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%%% Author: R. Aravire, R. Baeza
%%% Title:  The behavior of quadratic and differential forms under function
%%% field extensions in characteristic two
%%% Email: rbaeza@inst-mat.utalca.cl
%%% 2000 Mathematical Subject Classifications: 11E04, 11E81, 12E05, 12F20,
%%% 15A63, 15A75.
%%% Keywords and Phrases: quadratic forms,
%%% differential forms, bilinear forms, Witt-groups, function fields,
%%% generic
%%% splitting fields of quadratic forms, degree of quadratic forms.
%%% 
%%% Abstract Text:
% Let $F$ be a field of characteristic 2. Let $\Omega^n_F$ be the
% $F$-space of absolute differential forms over $F$. There is a
% homomorphism $\wp:\Omega^n_F \rightarrow \Omega^n_F/d\Omega^{n-1}_F$
% given by
% \begin{equation*}
% {\ \wp }(x\frac{dx_{1}}{x_{1}}\wedge \cdots \wedge \frac{dx_{n}}{x_{n}}
% )=(x^{2}-x)\frac{dx_{1}}{x_{1}}\wedge \cdots \wedge \frac{dx_{n}}{x_{n}}

% \;\;\;\; \mbox{ mod }d\Omega _{F}^{n-1}
% \end{equation*}
% Let $H^{n+1}(F)=\mbox{ Coker}(\wp)$.
% We study the behavior of $H^{n+1}(F)$ under the function field
% $F(\phi)/F$, where
% $\phi=\ll b_1,\ldots,b_n\gg$ is a $n$-fold Pfister form and $F(\phi)$ is
% the function field of the quadric $\phi=0$ over $F$.
% We show that
%  \begin{equation}
% \ker(H^{n+1} (F) \rightarrow H^{n+1} (F(\phi)) = \overline{F \cdot\frac{

% db_{1}}{b_{1}} \wedge\cdots\wedge\frac{db_{n}}{b_{n}}}\nonumber
% \end{equation}
% Using Kato's isomorphism of $H^{n+1}(F)$ with the quotient
% $I^nW_q(F)/$ $I^{n+1}W_q(F)$, where $W_q(F)$ is the Witt group of
% quadratic forms over $F$ and
% $I\subset W(F)$ the maximal ideal of even dimensional bilinear forms
% over $F$, we deduce from the above result the analogue in characteristic
% 2 of Knebusch's degree conjecture, i.e.
% $I^nW_q(F)$ is the set of all classes $\overline{q}$ with $\deg(q)\geq
% n$.

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\documentclass[12pt,leqno]{article}
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\begin{document}

\title{\vspace{4cm} The behavior of quadratic and differential forms under function
field extensions in characteristic two\vspace{2.5cm}}
\author{R. Aravire \and R. Baeza}
\date{}

\maketitle
\thispagestyle{empty}
\newpage

\tableofcontents
\thispagestyle{empty}
%\setlength{\baselineskip}{16pt} 
\thispagestyle{empty} 
\newpage

\pagenumbering{roman} \setcounter{page}{1}
\addtocontents{toc}{\protect\vspace{1cm}}
\addcontentsline{toc}{section}{\protect\numberline{ }{Abstract}}

\section*{Abstract}
{\small
\hspace{1.0cm}
\begin{minipage}[t]{12cm}
Let $F$ be a field of characteristic 2. Let $\Omega^n_F$ be the $F$-space of absolute differential forms over $F$. There is a homomorphism $\wp:\Omega^n_F \rightarrow \Omega^n_F/d\Omega^{n-1}_F$ given by
\begin{equation*}
{\ \wp }(x\frac{dx_{1}}{x_{1}}\wedge \cdots \wedge \frac{dx_{n}}{x_{n}}
)=(x^{2}-x)\frac{dx_{1}}{x_{1}}\wedge \cdots \wedge \frac{dx_{n}}{x_{n}}
\;\;\;\; \mbox{ mod }d\Omega _{F}^{n-1}
\end{equation*}
Let $H^{n+1}(F)=\mbox{ Coker}(\wp)$.
We study the behavior of $H^{n+1}(F)$ under the function field $F(\phi)/F$, where
$\phi=\ll b_1,\ldots,b_n\gg$ is a $n$-fold Pfister form and $F(\phi)$ is the function field of the quadric $\phi=0$ over $F$.
We show that
 \begin{equation}
\ker(H^{n+1} (F) \rightarrow H^{n+1} (F(\phi)) = \overline{F \cdot\frac{
db_{1}}{b_{1}} \wedge\cdots\wedge\frac{db_{n}}{b_{n}}}\nonumber
\end{equation}
Using Kato's isomorphism of $H^{n+1}(F)$ with the quotient
$I^nW_q(F)/$ $I^{n+1}W_q(F)$, where $W_q(F)$ is the Witt group of quadratic forms over $F$ and
$I\subset W(F)$ the maximal ideal of even dimensional bilinear forms over $F$, we deduce from the above result the analogue in characteristic 2 of Knebusch's degree conjecture, i.e.
$I^nW_q(F)$ is the set of all classes $\overline{q}$ with $\deg(q)\geq n$.
\end{minipage}
}

\vspace{1cm}\noindent{\textbf{Acknowledgments.}} This work has been
supported by Proyecto Fondecyt 197 0214 and Programa Formas Cuadr\'aticas,
Universidad de Talca (second author) and Proyecto Fondecyt 196 0880, 198
0681 (first author).

\vspace{1cm}\noindent {\textbf{Keywords and phrases:}} quadratic forms,
differential forms, bilinear forms, Witt-groups, function fields, generic
splitting fields of quadratic forms, degree of quadratic forms.

\vspace{1cm}\noindent {\textbf{1991 Mathematics Subject Classification.}}
11E04, 11E81, 12E05, 12F20, 15A63, 15A75.\newpage 

\pagenumbering{arabic} \addcontentsline{toc}{section}{\protect\numberline{
}{Introduction}}

\setcounter{page}{1}

\section*{Introduction}

Since Knebusch's seminal papers on generic splitting of quadratic forms
appeared (see \cite{Kn 1}, \cite{Kn 2}), few work until recently has been done on his
degree conjecture, which asserts that the $n$-th power $I^{n}$ of the ideal $%
I$ of even dimensional forms in the Witt-ring $W(F)$ of symmetric non
singular bilinear forms of a field $F$ with $2\neq 0$, coincides with the
ideal of forms of degree $\geq n$. Compare \cite{Kn 1}, \cite{Ar-Kn}, the 1998-preprint \cite{OVV} and \cite{A-Ba 2}. A similar
theory of generic splitting of quadratic forms over a field of
characteristic 2 can be developed (see section 6 of this work) and the
corresponding degree conjecture can be stated. The aim of this work is to
prove this analogue of Knebusch's conjecture for fields with 2=0.

The advantage of working with fields of characteristic 2 is the fact
discovered by K. Kato (see \cite{Ka 1}), that there is a strong relationship
between quadratic forms and differential forms defined over such fields (see
section 5 of this work). Thus many problems concerning quadratic forms in
characteristic 2 can be translated into the language of differential forms,
which are sometimes easier to handle, in particular choosing a suitable
2-basis of the ground field. Let us briefly recall Kato's above mentioned
correspondence. Let $F$ be a field of characteristic 2 and let $%
\Omega_{F}^{n}$ be the space of $n$-differential forms over $F$ (see \cite
{Ca}, \cite{Groth}). Let $d:\Omega _{F}^{n-1}\rightarrow \Omega _{F}^{n}$ be
the usual differential operator which extends $d:F\rightarrow \Omega _{F}^{1}
$, $a\mapsto da$. Then there is a well defined homomorphism $\wp :\Omega
_{F}^{n}\rightarrow \Omega _{F}^{n}/d\Omega _{F}^{n-1}$ given by 
\begin{equation*}
{\ \wp }(x\frac{dx_{1}}{x_{1}}\wedge \cdots \wedge \frac{dx_{n}}{x_{n}}
)=(x^{2}-x)\frac{dx_{1}}{x_{1}}\wedge \cdots \wedge \frac{dx_{n}}{x_{n}} %
\;\;\; \mbox{ mod }d\Omega _{F}^{n-1}
\end{equation*}
(see \cite{Ka 1}, \cite{Mi} or section 1 of this paper).

Choosing a 2-basis of $F$ one can easily lift $\wp$ to a homomorphism $%
\wp:\Omega_{F}^{n}\rightarrow \Omega_{F}^{n}$ which of course depends on the
2-basis ( see section 1). Let $\nu_{F}(n)=\ker(\wp)$ and $H^{n+1}(F)= %
\mbox{Coker}(\wp)$, so that there is a exact sequence 
\begin{equation*}
0 \rightarrow \nu_{F}(n)\rightarrow \Omega^n_F \rightarrow
\Omega^n_F/d\Omega^{n-1}_F \rightarrow H^{n+1}(F) \rightarrow 0.
\end{equation*}
In \cite{Ka 1} it is shown that there exists a natural isomorphism of groups 
\begin{equation*}
H^{n+1}(F)\overset{\sim}{\rightarrow}I^{n}W_{q}(F)/I^{n+1}W_{q}(F)
\end{equation*}
given by 
\begin{equation*}
\overline{b\frac{ba_{1}}{a_{1}}\wedge\cdots\wedge\frac{da_{n}}{a_{n}}}
\mapsto \overline{\ll a_{1},\ldots,a_{n};b]]}.
\end{equation*}
Here $W_{q}(F)$ denotes the $W(F)$-module of non singular quadratic forms
over $F$ and $\ll a_{1},\ldots,a_{n};b]]$ denotes the quadratic $n$-fold
Pfister form defined by $a_{1},\ldots,a_{n}\in F^{\ast}$, $b\in F$ (see \cite
{Ka 1}, \cite{A-Ba 1} or section 5 of this work).

The generic splitting theory developed in section 6 enables us to define the
degree $\deg(q)$ of a non singular quadratic form over a field $F$ with 2=0
along the same lines as Knebusch does for fields with $2\neq 0$. We show
also that $I^nW_q(F)\subseteq \{\overline{q}\in W_q(F) | \deg(q)\geq n\}$.
In \cite{A-Ba 2} we have shown that equality holds (i.e. the degree
conjecture holds) if one has the following equality:

\begin{equation*}
\ker\left[ \overline{I^{n} W_{q} (F)} \rightarrow \overline{I^{n} W_{q}
(F(\phi))}\right] = \overline{\phi\cdot W_{q} (F)}
\end{equation*}
for any anisotropic $n$-fold bilinear Pfister form $\phi$ over $F $. Here $%
\overline{I^{n} W_{q} (F)}$ means $I^{n} W_{q} (F)/I^{n+1} W_{q} (F)$ and $%
F(\phi) $ is the function field of $\phi $ over $F$.

This last equality is according to Kato's isomorphism equivalent with 
\begin{equation}
\ker\left[H^{n+1} (F) \rightarrow H^{n+1} (F(\phi)\right] = \overline{F \cdot\frac{
db_{1}}{b_{1}} \wedge\cdots\wedge\frac{db_{n}}{b_{n}}}  \tag{*}
\end{equation}
in $H^{n+1}(F)$, where $\phi=\ll b_{1}, \ldots, b_{n} \gg$. The proof of (*)
will be finished in section 4 of this work.
The corresponding result for fields with $2\neq 0$ has been announced in \cite{OVV}.

In section 1 we review some well known definitions and results concerning
differential forms over a field with $2=0$. We introduce the Cartier
operator $C$ and the $\wp $-homomorphism and we prove some technical results
about divisibility of forms by forms in the $F$-algebra $\Omega
_{F}=\bigoplus_{n=0}^{\infty }\Omega _{F}^{n}$ (see (1.16)). In
section 2 we start to study the behavior of $\Omega ^{n}$ and $H^{n+1}$
under field extensions. The computation of $\ker (\Omega _{F}^{n}\rightarrow
\Omega _{E}^{n})$ for some field extension $E/F$ is not so difficult if one
can choose a suitable 2-basis of $E$. We prove for example $\ker (\Omega
_{F}^{m}\rightarrow \Omega _{F(\phi )}^{m})=\Omega _{F}^{m-n}\wedge
db_{1}\wedge \cdots \wedge db_{n}$ if $m\geq n$ ( and = 0 otherwise). This
is the first evidence for the equality (*). The computation of $\ker
(H^{n+1}(F)\rightarrow H^{n+1}(E))$ is much harder, even for very simple
extensions $E/F$. We use frequently Kato's fundamental lemma (see \cite{Ka 2}
), which is stated here without proof as lemma (2.15). As an important
consequence of these computations we obtain that a form $w\in \Omega
_{F}^{n} $ belongs to $\ker (H^{n+1}(F)\rightarrow H^{n+1}(F(\phi )))$ if
and only if $w$ satisfies a certain ``differential equation'' in the space $%
\Omega _{L}^{n}$ where $L/F$ is a purely transcendental extension. In fact,
let $S_{n}$ be the set of all maps $\mu :\{1,\ldots ,n\}\rightarrow \{0,1\}$
with $\mu (i)=1$ for at least one $i$. Let $L=F(X_{\mu },\mu \in S_{n})$
where $X_{\mu }$ are independent variables. Let $M=F(X_{\mu }^{2},\mu \in
S_{n})\subset L$ and set $\Omega _{F}[M]$ for the sub-space of $\Omega _{L}$
(over $M$) generated by the forms $db,b\in F$, over $M$. For example if $%
T=\sum_{\mu \in S_{n}}b^{\mu }X_{\mu }^{2}$ with $b^{\mu
}=\prod_{i=1}^{n}b_{i}^{\mu (i)}$, then $T\in M$ and $dT\in \Omega _{F}[M]$.
Then we show that $w\in \Omega _{F}^{n}$ is contained in $\ker
(H^{n+1}(F)\rightarrow H^{n+1}(F(\phi )))$ if and only if $w$ satisfies an
equation in $\Omega _{L}$ of the form 
\begin{equation*}
w=\wp (u)+dv+\lambda \wedge dT
\end{equation*}
with $u$, $v$, $\lambda $ $\in \Omega _{F}[M]$ (see (2.25)). Section 3 is of
technical nature and prepares the way for the proof of our main result in
section 4. Section 4 is the heart of this work. We start with a relation $%
w=\wp (u)+dv+\lambda \wedge dT$ where $u$, $v$, $\lambda $ $\in \Omega
_{F}[M]$ and we develop a descent procedure to finally end with a relation $%
w=\wp (u_{0})+dv_{0}+\lambda _{0}\wedge db_{1}\wedge \cdots \wedge db_{n}$
where $u_{0}$, $v_{0}$, $\lambda _{0}$ $\in \Omega _{F}$. This is exactly
the content of (*). In section 5 we explain briefly the basic relations
between quadratic and differential forms. It is of expository character and
details can be found in \cite{Ka 1}, \cite{A-Ba 2}. Finally in section 6, as
mentioned above, we extend Knebusch's generic splitting theory to fields
with $2=0$ and prove the analogue of his degree conjecture. \newpage

\section{The algebra of differential forms}

We will consider in this paper only fields of characteristic 2. Let $F $ be
such a field. Let $\Omega_{F}^{1} $ be the $F$-vector space of absolute
differential 1-forms, i.e. $\Omega_{F}^{1} $ is the $F$-vector space
generated by the symbols $da$, $a \in F$, with the relations $d(a + b) = da
+ db$, $d(ab) = adb + bda $ for $a,b \in F$. In particular $d(F^{2}) = 0 $,
where $F^{2} = \{ a^{2} | a \in F \} $ and $d: F \rightarrow\Omega_{F}^{1} $
is a $F^{2}$-derivation.

Let us denote by $\Omega_{F}^{n}$ the $n$-exterior power $\bigwedge^{n}
(\Omega_{F}^{1})$. Thus $\Omega_{F}^{n}$ is a $F$-vector space generated by
the forms $da_{1}\wedge\cdots\wedge da_{n}$. The operator $d$ can be
extended to a $F^{2}$-linear map $d:\Omega_{F}^{n}\rightarrow
\Omega_{F}^{n+1} $ by $d(ada_{1}\wedge\cdots\wedge da_{n})=da\wedge
da_{1}\wedge\cdots\wedge da_{n}$. We will write $\Omega_{F}$ or $%
\Omega_{F}^{\ast}$ for the $F$-algebra $\oplus_{n=0}^{\infty}\Omega_{F}^{n}$
. We denote by $Z_{F}$ the $F^{2}$-sub algebra $\{w\in\Omega_{F}^{\ast}|dw=0
\} $. Since $d^{2}=0,$ we obtain the ideal $B_{F}=d\Omega_{F}^{\ast}$ in $%
Z_{F}$ of exact forms. Now let us fix a 2-basis $\mathcal{B}=\{b_{i},i\in
I\} $ of $F $ over $F^{2},$ i.e. if we take an ordering of $I,$ then the
monomials $b_{i_{1}}\cdots b_{i_{r}},i_{1} <\cdots<i_{r},$ form a $F^{2}$
-basis of $F$ (see \cite{Ca} or \cite{Groth} for details about $p$-basis).
Then it is well known that the (logarithmic) differential forms $\left\{ 
\frac{db_{i_{1}}}{b_{i_{n}}}\wedge\cdots \wedge\frac{db_{i_{n}}}{b_{i_{n}}}
,i_{1}<\cdots<i_{n}\right\} $ form a $F-$basis of $\Omega_{F}^{n}$. The
following fact is obvious. \newline

\bigskip\noindent{(1.1) \textbf{Lemma.}} \emph{Let $\eta\in\Omega_{F} $ be a
form which does not contain $db $ for some $b \in\mathcal{B}$, in its
expansion with respect to the above basis. Then $\eta\wedge db = 0 $
implies $\eta= 0$.}

\bigskip Let us denote by $\Omega_{F}^{[2]}$ the $F^{2}$-sub algebra of $%
\Omega_{F}^{\ast}$ generated by the logarithmic differential $db/b,b\in 
\mathcal{B}$. $\;\Omega_{F}^{[2]}$ depends on the choice of the 2-basis $%
\mathcal{B}$. Then a well known result of Cartier (see \cite{Ca}) asserts
that as an $F^{2}$- algebra we have 
\begin{equation}
Z_{F}=B_{F}\oplus\Omega_{F}^{[2]}  \tag{1.2}
\end{equation}

Moreover this decomposition is compatible with the graduation of $\Omega
_{F}^{*}$. Any $w \in Z_{F} $ can be written uniquely as 
\begin{equation*}
w=d\eta+\sum_{i_{1}<\cdots<i_{n}}a_{i_{1}\cdots i_{n}}^{2}\frac{db_{i_{1}}}{
b_{i_{1}}}\wedge\cdots\wedge\frac{db_{i_{n}}}{b_{i_{n}}}
\end{equation*}
so that we can define a homomorphism 
\begin{equation}
C:Z_{F}\rightarrow\Omega_{F}^{\ast}  \tag{1.3}
\end{equation}
by 
\begin{equation*}
C(w) =\sum_{i_{1} < \cdots< i_{n}} a_{i_{1} \cdots i_{n}} \frac{db_{i_{1}}}{
b_{i_{1}}} \wedge\cdots\wedge\frac{db_{i_{n}}}{b_{i_{n}}}
\end{equation*}
$C$ is the well known Cartier-operator (see \cite{Ca}) and it is uniquely
determined by the following properties 
\begin{equation}
C(a^{2}\frac{db}{b})=a\frac{db}{b}  \tag{1.4}
\end{equation}
\begin{equation}
C(d\eta)=0\mbox{\hspace{0.4cm }for \hspace{0.4cm}}\eta\in\Omega_{F}^{\cdot }
\tag{1.5}
\end{equation}
\begin{equation}
C(a^{2}w)=aC(w)\mbox{ , }a\in F\mbox{ , }w\in Z_{F}  \tag{1.6}
\end{equation}
\begin{equation}
C(w\wedge\lambda)=C(w)\wedge C(\lambda)  \tag{1.7}
\end{equation}
In particular $C$ does not depend on the choice of the $2-$basis and $%
\ker(C)=B_{F}$, Im$(C)=\Omega_{F}^{\ast}$. Thus we get a ring isomorphism
(compatible with the graduation) 
\begin{equation}
C:Z_{F}/B_{F}\overset{\sim}{\rightarrow}\Omega_{F}^{\ast}  \tag{1.8}
\end{equation}

For a fixed 2-basis $\mathcal{B} = \{ b_{i}, i \in I \} $ of $F $ we define
the square operator 
\begin{equation}
s:\Omega_{F}^{\ast}\rightarrow\Omega_{F}^{\ast}  \tag{1.9}
\end{equation}
by 
\begin{equation*}
s\left( \sum_{\sigma}a_{\sigma}\frac{db_{\sigma}}{b_{\sigma}}\right)
=\sum_{\sigma}a_{\sigma}^{2}\frac{db_{\sigma}}{b_{\sigma}}
\end{equation*}
where $\sigma$ runs over tuples of indices $i_{1}<\cdots<i_{p}$ and $%
b_{\sigma}=b_{i_{1}}\cdots b_{i_{p}}$, $db_{\sigma}=db_{i_{1}}\wedge
\cdots\wedge db_{i_{p}}$. Of course $s$ depends on the choice of $\mathcal{B}
$. We will write $w^{[2]}$ instead of $s(w)$ when we have a fixed 2-basis of 
$F$. Using (1.9) we also define the following operator 
\begin{equation}
\begin{array}{ll}
{\wp}: & \Omega_{F}^{\ast}\rightarrow\Omega_{F}^{\ast} \\ 
&  \\ 
& {\wp}(w)=w^{[2]}+w
\end{array}
\tag{1.10}
\end{equation}

Since $s$ is additive, ${\wp}$ is additive too and depends on the choice of
the 2-basis $\mathcal{B}$. But any other choice of a 2-basis changes $\wp(w)$
by an exact form, i.e. we get a well defined group-homomorphism 
\begin{equation*}
{\wp} : \Omega_{F}^{\ast} \rightarrow\Omega_{F}^{\ast} / d\Omega_{F}^{\ast}
\end{equation*}
which for every $n $ defines a homomorphism 
\begin{equation}
{\wp}:\Omega_{F}^{n}\rightarrow\Omega_{F}^{n}/d\Omega_{F}^{n-1}  \tag{1.11}
\end{equation}

Using (1.8) one easily checks that 
\begin{equation*}
{\wp} = C^{-1} - id
\end{equation*}

We now derive some useful properties of the Cartier operators, which will be
used frequently in the next sections. \newline

\noindent(1.12) \textbf{Proposition.} \emph{Let $\{ b_{1} , \ldots, b_{n} \} 
$ be elements of a 2-basis $\mathcal{B}$ of $F$. Let $\lambda\in\Omega_{F}^{
\ast} $ be such that $d( \lambda\wedge db_{1} \wedge\cdots\wedge db_{n} ) =
0 $. Then there is $\delta\in\Omega_{F}^{\ast} $ with 
\begin{equation*}
C (\lambda\wedge db_{1} \wedge\cdots\wedge db_{n} ) = \delta\wedge db_{1}
\wedge\cdots\wedge db_{n}.
\end{equation*}
}

\noindent\textbf{Proof.} We apply induction with respect to $n$. Assume
first $n=1$, i.e. let $u=\lambda\wedge db$ be a closed form and $b\in 
\mathcal{B}$. From $du=0$ we conclude from (1.2) that $u=dn+m^{[2]}$ with
some forms $m,n$. Write $m=m_{0}+m_{1}\wedge\frac{db}{b}$ where $m_{0},m_{1} 
$ are forms which do not contain $db$ in the basis representation with
respect to $\mathcal{B}$. We can also write $n=n_{0}+bn_{1}+n_{2}\wedge
db+bn_{3}\wedge db$ where $n_{0},n_{1},n_{2},n_{3}$ do not contain odd powers 
of $b$ in their coefficients nor $db$ in their 2-basis expansion. Thus 
\begin{equation*}
\begin{array}{lcl}
u & = & m_{0}^{[2]} + m_{1}^{[2]} \wedge\frac{db}{b} + dn_{0} + bdn_{1} +
n_{1} \wedge db \\ 
&  &  \\ 
& + & dn_{2} \wedge db + bdn_{3} \wedge db \\ 
&  &  \\ 
& = & \lambda\wedge db
\end{array}
\end{equation*}
Comparing terms with $db $ we obtain 
\begin{equation*}
\begin{array}{l}
m_{0}^{[2]} + dn_{0} + bdn_{1} = 0 \\ 
\\ 
u = \lambda\wedge db = m_{1}^{[2]} \wedge\frac{db}{b} + n_{1} \wedge db +
dn_{2} \wedge db + bdn_{3} \wedge db
\end{array}
\end{equation*}

The second equation and the assumption $du = 0 $ imply $d(n_{1} \wedge db) =
dn_{1} \wedge db = 0$.

But since $n_{1}$ does not contain $db$ as well $b$ as odd power in its
coefficients, we see that $dn_{1}$ does not contain $db$ in its basis
expansion. Thus (1.1) implies $dn_{1}=0$. Thus $n_{1}\wedge db=d(bn_{1})$,
and since $dn_{2}\wedge db=d(n_{2}\wedge db)$, $bdn_{3}\wedge
db=d(bn_{3}\wedge db)$ are exact, we conclude 
\begin{equation*}
u = m_{1}^{[2]} \wedge\frac{db}{b} + dv
\end{equation*}
with some form $v $. Applying the Cartier-operator to $u $ we obtain 
\begin{equation*}
C(u)=m_{1}\wedge\frac{db}{b}
\end{equation*}
and $\delta=b^{-1}m_{1}$ does the job. Let us now assume the proposition for
all integers less than $n$. Let $u=\lambda\wedge db_{1}\wedge\cdots\wedge
db_{n}$ with $du=0$ and $b_{1},\ldots,b_{n}\in\mathcal{B}$. By induction 
\begin{equation*}
C(u)=\delta\wedge db_{2}\wedge\cdots\wedge db_{n}=\mu\wedge db_{1}\wedge
db_{3}\wedge\cdots\wedge db_{n}
\end{equation*}
with some forms $\delta,\mu$. We can assume that $\delta$ is free from terms
containing $db_{2},\ldots,db_{n}$, respectively $\mu$ is free from terms
containing $db_{1},db_{3},\ldots,db_{n}$, in their basis expansion. Set $%
\delta=\delta_{0}+\delta_{1}\wedge db_{1}$ where $\delta_{0}$, $\delta_{1}$
are free from terms containing $db_{1}$. Then 
\begin{equation*}
\begin{array}{lcl}
\delta_{0}\wedge db_{2}\wedge\cdots\wedge db_{n} & + & \delta_{1}\wedge
db_{1}\wedge db_{2}\wedge\cdots\wedge db_{n} \\ 
&  &  \\ 
& = & \mu\wedge db_{1}\wedge db_{3}\wedge\cdots\wedge db_{n}.
\end{array}
\end{equation*}
Since $\delta_{0}\wedge db_{2}\wedge\cdots\wedge db_{n}$ is free from terms
containing $db_{1}$ we conclude $\delta_{0}\wedge db_{2}\wedge\cdots\wedge
db_{n}=0$ and $C(u)=\delta_{1}\wedge db_{1}\wedge db_{2}\wedge\cdots\wedge
db_{n}$. This proves the claim. $\square$\newline

\noindent(1.13) \textbf{Proposition.} \emph{Let $\{b_{1},\ldots,b_{n}\}$ be
elements belonging to a 2-basis $\mathcal{B}$ of $F$. Let $v\in\Omega_{F}$
be such that 
\begin{equation*}
dv \in\Omega_{F} \wedge db_{1} \wedge\cdots\wedge db_{n}.
\end{equation*}
Then there exist forms $z_{\mu}\in Z_{F} $ and $u \in\Omega_{F} $ with 
\begin{equation*}
dv=\left( \sum_{\mu\neq\mbox{\bf 1}}b^{\mu}z_{\mu}+b_{1}\cdots
b_{n}du\right) \wedge db_{1}\wedge\cdots\wedge db_{n}
\end{equation*}
where $\mu$ runs over all functions $\{1,\ldots,n\}\rightarrow\{0,1\}$
distinct from $\mbox{\bf 1}$ given by $\mbox{\bf 1}(i)=i$ for all $1\leq
i\leq n$. Here we have set $b^{\mu}=\prod_{i=1}^{n}b_{\mu(i)}$. Moreover the
forms $z_{\mu}$ and $u$ can be chosen free from terms containing $%
db_{1},\ldots,db_{n}$ as well as coefficients containing odd powers of $b_{1},\ldots,b_{n}$ in their 2-basis expansion.} \newline

\noindent\textbf{Proof.} We show the claim by induction on $n $. Let us
first assume $n = 1 $, i.e. $dv \in\Omega_{F} \wedge db $ with $b = b_{1}
\in \mathcal{B} $. We write 
\begin{equation*}
v=v_{0}+bv_{1}+v_{2}\wedge db+bv_{3}\wedge db
\end{equation*}
where $v_{0},v_{1},v_{2},v_{3}$ are free from terms containing $db$ and also
do not contain odd powers of $b$ in the expansion of their coefficients 
with respect to the 2-basis $\mathcal{B}$. Then we have 
\begin{equation*}
\begin{array}{lcl}
dv & = & dv_{0}+bdv_{1}+v_{1}\wedge db+dv_{2}\wedge db+bdv_{3}\wedge db \\ 
&  &  \\ 
& = & dv_{0}+bdv_{1}+(v_{1}+dv_{2}+bdv_{3})\wedge db
\end{array}
\end{equation*}

By the choice of $v_{0},v_{1}$ we see that $dv_{0}$, $dv_{1}$ do not contain
terms involving $db$ in their basis expansion with respect to $\mathcal{B}$.
But by hypothesis $dv\in\Omega_{F}\wedge db$, so that we get 
\begin{equation*}
dv_{0} = 0, \mbox{\hspace{0.2cm}}dv_{1} = 0
\end{equation*}
since $dv_{0} $ and $dv_{1} $ do not have coefficients containing odd powers of $b $ 
in their 2-basis expansion. Hence 
\begin{equation*}
dv = (v_{1} + dv_{2} + bdv_{3}) \wedge db
\end{equation*}
is of the desired form.

Let us assume the assertion for any integer less than $n$. Then from $dv
\in\Omega_{F} \wedge db_{1} \wedge\cdots\wedge db_{n} $ we conclude 
\begin{equation*}
dv=\left( \sum_{v\in S_{n-1},\nu\neq\mbox{\bf1}}b^{\nu}z_{\nu }+b_{1}\cdots
b_{n-1}du\right) \wedge db_{1}\wedge\cdots\wedge db_{n-1}
\end{equation*}
where $S_{r}=\{\nu:\{1,\ldots,r\}\rightarrow\{0,1\}\}$ is the set of all
functions $\{1,\ldots,r\}\rightarrow\{0,1\}$. Here all $z_{\nu}$ are closed
and the $z_{\nu}$, $u$ are free from $b_{1},\ldots,b_{n-1}$. We write 
\begin{equation*}
\begin{array}{rcl}
z_{\nu} & = & c_{\nu,0}+b_{n}c_{\nu,1}+c_{\nu,2}\wedge db_{n}+b_{n}c_{\nu
,3}\wedge db_{n} \\ 
&  &  \\ 
u & = & u_{0}+b_{n}u_{1}+u_{2}\wedge db_{n}+b_{n}u_{3}\wedge db_{n}
\end{array}
\end{equation*}
with $c_{\nu,i}$, $u_{i}$ free from $b_{n}$ (i.e. do not contain $db_{n}$ as
well as odd powers of $b_{n}$ in the 2-basis expansion of their coefficients). Since $dz_{\nu}=0 $,
we have 
\begin{equation*}
0=dc_{\nu,0}+b_{n}dc_{\nu,1}+c_{\nu,1}\wedge db_{n}+dc_{\nu,2}\wedge
db_{n}+b_{n}dc_{\nu,2}\wedge db_{n}
\end{equation*}
and the choice of the $c_{\nu,i}$ is imply 
\begin{equation*}
\begin{array}{c}
dc_{\nu, 0} = 0,\;\;\;\; dc_{\nu, 1} = 0 \\ 
\\ 
c_{\nu, 1} + dc_{\nu, 2} + b_{n} dc_{\nu, 3} = 0.
\end{array}
\end{equation*}

This last equation implies 
\begin{equation*}
c_{\nu, 1} = d c_{\nu, 2},\;\;\;\; dc_{\nu, 3} = 0.
\end{equation*}
Hence 
\begin{align*}
dv & = & \left( \sum_{\nu\in S_{n-1},\neq\mbox{\bf1}}b^{\nu}(c_{\nu
,0}+b_{n}dc_{\nu,2})+b^{\mbox{\bf1}}(du_{0}+b_{n}du_{1})\right) \wedge
db_{1}\wedge\cdots\wedge db_{n-1}+ \\
& & +\left( \sum_{\nu\in S_{n-1},\neq\mbox{\bf1}}b^{\nu}(c_{\nu
,2}+b_{n}c_{\nu,3})+b^{\mbox{\bf1}}(u_{1}+du_{2}+b_{n}du_{3})\right) \wedge
db_{1}\wedge\cdots\wedge db_{n}
\end{align*}
The expression in the first parenthesis does not contain $db_{1}$, $\ldots$, 
$db_{n}$ and since $dv\in\langle db_{1}\wedge\cdots\wedge db_{n}\rangle$, we
conclude (see (1.1)) 
\begin{equation*}
\sum_{v\in S_{n-1},\neq\mbox{\bf1}}b^{\nu}(c_{\nu,0}+b_{n}dc_{\nu,2})+b_{1}
\cdots b_{n-1}du_{0}+b_{1}\cdots b_{n}du_{1}=0
\end{equation*}

Since by hypothesis $c_{\nu,0}$, $c_{\nu,2}$, $u_{0}$, $u_{1}$ do not
contain coefficients with odd powers of $b_{n}$ in their 2-basis expansion, we conclude 
\begin{equation*}
\sum_{\nu\in S_{n-1},\neq\mbox{\bf1}}b^{\nu}dc_{\nu,2}+b_{1}\cdots
b_{n-1}du_{1}=0
\end{equation*}

By induction all $c_{\nu,2},u_{1}$ are free from $b_{1},\ldots,b_{n-1}$.
Then we obtain $dc_{\nu,2}=0$, $du_{1}=0$. Since 
\begin{equation*}
\begin{array}{lcl}
dv & = & \left( \sum_{\nu\in S_{n-1},\neq\mbox{\bf1}}b^{\nu}
c_{\nu,2}+\sum_{\nu\in S_{n-1},\neq\mbox{\bf1}}b_{n}b^{\nu} c_{\nu,3}\right.
\\ 
&  &  \\ 
&  & + \left. b^{\mbox{\bf1}}(u_{1}+du_{2})+b^{\mbox{\bf1}} b_{n}
du_{3}\right) \wedge db_{1}\wedge\cdots\wedge db_{n}
\end{array}
\end{equation*}
is of desired form, this concludes the proof of the lemma. $\square$ \newline

\noindent(1.14) \textbf{Remark.} Under the hypothesis of the lemma above,
one can give a more precise description of the form $v$. By the lemma we
have 
\begin{equation*}
dv=\left( \sum_{\mu\in S_{n},\neq\mbox{\bf 1}}b^{\mu}z_{\mu }+b^{\mbox{\bf 1}
}du\right) \wedge db_{1}\wedge\cdots\wedge db_{n}
\end{equation*}
where $b^{\mbox{\bf 1}}=b_{1}\cdots b_{n}$, $dz_{\mu}=0$ for all $\mu \neq %
\mbox{\bf 1}$. Let $\mu$ be such a function and let $1\leq i\leq n$ with $%
\mu(i)=0$. Then 
\begin{equation*}
\begin{array}{l}
d(b_{i} b^{\mu}z_{\mu}\wedge db_{1} \wedge\cdots\wedge db_{i-1} \wedge
db_{n}) \\ 
\\ 
= b^{\mu}z_{\mu}\wedge db_{1} \wedge\cdots\wedge db_{n}+ b_{i}
db^{\mu}\wedge z_{\mu}\wedge db_{1} \wedge\cdots\wedge db_{i-1} \wedge
db_{i+1} \wedge \cdots\wedge db_{n}
\end{array}
\end{equation*}

Since $b^{\mu}$ contains only $b_{j} $ with $j \neq i $, we see that the
last term is 0 and hence 
\begin{equation*}
d(b_{i}b^{\mu}z_{\mu}\wedge db_{1}\wedge\cdots\wedge
db_{i}\wedge\cdots\wedge db_{n})=b^{\mu}z_{\mu}\wedge
db_{1}\wedge\cdots\wedge db_{n}.
\end{equation*}

Also $b^{\mbox{\bf 1}} du \wedge db_{1} \wedge\cdots\wedge db_{n} = d(b^{%
\mbox{\bf 1}} u \wedge db_{1} \wedge\cdots\wedge db_{n}) $ and we conclude 
\begin{equation*}
dv=d\left( \sum\Sb{\mu\in S_{n},\neq\mbox{\bf1}\\ \mu(i)=0}b_{i}b^{\mu}z_{\mu}
\wedge db_{1}\wedge\cdots\wedge\hat{db_{i}}\wedge\cdots\wedge db_{n}+b^{%
\mbox{\bf 1}}u\wedge db_{1}\wedge\cdots\wedge db_{n}\right)
\end{equation*}
i.e.

\begin{equation}
v=\sum\Sb{\mu\in S_{n},\neq\mbox{\bf 1}\\ \mu(i)=0}b_{i}b^{\mu}z_{\mu }\wedge
db_{1}\wedge\cdots\wedge\hat{db_{i}}\wedge\cdots\wedge db_{n}+b^{\mbox{\bf 1}
}u\wedge db_{1}\wedge\cdots\wedge db_{n}+z  \tag{1.15}
\end{equation}
with $z\in Z_{F}$ a closed form.

\bigskip The next result characterizes divisibility by pure forms and will
be useful in the next sections. We say that a form $\omega\in\Omega_{F}$
divides the form $\lambda$ if $\lambda=\eta\wedge\omega$ with some form $%
\eta\in \Omega_{F}$. Then we have

\bigskip\noindent(1.16) \textbf{Proposition.} \emph{Let $b_{1},\ldots,b_{n}$
be elements of $F$ contained in a 2-basis $\mathcal{B}$ of $F$. Let $w\in
\Omega_{F}^{m}$ be a $m$-form. If $db_{1},\ldots,db_{n}$ divide $w$, then $%
db_{1}\wedge\cdots\wedge db_{n}$ divides $w$.} \newline

\noindent\textbf{Proof.} Let $\mathcal{B}=\{b_{1},\ldots,b_{n},\ldots\}$ be
the given 2-basis. We show the claim by induction on $n$. For $n=1$ the
claim is obvious. Assume the proposition for $n-1$. Then we have 
$w=\eta\wedge db_{1}\wedge\cdots\wedge db_{n-1}$. Write $\eta=\eta_{0}+\eta_{1}
\wedge db_{n}$ with forms $\eta_{0}$, $\eta_{1}$ which do not contain $db_{n}$
in their basis expansion. Then $w=\eta_{0}\wedge db_{1}\wedge\cdots\wedge
db_{n-1}+\eta _{1}\wedge db_{1}\wedge\cdots\wedge db_{n}$ is divisible by 
$db_{n}$ by hypothesis. Since $\eta_{0}\wedge db_{1}\wedge\cdots\wedge
db_{n-1}$ does not contains $db_{n}$, we conclude $\eta_{0}\wedge
db_{1}\wedge\cdots\wedge db_{n-1}=0$ and $w=\eta_{1}\wedge
db_{1}\wedge\cdots\wedge db_{n}$. $\square$

\bigskip\noindent(1.17) \textbf{Remark.} Let us consider a
rational function field $L=F(X_{\mu}; \mu\in A))$ with $A$ finite. Let 
$\mathcal{B}=\{b_{i}|i\in I\}$ be a 2-basis of $F$. Then $\mathcal{B}
\cup\{X_{\mu}, \mu\in A \} $ is a 2-basis of $L$. Let $B\subseteq A$ be a subset and $N=F(X_{\mu}, \mu\in B)$, respectively $M=F(X_{\mu}^2, \mu\in B)\subset N$. 
Let $X$ be any variable $X_{\mu}$ with $\mu\not\in B$. We will be later interested in forms contained in $\Omega_L$ which are generated over $M$ by the differentials $db$, $b\in F$. Thus we will define $\Omega_F M=\Omega_F\otimes M\subset \Omega_L$, respectively 
$\Omega_F M[X^2]=\Omega_F\otimes M[X^2]\subset \Omega_L$. This last set is the $M[X^2]$-submodule 
$$\bigoplus_{i_1<\cdots<i_m}M[X^2]db_{i_1}\wedge\cdots\wedge db_{i_m}$$
(for some ordering $<$ in $I$) of $\Omega_L$.
For every $p(X)\in N[X]$ irreducible and monic we set for every $n\geq 0$
\begin{equation}
p^{-\infty}\Omega^n_F M[X^2]=\left\{\frac{w}{p^s} | w\in \Omega^n_FM[X^2], s\geq 1, \deg_X w \leq s \deg_X p \right\}
\tag{1.18}
\end{equation}
if $p\in M[X^2]$, and 
\begin{equation}
p^{-\infty}\Omega^n_F M[X^2]=\left\{\frac{w}{p^{2s}} | w\in \Omega^n_FM[X^2], s\geq 1, \deg_X w \leq 2s \deg_X p \right\}
\tag{1.19}
\end{equation}
if $p\not\in M[X^2]$.

Here we have set $\deg_X w=2t$ whenever we have $w=w_0+w_1 X^2+\cdots+ w_t X^{2t}$ with 
$w_0,\ldots, w_t \in \Omega^n_F$, $w_t\neq 0$ for some $w\in \Omega^n_F M[X^2]$.

Now we have

\bigskip\noindent(1.20) \textbf{Lemma.} \emph{The sum 
$$\Omega_F^n M[X^2]+\sum_p p^{-\infty}\Omega^n_F M[X^2] \subseteq \Omega^n_L$$
is direct. Here $p$ runs over all irreducible polynomials contained in $N[X]$}. \newline

\noindent\textbf{Proof.} Let us assume
$$u_0+\sum_p \frac{u_p}{p^{s_p}} =0$$
in $\Omega_L$, with $u_0$, $u_p$ $\in \Omega_F M[X^2]$, $\deg_X u_p< s_p \deg_X p$ for all $p$. Thus
$$ \prod_p p^{s_p} u_o+\sum_p\left(\prod_{q\neq p}q^{s_q}\right)u_p=0$$
holds in $\Omega_F M[X^2]$. Recall that $s_p$ is even if $p\not\in M[X^2]$, and hence $p^{s_p}$ is contained always in $M[X^2]$. 

We fix now some $p_0$ and we get in $\Omega_F M[X^2]$
$$\left(\prod_p p^{s_p}\right) u_0= 
\left(\prod_{q\neq p_0} q^{s_q}\right) u_{p_0}+
\sum_{p\neq p_0}\left(\prod_{q\neq p} q^{s_q}\right) u_p.$$
This implies that $p_0^{s_{p_0}}$ divides the term $\left(\prod_{q\neq p_0} q^{s_q}\right) u_{p_0}$, and since $\Omega_F M[X^2]$ is a free module, $p_0^{s_{p_0}}$ divides $u_{p_0}$ in $\Omega_F M[X^2]$. Since $\deg_X u_{p_0}< s_{p_0} \deg_X p_0$, it follows $u_{p_0}=0$. This proves the lemma. $\square$

\bigskip The relevant point in the above decomposition is that the operators $\wp$ and $d$ respect this decompositions, whenever $\wp$ is defined with respect to a 2-basis containing a 2-basis of $F$. Thus we have:
\begin{equation}
\begin{array}{rcl}
d\Omega_F M[X^2] & \subseteq & \Omega_F M[X^2] \\
& & \\
\wp\Omega_F M[X^2] & \subseteq & \Omega_F M[X^2] \\
& & \\
d(p^{-\infty}\Omega_F M[X^2]) & \subseteq & p^{-\infty}\Omega_F M[X^2] \\
& & \\
\wp (p^{-\infty}\Omega_F M[X^2]) & \subseteq & p^{-\infty}\Omega_F M[X^2] \\
& & 
\end{array}
\tag{1.21}
\end{equation}

If we do not specify a particular 2-basis, the relations above must be understood as follows. For any field $F$ the maps $\Omega_F \rightarrow \Omega_F$ given by $w\mapsto w^{[2]}$ depends on the choice of a 2-basis. If we choose another 2-basis of $F$ and we denote by $w^{(2)}$ the same operation with respect to this new 2-basis, we have 
$$ w^{[2]}=w^{(2)}+dv$$ 
with some $v\in \Omega_F$. Therefore expressions of the form 
$ w^{[2]}+dv $, which will often occur in the sequel, make sense if the particular choice of the form $v$ does not matter. Thus, we will not make sometimes an explicit choice of a 2-basis when dealing with such expressions. Of course the same remark applies to expressions of the form $\wp w +d v$. For example if $u\in \Omega_F$ and $f\in F^{\ast}$, then for any choice of a 2-basis, the forms $\wp(u\wedge\frac{df}{f})$ and $\wp(u)\wedge \frac{df}{f}$ differ by an exact form. 

In the concrete situation above, if we choose $u\in \Omega_F M[X^2]$ and $f\in M[X^2]$, then there is some $v\in \Omega_F M[X^2]$ with 
$$
\wp\left( u\wedge \frac{df}{f}\right)=\wp( u)\wedge \frac{df}{f}+\frac{dv}{f^2}$$




\newpage

\section{The behavior of differential forms under some field extensions}

We continue in this section the algebraic study of differential forms by
considering their behavior under field extension. Any field extension $%
F\hookrightarrow L$ induces a natural homomorphism $\Omega_{F}^{n}
\rightarrow\Omega_{L}^{n}$ for all $n\geq 0$. We will denote by $%
\Omega^{n}(L/F)$ the kernel of this homomorphism. It is clear that $%
d(\Omega^{n}(L/F))$ is contained in $\Omega^{n+1}(L/F)$. Fixing any 2-basis
of $F$ we see that the same holds true for the operator ${\wp}$ and $s(=$
square). We are particularly interested in the following field extension of $%
F$. Let $\phi=\ll
b_{1},\ldots,b_{n}\gg=\langle1,b_{1}\rangle\otimes\cdots\otimes\langle
1,b_{n}\rangle$ be an anisotropic bilinear Pfister form (see \cite{Ba 1}, 
\cite{A-Ba 1}). The fact that $\phi$ is anisotropic means that $%
\{b_{1},\ldots,b_{n}\}$ are algebraically independent over $F^{2}$ and hence
can be chosen as part of a 2-basis of $F$. This will be always assumed in
the sequel. The function field $F(\phi)$ of the quadric $\phi=0$ is
constructed as follows. Let $S_{n}$ be the set of all maps $%
\mu:\{1,\ldots,n\}\rightarrow \{0,1\}$ with $\mu(i)=1$ for at least one
index $i$ and choose a variable $X_{\mu}$ for each $\mu\in S_{n}$. Let $%
L=F(X_{\mu},\mu\in S_{n})$ the rational function field over $F$ in the
variables $X_{\mu}$ and set $T=\sum_{\mu\in S_{n}}b^{\mu}X_{\mu}^{2}$, where 
$b^{\mu}=\prod_{i=1}^{n} b_{i}^{\mu(i)}$ ($T$ is the so called pure part of $%
\phi)$. Then the field 
\begin{equation}
F(\phi)=L(\sqrt{T})  \tag{2.1}
\end{equation}
is the function field of $\phi$. Obviously $\phi\otimes F(\phi)$ is
isotropic, although it is not necessarily hyperbolic, but metabolic. We are
interested in the behavior of $\Omega^{m}$ under the field extension $%
F\hookrightarrow F(\phi)$. To this end we will compute $\Omega^{m}(L/F)$, $%
\Omega^{m} (F(\phi)/L)$ and finally $\Omega^{m}(F(\phi)/F)$. We will denote
by $K$ the field $F(\phi)$ in what follows.

\bigskip\noindent(2.2) \textbf{Lemma.} \emph{Let $F(X)/F$ be a pure
transcendental extension of $F$. Then $\Omega^{m}(F(X)/F)=0$.} \newline

\noindent\textbf{Proof.} We may assume that $F(X)/F$ has transcendence
degree one. Let $\mathcal{B}=\{b_{i},i\in I\}$ be a 2-basis of $F$. Then $%
\mathcal{B}\cup\{X\}$ is a 2-basis of $F(X)$. If 
\begin{equation*}
w = \sum c_{i_{1}, \ldots, i_{m}} \frac{db_{i_{1}}}{b_{i_{m}}} \wedge
\cdots\wedge\frac{db_{i_{m}}}{b_{i_{m}}}
\end{equation*}
is in $\Omega^{m} (F(X)/F) $, we have $w = 0 $ in $\Omega_{F(X)}^{m} $. But
since 
\begin{equation*}
\left\{ \frac{db_{i_{1}}}{b_{i_{1}}}\wedge\cdots\wedge\frac{db_{i_{m}} }{
b_{i_{m}}},i_{1}<\cdots<i_{m}\right\}.
\end{equation*}
is part of a $F(X)$-basis of $\Omega_{F(X)}^{m}$, we conclude $%
c_{i_{1}\cdots i_{m}}=0$ for all $i_{1}<\cdots<i_{m}$, i.e. $w=0$ in $%
\Omega_{L}^{m}$. $\square$\newline

\noindent(2.3) \textbf{Remark.} If $\mathcal{B}$ is a 2-basis of $F$, then $%
\mathcal{B}\cup\{X_{\mu},\mu\in S_{n}\}$ is a 2-basis of $L$.\newline

\noindent(2.4) \textbf{Lemma. } \emph{Let $k $ be any field of
characteristic 2 and $t \in k \setminus k^{2}$. Then for $m \geq 1$ 
\begin{equation*}
\Omega^{m} (k (\sqrt{t})/k) = \Omega_{k}^{m-1} \wedge dt.
\end{equation*}
}
\noindent\textbf{Proof.} Since $t = (\sqrt{t})^{2} $ we have $dt = 0 $ in $%
\Omega_{k(\sqrt{t})}$, and therefore $\Omega_{k}^{m-1}\wedge dt
\subseteq\Omega^{m} (k(\sqrt{t})/k)$. Since $t \not \in k^{2}$, we can
choose a 2-basis of $k$, say $\mathcal{B}$, such that $t \in\mathcal{B}$,
i.e. $\mathcal{B} = \{t, c_{j}, j \in J \} $. Then $\{ \sqrt{t}, c_{j}, j
\in J\} $ is a 2-basis of $k(\sqrt{t})$. Let $w \in\Omega^{m} (k(\sqrt{t}
)/k) $. Then we have 
$$
w=\sum_{\sigma}a_{\sigma}dc_{\sigma}+(\sum b_{\tau}dc_{\tau})\wedge dt
$$
where $\sigma$ runs over all maps $\{1,\ldots,m\}\rightarrow J$ and $\tau$
over all maps $\{1,\ldots,m-1\}\rightarrow J$ which are monotone (i.e. $i<j$
implies $\sigma(i)<\sigma(j)$ in some ordering $<$ of $J)$. Moreover $%
dc_{\sigma}$ means $dc_{\sigma(1)}\wedge\cdots\wedge dc_{\sigma(m)}$, etc.
and $a_{\sigma},b_{\tau}\in k$. Then in $\Omega_{k(\sqrt{t})}^{m}$ we get $%
\sum_{\sigma}a_{\sigma}dc_{\sigma}=0$. Since all $c_{i}$, $i\in J$ are part
of a 2-basis of $k(\sqrt{t})$ we conclude $a_{\sigma}=0$ for all $\sigma$,
i.e. $w=(\sum b_{\tau}dc_{\tau})\wedge dt$. This proves the claim. $\square$ 
\newline

In particular we obtain \newline

\noindent(2.5) \textbf{Corollary. } \emph{$\Omega^{m}(K/L)=\Omega_{L}^{m-1}
\wedge dT$.} \newline

\noindent(2.6) \textbf{Remark. } In the case of a quadratic separable
extension of $k$, say $E=k+kz$ with $z^{2}+z=a$, we may assume
that $a$ is a square in $k$ and hence $dz=0$. In this case $z$ is a square in $%
E$ and a 2-basis of $k$ remains a 2-basis of $E$. Therefore the argument in
the proof of (2.4) shows $\Omega^{m}(E/k)=\{0\}$. \newline

\noindent(2.7) \textbf{Lemma.} \emph{For all $m \geq n $ 
\begin{equation*}
\Omega^{m} (F(\phi)/F)= \Omega_{F}^{m-n} \wedge db_{1} \wedge\cdots\wedge
db_{n}
\end{equation*}
Otherwise \mbox{ } $\Omega^{m}(F(\phi)/F)=0$.} \newline

\noindent\textbf{Proof.} We choose a 2-basis of $F $ containing $\{ b_{1},
\ldots, b_{n} \} $. Let us take $w$ in $\Omega^{m} (F(\phi)/F), $ i.e. $w
\in\Omega_{F}^{m} $ with $w = 0 $ in $\Omega_{F(\phi)}^{m}$. From (2.4) we
infer that $w = u \wedge dT $ with some $u \in\Omega_{L}^{m-1}$. Now $dT =
k_{1} db_{1}+ \cdots+ k_{n} db_{m} $ with some polynomials $k_{1},
\ldots,k_{n} $ (see (2.9) below). Thus replacing $db_{1} $ in $u $ we see
that one can assume that $u $ does not contain $db_{1} $ in its basis
expansion with respect to the 2-basis $\mathcal{B } \cup\{ X_{\mu}, \mu\in
S_{n} \}$. Let us write in $\Omega_{F}^{m} $ 
\begin{equation*}
w = w_{0} + w_{1} \wedge db_{1}
\end{equation*}
with forms $w_{0}, w_{1} $ not containing $db_{1} $. Then in $\Omega_{L}^{m} 
$ 
\begin{equation*}
\begin{array}{rcl}
w_{0} + w_{1} \wedge db_{1} & = & u \wedge k_{1} db_{1} + u \wedge(k_{2}
db_{2} + \cdots+ k_{n} db_{n}) \\ 
&  &  \\ 
(w_{1} + k_{1} u) \wedge db_{1} & = & w_{0} + u \wedge(k_{2} db_{2} +
\cdots+ k_{n} db_{n})
\end{array}
\end{equation*}

Since $b_{1} $ is part of the 2-basis of $L $ and the right hand side of
this equation does not contain $db_{1}$, we obtain $(w_{1} + k_{1} u) \wedge
db_{1} = 0 $ in $\Omega_{L}^{m} $. But $w_{1} + k_{1} u $ also does not
contain $db_{1} $, so that $w_{1} + k_{1} u = 0 $. Thus $u = k_{1}^{-1}
w_{1} $, and $w= k_{1}^{-1} w_{1} \wedge dT$. We get 
\begin{equation}
k_{1}w=w_{1}\wedge dT  \tag{2.8}
\end{equation}
which is a relation between forms all of whose coefficients are polynomials.
Now on such differential forms we can specialize the values of the variables 
$X_{\mu}$, obtaining forms defined over $F$. Letting $X_{\mu}\rightarrow 0$
for all $\mu\neq 1$ and $X_{1}\rightarrow 1$ and using $k_{1}\rightarrow 1$, 
$k_{i}\rightarrow 0$ for $i\geq 2$, we get from (2.8) in $\Omega_{F}^{m}$ 
\begin{equation*}
w=\bar{w}_{1}\wedge db_{1}
\end{equation*}
with some form $\bar{w}_{1}$. Thus $db_{1},\ldots,db_{n}$ divide the form $w$
. The lemma follows from (1.16). $\square$ \newline

\noindent(2.9) \textbf{Remark.} The differential of the polynomial $T$ is 
\begin{equation*}
dT = k_{1} db_{1} + \cdots+ k_{n} db_{n}
\end{equation*}
where $k_{1} , \ldots, k_{n} $ are polynomials in $F[X^2] $ given by 
\begin{equation*}
k_{i}=\sum_{\mu\in S_n , \mu(i)=1}b_i^{-1}b^{\mu}X^2_{\mu}, 
\mbox{\hspace{1cm}} 1\leq i \leq n.
\end{equation*}

\bigskip Let us recall that the Cartier-operator defines an isomorphism $C:
Z_{F}^{n} / B_{F}^{n} \rightarrow\Omega_{F}^{n} $ where $Z_{F}^{n} $ are the
closed forms in $\Omega_{F}^{n} $ and $B_{F}^{n} = d \Omega_{F}^{n-1} $ are
the exact forms. Following Kato (see \cite{Ka 1} and \cite{Mi}) we introduce
the operator 
\begin{equation}
{\wp}:\Omega_{F}^{n}\rightarrow\Omega_{F}^{n}/d\Omega_{F}^{n-1}  \tag{2.10}
\end{equation}
given by 
\begin{equation*}
{\wp} (w) = C^{-1} (w) - \overline{w}
\end{equation*}
i.e. ${\wp}= C^{-1} - \overline{id}$.

For example 
\begin{equation}
{\wp}\left( a\frac{db_{1}}{b_{1}}\wedge\cdots\wedge\frac{db_{n}}{b_{n} }
\right) =\overline{(a^{2}-a)\frac{db_{1}}{b_{1}}\wedge\cdots\wedge \frac{
db_{n}}{b_{n}}}  \tag{2.11}
\end{equation}

In order to complete (2.10) to an exact sequence we introduce the groups
(see \cite{Ka 1}) 
\begin{equation}
\nu_{F}(n)=\ker({\wp})  \tag{2.12}
\end{equation}
\begin{equation}
H^{n+1}(F)=\mbox{ Coker }({\wp})  \tag{2.13}
\end{equation}
and we have the exact sequence 
\begin{equation}
0\rightarrow\nu_{F}(n)\rightarrow\Omega_{F}^{n}\rightarrow\Omega_{F}
^{n}/d\Omega_{F}^{n-1}\rightarrow H^{n+1}(F)\rightarrow0.  \tag{2.14}
\end{equation}

\bigskip Taking a 2-basis $\mathcal{B} = \{ b_{i}, i \in I \} $ of $F $ we
get the square-operator $s: \Omega_{F}^{n} \rightarrow\Omega_{F}^{n} $ given
by $s: \left( \sum_{\sigma} a_{\sigma} \frac{db_{\sigma}}{b_{\sigma}}
\right) = \sum_{\sigma}a_{\sigma}^{2} \frac{db_{\sigma}}{b_{\sigma}} $ (see
section 1). We define the basis-dependent operator ${\wp}: \Omega_{F}^{n}
\rightarrow \Omega_{F}^{n} $ by ${\wp} (w) = s(w) - w $ and easily check
that 
\begin{equation*}
H^{n+1} (F) =\Omega_{F}^{n} /( d \Omega_{F}^{n-1} + {\wp} \Omega_{F}^{n}).
\end{equation*}

Another choice of a 2-basis would change ${\wp}(w)$ by an exact form. Thus $%
d\Omega_{F}^{n-1}+{\wp}\Omega_{F}^{n}$ is basis independent.

The next two propositions are due to Kato (see \cite{Ka 1}) and will be used
continuously in what follows.

Let $\{b_{i}\}_{i\in I}$ be a 2-basis of $F$ and endow $I$ with the
structure of a totally ordered set. For $j\in I$ let $F_{j}$ (resp. $F_{<j} $
) be the subfield of $F$ generated over $F^{2}$ by the elements $b_{i}$ with 
$i\leq j$ (resp. $i<j$). For fixed $n$ let $\sum_{n}$ be the set of all
functions $\alpha:\{1,\ldots,n\}\rightarrow I$ with $\alpha(i)<\alpha(j)$
whenever $1\leq i<j\leq n$. We endow $\sum_{n}$ with the lexicographic
ordering, namely $\alpha<\beta$ ( $\alpha$, $\beta\in\sum_{n})$ if and only
if there exists some $i$ such that $\alpha(i)<\beta(i)$ and $%
\alpha(j)\leq\beta(j)$ for all $j\leq i$. The $F$-vector space $%
\Omega_{F}^{n}$ has the basis $\{db_{\alpha (1)}\wedge\cdots\wedge
db_{\alpha(n)},\alpha\in\sum_{n}\}$ and we can introduce in $\Omega_{F}^{n}$
the following filtration: for $\alpha\in\sum_{n}$ let $\Omega_{F,\alpha}^{n}$,
(resp.$\Omega_{F,<\alpha}^{n})$, be the subspace of $\Omega_{F}^{n}$ generated
by the elements $db_{\beta(1)} \wedge\cdots\wedge db_{\beta(n)}$ with $\beta\leq\alpha$,
 (resp. $\beta<\alpha)$. Using this notation we formulate
the following basic result due to Kato (see also \cite{Ka 2}).

\bigskip\noindent(2.15) \textbf{Lemma.} \emph{Let $y \in F,
\alpha\in\sum_{n} $ and 
\begin{equation*}
w_{\alpha} = \frac{db_{\alpha(1) }}{b_{\alpha(1)}} \wedge\cdots\wedge \frac{
db_{\alpha(n)}}{b_{\alpha(n)}} \in\Omega_{F}^{n}
\end{equation*}
be such that 
\begin{equation*}
(y^{2} - y)w_{\alpha}\in\Omega_{F, < \alpha}^{n} + d \Omega_{F}^{n-1}
\end{equation*}
Then 
\begin{equation*}
yw_{\alpha}= v + \frac{da_{1}}{a_{1}} \wedge\cdots\wedge\frac{da_{n}}{a_{n}}
\end{equation*}
for some $v \in\Omega_{F, < \alpha}^{n} $ and some $a_{i} \in F_{\alpha
(i)}^{*}$, $1 \leq i \leq n $.}

\bigskip We will refer to this result in the sequel as Kato's lemma. An
immediate consequence of (2.15) is \newline

\noindent(2.16) \textbf{Corollary.} \emph{
\begin{equation*}
\nu_{F} (n) = \left\{ \sum_{i} \frac{da_{i_{1}}}{a_{i_{1}}} \wedge
\cdots\wedge\frac{da_{i_{n}}}{a_{i_{n}}}| a_{i_{k}} \in F^{*} \right\} .
\end{equation*}
}

\bigskip We will now study the behavior of $H^{n+1}$ under field extensions.
If $F\hookrightarrow L$ is a field extension we get the obvious maps $%
\nu_{F}(n)\rightarrow\nu_{L}(n)$ and $H^{n+1}(F)\rightarrow H^{n+1}(L)$. We
write $H^{n+1}(L/F)$ for the kernel of $H^{n+1}(F)\rightarrow H^{n+1}(L)$.
The main goal of this paper is the computation of $H^{n+1}(F(\phi)/F)$ for
an anisotropic bilinear $n$-fold Pfister form $\phi$ defined over $F$. This
will be done in section 4. We will now consider only some easier field
extensions. \newline

\noindent(2.17) \textbf{Lemma.} \emph{Let $F(X)/F$ be a pure transcendental
extension. Then} 
\begin{equation*}
H^{n+1} (F(X)/F) = 0
\end{equation*}

\noindent\textbf{Proof.} 
Let $\mathcal{B} = \{ b_{i},\; i \in I \} $ be a fixed 2-basis of $F$. 
We may assume that $F(X) / F $ has transcendence degree one. 
Then $\mathcal{B} \cup\{ X \} $ is a 2-basis of $F(X)$. 
We fix an ordering in $I $ (and hence in $\mathcal{B}) $ and we 
choose the ordering of $\mathcal{B}\cup\{ X \} $ with $X > b_{i} $ 
for all $i \in I$. 
Let $\bar{w } \in H^{n+1}(F) $ with $\bar{w} = 0 $ in $H^{n+1} (F(X))$. 
Thus in $\Omega_{F(X)}^{n} $ we have 
\begin{equation*}
w = {\wp} u + dv
\end{equation*}
for some $u \in\Omega_{F(X)}^{n} $ and $v \in\Omega_{F(X)}^{n-1}$. 
Here ${\wp }$ is defined with respect to the 2-basis 
$\mathcal{B} \cup\{ X \}$.
Hence ${\ \wp} u = w + dv $ with $w \in\Omega_{F}^{n}$.
Let $\alpha\in \Sigma_{n,F}\subset\Sigma_{n,F(X)}$ be the leading index of $w$, that is $w=\sum_{\gamma\in \Sigma_{n,F}} w_{\gamma}\frac{db_{\gamma}}{b_{\gamma}}$ with $w_{\alpha}\neq 0$ and $w_{\gamma}=0$ for all $\gamma>\alpha$, and $\beta\in \Sigma_{n,F(X)}$ the leading index of $u$. If $\beta>\alpha$, since 
$\wp(u_{\beta})\frac{db_{\gamma}}{b_{\gamma}}\in d\Omega^{n-1}_{F(X)}+\Omega^n_{F(X),<\beta}$, we apply Kato's lemma and conclude that 
$u_{\beta}\frac{db_{\gamma}}{b_{\gamma}}=u'+\frac{da_1}{a_1}\wedge\cdots\wedge\frac{da_n}{a_n}$, with $a_i\in F_{\beta(i)}$ and $u'\in \Omega^n_{F(X),<\beta}$. By this way we can replace $u$ by a differential form with lower leading index. This means that we may assume $\beta\leq\alpha$. Then we have
\begin{equation*}
\left(\wp(u_{\alpha})+w_{\alpha}\right)\frac{db_{\alpha}}{b_{\alpha}}=d(v) \mod \Omega^n_{F(X),<\alpha}
\end{equation*}
with $v\in \Omega^{n-1}_{F(X)}$. 
Since $b_{\alpha(i)}<X$ for all $i\in I$, we conclude that we may assume that the  coefficients of $d(v)$ are in $F(X^2)$. Because if we write $v=v_0+Xv_1+v_2\wedge dX+Xv_3\wedge dX$ where $v_0$, $v_1$, $v_2$, $v_3$ are generated over $F(X^2)$ by the differentials $db_i$, $i\in I$, we get
$$ dv=dv_0+Xdv_1+(v_1+dv_2+Xdv_3)\wedge dX$$
and hence $v_1+dv_2+Xdv_3=0$. By the choice of the forms $v_1$, $v_2$, $v_3$ we infer from the last relation that $v_1+dv_2=0$ and $dv_3=0$. Therefore inserting $v_1=dv_2$ into $v$ we obtain
$$ v=v_0+d(Xv_2)+Xd(Xv_3)$$
and since $d(Xd(Xv_3))=dX\wedge d(Xv_3)=0$, it follows $dv=dv_0$. Thus we may replace $v$ by $v_0$ and we can assume that $v$ is generated over $F(X^2)$ by the differentials $db_i$, $i\in I$. 
Then $u$ is also generated over $F(X^2)$ by the differentials $db_i$, $i\in I$.

Therefore the relation $ \wp(u)+w=d(v)$ holds in the subspace 
\begin{equation*}
\Omega_{F}^{n} [X^{2}] \oplus\bigoplus_{p} p^{-\infty}\Omega_{F}^{n} [X^{2}] %
\mbox{ of } \Omega_{F(X)}^{n},
\end{equation*}
where $p $ runs over all irreducible polynomials in $F[X^{2}]$. 
Let us write 
\begin{equation*}
u_{\alpha} = u_{\alpha,0} + \sum_{p} u_{\alpha,p} \;\;\mbox{ , }\;\; v = v_{0} + \sum_{p} v_{p}\;\;\mbox{ , }
\end{equation*}
with $u_{\alpha,0} \in F[X^{2}]$, $v_{0} \in\Omega_{F}^{n-1} [X^{2}] $,  
resp. $u_{p} \in p^{- \infty} F[X^{2}]$, $v_{p} \in p^{\infty}
\Omega_{F}^{n-1} [X^{2}]$. 
Then 
\begin{equation*}
w_{\alpha}\frac{db_{\alpha}}{b_{\alpha}} = 
\wp u_{\alpha,0}\frac{db_{\alpha}}{b_{\alpha}}+ dv_{0}+\sum_{p}\left(\wp u_{\alpha,p}\frac{db_{\alpha}}{b_{\alpha}}+dv_p\right) \mod \Omega^n_{F(X),<\alpha}
\end{equation*}
and since $\wp$, $d $ are compatible with the above direct sum, we see that 
\begin{equation*}
w_{\alpha}\frac{db_{\alpha}}{b_{\alpha}} = 
\wp u_{\alpha,0}\frac{db_{\alpha}}{b_{\alpha}}+ dv_{0} \mod \Omega^n_{F,<\alpha}[X^2]
\end{equation*}
holds in $\Omega_{F}^{n} [X^{2}]$. 
We write now $u_{\alpha,0} = \overline{u_{\alpha,0}} + X^{2s} u_{s}$, 
$v_{0} = \overline{v_{0}} + X^{2r} v_{r} $  with $u_{s}\in F$, 
$v_{r} \in\Omega_{F}^{n-1}$ and $2s =\deg(u_{\alpha,0})$, $2r = \deg(v_{0})$, 
$\deg(\overline{u_{\alpha,0}})< 2s$, $\deg( \overline{v_{0}}) < 2r$. 
Thus 
\begin{equation*}
w_{\alpha}\frac{db_{\alpha}}{b_{\alpha}}=\wp \overline{u_{\alpha,0}}\frac{db_{\alpha}}{b_{\alpha}} + d \overline{v_{0}}+ [X^{4s} u_{s}^{[2]} +
X^{2s} u_{s}]\frac{db_{\alpha}}{b_{\alpha}} + X^{2r} dv_{r} \mod \Omega^n_{F,<\alpha}[X^2].
\end{equation*}

Let us first assume $4s>2r$. 
Then comparing coefficients we get $u_{s}^{[2]}\frac{db_{\alpha}}{b_{\alpha}}\in\Omega^n_{F,<\alpha}$ and applying the 
Cartier-operator it follows $u_{s}=0$, which
is a contradiction. 
Similarly if $2r>4s$, then $dv_{r}\in\Omega^n_{F,<\alpha}$, including $X^{2r} dv_{r}$ in $\Omega^n_{F,<\alpha}$, we conclude that we can lower the degree of $v_0$. Thus we are lead to the case $4s=2r$. 
If $4s=2r>0$, then we get $u_{s}^{[2]}\frac{db_{\alpha}}{b_{\alpha}}=dv_{r} \mod \Omega^n_{F,<\alpha}$ and applying the Cartier 
operator we obtain $u_{s}\frac{db_{\alpha}}{b_{\alpha}}\in \Omega^n_{F,<\alpha}$, that is 
$u_{s}\frac{db_{\alpha}}{b_{\alpha}}=0$, which is a contradiction. 
Thus we have $s=r=0$, i.e. $w_{\alpha}\frac{db_{\alpha}}{b_{\alpha}}={\wp} u_{0}\frac{db_{\alpha}}{b_{\alpha}}+dv_{0} \mod \Omega^n_{F,<\alpha}$ with $u_{0}\in F$ and 
$v_{0} \in\Omega_{F}$.
Replacing in $w$, this see $\overline{w}$ can be represented by a differential form with lower leading index. This shows $\overline{w}=0$ in $H^{n+1}(F)$, and
concludes the proof of the lemma. $\square$

\bigskip In particular we have $H^{m+1} (L/F) = (0) $ for all $m $, where $L 
$ is the field extension $F(X_{\mu}, \mu\in S_{n}) $ introduced at the
beginning of this section.

\bigskip We want now to compute $H^{m+1}(F(\phi)/L)$, where $F(\phi)=L(\sqrt{
T})$ (see (2.1)). To this end we prove the following general fact. \newline

\noindent(2.18) \textbf{Lemma.} \emph{Let $k $ be field of characteristic 2,
and $b\in k\setminus k^{2}$. Then 
\begin{equation*}
H^{m+1} (k (\sqrt{b})/k ) = \overline{\Omega_{k}^{m-1} \wedge db}
\end{equation*}
}
\noindent\textbf{Proof. } Since $b \in k \setminus k^{2} $ we can take $b$
part of a 2-basis of $k $. Let $\mathcal{B} = \{ b_{1} = b, b_{2}, \ldots\}$
be a 2-basis of $k $. Then $\mathcal{B}^{\prime}= \{ \sqrt{b}, b_{2},
\ldots\}$ is a 2-basis of $k(\sqrt{b})$. Take now $\bar{w} \in H^{m+1} (k( 
\sqrt{b})/k) $, i.e. $\bar{w} \in H^{m+1} (k) $ with $\bar{w} = 0 $ in $%
H^{m+1} (k(\sqrt{b}))$. This means 
\begin{equation*}
w = {\wp} u + dv
\end{equation*}
with $u \in\Omega_{k(\sqrt{b})}^{m}$, $v \in\Omega_{k(\sqrt{b})}^{m-1}$. We
order the 2-basis $\mathcal{B}^{\prime}$ with $\sqrt{b} > b_{i} $ for all $i
\;\;(\neq 1)$. 
Let $\alpha\in\Sigma_{m,k}$, $\alpha(i)>1$ for all $i=1,\ldots,m$, be the leading index of $w$ and $\beta\in \Sigma_{m,k(\sqrt{b})}$ the leading index of $u$.
Thus we can assume, as in the proof of (2.17), $\beta\leq\alpha$. In this case we have
\begin{equation*}
\left(\wp(u_{\alpha})+w_{\alpha}\right)\frac{db_{\alpha}}{b_{\alpha}}=d(v) \mod \Omega^n_{k(\sqrt{b}),<\alpha}
\end{equation*}
with $v\in \Omega^{n-1}_{k(\sqrt{b})}$. 
Since $b_{\alpha(i)}<\sqrt{b}$ for all $i$, we conclude that the leading coefficient of $d(v)$ is in $k$, then $u_{\alpha}$ is also in $k$ and we may assume that $v\in \Omega^{m-1}_k$.


Since $\ker[\Omega^m_k \rightarrow \Omega^m_{k(\sqrt{b})}]=\Omega^{m-1}_k\wedge d(b)$, we conclude that
\begin{equation*}
w_{\alpha}\frac{db_{\alpha}}{b_{\alpha}}=\wp(u_{\alpha})\frac{db_{\alpha}}{b_{\alpha}}+d(v) \mod \Omega^n_{k,<\alpha}+\Omega^{m-1}_k\wedge d(b)
\end{equation*}
in $\Omega^m_k$. Replacing in $w$, this shows $\overline{w}$ can be represented by a differential form with lower leading index. This shows $\overline{w}\in \overline{\Omega^{m-1}_k\wedge db}$ in $H^{n+1}(k)$, and concludes the proof of the lemma.
$\square$
\newline

\noindent(2.19) \textbf{Corollary.} \emph{
\begin{equation*}
H^{m+1} (F(\phi)/L) = \overline{\Omega_{L}^{m-1} \wedge dT}
\end{equation*}
}

Let us close this section with some remarks concerning the computation of $%
H^{n+1} (F(\phi)/F)$. We write $y = \sqrt{T}, $ so that $K=F(\phi)= L[y] $.
Fix a 2-basis $\mathcal{B} = \{ b_{i}, i \in I \} $ of $F $, so that $%
\mathcal{B}\cup\{ X_{\mu},\mu\in S_{n} \}$ is a 2-basis of $L $. We order
the elements of $\mathcal{B} $ according to an order of $I $ and the
elements of $\{ X_{\mu}, \mu\in S_{n} \} $ for example using the
lexicographic ordering and we set $\mathcal{B} < \{ X_{\mu}, \mu\in S_{n} \} 
$ for an ordering in $\mathcal{B} \cup\{ X_{\mu}, \mu\in S_{n} \} $. Since $%
y^{2} = T = \sum_{\mu}b^{\mu}X_{\mu}^{2} $ in $K $, we see that the elements
of $\mathcal{B} $ are not 2-independent over $K^{2} $. Let us fix some $b =
b_{1} \in\mathcal{B} $. Then $\mathcal{B} \setminus\{ b_{1} \} \cup\{
X_{\mu}, \mu\in S_{n} \} \cup\{y \} $ is a 2-basis of $K $. We order this
basis such that $y $ is the maximal element. In particular we have the
operator ${\wp} $ on $\Omega_{K} $ defined with respect to this 2-basis.
Take now $\bar{w} \in H^{n+1} (K/F) $. Then 
\begin{equation}
w={\wp} u+dv  \tag{2.20}
\end{equation}
with $u\in\Omega_{K}^{n}$ and $v\in\Omega_{K}^{n-1}$. From ${\wp} u=w+dv$,
and using Kato's lemma with a filtration defined by the above ordering, we
see that one can assume in (2.20) that $u$ and $v$ are differential forms
generated over $K$ by the differential $db_{i}$, $i\in I$ $(i\neq1)$, i.e.
they do not contain differentials of the type $dX_{\mu}$ or $dy$. Looking at
the 2-basis expansion of both sides of (2.20) we easily conclude that $u$
and $v$ do not contain $y$ in their coefficients, i.e. they are contained in 
$L$. Therefore the forms $u$, $v$ (and $w$) are defined over $L$ and are
generated over $L$ by the differentials $db_{i}$, $i\in I$. From (2.4) we
conclude that $w+{\wp} u+dv\in\Omega^{n}(K/L)\subset\Omega_{L}^{n}$ as a
form in $\Omega_{L}^{n}$, and using (2.4) we obtain in $\Omega_{L}^{n}$ 
\begin{equation}
w={\wp} u+dv+\lambda\wedge dT  \tag{2.21}
\end{equation}
with $\lambda\in\Omega_{L}^{n-1}$. We will show below that the coefficients
of $u$, $v$, $\lambda$ in the 2-basis expansion of $L$ do not contain odd
powers of the variables $X_{\mu}$ and that $\lambda$ also is generated by
the differentials $db_{i}$, $i\in I$. Therefore if we define the subfield of 
$L$ 
\begin{equation*}
M = F(X_{\mu}^{2}, \mu\in S_{n} ) \subset L
\end{equation*}
we see that all forms $u$, $v$, $\lambda$, $dT $ and $w $ are generated over 
$M $ by the differentials $db_{i}$, $i \in I$, i.e. they are contained in
the subspace $\Omega_{F} \otimes M $ of $\Omega_{L}$. We show now this
assertion. The fact that $\lambda$ does not contain differentials $dX_{\mu}$, 
$\mu\in S_{n}$ follows from (2.21) and the fact that all other forms,
including $dT $, do not contain such forms. Let us write $u = u_{0} + Xu_{1}$
, $v = v_{0}+ Xv_{1}$, $\lambda= \lambda_{0} + X \lambda_{1} $, for some
fixed $X = X_{\mu} $, and where $u_{i}$, $v_{i}$, $\lambda_{i}$ $(i = 0, 1) $
do not contain odd powers of $X $ in their coefficients, with respect to their 2-basis
representation. 
Then in $\Omega_{L}^{n} $ 
$$
w  = \wp u_{0} + X^{2} u_{1}^{[2]} + Xu_{1} + dv_{0} + Xdv_{1} 
+ v_{1} \wedge dX + \lambda_{0} \wedge dT + X \lambda_{1} \wedge dT
$$
and this implies 
\begin{equation}
w={\wp} u_{0}+X^{2}u_{1}^{[2]}+dv_{0}+\lambda_{0}\wedge dT  \tag{2.22}
\end{equation}
\begin{equation}
u_{1}+dv_{1}+\lambda_{1}\wedge dT =0 \tag{2.23}
\end{equation}
\begin{equation}
 v_{1}\wedge dX=0.   \tag{2.24}
\end{equation}
Since $v_{1} $ does not contain $dX $, we obtain from (2.24) $v_{1} = 0 $
and therefore $u_{1} = \lambda_{1} \wedge dT $. Inserting this in (2.22) it
follows 
\begin{equation*}
w={\wp} u_{0}+dv_{0}+(\lambda_{0}+X^{2}T\lambda_{1}^{[2]})\wedge dT,
\end{equation*}
since from $u_{1}=T\lambda_{1}\wedge\frac{dT}{T}$ we get $u_{1}^{[2]}
=T^{2}\lambda_{1}^{2}\wedge\frac{dT}{T}=T\lambda_{1}^{[2]}\wedge dT$. The
form $\lambda_{0}+X^{2}T\lambda_{1}^{[2]}$ does not contain odd powers of $X$
, and the above equation therefore shows that we can eliminate all odd
powers of $X$ from $u$, $v$ and $\lambda$. Since $X=X_{\mu}$ was arbitrary,
this proves the claim. Therefore we have shown \newline

\noindent(2.25) \textbf{Proposition.} \emph{Let $\bar{w}\in H^{n+1}(K/F)$ and fix a 2-basis 
$\mathcal{B} = \{ b_{i}, i \in I \} $ of $F $. Then there exist forms $u$, $v$, $\lambda$ in $\Omega_{F}\otimes
M\subset\Omega _{L}$, i.e. generated over $M=F(X^2_{\mu}, \mu\in S_n)$ by the differentials $db_{i}$, $i\in I$ with 
\begin{equation*}
w = {\wp} u + dv + \lambda\wedge dT
\end{equation*}
where $\wp$ is defined with respect to any 2-basis of $L$ containing the basis $\mathcal{B} $
}

\bigskip The next two sections will be devoted to the study of this equation.

\bigskip \noindent (2.26) \textbf{Remark.} The behavior of $\nu _{F}(m)$
under field extensions is easier to handle than that of $H^{m+1}(F)$. For
example (2.7) implies $\Omega ^{m}(F(\phi )/F)=\Omega _{F}^{m-n}\wedge
db_{1}\wedge \cdots \wedge db_{n}$ if $\phi =\ll b_{1},\ldots ,b_{n}\gg $.
Hence $\ker (\nu _{F}(m)\rightarrow \nu _{F(\phi )}(m))=\Omega
_{F}^{m-n}\wedge db_{1}\wedge \cdots \wedge db_{n}\cap \nu _{F}(m)$. Thus
this kernel is determined by the following

\bigskip \noindent(2.27) \textbf{Lemma.} \emph{For any $a\in \Omega^{m-n}_F$
the following statements are equivalent: }

\bigskip\lefteqn{\mbox{(i)\hspace{1cm}} a\wedge\frac{db_1}{b_1}
\wedge\cdots\wedge\frac{db_n}{b_n}\in \nu_F(m)}

\bigskip\lefteqn{\mbox{(ii)
\hspace{1cm}} \wp(a)\in \sum_{\epsilon\in S_n, \epsilon\neq
0}b^{\epsilon}[\Omega_F^{m-n}]^{[2]}+d\Omega^{m-n-1}_F+\sum_{i=1}^n
\Omega^{m-n-1}_F\wedge db_i}

\emph{where $\epsilon=(\epsilon_1,\ldots,\epsilon_n)$ runs over all
sequences with $\epsilon_i=0$ or $1$. }

\bigskip\noindent\textbf{Proof. } Choose a 2-basis $\mathcal{B}$ of $F$ with 
$b_1,\ldots,b_n\in \mathcal{B}$. Without restriction we can assume $\mathcal{%
B}$ finite, i.e. $\mathcal{B}=\{b_1,\ldots,b_n,\ldots,b_N\}$ and let us
denote by $\eta$ the differential form $\frac{db_1}{b_1}\wedge\cdots\wedge%
\frac{db_n}{b_n}$.
\begin{enumerate}
\item  First we prove that (ii) implies (i). Let $a$ be an element in $%
\Omega _{F}^{m-n}$ such that $\wp (a)\in \sum_{\epsilon \in S_{n},\epsilon
\neq 0}b^{\epsilon }[\Omega _{F}^{m-n}]^{[2]}+d\Omega
_{F}^{m-n-1}+\sum_{i=1}^{n}\Omega _{F}^{m-n-1}\wedge db_{i}$, i.e. $a$ can
be written as 
\begin{equation*}
a=a^{[2]}+\sum_{\epsilon \neq 0}b^{\epsilon }A_{\epsilon
}^{[2]}+dB+\sum_{i=1}^{n}E_{i}\wedge db_{i}
\end{equation*}
with $A_{\epsilon }\in \Omega _{F}^{m-n}$, $B\in \Omega _{F}^{m-n-1}$, $%
E_{i}\in \Omega _{F}^{m-n-1}$. Since $(\sum_{\epsilon \neq 0}b^{\epsilon
}A_{\epsilon }^{[2]})\wedge \eta $ and $dB\wedge \eta $ are in $d(\Omega
_{F}^{m-1})$, and $E_{i}\wedge db_{i}\wedge \eta =0$, we obtain that 
\begin{equation*}
d(a\wedge \eta )=0
\end{equation*}
and 
\begin{eqnarray*}
C(a\wedge \eta ) &=&C((a^{[2]}+\sum_{\epsilon \neq 0}b^{\epsilon
}A_{\epsilon }^{[2]}+dB+\sum_{i=1}^{n}E_{i}\wedge db_{i})\wedge \eta ) \\
&=&C(a^{[2]}\wedge \eta )+\sum_{\epsilon \neq 0}C(b^{\epsilon }A_{\epsilon
}^{[2]}\wedge \eta )+C(dB\wedge \eta )=a\wedge \eta 
\end{eqnarray*}
This implies that $a\wedge \eta \in \nu (m)_{F}$ and proves (i).
\item  Now let $a$ be in $\Omega _{F}^{m-n}$ such that $a\wedge
\eta \in \nu _{F}(m)$. We write $a$ as 
\begin{equation*}
a=\sum_{\mu }c_{\mu }\frac{db_{\mu }}{b_{\mu }}
\end{equation*}
where $\frac{db_{\mu }}{b_{\mu }}=\frac{db_{\mu (1)}}{b_{\mu (1)}}\wedge
\cdots \wedge \frac{db_{\mu (m-n)}}{b_{\mu (m-n)}}$, and $c_{\mu }\in F$.
Note that $a\wedge \eta \in \nu _{F}(m)$
implies that $d(a\wedge \eta )=0$ and $C(a\wedge \eta )=a\wedge \eta $.

Be $k$ the maximal index $k>n$ such that 
\begin{equation*}
a=R_{0}+b_{k}R_{1}
\end{equation*}
where $R_{0}$, $R_{1}$ are differential forms generated by $\frac{db_{1}}{%
b_{1}}$, $\ldots $, $\frac{db_{N}}{b_{N}}$, with coefficients in $%
F^{2}(b_{1},\ldots ,b_{k-1})$. Also we decompose $R_{0}$ and $R_{1}$ as $%
R_{0}=M_{0}+M_{1}\wedge \frac{db_{k}}{b_{k}}$ and $R_{1}=M_{2}+M_{3}\wedge 
\frac{db_{k}}{b_{k}}$. Then $d(a\wedge \eta )=0$ implies that 
$$
\left[ d(M_{0})+b_{k}d(M_{1})+b_{k}M_{1}\wedge \frac{db_{k}}{b_{k}}
+(d(M_{2})+b_{k}d(M_{3}))\wedge 
\frac{db_{k}}{b_{k}}\right]\wedge \eta =0
$$
which means that 
\begin{eqnarray*}
d(M_{0})\wedge \eta  &=&0 \\
d(M_{1})\wedge \eta  &=&0 \\
d(M_{2})\wedge \eta  &=&0 \\
M_{1}\wedge \eta  &=&dM_{3}\wedge \eta .
\end{eqnarray*}
%%% FROM PROBLEM:
%>From the last relation we obtain $M_{1}=d(M_{3})+E$ where $E$ is in $\Omega
From the last relation we obtain $M_{1}=d(M_{3})+E$ where $E$ is in $\Omega
_{F}^{m-n-1}\wedge \frac{db_{1}}{b_{1}}+\cdots +\Omega _{F}^{m-n-1}\wedge 
\frac{db_{n}}{b_{n}}$. Replacing in the above decomposition of $a\wedge \eta 
$ we get 
\begin{equation*}
a=M_{0}+M_{2}\wedge \frac{db_{k}}{b_{k}}+d(b_{k}M_{3})+E.
\end{equation*}
Now we work with $a^{\prime }=M_{0}+M_{2}\wedge \frac{db_{k}}{b_{k}}$ which
is also generated by $\frac{db_{1}}{b_{1}}$, $\ldots $, $\frac{db_{N}}{b_{N}}
$, with coefficients in $F^{2}(b_{1},\ldots ,b_{k-1})$. Since $%
d(M_{0})\wedge \eta =0$ and $d(M_{2})\wedge \eta =0$ we have that $a^{\prime
}\wedge \eta =0$. By repeating of the above procedure on $a^{\prime }$ 
we conclude that $a$ can be written as 
\begin{equation*}
a=M+d(G)+H
\end{equation*}
where $M$ is generated by $\frac{db_{1}}{b_{1}},\ldots ,\frac{db_{N}}{b_{N}}$
with coefficients in $F^{2}(b_{1},\ldots ,b_{n})$, $G\in \Omega _{F}^{m-n-1}$
and $H$ in $\Omega _{F}^{m-n-1}\wedge \frac{db_{1}}{b_{1}}+\cdots +\Omega
_{F}^{m-n-1}\wedge \frac{db_{n}}{b_{n}}$. This means that $a$ can written as 
\begin{equation*}
a=\sum_{\mu \in \Sigma _{m-n}}c_{\mu }\frac{db_{\mu }}{b_{\mu }}+d(G)+H
\end{equation*}
where each $c_{\mu }$ is in $F^{2}(b_{1},\ldots ,b_{n})$, i.e. $c_{\mu
}=\sum_{\epsilon }c_{\mu ,\epsilon }^{2}b^{\epsilon }$ with $\epsilon
=(\epsilon _{1},\ldots ,\epsilon _{n})$ running over all sequences with $%
\epsilon _{i}=0$ or $1$. Reordering the above relation we obtain 
\begin{equation*}
a=\sum_{\epsilon }b^{\epsilon }\left( \sum_{\mu }c_{\mu ,\epsilon }^{2}\frac{%
db_{\mu }}{b_{\mu }}\right) +d(G)+H.
\end{equation*}
Note that each 
$\left( \sum_{\mu }c_{\mu ,\epsilon }^{2}\frac{db_{\mu }}{b_{\mu }}\right) $
 is in $\Omega _{F}^{[2]}$ so we will denote it by 
$A_{\epsilon }^{[2]}$. 
By this way we write $a$ as 
$$
a=\sum_{\epsilon }b^{\epsilon }A_{\epsilon }^{[2]}+d(G)+H
=A_{0}^{[2]}+\sum_{\epsilon \neq 0}b^{\epsilon }A_{\epsilon }^{[2]}+d(G)+H.
$$
Finally we compute $C(a\wedge \eta )$. Since $b^{\epsilon }A_{\epsilon
}^{[2]}\wedge \eta $, $d(G)\wedge \eta $ $\in d(\Omega _{F}^{m-1})$ and $%
H\wedge \eta =0$, we have 
$$
C(a\wedge \eta )=C(A_{0}^{[2]}\wedge \eta )=A_{0}\wedge \eta. 
$$
Using that $C(a\wedge \eta )=a\wedge \eta $, we obtain $a=A_{0}+H^{\prime }$
with $H^{\prime }$ in $\Omega _{F}^{m-n-1}\wedge \frac{db_{1}}{b_{1}}%
+\cdots +\Omega _{F}^{m-n-1}\wedge \frac{db_{n}}{b_{n}}$. This means that 
\begin{equation*}
a=a^{[2]}+\sum_{\epsilon \neq 0}b^{\epsilon }A_{\epsilon
}^{[2]}+d(G)+H^{\prime \prime }
\end{equation*}
where $H^{\prime \prime }=H+H^{\prime }$, completing the proof. $\square $
\end{enumerate}

\bigskip Considering the case $m=n$ we obtain

\bigskip\noindent(2.28) \textbf{Corollary.} 
\begin{equation*}
\ker(\nu_F(n)\rightarrow \nu_{F(\phi)}(n))=\left\{a\frac{db_1}{b_1}%
\wedge\cdots\wedge\frac{db_n}{b_n}| \wp(a)\in D_F(\phi^{\prime})\right\}
\end{equation*}

\newpage

\section{Some technical results}

Let us first recall some notation. Let $S_{n} $ be the set of maps $%
\mu:\{1,\ldots,n\}\rightarrow\{0,1\}$ with $\mu(i) = 1 $ for at least some
index $i $. If $\mu$ is defined by $\mu(i) = 1$, $\mu(j) = 0 $ for $j \neq i 
$ we write $i $ instead of $\mu$. Let $L = F(X_{\mu}, \mu\in S_{n} ) $ and
for a fixed $0 \leq s \leq n $ we set $N = F(X_{\mu}| \mu\neq0, \ldots, s )$
, $M = F(X_{\mu}^{2} | \mu\neq0, \ldots, s )$, (if $s = 0 $ we get $N = L $
and $M $ is the field introduced in the last section). \newline

\noindent(3.1) \textbf{Lemma.} \emph{Let $X=X_{s}$. If $f\in M[X^{2}]$ is
decomposed in monic irreducible polynomials in $M[X],$ say 
\begin{equation*}
f = a \prod_{p} p^{n_{p}} \hspace{2cm} (a \in M)
\end{equation*}
then for each $p $ holds $p \in M[X^{2}] $ or $n_{p} \equiv 0 $ (mod 2). } 
\newline

\noindent\textbf{Proof.} Let $p \in M[X] $ be an irreducible factor of $f $
and assume $p \not \in M[X^{2}]$. 
This means $D_{X}(p)\neq 0$, where $D_{X}(p)$ is the derivative of $p$ 
with respect to $X$. 
Let us assume $n_{p}= 2t + 1 $ for some $t \geq 0$. 
Since $f \in M[X^{2}] $ implies $D_{X}(f) =0 $, we obtain 
\begin{equation*}
D_{X} \left(a p^{2t+1} \prod_{q \neq p} q^{n_{q}}\right) = 0
\end{equation*}
\begin{equation*}
a p^{2t+1} D_{X} \left(\prod_{q \neq p}q^{n_{q}}\right) + a \left(\prod_{q
\neq p}q^{n_{q}}\right) p^{2t} D_{X} (p) = 0
\end{equation*}
\begin{equation*}
\left(\prod_{q \neq p}q^{n_{q}}\right) D_{X} (p) = pD_{X}\left(\prod_{q \neq
p}q^{n_{q}}\right)
\end{equation*}

But $\deg D_{X}(p)<\deg_{X}(p)$ implies that $p$ divides $\prod_{q\neq
p}q^{n_{q}}$ in $M[X]$, which is impossible. This proves the claim. $\square$
\newline

For any irreducible monic polynomial $p\in N[X](X=X_{s})$ we will write 
$N(p) $ for the quotient field $N[X]/(p)$. 
Let us fix a 2-basis $\mathcal{B}=\{b_{i},i\in I\}$ of $F,$ so that 
$\mathcal{B}\cup \{X_{\mu }\setminus \mu\neq 1,\ldots ,s\}$ is a 2-basis of $N$. 
If $p\in M[X^{2}]$, there is some $b_{i}\in \mathcal{B}$ such that 
$\mathcal{B}\setminus \{b_{i}\}\cup \{\bar{X}_{\mu },\mu \neq 1,\ldots ,s-1\}$ 
is a 2-basis of $N(p)$, where $\bar{X}_{\mu }$ denotes the image of 
$X_{\mu }$ in $N(p)$. 
If $p\not\in M[X^{2}]$ then $\mathcal{B}$ is part of a 2-basis of $N(p)$ 
and in fact there is some index $i_{0}\neq 1,\ldots ,s-1$ such that 
$\mathcal{B}\cup \{\bar{X}_{\mu},\mu \neq i_0,1,\ldots ,s-1\}$ is a 2-basis of $N(p)$. 
The natural map $N[X]\rightarrow N(p)$ induces a homomorphism 
$\Omega _{F}^{m}M[X^{2}]\rightarrow \Omega _{N(p)}^{m}$. \newline

\noindent(3.2) \textbf{Lemma.} \emph{a) If $p \in M[X^{2}] $ 
\begin{equation*}
\ker(\Omega_{F}^{m}M[X^{2}] \rightarrow\Omega_{N(p)}^{m} ) = 
p\Omega_{F}^{m}M[X^{2}] + \Omega_{F}^{m-1} M[X^{2}] \wedge dp
\end{equation*}
}

\emph{b) If $p \not \in M[X^{2}] $ 
\begin{equation*}
\ker(\Omega_{F}^{m} M[X^{2}] \rightarrow\Omega_{N(p)}^{m} ) = 
p^{2} \Omega_{F}^{m} M[X^{2}]
\end{equation*}
}

\noindent\textbf{Proof.} a) Since $p \in M[X^{2}]$, there is some $b_{k}\in 
\mathcal{B} $ such that $\overline{D_{b_{k}} (p)} \neq 0 $ in $N(p)$,
because otherwise one would infer that $p $ is a square in $M[X]$. But in $%
N(p) $ we have $0 = dp = \sum_{b_{j}} \overline{D_{b_{j}} (p)} db_{j}$,
where $\overline{D_{b_{k}}(p)} \neq 0 $ is the coefficient of $db_{k}$.
Choose $\Delta\in M[X^{2}] $ with $\Delta\cdot D_{b_{k}} (p) = 1 + p \cdot r 
$ in $M[X^{2}] $ and let $\bar{\Delta} $ be the image in $N(p)$. Then 
\begin{equation*}
db_{k} = \bar{\Delta} \sum_{i \neq k} \overline{D_{b_{i}}(p)} db_{i} 
\mbox{
in  } \Omega_{N(p)}
\end{equation*}

Let $w \in\Omega_{F}^{m} M[X^{2}] $ be in the kernel of $\Omega_{F}^{m}
M[X^{2}] \rightarrow\Omega_{N(p)}^{m}$, i.e. $w = 0 $ in $\Omega_{N(p)}$. We
have by definition of $\Omega_{F}^{m} M[X^{2}] $ 
\begin{equation*}
w = \sum_{ \gamma} a_{\gamma}db_{\gamma}+ ( \sum_{\delta}
c_{\delta}db_{\delta}) \wedge db_{k}
\end{equation*}
where $\gamma$ runs over all $\sum_{m} (I) $ with $k \not \in$ Im$(\gamma) $
and $\delta$ runs over $\sum_{m-1} (I) $ with $k \not \in $ Im$(\delta) $
and $a_{\gamma}$, $c_{\gamma}\in M[X^{2}]$. Recall that $\sum_{m} (I) $ are
all maps $\gamma: \{1, \ldots, m \} \rightarrow I $ with $\gamma(1) <
\cdots<\gamma(m) $ in a fixed ordering of $I$. Then in $\Omega_{N(p)}^{m} $ 
\begin{equation*}
\sum a_{ \gamma} db_{\gamma} + ( \sum c_{\delta} db_{\delta} ) \wedge
\Delta\sum_{i \neq k} D_{bi} (p) db_{i} = 0
\end{equation*}
(we omit the bars for simplicity).

Thus 
\begin{equation*}
\sum_{k \not \in\gamma} (a_{\gamma} + \Delta\sum_{\delta\cup j= \gamma}
c_{\delta} D_{b_{j}} (p)) db_{\gamma} = 0
\end{equation*}

Here $k\not\in\gamma$ means $\gamma\in \sum_m(I)$ with $k\not\in $ Im($\gamma$) and 
$\delta \cup j=\gamma$ means that $\delta\in \sum_{m-1}(I)$ can be extended
to $\gamma$ with $j\in $ Im($\gamma$).

Since $\mathcal{B} \setminus\{ b_{k} \} $ is part of a 2-basis of $N(p) $ it
follows that the $db_{\gamma}$, $k \not \in\gamma$ are linear independent
over $N(p) $ and hence in $N(p)$ 
\begin{equation*}
a_{\gamma}= \Delta\sum_{\delta\cup j= \gamma} c_{\delta} D_{b_{j}} (p).
\end{equation*}
Then in $N[X]$ we have 
\begin{equation*}
a_{\gamma}= \Delta\sum_{\delta\cup j= \gamma} c_{\gamma}D_{b_{j}} (p) +
t_{\gamma}\cdot p
\end{equation*}
for each $\gamma\in\sum_{m}$, $k \not \in\gamma, $ with $t_{\gamma}\in N[X]$
. It follows easily that $t_{\gamma}\in M[X^{2}] $ for all $\gamma$, i.e.
the above relation holds in $M[X^{2}]$. Inserting in $w $ we obtain 
\begin{equation*}
\begin{array}{lcl}
w & = & \sum_{k \not \in\gamma} ( \sum_{\delta\cup j= \gamma} \Delta
c_{\delta} D_{b_{j}} (p) + pt_{\gamma}) db_{\gamma}+(\sum_{k \not
\in\delta} c_{\delta} db_{\delta}) \wedge db_{k} \\ 
&  &  \\ 
 & = & \sum_{k \not \in\delta} (c_{\delta} db_{k} + \Delta\sum_{j \neq k}
c_{\delta} D_{b_{j}}(p) db_{j}) \wedge db_{\delta}+ p \sum_{k \not
\in\gamma} t_{\gamma}db_{\gamma} \\ 
&  &  \\ 
 & = & \sum_{k \not \in\delta} c_{\delta} (db_{k} + \Delta\sum_{j \neq k}
D_{b_{j}} (p) db_{j}) \wedge db_{\delta}+ p \sum_{k \not \in\gamma}
t_{\gamma }db_{\gamma}
\end{array}
\end{equation*}

Replacing the coefficient 1 of $db_{k} $ by $1 = \Delta D_{b_{k}} (p) + pr$,
we get 
\begin{equation*}
w = \sum_{k \not \in\delta} c_{\delta} \Delta dp \wedge db + p (\sum_{k \not
\in\gamma} t_{\gamma}db_{\gamma}+ r \sum_{k \not \in\delta} c_{\delta}
db_{k} \wedge db_{\delta})= w_{1} \wedge dp + p w_{2}
\end{equation*}
with $w_{1}, w_{2} $ in $\Omega_{F} M[X^{2}]$.

b) Let us assume $p \not \in M[X^{2}]$. Then $\mathcal{B} $ is part of a
2-basis of $N(p)$. Let $w \in\ker(\Omega_{F}^{m} M[X^{2}] \rightarrow
\Omega_{N(p)}^{m}) $ and set $w = \sum_{\gamma}a_{\gamma}db_{\gamma}\in
\Omega_{F}^{m} M[X^{2}]$, $\gamma\in\sum_{m}$, $a_{\gamma}\in M [X^{2}]$.
Thus $\sum\bar{a_{\gamma}}db_{\gamma}= 0 $ in $\Omega_{N}^{m} (p)$, and by
the remark above, we have $\bar{a_{\gamma}}= 0 $ in $N(p) $ for all $\gamma$. 
Thus $a_{\gamma}= p \cdot t_{\gamma}$, $t_{\gamma}\in N[X]$.

But $a_{\gamma}\in M [X^{2}] $ implies $D_{X_{\mu}} (a_{\gamma}) = 0 $ for
all $\mu\neq 1, \ldots, s-1$. Thus 
\begin{equation*}
D_{X_{\mu}} (p) \cdot t_{\gamma}+ p D_{X_{\mu}} (t_{\gamma}) = 0.
\end{equation*}
We have $p\not \in M[X^{2}]$, so that there is some $\mu\neq 1,\ldots,s-1$
with $D_{X_{\mu}}(p)\neq 0$. Choose this $\mu$ in the equation above. Then
it follows $p|D_{X_{\mu}}(p)t_{\gamma}$ in $N[X]$ for all $\gamma$. But $p$
does not divide $D_{X_{\mu}}(p)$, and hence $t_{\gamma}=ps_{\gamma}$, $%
s_{\gamma}\in N[X]$. Thus $a_{\gamma}=p^{2}s_{\gamma}$ for all $\gamma$, and
since $a_{\gamma}\in M[X^{2}]$, it follows $s_{\gamma}\in M[X^{2}]$ for all $%
\gamma$. Therefore $w=p^{2}s$ with some $s\in\Omega_{F}^{m}M[X^{2}]$. This
proves the lemma. $\square$\newline

Let $u\in \Omega_F M[X^2]$ be a form generated over $M[X^2]$ by forms defined over $F$.
The choice of any 2-basis $\mathcal B$ of $F$ enable us  to define $u^{[2]}$ (resp. $\wp(u)$) and this form is uniquely determined module $d\Omega_F M[X^2]$ (see remark (1.17)).
We are interested in the behavior of $u^{[2]}$ under the reduction homomorphism 
$\Omega_F M[X^2]\rightarrow \Omega_{N(p)}$, where $p$ is any irreducible polynomial in $N[X]$.
In particular we want to compare $\overline{u^{[2]}}$ with $\overline{u}^{[2]}$, where this last square is taken with respect to the 2-basis of $N(p)$ as defined at the beginning of this section.
In this case we have

\bigskip\noindent(3.3) \textbf{Lemma.} 
\emph{
$$\overline{u^{[2]}}-\overline{u}^{[2]}\in \overline{d\Omega_F M[X^2]}$$
}

\noindent(3.4) \textbf{Lemma.} 
\emph{Let $u,dv,\lambda\in\Omega_{F}^{\cdot}M[X^{2}]$ and $T\in M[X^{2}]$, defined by $T=\sum_{\mu\in
S_{n},\mu\neq1,\ldots,s-1}b^{\mu}X_{\mu}^{2}$. }

\emph{Assume 
\begin{equation*}
u^{[2]} + dv = \lambda\wedge db_{1} \wedge\cdots\wedge db_{r} \wedge d \bar{%
T }
\end{equation*}
in $\Omega_{N(p)}$, where $p \in N[X] $ is irreducible and monic and $b_{1},
\ldots, b_{r} \in F$. Then }

\begin{itemize}
\item[ a)]  \emph{If $p\in M[X^{2}]$, there exists $\delta ,u_{1},u_{2}\in
\Omega _{F}M[X^{2}]$ such that 
\begin{equation*}
u=\delta \wedge db_{1}\wedge \cdots \wedge db_{r}\wedge
dT+pu_{1}+u_{2}\wedge dp.
\end{equation*}
}

\item[ b)]  \emph{If $p\not\in M[X^{2}]$, there exists $\delta ,u_{1}\in
\Omega _{F}M[X^{2}]$ such that 
\begin{equation*}
u=\delta \wedge db_{1}\wedge \cdots \wedge db_{r}\wedge dT+p^{2}u_{1}.
\end{equation*}
}
\end{itemize}

\noindent\textbf{Proof.} Since $u^{[2]} + dv $ is closed, we can apply the
Cartier-operator to this form and we get $C(u^{[2]} + dv) = u $. Thus in $%
\Omega_{N (p)} $ 
\begin{equation*}
u = C( \lambda\wedge db_{1} \wedge\cdots\wedge db_{r} \wedge d \bar{T}) .
\end{equation*}
If $db_{1} \wedge\cdots\wedge db_{r} \wedge d \bar{T} = 0 $, then $u = 0 $
and the lemma follows from (3.2). Thus we can assume $db_{1}
\wedge\cdots\wedge db_{r} \wedge d \bar{T} \neq 0 $, in $\Omega_{N(p)} $,
and therefore we can take $\{ b_{1}, \ldots, b_{r}, \bar{T} \} $ as part of
a 2-basis of $N(p) $. Then (1.12) implies 
\begin{equation*}
u = \bar{\delta} \wedge db_{1} \wedge\cdots\wedge db_{r} \wedge d \bar{T}
\end{equation*}
with some $\bar{\delta} \in\Omega_{N(p)}$. We will show that $\bar{\delta}$
is contained in the image $\bar{\Omega}_{F}M[X^{2}] $ of $\Omega_{F}M[X^{2}]$
in $\Omega_{N(p)}$. It is clear that once we have this, the lemma follows
from (3.2).

a) Assume $p \in M[X^{2}] $. Then there is some $i_{0} \in I $ such that $%
\mathcal{B} \setminus\{ b_{i_{0}} \} \cup\{ \bar{X_{\mu}}$, $\mu\neq 1,
\ldots,s-1 \} $ is a 2-basis of $N(p) $ (we have chosen a 2-basis of $F$
including $b_{1}, \ldots, b_{r} ) $. We write in $\Omega_{N(p)} $ 
\begin{equation*}
\bar{\delta} = \bar{\delta}_{0} + \bar{\delta}_{1} \wedge d \bar{X}
\end{equation*}
where $\bar{\delta}_{0}$, $\bar{\delta}_{1}$ are forms not containing $d\bar{
X}= d\bar{X_{s}}$. Then 
\begin{equation*}
u = \delta_{0} \wedge db_{1} \wedge\cdots\wedge db_{r} \wedge d \bar{T} +
\delta_{1} \wedge d \bar{X} \wedge db_{1} \wedge\cdots\wedge db_{r} \wedge d 
\bar{T}
\end{equation*}
Since $u \in\Omega_{F} M[X^{2}] $ can not contain $d \bar{X} $ in its
expansion in this 2-basis of $N(p)$, we conclude that 
\begin{equation*}
u = \delta_{0} \wedge db_{1} \wedge\cdots\wedge db_{r} \wedge d \bar{T}
\end{equation*}
(Notice that the expansion of $db_{i_{0}} $ coming from $dp = 0 $ in $%
\Omega_{N(p)} $ does not contain $d \bar{X}$, because $p \in M[X^{2}]$).

We can proceed in the same way with the other variables $X _{\mu}, \mu\neq
1,\ldots,s-1$, and we finally obtain that $\delta_{0} $ is free from all
differentials $dX_{\mu}$, $\mu\neq 1, \ldots, s-1$. Thus $\delta_{0} $ is
generated over $N(p) $ by the differentials $db_{i}$, $i \in I \setminus\{
i_{0} \}$. We write now the coefficients of $\delta_{0} $ in the 2-basis
expansion. First we set $\delta_{0} = \delta^{\prime}_{0} + \bar{X}
\delta^{\prime\prime }_{0} $ where $\bar{X} $ appears in the coefficients of 
$\delta^{\prime}_{0}$, $\delta^{\prime\prime}_{0} $ only in even powers.
Then 
\begin{equation*}
u = \delta^{\prime}_{0} \wedge db_{1} \wedge\cdots\wedge db_{r} \wedge d 
\bar{T} + \bar{X} \delta^{\prime\prime}_{0} db_{1} \wedge\cdots\wedge db_{r}
\wedge d \bar{T}
\end{equation*}
The fact that $u $ is in $\bar{\Omega}_{F}M[X^{2}] $ implies that the
coefficients of $u $ (in the 2-basis expansion) do not contain odd power of $%
\bar{X}$. Comparing coefficients we obtain 
\begin{equation*}
u=\delta_{0}^{\prime}\wedge db_{1}\wedge\cdots\wedge db_{r}\wedge d\bar{T}
\end{equation*}
with $\delta_{0}^{\prime}$ free from odd powers of $\bar{X}$. Doing the same
with the other variables we finally conclude $u=\bar{\delta}\wedge
db_{1}\wedge\cdots\wedge db_{r}\wedge d\bar{T}$ with $\bar{\delta}\in \bar{
\Omega}_{F}M[X^{2}]$. From (3.2) follows the claim. The case (b) i.e. $p\not
\in M[X^{2}]$ can be treated in a similar way and we omit the proof. $%
\square $\newline

\noindent(3.5) \textbf{Lemma.} \emph{Let $p\in M[X^{2}]$ be irreducible and
monic (in $N[X])$. If 
\begin{equation*}
p u = v \wedge dp
\end{equation*}
in $\Omega_{F} M[X^{2}] $ with $u, v \in\Omega_{F} M[X^{2}] $, then there
exist $v_{1}, v_{2} \in\Omega_{F} M[X^{2}] $ with 
\begin{equation*}
v = pv_{1} + v_{2} \wedge dp.
\end{equation*}
}

\noindent\textbf{Proof.} Since $p \in M[X^{2}] $, we have $D_{X_{i}} (p) = 0 
$ for all $i \neq 1, \ldots, s-1 $. Thus 
\begin{equation*}
dp = \sum_{i \in I} D_{b_{i}} (p) db_{i}.
\end{equation*}
The fact that $p $ is irreducible implies $dp \neq 0 $ and hence there is
some $i_{0} \in I $ with $D_{b_{i_0}} (p) \neq 0 $. Let us write $%
p(X)=X^{2n}+\cdots=p_{0}+b_{i_0}p_{1},$ where $p_0,p_1$ are polynomials
whose coefficients do not contain odd powers of $b_{i_0}$ in the 2-basis
expansion. Hence $D_{b_{i_0}}(p) =p_{1}$ and $\deg_{X}p_1< \deg p$. In
particular $p \not| p_{1}$. Therefore one can find polynomials $\Delta, t
\in M[X^{2}] $ with $D_{b_{i_{0} }} (p) \Delta= 1 + p \cdot t $. Let us set $%
u = u_{0} + u_{1} \wedge db_{i_{0}}, v = v_{0} + v_{1} \wedge db_{i_{0}} $
with $u_{0}, u_{1}, v_{0}, v_{1} $ free from $db_{i_{0}}$. Thus 
\begin{equation*}
p( u_{0} + u_{1} \wedge db_{i_{0}}) = (v_{0} + v_{1} \wedge db_{i_{0}})
\wedge( D_{b_{i_{0}}}(p) db_{i_{0}} + \sum_{i \neq i_{0}} D_{b_{i}} (p)
db_{i})
\end{equation*}
implies 
\begin{eqnarray*}
pu_{0} & = & v_{0} \wedge\sum_{i\not=i_0} D_{b_{i}} (p) db_{i} \\ 
pu_{1} & = & D_{b_{i_{0}}}(p) v_{0} + v_{1} \wedge\sum_{i \neq i_{0}}
D_{b_{i}} (p) db_{i}.
\end{eqnarray*}
Taking modulo $p$ this equations, we obtain in $\Omega_{N(p)} $ 
\begin{equation*}
(D_{b_{i_{0}}}(p) v_{0} = v_{1} \wedge\sum_{i \neq i_{0}} D_{b_{i}} (p)
db_{i}
\end{equation*}
and since $\bar{\Delta} D_{b_{i_{0}}}(p) = 1$ in $N(p)$, it follows 
\begin{equation*}
v_{0} =\bar{\Delta} v_{1} \wedge\sum_{i \neq i_{0}} D_{b_{i}} (p) db_{i}.
\end{equation*}
But all these forms are contained in $\Omega_{F}M[X^2]$ so that (3.2) (a)
implies 
\begin{equation*}
v_{0} = \Delta v_{1} \wedge\sum_{i \neq i_{0}} D_{b_{i}} (p) db_{i}) +
pv_{3} + v_{4} \wedge dp
\end{equation*}
with $v_{3}$, $v_{4} \in\Omega_{F}M[X^{2}]$. Inserting $v_{0} $ in $v =
v_{0}+ v_{1} \wedge db_{i_{0}} $ we get 
\begin{eqnarray*}
v & = & \Delta v_{1} \wedge\sum_{i \neq i_{0}} D_{b_{i}} (p) db_{i} + pv_{3}
+ v_{4} \wedge dp + v_{1} \wedge db_{i_{0}} \\ 
 & = & v_{1} \wedge( \Delta\sum_{i \neq i_{0}} D_{b_{i}} (p) db_{i} +
db_{i_{0}}) + pv_{3} + v_{4} \wedge dp.
\end{eqnarray*}
Since $1 = \Delta Db_{i_{0}} (p) + p \cdot t $ in $M[X^{2}]$, we get 
\begin{eqnarray*}
v & = & v_{1}\wedge(\Delta dp+ptdb_{i_{0}})+pv_{3}+v_{4}\wedge dp \\ 
 & = & p(v_{1}\wedge tdb_{i_{0}}+v_{3})+(\Delta v_{1}+v_{4})\wedge dp
\end{eqnarray*}
which shows that $v$ has the desired form. $\square$ \newline

\noindent(3.6) \textbf{Lemma.} \emph{Let \;$dv$, $\lambda$ be forms in $%
\Omega_{F}M[X^{2}]$, $b_{1}, \ldots, b_{r} \in F $ be 2-independent and $T $
as before. Assume 
\begin{equation*}
dv = \lambda\wedge db_{1} \wedge\cdots\wedge db_{r} \wedge dT
\end{equation*}
in $\Omega_{N(p)}$, where $p $ is monic and irreducible. Then }
\begin{itemize}
\item[ a)]  \emph{If $p\in M[X^{2}]$, there exist $\delta $, $v_{1}$, $%
v_{2}\in \Omega _{F}M[X^{2}]$ such that 
\begin{equation*}
dv=\delta \wedge db_{1}\wedge \cdots \wedge db_{r}\wedge
dT+d(pv_{1}+v_{2}\wedge dp)
\end{equation*}
}
\item[ b)]  \emph{If $p\not\in M[X^{2}]$, there exist $\delta $, $v_{1}\in
\Omega _{F}M[X^{2}]$ such that 
\begin{equation*}
dv=\delta \wedge db_{1}\wedge \cdots \wedge db_{r}\wedge dT+p^{2}dv_{1}
\end{equation*}
}
\end{itemize}

\noindent\textbf{Proof.} If $db_{1}\wedge\cdots\wedge db_{r}\wedge dT=0$ in $%
\Omega_{N(p)}$, and therefore $dv=0$, we may use (3.2) to prove the lemma.
Hence we will assume $db_{1}\wedge\cdots\wedge db_{r}\wedge dT\neq 0 $ in $%
\Omega_{N(p)}$, and for the time being we set $T=b_{r+1}$. Thus $%
b_{1},\ldots,b_{r+1}$ can be chosen as part of a 2-basis of $N(p)$. From
(1.13) and (1.14) we infer in $N(p)$ 
\begin{equation*}
v=\sum_{\mu}b_{i}b^{\mu}z_{\mu}\wedge db_{1}\wedge\cdots\wedge\hat{db_{i} }
\wedge\cdots db_{r+1}+b^{\mbox{\bf 1}}u\wedge db_{1}\wedge\cdots\wedge
db_{r+1}+z
\end{equation*}
with $z_{\mu}$, $z$ closed forms in $\Omega_{N(p)}$, $\mu$ running over the
index set indicated by (1.13). We shall next prove that the form $z_{\mu}$, $%
u$, $z$ can be chosen in $\bar{\Omega}_F M[X^{2}]$. Let us first assume $p\in
M[X^{2}]$. We take a 2-basis of $N(p)$ of the form $\mathcal{B}
\setminus\{b_{i_{0}} \}\cup\{X_{\mu}\}$ where $X_{\mu}$ are all variable
involved $(X=X_{s})$. Let us fix some variable $X_{\mu}$ which we denote by $%
Y$. Thus in $N(p)$ we have 
\begin{equation*}
z_{\mu}= e_{\mu,1} + \bar{Y} e_{\mu, 2} + (e_{\mu, 3} + \bar{Y} e_{\mu, 4})
\wedge d \bar{Y}
\end{equation*}
with forms $e_{\mu, i} $ free from $d \bar{Y} $ and whose coefficients are
free from odd powers of $Y $. There are similar decompositions for $u, z $.
Since $z_{\mu}$ is closed we get 
\begin{equation*}
0 = d e_{\mu,1} + \bar{Y} d e_{\mu, 2} + e_{\mu, 2}\wedge d\bar{Y}+ d
e_{\mu, 3} \wedge d \bar{Y} + \bar{Y} d e_{\mu, 4} \wedge d \bar{Y}.
\end{equation*}
Thus we obtain 
\begin{equation*}
d e_{\mu,1} = d e_{\mu,4} = 0, \mbox{\hspace{0.5cm}} e_{\mu,2} = d e_{\mu,3}.
\end{equation*}
Inserting these expressions for $z_{\mu}$, $u $ and $z $ in the equation for 
$v, $ we get 
\begin{eqnarray*}
v & = & \sum_{\mu} b_{i} b^{\mu} z_{\mu, 1} \wedge db_{1} \wedge\cdots
\wedge \hat{db_{i}} \wedge\cdots\wedge db_{r+1} \\ 
& & +b^{\mbox{\bf 1}} u_{1} \wedge db_{1} \wedge\cdots\wedge db_{r+1} +
z_{1}
\end{eqnarray*}
when all $z_{\mu, 1} $ and $z_1$ are closed and moreover $z_{\mu, 1}, z_1 $
and $u_{1} $ are free from $d \bar{Y} $ and $\bar{Y}$. Doing the same with
the other variables, we finally conclude that the forms $z_{\mu}$, $u $ and $%
z $ can be taken in $\overline{\Omega_{F}}M[X^{2}] $.

Therefore we have $v - \sum_{\mu}b_{i} b^{\mu}z_{\mu}\wedge db_{1}
\wedge\cdots\wedge\hat{db_{i}} \wedge\cdots\wedge db_{r+1} - b^{%
\mbox
{\bf 1}} u \wedge db_{1} \wedge\cdots\wedge db_{r+1} - z \in\ker
(\Omega^{m}_F M[X^{2}] \rightarrow\Omega_{N(p)}^{m} ) $ and from (3.2) it
follows 
\begin{eqnarray*}
v & - & \sum_{\mu} b_{i} b^{\mu} z_{\mu} \wedge db_{1} \wedge\cdots\wedge 
\hat{db_{i}} \wedge\cdots\wedge db_{r+1}-b^{\mbox{\bf 1}}u \wedge db_{1}
\wedge\cdots\wedge db_{r+1}-z \\ 
& = & pv_{1} + v_{2} \wedge dp
\end{eqnarray*}
with some $v_{1}, v_{2} \in\Omega_F M[X^{2}] $. We apply now $d$ to this
relation and obtain (a).

Let us now assume $p\not \in M[X^{2}]$. Then these is some index $\mu_{0}$
such that $D_{X_{\mu_{0}}}(p)\neq 0$ and $\mathcal{B}\cup\{X_{\mu},\mu\neq
\mu_{0}\}\cup\{y\},$ where $y$ is the image of $X$ in $N(p)$, is a 2-basis
of $N(p)$. All variables $X_{\mu},\mu\neq\mu_{0}$ as well as $y$ can be
handled in the same way as in the first case, so that we are led to consider
only the variable $X_{\mu_{0}}$. Via the relation $dp=0$ we express $%
dX_{\mu_{0}}$ in terms of the other differentials, so that we may assume
that all $z_{\mu},u,z$ do not contain $dX_{\mu_{0}}$ too. Thus $\bar{X}
_{\mu_{0}}$ may appear in odd powers in the coefficients of $z_{\mu},u,z$.
Let us write $p=p_{0}+X_{\mu_{0}}p_{1}$with $p_{0},p_{1}$ not containing $%
X_{\mu_{0}}$ in odd powers. Thus in $N(p)$ we have $\bar{X}_{\mu_{0}}=\bar{p}
_{0}/\bar{p}_{1}$, so that replacing $\bar{X}_{\mu_{0}}$ by $\bar{p}_{0}/ 
\bar{p}_{1}$ in these coefficients, we get rid of the odd powers of $\bar{X}
_{\mu_{0}},$ but there appear again the variables $X_{\mu},\mu\neq\mu_{0},$
in these coefficients. We apply again the above procedure to get rid of the
odd powers of these variable in the forms $z_{\mu},u,z$. Therefore we may
assume $z_{\mu},u,z\in\Omega_{F}M[X^{2}]$. The assertion (b) follows again
from (3.2). $\square$\newline

\noindent(3.7) \textbf{Lemma}. \emph{Let $b_{1},\ldots,b_{r}\in F$ be
2-independent in $F$ and let $T$ be as before. Assume for a form $%
\lambda\in\Omega _{F}M[X^{2}]$ 
\begin{equation*}
\lambda\wedge db_{1} \wedge\cdots\wedge db_{r} \wedge dT = 0
\end{equation*}
in $\Omega_{N(p)}$. Then }

\begin{itemize}
\item[ a)]  \emph{If $p\in M[X^{2}]$ and $db_{1}\wedge \cdots \wedge
db_{r}\wedge dT\neq 0$ in $\Omega _{N(p)}$, there exist $\lambda
_{1},\lambda _{2}\in \Omega _{F}M[X^{2}]$ with 
\begin{equation*}
\lambda \wedge db_{1}\wedge \cdots \wedge db_{r}\wedge dT=(p\lambda
_{1}+\lambda _{2}\wedge dp)\wedge db_{1}\wedge \cdots \wedge db_{r}\wedge dT
\end{equation*}
in $\Omega _{F}M[X^{2}]$. }

\item[ b)]  \emph{If $p\not\in M[X^{2}]$ and $db_{1}\wedge \cdots \wedge
db_{r}\wedge dT\neq 0$ in $\Omega _{N(p)}$, there exist $\lambda _{1}\in
\Omega _{F}M[X^{2}]$ with 
\begin{equation*}
\lambda \wedge db_{1}\wedge \cdots \wedge db_{r}\wedge dT=p^{2}\lambda
_{1}\wedge db_{1}\wedge \cdots \wedge db_{r}\wedge dT
\end{equation*}
in $\Omega _{F}M[X^{2}]$. }

\item[ c)]  \emph{If $db_{1}\wedge \cdots \wedge db_{r}\wedge dT=0$ in $N(p)$,
there exists $t\in M[X^{2}]$ with $\deg _{X}t<\deg _{X}p$ and 
$db_{1}\wedge \cdots \wedge db_{r}\wedge dT=db_{1}\wedge \cdots \wedge
db_{r}\wedge d(pt)$ or $=db_{1}\wedge \cdots \wedge
d(pt)\wedge \cdots \wedge db_{r}\wedge dT$ for some $1\leq i\leq
r$, this case only occurs for $p\in M[X^2]$.}
\end{itemize}

\noindent\textbf{Proof.} Let us first assume $db_{1}\wedge\cdots\wedge
db_{r}\wedge dT\neq 0$ in $\Omega_{N(p)}$. 
Then $b_{1},\ldots,b_{r},\bar{T}$ can be chosen as part of a 2-basis of $N(p)$ 
and the assumption $\lambda\wedge db_{1}\wedge\cdots\wedge db_{r}\wedge d\bar{T}=0$ 
implies (see (1.1)) 
\begin{equation*}
\lambda=\sum_{1}^{r}\delta_{i}\wedge db_{i}+\delta\wedge d\bar{T}
\end{equation*}
with some forms $\delta_{i},\delta\in\Omega_{N(p)}$. 
Since $\lambda\in\bar{\Omega}M[X^{2}]$, one easily shows that the forms 
$\delta_{i}$, $\delta$ can be chosen in $\bar{\Omega_F}M[X^{2}]$ too. 
Then (3.2) implies 
\begin{equation*}
\lambda= \sum_{i=1}^{r} \delta_{i} \wedge db_{i} + \delta\wedge d {T} +
p \lambda_{1} + \lambda_{2} \wedge dp
\end{equation*}
if $p \in M[X^{2}] $ and $\lambda= \sum\delta_{i} \wedge db_{i} +
\delta\wedge dT + p^{2} \lambda_{2} $ if $p \not \in M[X^{2}]$, where 
$\lambda_{1}$, $\lambda_{2} \in\Omega_F M[X^{2}]$. 
Taking the product with $db_{1} \wedge \cdots\wedge db_{r} \wedge dT $ 
we obtain a) and b). 
Let us assume now $db_{1}\wedge\cdots\wedge db_{r} \wedge d \bar{T} = 0 $ in 
$\Omega_{N(p)}$. 
We write $b_{r+1} = \overline{T} $ for the time being, and we choose
(after reordering) a maximal 2-independent subset $\{ b_{1}, \ldots,
b_{j_{0}} \} $ of $\{ b_{1}, \ldots, b_{r+1} \}$. 
For example one could have a relation $\overline{T}=b_{r+1}= \sum_{\mu} p_{\mu}^{2} b_{\mu}$, where $\mu$ runs over the set of map 
$\mu: \{ 1, \ldots, j_{0} \} \rightarrow\{ 0, 1 \} $ and $p_{\mu}\in
N(p) $, $b_{\mu}= \prod_{i=1}^{j_{0}} b_{i}^{\mu(i)}$. 
Then we can write $T=\sum p_{\mu}^{2}b_{\mu}+p\cdot t$ in $N[X]$, 
with $p_{\mu},p,t\in N[X]$.
Since $T, p_{\mu}^{2} \in M[X^{2}] $, it follows $p \cdot t \in M[X^{2}]$,
and (3.1) implies $p, t \in M[X^{2}]$, or $t = p \cdot\ell$ with $\ell\in M[X^{2}]$. 
Thus we have 
\begin{equation*}
T = \sum p_{\mu}^{2} b_{\mu}+ p \cdot t, \,\,\, p,t \in M[X^{2}]
\end{equation*}
or 
\begin{equation*}
T = \sum p_{\mu}^{2} b_{\mu}+ p^{2} \cdot\ell, \,\,\, \ell\in M[X^{2}]
\end{equation*}
and $p \not \in M[X^{2}]$. But in the last case we see that $\deg_X(p^2 l)>\deg_X T$ and $\deg_X p_{\mu}^2 b_{\mu}$ (since $p_{\mu}$ can be chosen with $\deg_X p_{\mu}<\deg_X p$). This means that $p\not \in M[X^2]$ never happen. Therefore
\begin{equation*}
dT = \sum p_{\mu}^{2} db_{\mu}+ d (p t)
\end{equation*}
with $t \in M[X^{2}]$ and $\deg_X t < \deg_X p$. It
follows 
\begin{equation*}
db_{1}\wedge\cdots\wedge db_{r}\wedge dT=db_{1}\wedge\cdots\wedge
db_{r}\wedge d(pt).
\end{equation*}
The same argument applies for the case that some $b_{k}$ $(k\neq r+1)$ is
2-dependent of $\{b_{1},\ldots,b_{j_{0}}\}$. We omit the details. $\square$ 
\newline

\noindent(3.8) \textbf{Lemma.} \emph{Let $p \in M[X^{2}] $ be a monic
irreducible polynomial in $N[X]$. Let $u, v, \lambda\in\Omega_{F} M[X^{2}] $
be such that } 
\begin{equation*}
{\wp} u + dv = \lambda\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT
\mbox{\hspace{2cm} in } \Omega_{N(p)}
\end{equation*}
\emph{where $b_{1}, \ldots, b_{s-1} \in F^{*} $ and $T = b_{s} X^{2} + T^{\prime}$
as before $(\deg_{X} T^{\prime}= 0)$. 
Then there exist forms $u_1,u_2,\delta\in\Omega_FM[X^2]$, $f_{{ij}}$, 
$g_{{ij}}\in M[X^2]$ with $\deg(f_{ij}) < \deg(p) $ and $f_{{ij}} g_{{ij}} 
\equiv1 \;\;($ mod $p)$, such that }
\begin{eqnarray*}
u = \sum_{i} (g_{i1} df_{i1}) \wedge\cdots\wedge(g_{im} df_{im} ) & + & 
pu_{1} + u_{2} \wedge dp \\ 
& + & \delta\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT
\end{eqnarray*}
\emph{
holds in $\Omega_{F} M[X^{2}]$. } \newline

\noindent\textbf{Proof.} If $db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT=0$
in $\Omega_{N(p)}^{s}$, then $\wp u = dv$, and we conclude from (2.15)
\begin{equation*}
u = \sum_{i} \frac{df_{i1}}{f_{i1}} \wedge\cdots\wedge\frac{df_{im}}{f_{im}}
\end{equation*}
with certain $f_{ij} \in N(p)^{*}$.\newline
\noindent
Taking $g_{ij}\in N[X]$ with $g_{ij}f_{ij}\equiv 1\;\;($ mod $p)$, we obtain 
$u=\sum_{i}g_{i1}df_{i1}\wedge \cdots \wedge g_{is}df_{is}$. 
Since $u\in\Omega _{F}M[X^{2}]$ one can show that the $f_{ij}$ and 
$g_{ij}$ can be chosen in $M[X^{2}]$, and we can now apply (3.2). 
Thus we can assume $db_{1}\wedge \cdots \wedge db_{s-1}\wedge dT\neq 0$ in 
$\Omega _{N(p)}$, and hence $db_{1},\ldots ,db_{s-1},T$ are 2-independent in $N(p)$. 
We choose a 2-basis of $F$ which contains $\{b_{1},\ldots,b_{s-1}\}$ and we take the
constructed 2-basis of $N(p)$ for the case $p\in M[X^{2}]$ as indicated
after the proof of lemma (3.1). 
The excluded index $i_{0}$ can be chosen $\neq 1,\ldots ,s-1$. 
Moreover we replace the next element $b_{s}$ by the
image $\bar{T}$ of $T$ in $N(p)$ and we write $b_{s}$ for $\bar{T}$. 
Thus the new 2-basis of $N(p)$ is now $\{b_{1},\ldots ,b_{s-1}$, 
$b_{s}=\bar{T},\ldots ,\hat{b_{i_{0}}},\ldots ,b_{N},X_{\mu },\mu \neq 1,\ldots ,s-1\}$.
We order this basis such that all $b_{i}<X_{\mu }$ for all $\mu $ and all $i$. 
In $\Sigma _{m,N(p)}$ we choose the lexicographic ordering and with
respect to this ordering choose $\alpha $ minimal with $\alpha >\gamma $ for all 
$\gamma $ with $\gamma (1)=1,\ldots ,\gamma (s)=s$. 
Hence if $u\in \Omega_{N(p),<\alpha }^{m}$ then there exists some $\delta $ with 
$u=\delta \wedge db_{1}\wedge \cdots \wedge db_{s}$ and $\delta $ in the image of 
$\Omega _{F}M[X^{2}]$. 
We choose now $\beta \in \Sigma _{m,N(p)}$ minimal with the
property $u\in \Omega _{N(p),\beta }$. 
This means that in the representation of $u$ with respect to the above 2-basis of 
$N(p)$, $\beta $ corresponds to the leading index of this representation. 
Since $u$ comes from $\Omega_{F}M[X^{2}]$ we see that $b_{\beta (i)}\in M[X^{2}]$ 
for all $i$. 
Now we conclude that $N(p)_{\beta (i)}\subset \bar{M}[X^{2}]$.

If $\beta \leq \alpha $ we have $u=\delta \wedge db_{1}\wedge \cdots \wedge
db_{s}$ with $\delta $ in the image of $\Omega _{F}M[X^{2}]$, and we are
done by (3.2). 
Assume now $\alpha <\beta $. 
Then we have $\lambda \wedge db_{1}\wedge \cdots \wedge db_{s}\in 
\Omega _{N(p),<\beta }$ and hence 
\begin{equation*}
\wp (u)\in \Omega _{N(p),<\beta }+d\Omega _{N(p)}.
\end{equation*}
Applying Kato's lemma one gets 
\begin{equation*}
u=\frac{da_{1}}{a_{1}}\wedge \cdots \wedge \frac{da_{m}}{a_{m}}+u^{\prime }
\end{equation*}
in $\Omega _{N(p)}$ with $u^{\prime }\in \Omega _{N(p),<\beta }$, 
$a_{i}\in N(p)_{\beta (i)}\subset M[X^{2}]$. The form $u^{'}$ is contained in $\overline{\Omega_F} M[X^2]$ because $u$ and also $\frac{da_1}{a_1}\wedge \cdots\wedge \frac{da_n}{a_n}$ belong to $\overline{\Omega_F} M[X^2]$. We apply now the same procedure
to $u^{\prime }$ until we get 
\begin{equation*}
u=\sum \frac{df_{i,1}}{f_{i,1}}\wedge \cdots \wedge \frac{df_{i,m}}{f_{i,m}}
+v
\end{equation*}
with all $f_{i,j}\in \bar{M}[X^{2}]$ and $v\in \Omega _{N(p),<\alpha }$. By
the remark above we conclude $v=\delta \wedge db_{1}\wedge \cdots \wedge
db_{s}$ with $\delta $ in the image of $\Omega _{F}M[X^{2}]$. All $%
f_{i,j}\in M[X^{2}]$ can be chosen with $\deg _{X}f_{i,j}<\deg _{X}p$. Let $%
g_{i,j}\in M[X^{2}]$ with $f_{i,j}g_{i,j}\equiv 1(\mod p)$. Inserting in the
above equation and applying again lemma (3.2) we finally get the desired
result. $\square $\newpage

\section{The kernel $H^{n+1} (F (\protect\phi) /F)$}

In this section we will prove the main result of this paper, namely \newline

\noindent(4.1) \textbf{Theorem.} \emph{Let $\phi=\ll b_{1},\ldots,b_{n}\gg$
be an anisotropic bilinear Pfister form over $F$. Then for \;$m \geq n $ } 
\emph{
\begin{equation*}
H^{m+1} (F(\phi)/F) = \overline{\Omega_{F}^{m-n} \wedge db_{1} \wedge
\cdots\wedge db_{n} }.
\end{equation*}
If $m< n, $ then $H^{m+1} (F (\phi) / F) = 0$.}

\bigskip Let us write during this section $K $ for the function field $
F(\phi) $ of the conic $\phi= 0$. In section 2 we have shown that $w \in
H^{m+1} (F(\phi)/F) $ holds if and only if $w $ satisfies the following
equation in $\Omega_{L}^{m} $ 
\begin{equation}
w={\wp} u+dv+\lambda\wedge dT  \tag{4.2}
\end{equation}
with $u,v,\lambda\in\Omega_{F}M^{\prime}\subset\Omega_{L}$, where $%
M^{\prime} $ is the subfield $F(X_{\mu}^{2}|\mu\in S_{n})$ of $%
L=F(X_{\mu}|\mu\in S_{n}) $, and $\Omega_{F}M^{\prime}$ denotes the subspace
of $\Omega_{L}$ generated by the forms $db,b\in F$ over the field $%
M^{\prime} $. Recall that $T=\sum_{\mu\in S_{n}}b^{\mu}X_{\mu}^{2}$ and $%
S_{n}$ is the set of all maps $u:\{1,\ldots,n\}\rightarrow\{0,1\}$ with at
least one value $\mu(i)=1$. We will develop a descent procedure, which
starting from (4.2) will led us to an equation $w=\wp u+dv+\lambda\wedge
db_1\wedge\cdots\wedge db_n$ with $u,v,\lambda\in\Omega_F$ in $\Omega_F$.

Let us fix some integer $s $ with $1 \leq s \leq n$. Set $M = F
(X_{\mu}^{2}| \mu\in S_{n}, \mu\neq1, \ldots, s), X = X_{s} $ and let us
consider the equations in $\Omega_{L} $ 
\begin{equation}
w={\wp} u+dv+\lambda\wedge db_{1}\wedge\cdots\wedge db_{s-1}\wedge dT 
\tag{4.3}
\end{equation}
with $w\in\Omega_{F}$, $u$, $v$, $\lambda\in\Omega_{F}M(X^{2})$ and $%
T=b_{s}X^{2}+T^{\prime}$, $\deg_{X}T^{\prime}=0$ and $T^{\prime}$ is a
polynomial in $X_{\mu}^{2}$, $\mu\neq1,\ldots,s$ over $F$. The equation
(4.3) for $s=1$ is just (4.2). Our strategy is to start with (4.2) and to
push up the index $s$ until we get the factor $db_{1}\wedge\cdots\wedge
db_{n}$ and then to eliminate the rest of the variables until we get the
desired equation $w={\wp} u_{0}+dv_{0}+\lambda_{0}\wedge
db_{1}\wedge\cdots\wedge db_{n}$ in $\Omega_{F},$ which is obviously the
assertion of (4.1).

Any $u \in\Omega_{F} M(X^{2}) $ can be written in the form
\begin{equation}
u=u_{0}+\sum_{p}u_{p}  \tag{4.4}
\end{equation}
with $u_{0}\in\Omega_{F}M[X^{2}]$, $u_{p}\in p^{-\infty}\Omega_{F}M[X^{2}]$, 
$p$ running over all irreducible monic polynomials in $N[X]$, where $%
N=F(X_{\mu},\mu\neq1,\ldots,s)$. Recall that $p^{-\infty}\Omega_{F}M[X^{2}]$
denotes the space of forms $u/p^{r}$ with $u\in\Omega_{F}M[X^{2}]$ and $
\deg_{X}u<\deg_{X}(p^{r})$ (see section 2). Fixing a 2-basis of $F$ and $N$
we have seen that the operators ${\wp}$ and $d$ leave invariant the spaces $%
\Omega_{F}M[X^{2}]$, $p^{-\infty}\Omega_{F}M[X^{2}]$ (see (1.21)). Let us now insert in (4.3) the decompositions $u=u_{0}+\sum
u_{p},v=v_{0}+\sum v_{p}$ and $\lambda=\lambda_{0}+\sum\lambda_{p}$. The
terms $\lambda_{p}\wedge db_{1}\wedge\cdots\wedge db_{s-1}\wedge dT$ can
contribute eventually with an integral form (i.e. from $\Omega_{F}M[X^{2}])$
, which we will denote by $E_{p}$. Thus $\lambda_{p}\wedge
db_{1}\wedge\cdots\wedge db_{s-1}\wedge dT+E_{p}\in
p^{-\infty}\Omega_{F}M[X^{2}]$. Thus we conclude from (4.3), (1.20) and (1.21)

\begin{equation}
w=u_{0}^{[2]}+u_{0}+dv_{0}+\lambda_{0}\wedge db_{1}\wedge\cdots\wedge
db_{s-1}\wedge dT+\sum_{p}E_{p} \tag{4.5} \end{equation}
\begin{equation}
0=u_{p}^{[2]}+u_{p}+dv_{p}+\lambda_{p}\wedge db_{1}\wedge\cdots\wedge
db_{s-1}\wedge dT+E_{p}. \tag{4.6} 
\end{equation}

\bigskip\noindent(4.7) \textbf{Remark.}
As noticed in lemma (3.3), under the natural homomorphism
$$ \Omega_F M[X^2]\longrightarrow \Omega_{N(p)} $$
($p \in N[X]$ an irreducible polynomial), the operation $u^{[2]}$ behaves well, i.e., $\overline{u^{[2]}}- \overline{u}^{[2]}$ is contained in $\overline{d\Omega_F M[X^2]}$ provided the computations are done with respect to the 2-basis indicated there. In what follows we will frequently reduce modulo $p$ expressions of the form $u^{[2]}+dv$ and then we will lift back to $\Omega_F M[X^2]$. By the above remark, the differential form $dv$ will eventually change, but this will be of no importance for further computations. Because of this reason we will not explicitely mention these changes.

Our next goal is to study the forms $E_{p} \in\Omega_{F} M[X^{2}]$. To this
end we will distinguish three types of monic irreducible polynomials in $%
N[X] $, namely

\begin{itemize}
\item[ a)]  $p\not\in M[X^{2}]$ and $db_{1}\wedge \cdots \wedge
db_{s-1}\wedge dT\neq 0$ in $\Omega _{N(p)}$

\item[ b)]  $p\in M[X^{2}]$ and $db_{1}\wedge \cdots \wedge db_{s-1}\wedge
dT\neq 0$ in $\Omega _{N(p)}$

\item[ c)]  $db_{1}\wedge \cdots \wedge db_{s-1}\wedge dT=0$ in $\Omega
_{N(p)}$.
\end{itemize}

\noindent\textbf{Case (a):} We can write (see (4.6)) 
\begin{equation*}
E_{p} = \frac{\bar{u}^{[2]}}{p^{4r}} + \frac{\bar{u}}{p^{2r}} + \frac{d \bar{
v}}{p^{4r}} + \frac{\bar{\lambda}}{p^{4r}} \wedge db_{1} \wedge \cdots\wedge
db_{s-1} \wedge dT
\end{equation*}
with integral forms $\bar{u}, \bar{v}, \bar{\lambda} \in\Omega_{F} M[X^{2}] $
, and $r \geq0 $. 
For the time being, let us write $u, v, \lambda$ instead
of $\bar{u}, \bar{v}, \bar{\lambda} $. Then in $\Omega_{F} M[X^{2}] $ we get 
\begin{equation}
p^{4r}E_{p}=u^{[2]}+p^{2r}u+dv+\lambda\wedge db_{1}\wedge\cdots\wedge
db_{s-1}\wedge dT.  \tag{4.8}
\end{equation}

Next we show that the form $E_{p}$ can be absorbed by the first sum $%
u_{0}^{[2]}+u_{0}+dv_{0}+\lambda\wedge db_{1}\wedge\cdots\wedge
db_{s-1}\wedge dT$ in (4.5). 
Of course we can assume $r\geq 1$ in (4.8).
Taking (4.8) modulo $p$ we obtain in $\Omega_{N(p)}$ 
\begin{equation}
u^{[2]}+dv+\lambda\wedge db_{1}\wedge\cdots\wedge db_{s-1}\wedge dT=0. 
\tag{4.9}
\end{equation}

Since $p \not \in M[X^{2}]$, lemma (3.4) (b) implies 
\begin{equation*}
u = \delta\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT + p^{2} u_{1}
\end{equation*}
in $\Omega_{F} M[X^{2}]$. 
Inserting this expression in (4.8) we get 
\begin{equation}
p^{4r}E_{p}=p^{4}u_{1}^{[2]}+p^{2r+2}u_{1}+dv+\lambda\wedge db_{1}\wedge
\cdots\wedge db_{s-1}\wedge dT  \tag{4.10}
\end{equation}
with some new forms $v$ and $\lambda\in\Omega_{F}M[X^{2}]$. 
It follows $%
dv=\lambda\wedge db_{1}\wedge\cdots\wedge db_{s-1}\wedge dT$ in $%
\Omega_{N(p)}$ and lemma (3.5) implies 
\begin{equation*}
dv = \delta_{1} \wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT + p^{2}
dv_{1}
\end{equation*}
in $\Omega_{F} M[X^{2}]$. 
Thus (4.10) reads now 
\begin{equation*}
p^{4r} E_{p} = p^{4} u_{1}^{[2]} + p^{2r+2} u_{1} + p^{2} dv_{1} +
\delta^{\prime}\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT
\end{equation*}
and hence in $\Omega_{N(p)} $ we have 
\begin{equation*}
\delta^{\prime}\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT = 0.
\end{equation*}
%%% FROM PROBLEM:
%>From (3.6) (b) we conclude 
From (3.6) (b) we conclude 
\begin{equation*}
\delta^{\prime}\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT = p^{2}
\delta^{\prime\prime}\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT
\end{equation*}
with $\delta^{\prime\prime}\in\Omega_{F} M[X^{2}]$, since $db_{1}
\wedge\cdots\wedge db_{s-1} \wedge dT \neq0 $ in $\Omega_{N(p)} $. 
In this case we get 
\begin{equation*}
\begin{array}{rcl}
p^{4r} E_{p} & = & p^{4} u_{1}^{[2]} + p^{2r+2} u_{1} + p^{2} dv_{1} + p^{2}
\delta^{\prime\prime} \wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT \\ 
&  &  \\ 
p^{4r-2} E_{p} & = & p^{2} u_{1}^{[2]} + p^{2r} u_{1} + dv_{1} +
\delta^{\prime\prime} \wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT.
\end{array}
\end{equation*}
Again one obtains in $\Omega_{N(p)} $ 
\begin{equation*}
dv_{1} = \delta^{\prime\prime}\wedge db_{1} \wedge\cdots\wedge db_{s-1}
\wedge dT
\end{equation*}
and it follows $dv_{1} = \delta^{\prime\prime\prime}\wedge db_{1}
\wedge\cdots\wedge db_{s-1} \wedge dT + p^{2} dv_{2} $ with $%
\delta^{\prime\prime\prime}$, $v_{2} \in\Omega_{F} M[X^{2}]$. 
Repeating the last argument we finally obtain a relation 
\begin{equation*}
p^{4r-4} E_{p} = u_{1}^{[2]} + p^{2r-2} u_{1} + dv_{1} + \lambda_{1} \wedge
db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT
\end{equation*}
in $\Omega_{F} M[X^{2}]$, i.e. 
\begin{equation*}
E_{p} = {\wp} ( \frac{u_{1}}{p^{2r-2}}) + d (\frac{v_{1}}{p^{4r-4}}) + \frac{
\lambda_{1} }{p^{4r-4}} \wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT.
\end{equation*}
Thus we have reduced the number $r $ by one in (4.8). Iterating this process
we finally arrive at a relation with $r = 0$, i.e. 
\begin{equation*}
E_{p}={\wp}(u_{0}^{\prime})+dv_{0}^{\prime}+\lambda_{0}^{\prime}\wedge
db_{1}\wedge\cdots\wedge db_{s-1}\wedge dT
\end{equation*}
with $u_{0}^{\prime},v_{0}^{\prime},\lambda_{0}^{\prime}\in\Omega_{F}
M[X^{2}]$. 
This expression can be absorbed by the first integral part of
(4.5) and hence we have eliminated $E_{p}$ from this sum. \newline

\noindent\textbf{Case (b):} Thus we assume $p \in M[X^{2}] $ and $db_{1}
\wedge\cdots\wedge db_{s-1} \wedge dT \neq 0 $ in $\Omega_{N(p)}$. 
We can write 
\begin{equation*}
E_{p} = \frac{u^{[2]}}{p^{2r}} + \frac{u}{p^{r}} + \frac{dv}{p^{2r}} + \frac{
\lambda}{p^{2r}} \wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT
\end{equation*}
with some integer $r \geq 1 $. 
Thus 
\begin{equation}
p^{2r}E_{p}=u^{[2]}+p^{r}u+dv+\lambda\wedge db_{1}\wedge\cdots\wedge
db_{s-1}\wedge dT  \tag{4.11}
\end{equation}
with some forms $u,v,\lambda\in\Omega_{F}M[X^{2}]$. 
In $\Omega_{N(p)}$ we get 
\begin{equation*}
0 = u^{[2]} + dv + \lambda\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge
dT.
\end{equation*}
%%% FROM PROBLEM:
%>From lemma (3.4) (a) we conclude 
From lemma (3.4) (a) we conclude 
\begin{equation*}
u = \delta\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT + pu_{1} +
u_{2} \wedge dp
\end{equation*}
with $\delta, u_{1}, u_{2} \in\Omega_{F} M[X^{2}]$.

We insert now this expression in $E_{p} $ and obtain 
\begin{eqnarray*}
p^{2r} E_{p} & = & ( \delta\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge
dT)^{[2]} + p^{r} \delta\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT
\\ 
&  & +p^{2} u_{1}^{[2]} + p u_{2}^{[2]} \wedge dp + p^{r+1} u_{1} + p^{r}
u_{2} \wedge dp \\ 
&  & +dv + \lambda\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT \\ 
& = & p^{2} u_{1}^{[2]} + p u_{2}^{[2]} \wedge dp + p^{r+1}
u_{1} + p^{r} u_{2} \wedge dp \\ 
&  & +dv+ \lambda^{\prime}\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT
\end{eqnarray*}
with $\lambda^{\prime}= \lambda+ p^{r} \delta+ \delta^{\prime} $, where $%
\delta^{\prime}$ is some form in $\Omega_{F} M[X^{2}] $ with $( \delta\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT)^{[2]} = \delta^{\prime}\wedge
db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT $. 
Thus in $\Omega_{N(p)}$ it holds 
\begin{equation*}
dv = \lambda^{\prime}\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT
\end{equation*}
and lemma (3.5) implies 
\begin{equation*}
dv = \lambda^{\prime\prime}\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge
dT + d(pv_{1} + v_{2} \wedge dp)
\end{equation*}
with $\lambda^{\prime\prime}, v_{1}, v_{2} \in\Omega_{F} M[X^{2}]$.
Therefore 
\begin{eqnarray*}
p^{2r} E_{p} & = & p^{2} u_{1}^{[2]} + p u_{2}^{[2]} \wedge dp + p^{r} u_{2}
\wedge dp + p^{r+1} u_{1} \\ 
&  & +pdv_{3} + v_{3} \wedge dp + \lambda^{\prime\prime\prime}\wedge db_{1}
\wedge\cdots\wedge db_{s-1} \wedge dT
\end{eqnarray*}
where $v_{3} = v_{1} + dv_{2} $ and $\lambda^{\prime\prime\prime}$ is some
form in $\Omega_{F} M[X^{2}]$. 
Taking this equation modulo $p $ we obtain $%
\lambda^{\prime\prime\prime}\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge
dT = 0 $ in $\Omega_{N(p)}$. 
Since $db_{1} \wedge\cdots\wedge db_{s-1}
\wedge dT \neq0 $ in $\Omega_{N(p)}$, we obtain from lemma (3.6) (a) 
\begin{equation*}
\lambda^{\prime\prime\prime}\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge
dT = (p \lambda_{1} + \lambda_{2} \wedge dp) \wedge db_{1}
\wedge\cdots\wedge db_{s-1} \wedge dT
\end{equation*}
with $\lambda_{1}, \lambda_{2} \in\Omega_{F} M[X^{2}]$. 
Therefore 
\begin{eqnarray*}
p[p^{2r-1}E_{p}+pu_{1}^{[2]}&+&(u_{2}^{[2]}+ p^{r-1} u_{2}) \wedge
dp\\
& & + p^{r} u_{1} + dv_{3}+
\lambda_{1} \wedge db_{1} \wedge\cdots\wedge
db_{s-1} \wedge dT)] \\
&=& (v_{3} + \lambda_{2} \wedge db_{1} \wedge\cdots\wedge db_{s-1}\wedge
dT) \wedge dp.
\end{eqnarray*}
Lemma (3.4) implies 
\begin{equation*}
v_{3} + \lambda_{2} \wedge db_{1} \wedge\cdots\wedge db_{s-1}\wedge dT =
pv_{4} + v_{5} \wedge dp
\end{equation*}
with $v_{4}, v_{5} \in\Omega_{F} M[X^{2}]$. 
It follows 
\begin{eqnarray*}
p^{2r-1} E_{p} & + & p u_{1}^{[2]} + u_{2}^{[2]} \wedge dp + p^{r-1} u_{2}
\wedge dp + p^{r} u_{1} \\ 
& + & pdv_{4} + dv_{5} \wedge dp + \lambda_{3} \wedge db_{1} \wedge
\cdots\wedge db_{s-1} \wedge dT = 0
\end{eqnarray*}
where $\lambda_{3} = \lambda_{1} + d \lambda_{2}$. 
We get again in $\Omega_{N(p)}$ 
\begin{equation*}
\lambda_{3} \wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT = 0
\end{equation*}
and (3.6) (a) implies $\lambda_{3} \wedge db_{1} \wedge\cdots\wedge db_{s-1}
\wedge dT = ( p \lambda_{4} + \lambda_{5} \wedge dp) \wedge db_{1}
\wedge\cdots\wedge db_{s-1} \wedge dT $ in $\Omega_{F} M[X^{2}]$. 
Inserting this expression in the above equation we obtain 
\begin{equation}
\begin{array}{lcl}
p[p^{2r-2}E_{p} & + & u_{1}^{[2]}+p^{r-1}u_{1}+\lambda_{4}\wedge db_{1}
\wedge\cdots\wedge db_{s-1}\wedge dT+dv_{4}] \\ 
&  &  \\ 
& = & (u_{2}^{[2]}+p^{r-1}u_{2}+dv_{5}+\lambda_{5}\wedge db_{1}\wedge
\cdots\wedge db_{s-1}\wedge dT)\wedge dp
\end{array}
\tag{4.12}
\end{equation}
in $\Omega_{F}M[X^{2}]$. 
Lemma (3.4) implies 
\begin{equation*}
u_{2}^{[2]} + p^{r-1} u_{2} + dv_{5} + \lambda_{5} \wedge db_{1} \wedge
\cdots\wedge db_{s-1} \wedge dT = 0
\end{equation*}
in $\Omega_{N(p)}$.

We consider now two cases. \newline

\noindent\textbf{Case 1.} $r > 1 $. Then it holds in $\Omega_{N(p)} $, 
\begin{equation*}
u_{2}^{[2]} + dv_{5} + \lambda_{5} \wedge db_{1} \wedge\cdots\wedge db_{s-1}
\wedge dT=0
\end{equation*}
and we can apply (3.4) (a) to obtain 
\begin{equation*}
u_{2} = \lambda_{6} \wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT +
pu_{3} + u_{4} \wedge dp
\end{equation*}
in $\Omega_{F}M[X^{2}]$. Inserting this value of $u_{2} $ in the above
equation and using a similar argument as in case (a) we easily see that the
exponent of $p $ in (4.10) can be lowered. Therefore we are led to consider
the next case. \newline

\noindent\textbf{Case 2.} $r = 1 $. Then we have $\Omega_{N(p)}$, 
\begin{equation*}
u_{2}^{[2]} + u_{2} + dv_{5} + \lambda_{5} \wedge db_{1} \wedge\cdots\wedge
db_{s-1} \wedge dT = 0.
\end{equation*}

For the time being we will set $w_{s} = db_{1} \wedge\cdots\wedge db_{s-1}
\wedge dT$. From lemma (3.8) we conclude 
\begin{equation*}
u_{2} = \sum_{i} \bigwedge_{j} g_{ij} df_{ij} + pu_{3} + u_{4} \wedge dp +
\delta\wedge w_{s}
\end{equation*}
with $f_{ij}, g_{ij} \in M[X^{2}]$, $u_{3}, u_{4}, \delta\in\Omega_{F}
M[X^{2}] $ as indicated by the lemma. Therefore the right hand side of the
equation (4.12) reads now 
\begin{eqnarray*}
\left[ \sum_{i} \bigwedge_{j} (g_{ij} df_{ij})^{2} \right. \frac{df_{ij} }{
f_{ij}} & + & \sum_{i} \bigwedge_{j} g_{ij} f_{ij} \frac{df_{ij}}{f_{ij}} \\ 
& + & p^{2} u_{3}^{[2]} + pu_{3} + dv_{5} + \lambda_{0} \left. \wedge w_{s} 
\right] \wedge dp
\end{eqnarray*}
with certain form $\lambda_{0} \in\Omega_{F} M[X^{2}]$.

Since $g_{ij} f_{ij} \equiv 1 ($\;\; mod $p)$, it follows $\;\;\prod_{j} g_{ij}
f_{ij} = 1 + ph_{i} $ with some $h_{i} \in M[X^{2}] $ for all $i$. Thus
(4.12) implies 
\begin{eqnarray*}
p[E_{p} & + & u_{1}^{[2]} + u_{1} + \lambda_{4} \wedge w_{s} + dv_{4}] = [ (
\sum_{i} ph_{i} \frac{df_{i1}}{f_{i1}} \wedge\cdots\wedge\frac{df_{im}}{
f_{im}} )^{[2]} \\ 
& + & \sum ph_{i} \frac{df_{i1}}{f_{i1}} \wedge\cdots\wedge\frac{df_{im}}{
f_{im}} +\wp(pu_{3}) + dv_{5} + \lambda_{0} \wedge w_{s} ] \wedge dp.
\end{eqnarray*}

But one easily sees that 
\begin{eqnarray*}
{\wp}(pu_{3}) \wedge dp &\equiv& p {\wp} (u_{3} \wedge dp)\\
 & &  \\
{\wp}\left[\sum_{i} ph_{i} \wedge\frac{df_{i1}}{f_{i1}} \wedge\cdots\wedge 
\frac{df_{im}}{f_{im}}\right]\wedge dp 
&\equiv& p {\wp}\left[dp \wedge\sum_{i} h_{i} \frac{
df_{i1}}{f_{i1}} \wedge\cdots\wedge\frac{df_{im}}{f_{im}} \right]
\end{eqnarray*}
$(\mod d\Omega_F M[X^2]).$

We bring both terms to the left side of the above equation and we get 
\begin{eqnarray*}
p[E_{p} + {\wp} (u_{1} + u_{3} \wedge dp & + & \sum_{i} h_{i} \frac{df_{i1} 
}{f_{i1}} \wedge\cdots\wedge\frac{df_{im}}{f_{im}} \wedge dp) \\ 
& + & \lambda_{4} \wedge w_{s} + dv_{4}] = [dv_{5} + \lambda_{0} \wedge
w_{s}] \wedge dp.
\end{eqnarray*}
Thus 
\begin{eqnarray*}
E_{p} & = & {\wp} (u_{1} + u_{3} \wedge dp + \sum_{i} h_{i} \frac{df_{i1} }{
f_{i1}} \wedge\cdots\wedge\frac{df_{im}}{f_{im}} \wedge dp) \\ 
&  & + d(v_{4} + v_{5} \wedge\frac{dp}{p}) + ( \lambda_{4} + \lambda_{0}
\wedge\frac{dp}{p}) \wedge w_{s}.
\end{eqnarray*}
Notice that $u_{1} + u_{3} \wedge dp + \sum_{i} h_{i} \frac{df_{i1}}{f_{i1}}
\wedge\cdots\wedge\frac{df_{im}}{f_{im}} \wedge dp $ contains denominators
all of whose prime factors in $M[X^{2}] $ (and in $N[X]) $ are of degree $<
\deg(p)$. 
The other involved forms have at most $p $ in the denominator.
Using a similar argument as done at the beginning of this section we get 
\begin{equation*}
E_{p} = {\wp} (u^{\prime}_{0}) + dv^{\prime}_{0} + \lambda^{\prime}_{0}
\wedge w_{s} + G_{p} + \sum_{q \deg (q) < \deg(p)} G_{q}
\end{equation*}
with $u^{\prime}_{0}, v^{\prime}_{0}, \lambda^{\prime}_{0} \in\Omega_{F}
M[X^{2}] $ and integral forms $G_{p}, G_{q} $ of the type 
\begin{eqnarray*}
G_{p} & = & dv_{p} + \lambda_{p} \wedge w_{s} \\ 
G_{q} & = & {\wp} (u_{q}) + dv_{q} + \lambda_{q} \wedge w_{s}
\end{eqnarray*}
where $v_{p}, \lambda_{p} \in\frac{1}{p} \Omega_{F} M[X^{2}] $ and $u_{q},
v_{q}, \lambda_{q} \in q^{- \infty} \Omega_{F} M[X^{2}] $ if $\deg(q) <
\deg(p)$.

Since the forms $G_{q}$ have denominators of degree $<\deg(p)$ and are of
the same type as $E_{q},$ we can add them to the $E_{q}$'s, so that we
forget then now. 
Let us consider $G_{p}$. 
Let us write 
\begin{equation*}
v_{p} = \frac{v^{\prime}}{p^{2}} ,\mbox{\hspace{0.8cm}} \lambda_{p} = \frac{\lambda^{\prime}}{p} %
\mbox{\hspace{0.8cm} with \hspace{0.8cm} } v^{\prime},
\lambda^{\prime}\in\Omega_{F} M[X^{2}].
\end{equation*}

Thus $G_{p} = \frac{dv^{\prime}}{p^{2}} + \frac{\lambda^{\prime}}{p} \wedge
w_{s} $ and hence 
\begin{equation*}
p^{2}G_{p}=dv^{\prime }+p\lambda�\wedge w_{s}
\end{equation*}
holds in $\Omega _{F}M[X^{2}]$. 
Thus $dv^{\prime }=0$ in $\Omega _{N(p)}$.
Using Cartier's theorem (see \cite{Ca} or section 1) we can write $v^{\prime
}=A^{[2]}+dB$ in $\Omega _{N(p)}$. 
Since $v^{\prime }\in \bar{\Omega}
_{F}M[X^{2}]=\mbox{ Image of }\Omega _{F}M[X^{2}]$ in $\Omega _{N(p)}$, we
easily see that $B$ also can be chosen in $\bar{\Omega}_{F}M[X^{2}]$.
Therefore in $\Omega _{F}M[X^{2}]$ (see (3.2)) 
\begin{equation*}
v^{\prime}= A^{[2]} + dB + pv_{1} + v_{2} \wedge dp
\end{equation*}
with $v_{1}, v_{2} \in\Omega_{F}M[X^{2}]$. 
It follows $dv^{\prime}=
d(pv_{1} + v_{2} \wedge dp) = d(p (v_{1} + dv_{2})) = d(pv^{\prime}_{1}) $
where $v^{\prime}_{1} = v_{1} + dv_{2}$. Therefore in $\Omega_{F}M[X^{2}]$ 
\begin{equation*}
p[pG_{p} + dv^{\prime}_{1} + \lambda\wedge w_{s}] = v^{\prime}_{1} \wedge dp
\end{equation*}

Lemma (3.4) implies $v^{\prime}_{1} = pv_{2} + v_{3} \wedge dp $ in $%
\Omega_{F} M[X^{2}] $ and hence after replacing $v^{\prime}_{1} $ in the
above equation it follows 
\begin{equation*}
pG_{p} = pdv_{2} + dv_{3} \wedge dp + \lambda\wedge w_{s}.
\end{equation*}

Thus $\lambda\wedge w_{s} = 0 $ in $\Omega_{N(p)}$. Lemma (3.6) implies $%
\lambda\wedge w_{s} = (p \lambda_{1} + \lambda_{2} \wedge dp) \wedge w_{s} $
in $\Omega_{F} M[X^{2}] $ and therefore 
\begin{eqnarray*}
p G_{p} = pdv_{2} + dv_{3} \wedge dp + p \lambda_{1} \wedge w_{s} +
\lambda_{2} \wedge dp \wedge w_{s}\\
   \\
p [G_{p} + dv_{2} + \lambda_{1} \wedge w_{s}] = (dv_{3}+\lambda_{2}\wedge
w_{s})\wedge dp.
\end{eqnarray*}
We apply again (3.4) and obtain $dv_{3}+\lambda_{2}\wedge
w_{s}=pv_{4}+v_{5}\wedge dp$ in $\Omega_{F} M[X^{2}]$. Hence in $%
\Omega_{N(p)}$ it holds $dv_3=\lambda_2\wedge w_s$. Applying (3.5) we obtain 
$dv_3=\delta\wedge w_s+d(pv_6+v_7\wedge dp)$. 
This implies 
\begin{equation*}
p[G_{p}+dv_{2}+\lambda_{1}\wedge w_{s}]=pdv_{6} \wedge dp+\delta^{\prime
}\wedge w_{s}\wedge dp
\end{equation*}
with $\delta^{\prime}=\lambda_{2}+\delta$. Lemma (3.4) implies 
$\delta ^{\prime}\wedge w_{s} = p \lambda_{3} + \lambda_{4} \wedge dp $ in 
$\Omega_{p} M[X^{2}]$. 
Hence $\delta^{\prime}\wedge w_{s} \wedge dp = p\lambda_{3} \wedge dp$. 
But we are assuming that $w_{s} \neq 0 $ in $\Omega_{N(p)}$. 
Thus we easily conclude that there exist some $\lambda^{\prime}_{3} \in\Omega_{F} M[X^{2}] $ with $\delta^{\prime}\wedge
w_{s} \wedge dp = p \lambda^{\prime}_{3} \wedge w_{s} \wedge dp$. 
Inserting
this in the above relation we get 
\begin{eqnarray*}
p[G_{p} + dv_{2} + \lambda_{1} \wedge w_{s}] = pdv_{6} \wedge dp + p
\lambda^{\prime}_{3} \wedge w_{s} \wedge dp \\ 
\\
G_{p} = d(v_{2} + v_{6} \wedge dp) + ( \lambda_{1} +
\lambda^{\prime}_{3} \wedge dp) \wedge w_{s}
\end{eqnarray*}
in $\Omega_{F} M[X^{2}]$. 
Thus we also get rid of $G_{p} $ in the equation for $E_{p}$. \newline

\noindent\textbf{Case (c).} We assume $db_{1} \wedge\cdots\wedge db_{s-1}
\wedge dT = 0 $ in $\Omega_{N(p)}$. Thus we assume now $p\in M[X^{2}]$. 
In $\Omega_{F} M[X^{2}] $ we can write 
\begin{equation}
p^{2h}E_{p}=u^{[2]}+p^{h}u+dv+\lambda\wedge w_{s}  \tag{4.13}
\end{equation}
with $h\geq1,$ with $u,v,\lambda\in\Omega_{F}M[X^{2}]$ and $w_{s}=db_{1}
\wedge\cdots\wedge db_{s-1}\wedge dT$. We claim that there is $t\in M[X^{2}]$
, $\deg_{X}t<\deg_{X}p$ with 
\begin{equation}
t^{2}p^{2h-2}E_{p}=u^{\prime}{}^{[2]}+tp^{h-1}u^{\prime}+dv^{\prime}+
\lambda^{\prime}\wedge w_{s}  \tag{4.14}
\end{equation}
with $u^{\prime},v^{\prime},\lambda^{\prime}\in\Omega_{F}M[X^{2}]$.

%%% FROM PROBLEM:
%>From (4.13) it is clear that we have reduced the exponent of $p$ in the
From (4.13) it is clear that we have reduced the exponent of $p$ in the
denominators of $E_{p}$ at the cost that we increase the number of terms of
type $E_{q}$ but with $\deg _{X}(q)<\deg _{X}(p)$. Hence, iterating this
process and using partial fractions decompositions, we can finally eliminate 
$E_{p}$ from the original equation (4.5). Of course we have during this
process to modify the other terms $E_{q}$ for irreducibles $q$ with $\deg
_{X}(q)<\deg _{X}(p).$

The assumption $w_{s} = db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT = 0 $
in $\Omega_{N(p)}$ implies (see (3.7) (c)) 
\begin{equation*}
w_{s} = db_{1} \wedge\cdots\wedge db_{s-1} \wedge d(pt)
\end{equation*}
or 
\begin{equation*}
w_{s} = db_{1} \wedge\cdots\wedge db_{i-1} \wedge d(pt) \wedge db_{i+1}
\wedge\cdots\wedge db_{s-1} \wedge dT
\end{equation*}
with $t \in M[X^{2}]$, $\deg_{X} t < \deg_{X} p, $ and some $i$, $1 \leq i
\leq s-1$. 
Let us write $w_{s} = w_{s-1} \wedge d(pt) $ where $w_{s-1} $
denotes $db_{1} \wedge\cdots\wedge db_{s-1} $ or $db_{1} \wedge\cdots \wedge 
\hat{i} \wedge\cdots\wedge db_{s-1} \wedge dT$. 
We will assume $w_{s-1}\neq 0$  in $\Omega_{N(p)}$, and we will omit the proof in the case $w_{s-1}=0$, which can be treated similarly. 
Since we can assume $h \geq1$, we get from (4.13) 
\begin{equation*}
u^{[2]} + dv = 0
\end{equation*}
in $\Omega_{N(p)}$. It follows $u = 0 $ in $\Omega_{N(p)} $ and hence $u
=pu_{1} + u_{2} \wedge dp $ with $u_{1}, u_{2} \in\Omega_{F} M[X^{2}] $ (see
(3.2)). 
Then 
\begin{equation*}
p^{2h} E_{p} = p^{2} u_{1}^{[2]} + pu_{2}^{[2]} \wedge dp + p^{h+1} u_{1} +
p^{h} u_{1} \wedge dp + dv + \lambda\wedge w_{s}.
\end{equation*}

This implies $dv = 0 $ in $\Omega_{N(p)} $ and hence $dv = d(p v_{1}) $ with
some $v_{1} \in\Omega_{F} M[X^{2}] $ (see proof of case (2) above).
Replacing this value of $dv $ in the above equation it follows 
\begin{eqnarray*}
p[p^{2h-1} E_{p} + pu_{1}^{[2]} + u_{2}^{[2]} \wedge dp + p^{h} u_{1} +
p^{h-1} u_{2} \wedge dp \\ 
\\ 
+ dv_{1} + \lambda\wedge w_{s-1} \wedge dt] = (v_{1} + t \lambda\wedge
w_{s-1}) \wedge dp.
\end{eqnarray*}
Lemma (3.4) implies 
\begin{equation*}
v_{1} = t \lambda\wedge w_{s-1} + pv_{2} + v_{3} \wedge dp
\end{equation*}
with $v_2,v_3\in \Omega_FM[X^2]$. 
Thus 
\begin{eqnarray*}
p[p^{2h-2} E_{p} + u_{1}^{[2]} + p^{h-1} u_{1} + dv_{2}] = \\ 
\\ 
td \lambda\wedge w_{s-1} + [u_{2}^{[2]} + p^{h-1} u_{2} + dv_{3}] \wedge dp.
\end{eqnarray*}
Hence 
\begin{equation*}
t d \lambda\wedge w_{s-1} = 0
\end{equation*}
in $\Omega_{N(p)}$. 
Since $\deg_{X} t < \deg_{X} p, $ it follows $t \neq 0$
in $N(p)$ and we get 
\begin{equation*}
d \lambda\wedge w_{s-1} = 0.
\end{equation*}

This implies $d \lambda\wedge w_{s-1} = d (p \lambda_{1} ) \wedge w_{s-1} $
in $\Omega_{F} [X^{2}]$. 
Therefore 
\begin{eqnarray*}
p[p^{2h-2} E_{p} + u_{1}^{[2]} + p^{h-1} u_{1} + dv_{2} + t d \lambda_{1}
\wedge w_{s-1}] \\ 
= [ u_{2}^{[2]} + p^{h-1} u_{2} + dv_{3} + t \lambda_{1} \wedge w_{s-1}]
\wedge dp.
\end{eqnarray*}
Lemma (3.4) implies 
\begin{equation}
u_{2}^{[2]}+p^{h-1}u_{2}+dv_{3}=t\lambda_{1}\wedge w_{s-1}  \tag{4.15}
\end{equation}
in $\Omega_{N(p)}$. 
Thus we are led to consider two cases

\medskip\noindent a) $h > 1$. 
Then 
\begin{equation*}
u_{2}^{[2]} + dv_{3} = t \lambda_{1} \wedge w_{s-1}
\end{equation*}
implies (see (3.4)) 
\begin{equation*}
u_{2} = \delta\wedge w_{s-1} + pu_{3} + u_{4} \wedge dp
\end{equation*}
with $\delta, u_{3}, u_{4} \in\Omega_{F} M[X^{2}]$. 
Inserting this expression for $u_{2} $ in the equation for $E_{p} $ we obtain 
\begin{eqnarray*}
p[p^{2h-2} E_{p} + u_{1}^{[2]} + p^{h-1} u_{1} + dv_{2} + t d \lambda_{1}
\wedge w_{s-1} + pu_{3}^{[2]} \wedge dp + p^{h-1} u_{3} \wedge dp] \\ 
\\ 
= [(\delta\wedge w_{s-1})^{[2]} + p^{h-1} \delta\wedge w_{s-1} + dv_{3} + t
\lambda_{1} \wedge w_{s-1}] \wedge dp.
\end{eqnarray*}
It follows
\begin{equation*}
( \delta\wedge w_{s-1})^{[2]} + dv_{3} + t \lambda_{1} \wedge w_{s-1}\in \ker[\Omega^m_F M[X^2]\rightarrow  \Omega^m_{N(p)}]
\end{equation*}
which means that 
\begin{equation*}
dv_{3} = z \wedge w_{s-1}+pu_4+u_5\wedge dp
\end{equation*}
in $\Omega^m_F M[X^2]$. This relation says $dv_3\in \Omega^{m-s+1}_{N(p)}\wedge w_{s-1}$. 

We apply now lemma (1.13) and remark (1.14), and we obtain in $\Omega^m_F M[X^2]$
\begin{equation*}
dv_{3} = \left( \sum_{\mu\neq{\large 1}} b^{\mu}z_{\mu}+ b^{{\large 1}} dB
\right) \wedge w_{s-1} + d(pu_{4} + u_{5} \wedge dp)
\end{equation*}
with $z_{\mu}, B \in\Omega_{F} M[X^{2}], dz_{\mu}= 0 $ for all $\mu$. 
Notice that the form $( \sum_{\mu\neq 1} b^{\mu}z_{\mu}+ b^{1} dB) \wedge w_{s-1}$
is exact, i.e. equal to some $dH $ with $H \in\Omega_{F} M[X^{2}]$.
Inserting this in the above equation it follows 
\begin{eqnarray*}
p[p^{2h-2} E_{p} + u_{1}^{[2]} + p^{h-1} u_{1} + dv_{2} + t d \lambda_{1}
\wedge w_{s-1} + (pu_{3}^{[2]}+ p^{h-1} u_{3}+ du_{4})
\wedge dp] \\ 
\\
 = [ ( \delta\wedge w_{s-1})^{[2]} + p^{h-1} \delta\wedge w_{s-1} + t
\lambda_{1} \wedge w_{s-1} + dH] \wedge dp.
\end{eqnarray*}

Let us denote by $A$ the expression inside the parenthesis of the right hand
side of this equation, which is of the form $A = B \wedge w_{s-1}$, with
some $B \in\Omega_{F} M[X^{2}]$. 
The equation $p [\mbox{  } ] = B \wedge
w_{s-1}\wedge dp $ implies by lemma (3.4) $B \wedge w_{s-1} = 0 $ in $\Omega_{N(p)}$. 
Now we apply lemma (3.6) to conclude $B \wedge w_{s-1} = (p\mu_{1} + \mu_{2} \wedge dp) \wedge w_{s-1} $ in $\Omega_{F} M[X^{2}]$.
Inserting in the above relation we obtain then that we can assume $B\wedge w_{s-1} = p \mu_{1}\wedge w_{s-1} $ with $\mu_{1} \in\Omega_{F}M[X^{2}]$. 
Thus let us write $p B \wedge w_{s-1} $ instead of $B \wedge w_{s-1}$. 
Multiplying the above equation by $t$ and using $tdp=d(pt)+pdt$ as well as $w_{s-1}\wedge d(pt) = w_{s}$, we get 
\begin{eqnarray*}
p^{2h-2} E_{p} + u_{1}^{[2]} + p^{h-1} u_{1} + dv_{2} + t d \lambda_{1}
\wedge w_{s-1} + (pu_{3}^{[2]} + p^{h-1} u_{3} + du_{4})
\wedge dp \\
\\
= \frac{p A \wedge dt}{pt} + \frac{p B \wedge w_{s-1}}{pt} = A \wedge\frac
{dt}{t} + \frac{B}{t} \wedge w_{s-1}.
\end{eqnarray*}
This equation implies 
\begin{eqnarray*}
E_{p} & = & {\wp} \left( \frac{u_{1}}{p^{h-1}} \right) + {\wp} \left( \frac{
u_{3} \wedge dp}{p^{h-1}} \right) + d \left( \frac{v_{2} + u_{2} \wedge dp}{
p^{2(h-1)}} \right) \\
& & \\
& & +{\wp} \left( \frac{\delta\wedge w_{s-1}}{p^{h-1}} \wedge\frac{dt}{t}
\right) + d \left( \frac{H}{p^{2(h-1)}} \wedge\frac{dt}{t} \right) \\
& & \\
& & +\frac{td \lambda_{1} \wedge w_{s-1} + t \lambda_{1} \wedge w_{s-1}
\wedge\frac{dt}{t}}{p^{2h-2}} + \frac{1}{p^{2h-2}t} B \wedge w_{s}.
\end{eqnarray*}

Notice $t d \lambda_{1} \wedge w_{s-1} + \lambda_{1} \wedge w_{s-1} \wedge d
t = d ( t \lambda_{1} \wedge w_{s-1})$, so that 
\begin{eqnarray*}
E_{p} & = & {\wp} \left( \frac{u_{1}}{p^{h-1}} + \frac{u_{3} \wedge dp}{
p^{h-1}} + \frac{\delta\wedge w_{s-1} \wedge dt}{p^{h-1} t} \right) \\
& & \\
&  & + d \left( \frac{v_{2} + u_{4} \wedge dp}{p^{2h-2}} + \frac{(t
\lambda_{1}) \wedge w_{s-1}}{p^{2h-2}} + \frac{H \wedge dt}{p^{2h-2} t}
\right) \\
& & \\
&  & + \frac{B}{p^{2h-2} t} \wedge w_{s}.
\end{eqnarray*}
Thus 
\begin{equation*}
E_{p}={\wp}\left( \frac{u^{\prime}}{p^{h-1}t}\right) +d\left( \frac{
v^{\prime}}{p^{2(h-1)}t}\right) +\frac{\lambda^{\prime}}{tp^{2h-2}}\wedge
w_{s}
\end{equation*}
with forms $u^{\prime}$, $v^{\prime}$, $\lambda^{\prime}\in
\Omega_{F}M[X^{2}] $. Using partial fraction decomposition of the forms $%
u^{\prime}/(p^{h-1}t)$, $v^{\prime}/(p^{2(h-1)}t)$, $\lambda^{
\prime}/(p^{2h-2}t)$, we see that the exponent $h$ of $p$ in equation (4.13)
can be reduced by one, although expressions for polynomials $q$ of lower
degree of the same type can appear, which will be absorbed by the
corresponding $E_{q}$. Thus we are led to consider the case \newline

\noindent(b) $h = 1$. Then we have (see (4.15)) 
\begin{equation}
u_{2}^{[2]}+u_{2}+dv_{3}=t\lambda_{1}\wedge w_{s-1}  \tag{4.16}
\end{equation}
in $\Omega_{N(p)}$. We can now apply lemma (3.7) to conclude 
\begin{equation*}
u_{2} = \sum_{i} \bigwedge_{j} g_{ij} df_{ij} + pv_{1} + v_{2} \wedge dp +
\delta\wedge w_{s-1}
\end{equation*}
with $v_{1}, v_{2}, \delta\in\Omega_{F} M[X^{2}]$, $f_{ij}, g_{ij} \in
M[X^{2}]$, $\deg_{X} f_{ij} < \deg_{X} p$, $f_{ij} g_{ij} = 1 + h_{ij} p $
in $M[X^{2}]$. 
Next we compute ${\wp}(\sum_{i}\bigwedge_{j} g_{ij}df_{ij})$. 
We have 
\begin{eqnarray*}
{\wp}(\sum_{i} \bigwedge_{j} g_{ij} df_{ij}) & \equiv & \left[ \sum_{i}
\bigwedge_{j} g_{ij} f_{ij}) \frac{df_{ij}}{f_{ij}} \right] ^{[2]} +
\sum_{i} \bigwedge_{j} g_{ij} f_{ij} \frac{df_{ij}}{f_{ij}} \\ 
&  &  \\ 
& \equiv & \sum_{i} ( \prod_{j} (g_{ij} f_{ij})^{2} + \prod_{j} g_{ij} f_{ij})
\bigwedge_{j} \frac{df_{ij}}{f_{ij}} \\ 
&  &  \\ 
& \equiv & \sum_{i} ( \prod_{j} g_{ij}^2 f_{ij} + \prod_{j} g_{ij} )
\bigwedge_{j} df_{ij} \\ 
&  &  \\ 
& \equiv & \sum_{i} \prod_{j} g_{ij} (1 + \prod_{j} (1 + ph_{ij})) \bigwedge_{j}
df_{ij} \\ 
&  &  \\ 
& \equiv & \sum_{i} p \prod_{j} g_{ij} \cdot h_{i} \bigwedge_{j} df_{ij}
\end{eqnarray*}
$\mod d\Omega_F M[X^2]$, 
with certain $h_{i} \in M[X^{2}]$. Thus we have obtained 
\begin{equation*}
{\wp} ( \sum_{i} \bigwedge_{j} g_{ij} df_{ij}) = p \sum_{i} g_{i}
\bigwedge_{j} df_{ij}\mbox{\hspace{1.5cm}} \mod\;\;\; d\Omega_F M[X^2] 
\end{equation*}
with $g_{i} \in M[X^{2}]$. 
Let us insert $u_{2}$ in (4.15). According to the above computation we obtain in $\Omega_{F} M[X^2] $ 
\begin{equation*}
{\wp} ( \delta\wedge w_{s-1}) + dv_{3} + t \lambda_{1} \wedge w_{s-1}=pu_4+u_5\wedge dp 
\end{equation*}
and this implies 
\begin{equation*}
dv_{3} = \lambda\wedge w_{s-1}
\end{equation*}
in $\Omega_{N(p)}$, with some form $\lambda\in\bar{\Omega}_{F} M[X^{2}] \subset\Omega_{N(p)}$.
%%% FROM PROBLEM:
%>From lemma (1.13) and remark (1.14) we obtain in $\Omega_{F} M[X^{2}]$ 
From lemma (1.13) and remark (1.14) we obtain in $\Omega_{F} M[X^{2}]$ 
\begin{equation*}
dv_{3} = \left( \sum_{\mu\neq{\large 1}} b^{\mu}z_{\mu}+ b^{{\large 1}} dB
\right) \wedge w_{s-1} + d(pu_{4})
\end{equation*}
with $z_{\mu}, B \in\Omega_{F} M[X^{2}], dz_{\mu}= 0 $ for all $\mu$. 
Notice that the form $( \sum_{\mu\neq 1} b^{\mu}z_{\mu}+ b^{1} dB) \wedge w_{s-1}$
is exact, i.e. equal to some $dH $ with $H \in\Omega_{F} M[X^{2}]$.
Therefore
\begin{eqnarray*}
p[ E_{p} + \wp u_{1} + dv_{2} &+& t d \lambda_{1} \wedge w_{s-1}] = [ {\wp}
(\sum_{i} \bigwedge_{j} g_{ij} df_{ij} + pv_{1} +\delta\wedge w_{s-1})+\\
& & \left( \sum_{\mu\neq{\large 1}} b^{\mu}z_{\mu}+ b^{{\large 1}} dB
\right) \wedge w_{s-1} + d(pu_{4})+t\lambda_1\wedge w_{s-1}]\wedge dp.
\end{eqnarray*}

After dividing by $p$, we are led to consider the following expressions.
\begin{itemize}
\item[ a)]  ${\wp }(pv_1)\wedge \frac{dp}{p}={\wp }(pv_1\wedge \frac{dp}{p})=
{\wp }(v_1\wedge dp)\mbox{\hspace{1.5cm}} \mod\;\;\; d\Omega_F M[X^2]$

\item[ b)]
\begin{eqnarray*}
\wp\sum_i\bigwedge_{j}g_{ij}df_{ij}\wedge \frac{dp}{p}
&=&\wp\left(\sum_{i}\bigwedge_{j}g_{ij}f_{ij}\frac{df_{ij}}{f_{ij}}\wedge \frac{dp}{p}\right)\\
&& \\
&=&\wp\left(\sum_{i}\bigwedge_{j}(1+ph_{ij})\frac{df_{ij}}{f_{ij}}\wedge \frac{dp}{p}\right)\\
&& \\
&=&\wp\left(\sum_{i}\bigwedge_{j}\frac{df_{ij}}{f_{ij}}\wedge \frac{dp}{p}+
\sum_{i}H_i\bigwedge_{j}\frac{df_{ij}}{f_{ij}}\wedge dp\right)\\
&& \\
&=&\wp\left(\sum_{i}H_i\bigwedge_{j}\frac{df_{ij}}{f_{ij}}\wedge dp\right) \mod d\Omega_FM[X^2]
\end{eqnarray*}
with certain polynomials $H_i\in M[X^2]$.
\item[ c)]  ${\wp }(\delta \wedge w_{s-1})\wedge \frac{dp}{p}=
{\wp }(\delta \wedge w_{s-1}\wedge \frac{dt}{t})+{\wp }(\delta \wedge
w_{s-1}\wedge \frac{d(pt)}{pt})\mbox{\hspace{0.5cm}} \mod\; d\Omega_F M[X^2]$

\item[ d)]  $t\lambda_{1}\wedge w_{s-1}\wedge \frac{dp}{p}=\lambda_{1}\wedge w_{s-1}\wedge \frac{tdp}{p}=\lambda_{1}\wedge w_{s-1}\wedge \frac{d(tp)+pdt}{p}=
t\lambda_{1}\wedge w_{s-1}\wedge \frac{d(tp)}{tp}+\lambda_{1}\wedge w_{s-1}\wedge dt$ \;\;(recall $w_{s}=w_{s-1}\wedge d(pt))$

\item[ e)]  $\left( \sum_{\mu\neq{\large 1}} b^{\mu}z_{\mu}+ b^{{\large 1}} dB
\right) \wedge w_{s-1}\wedge\frac{dp}{p}=
\left( \sum_{\mu\neq{\large 1}} b^{\mu}z_{\mu}+ b^{{\large 1}} dB
\right) \wedge w_{s-1}\wedge\frac{d(pt)}{pt}+
\left( \sum_{\mu\neq{\large 1}} b^{\mu}z_{\mu}+ b^{{\large 1}} dB
\right) \wedge w_{s-1}\wedge\frac{d(t)}{t}$
\end{itemize}

Thus we get 
\begin{eqnarray*}
E_{p} & = & {\wp} (u_{1} + v_{1} \wedge dp) + d(v_{2} + u_{4} \wedge w_{s-1}
\wedge dp) \\
& & + {\wp} (\sum_{i} H_{i} \bigwedge_{j} \frac{df_{ij}}{f_{ij}} \wedge dp)
+ d(t \lambda_{1} \wedge w_{s-1}) \\
& & + {\wp} ( \delta\wedge w_{s-1}) \wedge\frac{dt}{t} + {\wp} (
\delta\wedge w_{s-1}) \wedge\frac{d(pt)}{pt} + \frac{\lambda_{1}}{p} \wedge
w_{s} \\
E_{p} & = & {\wp} \left(u_{1} + v_{1} \wedge dp + \sum_{i} H_{i} \bigwedge_{j} 
\frac{df_{ij}}{f_{ij}} \wedge dp + \delta\wedge w_{s-1} \wedge\frac{dt}{t}\right)
\\
& & + d(v_{2} + u_{4} \wedge w_{s-1} \wedge dp + t \lambda_{1} \wedge
w_{s-1}) + \alpha_{3} \wedge w_{s}
\end{eqnarray*}
where ${\wp} ( \delta\wedge w_{s-1}) \wedge\frac{d(pt)}{pt} + \frac
{\lambda_{1}}{p} \wedge w_{s} = \alpha_{3} \wedge w_{s}$, $\alpha_{3}$ being
a form of the type $\mu/pt $ with $\mu\in\Omega_{F} M[X^{2}]$, i.e. $%
\alpha_{3} $ has a denominator containing $p $ at most in the first power,
and all other prime factors are of degree (in $X) $ less than $\deg(p)$.
Thus we have 
\begin{equation*}
E_{p} = {\wp} (\alpha_{1} ) + d(\alpha_{2} ) + \alpha_{3} \wedge w_{s}
\end{equation*}
where $\alpha_{1}, \alpha_{2} $ have numerator in $\Omega_{F} M[X^{2}] $ and
denominators with prime factors of degree less than $\deg(p)$. 
In particular the above equation shows that $\alpha_{3} \wedge w_{s} $ is also a form of the type $\beta/ q $ where $q $ is a product of prime polynomials of degree
less than $p $ in $M[X^{2}]$. 
We claim 
\begin{equation}
\alpha_{3}\wedge w_{s}=\alpha\wedge w_{s}  \tag{4.17}
\end{equation}
where $\alpha$ is a form of the type $\gamma/q$ where $q$ is a product of
irreducible polynomials of degree $<\deg(p)$ contained in $M[X^{2}]$. 
Once we have shown this claim, we see that we can get rid of $E_{p}$ in the
equation (4.5). 
Thus we must show (after scaling (4.16) with a convenient polynomial $q)$. \newline

\noindent(4.18) \textbf{Lemma.} \emph{Let $\lambda$ be a form in $\frac{1}{p}
\Omega_{F} M[X^{2}] $ such that $\lambda\wedge w_{s} \in\Omega_{F}M[X^{2}]$.
Then there is some form $\lambda^{\prime}\in\Omega_{F} M[X^{2}] $ with } 
\emph{
\begin{equation*}
\lambda\wedge w_{s} = \lambda^{\prime}\wedge w_{s}.
\end{equation*}
}

\noindent\textbf{Proof.} Set $\lambda= p^{-1} \lambda_{0} $ with $\lambda_{0}\in\Omega_{F} M[X^{2}]$. 
We can obviously assume that $\deg_{X}
\lambda_{0} <\deg_{X} p$. We write $dT = k_{s} db_{s} + dT^{\prime}$, with $
\deg_{X}k_{s} = 2$, $\deg_{X} T^{\prime}= 0 $ and $dT^{\prime}$ not
containing $db_{s}$. 
Since $w_{s} = db_{1} \wedge\cdots\wedge db_{s-1}\wedge dT$, we may assume that $\lambda_{0}$ is only generated by forms 
$db_{j}, j \geq s$. 
Set $\lambda_{0} = \lambda_{s} \wedge db_{s} +\lambda^{\prime}_{s} $ with $\lambda_{s}, \lambda^{\prime}_{s} $ generated by forms $db_{j}, j > s$. 
Then
\begin{eqnarray*}
\lambda\wedge w_{s} & = & p^{-1}\lambda_{0}\wedge db_{1}\wedge\cdots\wedge
db_{s-1}\wedge(k_{s}db_{s}+dT^{\prime}) \\
& & \\
& = & p^{-1}db_{1}\wedge\cdots\wedge db_{s-1}\wedge db_{s}\wedge(\lambda
_{s}\wedge dT^{\prime}+k_{s}\lambda_{s}^{\prime}) \\
& & \\
& & + p^{-1}db_{1}\wedge\cdots\wedge db_{s-1}\wedge\lambda
_{s}^{\prime}\wedge dT^{\prime}.
\end{eqnarray*}
Because of the above choices, both summands do not interfere with each
other, so that both must be integral, since $\lambda\wedge w_{s}$ is
integral. 
In particular 
\begin{equation*}
p^{-1} db_{1} \wedge\cdots\wedge db_{s-1} \wedge\lambda^{\prime}_{s} \wedge
dT^{\prime }\in\Omega_{F} M[X^{2}].
\end{equation*}
Since $\deg_{X}\lambda^{\prime}_{s}<\deg_{X}p$ and $\deg_{X}T^{\prime}=0$,
it follows 
\begin{equation*}
db_{1} \wedge\cdots\wedge db_{s-1} \wedge\lambda^{\prime}_{s} \wedge
dT^{\prime}= 0.
\end{equation*}

But $\lambda^{\prime}_{s} \wedge dT^{\prime}$ is generated only by
differentials $db_{j}, j \geq s$, so that we obtain $\lambda^{\prime}_{s}
\wedge dT^{\prime}= 0$. 
It follows $\lambda^{\prime}_{s} = \delta\wedge
dT^{\prime}$ with some form $\delta\in\Omega_{F}M[X^{2}]$, and $%
\deg_{X}\delta< \deg_{X}p$. 
Therefore 
\begin{eqnarray*}
\lambda\wedge w_{s} & = & p^{-1} db_{1} \wedge\cdots\wedge db_{s}
\wedge(\lambda_{s} \wedge dT^{\prime}+ k_{s} \delta\wedge dT^{\prime}) \\
& & \\
& = & p^{-1} db_{1} \wedge\cdots\wedge db_{s} \wedge(\lambda_{s} + k_{s}
\delta) \wedge dT^{\prime} 
\end{eqnarray*}
Since $db_{1} \wedge\cdots\wedge db_{s} \wedge dT^{\prime}$ does not contain 
$X $ we easily see that $p $ divides $\lambda_{s} + k_{s} \delta$ i.e. $%
\lambda_{s} + k_{s} \delta= p \cdot\mu$ with $\mu\in\Omega_{F} M[X^{2}]$.
Hence
\begin{eqnarray*}
\lambda_{0} & = & (k_{s} \delta+ p \mu) \wedge db_{s} + \delta\wedge
dT^{\prime} \\
& & \\
& = & \delta\wedge dT + p \mu\wedge db_{s}.
\end{eqnarray*}
This implies 
\begin{eqnarray*}
\lambda\wedge w_{s} & = & p^{-1}\lambda_{0}\wedge db_{1}\wedge\cdots\wedge
db_{s-1}\wedge dT \\
& & \\
& = & p^{-1}p\mu\wedge db_{s}\wedge db_{1}\wedge\cdots\wedge db_{s-1}\wedge
dT \\
& & \\
& = & \mu\wedge w_{s}
\end{eqnarray*}
with $\mu\in\Omega_{F}M[X^{2}]$. This proves the claim. $\square$\newline

Therefore, using this descent procedure we have shown the following result. 
\newline

\noindent(4.19) \textbf{Proposition.} \emph{If \;$w \in\Omega_{F}^{n} $
satisfies an equation } \emph{
\begin{equation*}
w = {\wp} u + dv + \lambda\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT
\end{equation*}
with $u, v, \lambda\in\Omega_{F} M(X^{2}) $ and $T $ as before, then \;$w $
satisfies an equation } \emph{
\begin{equation*}
w={\wp} u^{\prime}+dv^{\prime}+\lambda^{\prime}\wedge
db_{1}\wedge\cdots\wedge db_{s-1}\wedge dT
\end{equation*}
with $u^{\prime},v^{\prime},\lambda^{\prime}\in\Omega_{F}M[X^{2}]$.} \newline

\bigskip\noindent(4.20) \textbf{Lemma.} \emph{Let $w \in\Omega_{F}^{n} $ be
such that there exist $u, v, \lambda\in\Omega_{F} M[X^{2}] $ with $w = {\wp}
u + dv + \lambda\wedge db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT$, $T =b_{s} X^{2}+ T^{\prime}$, $\deg_{X} T^{\prime}= 0$. 
Then there exist $u^{\prime},v^{\prime}, \lambda^{\prime}\in\Omega_{F} M $ with } \emph{
\begin{equation*}
w = {\wp} u^{\prime}+ dv^{\prime}+ \lambda^{\prime}\wedge db_{1}\wedge
\cdots\wedge db_{s-1} \wedge db_{s} \wedge dT^{\prime}.
\end{equation*}
}

\noindent\textbf{Proof.} Set $\deg_{X} u = 2h$, $\deg_{X} v = 2k$, $%
\deg_{X}\lambda= 2 \ell$. We consider the following cases: \newline

a) $4h \geq 2k$. Then the coefficient of $X^{4h} $ in ${\wp} u + dv
+\lambda\wedge w_{s}$ ($w_{s} = db_{1} \wedge\cdots\wedge db_{s-1} \wedge dT$
) is the form $u_{2h}^{[2]} + dv_{4h} + \lambda_{4h-2} \wedge w_{s-1} \wedge
db_{s}$, where $w_{s-1} = db_{1} \wedge\cdots\wedge db_{s-1}$. Then we must
have 
\begin{equation*}
u_{2h}^{[2]} + dv_{4h} + \lambda_{4h-2} \wedge w_{s-1} \wedge db_{s} = 0
\end{equation*}
if $h \geq 1$. Here we have used the following notation: $u = u_{0} + u_{2}
X^{2} + \cdots+ u_{2h} X^{2h}$, with $u_{i} \in\Omega_{F} M$. Then applying
the Cartier operator 
\begin{equation*}
0 = u_{2h} + C(\lambda_{4h-2} \wedge w_{s-1} \wedge db_{s}).
\end{equation*}
%%% FROM PROBLEM:
%>From lemma (1.12) we obtain in $\Omega_{F(X_{\mu})} $ 
From lemma (1.12) we obtain in $\Omega_{F(X_{\mu})} $ 
\begin{equation*}
u_{2h} = \mu\wedge w_{s-1} \wedge db_{s}.
\end{equation*}
Since $u_{2h} \in\Omega_{F} M$, it is easy to conclude that we can choose $%
\mu\in\Omega_{F} M$. Therefore
\begin{eqnarray*}
u & = & u_{0} + u_{2} X^{2} + \cdots + u_{2h} X^{2h} \\
& & \\
& = & u_{0} + \cdots+ \mu\wedge w_{s-1} \wedge db_{s} X^{2h} \\
& & \\
& = & u_{0} + \cdots+ \mu\wedge w_{s-1} \wedge(dT + dT^{\prime}) X^{2h-2} \\
& & \\
& = & u_{0} + \cdots+ u^{\prime}_{2h-2} X^{2h-2} + \mu X^{2h-2} \wedge
w_{s-1} \wedge dT
\end{eqnarray*}
where $u^{\prime}_{2h-2} = u_{2h-2} + \mu\wedge dT^{\prime}$. 
The term $\mu X^{2h-2} \wedge w_{s-1} \wedge dT $ can be added to $\lambda\wedge s_{s-1}\wedge dT $ replacing $\lambda$ by $\lambda+ \mu X^{2h-2}$. 
Thus we have lowered the degree of $u$. Iterating this procedure we are led to consider the following case \newline

b) $4h < 2k$. It follows $2 \ell\leq2k - 2$, and the coefficient of $X^{2k} $
is $dv_{2k} + \lambda_{2k-2} \wedge w_{s-1} \wedge db_{s}$. Therefore if $%
k\geq 1$, we must have 
\begin{equation*}
dv_{2k} + \lambda_{2k-2} \wedge w_{s-1} \wedge db_{s} = 0.
\end{equation*}

\noindent Then according (1.13) there exist forms $z_{\mu} \in\Omega_{F} M $ and $B\in\Omega_{F} M $ such that 
\begin{equation*}
dv_{2k} = \sum_{\mu\neq1_{s-1}} [b^{\mu} (z_{\mu, 2} + b_{s} z_{\mu, 3} ) +
b^{1_{s-1}} (z_{1} + b_{s} dB)] \wedge w_{s-1} \wedge db_{s}
\end{equation*}
with all $z_{\mu} $ closed, and $z_{\mu, 2}, z_{\mu, 3}, z_{1}, B $ not
containing $b_{s} $ in the 2-expansion of their coefficients. 
The term $b^{\mu}z_{\mu, 2} \wedge w_{s-1} $ is exact for all $\mu\neq1_{s-1} $ i.e. $b^{\mu} z_{\mu, 2} \wedge w_{s-1} \in d \Omega_{F} M$. 
Also $b_{s}b^{\mu}\wedge w_{s-1} \in b_{s} d \Omega_{F} M$. 
The form $b^{1_{s-1}} z_{1}\wedge w_{s-1} $ is closed and hence contained in $d \Omega_{F} M + (\Omega_{F}M)^{[2]} $ and $b_{s} b^{1_{s-1}} dB \wedge w_{s-1} \in b_{s} d\Omega_{F} M$.

Putting all this together, we obtain 
\begin{equation*}
dv_{2k} = (A^{[2]} + dC + b_{s} d E) \wedge db_{s}
\end{equation*}
with forms $A, C, E \in\Omega_{F} M$, which moreover are multiples of $w_{s-1}$. 
%%% FROM PROBLEM:
%>From 
From 
\begin{equation*}
\lambda_{2k-2} \wedge w_{s-1} \wedge db_{s} = (A^{[2]} + dC + b_{s} dE)
\wedge db_{s}
\end{equation*}
we infer 
\begin{equation*}
\lambda_{2k-2}\wedge w_{s-1}=A^{[2]}+dC+b_{s}dE+F\wedge db_{s}
\end{equation*}
with some $F\in\Omega_{F}M$, which can be chosen as a multiple of $w_{s-1}$,
since $F$ does not have terms containing $db_{s}$ and $F\wedge db_{s}
\in\langle w_{s-1}\rangle$. Therefore 
\begin{eqnarray*}
\lambda\wedge w_{s-1} & = & (\lambda_{0}+\lambda_{2}X^{2}+\cdots
+\lambda_{2k-2}X^{2k-2})\wedge w_{s-1} \\ 
&  &  \\ 
\lambda\wedge w_{s-1}\wedge dT & = & (\lambda_{0}+\lambda_{2}X^{2}+\cdots+
\lambda_{2k-4}X^{2k-4})\wedge w_{s-1}\wedge dT \\ 
&  &  \\ 
&  & + X^{2k-2}(A^{[2]}+dC+b_{s}dE+F\wedge db_s)\wedge dT.
\end{eqnarray*}
The terms $X^{2k-2}A^{[2]}\wedge dT$ and $X^{2k-2}dC\wedge dT$ are exact,
and hence can be absorbed by $dv$ without increasing the degree of $v$. 
Now 
\begin{eqnarray*}
X^{2k-2} b_{s} dE \wedge dT & = & X^{2k-4} (X^{2} b_{s}) dE \wedge dT \\ 
&  &  \\ 
& = & X^{2k-4} (T + T^{\prime})dE \wedge dT \\ 
&  &  \\ 
& = & d(X^{2k-4} TE \wedge dT) + X^{2k-4} T^{\prime}dE \wedge dT.
\end{eqnarray*}

The first summand can be added to $dv $ without increasing the degree of $v$.
Since $dE $ is a multiple of $w_{s-1}$, it follows that the second summand
in this expression is contained in $X^{2k-4} \Omega_{F} M \wedge w_{s-1}
\wedge dT $ and hence it can be added to the term $\lambda_{2k-4} \cdot
X^{2k-4} $ of degree $2k-4 $ in $\lambda\wedge w_{s-1} \wedge dT $. Finally the term $X^{2k-2} F \wedge db_{s}\wedge dT= X^{2k-4}F\wedge
dT^{\prime}\wedge dT$ is contained in $X^{2k-4} \Omega_{F} M \wedge w_{s-1}
\wedge dT = X^{2k-4} \Omega_{F} M \wedge w_{s}$, because $F $ is a multiple
of $w_{s-1}$, and therefore this term is absorbed by the term of degree $%
2k-4 $ in $\lambda\wedge w_{s}$.

This shows that can assume $\deg\lambda\leq2k - 4$. But this implies $dv_{2k}= 0, $ i.e. $\deg v < 2k$. 
Thus we have lowered the degree of $v$,
and iterating this procedure we come again to the case a).

Combining the procedures of cases a) and b) we finally arrive at an equation 
\begin{equation*}
w = u^{[2]} + u + dv + \lambda\wedge w_{s-1} \wedge dT
\end{equation*}
with $u, v, \lambda\in\Omega_{F} M$. 
Since $\lambda\wedge w_{s-1} \wedge dT
=X^{2} \lambda\wedge w_{s-1} \wedge db_{s} + \lambda\wedge w_{s-1} \wedge
dT^{\prime}$, we see that $X^{2} \lambda\wedge w_{s-1} \wedge db_{s}
\in\Omega_{F} M$, and this is possible only if $\lambda\wedge w_{s-1} \wedge
db_{s} = 0$. 
We have assumed that $\lambda$ does not contain terms with $db_{1}, \ldots, db_{s-1}$, so that it follows$\lambda= \lambda^{ \prime}\wedge db_{s} $ in $\Omega_{F} M$. Thus 
\begin{equation*}
w={\wp} u+dv+\lambda^{\prime}\wedge db_{1}\wedge\cdots\wedge db_{s}\wedge dT
\end{equation*}
and this proves the lemma. $\square$

\bigskip We are now ready to prove the main result of this paper. \newline

\noindent\textbf{Proof of Theorem (4.1)} Let us introduce the following
subfields of $L = F(X_{\mu} | \mu\in S_{n}) $ and polynomials

\begin{eqnarray*}
L_{0} & = & L = F (X_{\mu} | \mu\in S_{n} ) \\
& & \\
T_{0} & = & T = \sum_{\mu\in S_{n}} b^{\mu} X_{\mu}^{2} \\
& & \\
M_{0} & = & M = F (X_{\mu}^{2} | \mu\in S_{n} ) \\
& & \\
L_{1} & = & F (X_{\mu} | \mu\neq e_{1} ), e_{1} = (1, 0, \ldots, 0) \\
& & \\
T_{1} & = & \sum_{\mu\neq e_{1}} b^{\mu} X_{\mu}^{2} \\
& & \\
M_{1} & = & F (X_{\mu}^{2} | \mu\neq e_{1} ) \\
& & \\
\vdots & & \\
& & \\
L_{j} & = & F (X_{\mu} | \mu\neq e_{1}, \ldots, e_{j} ), e_{i} = (0,
\ldots1, \ldots, 0) \\
& & \\
T_{j} & = & \sum_{\mu\neq e_{1}, \ldots, e_{j}} b^{\mu} X_{\mu}^{2} \\
& & \\
M_{j} & = & F (X_{\mu}^{2} | \mu\neq e_{1}, \ldots, e_{j} )
\end{eqnarray*}
where $j = 1, 2, \ldots, n$. 
We have $L_{j-1}=L_{j}(X_{j})$,$T_{j+1} = T_{j}+b_{j+1}X_{j+1}^{2}\in M_{j+1}$.

Thus equation (4.2) corresponds to (4.3) with $s = 1$, i.e. $w = {\wp} u +
dv + \lambda\wedge dT_{0} $ with $w \in\Omega_{F}, u, v,
\lambda\in\Omega_{F} M_{0}$. 
The above process implies an equation $w = {\wp}u^{\prime}+ dv^{\prime}+ \lambda^{\prime}\wedge dT_{0} $ with $u^{\prime},v^{\prime},\lambda^{\prime} \in\Omega_{F} M_{1} [X_{1}^{2}]$. 
Lemma (4.19) implies now 
\begin{equation*}
w = {\wp} u^{\prime\prime}+ dv^{\prime\prime}+ \lambda^{\prime\prime}\wedge
db_{1} \wedge dT_{1}
\end{equation*}
with $u^{\prime\prime}, v^{\prime\prime},
\lambda^{\prime\prime}\in\Omega_{F} M_{1}$. 
We continue with this process
for $s = 2, \ldots, n-1 $ until we get forms $\bar{u}, \bar{v}, \bar{\lambda}
\in\Omega_{F} M_{n-1} $ such that 
\begin{equation*}
w = {\wp} \bar{u} + d \bar{v} + \bar{\lambda} \wedge db_{1}
\wedge\cdots\wedge db_{n-1} \wedge dT_{n-1}.
\end{equation*}

But we have 
\begin{equation*}
db_{1} \wedge\cdots\wedge db_{n-1} \wedge dT_{n-1} = db_{1}
\wedge\cdots\wedge db_{n-1} \wedge k_{n} db_{n}
\end{equation*}
where $k_{n} = b_{n}^{-1} \sum_{\mu(n) = 1} b^{\mu} X_{\mu}^{2}$. 
Therefore we obtain an equation 
\begin{equation}
w={\wp}\bar{u}+d\bar{v}+\bar{\lambda^{\prime}}\wedge
db_{1}\wedge\cdots\wedge db_{n}  \tag{4.21}
\end{equation}
in $\Omega_{F}M_{n-1}$. 
We can now get rid of the remaining variables just by comparing coefficients. Suppose namely that $k$ is a field, $K=k(X)$ a pure transcendental extension in one variable of $k,w\in\Omega_{k}$ and that there is a relation 
\begin{equation}
w={\wp} u+dv+\lambda\wedge w_{n}  \tag{4.22}
\end{equation}
in $\Omega_{k}(X^{2})$, where $w_{n}$ is now defined over $k$. 
We decompose $u,v,\lambda$ in partial fractions 
\begin{eqnarray*}
u & = & u_{0} + \sum u_{p} \\ 
&  &  \\ 
v & = & v_{0} + \sum v_{p} \\ 
&  &  \\ 
\lambda & = & \lambda_{0} + \sum\lambda_{p}
\end{eqnarray*}
with $u_{0}, v_{0}, \lambda_{0} \in\Omega_{k} [X^{2}],u_{p}, v_{p}, \lambda_{p} \in p^{-\infty} \Omega_{k} [X^{2}]$. 
Since the operators ${\wp}$ and $d $ respect these decomposition, we conclude from (4.21) that 
\begin{equation*}
w = {\wp} u_{0} + dv_{0} + \lambda_{0} \wedge db_{1} \wedge\cdots\wedge
db_{n}
\end{equation*}
in $\Omega_{k} [X^{2}]$. 
Letting $X = 0, $ we obtain $w = {\wp} \bar{u}_{0}+ d \bar{v}_{0} + \bar{\lambda}_{0} \wedge db_{1} \wedge\cdots\wedge db_{n} $
in $\Omega_{k}$. 
Applying this argument to the equation (4.21) we conclude 
\begin{equation*}
w={\wp}\alpha+d\beta+\gamma\wedge db_{1}\wedge\cdots\wedge db_{n}
\end{equation*}
with $\alpha,\beta,\gamma\in\Omega_{F}$. This finishes the proof of the
theorem. $\square$

\newpage

\section{Quadratic forms and differential forms}

We review briefly in this section for the sake of completeness some basic
notations and results from quadratic form theory and its relations with
differential forms over fields of characteristic two. Our basic references
will be \cite{A-Ba 1}, \cite{Ba 1}, \cite{Ka 1} and \cite{Sa}. Let $F$ be a
field of characteristic two. We denote by $[a,b]$ the binary nonsingular
quadratic form $ax^{2}+xy+by^{2}$ $(a,b\in F)$. Any non singular quadratic
form over $F$ is of the form $\perp _{i=1}^{n}[a_{i},b_{i}]$, ($%
a_{i},b_{i}\in F$), where $\perp $ means orthogonal sum. The form $[0,0]$ is
the hyperbolic plane and any orthogonal sum $\perp \lbrack 0,0]$ is called a
hyperbolic space. Two quadratic forms $q_{1}$, $q_{2}$ are called equivalent
($q_{1}\sim q_{2}$) if $H_{1}\perp q_{1}\simeq H_{2}\perp q_{2}$, where $%
H_{1}$, $H_{2}$ are hyperbolic spaces. A form $q$ is called isotropic if
there is a nonzero vector $x$ with $q(x)=0$, otherwise $q$ is called
anisotropic. The set of equivalence classes of anisotropic quadratic forms
over $F$ form the Witt-group $W_{q}(F)$ (with respect to orthogonal sums).
Respectively, we denote by $\langle a\rangle $ the one dimensional symmetric
bilinear form $axy$ ($a\in F^{\ast }$) and by $\langle a_{1},\ldots
,a_{n}\rangle $ the orthogonal sum $\langle a_{1}\rangle \perp \ldots \perp
\langle a_{n}\rangle $. Let $W(F)$ be the Witt ring of $F$, i.e. the ring of
classes of non singular symmetric bilinear forms over $F$. Then $W_{q}(F)$
is a $W(F)$-module via the operation $b\otimes q(x\otimes y)=b(x,x)\cdot
q(y) $ (see \cite{Ba 1}, \cite{Sa}). The maximal ideal $I\subset W(F)$ of
even dimensional bilinear forms is additively generated by the 1-Pfister
forms $\langle 1,a\rangle $, $a\in F^{\ast }$, so that the $n$-power $I^{n}$
is additively generated by the $n$-fold Pfister forms $\ll a_{1},\ldots
,a_{n}\gg =\langle 1,a_{1}\rangle \otimes \cdots \otimes \langle
1,a_{n}\rangle $. We get the submodules $I^{n}W_{q}(F)$ of $W_{q}(F)$, $%
n\geq 0$, which are additively generated by the quadratic $n$-fold Pfister
forms $\ll a_{1},\ldots ,a_{n};b]]=\ll a_{1},\ldots ,a_{n}\gg \otimes [1,b]$
, where $[1,b]=x^{2}+xy+by^{2}$ is a 0-fold Pfister form. Thus we have the
filtration $W(F)\supset I\supset I^{2}\supset \cdots $ and $W_{q}(F)\supset
IW_{q}(F)\supset I^{2}W_{q}(F)\supset \cdots $. In \cite{Ka 1} it is shown
that there are a natural isomorphisms (see section 2 for the definition of $%
\nu _{F}(n)$ and $H^{n+1}(F)$) 
\begin{equation}
\alpha:\nu_{F}(n)\overset{\sim}{\rightarrow}I_{F}^{n}/I_{F}^{n+1}  \tag{5.1}
\end{equation}
\begin{equation}
\beta:H^{n+1}(F)\overset{\sim}{\rightarrow}I^{n}W_{q}(F)/I^{n+1}W_{q}(F) 
\tag{5.2}
\end{equation}
given on generators by 
\begin{equation*}
\begin{array}{rcl}
\alpha\left( \frac{dx_{1}}{x_{1}} \wedge\cdots\wedge\frac{dx_{n}}{x_{n}}
\right) & = & \ll x_{1}, \ldots,x_{n} \gg \mod{ I^n} \\ 
&  &  \\ 
\beta\left( \overline{b \frac{dx_{1}}{x_{1}} \wedge\cdots\wedge\frac{dx_{n}}{
x_{n}}} \right) & = & \ll x_{1}, \ldots,x_{n} ; b ]] \mod{ I^nW_q(F)}.
\end{array}
\end{equation*}

If $L/F $ is any field extension, any quadratic form $q $ (or bilinear form $%
b$) over $F $ can be viewed as a form over $L $, which we denote by $q_{L} $
( or $b_{L} $). The natural homomorphisms $W_{q} (F) \rightarrow W_{q} (L)$,
resp. $W(F) \rightarrow W(L) $ are compatible with the above isomorphisms,
i.e.

\begin{picture}(300,80)(-180,-10)
\put(-30,0){\makebox(-30,0){$\nu_L (n)$}}
\put(30,0){\makebox(30,0){$I^n_L/I^{n+1}_L$}}
\put(0,0){\makebox(0,0){$\stackrel{\sim}{\longrightarrow}$}}

\put(-30,30){\makebox(-30,30){$\nu_F (n)$}}
\put(30,30){\makebox(30,30){$I^n_F/I^{n+1}_F$}}
\put(0,30){\makebox(0,30){$\stackrel{\sim}{\longrightarrow}$}}

\put(-40,30){\vector(0,-1){20}}
\put(40,30){\vector(0,-1){20}}

\end{picture}\newline

\noindent resp.

\begin{picture}(300,80)(-150,-10)
\put(-30,0){\makebox(-30,0){$H^{n+1}(L)$}}
\put(60,0){\makebox(30,0){$I^nW_q(L)/I^{n+1}W_q(L)$}}
\put(0,0){\makebox(0,0){$\stackrel{\sim}{\longrightarrow}$}}

\put(-30,30){\makebox(-30,30){$H^{n+1}(F)$}}
\put(60,30){\makebox(30,30){$I^nW_q(F)/I^{n+1}W_q(F)$}}
\put(0,30){\makebox(0,30){$\stackrel{\sim}{\longrightarrow}$}}

\put(-40,30){\vector(0,-1){20}}
\put(70,30){\vector(0,-1){20}}

\end{picture}\newline

In particular the main isomorphism (4.1) can be restated in terms of
quadratic forms as follows. \newline

\noindent(5.3) \textbf{Theorem.} \emph{Let $\phi=\ll b_{1},\ldots,b_{n}\gg$
be an anisotropic bilinear $n$-fold Pfister form over $F$. Then }

\emph{
\begin{equation*}
\ker\left[ \frac{ I^{n} W_{q} (F)}{ I^{n+1} W_{q} (F)} \rightarrow\frac{
I^{n} W_{q} (F(\phi))}{I^{n+1} W_{q} (F (\phi))} \right] = \{ \overline{
\phi \otimes[1, b]} | b \in F \}.
\end{equation*}
}

\bigskip Let $p=\ll b_{1},\ldots,b_{n};b]]$ be an anisotropic quadratic $n$
-fold Pfister form over $F$. Let $F(p)$ be the function field of the quadric 
$\{p=0\}$ over $F$. In \cite{A-Ba 2} we have shown that (5.3) implies the
following result, whose proof will be given here for the sake of
completeness. \newline

\noindent(5.4) \textbf{Theorem. } \emph{Let $p $ be as above. Then }

\emph{
\begin{equation*}
\ker\left[ \frac{I^{n} W_{q} (F)}{I^{n+1} W_{q} (F)} \rightarrow\frac{I^{n}
W_{q} (F(p))}{I^{n+1} W_{q} (F(p))} \right] = \{ 0, \bar{p} \}
\end{equation*}
}

\textbf{Proof. } Let $q\in I^{n}W_{q}(F)$ be such that $q_{F(p)}\in
I^{n+1}W_{q}(F(p))$. Set $q=\sum_{i=1}^{r}\phi_{i}[1,a_{i}]$, $\phi_{i}$ an $%
n$-fold bilinear Pfister form over $F$, $1\leq i\leq r$. If $r=1,$ i.e. $%
q=\phi\lbrack1,a]$, then the above assumption implies that $\phi\lbrack1,a]$
is hyperbolic over $F(p)$ (see \cite{Ba 2}) and then by the norm theorem (
see \cite{Ba 3}) we conclude $\phi\lbrack1,a]\simeq p$ over $F$, i.e. $\bar
{q}=\bar{p}$. Now assume $r>1$, and we will prove the assertion by induction
on $r$. Thus we assume the assertion true for any field and any form of
length less than $r$. Thus without restriction $\phi_{r}[1,a_{r}]$ is
anisotropic. Set $\psi=\phi_{r}$, and let $F(\psi)$ be its function field.
Then $q_{F(\psi)}=\sum_{1}^{r-1}\phi_{i}[1,a_{i}]\in I^{n}W_{q}(F(\psi))$
and over $F(\psi)(p)$ we get $q_{F(\psi)(p)}\in I^{n+1}W_{q}(F(\psi)(p))$.

Therefore by induction we obtain 
\begin{equation*}
q_{F( \psi)} \equiv\epsilon p_{F(\psi)} \mbox{  mod }I^{n+1} W_{q} (F(\psi))
\end{equation*}
with $\epsilon= 0 $ or 1. Thus 
\begin{equation*}
(q \perp\epsilon p)_{F(\psi)} \in I^{n+1} W_{q} (F(\psi)).
\end{equation*}

%%% FROM PROBLEM:
%>From (5.1) we conclude $q\perp\epsilon p\equiv\psi\lbrack1,c]$ mod $%
From (5.1) we conclude $q\perp\epsilon p\equiv\psi\lbrack1,c]$ mod $%
I^{n+1}W_{q}(F)$ with some $c\in F$. Since $q_{F(p)}\in I^{n+1}W_{q}(F(p)),$
it follows $\psi\lbrack1,c]_{F(p)}\in I^{n+1}W_{q}(F(p))$, i.e. $\psi
\lbrack1,c]$ is hyperbolic over $F(p)$, and hence by the norm theorem (see
loc. cit. ) $\psi\lbrack1,c]\equiv\eta p$ mod $I^{n+1}W_{q}(F)$ with $\eta=0$
or 1. Therefore $q\equiv(\epsilon+\eta)p$ mod $I^{n+1}W_{q}(F)$. This
conclude the proof. $\square$\newline

Using the isomorphism (5.2) we can restate the above result as follows.
\newpage

\noindent(5.5) \textbf{Theorem.} \emph{Let $p = \ll b_{1} , \ldots, b_{n}; b
]] $ be an anisotropic quadratic $n$-fold Pfister form over $F$. Then }

\emph{
\begin{equation*}
H^{n+1} (F(p) / F) = \left\{ 0, \overline{b \frac{db_{1}}{b_{1}}
\wedge\cdots \wedge\frac{db_{n}}{b_{n}}} \right\}.
\end{equation*}
} \newline

\noindent(5.6) \textbf{Remark.} The analogue of (5.3) for bilinear forms is
the following result \newline

\noindent(5.7) \textbf{Proposition.} \emph{\ Let $\phi=\ll b_1,\ldots,b_n\gg$
be an anisotropic bilinear $n$-fold Pfister form. Then for $m\geq n$} 
\begin{equation*}
\ker\left[\frac{I^m_F}{I^{m+1}_F}\rightarrow \frac{I^m_{F(\phi)}}{%
I^{m+1}_{F(\phi)}}\right]=\left\{ \sum I^{m-n}\overline{\ll x_1,\ldots,x_n\gg%
}| x_1,\ldots,x_n\in F^2(b_1,\ldots,b_n)\right\}.
\end{equation*}

\emph{In the case $m=n$, we have } 
\begin{equation*}
\ker\left[\frac{I^n_F}{I^{n+1}_F}\rightarrow \frac{I^n_{F(\phi)}}{%
I^{n+1}_{F(\phi)}}\right]=\left\{ \overline{\ll x_1,\ldots,x_n\gg}|
x_1,\ldots,x_n\in F^2(b_1,\ldots,b_n)\right\}.
\end{equation*}
\newline

We will show the special case $m=n$ for simplicity. \newline

If $x_1,\ldots,x_n\in F^2(b_1,\ldots,b_n)$, then $dx_i\in
Fdb_1\oplus\cdots\oplus Fdb_n$ and hence 
\begin{equation*}
\frac{dx_1}{x_1}\wedge\cdots\wedge\frac{dx_n}{x_n}=a\frac{db_1}{b_1}%
\wedge\cdots\wedge\frac{db_n}{b_n}
\end{equation*}
with some $a\in F^2(b_1,\ldots,b_n)$. Since $a\frac{db_1}{b_1}%
\wedge\cdots\wedge\frac{db_n}{b_n}\in \ker(\Omega^n_F\rightarrow
\Omega^n_{F(\phi)})$, it follows $\frac{dx_1}{x_1}\wedge\cdots\wedge\frac{dx_n%
}{x_n}\in \ker[\nu(n)_F\rightarrow \nu(n)_{F(\phi)}]$. Thus 
\begin{equation*}
\overline{\ll x_1,\ldots,x_n\gg}=\alpha(\frac{dx_1}{x_1}\wedge\cdots\wedge%
\frac{dx_n}{x_n})\in \ker\left[\frac{I^n_F}{I^{n+1}_F}\rightarrow \frac{%
I^n_{F(\phi)}}{I^{n+1}_{F(\phi)}}\right].
\end{equation*}
Conversely take any $\overline{\psi}=\alpha(w)\in \ker\left[\frac{I^n_F}{%
I^{n+1}_F}\rightarrow \frac{I^n_{F(\phi)}}{I^{n+1}_{F(\phi)}}\right]$, with $%
w\in \ker[\nu(n)_F\rightarrow \nu(n)_{F(\phi)}]$, i.e. $w=a\frac{db_1}{b_1}%
\wedge\cdots\wedge\frac{db_n}{b_n}$. Using Kato's lemma (see (2.15)) one
immediately obtains $w=\frac{dx_1}{x_1}\wedge\cdots\wedge\frac{dx_n}{x_n}$
with $x_1,\ldots,x_n\in F^2(b_1,\ldots,b_n)$ and this implies $\overline{\psi%
}=\overline{\ll x_1,\ldots,x_n\gg}$. \newline

In particular, since this kernel is a group, we obtain \newline

\noindent(5.8) \textbf{Corollary.} \emph{Given $x_1,\ldots,x_n,y_1,%
\ldots,y_n \in F^2(b_1,\ldots,b_n)$, then there exist $z_1,\ldots,z_n \in
F^2(b_1,\ldots,b_n)$ with} 
\begin{equation*}
\ll x_1,\ldots,x_n\gg+\ll y_1,\ldots,y_n\gg=\ll z_1,\ldots,z_n\gg %
\mbox{\hspace{1cm}mod }I^{n+1}_F
\end{equation*}
\newline

This corollary says that the field $F^2(b_1,\ldots,b_n)$ is $n$-linked
relative to $F$.

In particular if $\{b_1,\ldots,b_N\}$ is a 2-basis of the field $F$, i.e. $%
F=F^2(b_1,\ldots,b_N)$, then $I^{N+1}_F=0$ and it follows 
\begin{equation*}
I^N_F=\left\{\ll x_1,\ldots,x_N \gg | x_i\in F^{ \ast}\right\}.
\end{equation*}

\newpage

\section{Generic splitting of quadratic forms and the degree conjecture}

Since Knebusch's seminal papers on generic splitting of quadratic forms over
fields of characteristic $\neq2 $ (see \cite{Kn 1}, \cite{Kn 2}) appeared,
few work has been done on the subject (see \cite{Ar-Kn}, \cite{F}, ... ). In
particular Knebusch's degree conjecture $I^{n} = J_{n}$, where $J_{n} $ is
the ideal of $W(F) $ of forms of degree $\geq n $ remains still open. In
what follows we will briefly develop the analog of Knebusch's theory over
fields of characteristic 2 and using the results of section 5 we will show
that in this case the corresponding degree conjecture is true.

Our main reference will be Knebusch's paper \cite{Kn} on reduction theory of
quadratic and bilinear forms, which holds true for fields of any
characteristic, as well as his generic splitting papers cited above. Many of
the definitions and results of Knebusch's theory can be extended (using \cite
{Ba 2}, \cite{Ba 3}) mutatis mutandis to the case $2=0$, so that we will
often refer to the above papers for proofs.

%%% FROM PROBLEM:
%>From now on all fields have characteristic 2. The most basic notion in this
From now on all fields have characteristic 2. The most basic notion in this
theory is that of generic zero field of a quadratic form $q$ over $F$. A
field extension $L/F$ is a generic zero field of $q$ if $q_{L}$ is isotropic
and if $E/F$ is any extension with $q_{E}$ isotropic, then there exists a $F$
-place $\lambda :L\rightarrow E\cup \infty $ (see \cite{La}). One easily
checks that the function field $F(q)$ of $q$ is a generic field of $q$. If $%
q=\langle a_{1}\rangle \lbrack 1,b_{1}]\perp \cdots \perp \langle
a_{n}\rangle \lbrack 1,b_{n}]$, then $F(q)=F(x_{1},\ldots
,x_{n},y_{1},\ldots ,y_{n})$ with the single relation $%
\sum_{1}^{n}a_{i}(x_{i}^{2}+x_{i}y_{i}+b_{i}y_{i}^{2})=0$. Also the field $%
F(q)_{0}=F(u_{1},\ldots ,u_{n-1},v_{1},\ldots ,v_{n-1},z)$ with $%
a_{n}(z^{2}+z+b_{n})+
\sum_{1}^{n-1}a_{i}(u_{i}^{2}+u_{i}v_{i}+b_{i}v_{i}^{2})=0$ is a generic
zero field of $q$. $F(q)_{0}$ is purely transcendental over $F$ if and only
if $q$ is isotropic over $F$. Starting with a non singular quadratic form $q$
over $F$ we can define a field tower $F=F_{0}\subset F_{1}\subset \cdots
\subset F_{h}$ and forms $q=q_{0},q_{1},\ldots ,,q_{h}$ defined over $%
F_{0},F_{1},\ldots ,F_{h}$ respectively such that $q_{r}$ is anisotropic
over $F_{r},0\leq r\leq h-1$, $q_{h}$ is hyperbolic over $F_{h}$, and $%
q_{r-1}\otimes F_{r}\simeq q_{r}\perp i_{r}\times \lbrack 0,0]$ with some
integer $i_{r},F_{r}$ being a generic zero field of $q_{r-1}$ over $F_{r-1}$
. The sequence $(F_{r},q_{r},i_{r},0\leq r\leq h)$ is called a generic
splitting tower of $q$. Recall that two field extensions $L_{1}/F,L_{2}/F$
are called $F$-equivalent if there exist $F$-places $\lambda
_{1}:L_{1}\rightarrow L_{2}\cup \infty ,\lambda _{2}:L_{2}\rightarrow
L_{1}\cup \infty $. Then any field $L$ which is $F$-equivalent to $F_{h}$ is
called a generic splitting field of $q$ and any field $F$-equivalent to $%
F_{h-1}$ is called a leading field of $q$. A generic splitting tower of $q$
is essentially unique in the sense that if $(F_{r},q_{r},i_{r},0\leq r\leq
h) $, $(F_{s}^{\prime },q_{s}^{\prime },i_{s}^{\prime },0\leq s\leq
h^{\prime }) $ are two generic splitting towers of $q$, then $h=h^{\prime }$
, $i_{r}=i_{r}^{\prime }$, $0\leq r\leq h$, $F_{h}$ is equivalent with $%
F_{h}^{\prime }$. The number $h=h(q)$ is called the height of $q$. Obviously
any form $\langle a\rangle p$, $a\in F^{\ast }$ and $p$ a Pfister form, has
height 1. Conversely if $h(q)=1$, then $q_{F(q)}$ is hyperbolic and the norm
theorem proved in \cite{Ba 2} implies immediately that $q\simeq \langle
a\rangle p$ with $a\in F^{\ast }$ and $p$ a Pfister form. In particular for
any form $q$, the form $q_{h-1}$ is similar to a $n$-fold Pfister form over $%
F_{h-1}$. The degree $n$ of this form is uniquely determined and we will
call it the degree of $q$ and we denote it by $\deg (q)$. If $q$ is
hyperbolic we set $\deg (q)=\infty $. For any extension $L/F$ we have $\deg
(q_{L})\geq \deg (q)$ and $\deg $ is a well defined function on $W_{q}(F)$, $%
\deg :W_{q}(F)\rightarrow N\cup \{\infty \}$. We define for any $n\geq 0$ 
\begin{equation}
J(n)=\{\bar{q}\in W_{q}(F)|\deg(q)\geq n\}.  \tag{6.1}
\end{equation}

\medskip One easily checks that $J(0) = W_{q} (F)$, $J(1) = I W_{q} (F)$, $%
J(2) = I^{2} W_{q} (F) $ (see \cite{Ba 1}, \cite{A-Ba 1} ). First we show
that $J(n) $ is a $W(F)$ submodule of $W_{q} (F) $ and that $I^{n} W_{q} (F)
\subseteq J(n)$. The key fact is the following result (compare \cite{Kn 1}). 
\newline

\noindent(6.2) \textbf{Proposition.} \emph{Let $q=\langle a\rangle p\perp
q^{\prime }$ be a quadratic form over $F$, where $p$ is an anisotropic
Pfister form of degree $n\geq 1$, $a\in F^{\ast}$ and $\deg(q^{\prime})\geq
n+1$. Let $L$ be a leading field of $q$. Then }

\begin{itemize}
\item[ i)]  \emph{$\deg (q)=n$ }

\item[ ii)]  \emph{$p_{L}$ is a leading form of $q$. }

\item[ iii)]  \emph{If $\deg (q^{\prime })\geq n+2$, then $p_{L}$ is
anisotropic and $q_{L}$ is Witt-equivalent to $\langle a\rangle p_{L}$ with
some $a\in L^{\ast }$. }
\end{itemize}

\noindent\textbf{Proof.} We may assume that $q^{\prime}$ is not hyperbolic.
We will show that $\deg(q)=n$. Let $(L_{i},q_{i}^{\prime},0\leq i\leq e)$ be
a generic splitting tower of $q^{\prime}$. Then $p_{L_{e}}$ is anisotropic.
Otherwise $p_{L_{e}}$ is hyperbolic and we can choose $0\leq m\leq e$
maximal with $p_{L_{m}}$ anisotropic. Then $p$ is hyperbolic over $%
L_{m}(q_{m} ^{\prime})$ and the norm theorem (see \cite{Ba 2}) together with
the sub form theorem (see \cite{Ba 3}) show that $\langle a\rangle
q_{m}^{\prime}$ is a sub form of $p_{L_{m}}$ for some $b\in L_{m}^{\ast}$. In
particular $\deg(q^{\prime})=\deg(q^{\prime}_m)\leq n$ which is a
contradiction. Thus $p_{L_{e}}$ is anisotropic, and therefore $\langle
a\rangle p_{L_{e}}$ is the kernel form of $q_{L_{e}}$. This shows $%
\deg(q)\leq n$. If $\deg(q)=m<n$, let $L$ be a leading field of $q$ with $%
\ker(q_{L})=\langle c\rangle r$, $c\in L^{\ast}$ and $r$ a $m$-fold Pfister
form over $L$. Thus $q_{L}^{\prime}\sim\langle c\rangle r\perp\langle
a\rangle p_{L}$, and since $\dim(\langle c\rangle r\perp\langle a\rangle
p_{L})=2^{m}+2^{n}<2^{n+1}$, it follows $q_{L}^{\prime}\sim 0$ because $%
\deg(q^{\prime})\geq n+1$. Thus $\langle c\rangle r\sim\langle a\rangle
p_{L} $ over $L$, and this implies $p_{L}$ hyperbolic and hence $r$ is
hyperbolic, which is a contradiction. Thus $\deg(q)=n$. The rest of the
proposition follows easily and we omit the proof. $\square$\newline

\noindent(6.3) \textbf{Corollary.} \emph{For any two forms $q_{1}, q_{2} $
over $F$ } \emph{
\begin{equation*}
\deg(q_{1} \perp q_{2} ) \geq \min \{ \deg(q_{1}), \deg(q_{2}) \}
\end{equation*}
and if $\deg(q_{1}) \neq\deg(q_{2})$, the equality holds.}

\bigskip From these results we immediately obtain \newline

\noindent(6.4) \textbf{Theorem.}

\begin{itemize}
\item[(i)]  \emph{$J(n)$ is a $W(F)$-sub-module of $W_{q}(F)$ }

\item[(ii)]  \emph{$I^{n}W_{q}(F)\subset J(n)$ }

\item[(iii)]  \emph{$I^{m}J(n)\subset J(m+n)$. }
\end{itemize}

\noindent\textbf{Proof.} (i) The above corollary shows that $J(n)$ is
subgroup of $W_{q}(F)$. Since for any $a\in F^{\ast}$, $\deg(\langle
a\rangle q)=\deg(q)$, again the same corollary implies $\deg(\langle
a_{1},\ldots ,a_{m}\rangle q)\geq\deg(q)$ for any $a_{1},\ldots,a_{m}\in
F^{\ast}$, i.e. $J(n)$ is a $W(F)$-submodule of $W_{q}(F)$. This shows (i).
Since $I^{n} W_{q}(F)$ is additively generated by the $n$-fold Pfister forms 
$\ll a_{1},\ldots,a_{n};a]]$ of degree $n$, (ii) follows from (i), (iii) is
also an immediate corollary of (i). $\square$\newline

If $F $ is a field of characteristic different from 2, then one of the major
conjectures of Knebusch's generic splitting theory is the equality $J_{n} =
I^{n} $ in $W(F)$. In \cite{A-Ba 2} Theorem 1, we have shown that the
equality $I^{n} W_{q} (F) = J(n) $ for any field $F $ is equivalent with the
equality 
\begin{equation*}
\ker[I^{n} W_{q} (F) / I^{n+1} W_{q} (F) \rightarrow I^{n} W_{q} (F(p)) /
I^{n+1} W_{q} (F(p))] = \{ 0, \bar{p} \}
\end{equation*}
for any quadratic $n$-fold Pfister form $p $ over $F$. Now if $F $ is a
field of characteristic 2, theorem (5.4) asserts that this equality is true.
Thus combining these results we have the following. \newline

\noindent(6.5) \textbf{Main theorem.} \emph{Let $F $ be a field of
characteristic 2. Then for any $n \geq 0 $ }

\emph{
\begin{equation*}
I^{n} W_{q} (F) = J(n).
\end{equation*}
}

\noindent (6.6) \textbf{Remark.} Of course (6.5) implies (6.4), but the
above outlined proof of (6.4) is much more elementary.

\newpage

\addcontentsline{toc}{section}{\protect\numberline{ }{References}}

\begin{thebibliography}{ A-Ba 1}
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quadratischer Formen}. Math. Ann. 234, 167-192 (1978).

\bibitem[ A-Ba 1]{A-Ba 1}  Aravire, R., Baeza, R.: \emph{Milnor's K-theory
and quadratic forms over fields of characteristic two}. Comm. Algebra, 20
(4), 1087-1107 (1992).

\bibitem[ A-Ba 2]{A-Ba 2}  Aravire, R., Baeza, R.: \emph{A note on generic
splitting of quadratic forms}. Comm. Algebra, 27(7), 3473-3477 (1999).

\bibitem[ Ba 1]{Ba 1}  Baeza, R.: \emph{Quadratic forms over semi-local
rings }. Lecture Notes in Math. Vol. 655, Springer-Verlag (1978).

\bibitem[ Ba 2]{Ba 2}  Baeza, R.: \emph{The norm theorem for quadratic forms
over a field of characteristic two}. Comm. Algebra. 18(5), 1337-1348 (1990).

\bibitem[ Ba 3]{Ba 3}  Baeza, R.: \emph{Ein Teilformensatz f\"{u}r
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\bibitem[ Ca]{Ca}  Cartier, P.: \emph{Questions de rationalit\'{e} des
diviseurs en g\'{e}ometrie algebrique}. Bull. Soc. Math. I.H.E.S. Vol. 20
(1964).

\bibitem[ F]{F}  Fitzgerald, R. W.: \emph{Function fields of quadratic forms}
. Math. Z. 178, 63-76 (1981).

\bibitem[ Groth]{Groth}  Grothendieck, A.: \emph{El\'{e}ments de g\'{e}%
ometrie algebrique IV, premiere partie}. Publ. Math. I.H.E.S. Vol 20 (1964).

\bibitem[ Ka 1]{Ka 1}  Kato, K.: \emph{Symmetric bilinear forms, quadratic
forms and Milnor K-theory in characteristic two}. Inv. Math. 66, 493-510
(1982).

\bibitem[ Ka 2]{Ka 2}  Kato, K.: \emph{Galois cohomology of complete
discrete valuation fields}. Algebraic K-theory. Proc. Oberwolfach, Part II
(R.K. Dennis, Ed.) Lecture Notes in Math. Vol. 967, Springer-Verlag (1982).

\bibitem[ Kn]{Kn}  Knebusch, M.: \emph{Specialization of quadratic and
symmetric bilinear forms, and a norm theorem}. Acta Arith. 24, 279-299
(1973).

\bibitem[ Kn 1]{Kn 1}  Knebusch, M.: \emph{Generic splitting of quadratic
forms I}. Proc. London Math. Soc. (3) 33, 65-93 (1976).

\bibitem[ Kn 2]{Kn 2}  Knebusch, M.: \emph{Generic splitting of quadratic
forms II}. Proc. London Math. Soc. (3) 34, 1-31 (1977).

\bibitem[ La]{La}  Lang, S.: \emph{Algebra}. (Third Edition) Addison Wesley
(1983).

\bibitem[ Mi]{Mi}  Milne, J.S.: \emph{Duality in flat cohomology of a
surface }. Ann. Sc. Ec. N. Sup. 9, 171-202 (1976).

\bibitem[ OVV]{OVV}  Orlov, D., Vishik, A., Voevodsky, V.: \emph{Motivic cohomology of Pfister quadrics and Milnor's conjecture on quadratic forms }. Preprint, (1998).

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\end{thebibliography}


\bigskip\textsc{Roberto Aravire}

\textsc{Departamento de Ciencias F\'{\i}sicas y Matem\'{a}ticas }

\textsc{Universidad Arturo Prat, Casilla 121, Iquique, Chile }

\textsc{e-mail: raravire@cec.unap.cl. \newline
}

\textsc{Ricardo Baeza }

\textsc{Instituto de Matem\'{a}tica y F\'{\i}sica }

\textsc{Universidad de Talca, Casilla 747, Talca, Chile. }

\textsc{Departamento de Matem\'{a}ticas, Facultad de Ciencias }

\textsc{Universidad de Chile, Casilla 653, Santiago, Chile. }

\textsc{e-mail: rbaeza@inst-mat.utalca.cl. }

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