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\title{Harrison's criterion, Witt equivalence and reciprocity equivalence}
\author{N. Grenier-Boley\footnote{The author was supported by
the European research network HPRN-CT-2002-00287 \textquotedblleft K-theory, Linear Algebraic Groups and
Related Structures\textquotedblright}}

\begin{document}

\maketitle \noindent \textbf{Abstract.} Harrison's criterion
characterizes the isomorphy of the Witt rings of two fields in terms
of properties of these fields. In this article, we discuss about the
existence of such characterizations for the isomorphism of Witt
groups of hermitian forms over certain algebras with involution. In
the cases where we consider the Witt group of a quadratic extension
with its non-trivial automorphism or the Witt group of a quaternion
division algebra with its canonical involution, such criteria are
proved. In the framework of global fields, the first of these
criteria is reformulated in terms of
properties involving real places of the considered fields.\\

\noindent\textbf{Mathematics subject classification (2000):} Primary
11E39 ; Secondary 11E81\\

\noindent\textbf{Keywords:} Harrison's criterion, Witt equivalence, reciprocity equivalence,
algebras with involution.\\

\section{Introduction}
One of the basic questions in the algebraic theory
of quadratic forms is to give necessary and sufficient conditions for two fields $K_1$ and $K_2$ to have
isomorphic Witt rings: in this case, $K_1$ and $K_2$ are said to be
\emph{Witt equivalent}. In \cite{harrison}, Harrison expresses Witt
equivalence in the following terms:
\begin{thm}[Harrison]\label{harriso}
Let $K_1$ and $K_2$ be two fields of characteristic different from
$2$. Then the following
are equivalent:\\
$(1)$ $K_1$ and $K_2$ are Witt equivalent.\\
$(2)$ There is a group isomorphism $t: {K_1}^ {*}/{{K_1}^{*}}^{2}
\to {K_2}^{*}/{{K_2}^{*}}^{2}$ sending $-1$ to $-1$ such that the
quadratic form $\l x,y \r$ represents $1$ over $K_1$ if and only if
the quadratic form $\l t(x),t(y) \r$ represents $1$ over $K_2$ for
all $x, \ y \in {K_1}^*$.
\end{thm}\noindent
In the literature, the previous theorem is known as
\textquotedblleft Harrison's criterion". In \cite[Theorem
2.3]{cordes}, Cordes shows that the two conditions of Theorem
\ref{harriso} are equivalent to the following one,
 known as \textquotedblleft Harrison-Cordes condition":\\
$(3)$ There is a group isomorphism $t: {K_1}^ {*}/{{K_1}^{*}}^{2}
\to {K_2}^{*}/{{K_2}^{*}}^{2}$ sending $-1$ to $-1$ such that $t\pr
D_{K_1} (\l 1, a \r) \prt = D_{K_2}\pr\l 1 , t(a) \r\prt$ (if $K$ is
a field of characteristic different from $2$ and $q$ is a nondegenerate
quadratic form, recall that $D_K(q)$ denotes the set of nonzero square classes
represented by $q$). With this criterion, Cordes classifies Witt rings of order at most
$32$ up to Witt equivalence by their group structure: see
\cite[Theorem 7.1]{cordes}.

In \cite{baeza-moresi}, Baeza and Moresi study the possibilities to
extend Harrison's criterion to fields $K_1$ and $K_2$ of
characteristic $2$. On the one hand, they show that the bilinear
Witt rings $W(K_1)$ and $W(K_2)$ of $K_1$ and $K_2$ are isomorphic
if and only if $K_1$ and $K_2$ are isomorphic in the case where
$\dim_{{K_1}^2} K_1 = \dim_{{K_2}^2} K_2 > 2$, and they give a complete
treatment of the cases where $\dim_{{K_1}^2} K_1 = \dim_{{K_2}^2}
K_2 =1,\ 2$: see \cite[Theorem 2.9, Proposition 2.10]{baeza-moresi}.
On the other hand, in \cite[Theorem 3.1]{baeza-moresi}, they
characterize the isomorphy of the quadratic Witt modules $W_q (K_1)$
et $W_q(K_2)$ in the following way:
\begin{thm}[Baeza-Moresi]\label{Bmo}
Let $K_1$ and $K_2$ be two fields of characteristic $2$. Then the
following
are equivalent:\\
$(1)$ There exist a ring isomorphism $\Phi: W(K_1) \to W(K_2)$ and a
group isomorphism $\Psi: W_q (K_1) \to W_q(K_2)$ such that
$$\Psi (b.q) = \Phi (b) . \Psi(q)$$ for all $b \in W(K_1)$ and for all
$q \in W_q( K_1)$.\\
$(2)$ There exist groups isomorphisms
$$t_1:{K_1}^{*}/{{K_1}^{*}}^{2} \to {K_2}^{*}/{{K_2}^{*}}^{2}, \quad t_2:
K_1/\wp (K_1) \to K_2/\wp (K_2)$$ such that $$t_1
\prr D_{K_1}(\l 1, a \r)\prtt = D_{K_2} \prr\l 1, t_1 (a)
\r\prtt , \quad t_2 (D_{K_1} \lbrack 1, b\rbrack) = D_{K_2}
\lbrack 1, t_2 (b) \rbrack$$ for all $a \in {K_1}^*$ and for all $b
\in K_1$ \emph{(}where $\wp (K_i) = \lbrace a+a^2 \ | \ a\in
K_i \rbrace, \ i=1, \ 2$\emph{)}.
\end{thm}\noindent
Note that the first condition in $(2)$ is similar to
\textquotedblleft Harrison-Cordes condition".

Such criteria are very useful. For example, Harrison's criterion is
used by Min\'a\v{c} and Spira to connect the Witt equivalence of two
fields $K_1$ and $K_2$ to the isomorphy of some groups $G_{K_1}$ and
$G_{K_2}$ (called W-groups), $G_{K_i}$ being the Galois group of a
certain field extension ${K_i}^{(3)}$ of $K_i$ for $i=1, \ 2$: see
\cite{minac-spira}.

In this context, a natural question arises: is it possible to obtain
such criteria for the Witt group of a central simple algebra with
involution ? After recalling some notations and basic facts in
Section \ref{prel}, we explain how to obtain such criteria in two
particular cases in Section \ref{harris}. First, we obtain a
criterion similar to Theorem \ref{harriso} for the Witt group of a
quadratic field extension equipped with its nontrivial automorphism:
\begin{thm}\label{main6}
Let $K_1$ and $K_2$ be two fields of characteristic different from
$2$. Let $L_1=K_1(\sqrt{a})$ \emph{(}resp.
$L_2=K_2(\sqrt{b})$\emph{)} be a quadratic field extension of $K_1$
\emph{(}resp. $K_2$\emph{)} equipped with its non trivial
automorphism $\sigma_1$ \emph{(}resp. $\sigma_2$\emph{)}.
Then, the following are equivalent:\\
$(1)$ $W(L_1,\sigma_1)\simeq W(L_2,\sigma_2)$ as rings.\\
$(2)$ There is a group isomorphism $t: {K_1}^*/N_{L_1/K_1}({L_1}^*)
\to K_2^*/N_{L_2/K_2}({L_2}^*)$ sending $-1$ to $-1$ such that the
quadratic form $\l\l a,x,y \r\r$ is hyperbolic over $K_1$ if and
only if the quadratic form $\l \l b,t(x),t(y)\r \r$ is hyperbolic
over $K_2$ for all $x, \ y \in {K_1}^*$, where $N_{L_i/K_i}({L_i}^*)$
denotes the norm group of the extension $L_i/K_i$ for $i=1, \ 2$.
\end{thm}\noindent
Moreover, in Assertion $(2)$, we show that the condition
\textquotedblleft $t(-1)=-1$" is not a consequence of the two other
conditions: see Example \ref{exe}. Next, in the case of the Witt
group of a quaternion division algebra endowed with its canonical
involution, two examples point out that its structure of module
seems to be the right choice to obtain such a criterion: see
Examples \ref{exewitt}. In this direction, we obtain Theorem
\ref{main7} which is at the same time similar to Theorem
\ref{harriso} and to Theorem \ref{Bmo} (in some sense). At first
sight, Theorem \ref{main7} seems to be more analogous to Baeza and
Moresi's results, but the similarity with Harrison's criterion shows
up by taking $K_1=K_2=K$:
\begin{cor}\label{main8}
Let $Q_1$ (resp. $Q_2$) be quaternion division algebras over $K$ endowed
with its canonical involution $\gamma_1$ (resp. $\gamma_2$). Then, the following are equivalent:\\
$(1)$ $W(Q_1,\gamma_1)\simeq W(Q_2,\gamma_2)$ as $W(K)$-modules.\\
$(2)$ There is a group isomorphism $\tilde{t} :{K}^*/ {\Nrd}_{Q_1/K}({Q_1}^*)
\simeq {K}^*/ {\Nrd}_{Q_2/K}({Q_2}^*)$
with $\tilde{t}(-1)=-1$ such that the quadratic form $\l\l
a,b,u,v\r\r$ is hyperbolic over $K$ if and only if the quadratic
form $\l\l c,d,\tilde{t}(u),\tilde{t}(v)\r\r$ is hyperbolic over
$K$ for all $u, \ v \in {K}^*$.
\end{cor}

In this framework, another interesting problem is to give necessary
and sufficient conditions for two global fields to be Witt
equivalent. This problem is now entirely solved. The first step is
due to Baeza and Moresi who show that two global fields of
characteristic $2$ are always Witt equivalent: see
\cite{baeza-moresi} or \cite[\S 2]{pscl}. Besides, if two global
fields are Witt equivalent then either $\car (K_1), \ \car (K_2)
\neq 2$ or $\car(K_1)= \car(K_2) =2$: see \cite[\S 2]{pscl}. Lastly,
in \cite[\S 3, \S 4]{pscl}, Perlis, Szymiczek, Conner and Litherland
prove that two global fields $K_1$ and $K_2$ of characteristic different from $2$ are Witt equivalent
if and only if they are reciprocity equivalent (i.e. if there exist a
group isomorphism $t$ between their square class groups and a bijection $T$ between their
non trivial places such that the Hilbert symbols $(x,y)_P$ and $(t(x),t(y))_{T(P)}$ are equal
for any $x, \ y \in {K_1}^*/{K_1}^2$ and for any non trivial place $P$ over $K_1$).

In Section \ref{equiva}, we explain how to obtain such a classification for
the Witt group of a quadratic field extension endowed with its nontrivial automorphism.
For this purpose, we define the notion of $(a,b)$-quadratic
reciprocity equivalence between two global fields of characteristic different from $2$. This
notion is a natural adaptation of the reciprocity equivalence when considering norm class groups
of quadratic extensions instead of square class groups (see Definition \ref{eqrecq}).
\begin{thm}\label{main9}
Let $K_1$ and $K_2$ be two global fields of characteristic different
from $2$. Let $L_1=K_1(\sqrt{a})$ \emph{(}resp.
$L_2=K_2(\sqrt{b})$\emph{)} be a quadratic field extension of $K_1$
\emph{(}resp. $K_2$\emph{)} equipped with its nontrivial
automorphism $\sigma_1$ \emph{(}resp. $\sigma_2$\emph{)}. Then,
the following are equivalent:\\
$(1)$ $W(L_1,\sigma_1)\simeq W(L_2,\sigma_2)$ as rings.\\
$(2)$ There is an $(a,b)$-quadratic reciprocity equivalence between
$K_1$ and $K_2$.
\end{thm}

\section{Basic results and notations}\label{prel}

In this Section, we fix some notations and recall some basic definitions and results.

\subsection{Central simple algebras wth involution}

The general reference for the theory of central simple algebras
with involution is \cite{BOI}: see also \cite[Chapter 8]{scharlau}

In this Section, $K$ will denote a field of characteristic different
from $2$, $D$ will denote a finite-dimensional division algebra over
$K$. Then $\dim_K D = n^2$ for some $n\in \mathbb{N}$, and $n=\deg D$
is called the degree of $D$. Suppose that $D$ is endowed with an involution $\sigma$. The map
$\sigma$ restricts to an involution of the center $K$ and we can
distinguish two cases: if $\sigma|_K$ is the identity, we say that
$\sigma$ is of \emph{the first kind}, otherwise $\sigma|_K$ is an
automorphism of order two of $K$ and we say that $\sigma$ is of
\emph{the second kind}.

A central simple algebra $D$ of degree $2$ is called a
\emph{quaternion algebra}. As $\car(K) \neq 2$, every quaternion
algebra has a quaternion basis $\lbrace 1,i,j,k\rbrace$, that is a
basis of the $K$-algebra $Q$ subject to the relations
$$i^2=a \in {K}^*, \ j^2=b \in {K}^*, ij=k=-ji.$$
This algebra $Q$ is then denoted by $Q=(a,b)_{K}$. Note also that every
quaternion algebra has a canonical involution (usually denoted by
$\gamma$) which is of the first kind and defined as follows:
$$\gamma(i)=-i, \ \gamma(j)=-j.$$

\subsection{Hermitian forms}

The standard reference for the theory of hermitian forms is
\cite{scharlau}. All the vector spaces considered are supposed to be
finite dimensional right vector spaces.

A \emph{hermitian form} over $(D,\sigma)$ is a pair $(V,h)$ where
$V$ is a finite dimensional $D$-vector space and $h$ is a map $h :
V\times V \to D$ which is $\sigma$-sesquilinear in the first
argument, $K$-linear in the second argument and which satisfies
$$\sigma\pr h(x,y)\prt = h(y,x) \ \mathrm{for \ any} \ x, \ y \in V.$$
If $D=K$ and $\sigma=\id_{K}$ then a hermitian form is a symmetric
bilinear form which can be identified with a quadratic form as
$\car(K)\neq 2$. The integer $\dim_K V$ is called the dimension
(or the rank) of the hermitian form. Since $\car(K)\neq 2$ and $D$ is
division, a hermitian form over $(D,\sigma)$  may be given by a diagonalization
$\l a_1,\cdots, a_n\r$.

Note that the dual vector space $V^*:=\Hom_D(V,D)$ can be considered as a right
$D$-vector space by twisting the action of $D$ by $\sigma$. We
define the adjoint map of $(V,h)$ by the following morphism of right
$D$-vector spaces
$$\widehat{h} :\left\{
\begin{array}{cll}
V& \rightarrow & V^{*}\\
x& \mapsto & h_x
\end{array}\right.$$
where $h_{x}(y) = h(x,y)$ for all $y \in M$. If this map is an
isomorphism of right $D$-vector spaces, $h$ is said to be
\emph{nondegenerate}. We say that $h$ is \emph{isotropic} if there
exists an $x \in V \setminus\lbrace 0 \rbrace$ such that $h(x,x)=0$,
anisotropic otherwise. If $y$ is an element of $D$ such that
$h(x,x)=y$ for a certain $x \in V \setminus \lbrace 0 \rbrace$, then we
say that $h$ \emph{represents} $y$.

Let $(V,h)$ and $(V',h')$ be two hermitian forms over $(D,\sigma)$.
If these forms are isometric then we write $h \simeq h'$ for short.
Their \emph{orthogonal sum} is denoted by $h\bot h'$.

If $(V,h)$ is a hermitian form over $(D,\sigma)$, one can attach to
it a nondegenerate hermitian form over $(D,\sigma)$ denoted by $(V
\oplus V^*, \h_V)$ with
$$\h_V (v \oplus f, v' \oplus
g) = f(v') + \sigma \pr g(v)\prt$$ for all $f, \ g \in V^*$, $v, \
v' \in V$. A hermitian form over $(D,\sigma)$ which is isometric to
such a form is called a \emph{hyperbolic form}.

Witt's cancelation theorem says that if $(V,h)$, $(V',h')$,
$(V'',h'')$ are three hermitian forms over $(D,\sigma)$ and if
$h\bot h' \simeq h \bot h''$ then $h'\simeq h''$: see \cite[Chapter
5.11]{jacobson2}.

\subsection{The Witt group of a division algebra with involution}

We refer to \cite{scharlau} for more details about the Witt group.

Let $(V,h)$ be a nondegenerate hermitian form over $(D,\sigma)$. The
isometry class of this form is denoted by $[V,h]$ or $[h]$ for short
and $[0]$ denotes the zero-dimensional hermitian form which is supposed to be
nondegenerate by convention. Let  $S(D,\sigma)$ denote the set of
isometry classes of nondegenerate hermitian forms over $(D,\sigma)$.
This set is endowed with a structure of commutative mono\"{\i}d by the
law
$$[h]+[h']=[h\bot h'].$$
The \emph{Witt group} of $(D,\sigma)$ is the quotient group of the
Grothendieck group of $S(D,\sigma)$ by the subgroup generated by
hyperbolic forms and is denoted by $W(D,\sigma)$. In the case where
$D=K$, Kronecker's product can be used to define a structure of ring
on $W(K,\sigma)$. If moreover $\sigma=\id_K$, this ring is
called the \emph{Witt ring} of $K$ and is denoted by $W(K)$.

If $[V,h] \in W(K, \sigma|_K)$ and $[V',h'] \in W(D,\sigma)$, we
define $[V \otimes_K V', h \cdot h']\in W (D, \sigma)$ as the Witt
class of the following nondegenerate hermitian form
\begin{eqnarray*}
h\cdot h' :\left\{\begin{array}{cll}
 ( V \otimes_{K} V' ) \times (V
\otimes_{K} V') &
\rightarrow & D\\
(v_1 \otimes v_1', v_2\otimes v_2') & \mapsto & h(v_1,v_2) h'(v_1',v_2')
\end{array}.\right.
\end{eqnarray*}
It is easy to see that this action endows $W(D,\sigma)$ with a
structure of $W(K,\sigma|_K)$-module. The submodule generated by
nondegenerate hermitian forms of even rank is denoted by
$I_1(D,\sigma)$ or by $I(K)$ if $D=K$ and $\sigma=\id_K$. We write
$I^n (K)$ for $\pr I(K) \prt^n$. It is easy to see that $I^n(K)$ is
additively generated by the so-called Pfister forms
$$\l 1,-a_1 \r \otimes \cdots\otimes \l 1,-a_n \r:=\l\l a_1,\cdots,a_n\r\r,$$
for $a_1,\cdots,a_n \in K^*$.

\subsection{The (refined) discriminant of a hermitian form}

We refer to \cite[\S 2]{bayer-parimala1} and \cite[\S
3]{bayer-parimala2} for more general statements about this
invariant.

Let $(V,h)$ be a hermitian form over $(D,\sigma)$ and suppose first
that $\sigma$ is an involution of the first kind. Let $\lbrace
e_1,\cdots,e_n\rbrace$ be a $D$-basis of the right $D$-vector space
$V$. Let $M$ be the matrix of $h$ with respect to this basis,
$E=M_n(D)$ and $m=n \deg(D)$. We define the \emph{signed
discriminant} of $(V,h)$ by
$$\dpm(h)=(-1)^{\frac{m(m-1)}{2}}{\Nrd}_{E/K} (M) \in K^*,$$
where ${\Nrd}_{E/K}$ denotes the usual reduced norm map: see \cite[\S
22]{draxl}. One can show that $\dpm$ induces a well-defined group
homomorphism, again denoted by $\dpm$
$$\dpm :I_1 (D,\sigma) \rightarrow K^*/{K^*}^2.$$
If $\sigma$ is an involution of the second kind and if $F$ is the
fixed field of $\sigma$ in $K$, one can define the signed
discriminant of $(V,h)$ by the formula above and show that it
induces a group homomorphism
$$\dpm :I_1(D,\sigma) \rightarrow F^*/N_{K/F}(K^*),$$
where $N_{K/F}(K^*)$ is the group of norms of $K/F$. The kernel of
the signed discriminant is denoted by $I_2(D,\sigma)$.

In both cases, the signed discriminant can be refined as follows.
We will describe it only for $\sigma$ of the first kind.
Denote by $\Nrd_{D/K}(D^*)$ the group of reduced norms from $D$. With the same
formula as above, on can define a group homomorphism
$$\Disc :I_1(D,\sigma)\to{\Nrd}_{D/K}(D^*)/{\Nrd}_{D/K}(D^*)^2,$$
which is called the \emph{refined discriminant}.

\subsection{Chain equivalence}

Throughout this Subsection we will use the notations of
\cite[Chapter I, \S 5]{lam} and will refer to it for more general
statements.

Let $D$ be a division algebra over $K$ endowed with an involution
$\sigma$ (of arbitrary kind). Let $h=\l a_1,\cdots,a_n\r$ and $h'=\l
a'_1,\cdots,a'_n\r$ be two hermitian forms over $(D,\sigma)$. They
are said to be \emph{simply equivalent} if there exists indices $i, \ j
\in \lbrace 1,\cdots,n\rbrace$ such that $\l a_i,a_j\r \simeq \l
b_i,b_j\r$ and $a_k=b_k$ for every $k$ different from $i$ and $j$
(note that, if $i=j$, the expression $\l a_i, a_j \r$ is understood
to be $\l a_i\r$). Two (diagonalized) hermitian forms $h$ and $h'$ over $(D,\sigma)$
are \emph{chain equivalent} if there is a sequence of diagonalized
hermitian forms $f_0,\cdots,f_m$ over $(D,\sigma)$ such that
$h=f_0$, $h'=f_m$ and such that $f_i$ is simply equivalent to
$f_{i+1}$ for $0\leq i\leq m-1$. We immediately see that two chain
equivalent forms are isometric. In fact, the converse is also true
by \textquotedblleft Witt's Chain equivalence Theorem":
\begin{thm}[Witt]\label{eqch}
If $h$ and $h'$ are two \emph{(}diagonalized\emph{)} hermitian forms
over $(D,\sigma)$ and if $h$ is isometric to $h'$ then $h$ and $h'$
are chain equivalent.
\end{thm}
\begin{proof} The proof can be easily adapted from \cite[Chapter I, Theorem
5.2]{lam} by replacing usual squares by hermitian squares (that is,
elements of the form $\sigma(x) x$).
\end{proof}

\begin{rem} The above proposition is very useful for showing
general properties for quadratic (resp. hermitian) forms by reducing it to
properties for two-dimensional quadratic (resp. hermitian) forms.
See the proofs of Theorems \ref{harriso}, \ref{main6} and
\ref{main7}.
\end{rem}

\subsection{Further results}\label{jac}

Here we want to state two important results that are used several times in this paper.

The following result known as \textquotedblleft Arason-Pfister Hauptsatz" gives a
sufficient condition for a quadratic form to belong to $I^n(K)$.
\begin{thm}[Arason-Pfister]\label{ArPf}
Let $q$ be a positive-dimensional anisotropic quadratic form over $K$. If $q \in I^n (K)$,
then $\dim q \geq 2^n$.
\end{thm}
\begin{proof}
See \cite[Chapter X, Hauptsatz 5.1]{lam} or \cite[Chapter 4, Theorem 5.6]{scharlau}.
\end{proof}

Let $L/K$, $L=K(\sqrt{a})$, be a quadratic field extension endowed with its non trivial automorphism $-$
and $D=(a,b)_K$ be a quaternion algebra endowed with its canonical involution $\gamma$.
We define the following usual transfer maps
\begin{eqnarray*}
\begin{array}{cc}
\pi_L :\left\{
\begin{array}{cll}
W(L,-) & \rightarrow & W(K)\\
\lbrack h \rbrack    & \mapsto     & \lbrack x \mapsto h(x,x)\rbrack
\end{array},\right.
& \pi_D :\left\{ \begin{array}{cll}
W(D,\gamma) & \rightarrow & W(K)\\
\lbrack h \rbrack    & \mapsto     & \lbrack x \mapsto h(x,x)\rbrack
\end{array}.\right.
\end{array}
\end{eqnarray*}
\begin{thm}[Jacobson]\label{jacob}
With the above notations, the maps $\pi_L$ and $\pi_D$ are injective.
\end{thm}
\begin{proof}
See \cite[Chapter 10, 1.1, 1.2, 1.7]{scharlau} or
\cite{jacobson}.
\end{proof}\noindent
Moreover, $\im (\pi_L)=\l\l a \r\r W(K)$ and for any positive integer $n$, ${\pi_L}\pr {I_1(L,-)}^n\prt = \l\l a\r\r I^n(K)$.

\section{Analogues of Harrison's criterion}\label{harris}

In this Section, we state and prove results for quadratic field extension
with their nontrivial automorphism and for quaternion division
algebras with their canonical involutions in analogy with
Harrison's criterion \ref{harriso}.

\subsection{The case of quadratic field extensions}

First, we rephrase Theorem \ref{harriso} by introducing another
equivalent condition and it is this condition we will then generalize
to the setting of hermitian forms.
\begin{lem}
Let $K_1$ and $K_2$ be two fields of characterisic different from
$2$.
 Then the following are equivalent:\\
$(1)$ $K_1$ and $K_2$ are Witt equivalent.\\
$(2)$ There is a group isomorphism $t: {K_1}^ {*}/{{K_1}^{*}}^{2}
\to {K_2}^{*}/{{K_2}^{*}}^{2}$ sending $-1$ to $-1$ and such that
the quadratic form $\l\l x,y \r\r$ is hyperbolic over $K_1$ if and
only if the quadratic form $\l\l t(x),t(y) \r\r$ is hyperbolic over
$K_2$ for all $x, \ y \in {K_1}^*$.
\end{lem}
\begin{proof}
The quadratic form $\l x,y \r$ represents $1$ over $K_1$ if and only
if the quadratic form $\l 1, -x ,-y \r$ is isotropic over $K_1$ if
and only if the Pfister form $\l \l x,y \r\r$ is hyperbolic over
$K_1$. The equivalence then follows
from the second statement of Harrison's criterion \ref{harriso}.
\end{proof}

We now come to the proof of Theorem \ref{main6}:\\\\
\textbf{Theorem} \textit{Let $K_1$ and $K_2$ be two fields of
characteristic different from $2$. Let $L_1=K_1(\sqrt{a})$
\emph{(}resp. $L_2=K_2(\sqrt{b})$\emph{)} be a quadratic field
extension of $K_1$ \emph{(}resp. $K_2$\emph{)} equipped with its non
trivial automorphism $\sigma_1$ \emph{(}resp. $\sigma_2$\emph{)}.
Then, the following are equivalent:\\
$(1)$ $W(L_1,\sigma_1)\simeq W(L_2,\sigma_2)$ as rings.\\
$(2)$ There is a group isomorphism $t: {K_1}^*/N_{L_1/K_1}({L_1}^*)
\to K_2^*/N_{L_2/K_2}({L_2}^*)$ sending $-1$ to $-1$ such that the
quadratic form $\l\l a,x,y \r\r$ is hyperbolic over $K_1$ if and
only if the quadratic form $\l \l b,t(x),t(y)\r \r$ is hyperbolic
over $K_2$ for all $x, \ y \in {K_1}^*$.}\\\\
In the proof, we will need the following two lemmas.
\begin{lem}\label{isodisc}
For $i=1, \ 2$, the signed discriminant induces a group isomorphism
$$\dpm : I_1(L_i,\sigma_i) /I_2(L_i,\sigma_i) \simeq {K_i}^*/N_{L_i/K_i}({L_i}^*).$$
\end{lem}
\begin{proof} We know that
$$\dpm  :I_1(L_i,\sigma_i) \rightarrow {K_i}^*/N_{L_i/K_i}({L_i}^*)$$
is a group homomorphism whose kernel is $I_2(L_i,\sigma_i)$. If $b \in
{K_i}^*$ then
$$\dpm (\l 1,-b\r)=(-1)^1 {\Nrd}_{M_2(L_i)/L_i}(\l 1,-b\r)=-{\Nrd}_{L_i/K_i}(-b)=b \mod N_{L_i/K_i}({L_i}^*),$$
hence $\dpm$ is onto.
\end{proof}
We consider $W(L_i,\sigma_i)$ as a ring, $I_1(L_i,\sigma_i)$ and
$I_2(L_i,\sigma_i)$ as ideals for $i=1,\ 2$. As in the case of
quadratic forms, these two ideals are related in the following way:
\begin{lem}\label{Ideux}
For $i=1,\ 2$, $\prr I_1(L_i,\sigma_i)\prtt^2=I_2(L_i,\sigma_i)$.
\end{lem}
\begin{proof} By using the signed discriminant, it is easy
to show that $\pr I_1(L_i,\sigma_i)\prt^2 \subseteq
I_2(L_i,\sigma_i)$.

Conversely, if $\phi \in I_2(L_i,\sigma_i)$ with $\phi =\l
a_1,\cdots,a_{2s}\r$, $a_j \in {K_i}^*, \ j=1,\cdots,2s$, then
$$1=\dpm (\phi)= (-1)^{\frac{2s(2s-1)}{2}}\pr \prod_{j=1}^{2s} a_j \prt\in
{K_i}^*/N_{L_i/K_i}({L_i}^*).$$ We proceed by induction on $s$.

If $s=1$ and $\phi \simeq \l a,b\r$ then $-a=b \in
{K_i}^*/N_{L_i/K_i}({L_i}^*)$. Now $\phi \simeq\l a,-a\r$ and $\phi
\in \pr I_1(L_i,\sigma_i)\prt^2$.

If $s=2$ and $\phi\simeq \l a,b,c,d\r$ then $d=abc \in
{K_i}^*/N_{L_i/K_i}({L_i}^*)$ and
$$\phi \simeq \l a\r \otimes \l 1,ab,ac,bc\r \simeq \l a\r \otimes
\l 1, ab \r \otimes \l 1,ac \r.$$ We thus have $\phi \in \pr
I_1(L_i,\sigma_i)\prt^2$.

Suppose now $s \geq 3$. One can write $\phi = \l a,b,c\r \bot
\phi'$ and
$$\phi = \underbrace{\l a,b,c,abc\r}_{\alpha}\bot \underbrace
{(\phi' \bot \l -abc\r)}_{\beta} \in W(K,\sigma_a).$$ By assumption,
$\dpm(\phi)=1$ and $\dpm(\alpha)=1$ and this implies that
$\dpm(\beta)=1$. By induction, $\beta \in \pr
I_1(L_i,\sigma_i)\prt^2$ and $\phi \in \pr I_1(L_i,\sigma_i)\prt^2$,
thus finishing the proof.
\end{proof}\noindent
\textbf{Proof of Theorem \ref{main6}:} $(1) \Rightarrow (2):$ Let
$\Phi  :W(L_1,\sigma_1) \simeq W(L_2,\sigma_2)$ be a ring
isomorphism. The rank induces a group homomorphism, $\rank :
W(L_1,\sigma_1) \to \mathbb{Z}/2\mathbb{Z}$ and this shows that
$I_1(L_1,\sigma_1)$ is a maximal ideal of index $2$ in
$W(L_1,\sigma_1)$. In fact, it can be seen that $I_1(L_1, \sigma_1)$
is the unique ideal of index $2$ in $W(L_1,\sigma_1)$. As a
consequence, $\Phi$ induces the following group isomorphism
$$\Phi|_{I_1(L_1,\sigma_1)}
:I_1(L_1,\sigma_1) \rightarrow I_1(L_2,\sigma_2).$$ Hence, by Lemma
\ref{Ideux}, $\Phi\pr I_2(L_1,\sigma_1)\prt = I_2(L_2, \sigma_2)$
and we can define the following group isomorphism by means of Lemma
\ref{isodisc}:
\begin{eqnarray*}
t :\left\{
\begin{array}{cll}
{K_1}^*/N_{L_1/K_1}({L_1}^*) & \rightarrow & {K_2}^*/N_{L_2/K_2}({L_2}^*)\\
c & \mapsto & \dpm\pr \Phi(\l 1,-c\r)\prt
\end{array}.\right.
\end{eqnarray*}
Now, it is easy to see that $t(-1)=-1$.

$\Phi$ induces also the following group isomorphism
$$u :I_1(L_1,\sigma_1)^2/I_1(L_1,\sigma_1)^3 \simeq
I_1(L_2,\sigma_2)^2/I_1(L_2,\sigma_2)^3,$$ which can be used to
obtain the following commutative diagram:
$$\xymatrix@!C{\pr{K_1}^*/N_{L_1/K_1}({L_1}^*)\prt \times \pr{K_1}^*/N_{L_1/K_1}({L_1}^*)\prt\ar
[d]^{\normalsize\mbox{$(t,t)$}}
\ar[r]^{\normalsize\mbox{${\theta}_{L_1}$}} & \pr
I_1(L_1,\sigma_1)\prt^2/\pr
I_1(L_1,\sigma_1)\prt^3\ar[d]^{\normalsize\mbox{$u$}}
\\\pr{K_2}^*/N_{L_2/K_2}({L_2}^*)\prt \times
\pr{K_2}^*/N_{L_2/K_2}({L_2}^*)\prt
\ar[r]^{\normalsize\mbox{${\theta}_{L_2}$}} & \pr
I_1(L_2,\sigma_2)\prt^2/\pr I_1(L_2,\sigma_2)\prt^3 }$$ where
$$\theta_{L_i} (x,y) = \l 1,-x \r \otimes \l 1,-y \r \mod
I_1(L_i,\sigma_i)^3$$ for all $x,\ y \in {K_i}^*$ and for $i=1,\ 2$.
If the hermitian form $\l 1,-x,-y,xy\r$ is hyperbolic over
$(L_1,\sigma_1)$, then $\l 1,-x,-y,xy\r \in I_1(L_1,\sigma_1)^3$.
Conversely, let $\l 1,-x,-y,xy\r \in I_1(L_1,\sigma_1)^3$.
Using the notations of Subsection \ref{jac}, we know that
the map $\pi_{L_1}: W(L_1,\sigma_1) \rightarrow \l\l a\r\r W(K_1)$
is an additive group isomorphism and that
$$\Psi (\l 1,-x,-y,xy\r) \in I^4(K_1).$$
By Theorem \ref{ArPf}, the quadratic form $\Psi (\l 1,-x,-y,xy\r)$ is
hyperbolic over $K_1$ and it follows that the hermitian form
$\l1,-x,-y,xy\r$ is hyperbolic over $(L_1,\sigma_1)$.

Lastly, if the quadratic form $\l\l a, x ,y\r\r= \pi_{L_1}(\l 1,-x,-y,xy\r)$ is hyperbolic over
$K_1$ then, the hermitian form $\l 1,-x,-y,xy\r$
is hyperbolic over $(L_1,\sigma_1)$ and $\l 1,-x,-y,xy\r \in
I_1(L_1,\sigma_1)^3$. By commutativity of the previous diagram,
$$0=u \pr \theta_{L_1} (x,y) \prt= \theta_{L_2} \pr t(x),t(y)\prt.$$
We conclude that the hermitian form $\l 1,-t(x),-t(y),t(xy)\r \in
I_1(L_2,\sigma_2)^3$ which implies that the quadratic form $\l\l
b,t(x),t(y)\r\r=\pi_{L_2}(\l 1,-t(x),-t(y),t(xy)\r)$ is hyperbolic over $K_2$.
The converse is similar.\\
\noindent $(2) \Rightarrow (1):$ We define a map $\Phi$ on diagonal
forms by
$$\Phi(\l a_1,\cdots,a_n \r)=\l t(a_1),\cdots,t(a_n) \r.$$
We have to show that this definition does not depend on the chosen
diagonalization. If $n=1$, this is clear. If $n=2$, suppose that $\l
u,v \r \simeq \l u',v' \r$ as hermitian forms over $(L_1,\sigma_1)$.
By taking the signed discriminant on both sides, we obtain that
$uv=u'v' \in {K_1}^*/N_{L_1/K_1}({L_1}^*)$. If we let the
one-dimensional hermitian form $\l u \r$ act on both sides, it
follows that the hermitian form $\l 1,  -uu',-uv', u'v'\r$ is hyperbolic
over $(L_1,\sigma_1)$. As a consequence, the hermitian form $\l
1, -t(u)t(u'),- t(u)t(v'), t(u')t(v')\r$ is hyperbolic over $(L_2,\sigma_2)$ and,
multiplying by the form $\l t(u) \r$, it follows that the hermitian
forms $\l t(u),t(v) \r$ and $\l t(u'), t(v') \r$ are isometric over
$(L_2,\sigma_2)$. If $n>2$, the result comes from Theorem \ref{eqch}
and from the fact that the property holds for $n=2$. As $t(-1)=-1$,
$\Phi$ preserves hyperbolicity and induces a well-defined map
between $W(L_1,\sigma_1)$ and $W(L_2,\sigma_2)$. Besides, $\Phi$ is
additive and multiplicative ($\Phi$ being multiplicative over rank
one forms which generate additively $W(L_1,\sigma_1)$) and $t^{-1}$
provides an inverse for $\Phi$ which is thus a ring
isomorphism.\hfill$\Box$\bigskip
\begin{rems}
$(1)$ The previous proof is similar to the proof of Harrison's
criterion: see \cite{harrison} or \cite[\S 2]{pscl}.\\
$(2)$ As ${{K_i}^*}^2=D_{K_i} (\l 1 \r)$ and $N_{L_i/K_i}({L_i}^*)=
D_{K_i} (\l 1,-a\r)$ for $i=1, \ 2$, Theorem \ref{main6} is a
quadratic analogue of Theorem \ref{harriso}.
\end{rems}
In Theorem \ref{main6} we can show that the condition $t(-1)=-1$ is
not a consequence of the other two conditions of Assertion $(2)$:
\begin{ex}\label{exe}
Let $K_1=\mathbb{Q}_3$ and $K_2=\mathbb{Q}_5$. Then
$${K_1}^*/{{K_1}^*}^2=\lbrace 1,-1,3,-3 \rbrace, \ {K_2}^*/{{K_2}^*}^2= \lbrace 1,2,5,10 \rbrace, \
u(K_1)=u(K_2)=4,$$ and the unique anisotropic quadratic form of
dimension $4$ over $K_1$ (resp. over $K_2$) is $\l 1,1,-3,-3 \r$
(resp. $\l 1,-2,-5,10 \r$) (see \cite[Chapter 6, Theorem 2.2]{lam}).
Let $L_1=K_1(\sqrt{3})$ and $L_2=K_2(\sqrt{2})$. It is easy to show
that $| {K_1}^*/D_{K_1} (\l1, -3\r)|=2=| {K_2}^*/D_{K_2} (\l 1,
-2\r)|$ and that we have a group isomorphism defined by
\begin{eqnarray*}
t :{K_1}^*/D_{K_1} (\l 1, -3\r) & \rightarrow & {K_2}^*/D_{K_2} (\l 1,-2\r)\\
      1 & \mapsto & 1\\
      -1 & \mapsto & 5
\end{eqnarray*}
As $u(K_1)=u(K_2)=4$, the quadratic form $\l\l 3,x,y \r\r$ (resp.
$\l\l 2,t(x),t(y)\r\r$) is hyperbolic over $K_1$ (resp. over $K_2$)
for all $x, \ y \in {K_1}^*$. Finally, $\l 1,-2\r$ clearly represents
$-1$ over $K_2$ and
$$t(-1)\neq -1=1\in {K_2}^*/D_{K_2} (\l 1,-2\r).$$
\end{ex}

\subsection{The case of quaternion division algebras}

In this Subsection, $Q_1=(a,b)_{K_1}$ (resp. $Q_2=(c,d)_{K_2}$) will
denote a quaternion division algebra over $K_1$ (resp. over $K_2$)
with its canonical involution $\gamma_1$ (resp. $\gamma_2$).

First, we motivate our choice of the module structure in Theorem
\ref{main7} and Corollary \ref{main8} by giving two examples showing
that, for quadratic forms, the group structure of the Witt ring is
not sufficient to classify fields up to Witt equivalence as in
Theorem \ref{harriso}. In the first example, the cardinality of the
Witt ring is infinite and in the second, it is finite.
\begin{exs}\label{exewitt}
$(1)$ One can find this example in \cite[\S 7]{pscl}. If
$K_1=\mathbb{Q}(\root 3 \of {2})$ and $K_2=\mathbb{Q}$, one can show
that $W(K_1)\simeq W(K_2)$ as groups. But, by \cite[\S
4, Corollary 2]{pscl}, $W(K_1)$ and $W(K_2)$ are not isomorphic as rings.\\
$(2)$ One can find this example in \cite[Example 7.2]{cordes}. The
construction is based on \cite[\S II.1]{gross-fischer} which was
obtained in 1965 by Gross and Fischer. We choose $K_1=\mathbb{Q}_2
(\sqrt{d})$ where $d \in \mathbb{Q}_2^* \backslash \pm
{\mathbb{Q}_2^*}^2$. Then $K_1$ is a local field and we have
$$ |{K_1}^*/{{K_1}^*}^2|=16, \ s(K_1)=2, \ u(K_1)=4$$
(see \cite[Chapter VI, Corollary 2.24]{lam}, \cite[Chapter XI,
Examples 2.4(7), 6.2(4)]{lam}). Let $F$ be a field satisfying
$$ |F^*/{F^*}^2|=4, \ s(F)=2, \ u(F)=2$$
(there is such a field by Cordes' results in \cite{cordes}). Then
$K_2:=F((X))$ is such that
$$ |{K_2}^*/{{K_2}^*}^2|=8, \ s(K_2)=2, \ u(K_2)=4.$$
By \cite[Theorem 4.5]{cordes},
$$W(K_1) \simeq W(K_2) \simeq C_4 \times C_4\times C_2\times C_2$$
as groups. But $W(K_1)$ and $W(K_2)$ are not isomorphic as rings by
Harrison's criterion \ref{harriso} as we have
$|{K_1}^*/{{K_1}^*}^2|=16\neq 8=|{K_2}^*/{{K_2}^*}^2|$.
\end{exs}

In order to simplify the statement of Theorem \ref{main7}, we
define:
\begin{defi}\label{equ}
Two fields $K_1$ and $K_2$ whose characteristic is different from
$2$ are said to be $(Q_1,Q_2)$\emph{-equivalent} if there is a group
morphism $t: {K_1}^*/{{K_1}^*}^2 \rightarrow {K_2}^*/{{K_2}^*}^2$
sending $-1$ to $-1$ such that, if the quadratic form $\l\l x,y \r
\r$ is hyperbolic over $K_1$, then the quadratic form $\l\l
t(x),t(y)\r\r$ is hyperbolic over $K_2$ for all $x, \ y \in {K_1}^*$
and which induces a group isomorphism $\tilde{t}:
 {K_1}^*/D_{K_1} \pr\l\l a,b\r\r\prt \simeq {K_2}^*/D_{K_2} \pr\l\l c,d\r\r\prt$.
The pair $(t,\tilde{t})$ is called a \emph{$(Q_1,Q_2)$-equivalence}.
\end{defi}

\begin{rem}
If $A$ is a central simple algebra over a field $K$, let
$$\SH(A)= K^*/{\Nrd}_{A/K}(A^*).$$
In the literature, the group $\SH(A)$ is called the reduced norm
residue group of the algebra $A$. In particular, if $Q$ is a
quaternion division algebra with norm form $N_Q$,
$\SH(Q)={K}^*/D_{K} \pr N_Q\prt$ which allows us to translate
Definition \ref{equ} in terms of these groups.
\end{rem}

\begin{thm}\label{main7}
The following are equivalent:\\
$(1)$ There exist a ring homomorphism $\Phi : W(K_1) \to W(K_2)$
sending one-dimensional forms to one-dimensional forms and a group
isomorphism $\Psi :W(Q_1,\gamma_1) \to W(Q_2,\gamma_2)$ such that
$\Psi(\l 1\r) =\l 1\r$ and
$$\Psi(q.h)=\Phi(q).\Psi(h),$$
for all $q \in W(K_1)$, $h\in W(Q_1,\gamma_1)$.\\
$(2)$ There is a $(Q_1,Q_2)$-equivalence $(t,\tilde{t})$ between
$K_1$ and $K_2$ such that the hermitian forms $\l u,v \r$ and $\l
u',v' \r$ are isometric over $(Q_1,\gamma_1)$ if and only if the
hermitian forms $\l \tilde{t}(u),\tilde{t}(v) \r$ and
$\l \tilde{t}(u'),\tilde{t}(v') \r$ are isometric over $(Q_2,\gamma_2)$
for all $u, \ v, \ u', \ v' \in {K_1}^*$.\\
$(3)$ There is a $(Q_1,Q_2)$-equivalence $(t,\tilde{t})$ between
$K_1$ and $K_2$ such that the quadratic form $\l\l a,b,u,v\r\r$ is
hyperbolic over $K_1$ if and only if the quadratic form $\l\l
c,d,\tilde{t}(u),\tilde{t}(v)\r\r$ is hyperbolic over $K_2$ for all
$u, \ v \in {K_1}^*$.
\end{thm}\noindent
First, we need to prove the following lemma:
\begin{lem}\label{proo}
Let $Q$ be a quaternion division algebra over a field $K$ with norm
form $N_Q$. Let $u, \ v, \ u', \ v' \in K^*$. Suppose that the
quadratic form $N_Q \otimes \l u,v,-u',-v' \r$ is hyperbolic over
$K$. Then $uvu'v' \in D_K \prr N_Q \prtt$.
\end{lem}
\begin{proof}
As the quadratic forms $q=\l u,v,-u',-v'\r$ and $q'=\l 1,-uu'vv'\r$
have the same signed discriminant,  $q\bot(-q')$ belongs to $I^2
(K)$. Thus,
$$N_Q \otimes \l u,v,-u',-v'\r \equiv N_Q \otimes \l 1,-uu'vv'\r \mod \ I^4(K).$$
By assumption and by Theorem \ref{ArPf}, the quadratic
form $N_Q \otimes \l 1,-uu'vv' \r$ is hyperbolic over $K$ and it
follows that $uvu'v' \in D_K \pr N_Q \prt$.
\end{proof}\noindent
\textbf{Proof of Theorem \ref{main7}:}  $(3) \Rightarrow (2):$ Let
$u, \ v, \ u', \ v' \in {K_1}^*$ be such that
\begin{equation}\label{oo}
\l u,v \r \simeq \l u',v'\r
\end{equation}
as hermitian forms over $(Q_1,\gamma_1)$. We now use the notations
of Subsection \ref{jac}. By Theorem \ref{jacob},
$\pi_{Q_1}:W(Q_1,\gamma_1) \to W(K_1)$ is injective. The fact (\ref{oo}) is thus equivalent to the fact that
$$\l\l a,b \r \r \otimes \l u,v \r \simeq \l \l a,b \r \r \otimes
\l u',v' \r$$ as quadratic forms, which in turn is equivalent to the
hyperbolicity of the quadratic form $\l\l a,b\r\r \otimes \l
u,v,-u',-v' \r$ over $K_1$. By Lemma \ref{proo}, $uvu'v' \in D_{K_1}
\pr\l\l a,b\r\r\prt$ and $t(uvu'v') \in D_{K_2} \pr\l\l c,d
\r\r\prt$). Now we let the quadratic form $\l u \r$ act on (\ref{oo}) so we have $\l 1,
uv \r \simeq \l uu',uv'\r$ as hermitian forms over $(Q_1,\gamma_1)$
and, since $uvu'v' \in D_{K_1} \pr\l\l a,b\r\r\prt$, it follows that
$\l 1,u'v'\r \simeq \l vv',vu'\r$ as hermitian forms over
$(Q_1,\gamma_1)$. This is equivalent to the hyperbolicity of the
quadratic form $\l\l a,b, vv',vu'\r\r$ over $K_1$ and by Assertion
$(3)$, the quadratic form $\l\l c,d,
\tilde{t}(vv'),\tilde{t}(vu')\r\r$ is hyperbolic over $K_2$. Using
the fact that $t(uvu'v') \in D_{K_2} \pr\l\l c,d \r\r\prt$), we
deduce that the hermitian forms $\l \tilde{t}(u),\tilde{t}(v)\r$ and
$\l \tilde{t}(u'),\tilde{t}(v')\r$ are isometric over
$(Q_2,\gamma_2)$. The converse is similar.

\noindent$(2) \Rightarrow (1):$ let $(t,\tilde{t})$ be a
$(Q_1,Q_2)$-equivalence between $K_1$ and $K_2$ satisfying the
conditions of Assertion $(2)$. Mimicking the first part of the proof
of Theorem \ref{main6}, one can define a group homomorphism $\Phi :
W(K_1) \to W(K_2)$ sending a one-dimensional form to a
one-dimensional form. We define $\Psi$ in the following way
\begin{eqnarray*}
\Psi  :\left\{
\begin{array}{cll}
W(Q_1,\gamma_1) & \to & W(Q_2,\gamma_2)\\
\l a_1,\cdots, a_n\r   & \mapsto & \l
\tilde{t}(a_1),\cdots,\tilde{t}(a_n)\r
\end{array}.\right.
\end{eqnarray*}
As in the proof of Theorem \ref{main6}, by using Proposition
\ref{eqch}, we can show that $\Psi$ is a well-defined map which induces
a group homomorphism, and that the inverse of $\tilde{t}$ induces an
inverse for $\Psi$. Finally, the compatibility relation between
$\Phi$ and $\Psi$ is easily proved.

\noindent$(1) \Rightarrow (3):$ Let us suppose the existence of
$\Phi$ and $\Psi$ as in Assertion $(1)$. As $\Phi \pr I(K_1)\prt
\subset I(K_2)$, $\Phi$ induces the following group homomorphism
\begin{eqnarray*}
t :\left\{
\begin{array}{cll}
{K_1}^*/{{K_1}^*}^2 & \rightarrow & {K_2}^*/{{K_2}^*}^2\\
a & \mapsto & \dpm\pr \Phi(\l 1,-a\r)\prt
\end{array}\right.
\end{eqnarray*}
and $t$ satisfies the other properties stated in Definition
\ref{equ} by the implication $(1) \Rightarrow (2)$ in Harrison's criterion
\ref{harriso}. We are going to show that
\begin{equation}\label{ega}
D_{K_1} \pr\l\l a,b\r\r\prt/{{K_1}^*}^2 = t^{-1}\pr D_{K_2}\pr\l\l
c,d\r\r\prt/{{K_2}^*}^2\prt .
\end{equation}

Let $\overline{u} \in D_{K_1} \pr\l\l a,b \r\r\prt/{{K_1}^*}^2$.
Then $\Psi(\l u\r)=\Psi(\l 1\r)=\l 1 \r$ on the one hand, and
$\Psi(\l u\r)=\Phi(\l u \r ).\l 1\r$ on the other hand (note that
$\Phi(\l u \r)$ is a quadratic form over $K_2$ whereas $\Psi (\l u
\r)$ is a hermitian form over $(Q_2,\gamma_2)$). Letting $\Phi(\l u
\r ) =\l x\r$, we easily see that $t(\overline{u})=x$ and that $x
\in D_{K_2} \pr\l\l c,d\r\r\prt$. This implies that $t(\overline{u})
\in D_{K_2} \pr\l\l c,d\r\r\prt/{{K_2}^*}^2$ and finally $D_{K_1}
\pr\l\l a,b \r\r\prt/{{K_1}^*}^2 \subset t^{-1}\pr D_{K_2} \pr \l\l
c,d\r\r\prt/{{K_2}^*}^2\prt$.

Let $\overline{v} \in D_{K_2} \pr\l\l c,d \r\r\prt/{{K_2}^*}^2$ be
such that $t(\overline{y})=\overline{v}$ for a $y\in {K_1}^*$. As
$\Phi(\l y \r) =\l v\r$,
$$\Psi ( \l y \r) = \Phi(\l y \r).\l 1\r=\l v\r.\l 1 \r= \l 1
\r.$$ By injectivity of $\Psi$, it follows that $\l y \r \simeq \l 1
\r$ and (\ref{ega}) holds.

Hence, $t$ induces after factorization a unique injective group
homomorphism
\begin{eqnarray*}
\tilde{t} :\left\{
\begin{array}{cll}
{K_1}^*/D_{K_1} \pr\l\l a,b\r\r\prt & \rightarrow & {K_2}^*/D_{K_2} \pr\l\l c,d \r\r\prt\\
x & \mapsto & t(x)
\end{array}\right.
\end{eqnarray*}
Denote by $s_1 : {K_1}^*/{{K_1}^*}^2 \rightarrow {K_1}^*/ D_{K_1}
\pr\l\l a,b\r\r\prt$ and $s_2 :{K_2}^*/{{K_2}^*}^2 \rightarrow
{K_2}^*/ D_{K_2} \pr\l\l c,d\r\r\prt$ the two canonical surjections.
Then, the following diagram is commutative
$$\xymatrix@R=60pt@C=40pt{{K_1}^*/{{K_1}^*}^2 \ar[r]^{\normalsize\mbox{$t$}}
\ar[d]^{\normalsize\mbox{$s_1$}} & {K_2}^*/{{K_2}^*}^2
\ar[r]^{\normalsize\mbox{$s_2$}} & {K_2}^*/ D_{K_2} \pr\l\l c,d\r\r\prt\\
{K_1}^*/ D_{K_1} \pr\l\l
a,b\r\r\prt\ar[urr]^{\normalsize\mbox{$\tilde{t}$}}& &}$$

Now, we show that $s_2 \circ t$ is onto. Let $\overline{w} \in
{K_2}^*/D_{K_2} \pr\l\l c,d\r\r\prt$. $\Psi$ being surjective, there
is a hermitian form $h$ over $(Q_1,\gamma_1)$ such that $\Psi(h)=\l
w\r= \Phi(q).\l 1 \r$ where $h=q.\l 1 \r$ and $q$ is a quadratic form
over $K$. Without loss of generality, one can suppose
that $h = \l a_1,\cdots,a_n \r$ et $\Phi (q) = \l b_1,\cdots,b_n \r$
with $a_1,\cdots,a_n, \ b_1,\cdots,b_n\in {K_1}^*$ (note that $n$ is
odd). By taking the refined signed discriminant on both sides of the
previous equality, we obtain
$$\prod_{i=1}^n {b_i}^2 = w^2 \mod D_{K_2} \pr\l\l c,d\r\r\prt^2.$$
Consequently, there is a $\delta \in {Q_2}^*$ such that $\pr
\prod_{i=1}^n {b_i}^2\prt.{\Nrd}_{Q_2/K_2}(\delta)^2=w^2$ and $$w=\pm
\pr\prod_{i=1}^n {b_i}\prt.{\Nrd}_{Q_2/K_2}(\delta).$$ The element
$\pm \pr \prod_{i=1}^n a_i\prt $ is an antecedent
of $w$ via $s_2 \circ t$. Indeed,
\begin{eqnarray*}
t(\pm \prod_{i=1}^n a_i) & = & \pm \prod_{i=1}^{n} t(a_i)\\
                         & = & \pm \dpm \pr n\l 1 \r \bot -\Phi(\l a_1,\cdots,a_n \r)\prt\\
                         & = & \pm \dpm (n\l 1 \r \bot \l -b_1,\cdots,-b_n \r )\\
                         & = & \pm \prod_{i=1}^n {b_i} \in
                         {K_2}^*/{{K_2}^*}^2.
\end{eqnarray*}
So $$(s_2 \circ t)(\pm \prod_{i=1}^n a_i)= \pm \prod_{i=1}^n {b_i}
\mod D_{K_2} \pr\l\l c,d\r\r\prt= \overline{w}.$$ It follows that
$s_2 \circ t$ is surjective and that $\tilde{t}$ is a group
isomorphism.

Let $u, \ v \in {K_1}^*$ be such that the quadratic form $\l\l
a,b,u,v\r\r$ is hyperbolic over $K_1$. The hermitian form $\l
1,-u,-v,uv\r$ is also hyperbolic over $(Q_1,\gamma_1)$ by Theorem \ref{jacob}.
We obtain (in $W(Q_1,\gamma_1)$)
\begin{eqnarray*}
0 & = & \Psi(\l 1,-u,-v, uv \r)\\
  & = & \Phi(\l 1,-u, -v, uv \r).\l 1 \r\\
  & = & \pr \Phi(\l 1,-u\r )\otimes \Phi (\l 1, -v\r)\prt .\l 1 \r.
\end{eqnarray*}
By definition of $t$ and $\tilde{t}$, we then have
\begin{eqnarray*}
\Psi(\l 1,-u,-v,uv \r) & = & \l 1, -t(u) \r \otimes \l 1,-t(v) \r\\
                       & = & \l 1, -t(u), -t(v), t(u)t(v) \r\\
                       & = & \l 1, -\tilde{t}(u), -\tilde{t}(v),
                       \tilde{t}(u)\tilde{t}(v)\r.
\end{eqnarray*}
It follows that the quadratic form $\l\l c,d,
\tilde{t}(u),\tilde{t}(v)\r \r$ is hyperbolic over $K_2$. Conversely, if
the quadratic form $\l\l c,d, \tilde{t}(u),\tilde{t}(v)\r \r$ is hyperbolic over $K_2$ then
the quadratic form $\l\l a,b,u,v\r\r$ is hyperbolic over $K_1$ by Theorem \ref{jacob}
and by injectivity of $\Psi$.\hfill$\Box$\bigskip\\
In the particular case where $K_1=K_2=K$, Theorem \ref{main7} readily implies:\\\\
\textbf{Corollary} \textit{Then, the following are equivalent:\\
$(1)$ $W(Q_1,\gamma_1)\simeq W(Q_2,\gamma_2)$ as $W(K)$-modules.\\
$(2)$ There is a group isomorphism $\tilde{t} :{K}^*/ {\Nrd}_{Q_1/K}({Q_1}^*)
\simeq {K}^*/ {\Nrd}_{Q_2/K}({Q_2}^*)$
with $\tilde{t}(-1)=-1$ such that the quadratic form $\l\l
a,b,u,v\r\r$ is hyperbolic over $K$ if and only if the quadratic
form $\l\l c,d,\tilde{t}(u),\tilde{t}(v)\r\r$ is hyperbolic over
$K$ for all $u, \ v \in {K}^*$.}

\begin{rems}
$(1)$ The condition on the hyperbolicity of quadratic forms in
Theorem \ref{main7} can be replaced by a condition similar to the
\textquotedblleft Harrison-Cordes condition" (see Introduction).
Theorem \ref{main7} is thus analogous to Theorem \ref{Bmo} proved
by Baeza and Moresi. There are also some differences:\\
$-$ we do not suppose $\Phi$ to be bijective in Assertion $(1)$ ;\\
$-$ in addition, we suppose that $\Phi$ respects one-dimensional
forms and that $\Psi(\l1\r)=\l 1\r$ in Assertion $(1)$ ;\\
$(2)$ As ${{K_1}^*}^2=D_{K_1} (\l 1 \r)$ and
${\Nrd}_{Q_1/K_1}(Q_1^*)= D_{K_1} \pr\l\l a,b\r\r\prt$, Corollary
\ref{main8} is an analogue of Theorem \ref{harriso} for quaternion
division algebras (in the case where $K=L$).\\
$(3)$ In \cite{leep-marshall}, Leep and Marshall construct a
surjective map between $\Aut \pr W(K_1)\prt$ and the set of the
so-called \textquotedblleft Harrison maps" (i.e. satisfying
Assertion $(2)$ of Theorem \ref{harriso}) and describe the kernel of
this map. To prove their results, they used the fact that every
$\rho \in\Hom_{\ring}\pr W(K_1), W(K_2) \prt$ induces an element
$\overline{\rho} \in \Hom_{\ring}\pr W(K_1), W(K_2) \prt$ respecting
the dimension of quadratic forms and characterized by
$$\overline{\rho} (q) \equiv \rho(q) \mod \pr I(K_2)\prt^2$$
for all $q \in W(K_1)$. It might be interesting to check if such
properties hold for hermitian forms over a quaternion division
algebra.\\
$(4)$ In Assertion $(1)$ of Theorem \ref{main7}, the fact that
$\Psi(q.h)=\Phi(q).\Psi(h),$ for all $q \in W(K_1)$, $h\in
W(Q_1,\gamma_1)$ is equivalent to the commutativity of the following
diagram (with exact rows)
$$\xymatrix@!C{W(K_1) \ar[d]^{\normalsize\mbox{$\Phi$}} \ar[r]^{\normalsize\mbox{${\rho_1}$}}
& W(Q_1,\gamma_1) \ar[d]^{\normalsize\mbox{$\Psi$}}\ar[r] & 0\\
W(K_2) \ar[r]^{\normalsize\mbox{$\rho_2$}} & W(Q_2,\gamma_2)\ar[r] &
0}$$ where $\rho_i :W(K_i) \rightarrow W(Q_i,\gamma_i)$ is the
scalar extension map, for $i=1, \ 2$.
\end{rems}

\section{Reciprocity equivalence}\label{equiva}

In this Section we recall some basic results about global fields and
reciprocity equivalence: we refer to \cite{omeara} for more details
about quadratic forms over global fields and to \cite{pscl} for a
complete treatment of reciprocity equivalence. Finally, we define
the notion of quadratic reciprocity equivalence and prove Theorem
\ref{main9}.

\subsection{Preliminaries}

A \emph{global field} is either an algebraic number field (i.e. a
finite field extension of $\mathbb{Q}$) or an algebraic function
field (i.e. a finite field extension of a field of the form
$\mathbb{F}_q (X)$ where $\mathbb{F}_q$ is the finite field with $q$
elements for some prime power $q$ and $X$ is an indeterminate).

Let $K$ be a global field, $P$ be a nontrivial place of $K$ and
$K_P$ be a completion of $K$ at $P$. If $P$ is non archimedian then $P$
is discrete and $K_P$ is a local field whose residue field is
finite: we say that $P$ is a \emph{finite place}. If $P$ is
archimedian then there is a topological isomorphism between $K_P$
and $\mathbb{R}$ or $\mathbb{C}$: the place is called \emph{real} in
the first case, \emph{complex} in the second case. An archimedian
place is also called an \emph{infinite place}. If $K$ is an
algebraic function field, every place over $K$ is finite and if $K$
is an algebraic number field, there is a finite number of
archimedian places (more precisely $\lbrack K: \mathbb{Q}\rbrack=2r+s$
where $r$ is the number of complex places and $s$ is the number of real places).
We introduce the following notations:
\begin{eqnarray*}
\Omega_K & = & \lbrace \textrm{nontrivial places of } K\rbrace,\\
\Omega_K^f & = & \lbrace \textrm{finite places of } K\rbrace,\\
\Omega_K^{\infty} & = &\lbrace \textrm{infinite places of }
K\rbrace,\\
\Omega_K^{r} & = &\lbrace \textrm{real places of } K\rbrace=\lbrace
P\in \Omega_K^{\infty} \ | \ K_P \simeq\mathbb{R}\rbrace,\\
\Omega_K^{c} & = & \lbrace \textrm{complex places of }
K\rbrace=\lbrace P\in \Omega_K^{\infty}\ | \ K_P \simeq\mathbb{C}\rbrace.
\end{eqnarray*}
So we have the following disjoint unions: $\Omega_K= \Omega_K^f \cup
\Omega_K^{\infty}$ and $\Omega_K^{\infty}=\Omega_K^r \cup
\Omega_K^c$.

Let $K$ be a global field and suppose $P\in \Omega_K^r$. Then, there
is a topological isomorphism $\phi :K_P \simeq \mathbb{R}$. Via
$\phi$, $K_P$ is an ordered field, real closed and euclidian,
with unique ordering ${K_P}^2$ (see \cite[Chapter 3, Theorem
1.1.4]{scharlau}). We thus say that an element $a\in K^*$ is
positive (resp. negative) at $P$ if $a \in{K_P^*}^2$ (resp. $a
\in{-K_P^*}^2$), and we write $a \pos 0$ (resp. $a \nega 0$). If $a\in K^*$, we
introduce the notation
$$\Omega_{K}^a = \lbrace P \in \Omega_K^r\ | \ a \nega 0 \rbrace$$
to denote the set of real places at which $a$ is negative.

The following is immediate:
\begin{lem}\label{hypp}
Let $P \in \Omega_K^r$ and $q$ be a quadratic form over $K$. Then
$q$ is hyperbolic over $K_P$ if and only if $\sgn_P(q)=0$ where
$sgn_P$ denotes the usual signature at $P$ \emph{(}see \cite[Chapter
2, \S 4]{scharlau}\emph{)}.
\end{lem}\noindent
It is easy to calculate the signature of a $2$-fold Pfister form
$\l\l x,y\r\r$ at a real place $P$:
\begin{tabl}\label{signa}
\begin{center}
\begin{tabular}{|l|l|l|c|}   \hline
$x$ & $y$ & $xy$ & $\sgn_P \prr\l\l x,y\r\r\prtt$\\\hline
$+$ & $+$ & $+$ & $0$\\
$-$ & $+$ & $-$ & $0$\\
$+$ & $-$ & $-$ & $0$\\
$-$ & $-$ & $+$ & $4$\\
\hline
\end{tabular}
\end{center}
\end{tabl}\noindent
Finally, we get the following result which will be useful in
Subsection \ref{eqqua}:
\begin{lem}\label{refe}
Let $K$ be a global field and $P\in \Omega_K^r$. A $n$-fold Pfister
form $q=\l\l a_1,\cdots,a_n\r\r$ is anisotropic over $K_P$ if and
only if $a_i\nega 0$ for $i=1,\cdots,n$.
\end{lem}
\begin{proof}
The fact that $a_i \nega 0$ for all $i=1,\cdots,n$ is equivalent to
$\sgn_P(q)=2^n$ by mean of Table \ref{signa} and an induction on
$n$. This property is also equivalent to the fact that $q$ is
anisotropic over $K_P$ by Lemma \ref{hypp}, a Pfister form being
either anisotropic or hyperbolic.
\end{proof}

\subsection{Reciprocity equivalence}

Throughout this Subsection, $K_1$ and $K_2$ will denote global
fields of characteristic different from $2$. The notion
of reciprocity equivalence between such fields has been defined in  \cite[\S 1]{pscl}:
\begin{defi}\label{eqrec}
A \emph{reciprocity equivalence} between $K_1$ and $K_2$ is a pair
of maps $(t,T)$, where $t$ is a group isomorphism $t
:{K_1}^*/{{K_1}^*}^2\to {K_2}^*/{{K_2}^*}^2$ and $T$ is a
bijection $T :\Omega_{K_1} \to \Omega_{K_2}$ such that $(t,T)$
respects Hilbert symbols, i.e.
$$(x,y)_P=(tx,ty)_{TP}$$
for all $x, \ y \in {K_1}^*/{{K_1}^*}^2$ and for all $P \in
\Omega_{K_1}$.
\end{defi}
\begin{rem}\label{hilb}
As $(x,y)_P=1$ if and only if the $2$-fold Pfister form $\l\l x,y
\r\r$ is hyperbolic over $(K_1)_P$, one can replace the condition
concerning Hilbert symbols in Definition \ref{eqrec} by
$\l\l x,y\r\r$ being hyperbolic over $(K_1)_P$ if and only if the quadratic
form $\l\l t(x),t(y)\r\r$ is hyperbolic over $(K_2)_{T(P)}$.
\end{rem}\noindent
The main Theorem of \cite{pscl} says that
\begin{thm}\label{PSCLthm}
$K_1$ and $K_2$ are Witt equivalent
if and only if they are reciprocity equivalent.
\end{thm}\noindent
The proof of the \textquotedblleft if\textquotedblright
-part of this Theorem very much relies on Harrison's criterion \ref{harriso}. The proof of
the converse is much more difficult and is based on a description of
the $2$-torsion of the Brauer group by the set of subsets of
$\Omega_{K_1} \backslash \Omega_{K_1}^c$ of even order endowed with
symmetric difference. We refer to\cite[\S 3, 4]{pscl} for more
details.

\subsection{Quadratic Reciprocity equivalence}\label{eqqua}

The purpose of this Subsection is to give a proof of Theorem \ref{main9}.
Throughout, $K_1$ (resp. $K_2$) will denote a global field
of characteristic different from $2$, $L_1=K_1(\sqrt{a})$ (resp.
$L_2=K_2(\sqrt{b}$)) will denote a quadratic field extension of
$K_1$ (resp. $K_2$) and $\sigma_1$ (resp. $\sigma_2$) will denote
the nontrivial automorphism of $L_1$ (resp. $L_2$).
\begin{defi}\label{eqrecq}
An $(a,b)$-\emph{quadratic reciprocity equivalence} between $K_1$
and $K_2$ is a pair of maps $(t,T)$ where $t$ is a group isomorphism
$t :{K_1}^*/N_{L_1/K_1}({L_1}^*) \to {K_2}^*/N_{L_2/K_2}({L_2}^*)$
with $t(-1)=-1$ and where $T$ is a bijection $T :\Omega_{K_1}^a\to
\Omega_{K_2}^b$ such that the quadratic form $\l\l a,x,y\r\r$ is
hyperbolic over $(K_1)_P$ if and only if the quadratic form $\l\l
b,t(x),t(y)\r\r$ is hyperbolic over $(K_2)_{T(P)}$ for all $x, \ y
\in {K_1}^*/N_{L_1/K_1}({L_1}^*)$ and for all $P\in \Omega_{K_1}^a$.
\end{defi}\noindent
We now prove Theorem \ref{main9}:\\\\
\textbf{Theorem} \textit{The following are equivalent:\\
$(1)$ $W(L_1,\sigma_1)\simeq W(L_2,\sigma_2)$ as rings.\\
$(2)$ There is an $(a,b)$-quadratic reciprocity equivalence between
$K_1$ and $K_2$.}
\begin{proof} $(2)\Rightarrow (1):$
by Theorem \ref{main6}, it suffices to show that the quadratic form
$\l\l a,x,y\r\r$ is hyperbolic over $K_1$ if and only if the
quadratic form $\l\l b,t(x),t(y) \r\r$ is hyperbolic over $K_2$.
Note that, for all $P \in \Omega_{K_1} \backslash \Omega_{K_1}^a$
(resp. for all $Q \in \Omega_{K_2}\backslash \Omega_{K_2}^b$) and
for all $x, \ y\in {K_1}^*$, the quadratic form $\l\l a,x,y\r\r$
(resp. $\l\l b, t(x),t(y)\r\r$) is hyperbolic over $(K_1)_P$ (resp.
over $(K_2)_Q$). This fact is obvious if the place is complex or if
$a \pos 0$ (resp. if $b \underset{Q}{>} 0$). If $P$ (resp. $Q$) is
finite, it comes from the fact that $(K_1)_P$ (resp. $(K_2)_Q$) is a
local field with $u\pr (K_1)_P\prt=4=u\pr (K_2)_Q\prt$ (see
\cite[Chapter XI, Example 6.2(4)]{lam}). Suppose now that the
quadratic form $\l\l a,x,y\r\r$ is hyperbolic over $K_1$. Then
$\phi=\l\l b,t(x),t(y)\r\r$ is hyperbolic over $(K_2)_Q$ for all $Q
\in \Omega_{K_2}^b$ hence $\phi$ is hyperbolic over $K_2$ by the Hasse-Minkowski-Principle
(see \cite[Chapter VI, Hasse-Minkowski-Principle 3.1]{lam}).
The converse is similar.\\
\noindent $(1) \Rightarrow (2):$ by Theorem \ref{main6}, we only
have to find the map $T$ of Definition \ref{eqrecq}.

We first show that $K_1$ and $K_2$ have the same number of distinct
real places for which $a$ and $b$ are negative. First of all, $K_1$ has no real place where $a$
is negative if and only if $K_2$ has no real place where $b$ is
negative (this is a consequence of Theorem \ref{main6}, Lemma \ref{refe}
and the Hasse-Minkowski Principle applied to $\l\l a,-1,-1\r\r$ and to $\l\l b,t(-1),t(-1)\r\r
=\l\l b,-1,-1\r\r$). In this case, (for example, if $K_1$ and $K_2$ are
algebraic function fields), there is nothing to prove. Suppose now
that $\Omega_{K_1}^a=\lbrace P_1,\cdots,P_n \rbrace,$ $n \geq 1$. As
$K_1$ is a global field which has at least one real place, $K_1$ is
an SAP field (see \cite[\S 2]{lsu}). As a consequence, there exist
$x_1,\cdots,x_n \in {K_1}^*$ such that
$$ x_i \underset{P_i}{>}0, \quad x_i \underset{P_j}{<}0 \ \textrm{for all }
j\neq i,$$ for $i=1,\cdots ,n$. On the one hand, it is easy to see
that for each $i$,
$$(-1)^n x_1\cdots x_{i-1} x_{i+1} \cdots x_n \underset{P_i}{<}
0.$$ This implies, by Lemma \ref{refe}, that the quadratic form
$\l\l a, -1, (-1)^n x_1\cdots x_{i-1} x_{i+1} \cdots x_n\r\r$ is
anisotropic over $K_1$. Consequently, the quadratic form
$$\l\l b, -1, (-1)^n t(x_1)\cdots t(x_{i-1})
t(x_{i+1}) \cdots t(x_n)\r\r$$ is anisotropic over $K_2$ and there
is a place $Q_i$ (which has to be real) such that the quadratic form
$\l\l b, -1, (-1)^n t(x_1)\cdots t(x_{i-1}) t(x_{i+1}) \cdots
t(x_n)\r\r$ is anisotropic over $(K_2)_{Q_i}$ by the Hasse-Minkowski-Principle.
Applying Lemma \ref{refe}, we see that
$b\underset{Q_i}{<}0$ and
\begin{eqnarray}\label{equau}
(-1)^n t(x_1)\cdots t(x_{i-1}) t(x_{i+1}) \cdots t(x_n)
\underset{Q_i}{<} 0,
\end{eqnarray}
for all $i=1,\cdots,n$. On the other hand, one also has
$$ (-1)^n x_1\cdots x_n \underset{P_i}{<} 0$$
for all $i=1,\cdots,n$. This implies that the quadratic form
$\l\l a,-1, (-1)^n x_1\cdots x_n\r\r$ is anisotropic over $K_1$ and that
$\l\l b,-1, (-1)^n t(x_1)\cdots t(x_n)\r\r$  is anisotropic over $K_2$.
By the Hasse-Minkowski-Principle,
\begin{eqnarray}\label{equav}
(-1)^n t(x_1)\cdots  t(x_n) \underset{Q_i}{<} 0,
\end{eqnarray}
for all $i=1,\cdots,n$.
By combining (\ref{equau}) and (\ref{equav}), we get that $t(x_i)
\underset{Q_i}{>} 0$ for all $i=1,\cdots,n$. We now show that the
$n$ places $Q_i$ are distinct. If conversely, there are $i\neq j$
such that $Q=Q_i=Q_j$ then
$$t(x_i) \underset{Q}{>} 0, \quad t(x_j) \underset{Q}{>} 0$$
and the quadratic form $\l\l b,t(-x_i),t(-x_j)\r\r$ is anisotropic
over $K_2$. This implies that the quadratic form $\l\l
a,-x_i,-x_j\r\r$ is anisotropic over $K_1$. By the Hasse-Minkowski-Principle
and Lemma \ref{refe}, there is a place $P \in
\Omega_{K_1}^a$ for which
$$x_i\underset{P}{>} 0,\quad x_j \underset{P}{>} 0.$$
That is impossible, by construction of the $x_i$'s. We have thus shown that
there is an injection $T: \Omega_{K_1}^a \to \Omega_{K_2}^b$. By Symmetry, this is a bijection.

Finally, suppose that $\l\l b, t(x),t(y)\r\r$ is anisotropic over
$(K_2)_{Q_i}$. By Lemma \ref{refe}, $t(x)\underset{Q_i}{<} 0$ and
$t(y)\underset{Q_i}{<} 0$. As $t(x_i)\underset {Q_i}{>} 0$ by
definition, the quadratic forms $\l\l b,t(-x_i),t(x)\r\r$ and $\l\l
b,t(-x_i),t(y)\r\r$ are anisotropic over $(K_2)_{Q_i}$,
hence over $K_2$. As a consequence, the quadratic forms $\l\l a,-x_i,x\r\r$ and
$\l\l a,-x_i,y\r\r$ are anisotropic over $K_1$. By the Hasse-Minkowski-Principle,
there are $P,\ R \in \Omega_{K_1}^a$ such that
$$x_i \underset{P}{>} 0, \ x\underset{P}{<}0 \ \textrm{et} \
x_i \underset{R}{>} 0, \ y \underset{R}{<}0.$$ Thus $P=R=P_i$ and the
quadratic form $\l\l a,x,y\r\r$ is anisotropic over $(K_1)_{P_i}$.
The converse is similar.
\end{proof}

\begin{rems}
$(1)$ By means of Remark \ref{hilb}, one can say that Theorem \ref{PSCLthm}
and Theorem \ref{main9} are analogous. There are also
some major differences between the two results. In the proof of
Theorem \ref{PSCLthm}, the most important places are the finite
places whereas in Theorem \ref{main9}, they do not play a part, the
completion of a field at such places being local with $u$-invariant
$4$. Besides, in Theorem \ref{main9}, it seems that we have to
suppose that $t(-1)=-1$ which is not the case in Theorem
\ref{PSCLthm}. In fact, in this last case, this condition comes from
the \textquotedblleft Global Square Theorem" (see \cite[Chapter
VI, Theorem 3.7]{lam}) which has no analogue (as far as we know) in our
situation.\\
$(2)$ While proving $(1) \Rightarrow (2)$ in Theorem \ref{main9},
instead of saying that a certain field is SAP, we could have used
the Strong
Approximation Theorem for global fields: see \cite[Chapter II, Theorem 2.1.2]{omeara}.\\
$(3)$ In the case of algebraic function fields, the ring isomorphism $W(L_1,\sigma_1)\simeq
W(L_2,\sigma_2)$ is equivalent to having an isomorphism
${K_1}^*/N_{L_1/K_1}({L_1}^*)
\simeq K_2^*/N_{L_2/K_2}({L_2}^*)$ sending $-1$ to $-1$.
\end{rems}

Similarly, we can prove:
\begin{thm}
Let $K$ be a global field of characteristic different from $2$.
Let $Q_1=(a,b)_K$ (resp. $Q_2=(c,d)_K$) be a quaternion algebra over $K$
endowed with its canonical involution $\gamma_1$ (resp. $\gamma_2$). For $\alpha, \ \beta \in K^*$, denote
by $\Omega_{K}^{(\alpha ,\beta)}$ the set of real places at which $\alpha$ and $\beta$ are negative. Then, the following
are equivalent:\\
$(1)$ $W(Q_1,\gamma_1) \simeq W(Q_2,\gamma_2)$ as $W(K)$-modules.\\
$(2)$ There exists a pair of maps $(t,T)$ where $t$ is a group isomorphism
$t :{K}^*/ {\Nrd}_{Q_1/K}({Q_1}^*) \simeq {K}^*/ {\Nrd}_{Q_2/K}({Q_2}^*)$ with $t(-1)=-1$
and where $T$ is a bijection $T: \Omega_{K}^{(a ,b)} \to \Omega_{K}^{(c,d)}$
such that the quadratic form $\l\l a,b,x,y\r\r$ is hyperbolic over $K_P$ if and only
if the quadratic form $\l\l c,d, t(x), t(y)\r\r$ is hyperbolic over $K_{T(P)}$
for all $x, \ y \in {K}^*/ {\Nrd}_{Q_1/K}({Q_1}^*)$ and for all $P \in \Omega_{K}^{(a ,b)}$.
\end{thm}

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\begin{thebibliography}{1}

\bibitem{baeza-moresi} \textsc{R. Baeza}, \textsc{R. Moresi}: On
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\normalsize {\sc Division of Pure Mathematics, School of
Mathematical Sciences, University of
 Nottingham, University Park, Nottingham NG7 2RD, UK}

E-mail address:  {\tt grenier@math.univ-fcomte.fr}
\end{document}
