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{\Large{\textbf{On hermitian trace forms over hilbertian fields}}}
\\
\\
\textbf{Gr\'{e}gory Berhuy}
\\
\\
{\footnotesize{U.M.R.6623 du C.N.R.S., Labo de Maths, Bureau 401M, 16
route de Gray, \\ F-25030 Besan\c{c}on, France\\
(e-mail: berhuy@math.univ-fcomte.fr)}}
\\
\\
{\footnotesize{\textbf{Abstract: }Let $k$ be a field of characteristic
different from $2$. Let $E/k$ be a finite separable extension with
a $k$-linear involution $\sigma$. For every
$\sigma$-symmetric element $\mu\in E^*$, we define a
\textit{hermitian scaled trace form} by $x\in
E\mapsto\mathrm{Tr}_{E/k}(\mu x x^\sigma)$. If $\mu=1$, it is
called a \textit{hermitian trace form}. In the following, we
show that every even-dimensional quadratic form over a hilbertian
field, which is not isomorphic to the hyperbolic plane, 
is isomorphic to a hermitian scaled trace form. Then we give a
characterization of Witt classes of hermitian trace forms over
some hilbertian fields.}}
\begin{center}
\end{center}
\textbf{Introduction: }In this paper, the words ``quadratic
form'' are reserved to mean ``non-degenerate quadratic form''.
Let $k$ be a field of characteristic
different from $2$. If $E/k$ is a finite separable extension and
$\lambda\in k^*$, we can
define a quadratic form $E\to k,
x\mapsto\mathrm{Tr}_{E/k}(\lambda x^2)$,
denoted by $\mathrm{Tr}_{E/k}(<\lambda>)$. Such a form is called a \textit{scaled trace form.} 
If $\lambda=1$ this form is called \textit{ the trace form of $E/k$}. 
Recall that  a field $k$ is {hilbertian} if Hilbert's
irreducibility theorem holds. To simplify, $k$ is hilbertian if
for all $n,m\geq 1$ and for all irreducible polynomial $P\in
k(Y_1,\cdots,Y_m)[X_1,\cdots,X_n]$, there exist infinitely many specializations
$(y_1,\cdots,y_m)\in k^m$ such that $P(y_1,\cdots,y_m,X_1,\cdots,X_n)$
is still irreducible (for a precise statement of this
theorem, see [La] for example).
A natural problem is to know which
quadratic forms over $k$ are isomorphic, or more reasonably Witt-equivalent, to a (scaled)
trace form. No answer has been given in general, but Scharlau
and Waterhouse gave independently a characterization of scaled
trace forms over a hilbertian field (see Theorem 1 below).
The characterization of trace forms, which is much more
difficult, has been initiated by Conner and Perlis, who were
interested in the following question: which quadratic forms over
$\mathbb{Q}$ are Witt-equivalent to a trace form ? In [C-P], they
showed that such forms are precisely the positive quadratic forms
(Recall that a quadratic form over a field $k$ is called
\textit{positive} if all its signatures are non-negative).
In [S], Scharlau showed that the result is always true when $k$ is a number
field. Finally, Kr\"{u}skemper and Scharlau proved the validity
of this result for some hilbertian fields (see Theorem 3
below). Note that a characterization of isometry classes of trace
forms has been obtained by Epkenhans in [E] when $k$ is a number field.\\
Assume now that there exists a $k$-linear non-trivial involution
$\sigma$ on $E$. For each $\sigma$-symmetric element $\mu\in E^*$, we
can also define a quadratic form $E\to
k,x\mapsto\mathrm{Tr}_{E/k}(\mu x x^\sigma)$, denoted by $\mathrm{Tr}_{E/k}(<\mu>_\sigma)$.
Such a form is
called a \textit{hermitian scaled trace form}. If $\mu=1$, this
form is called \textit{the hermitian trace form of $E/k$ relative
to $\sigma$}. Note that the existence of $\sigma$ implies that
$[E:k]$ is even because the subfield $E_0$ fixed by the
involution verifies $[E:E_0]=2$. In this paper, we
give similar characterizations of hermitian trace forms and
hermitian scaled trace forms over hilbertian fields. 
A complete characterization of hermitian trace forms of number
fields can be found in [B1].
\\
\\
\textbf{A. Scaled trace forms and hermitian scaled trace forms
over hilbertian fields.}\\
\\
\textbf{1. Scaled trace forms over hilbertian fields.}\\
\\
In [S] and [W], Scharlau and Waterhouse respectively proved the
following theorem:\\
\\
\textbf{Theorem 1: }\textit{Let $k$ be a hilbertian field of
characteristic different from $2$. Then every quadratic form over
$k$ is isomorphic to a scaled trace form.} \\
\\
\\
\textbf{2. Hermitian scaled trace forms over hilbertian
fields.}\\
\\
In the following, we prove a similar theorem concerning hermitian
scaled trace forms. In fact, we show:\\
\\
\textbf{Theorem 2: }\textit{Let $k$ be a hilbertian field of
characteristic different from $2$. Then every 
even-dimensional quadratic form over $k$, which is not isomorphic
to the hyperbolic plane, is isomorphic to a hermitian scaled trace
form.}\\
\\
\\
\textbf{3. Proof of theorem 2.}\\
\\
The underlying idea of the proof is the following principle
from the general theory of hermitian forms used by Scharlau in [S] in order to prove
the theorem 1 above. An hermitian operator over an arbitrary field $k$ is a triple
$(V,b,u)$, where $(V,b)$ is a regular symmetric bilinear space and $u:V\to V$ is
a self-adjoint operator, that is, $b(ux,y)=b(x,uy)$ for all $x,y$ in $V$. Let us assume that the
characteristic polynomial $p$ of $u$ is separable and irreducible, and let $L=k[X]/(p)$.
Thus, $L$ is a field and $[L:k]=\mathrm{dim}_k(V)$. Then the law $L\times V\to V,
(\overline{R},v)\mapsto R(u)(v)$ endows $V$ with a structure of an $L$-vector space.
Moreover, $\mathrm{dim}_L(V)={\displaystyle{\frac{\mathrm{dim}_k(V)}{[L:k]}}}=1$, so we can identify $V$ and $L$.
Since $L/k$ is separable, the $k$-bilinear form $(x,y)\mapsto\mathrm{Tr}_{L/k}(xy)$ is non-degenerate,
so we get a $k$-isomorphism $L\to\mathrm{End}_k(L,k),s\mapsto (a\in L\mapsto\mathrm{Tr}_{L/k}(as))$.
Since $L\to k, a\mapsto b(ax,y)$ is $k$-linear, there exists a unique element
$B(x,y)$ of $L$ such that $\mathrm{Tr}_{L/k}(aB(x,y))=b(ax,y)$ for all $a\in L$. It is easy to see that
$B$ is symmetric and bilinear.
Writing $(V,B)=<\lambda>$, we get $\mathrm{Tr}_{L/k}(<\lambda>)\simeq (V,b)$.\\
Assume that dim$_k(V)=2n$ and that $p$ is even. Then $L=k(\alpha)$ has a subfield of
codimension $2$, namely $k(\alpha^2)$. Suppose that we have \\ $\alpha^2\not\equiv -1\pmod{k(\alpha^2)^{*2}}$,
and that $\lambda\in k(\alpha^2)$. One can easily check that $\mathrm{Tr}_{L/k}(<\lambda>)\simeq
\mathrm{Tr}_{k(\alpha^2)/k}(<2\lambda>)\perp\mathrm{Tr}_{k(\alpha^2)/k}(<2\lambda\alpha^2>)
\simeq\\ \mathrm{Tr}_{k(\sqrt{-\alpha^2})/k}(<\lambda>_\sigma)$, where $\sigma$ is the
$k$-linear involution of $L$ defined by \\ $\sigma(\alpha)=-\alpha$ and $\sigma\vert_{k(\alpha^2)}=\mathrm{Id}$.
\\
Let $q\simeq <s_1,\cdots,s_{2n}>$ be a non-degenerate quadratic form over $k$, and $D$ the
corresponding diagonal matrix. Notice that $U=BD$ is hermitian if and only if
$B$ is symmetric, that is $U^t D=DU$. In order to keep the notations of [B2], let $A=U^t$.
So we have $D^{-1}A D=A^t$. By the above arguments, it suffices to find a matrix $A$ with an even separable and
irreducible characteristic polynomial $p$ with a root $\alpha$ such that $\alpha^2\not\equiv -1\pmod{k(\alpha^2)^{*2}}$
and which verifies the above equality, and then to show that we have $\lambda\in k(\alpha^2)$.
Unfortunately, we are unable to show that $\lambda\in k(\alpha^2)$ using this point of view, so the proof of the
proposition below,
which is sufficient to prove theorem 2, uses another method to get the existence of $\lambda$, though
the underlying idea is the same.\\
\\
Before starting the proof, we recall the following lemma, proved
by O.Taussky in [T], first proof of theorem 1:\\
\\
\textbf{Lemma 1: }\textit{Let $k$ be a field of characteristic different
from $2$. Let $A\in M_n(k)$ with irreducible separable
characteristic polynomial. If $\alpha$ is an eigenvalue of $A$,
and $\mathbf{v}_\alpha=\left(\begin{array}{c}v_1\\ \vdots\\
v_n\end{array}\right)$ is a corresponding eigenvector, then 
$(v_1,\cdots,v_n)$ is a $k$-basis of $k(\alpha)$ and their exists
an eigenvector
$\mathbf{v}^\prime_\alpha=\left(\begin{array}{c}v^\prime_1\\
\vdots\\ v^\prime_n\end{array}\right)$ of $A^t$ corresponding to
$\alpha$ such that $(v^\prime_1,\cdots,v^\prime_n)$ is the dual
basis of $(v_1,\cdots,v_n)$ relative to
$\mathrm{Tr}_{k(\alpha)/k}.$}\\
\\
For the reader's convenience, we recall the sketch of the proof.
Denote by $v_j^{(i)}$ the $i^{th}$ conjugate of $v_j$.
Let $M=(v_j^{(i)})$ and let $c_{i1}$ be the cofactor of $v_1^{(i)}$. Then let
$v_i^\prime={\displaystyle{\frac{c_{i1}}{\mathrm{det}(M)}}}$. To be a little more
concrete, consider $M^t$. The columns of this matrix are the
eigenvectors for $A$, with $v_\alpha$ as the first column. Then
the entries of the corresponding $v_\alpha^\prime$ lie in the
first row of the inverse $(M^t)^{-1}$. It can be shown
that the coordinates of this vector belong to $k(\alpha)$. Using the properties of the determinant,
one can easily show that the vector defined by these $n$ elements is
an eigenvector of $A^t$ corresponding to $\alpha$. Moreover, the $j^{th}$
conjugate of $v_i^\prime$ is ${\displaystyle{\frac{c_{ij}}{\mathrm{det}(M)}}}$. Thus,
$\mathrm{Tr}_{k(\alpha)/k}(v_i v_k^\prime)=\underset{j=1}{\overset{n}{\sum}}v_i^{(j)}{\displaystyle{\frac{c_{kj}}{\mathrm{det}(M)}}}$,
and the relation $\mathrm{det}(M)I_n=(\mathrm{com}(M))^t M$
(where com$(M)$ is the matrix of cofactors associated to $M$) shows that the last sum is equal to
$\delta_{ik}$.\\
\\
In fact, O.Taussky proved this lemma for $k=\mathbb{Q}$ and $A\in
M_n(\mathbb{Z})$, but her proof is also valid under our
hypotheses. Indeed, during this part of the proof of Theorem 1,
she did not use the fact that the entries of $A$ are integers, and
she only used the separability of $\mathbb{Q}(\alpha)/\mathbb{Q}$.
\\
\\
\textbf{Proof of theorem 2: }An easy computation shows that any
hermitian scaled trace form of a quadratic extension of $k$ is
isomorphic to $<2\lambda,-2\lambda D>$, with $\lambda,D\in k^*$,
$D\not\in k^{*2}$. Comparing the discriminants shows that the
hyperbolic plane can't be realized as a hermitian scaled trace
form. Moreover, it is clear that any binary quadratic form, which
not isomorphic to the hyperbolic plane, is isomorphic to a
hermitian scaled trace form, so the case $n=1$ is completely
handled.
For the other cases, the comments given at the beginning imply that it
suffices to
prove the following proposition:
\\
\\
\textbf{Proposition 1: }\textit{Let $k$ be an hilbertian field of characteristic different from $2$.
Let $D$ be a diagonal matrix of $M_{2n}(k), n\geq 2$
with $\mathrm{det}(D)\neq 0$. Then
there exist an element $\alpha$ which is separable algebraic of
degree $2n$ over $k$, such that $\alpha^2\not\equiv -1
\pmod{k(\alpha^2)^{*2}}$, $\lambda
\in k(\alpha^2)$ and a basis of $(v_1,\cdots,v_{2n})$ of
$k(\alpha)$ such that $D=(\mathrm{Tr}_{k(\alpha)/k}(\lambda v_i
v_j))$.}\\
\\
\textbf{Proof of the proposition: }\\ $\bullet$ First, we show the
existence of a matrix $A_{2n}\in M_{2n}(k)$ with an even
irreducible and separable characteristic polynomial
$\chi_{A_{2n}}$ such that \\ $D^{-1}A_{2n}D=A_{2n}^t$. 
We construct a tridiagonal matrix as follows.\\
Let $T_4,T_6,\cdots,T_{2n}$ and $X$ be independent indeterminates
over $k$. \\ For $2\leq m\leq n$, let\\
$B_{2m}(T_4,T_6,\cdots,T_{2m})=\left(\begin{array}{ccccccccc}0&T_{2m}&
& & & & & & \\T_{2m}&0&1& & & & & & \\ &1&0&T_{2m-2}& & & & & \\ &
&T_{2m-2}&0&1& & & & \\ & & &1&0&\ddots& & & \\ & & &
&\ddots&\ddots&T_4& & \\ & & & & &T_4&0&1& \\ & & & & & &1&0&1\\ &
& & & & & &1&0\end{array}\right)$.\\
\\ If
$D=diag<s_{2n},\cdots,s_1>$, then set
$D_{2m}=diag<s_{2m},\cdots,s_1>$ for $2\leq m\leq n$. Now define
$\Delta_{2m}(T_4,\cdots,T_{2m},X)=\mathrm{det}(B_{2m}-XD^{-1}_{2m})$.
Since this tridiagonal matrix has zeros on the diagonal, its
characteristic polynomial is even (and so is $\Delta_{2m}$). Indeed, for any matrix $C\in M_{2m}(k)$ with
the previous properties, it is easy to see that $Q^{-1}CQ=-C$, where
\\ $Q=diag<1,-1,1,-1,\cdots,1,-1>$. Then we get
$\chi_{Q^{-1}CQ}(X)=\chi_C(X)=\chi_{-C}(X)
=\mathrm{det}(-C-XI_{2m})=\mathrm{det}(C+XI_{2m})=\chi_C(-X)$.  
It is shown in [B2] that $\Delta_{2n}$ is an irreducible
polynomial of $k(T_4,T_6,\cdots,T_{2n})[X]$ (In [B2], it is shown for $k=\mathbb{Q}$ but it is easy to see that this proof 
is still valid when $k$ is hilbertian). \\ Now, we prove that it is separable in $X$. It is well
known that an irreducible polynomial is separable if and only if
the derived polynomial is not zero. As a polynomial in $X$,
$\Delta_4(T_4,X)=(s_1 s_2 s_3 s_4)^{-1}X^4+$ (lower degree
terms). Since all $s_i\neq 0$ and $char(k)\neq 2$, the polynomial
${\displaystyle{\frac{\partial\Delta_4}{\partial X}}}(T_4,X)$ has nonzero
$X^3$-coefficient, so $\Delta_4$ is separable
in $X$. If $m>2$, let $P_{2m}$ be the determinant of the matrix
obtained by cancellation of the first row and the first column of
$B_{2m}-XD^{-1}_{2m}$. Thus, we have
$\Delta_{2m}=-s_1^{-1}XP_{2m}-T_{2m}^2\Delta_{2m-2}$. Since
$P_{2m}$ does not depend on $T_{2m}$, we get
${\displaystyle{\frac{\partial\Delta_{2m}}{\partial X}}}=U+
{\displaystyle{\frac{\partial\Delta_{2m-2}}{\partial
X}}}T_{2m}^2$, where $U$ is a polynomial which does not depend
on $T_{2m}$. The separability follows by induction. \\ By Hilbert's
irreducibility theorem, we can find $t_4,\cdots,t_{2m}\in k$ such
that $\Delta_{2n}(t_4,\cdots,t_{2n},X)$ is a separable irreducible
polynomial. Now $A_{2n}=DB_{2n}$ satisfies the required
conditions, because the characteristic polynomial of this matrix is a scalar
multiple of $\Delta_{2n}(t_4,\cdots,t_{2n},X)$.\\
\\
$\bullet$ Let $\alpha$ be an eigenvalue of $A_{2n}$. Since the
characteristic polynomial of this matrix has distinct roots, the
matrix $A_{2n}-\alpha I_{2n}$ has rank $2n-1$, so there exists some
non-zero cofactor. Let $v_j$ be the $j^{th}$
cofactor of $A-\alpha I_{2n}$ in this row and set 
$\mathbf{v}_\alpha=\left(\begin{array}{c}v_1\\
\vdots\\v_{2n}\end{array}\right)$. Then $\mathbf{v_\alpha}\neq
0$ and the properties of
the determinant show easily that
$\mathbf{v}_\alpha$ is an eigenvector corresponding to
$\alpha$. By the lemma, there exists an eigenvector
$\mathbf{v}_\alpha^\prime=\left(\begin{array}{c}v_1^\prime\\
\vdots\\v_{2n}^\prime\end{array}\right)$ of $A_{2n}^t$ which corresponds
to $\alpha$ such that $(v_1,\cdots,v_{2n})$ and
$(v_1^\prime,\cdots,v_{2n}^\prime)$ are dual bases with respect to
$\mathrm{Tr}_{k(\alpha)/k}$. The relation
$D^{-1}A_{2n}D=A_{2n}^t$ implies that $D^{-1}\mathbf{v}_\alpha$
is an eigenvector of $A_{2n}^t$ corresponding to $\alpha$. Since
the irreducible polynomials of $A_{2n}$ and its transpose are the
same, and since $\chi_{A_{2n}}$ is separable, the characteristic
spaces associated to $A_{2n}^t$ have dimension one. So there
exists $\lambda\in k(\alpha)$ such that $\lambda
D^{-1}\mathbf{v}_\alpha=\mathbf{v}_\alpha^\prime$. It is shown in
[B2] that $\lambda\in k(\alpha^2)$ for this particular choice of $\mathbf{v}_\alpha$.
Since we have $\lambda v_j=s_{2n-j+1}
v_j^\prime$ for all $j$, we get $\mathrm{Tr}_{k(\alpha)/k}(\lambda
v_i v_j)=\mathrm{Tr}_{k(\alpha)/k}(s_{2n_j+1} v_i
v_j^\prime)=\delta_{ij}s_{2n-j+1}$, that is
$D=(\mathrm{Tr}_{k(\alpha)/k}(\lambda v_i v_j))$.\\
\\
$\bullet$ Now we show that we can choose $\alpha$ such that
$\alpha^2\not\equiv -1\pmod{k(\alpha^2)^{*2}}$. Since
$\Delta_{2m}$ is even, we can write
$\Delta_{2m}(T_4,\cdots,T_{2m},X)=U_{2m}(T_4,\cdots,T_{2m},X^2)$.
We can show (as for $\Delta_{2m}$) by an inductive argument that the polynomial \\
$F_{2m}(T_4,\cdots,T_{2m},X)=U_{2m}(T_4,\cdots,T_{2m},-X^2)$
is irreducible in \\ $k(T_4,\cdots,T_{2m})[X]$ and
separable for all $m$. Indeed, if we replace $X^2$ by
$-X^2$ in the proof of the proposition in [Be2], we
obtain a relation between $F_{2m}$ and $F_{2m-1}$. Then we
can conclude using suitable valuations as in [Be2]. 
Then, by Hilbert's irreducibility theorem, we
can find a specialization of the $T_i$ such that the 
specialized polynomials $\Delta_{2n}(X)$ and $F_{2n}(X)$ are irreducible and separable. Let
$\alpha$ be a root of $\Delta_{2n}$. By definition,
$F_{2n}(\sqrt{-\alpha^2})=\Delta_{2n}(\alpha)=0$. Since $(-1)^n F_{2n}$
is monic and irreducible, we get $Irr(\sqrt{-\alpha^2},k)=(-1)^n F_{2n}$. So we
have $[k(\sqrt{-\alpha^2}):k]=2n$ and
$[k(\sqrt{-\alpha^2}):k(\alpha^2)]=2$, which means that
$-\alpha^2$ is not a square in $k(\alpha^2)$. This concludes the
proof of the proposition.
\\
\\
\\
\textbf{B. Witt classes of trace forms and hermitian trace forms
over hilbertian fields.}\\
\\
\textbf{1. Definitions and notation. }
\\
\\
We denote by $\sim$ the equivalence in $W(k)$ and by $\mathbb{H}$
the hyperbolic plane $<1,-1>$. 
\textit{The stability index} $st(k)$ of a field $k$ is the least integer
$s$ such that, for all $n>s$, and all $a_1,\cdots,a_n\in
k^*$, there exist $b_1,\cdots,b_s\in k^*$ such that the $n$-fold
Pfister forms $<<a_1,\cdots,a_n>>$ and
$<<b_1,\cdots,b_s,1,\cdots,1>>$ differ only by a torsion form
in $W(k)$ (cf.[K-S2]).  
We say that a quadratic form $\phi$ is
\textit{totally positive definite} if $\textrm{sign}_P
\phi=\textrm{dim}(\phi)$ for each ordering $P$ of $k$. This is
equivalent to say that there exists an isometry $\phi\simeq
<a_1,\cdots,a_n>$ where every $a_i$ is totally positive. 
Finally, a quadratic form is called \textit{algebraic}
(respectively \textit{h-algebraic}) if it is Witt-equivalent to a
trace form (respectively a hermitian trace form).\\
\\
\\
\textbf{2. Witt classes of trace forms over hilbertian fields. }
\\
\\
Kr\"{u}skemper and Scharlau proved the following result:\\
\\
\textbf{Theorem 3: }
\begin{enumerate}
\item[0.] \textit{Let $k$ be a hilbertian field. Then every $2$-dimensional
quadratic form is algebraic.}
\item[1.] \textit{Let $k$ be a hilbertian field, and let $\phi$ be a totally
positive definite quadratic form. Then $\phi$ is algebraic.}
\item[2.] \textit{Let $k$ be a non formally real hilbertian field. Then every
quadratic form is algebraic.}
\item[3.] \textit{Let $k=R(X)$, where $R$ is a real closed field. Then every
positive quadratic form is algebraic.}
\item[4.] \textit{More generally, let $k$ be a hilbertian field such that $st(k)\leq
2$. Then every positive quadratic form is algebraic.}
\end {enumerate}
\noindent
The fourth first points can be found in [K-S1], and the last one can
be obtained using theorems 2.1 and 2.3 of [K-S2].\\
\\
We could hope that every positive quadratic form over a
hilbertian field is algebraic, as it has been conjectured by Kr\"{u}skemper and Scharlau in
[K-S1]. In fact, this conjecture is false.
Indeed, it becomes false as soon as $st(k)\geq 4$ (cf.[K-S2], theorem 2.4).
\newpage
\noindent
\textbf{3. Witt classes of hermitian trace forms over hilbertian fields. }
\\
\\
In the following, we prove
\\
\\
\textbf{Theorem 4: }
\begin{enumerate}
\item[1.] \textit{Let $k$ be a hilbertian field, and let $\phi$ be an
even-dimensional quadratic form, which is totally
positive definite. Then
$\phi$ is $h$-algebraic.}
\item[2.] \textit{Let $k$ be a non formally real hilbertian field. Then every
even-dimensional quadratic form is $h$-algebraic.}
\item[3.] \textit{Let $k=R(X)$, where $R$ is a real closed field. Then every
even-dimensional positive quadratic form is $h$-algebraic.}
\item[4.] \textit{More generally, let $k$ be a hilbertian field such that $st(k)\leq 2$.
Then every even-dimensional positive quadratic form is $h$-algebraic.}\\
\end{enumerate}
\noindent
\textbf{4. Some useful results.}\\
\\
\textbf{Proposition 2 (cf.[K-S2], Theorem 2.1): }\textit{Let $k$ a
hilbertian field. Then a sum of algebraic forms is algebraic.}
\\
\\
\textbf{Proposition 3 (cf.[K-S1], Corollary 4): }\textit{Let $k$ be a field.
If every odd-dimensional positive quadratic form over $k$ is algebraic,
then every even-dimensional positive quadratic form
 $\phi\not\sim<-1,-1>$ is Witt-equivalent to a form 
$\mathrm{Tr}_{L(\sqrt{\lambda})/k}(<1>), \lambda\notin L^{*2}$.}
\\
\\
For the reader's convenience, we recall the sketch of the proof:\\
\\
By assumption, $\psi=<2>\phi\perp<1>$ is algebraic.
\\ Thus we have $\psi\sim\mathrm{Tr}_{L/k}(<1>)$. Then there exists
$\lambda\in L^*$ such that \\ $\mathrm{Tr_{L/k}}(<\lambda>)\sim <-1>$ (cf.[K-S1], Lemma
5).
$\lambda$ is not a square, otherwise an easy computation shows that
$\phi\sim <-1,-1>$. \\ Then,
$\mathrm{Tr}_{L(\sqrt{\lambda})/k}(<1>)\simeq
\mathrm{Tr}_{L/k}(<2,2\lambda>)\sim <2>\psi\perp<-2>\sim\phi$.
\\
\\
If we apply the previous proof to odd-dimensional quadratic
forms, which are totally positive definite, we get:\\
\\
\textbf{Proposition 4: }\textit{Let $k$ be a field. If every odd-dimensional
and totally positive definite
quadratic form is algebraic, then
every even-dimensional and totally positive definite quadratic form
$\phi\not\sim <-1,-1>$ is Witt-equivalent to
a form $\mathrm{Tr}_{L(\sqrt{b})/k}(<1>)$. In particular, it is
true over any hilbertian field.}
\\
\\
The last statement can be proved using proposition 2 and the fact that
every 1-dimensional positive form over a hilbertian field
is algebraic (cf.[K-S1], theorem 2).\\
\\
\\
We finish this section by a well-known lemma (cf. [K-S1], lemma 5
for example):\\
\\
\textbf{Lemma 2: }\textit{Let $L/k$ be a separable extension of
odd degree. For any $\delta\in k^*$ there exists a $\lambda\in L$ such
that $\mathrm{Tr}_{L/k}(<\lambda>)\sim <\delta>$.}
\\
\\
\textbf{5. Proof of theorem 4.}
\\
\begin{enumerate}
\item Let $\phi$ be a form which verifies the  hypotheses.\\ 
Suppose first that $\phi\sim 0$ (this case can occur if $k$ is
non formally real). Let $X,T$ be two indeterminates over $k$,
and let \\ $F(X,T)=X^4-(2X^2-1)T^2$. Clearly, $F$ is an irreducible
polynomial of $k(X)[T]$ (it suffices
to show that it has no roots in $k(X)$), so $F$ is irreducible in
$k[T][X]$. As a polynomial in $X$, $F$ is monic so $F$ is
irreducible in $k(T)[X]$. 
Since $k$ is hilbertian, there exists $t\in k$ such that $F(X,t)$
is still irreducible.
Let $\theta$ be a root of $X^4-2t^2X^2+t^2$ in an algebraic
closure of $k$.
Then $k(\theta)/k$ is of degree $4$, and $k(\theta^2)/k$ is of degree $2$.
Let $\sigma$ the $k$-linear involution of $k(\theta)$ defined by $\sigma(\theta)=-\theta$.
It is easy to see that the trace of $\theta,\theta^3$ and $\theta^5$
is equal to zero. Then, one can verify that $1,\theta,t^2-\theta^2,(2t^2-1)\theta-\theta^3$
is an orthogonal basis for $\mathrm{Tr}_{k(\theta)/k}(<1>_\sigma)$, and
that the corresponding diagonalization is \\
$<4,-4t^2,4t^2(t^2-1),(1-t^2)>$. Since the last quadratic form is
isomorphic to $2\mathbb{H}$, we get
$\phi\sim\mathrm{Tr}_{k(\theta)/k}(<1>_\sigma)$.\\
Assume now that $\phi\not\sim <-1,-1>$. Using proposition 4,
we get \\
$\phi\sim\mathrm{Tr}_{L(\sqrt{b})/k}(<1>)$. By the previous case,
we can assume that $\phi$ is
not hyperbolic. Then it is easy to show that $-b\notin L^{*2}$,
so we get \\ $\mathrm{Tr}_{L(\sqrt{-b})/k}(<1>_\sigma)\simeq
\mathrm{Tr}_{L(\sqrt{b})/k}(<1>)\sim\phi$, where $\sigma$
is the $k$-linear involution of $L(\sqrt{-b})$ defined by
$\sigma(\sqrt{-b})=-\sqrt{-b}$.
\\
Finally suppose that $\phi\sim <-1,-1>$. This implies that $k$ is
not formally real. If $-1\in k^{*2}$, then
$\phi\sim 0$ and we have finished. So we can suppose that
$-1\notin k^{*2}$. Since $k$ is non formally real, $<-2>$ is
algebraic by theorem 3. So we have $\mathrm{Tr}_{L/k}(<1>)\sim
<-2>$. Notice that $[L:k]$ is odd. By lemma 2, there exists a
$\lambda\in L$ such that $\mathrm{Tr}_{L/k}(<\lambda>)\sim <2>$.
Then $\lambda$ is not a square in $L$, otherwise we get
$<2>\sim\mathrm{Tr}_{L/k}(<\lambda>)\sim\mathrm{Tr}_{L/k}(<1>)\sim
<-2>$, that is\\ $<-2>\simeq <2>$, which implies that $-1\in k^{*2}$. Now set
$E=L(\sqrt{\lambda})$, and let $\sigma$ be the $k$-linear
involution of $E$ defined by
$\sigma(\sqrt{\lambda})=-\sqrt{\lambda}$ and
$\sigma\vert_L=\mathrm{Id}$. Then
$\mathrm{Tr}_{E/k}(<1>_\sigma)\simeq\mathrm{Tr}_{L/k}(<2,-2\lambda>)\sim
<-1,-1>$, 
and this concludes the proof.
\item It is a particular case of the previous point.
\item In the proof of [K-S1], Corollary 5,
it is shown that every odd-dimensional positive quadratic form is algebraic.
Let $\phi$ be an even-dimensional positive form. Since $k$ is an
ordered field in this case, $<-1,-1>$ is not positive,
so $\phi\not\sim <-1,-1>$. By proposition 3, we have \\
$\phi\sim\textrm{Tr}_{L(\sqrt{\lambda})/k}(<1>)$. If $-\lambda\in k^{*2}$,
$\phi$ is hyperbolic, and the argument used in the first point is
still valid, since $R(X)$ is hilbertian.
Otherwise, the isomorphism
$\textrm{Tr}_{L(\sqrt{\lambda})/k}(<1>)\simeq\textrm{Tr}_{L(\sqrt{-\lambda})/k}(
<1>_\sigma)$ gives the conclusion.
\item By the second point, we can assume that $k$ is formally
real. The fifth point of theorem 3 gives in particular that
every odd-dimensional positive quadratic form is algebraic. Now
conclude as previously.
\end{enumerate}
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\end{document}

