(Rx)^{\perp}}}$
with $u_{1}=\beta(x,x)\in R^{*}$ contradicting Lemma \ref{lem:rk_welldfn}

Repeat with $M_{1}$in place of $M_{0}$ to obtain $M=\left\langle u_{1}\right\rangle \perp\dots\perp\left\langle u_{k}\right\rangle \perp M_{k}$.
We can repeats as long as $\exists x\in M_{K}:\beta(x,x)\in R^{*}$.
But the procedure stops because $K>n$ impossible otherwise there
exists a surjective map $R^{n}\twoheadrightarrow M=\underset{R^{k}}{\underbrace{\left\langle u_{1}\right\rangle \perp\dots\perp\left\langle u_{n}\right\rangle }}\perp M_{k}\twoheadrightarrow R^{k}\Rightarrow n\geq k$ \end{proof}
\begin{rem*}
If $\beta(x,x)\in R\setminus R^{*}$ $\forall x\in N\neq0$ and $\beta$
is non-degenerate then $(N,\beta)$ cannot has an orthogonal basis
\begin{proof}
If $N=\langle u_{1}\rangle\perp\dots\perp\left\langle u_{n}\right\rangle $
and if $N$ is non-degenerate with respect to base $e_{1},\dots,e_{n}$
then $\left\langle u_{i}\right\rangle $ are non-degenerate $\Rightarrow u_{i}\in R^{*}$.
$\beta(e_{i},e_{j})=\begin{cases}
u_{i} & i=j\\
0 & i\neq j
\end{cases}$ in particular $\beta(e_{1},e_{1})=u_{1}\in R^{*}$
\end{proof}
\end{rem*}
\begin{thm}[Existence of orthogonal basis over fields of $\mbox{char}\neq2$]
Let $k$ be a field of $\mbox{char}\neq2$ and $(M,\beta)$ a finite
dimensional symmetric bilinear form. Then $M=\left\langle u_{1}\right\rangle \perp\dots\perp\left\langle u_{l}\right\rangle \perp N$
such that $\beta|_{N}=0$ ($\beta(x,y)=0\,\forall x,y\in N$). In
particular $(M,\beta)$ has an orthogonal basis, $e_{1},\dots,e_{l},e_{l+1},\dots e_{n}$
such that $\beta(e_{i},e_{j})=\begin{cases}
u_{i} & j=i=1,\dots l\\
0 & j=i=l+1,\dots n\\
0 & j\neq i
\end{cases}$\end{thm}
\begin{proof}
From corollary we have $(R=k)$ $M=\left\langle u_{1}\right\rangle \perp\dots\perp\left\langle u_{l}\right\rangle \perp N$
with $u_{i}\in R^{*},\beta(x,x)\in\underset{k\setminus k^{*}=0}{R\setminus R^{*}}\forall x\in N$.
Need to show $\beta|_{N}=0$. $\beta(x,x)=0\,\forall x\in N\Rightarrow$
associated quadratic form $q(x)=\beta(x,x)=0\underset{\mbox{char}\neq2}{\Rightarrow}\beta(x,y)=\frac{1}{2}(q(x+y)-q(x)-q(y))=0$ 
\end{proof}
Want to generalize theorem on existence of orthogonal basis to rings.
\begin{defn}
A ring $R$ is called \emph{local} if it has a unique maximal ideal
$m$.
\end{defn}
Note that $R/m$ is a field as $m\subset R$ is a maximal ideal.
\begin{notation*}
$(R,m,k)$ is a local ring if $k=R/m,m\subset R$ is the maximal ideal\end{notation*}
\begin{rem*}
In a local ring $(R,m,k)$ we have $R^{*}=R\setminus m$
\begin{proof}
Need to show $m=R\setminus R^{*}$. 
\begin{itemize}
\item We see that $m\cap R^{*}=\emptyset$ because $m\subsetneq R\Rightarrow m\subset R\setminus R^{*}$. 
\item If $a\in R\setminus R^{*}$ , then $(a)=Ra\subsetneq R$ (proper ideal
because $Ra=R\Rightarrow\exists b:ba=1$ contradicting $a\notin R^{*}$)
Every proper ideal is contained in a maximal ideal $\Rightarrow Ra\subset m\Rightarrow a\in m\Rightarrow R\setminus R^{*}\subset m$
\end{itemize}
\end{proof}
$\left(R,m,k\right)$ local then $A\in M_{n}(R)$ is invertible if
and only if $A\mod m\in M_{n}(k)$ is invertible because $A\in M_{n}(R)$
invertible $\iff\det A\in R^{*}=R\setminus m\iff\det(A\mod m)\neq0\in k=R/m\iff A\mod m\in M_{n}(k)$
is invertible.\end{rem*}
\begin{example*}
{}
\begin{itemize}
\item Fields are local rings with $m=0$
\item $\zz_{(p)}=\{\frac{a}{b}\in\qq:a,b\in\zz,p\nmid b\}$ where $p\in\zz$
is prime. This is a local ring with maximal ideal $m=\{\frac{a}{b}\in\qq:p\nmid b,p\mid a\}$.
Then $\zz_{(p)}/m=\mathbb{F}_{p}$
\item $k$ field $k[T]/T^{n}$ is a local ring with maximal ideal $(T)$
\end{itemize}
\end{example*}
\begin{defn}
A finitely generated $R$-module $P$ is called \emph{projective}
if it is a direct factor of some $R^{n},n\in\mathbb{N}$ i.e. $\exists R\mbox{-module }N:M\oplus N\cong R^{n}$\end{defn}
\begin{thm}
Let $(R,m,k)$ be a local ring and $M$ a finitely generated projective
$R$-module, then $M\cong R^{l}$ for some $l\in\mathbb{N}$\end{thm}
\begin{proof}
$M$ projective so $\exists N:M\oplus N\overset{f}{\underset{\leftarrow}{\cong}}R^{n}$
. Let $p:R^{n}\to R^{n}$ be the linear map $\overset{}{R^{n}\overset{f}{\to}M\oplus N\overset{q}{\underset{x,y\mapsto(x,0)}{\rightarrow}}M\oplus N\overset{f^{-1}}{\to}R^{n}}$
and let $p$ be the composition of all these maps. Note that $p^{2}=(f^{-1}qf)^{2}=f^{-1}qff^{-1}qf=f^{-1}\underset{=q}{\underbrace{q^{2}}}f=p$
and $\im p\overset{\overset{f}{\to}}{\underset{\underset{g}{\leftarrow}}{\cong}}M$
\begin{claim*}
$\exists$ basis of $R^{n}$ with respect to which $p$ has the form
$\begin{pmatrix}1_{l} & 0\\
0 & 0
\end{pmatrix}$ i.e.~$\exists U\in M_{n}(R)$ such that $p=U\begin{pmatrix}1_{l} & 0\\
0 & 0
\end{pmatrix}U^{-1}$.
\end{claim*}
Note that claim implies theorem because $\im p\overset{\cong}{\underset{U}{\leftarrow}}\im\begin{pmatrix}1_{l} & 0\\
0 & 0
\end{pmatrix}=R^{l}$

Claim is true over a field (i.e.~$\mod m$). $M/mM$ and $N/mN$
are finite dimensional $k$-vector spaces so $k^{l}\underset{\underset{g_{1}}{\to}}{\cong}M/mM$,
$k^{r}\underset{\underset{g_{2}}{\to}}{\cong}N/mN$
\[
\xymatrix{k^{l}\oplus k^{r}\ar^{g_{1}\oplus g_{2}}[d]\ar^{\begin{pmatrix}1_{l} & 0\\
0 & 0
\end{pmatrix}}[r] & k^{l}\oplus k^{r}\\
M/mM\oplus N/mN\ar_{f\mod m}^{\cong}[d]\ar^{q}[r] & M/m\oplus N/mN\ar^{g_{1}^{-1}\oplus g_{2}^{-1}}[u]\\
(R/m)^{n}\ar^{p}[r] & (R/m)^{n}\ar^{f^{-1}}[u]
}
\]
Then we get $\begin{pmatrix}1_{l} & 0\\
0 & 0
\end{pmatrix}=A^{-1}p\underset{A\in M_{n}(k)}{\underbrace{f(g_{1}\oplus g_{2})}}$, $p\mod m=A\begin{pmatrix}1_{l} & 0\\
0 & 0
\end{pmatrix}A^{-1}$, now lift all entries of $A$ to entries of $R$ (under the surjective
$R/m\to k$) to obtain a matrix $S\in M_{n}(R)$ such that $A=S\mod m$.
$A\in M_{n}(k)$ invertible $\Rightarrow S\in M_{n}(R)$ invertible.
$S^{-1}pS\mod m=\begin{pmatrix}1_{l} & 0\\
0 & 0
\end{pmatrix}\mod m$ so $S^{-1}pS=\begin{pmatrix}T & B\\
C & D
\end{pmatrix}$ with $T=1_{l}\mod m$ and $B,C,D=0\mod m$. $\Rightarrow$ after
base change (given by $S$) $p$ becomes $\begin{pmatrix}T & B\\
C & D
\end{pmatrix}$ as above. Idea: ``Want to perform row and column operation to make
$\begin{pmatrix}T & B\\
C & D
\end{pmatrix}$ into $\begin{pmatrix}1_{l} & 0\\
0 & 0
\end{pmatrix}$ after base changes''. Now, $T=1\mod m\Rightarrow T\in M_{n}(R)$
invertible. $p^{2}=p\Rightarrow\begin{pmatrix}T & B\\
C & D
\end{pmatrix}=\begin{pmatrix}T & B\\
C & D
\end{pmatrix}\begin{pmatrix}T & B\\
C & D
\end{pmatrix}=\begin{pmatrix}T^{2}+BC & *\\
* & *
\end{pmatrix}$. $T=T^{2}+BC\Rightarrow1=T+T^{-1}BC$. $\underset{X}{\underbrace{\begin{pmatrix}1 & T^{-1}B\\
0 & 1
\end{pmatrix}}}\begin{pmatrix}T & B\\
C & D
\end{pmatrix}\underset{X^{-1}}{\underbrace{\begin{pmatrix}1 & -T^{-1}B\\
0 & 1
\end{pmatrix}}}=\begin{pmatrix}1 & *\\
* & *
\end{pmatrix}$$\Rightarrow$after bases change $p$ becomes $\begin{pmatrix}1 & B\\
C & D
\end{pmatrix},B,C,D=0\mod m$. $p^{2}=p\Rightarrow\begin{pmatrix}1 & B\\
C & D
\end{pmatrix}=\begin{pmatrix}1 & B\\
C & D
\end{pmatrix}\begin{pmatrix}1 & B\\
C & D
\end{pmatrix}\Rightarrow BC=0,BD=0,DC=0,D=CB+D^{2}$. $\Rightarrow D^{2}=\underset{0}{\underbrace{DCB}}+D^{3}\Rightarrow D^{2}=D^{3}\Rightarrow D^{2}(1-D)=0\Rightarrow D^{2}=0\Rightarrow D=CB$.
The second to last implication is because $1-D$ is invertible as
$D\equiv0\mod m$. $\underset{Y^{-1}}{\underbrace{\begin{pmatrix}1 & 0\\
-C & 1
\end{pmatrix}}}\begin{pmatrix}1 & B\\
C & CB
\end{pmatrix}\underset{Y}{\underbrace{\begin{pmatrix}1 & 0\\
C & 1
\end{pmatrix}}}=\begin{pmatrix}1 & B\\
0 & 0
\end{pmatrix}$Then an other base change gives $\underset{Z}{\underbrace{\begin{pmatrix}1 & B\\
0 & 1
\end{pmatrix}}}\begin{pmatrix}1 & B\\
0 & 0
\end{pmatrix}\underset{Z^{-1}}{\underbrace{\begin{pmatrix}1 & -B\\
0 & 1
\end{pmatrix}}}=\begin{pmatrix}1 & 0\\
0 & 0
\end{pmatrix}$\end{proof}
\begin{rem*}
If $R$ is a Euclidean domain (e.g.~$R=\zz$) then every finitely
projective $R$-module is free. \end{rem*}
\begin{example*}
$R=\zz$ every finitely generated $\zz$-module is isomorphic to $\zz^{n}\oplus$finite
abelian group. If $P$ is finitely generated projective over $\zz$
then $P\subseteq\zz^{m}\Rightarrow P$ has no element of finite order.
$\Rightarrow P=Z^{n}\oplus\not\mbox{finite}$ \end{example*}
\begin{rem*}
For a general commutative ring, projective $R$-modules may not be
free\end{rem*}
\begin{example*}
$R=\zz[\sqrt{-5}]=\zz[T]/(T^{2}+5$) . Fact: $R$ is a Dedekind domain,
every ideal $I\subset R$ of a Dedekind domain is a projective $R$-module.
Let $I=(2,1+\sqrt{-5})\subset R$. From the fact $I$ is projective.
If $I$ was free then $I\cong R^{n}$, $n\neq0$ because $I\neq0$.
Let's compute $R/I$:
\begin{eqnarray*}
R/I & = & \zz[T]/(T^{2}+5,2,1+T)\\
 & = & \ff_{2}[T]/(T^{2}+5,T+1)\\
 & = & \ff_{2}[T]/(T^{2}+1,T+1)\\
 & = & \ff_{2}[T]/((T+1)^{2},T+1)\\
 & = & \ff_{2}[T]/(T+1)\\
 & \cong & \ff_{2}
\end{eqnarray*}
Assume $n=1$ then $I=Rt$ for some $t\in R$. Now $t$ is not a unit
because otherwise $0=R/Rt$ contradicting $R/I=\ff_{2}$. If $t\notin R^{*}:I=Rt\Rightarrow2=at\underset{2\mbox{ irreducible}}{\Rightarrow}a\in R^{*}\Rightarrow I=Rt=R\frac{1}{a}2=R\cdot2$
and 
\begin{eqnarray*}
R/I & = & \frac{\zz[T]}{T^{2}+5}/\underset{2R}{\underbrace{I}}\\
 & = & \zz[T]/(T^{2}+5,2)\\
 & = & \ff_{2}[T]/(T^{2}+1)\\
 & = & \ff_{2}[T]/(T+1)^{2}\\
 & \neq & \ff_{2}
\end{eqnarray*}
$\Rightarrow I\cong R^{n}\Rightarrow n\geq2$. So $R^{n}\cong I\subset R$,
let $F=$ field of fraction of $R$. Then $I\hookrightarrow I\otimes_{R}F$
and $R\hookrightarrow F=R\otimes_{R}F$ so we get $F^{n}=I\otimes_{R}F\subset F=R\otimes_{R}F$
(since $F$ is a localization of $R$) $\Rightarrow F^{n}\subset F$
contradiction so $n\not\geq2$.\end{example*}
\begin{defn}
An \emph{inner product space} is a non-degenerate bilinear form module
$(M,\beta)$ where $M$ is finitely generated and projective. \end{defn}
\begin{rem*}
Over a local ring any inner product space is free \end{rem*}
\begin{defn}
Let $R$ be a ring. The \emph{hyperbolic plane} $\mathbb{H}$is the
symmetric inner product space $\left\langle \left(\begin{matrix}0 & 1\\
1 & 0
\end{matrix}\right)\right\rangle $ i.e.~$\hh=(R^{2},\beta)$. 
\end{defn}
$\hh$ has basis $e_{1},e_{2}$ with respect to which we have $\beta(e_{i},e_{j})=\begin{cases}
0 & i=j\\
1 & i\neq j
\end{cases}$. $\hh$ is a symmetric space because $\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}^{T}=\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}$, it is non-degenerate because $\det\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}=-1\in R^{*}$. Does $\hh$ have an orthogonal basis? If $\frac{1}{2}\in R$ (i.e.
$2\in R^{*}$) then $\hh\cong\left\langle 1\right\rangle \perp\left\langle -1\right\rangle $
because $\begin{pmatrix}-1 & 0\\
0 & 1
\end{pmatrix}=\underset{A^{T}}{\underbrace{\begin{pmatrix}1 & -\frac{1}{2}\\
1 & \frac{1}{2}
\end{pmatrix}}}\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}\underset{=A}{\underbrace{\begin{pmatrix}1 & 1\\
-\frac{1}{2} & \frac{1}{2}
\end{pmatrix}}}$ and $\det A=1$ and thus $A$ is invertible. If $\frac{1}{2}\notin R$
($2\notin R^{*}$) then $\hh$ has no orthogonal basis. In $\hh\ni\begin{pmatrix}x\\
y
\end{pmatrix}$ then $\begin{pmatrix}x & y\end{pmatrix}\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}\begin{pmatrix}x\\
y
\end{pmatrix}=2xy\notin R^{*}\,\forall x,y\in\hh$ since $2\notin R^{*}$. If $(\hh,\beta)$ had an orthogonal basis
$e_{1},e_{2}$ then $\beta(e_{i},e_{i})\in R^{*}\Rightarrow\hh$ has
not orthogonal basis. 
\begin{example*}
If $(R,m,k)$ is a local ring with $\mbox{char}k=2$ then for all
$a,b\in m$ $\left\langle \begin{pmatrix}a & 1\\
1 & b
\end{pmatrix}\right\rangle $has no orthogonal basis (otherwise, any orthogonal basis would yield
an orthogonal basis mod $m$ but $\left\langle \begin{pmatrix}a & 1\\
1 & b
\end{pmatrix}\right\rangle \mod m=\left\langle \begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}\right\rangle =\hh$ has no orthogonal basis. \end{example*}
\begin{thm}
Let $(R,m,k)$ be a local ring and $M=(M,\beta)$ a symmetric inner
product space.
\begin{itemize}
\item If $\mbox{char}(k)\neq2$ then $M$ has an orthogonal basis
\item If $\mbox{char}(k)=2$ then $M=\left\langle u_{1}\right\rangle \perp\dots\perp\left\langle u_{l}\right\rangle \perp N_{1}\perp\dots\perp N_{r}$
where $u_{i}\in R^{*}$ and $N_{i}=\left\langle \begin{pmatrix}a_{i} & 1\\
1 & b_{i}
\end{pmatrix}\right\rangle ,a_{i},b_{i}\in m$
\end{itemize}
\end{thm}
\begin{proof}
Recall $M$ finitely generated, $\beta$ is symmetric $\Rightarrow M=\left\langle u_{1}\right\rangle \perp\dots\perp\left\langle u_{l}\right\rangle \perp N$
such that $u_{i}\in R^{*}$ and $\beta(x,x)\in R\setminus R^{*}=m\,\forall x\in N$. 

Recall $R$ local and $M$ finitely generated projective $\Rightarrow M\cong R^{n+l}$,
same for $N$ so $N\cong R^{n}$

If $n=0$ done. So assume $n\geq1$.Then $\beta$ non-degenerate $\Rightarrow$
for $\varphi:\underset{(x_{1},\dots,x_{n})\mapsto x_{1}}{R^{n}=N\to R}$
linear, $\exists x_{0}\in N:\beta(x_{0},y)=\varphi(y)\ \forall y\in N\Rightarrow\exists x,y\in N:\beta(x,y)=1$
($y=e_{1},x=x_{0}$). 

If $\mbox{char}(k)\neq2$ ($2\notin m=R\setminus R^{*}\Rightarrow2\in R^{*}$).
If $N\neq0\Rightarrow\exists x,y\in N,\beta(x,y)=1$. Then $x+y\in N$
so $\underset{\in m}{\underbrace{\beta(x+y,x+y)}}=\underset{\in m}{\underbrace{\beta(x,x)}}+\underset{=2}{\underbrace{2\beta(x,y)}}+\underset{\in m}{\underbrace{\beta(y,y)}}\Rightarrow2\in m\Rightarrow N=0$
(due to the contradiction of $\mbox{char}(k)\neq2$)

Now assume that $\mbox{char}(k)=2$. We are going to prove that $N=N_{1}\perp\dots\perp N_{r}$
with $N_{i}$ as in the theorem by induction on $n$ ($N\cong R^{n}$).
$n=0\Rightarrow N=0$ and we are done. $n=1\Rightarrow N\cong R$
then $\beta|_{N}$ is a non-degenerate symmetric form on $R$ but
any rank 1 inner product space is $\cong\left\langle u\right\rangle $
for $u\in R^{*}$ because $\beta:R\times R\to R,\beta(x,y)=xy\cdot\beta(1,1)$
and $\beta$ non-degenerated $\Rightarrow\beta(1,1)\in R^{*}$. This
contradict our assumption that $\beta(x,x)\in m\,\forall x\in N$.
So assume $n\geq2$: Since $\beta|_{N}$ is non degenerate and $N$
free (of rank $n$) $\Rightarrow\exists x,y\in N:\beta(x,y)=1$ (because
$N\cong R^{n},\varphi:R^{n}\to R$ by $x_{1},\dots,x_{n}\mapsto x_{1}\underset{\beta\mathrm{\, non-deg}}{\Rightarrow}\exists x:\varphi(y)=\beta(x,y)\,\forall y\in R^{n}$
so in particular $\exists x\in N\,1=\varphi(e_{1})=\beta(x,e_{1})$)
The subspace $Rx+Ry\subset N$ has bilinear form matrix with respect
to $\{x,y\}$ 
\[
\begin{pmatrix}\beta(x,x) & \beta(x,y)\\
\beta(y,x) & \beta(y,y)
\end{pmatrix}=\begin{pmatrix}a & 1\\
1 & b
\end{pmatrix}
\]
 and by assumption $a,b\in m$ because $\beta(z,z)\in\,\forall z\in N$.
This has determinant $\underset{\notin m}{\underbrace{\underset{\in m}{\underbrace{ab}}-\underset{\notin m}{\underbrace{1}}}}\in R^{*}\Rightarrow x,y$
linearly independent and $N_{1}:=Rx+Ry\subset N$ is isometric to
$\left\langle \begin{pmatrix}a & 1\\
1 & b
\end{pmatrix}\right\rangle $$\Rightarrow N=N_{1}\perp N_{1}^{\perp}$ and apply induction hypothesis
to $N_{1}^{\perp}$ and we are done\end{proof}
\begin{example*}
$(M,\beta)=\left\langle \begin{pmatrix}2 & 1\\
1 & 2
\end{pmatrix}\right\rangle $ over $\zz_{(2)}=\{\frac{p}{q}\in\qq|2\nmid q\}$. $\det\begin{pmatrix}2 & 1\\
1 & 2
\end{pmatrix}=3$ and $3\in(\zz_{(2)})^{*}$, hence $M$ is non-degenerate. But $M$
has no orthogonal basis because $\zz_{(2)}/2=\ff_{2}$ and any orthogonal
basis over $\zz_{(2)}$ induces a orthogonal basis over $\zz_{(2)}/2$
but $M=\left\langle \begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}\right\rangle $ over $\ff_{2}$ which we have seen has no orthogonal basis.

Over $R=\zz_{(p)}$ where $p\in\zz$ is prime, $p\neq3$ (otherwise
$M$ is degenerate as $3\notin(\zz_{(3)})^{*}$). Then $M$ is non-degenerate
since $\det M=3\in(\zz_{(p)})^{*}$. If furthermore $p\neq2$ then
by theorem $M$ has an orthogonal basis: For instance $x=\begin{pmatrix}1\\
0
\end{pmatrix}$ then $\beta(x,x)=2\in(\zz_{(p)})^{*}$ ($p\neq2$) $\Rightarrow Rx\subset M$
non-degenerate subspace so $M=Rx\perp(Rx)^{\perp}$. Now $(Rx)^{\perp}=\{y\in M:\beta(x,y)=0\}=\left\{ \begin{pmatrix}a\\
b
\end{pmatrix}\in R^{2}|\begin{pmatrix}1 & 0\end{pmatrix}\begin{pmatrix}2 & 1\\
1 & 2
\end{pmatrix}\begin{pmatrix}a\\
b
\end{pmatrix}=0\right\} =R\begin{pmatrix}1\\
-2
\end{pmatrix}\Rightarrow\left\{ \begin{pmatrix}1\\
0
\end{pmatrix},\begin{pmatrix}1\\
-2
\end{pmatrix}\right\} $ is an orthogonal basis of $M$ if $R=\zz_{(p)}$ $p\neq2,3$ (Also
works for $\qq,\rr,\cc$)\end{example*}
\begin{defn}
Let $(M,\beta)$ be a symplectic inner product space ($\beta$ symplectic
if $\beta(x,x)=0\,\forall x\in M$). A symplectic basis of $M$ is
a basis $x_{1},y_{2},x_{2},y_{x},\dots,x_{n},y_{n}$ such that $M=(Rx_{1}+Ry_{1})\perp\dots\perp(Rx_{n}+Ry_{n})$
and $\beta(x_{i},y_{i})=1\,\forall i$ ($\Rightarrow\beta(y_{i},x_{i})=-1$)
i.e. the bilinear form matrix of $\beta$ with respect to the basis
$x_{1},y_{1},...,x_{n},y_{n}$ is 
\[
\begin{pmatrix}\begin{matrix}0 & 1\\
-1 & 0
\end{matrix} &  &  & 0\\
 & \begin{matrix}0 & 1\\
-1 & 0
\end{matrix}\\
 &  & \ddots\\
0 &  &  & \begin{matrix}0 & 1\\
-1 & 0
\end{matrix}
\end{pmatrix}
\]
 \end{defn}
\begin{thm}
Let $(R,m,k)$ be a local ring and $(M,\beta)$ a symplectic inner
product space. Then $(M,\beta)$ has a symplectic basis\end{thm}
\begin{proof}
$(M,\beta)$ inner product space $\Rightarrow\beta$ non-degenerate,
$M$ projective $\Rightarrow$ free (since $R$ local) $\Rightarrow\exists x,y\in M:\beta(x,y)=1$.
So the inner product matrix of $\beta$ with respect to $\{x,y\}$
is $\begin{pmatrix}\beta(x,x) & \beta(x,y)\\
\beta(x,y) & \beta(y,y)
\end{pmatrix}\underset{\beta\mathrm{\, sympletic}}{=}\begin{pmatrix}0 & 1\\
-1 & 0
\end{pmatrix}$. Thus set $N_{1}=Rx+Ry\subset M$ is a non-degenerate free submodule
of rank 2 with symplectic basis $\{x,y\}$. $(N_{1},\beta|_{N_{1}})$
non-degenerate $\Rightarrow M=\underset{\left\langle \begin{pmatrix}0 & 1\\
-1 & 0
\end{pmatrix}\right\rangle }{\underbrace{N_{1}}}+N_{1}^{\perp}$ repeating the same argument with $N_{1}^{\perp}$ instead of $M$
we obtain $M=N_{1}\perp N_{2}\perp\dots\perp N_{n}$ where $N_{i}=\left\langle \begin{pmatrix}0 & 1\\
-1 & 0
\end{pmatrix}\right\rangle $\end{proof}
\begin{cor}
Over a local ring any symplectic inner product space has even dimension.
Furthermore any two symplectic inner product spaces are isometric
if and only if they have the same rank.
\end{cor}

\subsection{Witt Cancellation}

Motivation: Let $V_{1},V_{2}$ be finite dimensional vector spaces
over $k$, If $V_{1}\oplus W\cong V_{2}\oplus W$ for some finite
dimensional vector space $W$ then $V_{1}\cong V_{2}$ because $\dim V_{1}=\dim V_{1}\oplus W-\dim W=\dim V_{2}\oplus W-\dim W=\dim V_{2}$.
The same is true for free modules of finite rank (over a commutative
ring), and also over finitely generated projective modules over local
rings.

Question: If $V_{1},V_{2},W$ are symmetric inner product spaces does
$V_{1}\perp W\cong V_{2}\perp W\Rightarrow V_{1}\cong V_{2}$ ?
\begin{example*}
$R=\ff_{2}$ (or $R$ local with $\frac{1}{2}\notin R$) then $\left\langle -1\right\rangle \perp\left\langle -1\right\rangle \perp\left\langle -1\right\rangle \cong\left\langle -1\right\rangle \perp\hh$,
that is,

$\left\langle \begin{pmatrix}-1 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & -1
\end{pmatrix}\right\rangle \cong\left\langle \begin{pmatrix}-1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0
\end{pmatrix}\right\rangle $ but $\left\langle -1\right\rangle \perp\left\langle -1\right\rangle \ncong\hh$
because $\hh$ has no orthogonal basis over $\ff_{2}$. For the isometry
see the example below.\end{example*}
\begin{defn}
We say \emph{Witt cancellation} holds for a ring $R$ if $\forall M,N,P$
symmetric inner product spaces over $R$ $M\perp P\cong N\perp P\Rightarrow M\cong N$\end{defn}
\begin{example*}
Witt cancellation does not hold over fields of $\mbox{char}2$ (or
for local rings $R$ where $2\notin R^{*}$).\end{example*}
\begin{note*}
$\left\langle \begin{pmatrix}-1\\
 & -1\\
 &  & 1
\end{pmatrix}\right\rangle \cong\left\langle \begin{pmatrix}-1\\
 & 0 & 1\\
 & 1 & 0
\end{pmatrix}\right\rangle (*)$, because $\beta=\left\langle \begin{pmatrix}-1\\
 & -1\\
 &  & 1
\end{pmatrix}\right\rangle $ has orthogonal basis $e_{1},e_{2},e_{3}$ with $\beta(e_{i},e_{j})=0$
for $i\neq j$ and $\beta(e_{i},e_{i})=\begin{cases}
-1 & i=1,2\\
1 & i=3
\end{cases}$ . In the basis $e_{1}+e_{2}+e_{3},e_{1}+e_{3},e_{2}+e_{3},$ the
inner product $\beta$ has inner product matrix $\left\langle \begin{pmatrix}-1\\
 & 0 & 1\\
 & 1 & 0
\end{pmatrix}\right\rangle \Rightarrow(*)$. So If Witt cancellation holds then $\left\langle -1\right\rangle \perp\left\langle -1\right\rangle \cong\hh$
but this is not the case for field $\mbox{char}2$ (or local rings
$R$ with $2\notin R^{*}$)\end{note*}
\begin{defn}
Let $M$ be an symmetric inner product space and $N\subseteq M$ a
non-degenerate subspace then $M=N\perp N^{\perp}$ and \emph{the reflection
of $M$ at $N$} is the isometry 
\[
r_{N}:M=N\perp N^{\perp}\to N\perp N^{\perp}
\]
\[
(x,y)\mapsto(x,-y),\hspace{4ex}x\in N,y\in N^{\perp}
\]
 \end{defn}
\begin{rem*}
$r_{N}$ is $R$-linear, an isomorphism $(r_{N}\circ r_{N}=\id$)
and preserves inner product hence $r_{N}$ is an isometry\end{rem*}
\begin{lem}
\label{lem:reflection_map}Let $(M,\beta)$ be a symmetric inner product
space and $x,y\in M$ such that $\beta(x,x)=\beta(y,y)\in R^{*}$.
If $R$ is local with $\frac{1}{2}\in R$ then there is a reflection
$r$ of $M$ such that $r(x)=y$\end{lem}
\begin{proof}
Consider $u=x+y,v=x-y\in M$ then $x=\frac{1}{2}(u+v),y=\frac{1}{2}(u-v)$
\begin{itemize}
\item $u\perp v$: $\beta(u,v)=\beta(x+y,x-y)=\beta(x,x)-\beta(y,y)=0$
(Since by assumption $\beta(x,x)=\beta(y,y)$)
\item $\beta(u,u)$ or \textbf{$\beta(v,v)\in R^{*}$}: $4\beta(x,x)=\beta(2x,2x)=\beta(u+v,u+v)\in R^{*}$
(since $\beta(x,x)\in R^{*}$ and $2\in R^{*})$. By the first point
$\beta(u+v,u+v)=\beta(u,u)+\beta(v,v)$. If $\beta(u,u),\beta(v,v)\in m=$
maximal ideal of $R\Rightarrow\beta(u,u)+\beta(v,v)\in m$. Contradiction
hence $\beta(u,u)\mbox{ or }\beta(v,v)\in R^{*}$
\item If $\beta(u,u)\in R^{*}$ then $Ru\subseteq M$ non-degenerate subspace
$r_{Ru}(x)=r_{Ru}(\frac{u+v}{2})=\frac{1}{2}r_{Ru}(u+v)\underset{v\in(Ru)^{\perp}}{=}\frac{1}{2}(u-v)=y$\\
If $\beta(v,v)\in R^{*}$ then $Rv\subseteq M$ non-degenerate subspace
$r_{(Rv)^{\perp}}(x)=\frac{1}{2}r_{(Rv)^{\perp}}(u+v)\underset{\underset{v\in((Rv)^{\perp})^{\perp}=Rv}{u\in(Rv)^{\perp}}}{=}\frac{1}{2}(u-v)=y$
\end{itemize}
\end{proof}
\begin{thm}
Let $(R,m,k)$ be a local ring with $2\in R^{*}$. Then Witt cancellation
holds for $R$. That is $\forall M,N,P$ symmetric inner product space
over $R$ we have $M\perp P\cong N\perp P\Rightarrow M\cong N$.\end{thm}
\begin{proof}
Let $M,N,P$ be symmetric inner product spaces over $R$ such that
$M\perp P\cong N\perp P$. By our assumption $R$ local and $2\in R^{*}$
$\Rightarrow P=\left\langle u_{1}\right\rangle \perp\dots\perp\left\langle u_{r}\right\rangle $
for $u_{i}\in R^{*}$. Thus it suffices to show $M\perp\left\langle u\right\rangle \cong N\perp\left\langle u\right\rangle \Rightarrow M\cong N$.
Let $f:M\perp\left\langle u\right\rangle \overset{\cong}{\to}N\perp\left\langle u\right\rangle $
be an isometry and let $x\in M\perp\left\langle u\right\rangle $
and $y\in N\perp\left\langle u\right\rangle $ be a generator for
$\left\langle u\right\rangle $ i.e. $M\perp\left\langle u\right\rangle =M\perp Rx$
and $N\perp\left\langle u\right\rangle =N\perp Ry$. $\beta(x,x)_{M\perp\left\langle u\right\rangle }=u=\beta(y,y)_{N\perp\left\langle u\right\rangle }$.
$f$ isometry: $\beta(f(x),f(x))_{N\perp\left\langle u\right\rangle }=\beta(x,x)_{M\perp\left\langle u\right\rangle }=u=\beta(y,y)_{N\perp\left\langle u\right\rangle }\Rightarrow f(x),y\in N\perp\left\langle u\right\rangle $
satisfy hypotheses of lemma \ref{lem:reflection_map} $\Rightarrow\exists$
reflection $r:N\perp\left\langle u\right\rangle \to N\perp\left\langle u\right\rangle $
such that $r(f(x))=y$$\Rightarrow r\circ f:M\perp\left\langle u\right\rangle \to N\perp\left\langle u\right\rangle $
is an isometry such that $r\circ f:\underset{x\mapsto y}{Rx\overset{\cong}{\to}Ry}$.
$\Rightarrow r\circ f:\underset{M}{\underbrace{(Rx)^{\perp}}}\overset{\cong}{\to}\underset{N}{\underbrace{(Ry)^{\perp}}}\Rightarrow M\underset{r\circ f}{\overset{\cong}{\to}}N$
\end{proof}

\subsection{Symmetric Inner Product space over $\rr$}

Any symmetric inner product space $M$ over $\rr$ has an orthogonal
basis $M=\left\langle u_{1}\right\rangle \perp\dots\left\langle u_{n}\right\rangle ,n=\dim_{\rr}(M),u_{i}\in\rr^{*}=\rr\setminus\{0\}$.
If $u>0$ then $u=a^{2}$ for some $a\in\rr,\left\langle u\right\rangle \underset{\leftarrow a}{\cong}\left\langle 1\right\rangle $.
If $u<0$ then $u=-a^{2}$and $\left\langle u\right\rangle \underset{\leftarrow a}{\cong}\left\langle -1\right\rangle $.
So $M=r\left\langle 1\right\rangle \perp s\left\langle -1\right\rangle $
(and $r+s=\dim_{\rr}(M)$)
\begin{prop}[Inertia Theorem]
Over $\rr$ we have $r\left\langle 1\right\rangle \perp s\left\langle -1\right\rangle \cong m\left\langle 1\right\rangle \perp n\left\langle -1\right\rangle \Rightarrow r=m$
and $s=n$\end{prop}
\begin{proof}
The equation implies that $r+s=\dim_{\rr}(M)=n+m$. Assume without
loss of generality that $r\leq m$ then $n\leq s$. Witt cancellation
tells us that $(s-n)\left\langle -1\right\rangle \cong(m-r)\left\langle 1\right\rangle $.
Note that if $m-r=s-n\neq0$ then $\forall x\neq0\in(s-n)\left\langle -1\right\rangle $
we have $\beta(x,x)=-\sum x_{i}^{2}<0$. However $\forall x\neq0\in(m-r)\left\langle 1\right\rangle $
we have $\beta(x,x)=\sum x_{i}^{2}>0$. Contradiction. Hence $s-n=m-r=0$\end{proof}
\begin{cor}
The numbers $r,s$ in $M\cong r\left\langle 1\right\rangle +s\left\langle -1\right\rangle $
do not depend on the choice of an orthogonal basis for $M$\end{cor}
\begin{defn}
If $M\cong r\left\langle 1\right\rangle \perp s\left\langle -1\right\rangle $
over $\rr$ then $r=i^{+}M$ is called the \emph{positive index of
$M$. }$s=i^{-}M$ is called the \emph{negative index of $M$ }and
$i^{+}M-i^{-}M=r-s=\mbox{sgn}(M)$ is called the \emph{signature of
$M$}
\end{defn}
We have showed that if (over $\rr$) $M\cong N$ then $i^{+}N=i^{+}M,i^{-}N=i^{-}M$
and $\sgn(N)=\sgn(M)$
\begin{cor}
Two symmetric inner product-spaces $M,N$ over $\rr$ are isometric
$M\cong N\iff i^{+}M=i^{+}N,i^{-}M=i^{-}N\iff\mbox{rank }M=\mbox{rank }N,\sgn M=\sgn N$
\end{cor}

\subsection{Witt chain equivalence theorem}
\begin{notation*}
Let $u_{1},\dots,u_{l}\in R^{*}$ write $\left\langle u_{1},\dots,u_{l}\right\rangle $
for $\left\langle u_{1}\right\rangle \perp\dots\perp\left\langle u_{l}\right\rangle =\left\langle \begin{pmatrix}u_{1} &  & 0\\
 & \ddots\\
0 &  & u_{l}
\end{pmatrix}\right\rangle $. We say $\left\langle u_{1},\dots,u_{l}\right\rangle $ is a diagonal
form\end{notation*}
\begin{defn}
We say $(M,\beta)$ represent $a\in R$ if $\exists x\in M$ such
that $\beta(x,x)=a$\end{defn}
\begin{example*}
A diagonal form $\left\langle u_{1},\dots,u_{l}\right\rangle $ represents
$u_{1},\dots,u_{l},u_{1}+u_{2},\dots$. The equation $a=u_{1}x_{1}^{2}+\dots+u_{l}x_{l}^{2}$
has a solution $x_{1},\dots,x_{l}\in R\iff a$ is represented by $\left\langle u_{1},\dots,u_{l}\right\rangle $\end{example*}
\begin{lem}
\label{lem:ClassfcnBinaryQf}Let $R$ be a local ring (or a ring in
which every direct summand of a finitely generated free module is
free) Let $\left\langle a,b\right\rangle $ and $\left\langle c,d\right\rangle $
be non-degenerate diagonal forms ($a,b,c,d\in R^{*}$). Then $\left\langle a,b\right\rangle \cong\left\langle c,d\right\rangle \iff ab=cd\in R^{*}/(R^{*})^{2}$
and $\exists e\in R^{*}$ which represent $\left\langle a,b\right\rangle $
and $\left\langle c,d\right\rangle $\end{lem}
\begin{proof}
``$\Rightarrow$'': We've already done. (They obviously need the
same determinant modulo squares, and need to represent the same numbers)

``$\Leftarrow$'': $e\in R^{*}$ represents $\left\langle a,b\right\rangle $
and $\left\langle c,d\right\rangle \Rightarrow\exists x,y\in R^{2}$
such that $\beta(x,x)_{\left\langle a,b\right\rangle }=e=\beta(y,y)_{\left\langle c,d\right\rangle }\Rightarrow\left\langle a,b\right\rangle =\underset{\left\langle e\right\rangle \perp\left\langle u_{1}\right\rangle }{\underbrace{\underset{\mathrm{non-degenerat}}{\underbrace{Rx}}\perp\underset{\mathrm{rank}1}{\underbrace{(Rx)^{\perp}}}}}$
and $\left\langle c,d\right\rangle =\underset{\left\langle e\right\rangle \perp\left\langle u_{2}\right\rangle }{\underbrace{\underset{\mathrm{non-degenerat}}{\underbrace{Ry}}\perp\underset{\mathrm{rank}1}{\underbrace{(Ry)^{\perp}}}}}$
with $u_{1},u_{2}\in R^{*}$. Now $e\cdot u_{1}=\det(\left\langle e\right\rangle \perp\left\langle u_{1}\right\rangle )=\det\left\langle a,b\right\rangle =\det\left\langle c,d\right\rangle =\det\left\langle e,u_{2}\right\rangle =eu_{2}\in R^{*}/(R^{*})^{2}\Rightarrow eu_{1}=eu_{2}g^{2}$
for some $g\in R^{*}$ $\Rightarrow u_{1}=u_{2}g^{2}$ for some $g\in R^{*}$,
$\left\langle u_{1}\right\rangle \cong\left\langle u_{2}\right\rangle \Rightarrow\left\langle a,b\right\rangle \cong\left\langle e,u_{1}\right\rangle \cong\left\langle e,u_{2}\right\rangle \cong\left\langle c,d\right\rangle $\end{proof}
\begin{defn}
Two non-degenerate diagonal forms $\left\langle a_{1},\dots,a_{n}\right\rangle $
and $\left\langle b_{1},\dots,b_{n}\right\rangle $ (of the same rank
$n$) are called \emph{simply (chain) equivalent }(with notation $\approx_{s}$)
if either 
\begin{itemize}
\item $n\geq2$ and $\exists1\leq i<j\leq n$ such that $\left\langle a_{i},a_{j}\right\rangle \cong\left\langle b_{i},b_{j}\right\rangle $
and $a_{l}=b_{l}\,\forall l\neq i,j$ 
\item or $n=1$ $\left\langle a_{1}\right\rangle \cong\left\langle b_{1}\right\rangle $
\end{itemize}
Two non-degenerate diagonal forms $\left\langle a_{1},\dots,a_{n}\right\rangle $
and $\left\langle b_{1},\dots,b_{n}\right\rangle $ are \emph{chain
equivalent (with notation $\approx)$} if $\exists M_{1},\dots,M_{r}$
non degenerated diagonal forms of rank $n$ such that $M_{1}=\left\langle a_{1},\dots,a_{n}\right\rangle ,M_{r}=\left\langle b_{1},\dots,b_{n}\right\rangle $
and $M_{1}\approx_{s}M_{2}\approx_{s}\dots\approx_{s}M_{r}$\end{defn}
\begin{rem*}
$\left\langle a_{1},\dots,a_{n}\right\rangle \approx\left\langle b_{1},...,b_{n}\right\rangle \Rightarrow\left\langle a_{1},\dots,a_{n}\right\rangle \cong\left\langle b_{1},\dots,b_{n}\right\rangle $\end{rem*}
\begin{example*}
$\sigma\in\sum_{n}=$ permutation group on $n$ letters. $\left\langle a_{1},\dots,a_{n}\right\rangle \approx\left\langle a_{\sigma(1)},\dots a_{\sigma(n)}\right\rangle $
because true for transpositions because $\underset{\left\langle a\right\rangle \perp\left\langle b\right\rangle \cong\left\langle b\right\rangle \perp\left\langle a\right\rangle }{\underbrace{\left\langle \begin{pmatrix}a & 0\\
0 & b
\end{pmatrix}\right\rangle \cong\left\langle \begin{pmatrix}b & 0\\
0 & a
\end{pmatrix}\right\rangle }}\Rightarrow$ true for all $\sigma\in\sum_{n}$ because $\sum_{n}$ is generated
by transpositions.\end{example*}
\begin{namedthm}[Witt's Chain Equivalence Theorem]
\label{Chain equivalence form} Let $\left\langle a_{1},\dots,a_{n}\right\rangle $
and $\left\langle b_{1},\dots,b_{n}\right\rangle $ be non-degenerate
diagonal forms over a local ring $R$ with $2\in R^{*}$ then $\left\langle a_{1},\dots a_{n}\right\rangle \approx\left\langle b_{1},\dots,b_{n}\right\rangle \iff\left\langle a_{1},\dots,a_{n}\right\rangle \cong\left\langle b_{1},\dots,b_{n}\right\rangle $\end{namedthm}
\begin{proof}
``$\Rightarrow$'': We seen this in the remark

``$\Leftarrow$'': We use induction on $n$. For $n=0$ there is
nothing to say. $n=1,n=2$ is true by definition of chain equivalence.

Assume $n\geq3$ and $\left\langle a_{1},\dots,a_{n}\right\rangle \cong\left\langle b_{1},\dots,b_{n}\right\rangle $

Claim: $\exists$ non-degenerate diagonal form $\left\langle c_{1},\dots,c_{n}\right\rangle \approx\left\langle a_{1},\dots,a_{n}\right\rangle $
with $c_{1}=b_{1}$.

Note that the claim implies the theorem because $\left\langle a_{1},\dots,a_{n}\right\rangle \approx\left\langle c_{1},\dots,c_{n}\right\rangle ,\left\langle c_{1},\dots,c_{n}\right\rangle =\left\langle b_{1},c_{2},\dots,c_{n}\right\rangle \cong\left\langle b_{1},\dots,b_{n}\right\rangle \underset{\mathrm{Witt\, cancellation}}{\Rightarrow}\left\langle c_{2},\dots,c_{n}\right\rangle \cong\left\langle b_{2},\dots b_{n}\right\rangle $.
So by hypothesis $\Rightarrow\left\langle c_{2},\dots,c_{n}\right\rangle \approx\left\langle b_{2},\dots,b_{n}\right\rangle $
hence $\left\langle a_{1},\dots,a_{n}\right\rangle \approx\left\langle \underset{b_{1}}{\underbrace{c_{1}}},c_{2},\dots,c_{n}\right\rangle \approx\left\langle b_{1},\dots,b_{n}\right\rangle $.

Proof of claim: Let $P=\{(c,p)|c=\left\langle c_{1},\dots,c_{n}\right\rangle ,1\leq p\leq n$
such that $\left\langle c_{1},\dots,c_{p}\right\rangle $ represents
$b_{1}\}$. Note that $(\left\langle a_{1},\dots,a_{n}\right\rangle ,n)\in P$
so $P\neq\emptyset$. Let $p=\min\{l|\exists(c,l)\in P\}$ well defined
and $1\leq p\leq n$ because $P\neq\emptyset$. Choose $(c,p)$ with
$p$ minimal as above. $c=\left\langle c_{1},\dots,c_{n}\right\rangle $
has property that $\left\langle c_{1},\dots,c_{p}\right\rangle $
represent $b_{1}$$\Rightarrow\exists x_{1},\dots x_{p}\in R$ such
that $b_{1}=c_{1}x_{1}^{2}+\dots+c_{p}x_{p}^{2}\in R^{*}$. Assume
that $p\geq2$. If $\forall i\neq j$ $c_{i}x_{i}^{2}+c_{j}x_{j}^{2}\in m=R\setminus R^{*}$
then $\underset{\in m}{\underbrace{(c_{1}x_{1}^{2}+c_{2}x_{2}^{x})}}+\underset{\in m}{\underbrace{(c_{2}x_{2}^{2}+c_{3}x_{3}^{2})}}+\dots+\underset{\in m}{\underbrace{(c_{p}x_{p}^{2}+c_{1}x_{1}^{2})}}=2b_{1}$
which is a contradiction as $2\in R^{*}$ and $b_{1}\in R^{*}\Rightarrow\exists i<j$
such that $c_{i}x_{i}^{2}+c_{j}x_{j}^{2}\in R^{*}$. Since $\left\langle c_{1},\dots,c_{p}\right\rangle \approx\left\langle c_{\sigma(1)},\dots,c_{\sigma(p)}\right\rangle \,\forall\sigma\in\sum_{p}$
we can assume $d=c_{1}x_{1}^{2}+c_{2}x_{2}^{2}\in R^{*}$. Then $\left\langle c_{1},c_{2}\right\rangle \cong\left\langle d,dc_{1}c_{2}\right\rangle $
because both represent $d\in R^{*}$ and both have the same determinant
(in $R^{*}/R^{2*}$), $\Rightarrow\left\langle c_{1},\dots,c_{p}\right\rangle \approx\left\langle d,dc_{1}c_{2},c_{3},\dots,c_{p}\right\rangle \approx\left\langle d,c_{3},\dots,c_{p},dc_{1}c_{2}\right\rangle $
but $\left\langle d,c_{3},\dots,c_{p}\right\rangle $represent $b_{1}$
because $b_{1}=\underset{d\cdot1^{2}}{\underbrace{c_{1}x_{1}^{2}+c_{2}x_{2}^{2}}}+\dots+c_{p}x_{p}^{2}$$\Rightarrow(\left\langle d,c_{1},\dots,c_{p},dc_{1}c_{2},c_{p+1},\dots,c_{n}\right\rangle ,p-1)\in P$
which contradicts minimality of $p$$\Rightarrow p=1\Rightarrow\exists\left\langle c_{1},\dots,c_{p}\right\rangle \approx\left\langle a_{1},\dots,a_{n}\right\rangle $and
$\left\langle c_{1}\right\rangle $ represents $b_{1}$, i.e., $b_{1}=c_{1}x^{2}\Rightarrow\left\langle b_{1}\right\rangle \cong\left\langle c_{1}\right\rangle $$\Rightarrow\exists\left\langle b_{1},c_{2},\dots,c_{p}\right\rangle \approx\left\langle a_{1},\dots,a_{n}\right\rangle $
\end{proof}

\subsection{Witt Groups:}

Goal: Define $W(R)=$ abelian group to be $\{$isometry classes of
symmetric inner product space over $R\}$/metabolic forms ($=$hyperbolic
if $\frac{1}{2}\in R$) with group operation given by $\perp$
\begin{defn}
A symmetric inner product space $(M,\beta$) is called \emph{metabolic}
(or split) if $\exists$ direct summand $N\subseteq M$ such that
$N=N^{\perp}$. Such a direct summand $N$ is called \emph{Lagrangian.}\end{defn}
\begin{rem*}
$N\subseteq M$ is a direct summand if $\exists P\subseteq M$ such
that $N\oplus P=M$ (i.e. $N+P=M$ and $N\cap P=0$)\end{rem*}
\begin{example*}
{}
\begin{itemize}
\item $\hh=\left\langle \begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}\right\rangle $ is metabolic with Lagrangian $\begin{pmatrix}1\\
0
\end{pmatrix}:R\to R^{2}$ defined by $x\mapsto\begin{pmatrix}x\\
0
\end{pmatrix}$. i.e. $L=\{(x,0)\in R^{2}|x\in R\}\subset\hh$ is a Lagrangian. Because 

\begin{itemize}
\item $L$ is a direct summand with complement $P=\{(0,y)|y\in R\}\subseteq\hh$,
\item $L^{\perp}=\{(x,y)\in R^{2}|\underset{zy=0\,\forall z\iff y=0}{\underbrace{\underset{\in L}{\underbrace{(z,0})}\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}\begin{pmatrix}x\\
y
\end{pmatrix}=0}}\,\forall z\in R\}=\{(x,y)\in R^{2}|y=0\}=L$
\end{itemize}
\item $I_{n}$ is the identity matrix in $M_{n}(R)$. $A\in M_{n}(R)$ then
$\left\langle \begin{pmatrix}0 & I_{n}\\
I_{n} & A_{n}
\end{pmatrix}\right\rangle $ is metabolic with Lagrangian the image of the map $R^{n}\overset{\begin{pmatrix}I_{n}\\
0
\end{pmatrix}}{\longrightarrow}R^{n}\oplus R^{n}$ (proof is the same as above) 
\item {}

\begin{lem}
Let $(M,\beta)$ be a symmetric inner product space then $(M,\beta)\perp(M,-\beta)$
is metabolic. \end{lem}
\begin{proof}
The submodule $L=\{(x,x)\in M\oplus M|x\in M\}\subseteq M\perp M$
is a Lagrangian for $(M.\beta)\perp(M,-\beta)$ because
\begin{itemize}
\item $L$ is a direct summand with complement $P=\{(y,0)\in M^{2}|y\in M\}$
as $L\cap P=0$ and every element $(a,b)\in M^{2}$ is $(a,b)=\underset{\in L}{\underbrace{(b,b)}}+\underset{\in P}{\underbrace{(a-b,0)}}$
so $L\oplus P=L+P=M^{2}$
\item $L=L^{\perp}$ because let $\begin{pmatrix}a\\
b
\end{pmatrix}\in L^{\perp}\subseteq M^{2}\iff\beta(a,x)-\beta(b,x)=0\,\forall x\in M\iff\beta(a-b,x)=0\,\forall x\in M\underset{\beta\mathrm{\, non\, degenerate}}{\iff}a-b=0\Rightarrow a=b\Rightarrow\begin{pmatrix}a\\
b
\end{pmatrix}\in L$. Hence $L^{\perp}=L$
\end{itemize}
\end{proof}
\end{itemize}
\end{example*}
\begin{defn}
A free symmetric inner product space is called \emph{hyperbolic} if
it is isometric to $\hh^{n}$\end{defn}
\begin{note*}
$M,N$ are metabolic (or hyperbolic) then so is $M\perp N$. If $M,N$
are metabolic with Lagrangian $L_{1}\subseteq M,L_{2}\subseteq N$
then $M\perp N$ has Lagrangian $L_{1}\perp L_{2}\subseteq M\oplus N$.

$\left\langle \begin{pmatrix}0 & I_{n}\\
I_{n} & 0
\end{pmatrix}\right\rangle \cong\hh^{n}$ (by change of basis). Let the basis of the first one to be $e_{1},\dots,e_{n},e_{n+1},\dots,e_{2n}$
then the basis of the second one is $(e_{1},e_{n+1}),(e_{2},e_{n+2}),\dots(e_{n},e_{2n})$
where each pairs gives a copy of $\hh$\end{note*}
\begin{lem}
If $2\in R^{*}$ then for all $A\in M_{n}(R)$, $\left\langle \begin{pmatrix}0 & I_{n}\\
I_{n} & A
\end{pmatrix}\right\rangle \cong\left\langle \begin{pmatrix}0 & I_{n}\\
I_{n} & 0
\end{pmatrix}\right\rangle \cong\hh^{n}$\end{lem}
\begin{proof}
We need to find a base change, that is $\exists X\in M_{n}(R)$ which
is invertible such that $X\begin{pmatrix}0 & I_{n}\\
I_{n} & A
\end{pmatrix}X^{T}=\begin{pmatrix}0 & I_{n}\\
I_{n} & 0
\end{pmatrix}$. Take $X=\begin{pmatrix}I_{n} & 0\\
-\frac{1}{2}A & I_{n}
\end{pmatrix}$ which is invertible with inverse $X^{-1}=\begin{pmatrix}I_{n} & 0\\
\frac{1}{2}A & I_{n}
\end{pmatrix}$\end{proof}
\begin{lem}
Let $R$ be a ring for which all finitely generated projective $R$-module
are free (e.g.., $R$ local or $R=\zz$) then any metabolic inner
product space $(M,\beta)$ is isometric to $\left\langle \begin{pmatrix}0 & I_{n}\\
I_{n} & A
\end{pmatrix}\right\rangle $ for some $n\in\nn,\mbox{ and }A\in M_{n}(R)$. If moreover, $2\in R^{*}$
then every metabolic space is hyperbolic.\end{lem}
\begin{proof}
Let $(M,\beta)$ be metabolic with Lagrangian $L\subseteq M$. $L\subseteq M$
being a direct summand $\Rightarrow P\subseteq M$ such that $L\cap P=0$
and $L+P=M$. By assumption $M$ projective $\Rightarrow P,L$ projective
$\underset{\mathrm{assumption\, on\,}R}{\Rightarrow}P,L$ are free.
In a basis for $L$, and $P$, the inner product space $\beta$ has
inner product matrix $\begin{pmatrix}0 & B\\
B^{T} & C
\end{pmatrix}$, with $C=C^{T}$. The upper left corner is $0$ because $L=L^{\perp}$
we have $\beta(x,x)=0\,\forall x\in L$.

Claim: The matrix $B$ is invertible. 

Proof of claim: $B$ is the matrix of the linear map $\underset{x\mapsto\beta(x,-)}{P\to L^{*}}=\Hom_{R}(L,R)$
with respect to the basis of $P$ and the dual basis of $L$. Need
to show $P\to L^{*}$ defined by $x\mapsto\beta(x,-)$ is an isomorphism.

Injectivitiy: $x\in P:\beta(x,y)=0\,\forall y\in L\Rightarrow x\in L^{\perp}=L\Rightarrow x\in L\cap P=0\Rightarrow x=0$

Surjectivity: Let $\phi\in L^{*},\phi:L\to R$. Define $\bar{\phi}:M=L\oplus P\to P$
by $(x,y)\mapsto\phi(x)$. Now $\beta$ is non-degenerate $\Rightarrow\exists\underset{\in L}{a},\underset{\in P}{b}\in M=L\oplus P$
such that $\bar{\phi}(x,y)=\beta(a+b,x+y)\,\forall x\in L,y\in P$$\Rightarrow\phi(x)=\bar{\phi}(x,0)=\beta(a+b,x)=\underset{=0\mathrm{\, since\,}a,x\in L=L^{\perp}}{\underbrace{\beta(a,x)}}+\beta(b,x)=\beta(b,x)\,\forall x\in L$.
So $b\in P$ is sent to $\phi$ under the map $P\to L^{*}$. This
shows surjectivity. 

Notice that $\begin{pmatrix}0 & I_{n}\\
I_{n} & (B^{T})^{-1}CB^{-1}
\end{pmatrix}=\begin{pmatrix}1 & 0\\
0 & (B^{T})^{-1}
\end{pmatrix}\begin{pmatrix}0 & B\\
B^{T} & C
\end{pmatrix}\begin{pmatrix}1 & 0\\
0 & B^{-1}
\end{pmatrix}\Rightarrow M\cong\left\langle \begin{pmatrix}0 & B\\
B^{T} & C
\end{pmatrix}\right\rangle \cong\left\langle \begin{pmatrix}0 & I_{n}\\
I_{n} & A
\end{pmatrix}\right\rangle $. For $A=(B^{T})^{-1}CB^{-1}$\end{proof}
\begin{note*}
If $2\in R^{*}$ then $\left\langle \begin{pmatrix}0 & I_{n}\\
I_{n} & A
\end{pmatrix}\right\rangle \cong\hh^{n}$\end{note*}
\begin{cor}
Over a local ring $R$, every metabolic space has even dimension and
if $2\in R^{*}$, $R$ local, then every metabolic space is hyperbolic\end{cor}
\begin{defn}
Let $M,N$ be symmetric inner product spaces over $R$ then $M$ are
$N$ are called \emph{Witt Equivalent} ($M\sim N$) if $\exists$
metabolic spaces $P,Q$ such that $M\perp P\cong N\perp Q$. Denote
by $W(R)$ be the set of Witt equivalence classes $[M]$ of symmetric
inner product spaces $M$ over $R$\end{defn}
\begin{lem}[\textbf{Definition}]
 Orthogonal sum $\perp$ makes $W(R)$ into an abelian group with
$0=[0],[M]+[N]=[M\perp N]$ and $-[M,\beta]=[M,-\beta]$. $W(R)$
is called the \emph{Witt group of $R$.}\end{lem}
\begin{proof}
{}
\begin{itemize}
\item ``$+$'' is well defined because if $M\sim M',N\sim N'$ then $\exists P,P',Q,Q'$
metabolic such that $M\perp P\cong M'\perp P',N\perp Q\cong N'\perp Q'$.
Then $(M\perp N)\perp\underset{\mathrm{metabolic}}{\underbrace{(P\perp Q)}}\cong(M'\perp N')\perp\underset{\mathrm{metabolic}}{\underbrace{(P'\perp Q')}}\Rightarrow M\perp N\sim M'\perp N'$
\item We have $[M]+[N]=[M\perp N]=[N\perp M]=[N]+[M]$ (since $M\perp N\cong N\perp M$)
and the group law is commutative
\item $[0]+[M]=[0+M]=[M]$ because $0\perp M\cong M$
\item $[M,\beta]+[M,-\beta]=[(M,\beta)\perp(M,-\beta)]=0$ because $(M,\beta)\perp(M,-\beta)$
is metabolic for any inner product space $(M,\beta)$.
\end{itemize}
\end{proof}
\begin{rem*}
$W:$ (commutative) rings $\to$ abelian groups, defined by $R\mapsto W(R)$
is a functor. For $f:R\to S$ a ring homomorphism, we define a map
of abelian groups $W(f):W(R)\to W(S)$ by $[M,\beta]\mapsto[M_{S},\beta_{S}]$
where $M_{S}=S\otimes_{R}M$ and $\beta_{S}:M_{S}\times M_{S}\to S$
is defined $s_{1}\otimes x_{1},s_{2}\otimes x_{2}\mapsto s_{1}s_{2}\beta(x_{1},x_{2})$.
Note that if $(M,\beta)$ is non-degenerate then so is $(M_{S},\beta_{S}):$
if $M$ is free choose an $R$-basis of $M$, say $x_{1},\dots,x_{n}\in M$
then $M_{S}$ is free with $S$-basis $1\otimes x_{1},\dots,1\otimes x_{n}$.
Then $(M,\beta)$ non-degenerate $\iff(\beta(x_{i},x_{j}))\in M_{n}(R)$
is invertible $\underset{f:R^{*}\to S^{*}}{\Rightarrow}(f(\beta(x_{i},x_{j}))=(\beta_{S}(1\otimes x_{i},1\otimes x_{j}))\in M_{n}(S)\mathrm{\, is\, invertible}\iff$$(M_{S},\beta_{S})$
is non-degenerate. In the case $(M,\beta)$ is projective do it as
an exercise
\end{rem*}
For $R\overset{g}{\to}S\overset{f}{\to}T$ ring homomorphism, note
that $W(f)\circ W(g)=W(f\circ g)$ because $T\otimes_{S}(S\otimes_{R}M)\cong\underset{\underset{x\otimes y\mapsto xf(y)}{\cong T}}{\underbrace{(T\otimes_{S}S)}}\otimes_{R}M\cong T\otimes_{R}M$
\begin{prop}
Let $R$ be a local ring with $2\in R^{*}$. Then two symmetric inner
product spaces $M,N$ are isometric if and only if $\mbox{rank }M=\mbox{rank }N$
and $[M]=[N]\in W(R)$\end{prop}
\begin{proof}
``$\Rightarrow$'': Obvious

``$\Leftarrow$'': $[M]=[N]\in W(R)\Rightarrow M\sim N\Rightarrow\exists$
metabolic $P,Q$ such that $M\perp P\cong N\perp Q$. $R$ local,
$2\in R^{*}\Rightarrow$ metabolic $=$ hyperbolic so $P\cong\hh^{p},Q\cong\hh^{q}$.
Now $\mbox{rank }M=\mbox{rank }N$ and $\mbox{rank }M\perp P=\mbox{rank }N\perp Q\Rightarrow p=q\Rightarrow M\perp\hh^{p}\cong N\perp\hh^{p}$
so by Witt cancellation $\Rightarrow M\cong N$.\end{proof}
\begin{defn}
Let $ $$R$ be a local ring. The rank homomorphism $\mbox{rk}:W(R)\to\zz/2\zz$
is defined by $[M]\mapsto\mbox{rank }M$. Note this map is well defined
as $M\sim0\iff\exists P$ metabolic such that $M\perp P$ is metabolic$\Rightarrow M\perp P$
and $P$ have even rank $\Rightarrow\rk M=0\in\zz/2\zz$. The rank
map is surjective because $\rk(\left\langle 1\right\rangle )=1$. 

Let $I(R)=\ker(\rk)=$set of equivalence classes of even rank inner
product spaces 

The \emph{discriminant} is the homomorphism $\mbox{disc: }I(R)\to R^{*}/R^{2*}$
defined by $[M]\mapsto(-1)^{\frac{\rk M}{2}}\det M$ which is well
defined because $\mbox{disc }P=1$ for $P$ metabolic as $\det\begin{pmatrix}0 & I_{n}\\
I_{n} & A
\end{pmatrix}=(-1)^{n}$
\end{defn}
Note that $\mbox{disc}$ map is surjective because $\mbox{disc}(\left\langle u,-1\right\rangle )=u\,\forall u\in R^{*}$
\begin{prop}
Let $F$ be a field in which every element is a square \emph{(}e.g.
$F$ algebraic closed, or $\mbox{char }F=2$ and $F$ perfect, e.g.,
finite and $\mbox{char }F=2$\emph{)} then $\rk:W(F)\overset{\cong}{\to}\zz/2\zz$
is an isomorphism.\end{prop}
\begin{proof}
$\rk:W(F)\to\zz/2\zz$ is surjective (for any commutative ring) since
$\left\langle 1\right\rangle \mapsto1$. Recall that every symmetric
inner product space over $F$ a field is isometric to $\left\langle u_{1}\right\rangle \perp\dots\perp\left\langle u_{l}\right\rangle \perp N_{1}\perp\dots\perp N_{r}$
with $N_{i}=\hh$ (in the case of a field). So $W(F)$ is generated
by $\left\langle u\right\rangle ,u\in F^{*}/F^{2*}$ as an abelian
group. Consider the map $\zz\to W(F)$ defined by $1\mapsto\left\langle 1\right\rangle $.
Since $\left\langle 1\right\rangle +\left\langle 1\right\rangle =\left\langle 1\right\rangle +\left\langle -1\right\rangle $
(as $-1\in F^{2*}$), we have $\left\langle 1\right\rangle +\left\langle 1\right\rangle =0\Rightarrow$
This map factors as $\zz/2\zz\to W(F)$ with $1\mapsto\left\langle 1\right\rangle $.
As $W(F)$ is generated by $\left\langle u\right\rangle ,u\in F^{*}/F^{2*}=\{1\}$
this means that the map $\zz/2\zz\to W(F)$ is surjective and it is
injective as $\zz/2\zz\twoheadrightarrow W(F)\overset{\rk}{\twoheadrightarrow}\zz/2\zz$
sends $\underset{\id}{\underbrace{1\mapsto\left\langle 1\right\rangle \mapsto1}}\Rightarrow\zz/\zz2\overset{\cong}{\to}W(F)\Rightarrow W(F)\overset{\rk}{\underset{\cong}{\to}}\zz/2\zz$\end{proof}
\begin{cor}
$W(\ff_{q})=\zz/2\zz$ for $q$ even, $\rk:W(\cc)\overset{\cong}{\to}\zz/2\zz$\end{cor}
\begin{example*}
$W(\rr)\to\zz$ defined by $[M]\mapsto\sgn M$ is well defined because
$\sgn(\hh)=0$

Claim: $\sgn:W(\rr)\overset{\cong}{\to}\zz$ is an isomorphism

Surjective: $\sgn(n\left\langle 1\right\rangle )=n\sgn(\left\langle 1\right\rangle )=n\cdot1=n$

Injectivitiy: Every symmetric inner product space over $\rr$ is $M\cong n\left\langle 1\right\rangle +m\left\langle -1\right\rangle $.
If $\sgn M=\sgn(n\left\langle 1\right\rangle +m\left\langle -1\right\rangle )=n-m$
then $n=m\Rightarrow M=n(\left\langle 1\right\rangle +\left\langle -1\right\rangle )=0\in W(\rr)$
\end{example*}
Recall:{}
\begin{itemize}
\item $I(F)=\ker(W(F)\overset{\rk}{\to}\zz/2\zz)=$ ``\emph{fundamental
ideal}''
\item $I(F)\overset{\mbox{disc}}{\to}F^{*}/F^{2*}$ map of abelian groups
defined by $M\mapsto(-1)^{\frac{\rk M}{2}}\det M$\end{itemize}
\begin{note*}
The disc map extends to all of $W(F)$ by $W(F)\to F^{*}/F^{2*}$
defined by $M\mapsto(-1)^{\frac{r(r-1)}{2}}\det M$, where $r=\rk M$,
but, in general, this is not a map of abelian groups so we don't often
use this.\end{note*}
\begin{prop}
Let $F$ be a finite field. Then the discriminant map is an isomorphism:
$I(F)\overset{\cong}{\to}F^{*}/F^{2*}$\end{prop}
\begin{proof}
If $\mbox{char }F=2$ this is true because both sides are equal to
$0$. (Since $F$ finite and $\mbox{char }F=2\Rightarrow F^{*}/F^{2*}=0$)

So assume $\mbox{char }F$ is odd. We have to prove the following
special case

Claim: $\left\langle a,b\right\rangle \cong\left\langle ab,1\right\rangle $

The claim implies the proposition: define the map $\rho:F^{*}/F^{2*}\to I(F)$
by $a\mapsto\left\langle a,-1\right\rangle $ this is easily seen
to be a well defined map of \uline{sets}. This is a map of abelian
groups because $\rho(ab)=\left\langle ab,-1\right\rangle =\left\langle ab\right\rangle +\left\langle -1\right\rangle =\left\langle ab\right\rangle +\left\langle 1\right\rangle +\left\langle -1\right\rangle +\left\langle -1\right\rangle =\left\langle ab,1\right\rangle +\left\langle -1\right\rangle +\left\langle -1\right\rangle \underset{\mathrm{claim}}{=}\left\langle a,b\right\rangle +2\left\langle -1\right\rangle =\left\langle a,-1\right\rangle +\left\langle b,-1\right\rangle =\rho(a)+\rho(b)$.
The maps is surjective because every $\left\langle a_{1},\dots,a_{2n}\right\rangle \in I(F)$
is $\left\langle a_{1},\dots,a_{2n}\right\rangle =\left\langle a_{1}\dots a_{2n},1,1,\dots,1\right\rangle \underset{\mathrm{claim}}{=}\left\langle a_{1}\dots a_{2n},-1\right\rangle +2n\left\langle 1\right\rangle =\rho(a_{1}\dots a_{2n})+n\rho(-1)$
because $\rho(-1)=\left\langle -1,-1\right\rangle \underset{\mathrm{claim}}{=}\left\langle 1,1\right\rangle $.
The map $\rho$ is injective because $F^{*}/F^{2*}\overset{\rho}{\to}I(F)\overset{\mbox{disc }}{\to}F^{*}/F^{2*}$
defined by $\underset{\id}{\underbrace{a\mapsto\left\langle a,-1\right\rangle \mapsto a}}$,
hence we are done.

Proof of claim: Recall: $\left\langle a,b\right\rangle \cong\left\langle c,d\right\rangle \iff ab=cd\in F^{*}/F^{2*}$
and $\exists e\in F$ such that $e$ is represented by both forms.
Obviously $\left\langle a,b\right\rangle \cong\left\langle ab,1\right\rangle $
has the same determinant, so the claim is equivalence to the fact,
since $\left\langle ab,1\right\rangle $represent $1$, that $\left\langle a,b\right\rangle $
represent 1, i.e., $\exists x,y\in F$ such that $1=ax^{2}+by^{2}$.
This follows from the following lemma
\begin{lem}
Let $F$ be a finite field of order $q=$odd. Then $\forall a,b\in F^{*}$
the equation $1=ax^{2}+by^{2}$ has a solution $x,y\in F$\end{lem}
\begin{proof}
Need to find $x,y$ such that $1-by^{2}=ax^{2}$. We use the pigeon
hole principle.

Let $\varphi:F^{*}\to F^{*}:x\mapsto x^{2}$ hence $F^{2*}=\im\varphi=F^{*}/\ker(\varphi)=F^{*}/\{\pm1\}$
(The last equality holds since $\mbox{char }F\neq2$)$\Rightarrow|(F^{2*})|=\frac{|F^{*}|}{2}=\frac{q-1}{2}\Rightarrow$
number of square in $F=|F^{2*}|+1$ (for zero) $=\frac{q+1}{2}$.
Hence $n_{1}=|\{ax^{2}|x\in F\}|=|\{x^{2}|x\in F\}=\frac{q+1}{2}$.
Similarly $n_{2}=|\{1-by^{2}|y\in F\}|=|\{y^{2}|y\in F\}=\frac{q+1}{2}$.
Hence $n_{1}+n_{2}=q+1>|F|\Rightarrow\{ax^{2}|x\in F\}\cap\{1-by^{2}|y\in F\}\neq\emptyset\Rightarrow1-by^{2}=ax^{2}$
has a solution.
\end{proof}
{}\end{proof}
\begin{thm}
\label{thm:Char of finite field}Let $F$ be a finite field with $q$
elements then 
\[
W(F)=\begin{cases}
\zz/2\zz & \mbox{char }F=2\\
\zz/2\zz\oplus\zz/2\zz & q\equiv1\mod4\;(\iff-1\in F^{2*})\\
\zz/4\zz & q\equiv3\mod4\;(\iff-1\notin F^{2*})
\end{cases}
\]
 \end{thm}
\begin{proof}
$\mbox{char }F=2$ we have already done (in this case $\rk:W(F)\overset{\cong}{\to}\zz/2\zz$)

Assume $\mbox{char }F$ odd, so $q$ odd. We have an exact sequence
\[
0\to\underset{\cong F^{*}/F^{2*}}{I(F)}\to W(F)\overset{\rk}{\to}\zz/2\zz\to0
\]
Since $q$ is odd we have $|F^{*}/F^{2*}|=2$ because $F^{2*}=\im(\underset{x\mapsto x^{2}}{F^{*}\overset{.^{2}}{\to}F^{*}})$
and $\ker(F^{*}\overset{.^{2}}{\to}F^{*})=\{\pm1\}$. $\Rightarrow|W(F)|=4\Rightarrow$(by
the structure theorem of finite groups) $W(F)=\zz/2\zz\oplus\zz/2\zz$
or $W(F)=\zz/4\zz$. 

If $-1\in F^{2*},(-1=a^{2}),\Rightarrow2\left\langle u\right\rangle =\left\langle u\right\rangle +\left\langle u\right\rangle =\left\langle u\right\rangle +\left\langle a^{2}u\right\rangle =\left\langle u\right\rangle +\left\langle -u\right\rangle =0\in W(F)\,\forall u\in F^{*}$.
$W(F)$ generated as an abelian group by $\left\langle u\right\rangle ,u\in F^{*}\Rightarrow$
every element in $W(F)$ has order $\leq2$. $\Rightarrow W(F)\neq\zz/4\zz\Rightarrow W(F)=\zz/2\zz\oplus\zz/2\zz$.

If $-1\notin F^{2*}\Rightarrow$ if $2\left\langle 1\right\rangle =0\in W(F)$
then $\left\langle 1\right\rangle +\left\langle 1\right\rangle =\left\langle 1\right\rangle +\left\langle -1\right\rangle \in W(F)\underset{\mbox{char }F\mbox{ odd}}{\Rightarrow}\left\langle 1\right\rangle +\left\langle 1\right\rangle \cong\left\langle 1\right\rangle +\left\langle -1\right\rangle \underset{\mathrm{Witt\, Cancellation}}{\Rightarrow}\left\langle 1\right\rangle \cong\left\langle -1\right\rangle \Rightarrow1=\det\left\langle 1\right\rangle =\det\left\langle -1\right\rangle =-1\in F^{*}/F^{2*}\Rightarrow-1\in F^{2*}$
which is a contradiction to the assumption $-1\notin F^{*2}\Rightarrow2\left\langle 1\right\rangle \neq0\Rightarrow W(F)\neq\zz/2\zz\oplus\zz/2\zz\Rightarrow W(F)=\zz/4\zz$.
(Then the theorem follows from the following lemma)\end{proof}
\begin{lem}
Let $F$ be a finite field of odd characteristic, with $q=|F|$ elements.
Then $-1\in F^{2*}\iff q\equiv1\mod4$.\end{lem}
\begin{proof}
$-1\in F^{*}\cong\zz/(q-1)\zz$ is the only element of order $2$.
$\Rightarrow-1\in F^{2*}\iff F^{*}=\zz/(q-1)\zz$ has an element of
order 4 $\iff4|(q-1)\iff q\equiv1\mod4$.\end{proof}
\begin{rem*}
$p\in\zz$ is an odd prime, then $p$ can be written as $p=a^{2}+b^{2}$
with $a,b\in\zz$ $\iff-1$ is a square in $\ff_{p}$ ($\iff p\equiv1\mod4$)\end{rem*}
\begin{proof}[To see this:]
 $a^{2}+b^{2}=(a+ib)(a-ib)\in\zz[i]$. Recall that $\zz[i]$ is a
Euclidean domain, hence a UFD (unique factorization domain) and thus,
irreducible elements and prime elements are the same. If $a^{2}+b^{2}=(a+ib)(a-ib)\in\zz[i]$,
we have $p=a^{2}+b^{2}$$\Rightarrow p$ not prime in $\zz[i]$. The
converse also holds: if $p$ is not a prime in $\zz[i]$ then $p=xy\in\zz[i]$
for non-units $x,y\in\zz[i]$. But then (if $x=a+ib$) $N(x)=a^{2}+b^{2}$
has the properties $N(xy)=N(x)N(y),N(x)=1\iff x$ unit. So $p=xy\Rightarrow\underset{=p^{2}}{\underbrace{N(p)}}=N(x)N(y)\Rightarrow N(x)=p=N(y)\Rightarrow a^{2}+b^{2}=N(x)=p$.
So $p$ can be written as $p=a^{2}+b^{2}$ with $a,b\in\zz$ $\iff p$
is not a prime in $\zz[i]$. But $p$ is a prime in $\zz[i]\iff\zz[i]/p$
is a domain. But $\zz[i]=\zz[t]/(t^{2}+1)$ so $\zz[i]/p=\zz[t]/(t^{2}+1,p)=\frac{\zz}{p}[t]/(t^{2}+1)=\ff_{p}[t]/(t^{2}+1)$.
Now $\ff_{p}[t]/(t^{2}+1)$ is a field $\iff t^{2}+1$ irreducible
$\iff t^{2}+1$ has no solution in $\ff_{p}\iff-1\notin\ff_{p}^{2*}$.
On the other hand $t^{2}+1$ reducible $\iff-1\in\ff_{p}^{2*},-1=a^{2},t^{2}+1=(t+a)(t-a)$.
Then $\ff[t]/(t^{2}+1)=\ff_{p}[t]/((t-a)(t+a))\underset{\mathrm{CRT}}{=}\ff_{p}[t]/(t+a)\times\ff_{p}[t]/(t-a)=\ff_{p}\times\ff_{p}$
not a domain.

Hence $p=a^{2}+b^{2}\iff p$ not a prime in $\zz[i]\iff\zz[i]/p=\ff_{q}[t]/(t^{2}+1)$
not a domain $\iff-1\in\ff_{p}^{2*}\iff p\equiv1\mod4$\end{proof}
\begin{cor}[of Theorem \ref{thm:Char of finite field} ]
 Two symmetric inner product spaces over a finite field of odd characteristic
are isometric if and only if they have the same rank and the same
determinant $(\in F^{*}/F^{2*})$.\end{cor}
\begin{thm}
Let $F$ be a field then $W(F)$ is generated as an abelian group
by $\left\langle u\right\rangle ,u\in F^{*}$ subject to the relations:
\begin{enumerate}
\item $\left\langle u\right\rangle =\left\langle a^{2}u\right\rangle \,\forall a,u\in F^{*}$
\item $\left\langle u\right\rangle +\left\langle -u\right\rangle =0\,\forall u\in F^{*}$
\item $\left\langle u\right\rangle +\left\langle v\right\rangle =\left\langle u+v\right\rangle +\left\langle uv(u+v)\right\rangle \,\forall u,v\in F^{*},u+v\in F^{*}$
\end{enumerate}
\end{thm}
\begin{rem*}
The theorem asserts that

\[
\frac{\oplus_{a\in F^{*}}\overset{\mathrm{rank\,1\, free\, abelian\, group\, with\, basis\,}\{a\}}{\overbrace{\zz\{a\}}}}{\{u\}-\{a^{2}u\},\{u\}+\{-u\},\{u\}+\{v\}-\{u+v\}-\{uv(u+v)\}}\overset{\cong}{\to}W(F)
\]
defined by $\{a\}\mapsto\left\langle a\right\rangle $ is an isomorphism.
(Here $\zz\{a\}\cong\zz$ denotes the free $\zz$-module of rank $1$
with basis $\{a\}$.)\end{rem*}
\begin{proof}
We already know that $W(F)$ is generated by $\left\langle u\right\rangle ,u\in F^{*}$
and that $1,2,3$ holds in $W(F)\Rightarrow$
\[
\rho:\frac{\oplus_{a\in F^{*}}\overset{\mathrm{rank\,1\, f.a.g.w/\, basis\,}\{a\}}{\overbrace{\zz\{a\}}}}{\{u\}-\{a^{2}u\},\{u\}+\{-u\},\{u\}+\{v\}-\{u+v\}-\{uv(u+v)\}}\to W(F)
\]
 is a well defined surjective map of abelian groups. So we need to
check that $\rho$ is injective. Will give a proof when $\mbox{char }F\neq2$
(the $\mbox{char }F=2$ case needs a different, longer proof)

Using relation 2. ($\{u\}=-\{-u\}$) we can write every element in
LHS as $\sum_{i=1}^{n}\{u_{i}\}$. Given $U=\sum_{i=1}^{n}\{u_{i}\}$
and $V=\sum_{j=1}^{m}\{v_{j}\}$ in the LHS such that $\rho(U)=\rho(V)\in W(F)\Rightarrow n=\rk\,\rho(u)=\rk\,\rho(V)=m\in\zz/2\zz$.
So $m\equiv n\mod2$ and without loss of generality say $m\geq n$
so $m-n=2k,k\geq0$. Then $U=U+k\underset{=0\mathrm{\, by\,}2.}{\underbrace{(\{1\}+\{-1\}}})\in$
LHS. Replacing $U\in$ LHS with $U+k(\{1\}+\{-1\})$ we can assume
that $m=n$. Then $\rho(U)=\rho(V)\in W(F)$ and $\rk\rho(U)=\rk\rho(V)$.
$\underset{\frac{1}{2}\in F^{*}}{\Rightarrow}\left\langle u_{1},\dots,u_{n}\right\rangle \cong\left\langle v_{1},\dots,v_{n}\right\rangle \underset{\mathrm{chain\, equivalence\, thm}}{\Rightarrow}\left\langle u_{1},\dots,u_{n}\right\rangle \approx\left\langle v_{1},\dots,v_{n}\right\rangle \Rightarrow\exists$
diagonal forms $c_{1},\dots,c_{l}$ such that $\left\langle u_{1},\dots u_{n}\right\rangle \approx_{S}c_{1}\approx_{S}\dots\approx_{S}c_{l}\approx_{S}\left\langle v_{1},\dots,v_{n}\right\rangle \Rightarrow$
it suffices to show that $\{u_{1}\}+\dots+\{u_{n}\}=\{v_{1}\}+\dots+\{v_{n}\}$
in the case $\left\langle u_{1},\dots,u_{n}\right\rangle ,\left\langle v_{1},\dots,v_{n}\right\rangle $
are simply chain equivalence (i.e., they differ in two places). So,
we can assume $n=2$, we need to show that $\left\langle u_{1},u_{2}\right\rangle \cong\left\langle v_{1},v_{2}\right\rangle $
then $\{u_{1}\}+\{u_{2}\}=\{v_{1}\}+\{v_{2}\}$ in LHS. Assume $\left\langle u_{1},u_{2}\right\rangle \cong\left\langle v_{1},v_{2}\right\rangle \Rightarrow u_{1}u_{2}=v_{1}v_{2}a^{2}$
for some $a\in F^{*}$ and $u_{1}=v_{1}x^{2}+v_{2}y^{2}$ for some
$x,y\in F$. If $x,y\neq0$ then $\{v_{1}\}+\{v_{2}\}\underset{1.}{=}\{v_{1}x^{2}\}+\{v_{2}y^{2}\}\underset{3.}{=}\{v_{1}x^{2}+v_{2}y^{2}\}+\{v_{1}v_{2}x^{2}y^{2}(v_{1}x^{2}+v_{2}y^{2})\}=\{u_{1}\}+\{\frac{1}{a^{2}}u_{1}u_{2}x^{2}y^{2}u_{1}\}\underset{1}{=}\{u_{1}\}+\{u_{2}\}\in$
LHS. If $x$ or $y=0$, say $x=0$ then $y\neq0$ since $u_{1}\in F^{*}$,
then we get $u_{1}=v_{2}y^{2}$ and $v_{1}v_{2}a^{2}=v_{2}y^{2}u_{2}\Rightarrow v_{1}(\frac{a}{y})^{2}=u_{2}$.
Then $\{u_{1}\}+\{u_{2}\}=\{v_{2}y^{2}\}+\{v_{1}(\frac{a}{y})^{2}\}\underset{1.}{=}\{v_{2}\}+\{v_{1}\}\in$
LHS.

So $\rho(U)=\rho(V)\in W(F)\Rightarrow U=V\in$ LHS$\Rightarrow\rho$
injective
\end{proof}

\subsection{Second Residue Homomorphism}

For a DVR (\emph{Discrete valuation ring}) $R$ with field of fraction
$F$, residue field $k=R/m$ and uniformizing element $\pi\in R$,
we will construct maps $\partial_{\pi}:W(F)\to W(k)$ which will help
compute $W(\qq),W(\zz),W(\qq_{p})\dots$
\begin{defn}
A \emph{discrete valuation ring} (DVR) is a local ring $(R,m,k)$
which is:
\begin{itemize}
\item Noetherian
\item A domain ($ab=0\in R\Rightarrow a=0$ or $b=0$)
\item $m\neq0$ is a principal ideal $(m=\pi R$ for some $\pi\in R$)
\end{itemize}
\end{defn}
There are other (equivalent) characterizations of DVR (which we won't
need):
\begin{itemize}
\item $R$ is a DVR $\iff$ local 1-dimensional integrally closed noetherian
domain
\item $\iff$ Local $1$-dimensional noetherian regular domain
\item $\iff$ Local PID
\item $\iff$ Local domain with principal $m$ such that $\cap_{n\geq0}m^{n}=0$
\item $\iff$ valuation ring of a discrete valuation on a field\end{itemize}
\begin{example*}[Of DVR]
{}
\begin{itemize}
\item $\zz_{(p)}=\{\frac{a}{b}\in\qq|p\nmid a\}$, $p\in\zz$ prime. $m=p\zz_{(p)},k=\ff_{p}$.
Fraction field $\qq$.
\item The $p$-adic integers $\zz_{p}=\underset{\underset{n\to\infty}{\leftarrow}}{\lim}\zz/p^{n}\zz=\{(x_{n})_{n\in\nn_{\geq1}},x_{n}\in\zz/p^{n}\zz:x_{n+1}=x_{n}\mod p^{n}\}$,
$m=p\zz_{p},k=\zz_{p}/p\zz_{p}=\ff_{p}$ and field of fractions $\zz_{p}=\qq_{p}$
\item $D$ = Dedekind domain, $p\subset D$ a prime ideal then $D_{p}=\{\frac{a}{b}\in\mbox{Frac}D|b\notin p\}$,
$m=pD_{p},k=D/p$
\item $K$ is a field, $f\in K[T]$ is irreducible. $K[T]_{(f)}=\{\frac{a}{b}\in\mbox{Frac}(K[T])=K(T)|f\nmid b\}$,
$m=fK[T]_{(f)},k=K[T]/f$.
\item $R$ is a UFD, $f\in R$ an irreducible element (=prime element) $R_{(f)}=\{\frac{a}{b}\in\mbox{Frac }R|f\nmid b\}$
\end{itemize}
\end{example*}
\begin{defn}
Let $(R,m,k)$ be a DVR, a \emph{uniformizing element }of $R$ is
a choice $\pi\in m\subset R$ generating $m$, i.e. $\pi R=m$\end{defn}
\begin{lem}
Let $R$ be a DVR with uniformizing element $\pi\in R$, then every
element $a\in R,a\neq0$ can be written uniquely as $a=\pi^{n}u$
for some $n\in\nn$ and $u\in R^{*}$\end{lem}
\begin{proof}
\uline{Uniqueness }:Assume $\pi^{n}u=\pi^{m}v$ with $u,v\in R^{*}$.
Without loss of generality assume $m\geq n$. $R$ domain $\Rightarrow\pi^{m-n}=vu^{-1}\in R$.
If $m\neq n\Rightarrow\pi^{n-m}\in\pi R=m$ but $vu^{-1}\in R^{*}=R\setminus m$
which is a contradiction $\Rightarrow n=m\Rightarrow1=vu^{-1}\Rightarrow u=v$

\uline{Existence }:Let $a\in R$, $a\neq0$. If $a\in\cap_{n\geq0}m^{n}=\cap_{n\geq0}\pi^{m}R$
then $a=\pi^{n}b_{n}\,\forall n$. $\underset{R\mathrm{\, domain}}{\Rightarrow}b_{n}=\pi b_{n+1}\Rightarrow(b_{n})\subset(b_{n+1})\subset(b_{n+2})\subset\dots R$,
is an ascending chain of ideals which has to stop because $R$ is
noetherian. $\exists n$ such that $(b_{n})=(b_{n+1})$ in particular
$b_{n+1}\in(b_{n})\Rightarrow b_{n+1}=cb_{n}$ but $b_{n}=\pi b_{n+1}\underset{R\mathrm{\, domain}}{\Rightarrow}1=c\pi\Rightarrow\pi\in R^{*}=R\setminus m$
which contradicts the fact that $\pi\in m$. Hence $a=0\Rightarrow\cap_{n\geq0}\pi^{n}R=0$.
Hence $\exists n$ such that $a\in\pi^{n}R$ but $a\notin\pi^{n+1}R\Rightarrow a=\pi^{n}u,u\notin\pi R=m$
hence $u\in R^{*}$.\end{proof}
\begin{rem*}
$\cap_{n\geq0}\pi^{n}R=\cap_{n\geq0}m^{n}$. For all Noetherian $R:\cap_{n\geq0}m^{n}=0$\end{rem*}
\begin{cor}
Let $R$ be a DVR with uniformizing element $\pi$ and $F$ its field
of fractions, then every $a\in F,a\neq0$ can be written uniquely
as $a=\pi^{n}u$ where $u\in R^{*}$.
\end{cor}
So we can define a function $\nu:F^{*}\to\zz$ defined by $a=\pi^{n}u\mapsto n=\nu(a)$
(with $u\in R^{*}$) with the properties
\begin{enumerate}
\item $\nu(ab)=\nu(a)+\nu(b)$
\item $\nu(a+b)\geq\min(\nu(a),\nu(a))$
\item Setting $\nu(0)=\infty$ we have $R=\{a\in F:\nu(a)\geq0\}$, $R^{*}=\{a\in F|\nu(a)=0\}$,
$m=\{a\in F|\nu(a)>0\}$\end{enumerate}
\begin{defn}
A \emph{discrete valuation} on a field $F$ is a function $\nu:F^{*}\to\zz$
satisfying 1., 2.~above.

The valuation ring of $\nu$ is the ring $R=\{a\in F|\nu(a)\geq0\}$
where $\nu(0)=\infty$.
\end{defn}

\begin{defn}
Let $(F,\nu)$ be a discrete valuation on a field $F$ with associated
DVR $R$ and choice of uniformizing element $\pi\in R$. The \emph{second
residue homomorphism} is the map $\partial_{\pi}:W(F)\to W(R/\pi)$
defined by 
\[
\left\langle a\right\rangle \mapsto\begin{cases}
\left\langle \underline{u}\right\rangle  & n=\nu(a)\,\mbox{odd}\\
0 & n=\nu(a)\,\mbox{even}
\end{cases}
\]
where $a\in F^{*}$, $a=\pi^{n}u,n=\nu(a),u\in R^{*},\underline{u}=u\mod m=\pi R$,
$u\in R/\pi$\end{defn}
\begin{note*}
$\partial$ depends on the choice of the uniformizing element $\pi$.\end{note*}
\begin{lem}
The second residue homomorphism is well defined\end{lem}
\begin{proof}
Recall that $W(F)$ is generated by $\left\langle a\right\rangle $,
$a\in F^{*}$ subject to:
\begin{enumerate}
\item $\left\langle u\right\rangle =\left\langle x^{2}u\right\rangle $,
$u,x\in F^{*}$
\item $\left\langle u\right\rangle +\left\langle -u\right\rangle =0$
\item $\left\langle u\right\rangle +\left\langle v\right\rangle =\left\langle u+v\right\rangle +\left\langle uv(u+v)\right\rangle $,
$u,v,u+v\in F^{*}$
\end{enumerate}
We need to check that $\partial_{\pi}$ preserves these relations.
First define $\epsilon_{i}=\begin{cases}
0 & i\,\mbox{even}\\
1 & i\,\mbox{odd}
\end{cases}$, so that we can write $\partial\left\langle u\right\rangle =\epsilon_{\nu(u)}\left\langle \underline{\phi}\right\rangle $
where $u=\pi^{\nu(u)}\phi,\phi\in R^{*},\underline{\phi}=\phi\mod\pi R$.
Then:
\begin{enumerate}
\item Let $u=\pi^{n}\phi,x=\pi^{m}\psi$ where $\phi,\psi\in R^{*}$. Then
$x^{2}u=\pi^{2m+n}\phi\psi^{2}$ so $\partial\left\langle x^{2}u\right\rangle =\epsilon_{2m+n}\left\langle \underline{\phi\psi^{2}}\right\rangle =\epsilon_{n}\left\langle \underline{\phi}\right\rangle \in W(R/\pi)$
as required $\partial\left\langle u\right\rangle =\epsilon_{n}\left\langle \underline{\phi}\right\rangle $.
\item Let $u=\pi^{n}\phi,-u=\pi^{n}(-\phi)$ with $\phi\in R^{*}$$ $.
Then $\partial\left\langle u\right\rangle +\partial\left\langle -u\right\rangle =\epsilon_{n}\left\langle \underline{\phi}\right\rangle +\epsilon_{n}\left\langle \underline{-\phi}\right\rangle =\epsilon_{n}(\underset{=0}{\underbrace{\left\langle \underline{\phi}\right\rangle +\left\langle \underline{-\phi}\right\rangle }})=0\in W(R/\pi)$
\item Let $u=\pi^{n}\phi,v=\pi^{m}\psi$, without loss of generality assume
$n\geq m$. $u+v=\pi^{n}\phi+\pi^{m}\psi=\pi^{m}(\pi^{n-m}\phi+\psi)$

\begin{casenv}
\item $n>m:\mbox{ Then }n-m>0\Rightarrow t=\underset{\in m}{\underbrace{\pi^{n-m}\phi}}+\underset{\notin m}{\underline{\psi}}\in R^{*},u+v=\pi^{m}t$,
note that $t\equiv\psi\mod\pi R$. Now $uv(u+v)=\pi^{n+2m}\underset{\in R^{*}}{\underbrace{\phi\psi t}}$.
So $\partial\left\langle u+v\right\rangle +\partial\left\langle uv(u+v)\right\rangle =\epsilon_{m}\left\langle \underline{t}\right\rangle +\epsilon_{n+2m}\left\langle \underline{\phi\psi t}\right\rangle =\epsilon_{m}\left\langle \underline{\psi}\right\rangle +\epsilon_{n}\left\langle \underline{\phi\psi^{2}}\right\rangle =\partial\left\langle v\right\rangle +\partial\left\langle u\right\rangle $
\item $n=m$: Now $u+v=\pi^{n}(\phi+\psi)$ and $\phi+\psi=\pi^{l}t$ where
$t\in R^{*}$.

\begin{casenv}
\item $l=0$: $\phi+\psi=t\in R^{*}$. Now $u+v=\pi^{n}t$ and $uv(u+v)=\pi^{3n}\underset{\in R^{*}}{\underbrace{\phi\psi t}}$.
Then $\partial\left\langle u+v\right\rangle +\partial\left\langle uv(u+v)\right\rangle =\epsilon_{n}\left\langle \underline{t}\right\rangle +\epsilon_{3n}\left\langle \underline{\phi\psi t}\right\rangle =\epsilon_{n}\left\langle \underline{\phi}+\underline{\psi}\right\rangle +\epsilon_{n}\left\langle \underline{\phi\psi}(\underline{\phi}+\underline{\psi})\right\rangle =\epsilon_{n}(\left\langle \underline{\psi}\right\rangle +\left\langle \underline{\phi}\right\rangle )\in W(R/\pi)$
which is what we wanted.
\item $l>0$: $u+v=\pi^{l+n}t\in\pi R$. In particular $\underline{\psi}+\underline{\phi}=0$
so $\underline{\psi}=-\underline{\phi}\in R/\pi R$. So $uv(u+v)=\pi^{3n+l}\phi\psi t$.
Then $\partial\left\langle u+v\right\rangle +\partial\left\langle uv(u+v)\right\rangle =\epsilon_{n+l}\left\langle \underline{t}\right\rangle +\epsilon_{3n+l}\left\langle \underline{\phi\psi t}\right\rangle =\epsilon_{n+t}\left\langle \underline{t}\right\rangle +\epsilon_{n+l}\partial\left\langle -\underline{\psi^{2}}t\right\rangle =\epsilon_{n+l}(\left\langle \underline{t}\right\rangle +\left\langle \underline{-t}\right\rangle )=0=\epsilon_{n}(\left\langle -\underline{\psi}\right\rangle +\left\langle \underline{\psi}\right\rangle )=\epsilon_{n}\left\langle \underline{\phi}\right\rangle +\epsilon_{n}\left\langle \underline{\psi}\right\rangle =\partial\left\langle u\right\rangle +\partial\left\langle v\right\rangle $ 
\end{casenv}
\end{casenv}
\end{enumerate}
\end{proof}
\begin{thm}
Let $D$ be a Dedekind domain with field of fractions $F$. Then the
sequence of abelian group 
\[
0\longrightarrow W(D)\longrightarrow W(F)\hspace{1ex}\overset{\bigoplus\partial_{\wp}}{\longrightarrow}\bigoplus_{\begin{array}{c}
\wp\subset D\\
\text{{\rm max. ideal}}
\end{array}}W(D/\wp)
\]
is exact.
\end{thm}
We will prove the above theorem in the special cases: $D=\zz,$ DVR,
$k[T]$.
\begin{lem}
Let $(R,m,k)$ be a DVR with field of fraction $k$ and uniformizing
element $\pi\in m\subset R$. Then the composition
\[
\xymatrix{W(R)\ar[r] & W(F)\ar^{\partial_{\pi}}[r] & W(R/\pi)}
\]
 is zero\end{lem}
\begin{proof}
$R$ local $\Rightarrow W(R)$ is generated by $\left\langle u\right\rangle ,u\in R^{*}$
and $\left\langle \begin{pmatrix}a & 1\\
1 & b
\end{pmatrix}\right\rangle ,a,b\in m$. We have $\partial\left\langle u\right\rangle =0$ by definition
of $\partial$: $\partial\left\langle u\right\rangle =\begin{cases}
0 & \nu(u)\,\mbox{even}\\
\left\langle \underline{v}\right\rangle  & \nu(u)\,\mbox{odd}
\end{cases}$ where $u=\pi^{\nu(u)}v,v\in R^{*}$. 

If $a=0$ then $\left\langle \begin{pmatrix}0 & 1\\
1 & b
\end{pmatrix}\right\rangle $ is metabolic so, $W(F)\ni\left\langle \begin{pmatrix}a & 1\\
1 & b
\end{pmatrix}\right\rangle =0\Rightarrow\partial\left\langle \begin{pmatrix}a & 1\\
1 & b
\end{pmatrix}\right\rangle =0$

If $a\neq0$ then $a\in F^{*}$ so $\left\langle \begin{pmatrix}a & 1\\
1 & b
\end{pmatrix}\right\rangle =\left\langle a\right\rangle +\left\langle a(ab-1)\right\rangle ,a=\pi^{\nu(a)}v,v\in F^{*},a(ab-1)=\pi^{\nu(a)}v\underset{\in R^{*}}{\underbrace{(ab-1})},ab\in m\Rightarrow\partial\left\langle \begin{pmatrix}a & 1\\
1 & b
\end{pmatrix}\right\rangle =\partial\left\langle a\right\rangle +\partial\left\langle a(ab-1)\right\rangle =\epsilon_{\nu(a)}\left\langle \underline{v}\right\rangle +\epsilon_{\nu(a)}\left\langle \underline{v}\underset{-1\mod m}{\underbrace{(ab-1)}}\right\rangle =\epsilon_{\nu(a)}(\left\langle \underline{v}\right\rangle +\left\langle -\underline{v}\right\rangle )=0$\end{proof}
\begin{cor}
The composition $W(\zz)\to W(\qq)\overset{\partial_{p}}{\to}\oplus_{p\in\zz}W(\zz/p\zz)$
($p$ prime) is zero, where $W(\qq)\overset{\partial_{p}}{\to}W(\zz/p\zz)$
is the 2nd residue homomorphism associated with the $p$-adic valuation
on $\qq$ which has valuation ring $\zz_{(p)}=\{\frac{a}{b}\in\qq|p\nmid b\}$\end{cor}
\begin{proof}
We have defined
\[
\xyC{3pc}\xymatrix{W(\qq)\ar^{\prod\partial_{p}}[r]\ar_{?}@{-->}[dr] & \prod W(\zz/p\zz)\\
 & \oplus W(\zz/p\zz)\ar@{^{(}->}[u]
}
\]
Need to see that $\prod\partial_{p}$ has image in $\oplus_{p}W(\zz/p\zz)$.
This is the case because if $\left\langle u\right\rangle \in W(\qq),u\in\qq^{*}$
then $u=\frac{a}{b}$ and $\{p\in\zz$ prime $|\nu(u)\ne0\}\subset\{$primes
in the factorization of $a,b\}$ finite $\Rightarrow\forall\xi\in W(\qq),\partial_{p}\xi=0$
for all but finitely many $p\in\zz$ prime. For the composition to
be zero, it suffices to check that the composition $W(\zz)\to W(\qq)\overset{\partial_{p}}{\to}W(\zz/p\zz)$
is zero for all $p$. The composition is zero because it factors as
$W(\zz)\to\underset{0\mbox{}}{\underbrace{W(\zz_{(p)})\to W(\qq)\overset{\partial_{\pi}}{\to}W(\zz/p\zz)}}$\end{proof}
\begin{defn}
\emph{A principal ideal domain (PID)} is a commutative ring $R$ which
is a domain ($ab=0\Rightarrow a$ or $b=0$) and for which every ideal
is a principal ideal ($I\subset R\Rightarrow I=Rx$ for some $x\in I$)\end{defn}
\begin{example*}
$\zz,k[T]$ ($k$ a field) are PIDs (Euclidean domain$\Rightarrow$PID)

A DVR $R$ is a PID: Let $\pi$ be a uniformizing element, so $m=\pi R$
and let $I\subset R$ be any ideal. $I=0$ is principal, so assume
$I\neq0$. Let $n=\min\{\nu(a)|a\in I,a\neq0\}\in\nn_{\geq0}$. Then
$I=\pi^{n}R$ because $I\subset\pi^{n}R$ since if $a\in I,a\neq0,a=\pi^{\nu(a)}u=\pi^{n}(\underset{\in R}{\underbrace{\pi^{\overset{\geq0}{\overbrace{\nu(a)-n}}}u}})\in\pi^{n}R$,
and $\pi^{n}R\subset I$ since $\exists a\in I,a\neq0$, $\nu(a)=n$
so $a=\pi^{n}u,u\in R^{*},a\in I\Rightarrow\pi^{n}R=\pi^{n}uR=aR\subset I$.
Hence $I=\pi^{n}R$ is principal.\end{example*}
\begin{rem*}
A PID is noetherian
\begin{proof}
Let $R$ be a PID, $I_{1}\subset I_{2}\subset\dots\subset I_{2}\subset\dots\subset R$
be an ascending chain of ideals. Then $I=\cup I_{n}\subset R$ is
an ideal $\Rightarrow I=xR$ for some $x\in I\Rightarrow x\in I_{n}$
for some $n\Rightarrow I=Rx\subset I_{n}\subset I\Rightarrow I_{n}=I_{m}=I\,\forall m\geq n$
\end{proof}
\end{rem*}
\begin{defn}
An $R$-module $M$ is called \emph{cyclic} if $M\cong R/I$ for some
ideal $I\subset R$.\end{defn}
\begin{fact*}
Every finitely generated module $M$ over a PID $R$ is a finite direct
sum of cyclic $R$-modules, that is, $M\cong\oplus_{i=1}^{n}R/a_{i}$
for some non-units $a_{1},\dots,a_{n}\in R$\end{fact*}
\begin{cor}
Let $R$ be a PID with field of fractions $F$
\begin{enumerate}
\item Every submodule $M$ of a finitely generated free $R$-module is free
\item Every finitely generated $R$-submodule $M\subset F^{n}$ is also
free
\end{enumerate}
\end{cor}
\begin{proof}
$R$ is a PID then $R$ is noetherian, $M\subset R^{n}\Rightarrow M$
is finitely generated so in $1$ and 2 the module $M$ is finitely
generated $\Rightarrow M=\oplus_{i=1}^{n}R/a_{i}$ for some non-unit
$a_{i}$. But $R/a\subset R^{n}$ or $R/a\subset F^{n}\Rightarrow a=0$
because:

If $a\neq0$ and $a\notin R^{*}$ then the composition $R/a\subset R\overset{a}{\hookrightarrow}R$
is injective and $R/a\overset{a=0}{\to}R/a\hookrightarrow R$ zero
at the same time (a contradiction). Similarly the composition $R/a\subset F^{n}\overset{a}{\to}F^{n}$
is injective (if $a\neq0,a\notin R^{*}$) and equals $R/a\overset{a=0}{\to}R/a\hookrightarrow F^{n}$
zero, again a contradiction\end{proof}
\begin{cor}
Every inner product space over a PID is free\end{cor}
\begin{lem}
Let $R$ be a PID with field of fractions $F$ then the map $W(R)\to W(F)$
defined by $[M,\beta]\mapsto(M_{F},\beta_{F})$ is injective, where
$M_{F}=M\otimes_{R}F$ and $\beta_{F}(x\otimes a,y\otimes b)=ab\beta(x,y)$
for $a,b\in F,x,y\in M$.\end{lem}
\begin{proof}
Assume $[M_{F},\beta_{F}]=0\in W(F)$ then $(M_{F},\beta_{F})\sim0\Rightarrow$
$\exists$ metabolic $V=\left\langle \begin{pmatrix}0 & 1\\
1 & A
\end{pmatrix}\right\rangle $ with $A\in M_{n}(F),A^{T}=A$ such that $(M_{F},\beta_{F})\perp V$
is metabolic. There exist $d\in R$ such that $dA\in M_{n}(R)$. Then
\[
\begin{pmatrix}d^{-1} & 0\\
0 & d
\end{pmatrix}\begin{pmatrix}0 & 1\\
1 & A
\end{pmatrix}\begin{pmatrix}d & 0\\
0 & d^{-1}
\end{pmatrix}=\begin{pmatrix}0 & 1\\
1 & d^{2}A
\end{pmatrix}\Rightarrow\left\langle \begin{pmatrix}0 & 1\\
1 & A
\end{pmatrix}\right\rangle \cong\left\langle \begin{pmatrix}0 & 1\\
1 & d^{2}A
\end{pmatrix}\right\rangle \Rightarrow(M,\beta)\perp\underset{\mathrm{metabolic\, over\,}R}{\underbrace{\left\langle \begin{pmatrix}0 & 1\\
1 & d^{2}A
\end{pmatrix}\right\rangle }}
\]
 is metabolic over $F$. Hence can assume $(M,\beta)$ to be metabolic
over $F$.

Let $(M,\beta)$ be a symmetric inner product space over $R$ such
that $(M,\beta)_{F}=(M_{F},\beta_{F})$ is metabolic over $F$. $M_{F}=M\otimes_{R}F$,
$\beta_{F}(x\otimes a,y\otimes b)=ab\beta(x,y)$, $x,y\in M,a,b\in F$.
Note $M\subset M_{F}$ (as $R\subset F,M\cong R^{n},M_{F}\cong F^{n})$.
Now $(M_{F},\beta_{F})$ metabolic$\Rightarrow\exists N\subset M_{F}$
Lagrangian

\uline{Claim}: $M\cap N\subset M$ is a Lagrangian for $(M,\beta)$

$M\cap N$ is a direct summand of $M$ because $M/M\cap N\subset M_{F}/N\cong F^{m}$
is a finitely generated $R$-submodule of $F^{m}$ $\underset{\mathrm{PID}}{\Rightarrow}M/M\cap N$
is finitely generated free $R$-module, so $M/M\cap N\cong R^{l}$.
Any section $s:M/M\cap N\to M$ of $g:M\to M/M\cap N\cong R^{l}$
(that is $gs=1$) yields a direct sum decomposition $(M\cap N)\oplus\im(s)=M\Rightarrow M\cap N\subset M$
is a direct summand. We now need to check that $(M\cap N)^{\perp}=M\cap N$.
Let $x\in M$, then $x\in(M\cap N)^{\perp}\iff\beta(x,y)=0\,\forall y\in M\cap N\iff\beta(x,y)=0\,\forall y\in N$
(because $\forall t\in M_{F}=M\otimes_{R}F$ $\exists a\in R,a\neq0$
such that $at\in M$, in particular $\forall y\in N,\exists a\in R,a\neq0,ay\in M\cap N$,
$\beta(x,y)=0\iff\beta(x,ay)\underset{a\neq0}{=}0$). But $\beta(x,y)=0\,\forall y\in N\underset{N=N^{\perp}}{\iff}x\in N^{\perp}=N\underset{x\in M}{\iff}x\in M\cap N$.
Hence $M\cap N\subset M$ is a Lagrangian $\Rightarrow(M,\beta)$
is metabolic $\Rightarrow[M,\beta]=0\in W(R)$

To finish the proof, take $[M,\beta]\in W(R)$ such that $(M,\beta)_{F}=0\in W(F)\Rightarrow\exists V$
metabolic symmetric inner product space over $R$ such that $(M\perp V)_{F}$
is metabolic over $F$$\Rightarrow M\perp V$ metabolic over $R\Rightarrow[M]=[M]+[V]=[M\perp V]=0\in W(R)$.\end{proof}
\begin{thm}
The sequence of abelian group
\[
0\to W(\zz)\to W(\qq)\overset{\oplus_{p}\partial_{p}}{\longrightarrow}\bigoplus_{p\in\zz_{\geq2}\mathrm{prime}}W(\ff_{p})\to0
\]
is exact and the map $W(\zz)\to W(\rr)$ defined by $M\mapsto M\otimes_{\zz}\rr$
is an isomorphism\end{thm}
\begin{proof}
We have already proved that $W(\zz)\to W(\qq)$ is injective since
$\zz$ is a PID and the composition $W(\zz)\to W(\qq)\to\oplus_{p}W(\ff_{p})$
is zero.

For $n\in\zz_{\geq1},$ let $\mathscr{P}_{n}$ be the set $\mathscr{P}_{n}=\{a\in\zz\setminus\{0\}|\forall\mbox{ prime }p:p|a\Rightarrow p\leq n\}$
(e.g. $\mathscr{P}_{1}=\{+1,-1\},\mathscr{P}_{2}=\{+2^{n},-2^{n}\},\mathscr{P}_{n}\subset\mathscr{P}_{n+1}$).
Note that $\mathscr{P}_{n-1}=\mathscr{P}_{n}$ unless $n$ is prime.
Let $L_{n}\subset W(\qq)$ be the subgroup generated by $\left\langle a\right\rangle $
with $a\in\mathscr{P}_{n}$. So $L_{n-1}\subset L_{n}$ and $L_{n-1}=L_{n}$
unless $n$ is prime. The composition $L_{p-1}\subset L_{p}\overset{\partial_{p}}{\to}W(\ff_{p})$
is zero because, for $a\in\mathscr{P}_{p-1},\nu_{p}(a)=0$ so $\partial_{p}(a)=0$.
Hence we get a map of abelian groups $L_{p}/L_{p-1}\overset{\partial_{p}}{\to}W(\ff_{p})$.

\uline{Claim:} $L_{p}/L_{p-1}\overset{\partial_{p}}{\to}W(\ff_{p})$
is an isomorphism for all primes $p\in\zz_{\geq2}$.

The claim will follow from:
\begin{lem}
If $0<|n|,|n_{1}|,\dots,|n_{k}|<p,\mbox{ and }n\equiv n_{1}\dots n_{k}\mod p$,
then $\left\langle pn\right\rangle =\left\langle pn_{1}\dots n_{k}\right\rangle \in L_{p}/L_{p-1}$
\begin{proof}
We use induction on $k$. The case $k=1$ follows from $k=2$ with
$n_{2}=1$.

Assume $k\geq2$: Write $n_{1}n_{2}=pl+r,0\neq|r|<p$. $|pl|=|n_{1}n_{2}-r|\leq|n_{1}||n_{2}|+|r|\leq(p-1)(p-1)+p-1=p^{2}-p\leq p^{2}\Rightarrow|l|<p$.
We will show that $\left\langle pn_{1}\dots n_{k}\right\rangle =\left\langle prn_{3}\dots n_{k}\right\rangle \in L_{p}/L_{p-1}$.
This is clear for $l=0$ as then both sides are the same. Assume $l\neq0$
and write $m=n_{3}\dots n_{k}$. 
\begin{eqnarray*}
\left\langle pn_{1}\dots n_{k}\right\rangle -\left\langle prn_{3}\dots n_{k}\right\rangle  & = & \left\langle pn_{1}n_{2}m\right\rangle -\left\langle prm\right\rangle \\
 & = & \left\langle pn_{1}n_{2}m\right\rangle -\left\langle lm\right\rangle -\left\langle prm\right\rangle \mod L_{p-1}\\
 & = & \left\langle pn_{1}n_{2}m\right\rangle -\left\langle \underset{=:v}{\underbrace{p^{2}lm}}\right\rangle -\left\langle \underset{=:u}{\underbrace{prm}}\right\rangle \mod L_{p-1}\\
 & = & \left\langle pn_{1}n_{2}m\right\rangle -\left\langle u+v\right\rangle -\left\langle uv(u+v)\right\rangle \\
 & = & \left\langle pn_{1}n_{2}m\right\rangle -\left\langle pm(pl+r)\right\rangle -\left\langle p^{4}m^{3}lr(pl+r)\right\rangle \\
 & = & \left\langle pn_{1}n_{2}m\right\rangle -\left\langle pmn_{1}n_{2}\right\rangle -\left\langle p^{4}m^{3}lrn_{1}n_{2}\right\rangle \\
 & = & -\left\langle mlrn_{1}n_{2}\right\rangle =0\mod L_{p-1}
\end{eqnarray*}
Hence $\left\langle pn_{1}n_{2}\dots n_{k}\right\rangle =\left\langle prn_{3}\dots n_{k}\right\rangle $
where $n_{1}n_{2}=pl+r,|r|<p$. This proves the case $k=2$ (and hence
$k=1$). Now, the product $rn_{3},...,n_{k}$ has $k-1$ factors,
and we can apply the induction hypothesis. 
\end{proof}
\end{lem}
We now construct an inverse of the map in the claim. We define the
map $\phi:\oplus_{u\in\ff_{p}^{*}}\zz\{u\}\to L_{p}/L_{p-1}$ by $\{u\}\mapsto\left\langle pn\right\rangle $
where $n\in\zz\setminus\{0\}$, $|n|<p,u\equiv n\mod p$. Note that
$\phi$ is well defined by the lemma, that is our choice of $n$ does
not matter. Need to check that $\phi$ preserves the three relations
for $W(\ff_{p})$
\begin{enumerate}
\item $\left\langle u\right\rangle =\left\langle a^{2}u\right\rangle ,a,u\in\ff_{p}^{*}$.
Choose $a_{0},u_{0},n_{o}\in\zz\setminus\{0\}$ such that $a_{0}=a,u_{0}=u,n_{0}=a^{2}u\in\ff_{p},|a_{0}|,|u_{0}|,|n_{0}|<p$.
Then $\{u\}-\{a^{2}u\}\overset{\phi}{\mapsto}\left\langle pu_{0}\right\rangle -\left\langle pn_{0}\right\rangle \underset{\mathrm{lemma}}{=}\left\langle pu_{0}\right\rangle -\left\langle pa_{0}^{2}u_{0}\right\rangle =0\in L_{p}/L_{p-1}$
\item $\left\langle u\right\rangle +\left\langle -u\right\rangle =0\in W(\ff_{p})$.
Choose $u_{0}\in\zz\setminus\{0\}$, $|u_{0}|<p,u_{0}=u\in\ff_{p}$.
Then $\{u\}+\{-u\}\overset{\phi}{\mapsto}\left\langle pu_{0}\right\rangle +\left\langle -pu_{0}\right\rangle =0\in L_{p}/L_{p-1}$
\item $\left\langle u\right\rangle +\left\langle v\right\rangle =\left\langle u+v\right\rangle +\left\langle uv(u+v)\right\rangle \in W(\mathbb{F}_{p}),u,v,u+v\in\ff_{p}^{*}$.
Choose $-p<u_{0}<0<v_{0}<p,$ (then $|u_{0}+v_{0}|<p,|n_{0}|<p$),
$|n_{0}|<p$ such that $u_{0},v_{0},n_{0}\in\zz\setminus\{0\}$, $u_{0}=u,v_{0}=v,n_{0}=uv(u+v)\in\ff_{p}$.
Then $\{u\}+\{v\}-\{u+v\}-\{uv(u+v)\}\overset{\phi}{\mapsto}\left\langle pu_{0}\right\rangle +\left\langle pv_{0}\right\rangle -\left\langle p(u_{0}+v_{0})\right\rangle -\left\langle pn_{0}\right\rangle \underset{\mathrm{lemma}}{=}\left\langle pu_{0}\right\rangle +\left\langle pv_{0}\right\rangle -\left\langle pu_{0}+pv_{0}\right\rangle -\left\langle pu_{0}pv_{0}(pu_{0}+pv_{0})\right\rangle =0\in L_{p}/L_{p-1}$
\end{enumerate}
From this it follows that $\phi$ induces a well defined map of abelian
groups $\bar{\phi}:W(\ff_{p})\to L_{p}/L_{p-1}$. The map $\bar{\phi}$
is surjective because $L_{p}/L_{p-1}$ generated by $\left\langle pm\right\rangle $,
all prime divisors $q$ of $m$ are $q<p$. By the lemma $\left\langle pm\right\rangle \in\im(\bar{\phi})$.
It is injective because $\underset{\left\langle u\right\rangle \mapsto\left\langle pn\right\rangle \mapsto\left\langle \underline{n}\right\rangle =\left\langle u\right\rangle }{W(\ff_{p})\to L_{p}/L_{p-1}\overset{\partial_{p}}{\to}W(\ff_{p})}$
is the identity $(*)$. Hence $\bar{\phi}$ is an isomorphism$\underset{(*)}{\Rightarrow}\partial_{p}:L_{p}/L_{p-1}\overset{\cong}{\to}W(\ff_{p})$
is an isomorphism. This finishes the claim.

We prove by induction on $n\geq1$ that $L_{n}/L_{1}\to\oplus_{p\leq n}W(\ff_{p})$
is an isomorphism. 

The case $n=1$ is clear as both sides are 0

The case $n=2$ is true by the claim

$n-1$ to $n$: If $n$ is not a prime then LHS$_{n}$=LHS$_{n-1}$=RHS$_{n-1}$=RHS$_{n}$

If $n$ is a prime, we have a map of short exact sequences: 
\[
\xymatrix{0\ar[r] & L_{n-1}/L_{1}\ar[r]\ar^{\cong\mathrm{\, by\, induction}}[d] & L_{n}/L_{1}\ar[r]\ar[d] & L_{n}/L_{n-1}\ar^{\cong\mathrm{\, by\, claim}}[d]\ar[r] & 0\\
0\ar[r] & \oplus_{p\leq n-1}W(\ff_{p})\ar[r] & \oplus_{p\leq n}W(\ff_{p})\ar[r] & W(\ff_{n})\ar[r] & 0
}
\]
By the five lemma we have that $L_{n}/L_{1}\overset{\partial_{p}}{\to}\oplus_{p\leq n}W(\ff_{p})$
is also an isomorphism. 

Hence $W(\qq)/L_{1}=\cup_{n\geq1}L_{n}/L_{1}\overset{\cong}{\to}\cup_{n\geq1}\oplus_{p\leq n}W(\ff_{p})=\oplus_{p\geq1,\mbox{\ensuremath{\mathrm{prime}}}}W(\ff_{p})$.
So we get the exact sequence
\[
\xymatrix{0\ar[r] & L_{1}\ar[r] & W(\qq)\ar[r] & \oplus_{p\in\zz\geq2\mathrm{\, prime}}W(\ff_{p})\ar[r] & 0\\
 & W(\zz)\ar@{^{(}->}[ur]\ar_{0}[urr]
}
\]
Since $W(\zz)\subset W(\qq)$ and $(\oplus\partial_{p})W(\zz)=0\Rightarrow W(\zz)\subset L_{1}$.
But $L_{1}\subset W(\zz)$ because $L_{1}$ is generated by $\left\langle 1\right\rangle ,\left\langle -1\right\rangle \Rightarrow W(\zz)=L_{1}$,
and we have exactness of $0\to W(\zz)\to W(\qq)\to\oplus_{p}W(\ff_{p})\to0$.
Finally the map $W(\zz)\to W(\rr)\underset{\sgn}{\cong}\zz$ is an
isomorphism. This is due to the fact it is surjective since if $U\in W(\rr)$
then $U=n\left\langle 1\right\rangle +m\left\langle -1\right\rangle $
but $n\left\langle 1\right\rangle +m\left\langle -1\right\rangle \in W(\zz)$.
It is also injective since $W(\zz)=L_{1}$ is generated by $\left\langle 1\right\rangle ,\left\langle -1\right\rangle $,
so every element $V$ of $W(\zz)$ has the form $V=n\left\langle 1\right\rangle +m\left\langle -1\right\rangle $
which is zero in $W(\rr)\iff n-m=\sgn(n\left\langle 1\right\rangle +m\left\langle -1\right\rangle )=0\in\zz\iff n=m\iff V=n\left\langle 1\right\rangle +n\left\langle -1\right\rangle =n(\left\langle 1\right\rangle +\left\langle -1\right\rangle )=0\in W(\zz)$.
Hence $W(\zz)\to W(\rr)$ is an isomorphism\end{proof}
\begin{cor}
The map 
\[
W(\qq)\to W(\rr)\oplus\bigoplus_{p\in\zz_{\geq2}\mathrm{\, prime}}W(\ff_{p})
\]
 defined by $M\mapsto(M\otimes_{\qq}\rr,\sum_{p}\partial_{p}M)$ is
an isomorphism.\end{cor}
\begin{proof}
This follows from the exact sequence $0\to W(\zz)\to W(\qq)\to\oplus_{p}W(\ff_{p})\to0$,
which is split exact via 
\[
\xymatrix{W(\zz)\ar@{^{(}->}[r]\ar_{\cong}[dr] & W(\qq)\ar[d]\\
 & W(\rr)
}
\]
\end{proof}
\begin{cor}
Two symmetric inner product spaces $M,N$ over $\qq$ are isometric
$\iff\sgn M=\sgn N,\rk M=\rk N$, $\partial_{p}M=\partial_{p}N\in W(\ff_{p})\,\forall p\in\zz$
prime. (In terms of quadratic forms, any two regular quadratic forms
are equivalent over $\qq$ if and only if the previous condition are
fulfilled)\end{cor}
\begin{proof}
$M\cong N\iff\rk M=\rk N$ and $[M]=[N]\in W(\qq)\iff\rk M=\rk N$
and $\underset{\sgn M=\sgn N}{\underbrace{[M]=[N]\in W(\rr)}}$ and
$\partial_{p}M=\partial_{p}N$\end{proof}
\begin{cor}[Weak Hasse Principle]
 The map%
\footnote{In the lectures I carelessly wrote $\bigoplus_{p}$ instead of $\prod_{p}$
but the image of $W(\qq)$ does not lie in $\bigoplus_{p}$, otherwise,
what is the image of $\langle1\rangle$ which is $\neq0\in W(F)$
for any field $F$?%
} 
\[
W(\qq)\hookrightarrow W(\rr)\oplus\prod_{\zz\ni p\,\mathrm{prime}}W(\qq_{p})
\]
 is injective. In particular two inner product spaces $M,N$ over
$\qq$ are isometric over $\qq$ if and only if $M$ and $N$ are
isometric over $\rr$ and $\qq_{p}$ for all $p\in\zz$ prime.\end{cor}
\begin{proof}
We have the following commutative diagram by definition of $\partial_{p}$
\[
\xyC{1.5pc}\xyR{2.5pc}\xymatrix{W(\qq)\ar_{(b)}[d]\ar_{(a)}^{\cong}[rr] &  & W(\rr)\oplus\bigoplus_{p\,\mathrm{prime}}W(\ff_{p})\ar[d]^{(c)}\\
W(\rr)\oplus\prod_{p\,\mathrm{prime}}W(\qq_{p})\ar_{\id_{W(\rr)}\oplus\prod\partial_{p}}[rr] &  & W(\rr)\oplus\prod_{p\,\mathrm{prime}}W(\ff_{p})
}
\]
Now $(a)$ isomorphism and $(c)$ injective implies $(b)$ injective.

$M\cong_{\qq}N\Rightarrow M\cong_{\rr}N$ and $M\cong_{\qq_{p}}N$
for all $p\in\zz$ prime.

Assume $M\cong_{\rr}N$ and $M\cong_{\qq_{p}}N$ for all $p\in\zz$
prime. Then $\rk M=\rk N$ and $[M]=[N]\in W(\rr)$ and $[M]=[N]\in W(\qq_{p})$
for all $p$. But $(b)$ injective $\Rightarrow\rk M=\rk N,[M]=[N]\in W(\qq)\underset{\mathrm{char\,}\qq\neq2}{\iff}M\cong N$\end{proof}
\begin{example*}
{}We start with two side remarks: The quadratic form $q=\sum_{i=1}^{n}a_{i}x_{i}^{2}+\sum_{i<j}a_{ij}x_{i}x_{j}$
has associated form matrix $B=(\beta_{ij})$ (with respect to $e_{1},e_{2},e_{3},\dots,e_{n}$)
where $\beta_{ij}=q(e_{i}+e_{j})-q(e_{i})-q(e_{j})=\begin{cases}
a_{ij} & i<j\\
a_{ij} & j<i\\
2a_{i} & i=j
\end{cases}$

That is, $B=\begin{pmatrix}2a_{1} & a_{12} & \dots & a_{1n}\\
a_{12} & 2a_{2}\\
\vdots &  & \ddots\\
a_{1n} &  &  & 2a_{n}
\end{pmatrix}$.

The diagonalisation of a symmetric matrix $B=\left(\begin{smallmatrix}a_{1} & a_{12} & \dots & a_{1n}\\
a_{12} & a_{2}\\
\vdots &  & \ddots\\
a_{1n} &  &  & a_{n}
\end{smallmatrix}\right)$ is $\left\langle B\right\rangle =\left\langle d_{1},\frac{d_{2}}{d_{1}},\dots,\frac{d_{n}}{d_{n-1}}\right\rangle $
where $d_{i}=$ determinant of the upper left corner of size $i\times i$
of $B$, provided $d_{1},\dots,d_{n-1}\neq0$.
\begin{enumerate}
\item Does $15=x^{2}+2xy+3y^{2}-4yz$ have a solution $x,y,z\in\qq$?


Solution: Let $q=x^{2}+2xy+3y^{2}-4yz$. Does $q$ represent $15$?
The associated symmetric bilinear form $\beta(u,v)=q(u+v)-q(u)-q(v)$,
$u,v\in\qq^{3}$ has form matrix 
\[
B=\begin{pmatrix}2 & 2 & 0\\
2 & 6 & -4\\
0 & -4 & 0
\end{pmatrix}
\]
 which has determinant $-32$ hence it is non-degenerate. It has diagonalisation
$\left\langle B\right\rangle \cong\left\langle 2,\frac{8}{2},\frac{-32}{8}\right\rangle \cong\left\langle 2,\underset{\hh}{\underbrace{1,-1}}\right\rangle $$\underset{\mathrm{exercise}}{\Rightarrow}q$
isotropic and represent any rational number. In particular there exists
$x,y,z$ such that $q(x,y,z)=15$


Note that $\left\langle B\right\rangle \cong\left\langle 2,4,-4\right\rangle \Rightarrow q\cong x^{2}+2y^{2}-2z^{2}\cong x^{2}+yz$

\item Does $15=x^{2}+4xy-2xz+7y^{2}-4yz+z^{2}=:q$ has a solution $x,y,z\in\qq$.


Solution: The associated bilinear form $\beta$ of $q$ has matrix
form
\[
B=\begin{pmatrix}2 & 4 & -2\\
4 & 14 & -4\\
-2 & -4 & 2
\end{pmatrix}
\]
with determinant $=0$ (since $Be_{3}=-Be_{1}$). So $q$ is degenerate,
and we can eliminate a variable as follows: The inner product space
$\left\langle B\right\rangle $ has diagonalisation $\left\langle B\right\rangle \cong\underset{\mathrm{non-degenrate\, as\,}\det\left(\begin{smallmatrix}2 & 4\\
4 & 14
\end{smallmatrix}\right)=12\neq0}{\underbrace{(\qq e_{1}+\qq e_{2})}}+\underset{\mathrm{dim}=1,\mathrm{degenerate\, as\,}\det B=0}{\underbrace{(\qq e_{1}+\qq e_{2})^{\perp}}}\cong\left\langle \begin{pmatrix}2 & 4\\
4 & 14
\end{pmatrix}\right\rangle \perp\left\langle 0\right\rangle \cong\left\langle 2,\frac{12}{2}\right\rangle \perp\left\langle 0\right\rangle \cong\left\langle 2,6\right\rangle \perp\left\langle 0\right\rangle $. This means that $q\cong x^{2}+3y^{2}$, so does this represent 15?
This is equivalent to asking $\left\langle 2,6\right\rangle \cong\left\langle 30,a\right\rangle $
for some $a\in\qq^{*}$. Then $\det$ LHS $=\det$ RHS modulo square
units $\iff\left\langle 1,3\right\rangle \cong\left\langle 15,5\right\rangle \iff\left\langle 1,3\right\rangle =\left\langle 15,5\right\rangle \in W(\qq)$
(because $\left\langle 1,3\right\rangle $ and $\left\langle 15,5\right\rangle $
have the same rank) $\iff\left\langle 1,3\right\rangle \cong\left\langle 15,5\right\rangle \in W(\rr)$
and $\partial_{p}\left\langle 1,3\right\rangle \cong\partial_{p}\left\langle 15,5\right\rangle \in W(\ff_{p})$
for all $p$ prime. 
\begin{itemize}
\item If $p\neq3$ or $5$ then $\partial_{p}\left\langle 1,3\right\rangle =0=\partial_{p}\left\langle 15,5\right\rangle $
\item if $p=3$ then $\partial_{3}\left\langle 1,3\right\rangle =\partial_{3}\left\langle 1\right\rangle +\partial_{3}\left\langle 3\right\rangle =0+\left\langle 1\right\rangle $,
and $ $$\partial_{3}\left\langle 15,5\right\rangle =\partial_{3}\left\langle 15\right\rangle +\partial_{3}\left\langle 5\right\rangle =\left\langle 5\right\rangle +0=\left\langle -1\right\rangle $.
Do they agree in $W(\ff_{3})$? No because $\left\langle 1\right\rangle \neq\left\langle -1\right\rangle \in W(\ff_{3})\cong\zz/4\zz$
generated by $\left\langle 1\right\rangle $$\Rightarrow\left\langle 1\right\rangle -\left\langle -1\right\rangle =2\left\langle 1\right\rangle \neq0\in W(\ff_{3})$
\end{itemize}

We have showed that $\left\langle 1,3\right\rangle \neq\left\langle 15,5\right\rangle \in W(\qq)\Rightarrow q$
does not represent 15 and the equation has no solution in $x,y,z\in\qq$

\end{enumerate}
\end{example*}

\subsection{The Brauer Group and the Hasse Invariant}

Recall from MA377 (Rings and Modules):
\begin{defn}
Let $k$ be a field, a $k$-algebra $A$ is called:
\begin{itemize}
\item \emph{central:} if $k\overset{\cong}{\to}Z(A)$, where $Z(A)$ denotes
the center of $A$
\item \emph{simple:} if $A\neq0$ and the only ideals of $A$ are $0$ and
$A$
\item \emph{finite dimensional}: if $\dim_{k}A<\infty$
\end{itemize}
\end{defn}
\begin{fact*}
Let $A,B$ be finite dimensional central simple $k$-algebras. Then:
\begin{enumerate}
\item $A=M_{n}(D)$ where $D$ is a finite dimensional division $k$-algebra
\item $A\otimes_{k}B$ is also a finite dimensional central simple $k$-algebra
\item $A\otimes_{k}A^{op}\cong M_{n}(k)$ where $n=\dim_{k}A$
\end{enumerate}
\end{fact*}
\begin{defn}
Let $F$ be a field. The \emph{Brauer group, $\mbox{Br}(F)$, }is
the set of Brauer equivalence classes $[A]$ of finite dimensional
central simple $F$-algebras $A$, where $A\sim B$ ($A$ is \emph{Brauer
equivalent }to $B$) if $M_{m}(A)\cong M_{n}(B)$ as $F$-algebras
for some $m,n\in\nn_{\geq1}$.
\end{defn}
$\Br(F)$ is a group with group law: $[A][B]:=[A\otimes_{F}B]$, with
$1=[F]$ and $[A]^{-1}=[A^{op}]$. Indeed $\Br(F)$ is an abelian
group: $[A][B]=[A\otimes_{F}B]=[B\otimes_{F}A]=[B][A]$, $1[A]=[F][A]=[F\otimes_{F}A]=[A]$,
$[A][A^{op}]=[A\otimes_{F}A^{op}]\underset{\mathrm{Fact}}{=}[M_{n}(F)]=[F]=1$
\begin{example*}
(From MA377)\end{example*}
\begin{itemize}
\item $\Br(\cc)=\Br(F)=\{F\}=0$ where $F=\bar{F}$ is algebraically closed
\item $\Br(F)=0$ if $F$ is a finite field
\item $\Br(\rr)=\{\rr,\hh\}=\zz/2$\end{itemize}
\begin{defn}
Let $F$ be a field with $\mbox{char }F\neq2$ and $a,b\in F^{*}$.
Let $(\frac{a,b}{F})$ be the $4$-dimensional $F$-algebra with basis,
$1,i,j,k$ such that $i^{2}=a,j^{2}=b,k=ij=-ji$\end{defn}
\begin{note*}
$k^{2}=ij(-ji)=-ab$\end{note*}
\begin{fact*}
For $a,b\in F^{*}$, $(\frac{a,b}{F})$ is a $4$-dimensional central
simple $F$-algebra.\end{fact*}
\begin{defn}
An $F$-algebra which is $F$-algebra isomorphic to $(\frac{a,b}{F})$
for some $a,b\in F^{*}$ is called (generalized) \emph{quaternion
algebra} (over $F$)\end{defn}
\begin{namedthm}[Structure theorem for Quaternion algebras]
 Let $F$ be a field with $\mbox{char }F\neq2$. Then $(\frac{a,b}{F})\cong(\frac{c,d}{F})\iff\left\langle a,b,-ab\right\rangle =\left\langle c,d,-cd\right\rangle \in W(F)$\end{namedthm}
\begin{rem*}
Let $A,B$ be finite dimensional central simple $F$-algebras. Then
$A\cong B\iff\dim_{F}A=\dim_{F}B$ and $[A]=[B]\in\Br(F)$.

In particular $[(\frac{a,b}{F})]=[(\frac{c,d}{F})]\in\Br(F)\iff\left\langle a,b,-ab\right\rangle =\left\langle c,d,-cd\right\rangle \in W(F)$\end{rem*}
\begin{example*}
$(\frac{1,1}{F})\cong M_{2}(F)$ by $i\mapsto\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix},j\mapsto\begin{pmatrix}1 & 0\\
0 & -1
\end{pmatrix}$

$(\frac{-1,-1}{\rr})=$ Real quaternion algebra.

$M_{2}(\rr)=(\frac{1,1}{\rr})\ncong(\frac{-1,-1}{\rr})$ because $\underset{\sgn=1}{\underbrace{\left\langle 1,1,-1\right\rangle }}\neq\underset{\sgn=-3}{\underbrace{\left\langle -1,-1,-1\right\rangle }}\in W(\rr)$\end{example*}
\begin{rem*}
$(\frac{a,b}{F})$ is a division algebra $\iff$$(\frac{a,b}{F})\ncong M_{2}(F)$$\iff\left\langle a,b,-ab\right\rangle \ncong\left\langle 1,1,-1\right\rangle \iff\left\langle a,b,-ab\right\rangle $
is an isotropic (i.e., does not represent $0$)
\end{rem*}

\begin{rem*}
$(\frac{a,b}{F})\cong(\frac{a,b}{F})^{op}$ by $1\mapsto1,i\mapsto-i,j\mapsto-j,k\mapsto-k$.
This means that $[(\frac{a,b}{2})]$ has order $2$ in $\Br(F)$ because
$(\frac{a,b}{F})\otimes_{F}(\frac{a,b}{F})\cong(\frac{a,b}{F})\otimes_{F}(\frac{a,b}{F})^{op}\cong M_{4}(F)\Rightarrow[(\frac{a,b}{F})][(\frac{a,b}{F})]=[M_{4}(F)]=[F]=1\in\Br(F)$.
Hence $[(\frac{a,b}{F})]\in\,_{2}\Br(F)$, where for an abelian group
$G$ we denote $_{2}G=\{x\in G|x^{2}=1\}$\end{rem*}
\begin{lem}
\label{lem:Prodict of (1,1/F)}Let $F$ be a field with $\mbox{char }F\neq2$.
Then:
\begin{enumerate}
\item $(\frac{a,b}{F})\cong(\frac{a,-ab}{F})\cong(\frac{b,-ab}{F})$
\item $(\frac{a,b}{F})\otimes_{F}(\frac{a,c}{F})\cong(\frac{a,bc}{F})\otimes_{F}M_{2}(F)$
\end{enumerate}
\end{lem}
\begin{proof}
{}
\begin{enumerate}
\item $(\frac{a,b}{F})\cong(\frac{a,-ab}{F})$ because $\left\langle a,b,-ab\right\rangle \cong\left\langle a,-ab,a^{2}b\right\rangle $
\item Let $A=(\frac{a,b}{F}),B=(\frac{a,c}{F})$ have basis $\mathscr{B}_{A}=\{1,i_{A},j_{A},k_{A}\}$
and $\mathscr{B}_{B}=\{1,i_{B},j_{B},k_{B}\}$ respectively. Then
$A\otimes_{F}B$ has basis $\{u\otimes v|u\in\mathscr{B}_{A},v\in\mathscr{B}_{B}\}$.
Let $\Sigma_{A}=\{1\otimes1,i_{A}\otimes1,j_{A}\otimes j_{B},k_{A}\otimes j_{B}\}\subset A\otimes_{F}B$
and $\Sigma_{B}=\{1\otimes1,1\otimes j_{B},i_{A}\otimes k_{B},-ci_{A}\otimes i_{B}\}$.
Then $\Sigma_{A},\Sigma_{B}$are the basis of $A',B'\subset A\otimes_{F}B$-subalgebras
with $A'\cong(\frac{a,bc}{F})$ and $B'\cong(\frac{c,-a^{2}c}{F})$
because

\begin{itemize}
\item $(i_{A}\otimes1)^{2}=i_{A}^{2}\otimes1=a(1\otimes1)=a$
\item $(j_{A}\otimes j_{B})^{2}=j_{A}^{2}\otimes j_{B}^{2}=b\otimes c=bc(1\otimes1)=bc$
\item $(1_{A}\otimes1)(j_{A}\otimes j_{B})=k_{A}\otimes j_{B}=-(j_{A}\otimes j_{B})(i_{A}\otimes1)$
\end{itemize}

and
\begin{itemize}
\item $(1\otimes j_{B})^{2}=1\otimes j_{B}^{2}=c$
\item $(i_{A}\otimes k_{B})^{2}=i_{A}^{2}\otimes k_{B}^{2}=a\cdot(-ac)=-a^{2}c$
\item $(1\otimes j_{B})(i_{A}\otimes k_{B})=-(i_{A}\otimes k_{B})(1\otimes j_{B})=i_{A}\otimes-ci_{B}=-ci_{A}\otimes i_{B}$
\end{itemize}

But $(\frac{c,-a^{2}c}{F})\cong(\frac{1,1}{F})=M_{2}(F)$ because
$\left\langle c,-a^{2}c,a^{2}c^{2}\right\rangle \cong\left\langle c,-c,1\right\rangle \cong\left\langle 1,1,-1\right\rangle $.
Every element of $A'$ commutes with every element of $B'$ (one checks
that $\Sigma_{A}$ commutes with $\Sigma_{B}$)$\Rightarrow$ The
map $\phi:$$A'\otimes_{F}B'\to A\otimes_{F}B$ defined by $x\otimes y\mapsto xy$
is a well defined map of $F$-algebras. The elements $\{xy|x\in\Sigma_{A},y\in\Sigma_{B}\}$
are linearly independent in $A\otimes_{F}B$ (check!) $\Rightarrow\phi$
is injective. Since $\dim_{F}A'\otimes_{F}B'=8=\dim_{F}A\otimes_{F}B$,
this means that $\phi$ is an isomorphism $\Rightarrow(\frac{a,bc}{F})\otimes_{F}M_{2}(F)\cong A'\otimes_{F}B'\cong A\otimes_{F}B\cong(\frac{a,b}{F})\otimes_{F}(\frac{a,c}{F})$

\end{enumerate}
\end{proof}
\begin{defn}
Let $F$ be a field with $\mbox{char }F\neq2$, and $V$ be a symmetric
inner product space over $F$ with diagonalisation $V\cong\left\langle a_{1},\dots,a_{n}\right\rangle $.
The \emph{Hasse invariant} of $V$ is the algebra 
\[
\Hasse(V)=\prod_{1\leq i<j\leq n}\left(\frac{a_{i},a_{j}}{F}\right)\in\,_{2}\Br(F)
\]
\end{defn}
\begin{lem}
$\Hasse(V)\in_{2}\Br(F)$ does not depend on diagonalisation $\left\langle a_{1},\dots,a_{n}\right\rangle $
of $V$ used to define $\Hasse(V)$\end{lem}
\begin{proof}
{}
\begin{enumerate}
\item $\Hasse(\left\langle a_{1},\dots a_{i},a_{i+1},\dots,a_{n}\right\rangle )=\Hasse(\left\langle a_{1},\dots,a_{i-1},a_{i+1},a_{i},a_{i+2},\dots,a_{n}\right\rangle )$.
This is because 
\begin{eqnarray*}
\mbox{LHS} & = & \left(\prod_{r<s,\{r,s\}\neq\{i,i+1\}}\left(\frac{a_{r},a_{s}}{F}\right)\right)\left(\frac{a_{i},a_{i+1}}{F}\right)\\
\mbox{RHS} & = & \left(\prod_{r<s,\{r,s\}\neq\{i,i+1\}}\left(\frac{a_{r},a_{s}}{F}\right)\right)\left(\frac{a_{i+1},a_{i}}{F}\right)
\end{eqnarray*}
Since $(\frac{a,b}{F})\cong\left(\frac{b,a}{F}\right)\Rightarrow$
LHS = RHS. Hence for all $\sigma\in\Sigma_{n}=$ permutation group,
we have $\Hasse(\left\langle a_{1},\dots,a_{n}\right\rangle )=\Hasse(\left\langle a_{\sigma(1)},\dots,a_{\sigma(n)}\right\rangle )$
\item $\mbox{char}\neq2$, if $\left\langle a_{1},\dots,a_{n}\right\rangle $
and $\left\langle b_{1},\dots,b_{n}\right\rangle $ are diagonalisation
of $V$ then $\left\langle a_{1},\dots,a_{n}\right\rangle \approx\left\langle b_{1},\dots,b_{n}\right\rangle $
(Chain equivalence Theorem \vpageref{Chain equivalence form}). Hence
it suffices to show that $\Hasse(\left\langle a_{1},\dots,a_{n}\right\rangle )=\Hasse(\left\langle b_{1},\dots,b_{n}\right\rangle )$
for $\left\langle a_{1},\dots,a_{n}\right\rangle \approx_{s}\left\langle b_{1},\dots,b_{n}\right\rangle $
simply chain equivalent. By 1. it suffices to show $\Hasse(\left\langle a,b,e_{1},\dots,e_{n}\right\rangle )=\Hasse(\left\langle c,d,e_{1},\dots,e_{n}\right\rangle )$
where $\left\langle a,b\right\rangle \cong\left\langle c,d\right\rangle $.
Recall that $\left\langle a,b\right\rangle \cong\left\langle c,d\right\rangle \iff ab=cd\cdot x^{2},a=cy^{2}+dz^{2}$
for some $x,y,z\in F$. Now
\begin{eqnarray*}
\mbox{LHS} & = & \left(\frac{a,b}{F}\right)\prod_{i=1}^{n}\left(\frac{a,e_{i}}{F}\right)\left(\frac{b,e_{i}}{F}\right)\Hasse(\left\langle e_{1},\dots,e_{n}\right\rangle )\\
\mbox{RHS} & = & \left(\frac{c,d}{F}\right)\prod_{i=1}^{n}\left(\frac{c,e_{i}}{F}\right)\left(\frac{d,e_{i}}{F}\right)\Hasse(\left\langle e_{1},\dots,e_{n}\right\rangle )
\end{eqnarray*}
Note that 
\begin{eqnarray*}
\left(\frac{a,e}{F}\right)\left(\frac{b,e}{F}\right) & = & \left(\frac{ab,e}{F}\right)\in\Br(F)\\
 & = & \left(\frac{cd,e}{F}\right)\mbox{ since }ab=cdx^{2}\\
 & = & \left(\frac{c,e}{F}\right)\left(\frac{d,e}{F}\right)
\end{eqnarray*}
and $(\frac{a,b}{F})\cong(\frac{c,d}{F})$ because $\left\langle a,b,-ab\right\rangle =\left\langle a,b\right\rangle +\left\langle -ab\right\rangle =\left\langle c,d\right\rangle +\left\langle -cd\right\rangle =\left\langle c,d,-cd\right\rangle $
\end{enumerate}
\end{proof}
\begin{lem}
$\Hasse(V\perp W)=\Hasse(V)\Hasse(W)\cdot(\frac{\det V,\det W}{F})$\end{lem}
\begin{proof}
Exercise
\end{proof}
So $\Hasse(-)$ does not define a group homomorphism $W(F)\to\,_{2}\Br(F)$.
$\Hasse(\hh)=\Hasse(\left\langle 1,-1\right\rangle )=(\frac{1,-1}{F})\cong(\frac{1,1}{F})=M_{2}(F)$
because $\left\langle 1,-1,1\right\rangle \cong\left\langle 1,1,-1\right\rangle $.
But $\Hasse(\hh^{2})=\Hasse(\hh)\Hasse(\hh)\cdot(\frac{\det\hh,\det\hh}{F})=(\frac{-1,-1}{F})\neq(\frac{1,1}{F})=M_{2}(F)=F\in\Br$
in general. 


\subsection{Tensor Product of Inner Product Spaces}
\begin{defn}
Let $(M,\beta),(B,\gamma)$ be symmetric bilinear forms over $R$.
We define $(M\otimes_{R}N,\beta\otimes_{R}\gamma)$ to be the bilinear
form $\beta\otimes\gamma:M\otimes_{R}N\times M\otimes_{R}N\to R$
defined by $(x\otimes u,y\otimes v)\mapsto\beta(x,y)\cdot\gamma(u,v)$,
which is symmetric: $\beta\otimes\gamma(x\otimes u,y\otimes v)=\beta(x,y)\gamma(u,v)=\beta(y,x)\gamma(v,u)=\beta\otimes\gamma(y\otimes v,x\otimes u)$ \end{defn}
\begin{lem}
Let $P,Q$ be finitely generated $R$-module then the following map
$\phi:\Hom_{R}(P,R)\otimes_{R}\Hom_{R}(Q,R)\to\Hom_{R}(P\otimes_{R}Q,R)$
defined by $f\otimes g\mapsto f\cdot g$ where $(f\cdot g)(x\otimes u)=f(x)g(u)$,
is an isomorphism\end{lem}
\begin{proof}
$\phi$ is an isomorphism for $(P,Q)=(R,R)$.

If $\phi$ is an isomorphism for $(P_{1},Q)$ and $(P_{2},Q)$ then
$\phi$ is an isomorphism for $(P_{1}\oplus P_{2},Q)$ because $(P_{1}\oplus P_{2})\otimes Q=P_{1}\otimes Q\oplus P_{2}\otimes Q$,
$\Hom(P_{1}\oplus P_{2},R)=\Hom(P_{1},R)\oplus\Hom(P_{2},R)$ and
$\phi_{1}\oplus\phi_{2}$ is an isomorphism if and only if $\phi_{1}$
and $\phi_{2}$ are isomorphisms. $\Rightarrow\phi$ is isomorphism
for $(P,Q)=(R^{m},R^{n})$ $m,n\in\zz_{\geq0}$.

A finitely generated projective module is a direct factor of $R^{n}$
for some $n$. If $\phi_{1}$ is a direct summand of a map $\phi$
which is an isomorphism then $\phi_{1}$ is an isomorphism $\Rightarrow\phi$
is an isomorphism for $P,Q$ finitely generated projective modules.\end{proof}
\begin{lem}
Let $(M,\beta),(N,\gamma)$ be symmetric inner product spaces over
$R$. Then $(M\otimes_{R}N,\beta\otimes\gamma)$ is an inner product
space over $R$\end{lem}
\begin{proof}
$M,N$ is finitely generated projective $\Rightarrow$ $M\otimes_{R}N$
is finitely generated projective. We need to show $\beta\otimes\gamma$
is non-degenerate. Now $\beta,\gamma$ non-degenerated $\iff M\to\Hom_{R}(M,R)$
defined by $x\mapsto\beta(x,-)$ and $N\to\Hom_{R}(N,R)$ defined
by $y\mapsto\gamma(y,-)$ are isomorphisms. $\beta\otimes\gamma$
is non-degenerate $\iff M\otimes_{R}N\to\Hom_{R}(M\otimes_{R}N,R)$
defined by $x\otimes y\mapsto\beta(x,-)\gamma(y,-)$ is an isomorphism,
but this map is the composition of the following two maps
\[
M\otimes_{R}N\overset{\cong}{\longrightarrow}\Hom_{R}(M,R)\otimes\Hom_{R}(N,R)\overset{\cong}{\underset{\mathrm{Lemma}}{\longrightarrow}}\Hom_{R}(M\otimes_{R}N,R)
\]
\[
x\otimes y\mapsto\beta(x,-)\otimes\gamma(y,-)\mapsto\beta(x,-)\cdot\gamma(y,-)
\]
\end{proof}
\begin{lem}[\textbf{Definition}]
 The Wilt group $W(R)$ of a a commutative ring $R$ is a commutative
ring with multiplication $[M,\beta]\cdot[N,\gamma]=[(M,\beta)\otimes_{R}(N,\gamma)]$
and unit $\left\langle 1\right\rangle $. $W(R)$ is called the \emph{Witt
ring of $R$.}\end{lem}
\begin{proof}
We need to show that if $(M,\beta)$ is metabolic and $(N,\gamma)$
arbitrary then $(M,\beta)\otimes(N,\gamma)$ is metabolic. But a Lagrangian
$L\subset M$ of $(M,\beta)$ defines a Lagrangian $L\otimes N\subset M\otimes N$
of $(M\otimes N,\beta\otimes\gamma)$ (exercise)\end{proof}
\begin{rem*}
$\left\langle u\right\rangle \cdot\left\langle v\right\rangle =\left\langle uv\right\rangle \in W(R)$\end{rem*}
\begin{defn}
Let $R$ be a local ring then the rank map $W(R)\to\zz/2$ defined
by $M\mapsto\rk M$ is a ring homomorphism. The kernel $\ker(\rk)$
is an ideal $I(R)$ which is called \emph{the fundamental ideal}. \end{defn}
\begin{rem*}
$I(F)$ is generated by even dimensional forms, hence additively generated
by 2 dimensional forms $\left\langle a,b\right\rangle =\left\langle a,1\right\rangle -\left\langle -b,1\right\rangle \Rightarrow I(F)$
is additively generated by $\left\langle a,1\right\rangle ,a\in F^{*}$$\Rightarrow I^{2}(F)$
is additively generated by $\left\langle a,1\right\rangle \otimes\left\langle b,1\right\rangle =\left\langle ab,a,b,1\right\rangle $,
the discriminant map, $\mbox{disc}:I(F)\to F^{*}/F^{2*}$ defined
by $V\mapsto(-1)^{\frac{\dim V}{2}}\det V$, in our case we have $\left\langle ab,a,b,1\right\rangle \mapsto a^{2}b^{2}=1\in F^{*}/F^{2*}$
hence $\mbox{disc}(I^{2})=0$ and $I(F)/I^{2}(F)\to F^{*}/F^{2*}$
well defined surjective map of abelian groups\end{rem*}
\begin{thm}[Pfister]
 The map $I(F)/I^{2}(F)\to F^{*}/F^{2*}$ is an isomorphism for all
fields $F$.\end{thm}
\begin{proof}
The map is surjective because $I(F)\overset{\mbox{disc}}{\to}F^{*}/F^{2*}$
sends $\left\langle a,-1\right\rangle $ to $a$ for $a\in F^{*}$.

In $W(F)/I^{2}$ we have:
\begin{enumerate}
\item $\left\langle a\right\rangle +\left\langle b\right\rangle =\left\langle -ab\right\rangle +\left\langle -1\right\rangle $
because $\left\langle ab,a,b,1\right\rangle \in I^{2}$
\item $3\left\langle -1\right\rangle =\left\langle 1\right\rangle $ because
$\left\langle 1,1,1,1\right\rangle =\left\langle 1,1\right\rangle \otimes\left\langle 1,1\right\rangle \in I^{2}$,
hence $4\left\langle -1\right\rangle =0$.
\end{enumerate}
If $\xi=\left\langle u_{1},\dots,u_{n}\right\rangle \in I/I^{2}$
then $n=2m$. For $u=\mbox{dics}\xi$ we have 
\begin{eqnarray*}
\xi=\left\langle u_{1},\dots,u_{2m}\right\rangle  & \underset{1.}{=} & \langle-(-1)^{m}u,\underset{2m-1}{\underbrace{-1,\dots,-1}}\rangle\mbox{ in }I/I^{2}\\
 & \underset{2.}{=} & \left\{ \begin{array}{cl}
\langle u,-1,-1,-1\rangle & m\ \text{even}\\
\langle u,-1\rangle & m\ \text{odd}
\end{array}\right.\\
 & = & \left\langle -u,1\right\rangle 
\end{eqnarray*}
where the last equation follows from 2 when $m$ is even, and when
$m$ is odd we have $\langle u,-1\rangle=\langle-u,1\rangle$ because
$\langle u,u,-1,-1\rangle=\langle-u^{2},-1,-1,-1\rangle=\langle-1,-1,-1,-1\rangle=0\in I/I^{2}$,
by 1 and 2. Thus, if $\xi$ is in the kernel of the discriminant map
then $1=\mbox{disc}(\xi)=\mbox{disc}(-u,1)=u\Rightarrow u=1\in F^{*}/F^{2*}\Rightarrow\xi=\left\langle -u,1\right\rangle =\left\langle -1,1\right\rangle =0\in I/I^{2}\Rightarrow$
$\xi=0\in I/I^{2}$ and the map $I/I^{2}\to F^{*}/F^{2*}$ is injective.\end{proof}
\begin{example*}
{}
\begin{itemize}
\item $I^{2}(\ff_{q})=0$ because $\mbox{disc}:I(\ff_{q})\overset{\cong}{\to}\ff_{q}^{*}/\ff_{q}^{2*}$
is an isomorphism. 
\item $I^{2}(F)=I(F)=0$ for any algebraically closed field $F$ because
$\rk:W(F)\overset{\cong}{\to}\zz/2\zz$
\item $\xymatrix{I(\rr)\ar@{=}[d]\ar@{^{(}->}[r] & W(\rr)\ar^{\rk}[r]\ar_{\sgn}@{=}[d] & \zz/2\zz\\
2\zz\subset & \zz\ar_{\mathrm{reduction\, mod\,2}}[ur]
}
$
\item We'll see later: $I^{2}(\qq_{p})=\zz/2\zz$
\end{itemize}
\end{example*}
\begin{defn}
Let $V$ be a $4k$-dimensional symmetric inner product space over
$F$ with $\mbox{disc}V$ ($=\det V$) $=1$. \emph{The Signed Hasse
Invariant }is $s(V)=(\frac{-1,-1}{F})^{k}\Hasse(V)=(\frac{(-1)^{k}-1}{F})\Hasse(V)$\end{defn}
\begin{note*}
If $V,W$ have dimension divisible by 4 and $\mbox{disc}V=\mbox{disc}W=1$
then 
\begin{eqnarray*}
s(V\perp W) & = & (\frac{-1,-1}{F})^{\frac{\dim V+\dim W}{4}}\Hasse(V\perp W)\\
 & = & \left(\frac{-1,-1}{F}\right)^{\frac{\dim V}{4}}\left(\frac{-1,-1}{F}\right)^{\frac{\dim W}{4}}\Hasse(V)\Hasse(W)\underset{\left(\frac{1,1}{F}\right)}{\underbrace{\left(\frac{\det V,\det W}{F}\right)}}\\
 & = & s(V)s(W)
\end{eqnarray*}
and $s(\hh^{2})=(\frac{-1,-1}{2})\Hasse(\hh^{2})=(\frac{-1,-1}{F})(\frac{-1,-1}{F})=[F]\in\Br(F)\Rightarrow s(\hh^{2k})=[F]\in\Br(F)$.
Hence $s:I^{2}(F)\to\,_{2}\Br(F)$ is a well defined map of abelian
groups. (as $I^{2}$ is generated by 4-dimesional spaces)\end{note*}
\begin{lem}
$s(I^{3}F)=0$ for every field $F$ of $\mbox{char}\neq2$\end{lem}
\begin{proof}
$I$ is generated by $\left\langle 1,a\right\rangle \Rightarrow I^{3}$
is generated by $\left\langle 1,a\right\rangle \otimes\left\langle 1,b\right\rangle \otimes\left\langle 1,c\right\rangle =\left\langle 1,a,b,c,ab,ac,bc,abc\right\rangle =\left\langle a_{1},\dots,a_{8}\right\rangle $.
So, 
\begin{eqnarray*}
s(\left\langle a_{1},\dots,a_{8}\right\rangle  & = & (-1)^{\frac{8}{4}}\Hasse(\left\langle a_{1},\dots,a_{8}\right\rangle )\\
 & = & \prod_{i\leq i<j\leq8}\left(\frac{a_{i},a_{j}}{F}\right)\\
 & = & \prod_{1\leq i\leq7}\left(\frac{a_{i}\prod_{i<j\leq8}a_{j}}{F}\right)\mbox{ by Lemma }\ref{lem:Prodict of (1,1/F)}\\
 & = & \left(\frac{1,a^{4}b^{4}c^{4}}{F}\right)\left(\frac{a,a^{3}b^{4}c^{4}}{F}\right)\left(\frac{b,a^{3}b^{3}c^{4}}{F}\right)\left(\frac{c,a^{3}b^{3}c^{3}}{F}\right)\left(\frac{ab,a^{2}b^{2}c^{3}}{F}\right)\left(\frac{ac,ab^{2}c^{2}}{F}\right)\left(\frac{bc,abc}{F}\right)\\
 & = & \left(\frac{1,1}{F}\right)\left(\frac{a,a}{F}\right)\left(\frac{b,ab}{F}\right)\left(\frac{c,abc}{F}\right)\left(\frac{ab,c}{F}\right)\left(\frac{ac,a}{F}\right)\left(\frac{bc,abc}{F}\right)\mbox{ removing powers of }2\\
 & = & \left(\frac{a,a}{F}\right)\left(\frac{b,-a}{F}\right)\left(\frac{c,-ab}{F}\right)\left(\frac{ab,c}{F}\right)\left(\frac{a,-c}{F}\right)\left(\frac{bc,-a}{F}\right)\mbox{ by using the relation }\left(\frac{a,b}{F}\right)=\left(\frac{a,-ab}{F}\right)\\
 & = & \left(\frac{a,a}{F}\right)\left(\frac{a,-c}{F}\right)\left(\frac{b,-a}{F}\right)\left(\frac{bc,-a}{F}\right)\left(\frac{c,-ab}{F}\right)\left(\frac{c,ab}{F}\right)\mbox{ by rearranging}\\
 & = & \left(\frac{a,-ac}{F}\right)\left(\frac{c,-a}{F}\right)\left(\frac{c,-1}{F}\right)\mbox{pairing off and Lemma \ref{lem:Prodict of (1,1/F)}}\\
 & = & \left(\frac{a,-ac}{F}\right)\left(\frac{c,a}{F}\right)\mbox{ Lemma \ref{lem:Prodict of (1,1/F)} on the last two pairs}\\
 & = & \left(\frac{a,-ac^{2}}{F}\right)\mbox{ Lemma \ref{lem:Prodict of (1,1/F)}}\\
 & = & \left(\frac{a,-a}{F}\right)\mbox{ removing powers of }2\\
 & = & \left(\frac{1,1}{F}\right)\mbox{ because }\left\langle a,-a,a^{2}\right\rangle =\left\langle 1,1,-1\right\rangle \\
 & = & 0
\end{eqnarray*}
\end{proof}
\begin{cor}
The signed Hasse invariant gives a well defined map of abelian groups
$I^{2}(F)/I^{3}(F)\to\,_{2}\Br(F)$\end{cor}
\begin{thm}[Merkurev, 1981]
The map $I^{2}F/I^{3}F\overset{\cong}{\to}\,_{2}\Br(F)$ is an isomorphism
($\mbox{char}(F)\neq2$)\end{thm}
\begin{rem*}
$I^{0}/I^{2}=W(F)/I=\zz/2\zz$, $I/I^{2}=F^{*}/F^{2*},I^{2}/I^{3}=\,_{2}\Br(F)$,
what about $I^{k}/I^{k+1}=?$
\end{rem*}
For any field $F$ there are defined cohomology groups $H^{n}(F,\zz/2\zz)$,
sometimes called ``Galois cohomology groups'', which satisfy $H^{0}(F,\zz/2\zz)=\zz/2\zz$,
$H^{1}(F,\zz/2\zz)=F^{*}/F^{2*}$ and $H^{2}(F,\zz/2\zz)=\,_{2}\Br(F)$
for any field $F$ of characteristic $\neq2$. %These identification are easy to prove once one knows the definition of Galois cohomology.
This makes the statement of the following theorem plausible. For its
proof and the development of the tools needed in the proof (motivic
cohomology and motivic homotopy theory), Voevodsky was awarded the
fields medal in 2002.
\begin{thm}[Voevodsky, conjectured by Milnor]
Let $F$ be a field of $\mbox{char}\neq2$ then 
\[
I^{n}(F)/I^{n+1}(F)\cong H^{n}(F,\zz/2\zz)
\]
\end{thm}
\begin{lem}
Let $F$ be a field with $\mbox{char}F\neq2$ and $V,W$ symmetric
inner product spaces over $F$ of dimension $\leq3$. Then $V\cong W$$\iff\dim V=\dim W,\det V=\det W$
$\in F^{*}/F^{2*}$ and $\Hasse V=\Hasse W\in\Br(F)$\end{lem}
\begin{proof}
``$\Rightarrow$'' is clear

``$\Leftarrow$'': $\dim V=\dim W=1$: $V\cong\left\langle \det V\right\rangle =\left\langle \det W\right\rangle \cong W$,
hence we are done.

$\dim V=\dim W=2$: Then $V\cong\left\langle a,b\right\rangle ,W\cong\left\langle c,d\right\rangle $
($\mbox{char}F\neq2$). $(\frac{a,b}{F})=\Hasse(V)=\Hasse(W)=(\frac{c,d}{F})\Rightarrow\left\langle a,b,-ab\right\rangle \cong\left\langle c,d,-cd\right\rangle \underset{ab=cd\in F^{*}/F^{2*}}{\Rightarrow}\left\langle a,b\right\rangle \cong\left\langle c,d\right\rangle $
(Witt cancellation)

$\dim V=\dim W=3$: $V\cong\left\langle a,b,c\right\rangle ,W\cong\left\langle x,y,z\right\rangle ,a,b,c,x,y,z\in F^{*}$.
$\Hasse(\left\langle a,b,c\right\rangle )=(\frac{-abc,-1}{F})\Hasse(\left\langle -ab,-ac,-bc\right\rangle )\,(*)$
(Exercise). 
\begin{eqnarray*}
\Hasse(V)=\Hasse(W) & , & abc=\det V=\det W=xyz\\
\underset{(*)}{\Rightarrow}\Hasse(\left\langle -ab,-ac,-bc\right\rangle ) & = & \Hasse(\left\langle -xy,-xz,-yz\right\rangle )\\
\Rightarrow\left(\frac{-ab,-ac}{F}\right)\left(\frac{-ab,-bc}{F}\right)\left(\frac{-ac,-bc}{F}\right) & = & \left(\frac{-xy,-xz}{F}\right)\left(\frac{-xz,-yz}{F}\right)\left(\frac{-xz,-yz}{F}\right)\\
\Rightarrow\left(\frac{-ab,ab}{F}\right)\left(\frac{-ac,-bc}{F}\right) & = & \left(\frac{-xy,xy}{F}\right)\left(\frac{-xz,-yz}{F}\right)
\end{eqnarray*}
but $(\frac{-ab,ab}{F})=(\frac{1,1}{F})$ because $\left\langle -ab,ab,1\right\rangle \cong\left\langle 1,1,-1\right\rangle $
\begin{eqnarray*}
\Rightarrow\left(\frac{-ac,-bc}{F}\right) & \cong & \left(\frac{-xz,-yz}{F}\right)\\
\Rightarrow\left\langle -ac,-bc,-abc^{2}\right\rangle  & \cong & \left\langle -xz,-yz,-xyz^{2}\right\rangle \\
\Rightarrow\left\langle -abc\right\rangle \otimes\left\langle -ac,-bc,-ab\right\rangle  & \cong & \left\langle -xyz\right\rangle \otimes\left\langle -xz,-yz,-xy\right\rangle \mbox{ as }\left\langle -\det V\right\rangle =\left\langle -\det W\right\rangle \\
\Rightarrow\left\langle b,a,c\right\rangle  & \cong & \left\langle y,x,z\right\rangle 
\end{eqnarray*}
\end{proof}
\begin{prop}
Let $F$ be a field with $\mbox{char}F\neq2$. Assume that every 5-dimensional
symmetric inner product space is isotropic, i.e., represent $0$ non-trivially.
Then for symmetric inner product spaces $V,W$ over $F$, $V\cong W\iff\dim V=\dim W,\det V=\det W\in F^{*}/F^{2*},\Hasse(V)=\Hasse(W)\in\Br(F)$\end{prop}
\begin{rem*}
Proposition applies when $F=\qq_{p}$ (See below). (Also if $F=$
any local field, or non-real number field)\end{rem*}
\begin{proof}
Induction on $n=\dim V=\dim W$

$n\leq3$: This case is the previous lemma

Assume $n\geq4$. $V\perp\left\langle -1\right\rangle $ has dimension
$\geq5$ hence it is isotropic. $\Rightarrow V\perp\left\langle -1\right\rangle \cong V_{0}\perp\left\langle 1,-1\right\rangle \Rightarrow V\cong V_{0}\perp\left\langle 1\right\rangle $.
Similarly $W\cong W_{0}\perp\left\langle 1\right\rangle $. Now
\begin{itemize}
\item $\dim V_{0}=\dim W_{0}=n-1$.
\item $\det V_{0}=\det V_{0}\cdot\det\left\langle 1\right\rangle =\det V=\det W=\det W_{0}$
\item $\Hasse(V_{0}\perp\left\langle 1\right\rangle )=\Hasse(V)=\Hasse(W)=\Hasse(W_{0}\perp\left\langle 1\right\rangle )\Rightarrow\Hasse(V_{0})\cdot\Hasse(\left\langle 1\right\rangle )\cdot(\frac{\det V_{0},1}{F})=\Hasse(W_{0})\cdot\Hasse(\left\langle 1\right\rangle )\cdot(\frac{\det W_{0},1}{F})\underset{\det V_{0}=\det W_{0}}{\Rightarrow}\Hasse(V_{0})=\Hasse(W_{0})$
\end{itemize}
So by induction hypothesis $V_{0}\cong W_{0}\Rightarrow V=V_{0}\perp\left\langle 1\right\rangle \cong W_{0}\perp\left\langle 1\right\rangle =W$\end{proof}
\begin{cor}
\label{cor:Isotorpic 5-0dim}Let $F$ be a field with $\mbox{char}F\neq2$
for which every 5-dimensional form is isotropic. Then $I^{3}F=0$\end{cor}
\begin{proof}
Let $V$ be a symmetric inner product space over $F$, $[V]\in I^{3}F\subset I(F)\Rightarrow\dim_{F}V=2k$.
If $4\nmid\dim V$ replace $V$ with $V\perp\hh$, this doesn't change
$[V]=[V\perp\hh]$. Hence we can assume $\dim V=4\cdot l$ for some
$l\in\nn$. Now
\begin{itemize}
\item $\dim V=4l=\dim\hh^{2l}$
\item $\det V=(-1)^{\frac{\dim V}{2}}\mbox{disc}V=1$ because $[V]\in I^{2}$
and $(-1)^{\frac{\dim V}{2}}=(-1)^{2l}=1$. But $\det\hh^{2l}=1$
\item $\Hasse(V)=(\frac{(-1)^{l},-1}{F})\underset{[F]\in\Br}{\underbrace{s(V)}}=(\frac{(-1)^{l},-1}{F})$,
since $[V]\in I^{3}\subset\ker(s:I^{2}\to\Br)$. But $\Hasse\hh^{2l}=(\frac{(-1)^{l},-1}{F})$
\end{itemize}
So by the proposition we have $V\cong\hh^{2l}\Rightarrow[V]=0\in W(F)$
\end{proof}

\subsection{Quadratic Forms over $p$-adic numbers}
\begin{defn}
The $p$-adic integers $\zz_{p}$ are ($p\in\zz$ prime) 
\begin{eqnarray*}
\zz_{p} & = & \lim_{n\to\infty}\zz/p^{n}\zz\\
 & = & \{(x_{n})_{n\in\nn_{\geq1}}|x_{n}\in\zz/p^{n}\zz,x_{n+1}\equiv x_{n}\mod p^{n}\}\\
 & = & \{\sum_{i=0}^{\infty}a_{i}p^{i}|a_{i}\in\{0,\dots,p-1\}\}\\
 & = & \mbox{completion of }\zz\mbox{ with repsect to }||a||_{p}=p^{-\nu_{p}(a)}
\end{eqnarray*}
 
\end{defn}
$\zz_{p}$ is a Discrete Valuation Ring with maximal ideal $p\zz_{p}$
and residue field $\zz_{p}/p\zz_{p}=\ff_{p}$
\begin{defn}
The $p$-adic rational numbers $\qq_{p}$ are 
\begin{eqnarray*}
\qq_{p} & = & \mbox{field of fractions of }\zz_{p}\\
 & = & \mbox{completion of }\qq\mbox{ with respect to }||a||_{p}=p^{-\nu_{p}(a)}\\
 & = & \{\sum_{i=N}^{\infty}a_{i}p^{i}|a_{i}\in\{0,\dots,p-1\},N\in\zz\}
\end{eqnarray*}

\end{defn}
We have the surjective ring homomorphism $\zz_{p}\twoheadrightarrow\zz/p^{n}\zz$
by $\sum_{i=1}^{\infty}a_{i}p^{i}\mapsto\sum_{i=1}^{k}a_{i}p^{i}\mod p^{n}$
($k\geq n-1)$. $x\in\zz_{p}$ is a unit $\iff x\in\zz_{p}/p\zz_{p}=\ff_{p}$
is a unit ($\zz_{p}$ local). So $\sum_{i=0}^{\infty}a_{i}p^{i}\in\zz_{p}$
is a unit $\iff a_{0}\neq0$ in $\ff_{p}$.

We want to understand $\qq_{p}^{*}/\qq_{p}^{2*}$. If $p$ is odd
this is an exercise. For $p=2$ we first look at $\zz_{2}^{*}/\zz_{2}^{2*}$.
We have a ring homomorphism $\zz_{2}\to\zz/8\zz$ defined by $\sum_{i=0}^{\infty}a_{i}2^{i}\mapsto a_{0}+a_{1}2+a_{2}4$.
Therefore $(\zz_{2})^{*}\twoheadrightarrow(\zz/8\zz)^{*}$ is surjective
by the map $1+\sum_{i=1}^{\infty}a_{i}2^{i}\mapsto1+a_{1}2+a_{2}4$.
Now $(\zz_{2})^{2*}\to(\zz/8\zz)^{2*}=\{1^{2},3^{2},5^{2},7^{2}\}=\{1\}$,
so we have a well defined group homomorphism $(\zz_{2})^{*}/(\zz_{2})^{2*}\twoheadrightarrow(\zz/8\zz)^{*}$.
\begin{prop}
The map $\zz_{2}^{*}/\zz_{2}^{2*}\to(\zz/8\zz)^{*}$ defined by $\sum_{i=0}^{\infty}a_{i}z^{i}\mapsto a_{0}+a_{1}2+a_{2}4$
is an isomorphism.\end{prop}
\begin{proof}
We already know that the map is surjective. $z=1+\sum_{i=1}^{\infty}a_{i}2^{i}\in$
kernel of the map $\iff a_{1},a_{2}=0\iff x=1+8y$ for some $y\in\zz_{2}$.
We need to sow that $x$ is a square in $\zz_{2}$. 
\begin{eqnarray*}
z & = & (1+8y)^{1/2}\\
 & := & \sum_{k=0}^{\infty}{\nicefrac{1}{2} \choose k}(8y)^{k}\\
 & = & \sum_{k=0}^{\infty}{\nicefrac{1}{2} \choose k}4^{k}(2y)^{k}\\
 & = & \sum_{k=0}^{\infty}b_{k}(2y)^{k}
\end{eqnarray*}
where
\begin{eqnarray*}
b_{k} & = & {\nicefrac{1}{2} \choose k}4^{k}\\
 & = & \frac{\nicefrac{1}{2}(\nicefrac{1}{2}-1)\cdot\dots\cdot(\nicefrac{1}{2}-k+1)}{k!}4^{k}\\
 & = & \frac{(\nicefrac{1}{2})^{k}\cdot1\cdot(-1)\cdot\dots\cdot(-2k+3)}{k!}4^{k}\\
 & = & (-1)^{k-1}\cdot1\cdot3\cdot\dots\cdot(2k-3)\cdot\frac{2^{k}}{k!}.
\end{eqnarray*}
Now $k=\nu_{2}(2^{k})\geq\nu_{2}(k!)$ since $\nu_{2}(k!)\leq$ (number
of even number $\leq k$) + (number of number divisible by $4\leq k$)
$+\dots\leq\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k}{4}\rfloor+\dots\leq\sum_{i=1}^{\infty}\frac{k}{2^{i}}=\frac{k}{2}\sum_{i=0}^{\infty}\frac{1}{2^{i}}=\frac{k}{2}\frac{1}{1-\frac{1}{2}}=\frac{k}{2}2=k$.
Hence $\nu_{2}(b_{k})\geq0$ and since $\zz_{2}=\{t\in\qq_{2}|\nu_{2}(t)\geq0\}$
we have that $b_{k}\in\zz_{2}$. 
\begin{eqnarray*}
\left\Vert \sum_{k=n}^{m}\underset{\in\zz_{2}}{\underbrace{b_{k}y^{k}}}\cdot2^{k}\right\Vert _{2} & = & \left\Vert \underset{\in\zz_{2}}{\underbrace{\sum_{k=n}^{\infty}b_{k}y^{k}2^{k-n}}}\right\Vert _{2}\left\Vert 2^{n}\right\Vert _{2}\\
 & \leq & 1\cdot2^{-n}
\end{eqnarray*}
where $||a||_{2}=2^{-\nu_{2}(a)}\leq1$ for all $a\in\zz_{2}$. Hence
$m\mapsto\sum_{k=0}^{m}b_{k}y^{k}2^{k}$ is a Cauchy sequence $\Rightarrow z:=\sum_{k=0}^{\infty}b_{k}y^{k}2^{k}$
defines an element in $\zz_{2}$. Then $z^{2}=x$.\end{proof}
\begin{rem*}
For $p$ odd $\zz_{p}^{*}/\zz_{p}^{2*}\overset{\cong}{\to}\ff_{p}^{*}/\ff_{p}^{2*}$
(reduction mod $p$) is an isomorphism (exercise)\end{rem*}
\begin{cor}
The map 
\[
\qq_{p}^{*}/\qq_{p}^{2*}\overset{\cong}{\to}\zz/2\zz\times\zz_{2}^{*}/\zz_{p}^{2*}=\begin{cases}
\zz/2\zz\times\ff_{p}^{*}/\ff_{p}^{2*} & p\,\mbox{odd}\\
\zz/2\zz\times(\zz/8\zz)^{*}=\zz/2\zz\times(\zz/2\zz)^{2}=(\zz/2\zz)^{3} & p=2
\end{cases}
\]
 defined by $p^{\nu}a\mapsto\nu,a$ where $a\in\zz_{p}^{*}$is an
isomorphism.\end{cor}
\begin{proof}
For any Discrete Valuation Ring $R$ with field of fractions $F$,
the map $F^{*}\overset{\cong}{\to}\zz\times R^{*}$ defined by $p^{\nu}a\mapsto\nu,a$
where $a\in R^{*}$ is a isomorphism. Hence $F^{*}/F^{2*}\overset{\cong}{\to}\zz/2\zz\times R^{*}/R^{2*}$.
Now $(\zz/8\zz)^{*}$ is generated by $3,5$ and $3^{2}=5^{2}=1\mod8$,
hence $(\zz/8\zz)\cong\zz/2\zz\oplus\zz/2\zz$.\end{proof}
\begin{cor}
$\zz_{2}^{*}/\zz_{2}^{2*}=\{1,3,5,7\}$ and $\qq_{2}^{*}/\qq_{2}^{2*}=\{1,3,5,7,2,6,10,14\}$.\end{cor}
\begin{prop}
Let $p\in\zz$ be a prime. Then there is, up to isometry, a unique
anisotropic $4$-dimensional regular quadratic form over $\qq_{p}$.
This form has determinant $1$ and represents all of $\qq_{p}^{*}/\qq_{p}^{2*}$.\end{prop}
\begin{proof}
$p=$ odd (exercise)

$p=2$: Consider all possible 2-dimensioanl forms $\left\langle 1,a\right\rangle $
where $a\in\qq_{2}^{*}/\qq_{2}^{2*}$. Set $D_{a}=\{t\in\qq_{2}^{*}/\qq_{2}^{2*}|t$
represent $\left\langle 1,a\right\rangle \}$

\begin{tabular}{|c|c|}
\hline 
$\left\langle 1,a\right\rangle $ & $D_{a}\subset\qq_{2}^{*}/\qq_{2}^{2*}=\{1,3,5,7,2,6,10,14\}$\tabularnewline
\hline 
\hline 
$\left\langle 1,1\right\rangle $ & $1,2,5,10$\tabularnewline
\hline 
$\left\langle 1,2\right\rangle $ & $1,2,3,6$\tabularnewline
\hline 
$\left\langle 1,3\right\rangle $ & $1,3,5,7$\tabularnewline
\hline 
$\left\langle 1,5\right\rangle $ & $1,5,6,14$\tabularnewline
\hline 
$\left\langle 1,6\right\rangle $ & $1,6,7,10$\tabularnewline
\hline 
$\left\langle 1,7\right\rangle $ & Hyperbolic\tabularnewline
\hline 
$\left\langle 1,10\right\rangle $ & $1,3,10,14$\tabularnewline
\hline 
$\left\langle 1,14\right\rangle $ & $1,2,7,14$\tabularnewline
\hline 
\end{tabular}

We check this table for $\left\langle 1,1\right\rangle $: This represent
$1,2,5,10$ because $1=1\cdot1^{2}+1\cdot0^{2}$, $2=1^{2}+1^{2},5=2^{2}+1^{2}$
and $10=3^{2}+1^{2}$, and it does not represent $3,7,6,14$ because
$x^{2}+y^{2}\in\{3,7,6,14\}$ has no solution in $\zz_{2}$ since
it has no solution $\mod8$ as $x^{2}+y^{2}\in\{0,1,2\}\mod8$ since
$x^{2},y^{2}\in\{0,1\}\mod8$. If $x^{2}+y^{2}=a\in\{3,7,6,14\}$
has a solution in $\qq_{2}$ clearing denominators (multiplying with
respect to $2^{n}$) $(*)\, x^{2}+y^{2}=at^{2}$ has a solution in
$\zz_{2}$ and not all of $x,y,t$ are divisible by $2$.
\begin{casenv}
\item $2\nmid t$ then $t\in\zz_{2}^{*}\Rightarrow t^{2}=1\mod8$ and $(*)$
has no solution $\mod8$
\item $t=2u$ and $2\nmid x$ then $x^{2}=1\mod8$, $\underset{1}{\underbrace{x^{2}}}+y^{2}=4u^{2}a$
has no solution $\mod8$ as $y^{2}\in\{1,0\}\mod8$.
\end{casenv}
Hence $\left\langle 1,1\right\rangle $ does not represent $3,7,6,14$.

We also can check that $\left\langle 1,1\right\rangle \cong\left\langle 2,2\right\rangle \cong\left\langle 5,5\right\rangle \cong\left\langle 10,10\right\rangle \ncong\left\langle 3,3\right\rangle \cong\left\langle 7,7\right\rangle \cong\left\langle 6,6\right\rangle \cong\left\langle 14,14\right\rangle $.
e.g., $\left\langle 1,1\right\rangle \cong\left\langle 2,2\right\rangle \cong\left\langle 5,5\right\rangle \cong\left\langle 10,10\right\rangle $
since $\left\langle 1,1\right\rangle $ represents 2,5 and 10 and
the all have the same determinant. Now $\left\langle 1,1\right\rangle \ncong\left\langle 3,3\right\rangle =\left\langle 3\right\rangle \cdot\left\langle 1,1\right\rangle $
because $\left\langle 1,1\right\rangle $ represent $1,2,5,10$ but
$\left\langle 3\right\rangle \cdot\left\langle 1,1\right\rangle $
represents $3,6,15=7,30=14\in\qq_{2}^{*}/\qq_{2}^{2*}=\zz/2\zz\times(\zz/8\zz)^{*}$.

Let $\phi$ be a $4$-dimensional anisotropic form over $\qq_{2}$,
$\phi=\left\langle d,\dots\right\rangle $, then $\psi=\left\langle d\right\rangle \phi$
is also anisotropic and represent $d^{2}=1\in\qq_{2}^{*}/\qq_{2}^{2*}$.
So $\psi=\left\langle 1,a,-b,-bc\right\rangle $ for some $a,b,c\in\qq_{2}^{*}$.
Rewrite this as $\psi=\left\langle 1,a\right\rangle \perp\left\langle -b\right\rangle \cdot\left\langle 1,c\right\rangle $.
If $\left\langle 1,a\right\rangle $ and $\left\langle b\right\rangle \cdot\left\langle 1,c\right\rangle $
represent a common element, then $\psi$ represent $0$ which contradicts
the fact that $\psi$ is anisotropic. Note also that $a,c\neq7$ because
$\left\langle 1,a\right\rangle $ and $\left\langle 1,c\right\rangle $
are not hyperbolic. Therefore $D_{a}\cap bD_{c}=\emptyset\underset{|D_{a}|=|D_{c}|}{\Rightarrow}D_{a}\sqcup bD_{c}=\qq_{2}^{*}/\qq_{2}^{2*}$.
We can use the table to see $D_{a}\subset\qq_{2}^{*}/\qq_{2}^{2*}$
is a subgroup. Now $1\in D_{a},D_{c}$, $1\notin bD_{c}\underset{D_{c}\mathrm{\, subgroup}}{\Rightarrow}bD_{c}\cap D_{c}=\emptyset\Rightarrow D_{c}\sqcup bD_{c}=D_{a}\sqcup bD_{c}=\qq_{2}^{*}/\qq_{2}^{2*}\Rightarrow D_{a}=D_{c}\underset{\mathrm{table}}{\Rightarrow}a=c\in\qq_{2}^{*}/\qq_{2}^{2*}\Rightarrow\psi=\left\langle 1,a,-b,-ab\right\rangle \Rightarrow\det\psi=1$.
Now $\phi=\left\langle d^{2}\right\rangle \cdot\psi=\left\langle d\right\rangle \psi=\left\langle d,da,-db,-dab\right\rangle $
has determinant $=1$$\Rightarrow$ every anisotropic $4$-dimensional
form has determinant 1. In particular $\left\langle -1,a,-b,-ab\right\rangle $
is isotropic as it has determinant $-1\neq1\in\qq_{2}^{*}/\qq_{2}^{2*}\Rightarrow\left\langle -1,a,-b,-ab\right\rangle =\left\langle -1,1,\dots\right\rangle \Rightarrow\left\langle a,-b,-ab\right\rangle $
represent $1$. Hence $\psi=\left\langle 1,a,-b,-ab\right\rangle =\left\langle 1,1,e,e\right\rangle $
since $\det\psi=1$. But $e\notin\{3,7,6,14\}$ because otherwise
$\left\langle e,e\right\rangle \underset{\mathrm{table}}{=}\left\langle 7,7\right\rangle =\left\langle -1,-1\right\rangle $and
$\psi$ isotropic $\Rightarrow e\in\{1,2,5,10\}$, $\left\langle e,e\right\rangle \underset{\mathrm{table}}{\cong}\left\langle 1,1\right\rangle \Rightarrow\psi=\left\langle 1,1,1,1\right\rangle $.

Let us check $\psi=\left\langle 1,1,1,1\right\rangle $ is indeed
anisotropic because otherwise $\left\langle 1,1,1,1\right\rangle \cong\left\langle 1,-1,\_,\_\right\rangle \Rightarrow\left\langle 1,1,1\right\rangle $
represent $-1=7\in\qq_{2}^{*}/\qq_{2}^{2*}$ but $x^{2}+y^{2}+z^{2}=7$
has no solution in $\qq_{2}^{*}$ because $x^{2}+y^{2}+z^{2}$ has
no solution in $\zz_{2}$ (since no solution $\mod8$). If $x^{2}+y^{2}+z^{2}=7$
has a solution in $\qq_{2}$ then there exists $x^{2}+y^{2}+z^{2}=7t^{2}$
for some $x,y,z,t\in\zz_{2}$ and not all of $x,y,z,t$ are divisible
by $2$.
\begin{casenv}
\item If $2\nmid t$ then $t\in\zz_{2}^{*}\Rightarrow t^{2}=1\mod8$ contradiction
since $x^{2}+y^{2}+z^{2}=7$ has no solution $\mod8$
\item If $2|t\Rightarrow t=2u,u\in\zz_{2}$ and one of $x,y,z$ is not divisible
by $2$, say $2\nmid x$$\Rightarrow x^{2}+y^{2}+z^{2}=4\cdot7\cdot u^{2}$
has no solution $\mod8$ since $x^{2}=1\mod8$ and $y^{2},z^{2}\in\{1,0\}\mod8$
while $4\cdot7u^{2}\in\{0,4\}\mod8$.
\end{casenv}
Hence $\psi=\left\langle 1,1,1,1\right\rangle $ is anisotropic.Now
$\left\langle 1,1\right\rangle $, hence $\psi$, represents $1,2,5,10$
and $\psi$ also represents $-1=7=2^{2}+1^{2}+1^{2}+1^{2}\in\qq_{2}^{*}/\qq_{2}^{2*}$.
$\left\langle -1\right\rangle \cdot\psi$ represents 1 and is anisotropic
$\Rightarrow\left\langle -1\right\rangle \cdot\psi=\psi\Rightarrow\psi\cong\left\langle -1,-1,-1,-1\right\rangle \equiv\left\langle 7,7,7,7\right\rangle \underset{\mathrm{table}}{\Rightarrow}\psi\cong\left\langle d\right\rangle \psi=\phi\,\forall d\in\qq_{2}^{*}/\qq_{2}^{2*}$\end{proof}
\begin{thm}
Let $p\in\zz$ be a prime then:
\begin{enumerate}
\item Every $5$-dimensional inner product space over $\qq_{p}$ is isotropic
\item $I^{3}(\qq_{p})=0$, $I^{2}(\qq_{p})=\zz/2\zz$ generated by the unique
anisotropic form of dimension $4$. $I/I^{2}(\qq_{p})=\qq_{p}^{*}/\qq_{p}^{2*}$,
$\frac{W(\qq_{p})}{I(\qq_{p})}=\zz/2\zz$
\end{enumerate}
\end{thm}
\begin{proof}
{}
\begin{enumerate}
\item $\left\langle a_{1},\dots,a_{5}\right\rangle $ anisotropic $\Rightarrow\left\langle a_{1},\dots,a_{4}\right\rangle $
is anisotropic hence is the unique $4$-dimensional anisotropic form
representing all of $\qq_{p}^{*}/\qq_{p}^{2*}$, in particular $\left\langle a_{1},\dots,a_{4}\right\rangle $
represents $-a_{5}\Rightarrow\left\langle a_{1},\dots,a_{5}\right\rangle $
isotropic
\item Now 1. $\Rightarrow I^{3}(\qq_{p})=0$ by Corollary \ref{cor:Isotorpic 5-0dim}.
$I^{2}(\qq_{p})=\zz/2\zz$ because let $\phi$ be the unique $4$-dimensional
anisotropic form over $\qq_{p}$ then $\phi\in I$ because $\dim\phi=4=0\in\zz/2\zz$
and $0\neq\phi\in I^{2}$ because $\mbox{disc}\phi=\det\phi=1\Rightarrow0\neq\phi\in\ker(\mbox{disc})=I^{2}$.
If $0\neq\psi\in I^{2}$ is anisotropic, $\phi\neq\psi\Rightarrow\dim\psi<4,\dim\psi=0\mod2$
since $\psi\in I,\Rightarrow\dim\psi=2\Rightarrow\psi=\left\langle a,b\right\rangle $
but $1=\mbox{disc}\psi=-ab$ since $\mbox{disc}I^{2}=1\Rightarrow\psi=\left\langle a,-a\right\rangle $
is hyperbolic, in particular not anisotropic $\Rightarrow\psi=0\in W(\qq_{p})$.
Hence, $I^{2}=\{0,\phi\}=\zz/2\zz$. The rest is true for any field
$F$ with $\mbox{char}F\neq2$
\end{enumerate}
\end{proof}
\begin{thm}
Let $p\in\zz$ be a prime. Then the Witt groups of $\zz_{p}$ and
$\qq_{p}$ are:\end{thm}
\begin{casenv}
\item $p$ odd: 
\[
\xyR{0.5pc}\xyC{4.5pc}\xymatrix{W(\zz_{p})\ar^{\cong}[r] & W(\ff_{p}) & \hspace{-10ex}\text{(reduction \ensuremath{\mod p})}\\
W(\qq_{p})\ar_{\cong}^{\partial^{1},\partial^{2}}[r] & W(\ff_{p})\oplus W(\ff_{p})
}
\]
where $ $$\partial^{1},\partial^{2}$ are the first and second residue
homomorphism ($\partial^{1}(\zeta)=\partial^{2}(\left\langle p\right\rangle \otimes\zeta)$)
\item $p=2$
\[
\xymatrix{W(\zz_{2})\ar^{\cong}[r] & \zz/8\zz\times\zz/2\zz\\
W(\qq_{2})\ar^{\cong}[r] & \zz/8\zz\times(\zz/2\zz)^{2}
}
\]
\end{casenv}
\begin{proof}
The case $p$ is odd is left as an exercise.

$p=2$: $I^{3}(\qq_{2})=0,I^{2}(\qq_{2})=\zz/2\zz,I/I^{2}(\qq_{2})=\qq_{2}^{*}/\qq_{2}^{2*}\cong(\zz/8\zz)^{*}\times(\zz/2\zz)=(\zz/2\zz)^{3},W(\qq)/I=\zz/2\zz$.
$0=I^{3}\subset I^{2}\subset I\subset W(\qq_{2})$. $|W(\qq_{2})|=|W/I|\cdot|I/I^{2}|\cdot|I^{2}|=2\cdot8\cdot2=32\Rightarrow$
every element of $W(\qq_{2})$ has order a power of $2$. We have:
\begin{itemize}
\item $\left\langle 1\right\rangle \in W(\qq_{2})$ has order $8$ because
$0\neq4\left\langle 1\right\rangle =\left\langle 1,1,1,1\right\rangle $
as it is a generator of $I^{2}(\qq_{2})=\zz/2\zz$ and $8\left\langle 1\right\rangle =4\left\langle 1\right\rangle +4\left\langle 1\right\rangle =4\left\langle 1\right\rangle +4\left\langle -1\right\rangle =0$
because $\left\langle 1,1,1,1\right\rangle =\left\langle -1,-1,-1,-1\right\rangle =\left\langle -1\right\rangle \otimes\left\langle 1,1,1,1\right\rangle $
(both are anisotropic and there exists a unique anisotropic form of
dimension 4) over $\qq_{2}$. 
\item $\left\langle 1,3\right\rangle \in W(\qq_{2})$ has order $2$ because
it represents $-1$ as $1+3\cdot3^{2}=28=2^{2}\cdot7=7=-1\in\qq_{2}^{*}/\qq_{2}^{2*}$.
So $\left\langle 1,3\right\rangle \cong\left\langle -1,-3\right\rangle $
and $\left\langle 1,3\right\rangle +\left\langle 1,3\right\rangle =\left\langle 1,-1\right\rangle +\underset{\mathrm{hyperbolic}}{\underbrace{\left\langle 3,-3\right\rangle }}=0\in W(\qq_{2})$
and $\left\langle 1,3\right\rangle \neq0\in W(\qq_{2})$ since $\mbox{disc}\left\langle 1,3\right\rangle =-3=5\neq1\in\qq_{2}^{*}/\qq_{2}^{2*}$
\item $\left\langle 1,6\right\rangle \in W(\qq_{2})$ has order 2 because
it represent $-1$ as $-1=7=1\cdot1^{2}+6\cdot1^{2}\in\qq_{2}^{*}/\qq_{2}^{2*}$.
So $\left\langle 1,6\right\rangle \cong\left\langle -1,-6\right\rangle \Rightarrow\left\langle 1,6\right\rangle +\left\langle 1,6\right\rangle =\left\langle -1,-6\right\rangle +\left\langle 1,6\right\rangle =\left\langle 1,-1\right\rangle +\left\langle 6,-6\right\rangle =0\in W(\qq_{2})$
and $0\neq\left\langle 1,6\right\rangle \in W(\qq_{2})$ because $\mbox{disc}\left\langle 1,6\right\rangle =-6\neq1\in\qq_{2}^{*}/\qq_{2}^{2*}$
\end{itemize}
Hence the map $\zz/8\zz\oplus\zz/2\zz\oplus\zz/2\zz\to W(\qq_{2})$
defined by $a,b,c\mapsto a\left\langle 1\right\rangle +b\left\langle 1,3\right\rangle +c\left\langle 1,6\right\rangle $
is well defined. Both groups have order $32$. In order to show that
the map is an isomorphism it suffices to show that it is injective.
Assume $0=a\left\langle 1\right\rangle +b\left\langle 1,3\right\rangle +c\left\langle 1,6\right\rangle \in W(\qq_{2})$.
Now $b\left\langle 1,3\right\rangle +c\left\langle 1,6\right\rangle $
has order $\leq2\Rightarrow a\left\langle 1\right\rangle $ has order
$\leq2$ $\Rightarrow a=4\mod8\Rightarrow a\left\langle 1\right\rangle =a'\left\langle 1,1,1,1\right\rangle $
with $a'\in\zz/2\zz.$ We compute the discriminant 
\begin{eqnarray*}
\mbox{disc}(a'\left\langle 1,1,1,1\right\rangle +b\left\langle 1,3\right\rangle +c\left\langle 1,6\right\rangle ) & = & (\mbox{disc}\left\langle 1,1,1,1\right\rangle )^{a'}\cdot\mbox{disc}(\left\langle 1,3\right\rangle )^{b}\cdot\mbox{disc}(\left\langle 1,6\right\rangle )^{c}\in\qq_{2}^{*}/\qq_{2}^{2*}\\
 & = & 1\cdot(-3)^{3}(-6)^{c}=5^{b}\cdot10^{c}\in\qq_{2}^{*}/\qq_{2}^{2*}.
\end{eqnarray*}
Now $5\neq10\in\qq_{2}^{*}/\qq_{2}^{2*}$, hence, they linearly independent
in the $\ff_{2}$-vector space in $\qq_{2}^{*}/\qq_{2}^{2*}=(\zz/2\zz)^{3}\Rightarrow b,c=0\in\zz/2\zz\Rightarrow a'\left\langle 1,1,1,1\right\rangle =0\Rightarrow a'=0$
since $\left\langle 1,1,1,1\right\rangle \neq0\in W(\qq_{2})\Rightarrow$
the map is injective. Hence $W(\qq_{2})\cong\zz/8\zz\oplus\zz/2\zz\oplus\zz/2\zz$.

$\left\langle 1\right\rangle ,\left\langle 1,3\right\rangle \in W(\zz_{2})$
since $1,3\in\zz_{2}^{*}$, $W(\zz_{2})\hookrightarrow W(\qq_{2})$
is injective, it follows that $(\zz/8\zz)\left\langle 1\right\rangle \oplus(\zz/2\zz)\left\langle 1,3\right\rangle \subset W(\zz_{2})\Rightarrow|W(\zz_{2})|\geq8\cdot2=16$.
Also $W(\zz_{2})\to W(\qq_{2})\overset{\partial^{2}}{\twoheadrightarrow}W(\ff_{2})$
is zero. $W(\zz_{2})\subset\ker(\partial^{2})\Rightarrow|W(\zz_{2})|\leq|\ker(\partial^{2})|=\frac{|W(\qq_{2})|}{|W(\ff_{2})|}=\frac{32}{2}=16\Rightarrow|W(\zz_{2})|=16$.
Hence $(\zz/8\zz)\left\langle 1\right\rangle \oplus(\zz/2\zz)\left\langle 1,3\right\rangle =W(\zz_{2})$.\end{proof}
\begin{lem}[\textbf{Definition}]
 Set $\qq_{\infty}=\rr$, $p=\infty=$ ``infinite prime''. For
$p\in\zz\cup\{\infty\}$ prime there is a unique quaternion algebra
over $\qq_{p}$ that doesn't split (i.e., $\ncong M_{2}(\qq_{p})$).
Therefore, $\Hasse(V)=\begin{cases}
[A] & \in\Br(\qq_{p})\\
{}[\qq_{p}] & \in\Br(\qq_{p})
\end{cases}$, where $A$ is a division quaternion algebra. The \emph{Hasse symbol
$h_{p}(V)$ }for a symmetric inner product space $V$ over $\qq_{p}$
is defined by
\[
h_{p}(V)=\begin{cases}
-1 & \mathrm{if}\,\Hasse(V)\,\mathrm{does\, not\, split\,}(\neq[\qq_{p}]\in\Br(\qq_{p}))\\
1 & \mathrm{if\,}\Hasse(V)=[\qq_{p}]\in\Br(\qq_{p})
\end{cases}
\]
For $a,b\in\qq_{p}^{*}$, the \emph{Hilbert Symbol }is:
\[
(a,b)_{p}=h_{p}(\left\langle a,b\right\rangle )=\begin{cases}
1 & \mbox{if }\left(\frac{a,b}{\qq_{p}}\right)\mathrm{\, splits}\\
-1 & \mathrm{if\,}\left(\frac{a,b}{\qq_{p}}\right)\,\mathrm{does\, not\, split}
\end{cases}
\]
\end{lem}
\begin{proof}
We need to justify that there exists a unique non-split quaternion
algebra over $\qq_{p}$. If $p=\infty$ then $\Br(\qq_{\infty})=\Br(\rr)=\{\rr,\hh\}$,
so $\hh$ is the unique non-split quaternion algebra over $\rr$.

If $p<\infty$: $(\frac{a,b}{\qq_{p}}),(\frac{c,d}{\qq_{p}})\ncong M_{2}(\qq_{2})\iff\left\langle a,b,-ab,-1\right\rangle ,\left\langle c,d,-cd,-1\right\rangle \ncong\left\langle 1,1,-1,-1\right\rangle $
(all forms are in $I^{2}(\qq_{p})\cong\zz/2\zz$ as $\mbox{disc}=0$
for them) $\iff\left\langle a,b,-ab,-1\right\rangle ,\left\langle c,d,-cd,-1\right\rangle $
are both the unique anisotropic 4-dimensional form over $\qq_{p}$$\iff\left\langle a,b,-ab,-1\right\rangle \cong\left\langle c,d,-cd,-1\right\rangle \ncong\left\langle 1,1,-1,-1\right\rangle \iff(\frac{a,b}{\qq_{p}})\cong(\frac{c,d}{\qq_{p}})\ncong(\frac{1,1}{\qq_{p}})=M_{2}(\qq_{p})$.\end{proof}
\begin{namedthm}[Hilbert Reciprocity Law]
 Let $V$ be a symmetric inner product space over $\qq$. Then $h_{p}(V)=1$
for all but finitely many primes $p\in\zz\cup\{\infty\}$. And $\prod_{p\in\zz\cup\{\infty\}\mathrm{\, prime}}h_{p}(V)=1$\end{namedthm}
\begin{proof}
Since $h_{p}(V)$ is a product of Hilbert Symbols $(a,b)_{p}$, it
suffices to show claim for $V=\left\langle a,b\right\rangle $ and
thus $\prod_{p\in\zz\cup\{\infty\}}(a,b)_{p}=1\,\forall a,b\in\qq_{p}^{*}$.
To show $\prod(a,b)_{p}=1$, using bilinearity of Hilbert symbol $(ab,c)=(a,c)_{p}(b,c)_{p}$,
we just need to show $\prod(a,b)_{p}=1$ for $a,b$ prime or $\pm1$.
In this case, express $(a,b)_{p}$ in terms of Legendre symbol which
mean the proof is a consequence of Quadratic Reciprocity. (Details
are left as an exercise)\end{proof}
\begin{cor}
Let $V,W$ be inner product spaces over $\qq$. Let $q\in\zz\cup\{\infty\}$
be a prime. If $h_{p}(V)=h_{p}(W)\,\forall p\in\zz\cup\{\infty\}$
prime, $p\neq q$. Then $h_{q}(V)=h_{q}(W)$\end{cor}
\begin{proof}
$\prod_{p\in\zz\cup\{\infty\}}h_{p}(V)=1=\prod_{p\in\zz\cup\{\infty\}}h_{p}(W)$
\end{proof}
We will need this theorem:
\begin{thm}[Dirichlet]
 Let $a,b\in\zz$ be integers with $\gcd(a,b)=1$, then the set of
integers of the form $a+nb$, $n\in\zz$, contains infinitely many
primes.\end{thm}
\begin{proof}
This theorem is beyond the scope of this module\end{proof}
\begin{namedthm}[Strong Hasse principle for quadratic forms over $\qq$]
 A symmetric inner product space $V$ over $\qq$ is isotropic if
and only if $V$ is isotropic over $\rr$ and $\qq_{p}\,\forall p\in\zz$
prime.\end{namedthm}
\begin{rem*}
The Theorem says: A homogeneous quadratic polynomial has a non-trivial
zero in $\qq$ if and only if it has a non-trivial zero in $\rr$
and $\qq_{p}\,\forall p\in\zz$ prime.\end{rem*}
\begin{proof}
{}
\begin{description}
\item [{``$\Rightarrow$'':}] Is clear.
\item [{``$\Leftarrow$'':}] We assume Dirichlet Theorem. We use induction
on $n=\dim_{\qq}V$

\begin{description}
\item [{$n=1$:}] Every 1-dimensional inner space is anisotropic (over
any field)
\item [{$n=2$:}] A $2$-dimensional form $V$ is isotropic over $\qq$
(any field of characteristic not 2) $\iff$ $V$ hyperbolic over $\qq$,
i.e., $V\cong\hh\iff V\cong\hh$ over $\rr$ and $\qq_{p}\,\forall p\in\zz$
prime$\iff V$ is isotropic over $\rr$ and $\qq_{p}\,\forall p\in\zz$
prime
\item [{$n=3:$}] $V$ isotropic over $\qq\iff V\cong\left\langle 1,-1,-\det V\right\rangle $\noun{
}over $\qq$$\iff V\cong\left\langle 1,-1,-\det V\right\rangle $
over $\rr$ and $\qq_{p}\,\forall p$ prime (Weak Hasse Principle)
$\iff V$ isotropic over $\rr$ and $\qq_{p}\,\forall p$ primes.
\item [{$n=4$:}] Write $V=\left\langle d_{1},d_{2},d_{3},d_{4}\right\rangle $
with $d_{i}\in\zz\setminus\{0\}$ square free, $d=\det V$ square
free. Let $\mathscr{P}=\{2\}\cup\{p\in\zz\mbox{ prime }:p|d_{1}\dots d_{4}\}<\infty$.
Write $V_{p}$ for $V\otimes_{\qq}\qq_{p}$. Now $V_{p}$ is isotropic
by assumption, $\Rightarrow V_{p}\cong\left\langle 1,-1\right\rangle \perp\left\langle a_{p},-a_{p}d\right\rangle $
(over $\qq_{p}$) with $a_{p}\in\zz\setminus\{0\}$ square free. 

\begin{itemize}
\item If $p\notin\mathscr{P}$ we can assume that $a_{p}\in\zz_{p}^{*}$
and $a_{\infty}=1$. Otherwise if $p=\infty$ replace $(V,\beta)$
with $(V,-\beta)$, and if $\infty\neq p\notin\mathscr{P}$ we would
have $a_{p}=pb_{p}$ (as $a_{p}\in\zz_{p}\setminus\{0\}$ square free).
Therefore, $0\underset{p\nmid d_{1}\dots d_{4}}{=}\partial_{p}^{2}V=\partial_{p}^{2}\left\langle 1,-1,pb_{p},-pb_{p}d\right\rangle =\left\langle b_{p},-b_{p}d\right\rangle \Rightarrow\mbox{disc}\left\langle b_{p},-b_{p}d\right\rangle =d=1\in\ff_{p}^{*}/\ff_{p}^{2*}\cong\zz_{p}^{*}/\zz_{p}^{2*}$
and $d$ is a square in $\qq_{p}$ $\Rightarrow\left\langle a_{p},-a_{p}d\right\rangle \cong\left\langle a_{p},-a_{p}\right\rangle \cong\left\langle 1,-1\right\rangle $
over $\qq_{p}$ and we can even assume $a_{p}=1$ 
\item There exists $q\in\zz$ prime such that $a:=q\pi=a_{p}\in\qq_{p}^{*}/\qq_{p}^{2*}\,\forall p\in\mathscr{P}$
where $\pi=\prod_{p\in\mathscr{P},\nu(a_{p})=1}p$ (Note $\nu(a_{p})=0$
or $1$ as $a_{p}\in\zz_{p}$ is square free). To justify existence
of $q$ note that $a_{p}=\pi u_{p},u_{p}\in\zz_{p}^{*}$. By the Chinese
Remainder Theorem $\zz\twoheadrightarrow\zz/8\zz\times\prod_{p\in\mathscr{P},p\neq2}\zz/p\zz$
is surjective. So there exists an integer $r$ such that $r=a_{2}\in(\zz/8\zz)^{*}\subset\zz/8\zz$
and $r=u_{p}\in\zz/p\zz$ for $p\in\mathscr{P}\setminus\{2\}$. In
fact, any integer of the form $r+ns$ with $s=2^{3}\prod_{p\in\mathscr{P}\setminus\{2\}}p$
can be chosen instead of $r$. Since the $a_{p}$'s are units, it
follows that $s$ and $r$ are relatively prime. By Dirichlet's theorem
on existence on infinitely many primes in an arithmetic progression,
we can choose $r=q$ a prime. By construction $a=q\pi=a_{p}\in\qq_{p}^{*}/\qq_{p}^{2*}$. 
\end{itemize}

\uline{Claim: }$V\cong\left\langle 1,-1\right\rangle \perp\left\langle a,-ad\right\rangle $
over $\qq$ (in particular, $V$ isotropic over $\qq$ as it contains
$\hh$)


Proof of claim: By the weak Hasse principle it suffices to show that
$(V_{p}=)\ \left\langle 1,-1,a_{p},-a_{p}d\right\rangle =\left\langle 1,-1,a,-ad\right\rangle $
over $\qq_{p}$ $\forall p\in\zz\cup\{\infty\}$ prime. 
\begin{casenv}
\item $p\in\mathscr{P}$: We have $\left\langle a_{p},-a_{p}d\right\rangle \cong\left\langle a,-ad\right\rangle $
since, by construction of $a$, we have $a=a_{p}\in\qq_{p}^{*}/\qq_{p}^{2*}$
for $p\in\mathscr{P}$. 
\item $p\notin\mathscr{P}$ and $p\neq q,\infty$: One checks that $\partial^{1}$
and $\partial^{2}$ agree: $\partial^{1}\langle a_{p},-a_{p}d\rangle=\langle a_{p},-a_{p}d\rangle=\langle a,-ad\rangle=\partial^{1}\langle a,-ad\rangle\in W(\ff_{p})$
because $p$ does not divide $a,a_{p},d$ and over $\ff_{p}$ quadratic
forms are classified by rank and determinant. Further, we have $\partial^{2}\langle a_{p},-a_{p}d\rangle=0=\partial^{2}\langle a,-ad\rangle\in W(\ff_{p})$
because $p$ does not divide $a,a_{p},d$. And so $\underset{p\neq2}{\Rightarrow}\left\langle a,-ad\right\rangle \cong\left\langle a_{p},-a_{p}d\right\rangle $
over $\qq_{p}$. 
\item $p=\infty$: $\langle a,-ad\rangle=\langle a_{\infty},-a_{\infty}d\rangle$
over $\rr=\qq_{\infty}$ because $a_{\infty}=1$ and $a>0$. 
\item $q$: Over $\qq_{q}$ the forms $\left\langle 1,-1,a,-ad\right\rangle $
and $V$ have the same rank ($=4$), determinant $d$ and Hasse invariant
(by Hilbert reciprocity, as both are isometric over $\qq_{p}$, $p\neq q$,
and thus have same Hasse symbol over $\qq_{p}$, $p\neq q$) $\Rightarrow$
$\left\langle 1,-1,a,-ad\right\rangle \cong V$ over $\qq_{q}$. 
\end{casenv}
\item [{$n\geq5$:}] Choose an orthogonal sum decomposition $V\cong U\perp W$
with $\dim U=2$ and $\dim W=n-2\geq3$. Want to find a non-degenerate
subspace of $V$ of dimension less than $n$ which is isotropic over
$\rr$ and $\qq_{p}\,\forall p$. Then by induction hypothesis, this
subspace is isotropic over $\qq$, hence $V$ is isotropic over $\qq$.
If ($U$ or) $W$ is isotropic over $\rr$ and $\qq_{p}\,\forall p$
then by induction ($U$ or)$W$ is isotropic over $\qq$ (then so
is $V$ and we are done). Hence suppose $W$ anisotropic over some
$\qq_{p}$. Let $\mathscr{P}=\{p\in\zz\cup\{\infty\}\mathrm{\, prime}|W\mbox{ anisotropic over }\qq_{p}\}$
($\neq\emptyset$). $\mathscr{P}$ is a finite set because $\dim W\geq3$
and $\left\langle a,b,c\right\rangle $ ($a,b,c\in\zz$) is isotropic
over $\qq_{p}$ ($p\neq2$) if and only if $\left\langle a,b,c\right\rangle \cong\left\langle 1,-1,-abc\right\rangle $
over $\qq_{p}$, but if $p\nmid a,b,c$ and $p\neq2,\infty$ then
$\left\langle a,b,c\right\rangle \cong\left\langle 1,-1,-abc\right\rangle $
over $\qq_{p}$. (This holds because if $p\neq2,\infty$ then $\partial^{2}$LHS
$=0=\partial^{2}$RHS, $\partial^{1}$LHS$=\left\langle a,b,c\right\rangle =\left\langle 1,-1,-abc\right\rangle =\partial^{1}$RHS,
recall $W(\qq_{p})\underset{\partial^{1},\partial^{2}}{\cong}W(\ff_{p})\oplus W(\ff_{p})$
and over $\ff_{p}$ quadratic forms are classified by rank and determinant)


Let $q$ be the quadratic form of $(V,\beta),$ $q(x)=\beta(x,x)$,
$q$ isotropic over $\qq_{p}\,\forall p\in\zz\cup\{\infty\}$ prime.
Hence $\forall p\in\zz\cup\{\infty\}$ there exists $0\neq u_{p}\in U\otimes\qq_{p}$
and $0\neq w_{p}\in W\otimes\qq_{p}$ such that $q(u_{p})+q(w_{p})=0$


\uline{Claim: }There exists $u\in\zz\setminus\{0\}$ such that
$q(u)=q(u_{p})\in\qq_{p}^{*}/\qq_{p}^{2*}$ for all $p\in\mathscr{P}$


Then the claim $\Rightarrow$ $\qq u\perp W\subset V$ has dimension
$n-1$ and is isotropic over $\rr$ and $\qq_{p}\,\forall p$ prime
because $W$ isotropic over $\qq_{p}\,\forall p\notin\mathscr{P}$
and $Qu\perp W$ isotropic over $p\in\mathscr{P}$ as $q(u)+q(w_{p})=0\,\forall p\in\mathscr{P}$.
So by induction hypothesis the subspace $\qq u\perp W$ is isotropic
over $\qq\Rightarrow V$ is isotropic over $\qq$


Justification of the claim: Now $q=ax^{2}+by^{2}$ with $a,b\in\zz\setminus\{0\}$
so $u_{p}=(x_{p},y_{p})\in\zz_{p}\times\zz_{p}$. 
\begin{casenv}
\item Assume first $\infty\notin\mathscr{P}$, then $p^{l_{p}}\xi_{p}=q(u_{p})=ax_{p}^{2}+by_{p}^{2}\in\zz_{p}\setminus\{0\}$
where $\xi_{p}\in\zz_{p}^{*}$ and $l_{p}\in\zz_{\geq0}$. By the
Chinese Remainder Theorem the map $\zz\twoheadrightarrow\prod_{p\in\mathscr{P}}\zz/p^{l_{p}+3}=\prod_{p\in\mathscr{P}}\zz_{p}/p^{l_{p}+3}$
is surjective. Hence there exists $x,y\in\zz$ such that $x=x_{p},\ y=y_{p}\mod p^{l_{p}+3}$
for $p\in\mathscr{P}$. and Set $u=(x,y)$. Then $q(u)=ax^{2}+by^{2}=ax_{p}^{2}+by_{p}^{2}\in\zz_{p}/p^{l_{p}+3}$
implies that $ax^{2}+by^{2}=p^{l_{p}}e_{p}$ with $e_{p}=\xi_{p}\mod p^{3}$.
Now $\xi_{p}\in\zz_{p}^{*}$ implies that $\xi_{p}=e_{p}\in\qq_{p}^{*}/\qq_{p}^{2*}$
due to the fact that

\begin{itemize}
\item $e_{p}=\xi_{p}\mod p^{3}$
\item $\zz_{p}^{*}/\zz_{p}^{2*}=\ff_{p}^{*}/\ff_{p}^{2*}\mod p$ ($p$ odd)
\item $\zz_{2}^{*}/\zz_{2}^{2*}=(\zz/2^{3}\zz)^{*}$
\end{itemize}

Now $\xi_{p}=e_{p}\in\qq_{p}^{*}/\qq_{p}^{2*}\Rightarrow p^{l_{p}}\xi_{p}=p^{l_{p}}e_{p}\in\qq_{p}^{*}/\qq_{p}^{2*}\,\forall p\in\mathscr{P}\Rightarrow ax^{2}+by^{2}=ax_{p}^{2}+by_{p}^{2}\in\qq_{p}^{*}/\qq_{p}^{2*}\Rightarrow q(u)=q(u_{p})\in\qq_{p}^{*}/\qq_{p}^{2*}\,\forall p\in\mathscr{P}$

\item If $\infty\in\mathscr{P}$, $\qq_{\infty}=\rr$. Then $q(u_{\infty})\in\rr^{*}/\rr^{2*}$
either $>0$ or $<0$, by replacing $q$ with $-q$ we can assume
$q(u_{\infty})>0$. Let $u_{\infty}=(x_{\infty},y_{\infty})\in\rr\times\rr$,
$q(u_{\infty})=ax^{2}+by^{2}$ not both $a,b<0$ since $q(u_{\infty})>0$,
so without loss of generality we can assume $a>0$. Choose $x,y$
as in the first case such that moreover $x^{2}>-\frac{b}{a}y^{2}$
then $q(u)=q(u_{p})\in\qq_{p}^{*}/\qq_{p}^{2*}\,\forall p\in\mathscr{P}\setminus\{\infty\}$
as above. Furthermore $q(u)=ax^{2}+by^{2}>0\Rightarrow q(u)=q(u_{\infty})\in\rr^{2}/\rr^{2*}$.
\end{casenv}

This ends the proof of the claim.

\end{description}
\end{description}
\end{proof}
\begin{defn}
Let $q$ be a rational (or integral) quadratic form. Then $q$ is
said to be
\begin{itemize}
\item \emph{positive definite} if $q(x)>0\,\forall x\neq0$. 
\item \emph{negative definite} if $q(x)<0\,\forall x\neq0$.
\item \emph{indefinite} if $q$ is neither positive nor negative definite.
\end{itemize}
\end{defn}
\begin{cor}
Let $q$ be a rational quadratic form of dimension $\geq5$. If $q$
is indefinite, then it represents $0$ over $\qq$.\end{cor}
\begin{proof}
By the strong Hasse principle, we nee to see that $q$ represents
$0$ over $\rr$ and $\qq_{p}$ for all $p\in\zz$ prime. Since $q$
indefinite $\Rightarrow q_{\rr}=\left\langle 1,-1\right\rangle \perp\dots$
so $q$ represent $0$ over $\rr$. Since dimension of $q$$\geq5\Rightarrow q$
represent $0$ over $\qq_{p}$ for all $p$ prime because every $5$
dimensional form over $\qq_{p}$ is isotropic.
\end{proof}

\subsection{Integral quadratic forms}

Recall $W(\zz)\underset{\sgn}{\overset{\cong}{\to}}\zz$. 
\begin{defn}
A symmetric inner product space over $\zz$, $(V,\beta)$ is called:
\begin{itemize}
\item \emph{even} (or of type II) if $\beta(x,x)\in\zz$ is even $\forall x\in V$
\item \emph{odd} (or of type I) if $\exists x\in V$ such that $\beta(x,x)\in\zz$
is odd.
\end{itemize}
\end{defn}
\begin{rem*}
{}
\begin{itemize}
\item A symmetric inner product space over $\zz$ is also called unimodular
lattice
\item If $q$ is a quadratic form over $\zz$ then $\beta(x,y)=q(x+y)-q(x)-q(y)$
and $\beta(x,x)=2q(X)$. So the even symmetric inner product spaces
over $\zz$ are precisely the inner product spaces that come from
regular quadratic forms. 
\end{itemize}
\end{rem*}
\begin{lem}
Let $(V,\beta)$ be an indefinite symmetric inner product space over
$\zz$ then there exists $x\in V,x\neq0$ such that $\beta(x,x)=0$.\end{lem}
\begin{proof}
Recall that the image of $W(\zz)\hookrightarrow W(\qq)$ is generated
by $\left\langle 1\right\rangle $ and $\left\langle -1\right\rangle $.
$(V,\beta)$ indefinite $\Rightarrow V_{\qq}=V\otimes_{\zz}\qq=\left\langle 1,-1\right\rangle \perp\dots$
isotropic $\Rightarrow\exists x\in V_{\qq}$ , $x\neq0$ such that
$\beta(x,x)=0$. But $V\subset V_{\qq},x=\frac{y}{n}$ for some $n\in\zz\setminus\{0\}$,
$y\in V\Rightarrow\beta(y,y)=\beta(nx,nx)=n^{2}\beta(x,x)=0$ and
$0\neq y\in V$\end{proof}
\begin{thm}
\label{thm:typeIindefOrthbasis} Let $(V,\beta)$ be an odd (i.e.,
type I) indefinite symmetric inner product space over $\zz$. Then
$(V,\beta)$ has an orthogonal basis. In particular $(V,\beta)\cong m\left\langle 1\right\rangle \perp n\left\langle -1\right\rangle $
over $\zz$.\end{thm}
\begin{proof}
\uline{Claim:} $(V,\beta)\cong\left\langle \begin{pmatrix}0 & 1\\
1 & \mbox{odd}
\end{pmatrix}\right\rangle \perp(V',\beta')$.

The theorem follows from claim by induction on dimension $V$: For
$k\in\zz$ we have 
\[
\begin{pmatrix}0 & 1\\
1 & 2k+1
\end{pmatrix}=\begin{pmatrix}1 & 1\\
k+1 & k
\end{pmatrix}\begin{pmatrix}1 & 0\\
0 & -1
\end{pmatrix}\underset{\det=-1}{\underbrace{\begin{pmatrix}1 & k+1\\
1 & k
\end{pmatrix}}}\Rightarrow\left\langle \begin{pmatrix}0 & 1\\
1 & 2k+1
\end{pmatrix}\right\rangle \cong\left\langle 1,-1\right\rangle .
\]
 From this the theorem follows as we have $\left\langle 1\right\rangle \perp V'$
or $\left\langle -1\right\rangle \perp V'$ is indefinite and both
are odd and have dimension less that $V$, and $V=\left\langle \pm1\right\rangle \perp(\left\langle \mp1\right\rangle \perp V')$

To prove the claim: We know $(V,\beta)$ indefinite $\underset{\mathrm{previous\, lemma}}{\Rightarrow}\exists x\in V$
with $x\neq0$ and $\beta(x,x)=0$. Now $(V,\beta)$ inner product
space over $\zz\Rightarrow V=\zz^{n}$. So $x=(a_{1},\dots,a_{n})\in\zz^{n}$,
we can assume $d=\gcd(a_{1},\dots,a_{n})=1$ (otherwise replace $x$
with $\frac{x}{d}$). We can extend $x$ to a $\zz$ basis $x_{1}=x,x_{2},\dots,x_{n}$
of $V=\zz^{n}$ because $\phi:V/\zz x\to(V\otimes_{\zz}\qq)/\qq x$
is injective as $y=(c_{1},\dots,c_{n})\in\ker\phi$ then $\exists r,s\in\zz$
with $r\neq0$ and $ry=sx\Rightarrow rc_{i}=sa_{i}$ for all $i=1,\dots,n$,
since $\gcd(a_{1},\dots,a_{n})=1$ there exists $b_{1},\dots,b_{n}\in\zz$
such that $\sum a_{i}b_{i}=1$. So $r\sum c_{i}b_{i}=s\sum a_{i}b_{i}=s$
$\Rightarrow s=rt$ where $t=\sum c_{i}b_{i}$, hence $ry=sx\Rightarrow ry=rtx\underset{r\neq0}{\Rightarrow}y=tx\Rightarrow y=0\in V/\zz x$.
Hence our map is indeed injective. Now $V/\zz x$ is a finitely generated
$\zz$-module, submodule of $\frac{V\otimes_{\zz}\qq}{\qq x}=\qq^{n-1}\Rightarrow V/x$
is a free $\zz$-module $\Rightarrow V\overset{p}{\twoheadrightarrow}V/x$
has a section $\sigma:V/\zz x\to V$ ($p\sigma=1$)$\Rightarrow V=x\oplus\underset{\cong V/x\cong\zz^{n-1}}{\underbrace{\im\sigma}}$.
Hence $x$ can be extended to a $\zz$-basis $x_{1}=x,x_{2},\dots,x_{n}$
of $V$. 

Let $y_{1},\dots,y_{n}$ be the dual basis of $x_{1},\dots,x_{n}$,
i.e., $\beta(x_{i},y_{j})=\begin{cases}
1 & i=j\\
0 & i\neq j
\end{cases}$ which exists because $\beta$ is non-degenerated $\Rightarrow$ $\beta:V\overset{\cong}{\to}\Hom(V,\zz)\cong\zz^{n}$
has a $\zz$ basis $e_{1},\dots,e_{n}$, where $(e_{i})(\sum\alpha_{j}x_{j})=\alpha_{i}$
, $y_{i}\overset{\beta}{\leftrightarrow e_{i}}.$ Now $(V,\beta)$
odd $\Rightarrow$ there exists $k\in\{1,\dots,n\}$ such that $\beta(y_{k},y_{k})$
is odd. If $\beta(y_{1},y_{1})$ is odd then $\beta|_{(\zz x_{1}+\zz y_{1})}=\left\langle \begin{pmatrix}0 & 1\\
1 & \mathrm{odd}
\end{pmatrix}\right\rangle \Rightarrow(V,\beta)\cong\left\langle \begin{pmatrix}0 & 1\\
1 & \mathrm{odd}
\end{pmatrix}\right\rangle \overset{\mathrm{non-degenerate}}{\perp}(\zz x_{1}\oplus\zz y_{1})^{\perp}$. If $\beta(y_{1},y_{1})$ even and $\beta(y_{k},y_{k})$ odd for
$k\neq1$ then $\beta|_{(\zz x_{1}+\zz(y_{1}+y_{k}))}=\left\langle \begin{pmatrix}0 & 1\\
1 & \mathrm{odd}
\end{pmatrix}\right\rangle \Rightarrow(V,\beta)\cong\left\langle \begin{pmatrix}0 & 1\\
1 & \mathrm{odd}
\end{pmatrix}\right\rangle \overset{\mathrm{non-degenerate}}{\perp}(\zz x_{1}\oplus\zz(y_{1}+y_{k}))^{\perp}$\end{proof}
\begin{thm}
If $(V,\beta)$ is an even symmetric inner product space over $\zz$
then its signature is divisible by $8$.\end{thm}
\begin{proof}
Let $(V,\beta)$ be an arbitary symmetric inner product space over
$\zz$. Then $V/2V=V\otimes_{\zz}\ff_{2}$ is a symmetric inner product
space over $\ff_{2}$. But over $\ff_{2}$ the map $x\mapsto\beta(x,x)$
is linear because $\beta(x+y,x+y)=\beta(x,x)+\underset{=0}{\underbrace{2\beta(x,y)}}+\beta(y,y)\in\ff_{2}$.
As $V/2V$ is non degenerate, there exists a unique $\overline{u}\in V/2V$
such that $\beta(\overline{u},x)=\beta(x,x)\mod2\,\forall x\in V.$
If $u,u'\in V$ are two lifts of $\overline{u}\in V/2V$ then $u'=u+2v$
for some $v\in V$, and $\beta(u',u')=\beta(u,u)+\underset{=0\mod2}{4(\underbrace{\beta(u,v)+\beta(v,v)})}=\beta(u,u)\in\zz/8\zz$
because $\beta(u,v)=\beta(v,v)\in\ff_{2}$ by definition of $u$.
Set $\phi(V):=\beta(u,u)\in\zz/8\zz$ for any lift $u$ of $\overline{u}\in V/2V$.
We have seen that $\phi(V)$ does not depend on the lift $u$ of $\overline{u}$.
From the definition of $\phi$ we have $\phi(V\perp W)=\phi(V)+\phi(W)$
and $\phi(\left\langle 1\right\rangle )=1$, $\phi(\left\langle -1\right\rangle )=-1$,
so $\phi:W(\zz)\to\zz/8\zz:V\mapsto\phi(V)$ is a well defined map.
If $(V,\beta)$ is even then $\beta(x,x)=0\mod2\,\forall x\in V$
and we can choose $\overline{u}=0\Rightarrow u=0\Rightarrow\phi(V)=0\in\zz/8\zz$.
Since $\phi(\left\langle 1\right\rangle )=1\Rightarrow\phi:\zz\cdot\langle1\rangle=W(\zz)\to\zz/8\zz$
is the signature. Now, if $(V,\beta)$ is even then $\beta(x,x)=0\mod2\,\forall x\in V$
and $\overline{u}=0\Rightarrow\text{we can choose }u=0\Rightarrow\phi(V)=0\in\zz/8\zz$.
This implies that signature of any even symmetric inner product space
over $\mbox{\ensuremath{\zz}}$ is divisible by $8$.\end{proof}
\begin{cor}
Every even positive definite inner product space over $\zz$ has rank
divisible by $8$.\end{cor}
\begin{proof}
If $M$ is positive definite then $\rk M=\sgn M$.\end{proof}
\begin{thm}
Let $M,N$ be indefinite symmetric inner product spaces over $\zz$.
Then $M\cong N$ $\iff M,N$ have the same rank, signature and type
(odd or even).\end{thm}
\begin{proof}
If $M,N$ are odd then $M,N$ have orthogonal basis, by Theorem \ref{thm:typeIindefOrthbasis},then
the theorem follows.

If $M,N$ are even, we do not have the time to prove this in this
course.\end{proof}
\begin{example*}
Of even positive definite inner product spaces over $\zz$.

\uline{General Remark}: Let $\rr^{n}$ be equipped with standard
Euclidean inner product $\left\langle (x_{1},\dots,x_{n}),(y_{1},\dots,y_{n})\right\rangle =\sum_{i=1}^{n}x_{i}y_{i}$.
If $M\subseteq\rr^{n}$ is a finitely generated $\zz$-submodule,
then $M\cong\zz^{k}$ for some $k\leq n$. Restricting $\left\langle ,\right\rangle _{\rr^{n}}$
to $M$ defines a symmetric bilinear form $\beta(x,y)=\left\langle x,y\right\rangle \in\rr$
on $M$ with values in $\rr$. Assume $\rk_{\zz}M=n=\dim_{\rr}\rr^{n}\Rightarrow\rr^{n}/M$
compact Riemanian manifold. $\mbox{Vol}(\rr^{n}/M)=$ volume of parallelepiped
spanned by a $\zz$-basis of $M$. If we let $A=(b_{1},\dots,b_{n})$
where $b_{1},\dots,b_{n}$ is a $\zz$-basis of $M$ then $\mbox{Vol}(\rr^{n}/M)=|\det A|=\sqrt{\det(A^{T}A)}=\sqrt{\det\underset{\mathrm{biliner\, form\, matrix\, of\,}M}{\underbrace{(\left\langle b_{i},b_{j}\right\rangle )_{i,j=1,\dots,n}}}}\Rightarrow$
a finitely generated $M\subset\rr^{n}$ of $\rk_{\zz}M=n$ defines
a (positive definite) inner product space over $\zz$ if and only
if:
\begin{itemize}
\item $\left\langle x,y\right\rangle \in\zz\,\forall x,y\in M$, and
\item $\mbox{Vol}(\rr^{n}/M)=1$
\end{itemize}

In fact every possible definite inner product space $(M,\beta)$ over
$\zz$ arises in that way, because $\rr^{n}\cong M\otimes_{\zz}\rr\supset M$
and $\beta_{\rr}\cong\underset{n}{\underbrace{\left\langle 1,\dots,1\right\rangle }}$
over $\rr$ since $(M,\beta)$ is positive definite. 
\begin{lem}
Let $E_{4m}\subseteq\rr^{4m}$ be the $\zz$-submodule ($m\in\zz_{\geq1}$)
generated by $e_{i}+e_{j}$ ($i,j=1,\dots,4m)$ and $\frac{1}{2}(e_{1}+e_{2}+\dots+e_{4m})$
where $e_{i}=(0,\dots,0,\underset{\underset{i}{\uparrow}}{1},0,\dots,0)$
is the standard basis vector of $\rr^{4m}$. Then $E_{4m}$ is a symmetric
inner product space over $\zz$ of rank $4m$ which is even (respectively
odd) if $m$ is even (respectively odd)\end{lem}
\begin{proof}
{}
\begin{itemize}
\item $E_{4m}\subset\rr^{4m}$ finitely generated $\zz$-submodule $\Rightarrow E_{4m}$
free $\zz$-module, i.e., $E_{4m}\cong\zz^{k}$. Now $\rk_{\zz}E_{4m}=\dim_{\qq}E_{4m}\otimes_{\zz}\qq=4m$
because $e_{i}+e_{j},i,j=1,\dots,4m$ and $\frac{1}{2}(e_{1}+\dots+e_{4m})$
span $\qq^{n}$. (Note that this contains $2e_{i},i=1,\dots,m$ by
setting $i=j$).
\item $\left\langle x,y\right\rangle \in\zz\,\forall x,y\in E_{4m}$. (check
for $x,y$ generators of $E_{4m}$). E.g., $\left\langle e_{i}+e_{j},e_{i}+e_{j}\right\rangle =\begin{cases}
2 & i\neq j\\
4 & i=j
\end{cases}$, \\
$\left\langle \frac{1}{2}(e_{1}+\dots+e_{4m}),\frac{1}{2}(e_{1}+\dots+e_{4m})\right\rangle =\frac{1}{4}4m=m$
$\Rightarrow E_{4m}$ is even if and only if $m$ even.
\item We are left to check it is non-degenerate. We will use the following
trick: If $M\subset N\subset\rr^{n}$ of rank $n$ $\zz$-submodule,
then $\rr^{n}/M\twoheadrightarrow\rr^{n}/N$ covering with $|N/M|$
sheets because $N/M$ acts freely on $\rr^{n}/M$ with quotient $\rr^{n}/N$.
So $|N/M|\cdot\mbox{Vol}(\rr^{n}/N)=\mbox{Vol}(\rr^{n}/M)\,\forall M\subset N\subset\rr^{n}$
$\rk=n$ $\zz$-submodules.


Now we prove $E_{4m}$ is non-degenerate, i.e., $\mbox{Vol}(\rr^{4m}/E_{4m})=1$.
Let $E^{0}\subset E_{4m}$ be the $\zz$-submodule generated by $e_{i}+e_{j},i,j=1,\dots4m$.
Then $E_{4m}/E^{0}$ is generated by $\xi=\frac{1}{2}(e_{1}+\dots+e_{4m})\notin E^{0}$,
and $2\xi\in E^{0}$ so $2\xi=0\in E_{4m}/E^{0}$. Therefore $E_{4m}/E^{0}=\zz/2\zz\Rightarrow2\mbox{Vol}(\rr^{4m}/E_{4m})=\mbox{Vol}(\rr^{4m}/E^{0})$.
But notice $E^{0}\subset\zz^{4m}$ where $\zz^{4m}$ is generated
by $e_{1},\dots,e_{4m}$. Now $\zz^{4m}/E^{0}$ is generated by $e_{1}$
because $e_{i}=e_{i}+e_{1}-e_{1}\,\forall i=2,\dots,4m$. Now $e_{1}\notin E^{0}$
but $2e_{1}=e_{1}+e_{1}\in E^{0}\Rightarrow\zz^{4m}/E^{0}=\zz/2\zz\Rightarrow\mbox{Vol}(\rr^{4m}/E^{0})=2\mbox{Vol}(\rr^{4m}/\zz^{4m})\Rightarrow\mbox{Vol}(\rr^{4m}/E^{4m})=\mbox{Vol}(\rr^{4m}/\zz^{4m})=1\Rightarrow E_{4m}$
is non-degenerate.

\end{itemize}
\end{proof}
\begin{cor}
$E_{8m}$ is an even positive definite symmetric inner product space
of rank $8m$\end{cor}
\begin{fact*}
For all $n\in\zz_{\geq0}$, $\{$symmetric inner product spaces over
$\zz$ of given rank $n\}/$isometry is a finite set.\end{fact*}
\begin{example*}
~

\begin{tabular}{|c|c|c|c|c|}
\hline 
rank$ $$n$  & 8  & 16  & 24  & 32\tabularnewline
\hline 
\hline 
number of even positive definite inner products space over $\zz$  & 1  & 2  & 24  & Unknown $\geq10^{7}$\tabularnewline
\hline 
representative  & $E_{8}$  & $E_{16},E_{8}\perp E_{8}$  & %
\begin{tabular}{cc}
Niemeier & \tabularnewline
(1968) & \tabularnewline
\end{tabular} & \tabularnewline
\hline 
\end{tabular}\end{example*}
\end{example*}

\end{document}
