\documentclass{article}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{amsopn}
\usepackage[all]{xy}
\usepackage{amsthm}
\usepackage{amscd}

\swapnumbers
\theoremstyle{plain}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lemma}[thm]{Lemma}
\newtheorem{sublemma}[thm]{Sublemma}
\newtheorem{cor}[thm]{Corollary}

\theoremstyle{definition}
\newtheorem{rem}[thm]{Remark}
\newtheorem{ntt}[thm]{ }


\newcommand{\ra}{\rightarrow}
\DeclareMathOperator{\spec}{Spec}

\newcommand{\FF}{{\mathbb F}}
\newcommand{\AS}{{\mathbb A}}
\newcommand{\mm}{{\mathcal M}}
\newcommand{\Bb}{{\mathcal B}}


\title{On Knebusch's Norm Principle for quadratic forms over semi-local rings}
\author{K.~Zainoulline}
\date{kirill@mathematik.uni-bielefeld.de}


\begin{document}

\maketitle

\begin{abstract}
We prove Knebusch's
Norm Principle for finite extensions of semi-local regular
rings containing a field of characteristic $0$.
As an application we prove
the version of Grothendieck-Serre's conjecture 
on principal homogeneous spaces for the split case of the spinor group.
\paragraph{MSC2000:} 13H05, 11E08, 14M17
\paragraph{Key Words:} local regular ring, quadratic form, norm principle
\end{abstract}

\section{Introduction}

Let $L/K$ be a finite field extension
and $q$ be a regular quadratic form over $K$. 
Let $D(q_L)\subset L^*$ be the subgroup generated
by the set of non-zero elements of the field $L$ represented by the form $q$.
The well-known Knebusch's Norm Principle for quadratic forms 
over fields
\cite{KN}, \cite[VII.5.1]{La}
says there is an inclusion $N^L_K(D(q_L))\subset D(q_K)$
between the subgroups of $K^*$, where 
$N^L_K$ is the norm map.

The present paper is devoted to the proof of
Knebusch's Norm Principle for quadratic forms 
over semi-local regular rings.
Namely, we want to prove the following

\begin{thm}\label{mainthm}
Let $S/R$ be a finite extension 
of semi-local regular rings containing 
a field $k$ of characteristic $0$.
Let $q$ be a regular quadratic form over $R$ of rank $m$
and $q_S=q\otimes_R S$ be it's base change.
Then we have the following inclusion between the subgroups of the group
of invertible elements $R^*$
$$
N^S_R(D^0(q_S)) \subset D^0(q_R),
$$
where $N^S_R:S^*\ra R^*$ is the norm map and the subgroup
$D^0(q_S)\subset S^*$ is generated
by all products of two elements represented by $q_S$, i.e.,
$D^0(q_S)=\langle q_S(x_1)q_S(x_2)\mid x_1,\, x_2 \in S^m \rangle$.
In particular, the following inclusion holds
$$
N^S_R(D(q_S)) \subset D(q_R),
$$
where the subgroup
$D(q_S)\subset S^*$ is generated by all invertible
elements of $S$ that are represented by the quadratic form $q_S$, i.e.,
$D(q_S)=\langle q_S(x),\; x\in S^m \rangle$.
\end{thm}

In order to prove \ref{mainthm}, first, we prove Norm Principle for
simple extensions of semi-local domains (Theorem \ref{normpr}).
Briefly speaking, a finite extension $S/R$ is called simple, 
where $R$ is a semi-local domain, 
if $S=R[t]/(p(t))$ is 
the quotient of the polynomial ring $R[t]$ by some monic polynomial $p$. 
A finite etale extension (or separable in the case of fields)
is an example of a simple extension (see section 2).
The proof of \ref{normpr} proceeds by induction
on degree of extension. 
In order to make the induction step
we use general position arguments of section 3
(Theorem \ref{genpos}).
Then, using \ref{normpr}
we prove the version of
Grothendieck-Serre's conjecture
on principal homogeneous spaces \cite{Gr} for the case of the spinor group 
(Theorem \ref{groth}).
The conjecture states
the canonical map 
$$
i_R: H_{et}^1(R,Spin_q) \ra H^1_{et}(K,Spin_q)
$$
has trivial kernel, where $R$ is a semi-local regular ring and
$K$ is it's quotient field. We prove the map $i_R$ has the trivial
kernel if $R$ is a semi-local regular ring containing
a field of characteristic $0$. The main tool of the proof
is the injectivity theorem of section 2 of \cite{Za} together
with the etale version of Geometric Presentation Lemma \cite[6.1]{Za1}. 
We finish the proof of \ref{mainthm} by some diagram chase,
where Knebusch's Norm Principle over fields and Grothendieck-Serre's
conjecture over rings play the crucial role.

\paragraph{Agreements and Notations}
All rings are assumed to be commutative with units.
By $R$ and $S$ we denote semi-local domains.
$k$ means the residue field (sections 2, 3 and 4) or 
the base field (section 5).
By ``bar'' we mean the reduction modulo the maximal ideal 
(radical) of
a (semi-)local ring. 
By a ring extension $S/R$ we mean a ring $S$ together
with a ring monomorphism $R\ra S$.
By $S^*$ we denote the group of invertible elements of $S$.

For simplicity all the proofs and definitions 
of sections 2, 3 and 4
are given for local rings only. 
In order to pass from the local case
to the semi-local case, one has to replace
the words ``maximal ideal'' by the words ``radical ideal''.

Let $S/R$ be a ring extension such that $S$ is free as the $R$-module.
Then there is the norm map denoted by $N^S_R$ that is given as follows.
For any $c\in S$ let $l_c:S\ra S$ be the endomorphism of the free
$R$-module $S$ given by the left
multiplication by $c$. Then we set $N^S_R(c)=\det(l_c)$, where $l_c$
is the respective matrix.

We say a quadratic form $q$ over $R$ is regular if
the determinant of the respective symmetric matrix is invertible in $R$.
Observe that any regular 
quadratic form over a semi-local ring of characteristic
different from 2 is diagonalizable, i.e., $q$ can be represented
as a sum of squares with coefficients from $R^*$.



\ 

The author is grateful to Max-Planck-Institute f\"ur Mathematik in Bonn 
and European Post-Doctoral Institute for hospitality and financial support.
The author is also grateful to the RTN-Network HPRN-CT-2002-00287 and 
INTAS-Project 99-00817 for the travel support during the work. 


\section{Simple extensions of local rings}

\begin{ntt}
We say a ring extension $S/R$ is a simple extension of degree $n$ 
if there exists an element $c\in S^*$ such that $S$ is free as the $R$-module
with the basis $\{1, c,\ldots , c^{n-1}\}$ denoted by $\Bb(c)$. The element $c$
is called a primitive element.
Let $c^n=a_0+a_1c+\ldots+ a_{n-1}c^{n-1}$ be the unique presentation
of the element $c^n$ in the basis $\Bb(c)$.
The polynomial $p_c(t)=t^n-a_{n-1}t^{n-1}-\ldots -a_0$ is called
a minimal polynomial for the primitive element $c$.
It has the property $p_c(c)=0$. Observe that $S$ can be identified
with the quotient $R[t]/p_c(t)$ of the polynomial ring $R[t]$
modulo the principal ideal generated by $p_c(t)$.
Since $c$ is invertible, it's norm $N^S_R(c)=(-1)^{n}p_c(0)=(-1)^{n-1}a_0$ 
is invertible as well.

To the opposite direction, for a given monic 
polynomial $p(t)\in R[t]$ such that $p(0)\in R^*$ 
the ring extension $S=R[t]/p(t)$ over $R$ 
is a simple extension of degree $n=\deg p$.
The image of $t$ by means of the canonical map $R[t]\ra R[t]/p(t)$
gives the respective primitive element of the extension $S/R$. 
\end{ntt}

\begin{ntt}
Clearly, if $c\in S^*$ is a primitive element of 
a simple extension $S/R$,
then it's inverse $c^{-1}$ is primitive and
$r_1c+r_0$ is primitive for any $r_1\in R^*$ and 
$r_0\in R$ such that $p_c(-r_0/r_1)\in R^*$. 
\end{ntt}

\begin{ntt}\label{primelts}
Let $S/R$ be a simple extension with the primitive element $c$,
then $\bar{S}/\bar{R}$ is a simple extension of the same degree
with the primitive element $\bar{c}$. Moreover, if $b$ is a primitive
element of $\bar{S}$, then by Nakayama's lemma 
the preimage $c\in \rho^{-1}(b)$ is primitive as well,
where $\rho: S \ra \bar{S}$ is the reduction map. 
In other words, if $S^*_{prim}$ denotes the subset of primitive elements of 
$S$ and $\bar{S}^*_{prim}$ 
denotes the subset of primitive elements of $\bar{S}$,
then we have 
$S^*_{prim}=\rho^{-1}(\bar{S}^*_{prim})$. 
\end{ntt}

\begin{rem} \label{algcl} Let $S/R$ be a simple extension.
In the case $R=k$ is a field the algebra $S$ can be
viewed as the product of local Artinian algebras over $k$.
For instance, if $k$ is algebraically closed, the algebra $S$ 
is isomorphic to the product of algebras of the kind $k[t]/t^m$, $m\geq 1$.
In the case $S$ is a field 
we get a simple field extension $S/k$. Recall 
(by the Primitive Element Theorem) that any finite separable field
extension is simple but not in the other direction. There are examples of
finite field extensions which are not simple.
\end{rem}

\section{Some general position arguments}

In the present section $S$ will be a simple extension
of an infinite field $k$.

\begin{ntt}\label{topol}
Let $S$ be a simple extension of $k$ of degree $n$.
The $k$-algebra $S$ can be viewed as the $k$-vector space of dimension $n$
and, thus, can be identified with the set of rational points of the
affine space $\AS^n$ over $k$. From this point on we assume
$S=\AS^n(k)$ is the topological space by means of Zariski topology structure.

For example, any map $S\ra S$ given by $b\mapsto f(b)$ is continuous,
where $f(t)\in S[t]$ is a polynomial with coefficients in $S$.
And the set $S^*$ of invertible elements 
is open in $S$, since it is given by the equation $N^S_k(x)\neq 0$,
where $N^S_k$ is the norm map.

In the case of an infinite field $k$ 
this topology has the important property 
-- the intersection
of any two open subsets is non-empty.
\end{ntt}

\begin{lemma}
Let $S$ be a simple extension of a field $k$. 
Then the subset of primitive elements
$S^*_{prim}$ is non-empty and open in $S^*$. 
\end{lemma}

\begin{proof}
We fix some primitive element $c$ of $S$. An element $b\in S^*$
is primitive iff the matrix $(b_{i,j})_{i,j=0}^{n-1}$ 
has non-zero determinant, where $b_{i,j}$ is the $i$-th
coefficient in the presentation of the $j$-th power $b^j$ of the element $b$
in the basis $\Bb(c)$. 
Observe that $b_{i,1}=b_i$ are the coefficients of the presentation 
of the element $b$ in the basis $\Bb(c)$, i.e,
$b=b_0+b_1c+\ldots + b_{n-1}c^{n-1}$.
Clearly, the determinant $\det(b_{i,j})$ can be viewed as the polynomial
in $n$ variables $b_0,\ldots,b_{n-1}$ and, thus,
the subset 
$$
S^*_{prim}=\{b=(b_0,\ldots,b_{n-1})\in S^*\mid\det(b_{i,j})\neq 0\}
$$ 
is open in $S^*$.
\end{proof}

\begin{lemma}\label{cprim}
Let $S$ be a simple extension of an infinite field $k$ 
of characteristic different from 2 and 
let $c\in S^*$ be an invertible element.
Then the subset 
$V_c=\{b\in S^*\mid cb^2\; \text{is primitive} \}$ is non-empty and open in $S^*$.
\end{lemma}

\begin{proof}
Since the map $S^*\ra S^*$ given by $b\mapsto cb^2$ is continuous, 
it is enough to show that the image of the map 
$f:S^*\ra S^*$ given by $b\mapsto b^2$ is dense.
(observe that the multiplication by $c$ is the homeomorphism).
The algebra $S$ splits as the product of local Artinian algebras and
the image of $f$ is dense if the image of the restriction
of $f$ to the each component of this product is dense.
Thus, we may assume $S$ is a local Artinian algebra over $k$ and,
hence, it is irreducible.
 
Assume the image of $f$ is not dense. 
Then the closure of the image of $f$ in $S$
must have the dimension strictly less than the dimension of $S$ (considered
as the affine space over $k$). It means that $f$ induces the regular
map between two affine spaces such that the dimension of the target
space is strictly less than the dimension of the origin space $S$.
In particular, there exists an element $u\in S^*$ such that
the equation $b^2=u$ has infinite number of solutions
(the dimension of the fiber of $f$ over $u$ is $\geq 1$ and $k$ is infinite).
Hence, we get contradiction by Lemma \ref{densit}. 
\end{proof}

\begin{lemma}\label{densit}
Let $S$ be a simple extension over an infinite field $k$
of characteristic different from 2. 
Let $u$ be an invertible element of $S$.
Then the number of solution of the equation $b^2=u$ is finite (or empty).
\end{lemma}

\begin{proof}
Let $k'$ be the algebraic closure of the field $k$.
Let $S'=S\otimes_k k'$ be the base change of $S$.
Clearly, $S'$ is the simple extension of $k'$ of the same degree.
The number of solutions of $b^2=u$ over $S$ is finite
if it is finite over $S'$. 

Since the algebra $S'$ splits as the finite product of algebras of the kind
$A_m=k'[t]/t^m$, $m\geq 1$, it is enough to show that the number
of solutions of $b^2=u$ is finite in $A_m$ for any $m$.

The case $m=1$ is trivial, since $A_1=k'$ is a field.
Let $m>1$.
In the basis $\Bb(t)$ of $A_m$ our equation can be written as:
$$
(b_0+b_1t+\ldots+b_{m-1}t^{m-1})^2=u_0+u_1t+\ldots + u_{m-1}t^{m-1}.
$$
Hence, we get the system of $m$ quadratic equations over $k'$
\begin{equation}
\tag{*}
b_0^2=u_0,\; 2b_0b_1=u_1,\; 2(b_0b_2+b_1^2)=u_2,\; 2(b_3b_0+b_2b_1)=u_3,\ldots
\end{equation}
which has the property that any element $b_j$ is the solution of
the quadratic or linear equation over $k'$ (precisely the $j+1$-th equation) 
with coefficients $b_i$, $i<j$, and $u_i$, $i\leq j$.
Then it follows immediately that the number of solutions of (*) is finite.
\end{proof}

\begin{rem}
The assumption that the characteristic of $k$ is different from 2 is essential.
Take $k$ to be the algebraic closure of the finite field ${\mathbb F}_2$
then the algebra $S=k[t]/t^2$ is the simple extension of $k$.
But it easy to see that the image of the map $b\ra b^2$ coincides
with the subspace $k\cdot 1$ in $S=k\cdot1 \oplus k\cdot t$ 
which consists of all non-primitive
elements of $S$..
\end{rem}

\begin{ntt}
Let $c\in S^*$ be a primitive element of 
a simple extension $S/k$ of degree $n$.
Let $x$ be an element of $S$. 
By the symbol $\{x,c\}$
we denote the $n$-th coefficient
of the presentation of $x$
in the basis $\Bb(c)$, i.e., $\{x,c\}=x_{n-1}$, where 
$\sum_{i=0}^{n-1} x_ic^i=x$.
\end{ntt}


\begin{lemma}\label{cxprim}
Let $S$ be a simple extension of degree $n$ of a field $k$ of
characteristic 0.
Let $c\in S^*_{prim}$ be a primitive element and let $x$ be a non-zero element
of $S$.
Then the subset $W_{c,x}=\{b\in V_c\mid\{xb^{-1},cb^2\}\neq 0\}$  
is non-empty and open in $S^*$, where $V_c$ is the non-empty open subset
from Lemma \ref{cprim}. 
\end{lemma}

\begin{rem}\label{charrem} The assumption that $k$ has characteristic 0 is essential.
Take $k$ to be the algebraic closure of the finite field ${\mathbb F}_3$.
Consider the algebra $S=k[t]/t^3-1$. Take $x=c=t$. It is easy to see
that $\{xb^{-1},cb^2\}=0$ for all $b\in S^*$.
\end{rem}

\begin{proof}
Let 
$
xb^{-1}=v_0+v_1(cb^2)+\ldots + v_{n-1}(cb^2)^{n-1}
$ 
be the presentation
of the element $xb^{-1}$ in the basis $\Bb(cb^2)$.
Our goal is to show that the subset of the elements 
$b\in V_c$ with $v_{n-1}\neq 0$ 
is non-empty and open in $S^*$.

Multiplying the presentation of $xb^{-1}$ by $b$ we get the following equation:
\begin{equation}\tag{*}
x=v_0b+v_1(cb^3)+\ldots + v_{n-1}c^{n-1}b^{2n-1}.
\end{equation}
Now consider the primitive element $c$ of $S$.
Let $x=x_0+x_1c+\ldots +x_{n-1}c^{n-1}$ 
and 
$b=b_0+b_1c+\ldots+b_{n-1}c^{n-1}$ be the presentations of the
elements $x$ and $b$ in the basis $\Bb(c)$.
Consider the equation (*) in terms of the basis $\Bb(c)$.
We get the system of $n$ linear equations in $n$ variables 
$v_0,\ldots,v_{n-1}$: 
$$
\begin{pmatrix}
x_0\cr
x_1\cr
\vdots\cr
x_{n-1}\cr
\end{pmatrix}
=v_0\cdot 
\begin{pmatrix}
h_{0,0}(b_0,\ldots,b_{n-1})\cr
h_{1,0}(b_0,\ldots,b_{n-1})\cr
\vdots\cr
h_{n-1,0}(b_0,\ldots,b_{n-1})\cr
\end{pmatrix} 
 + \ldots v_{n-1}\cdot 
\begin{pmatrix}
h_{0,n-1}(b_0,\ldots,b_{n-1})\cr
h_{1,n-1}(b_0,\ldots,b_{n-1})\cr
\vdots\cr
h_{n-1,n-1}(b_0,\ldots,b_{n-1})\cr
\end{pmatrix},
$$
where $h_{i,j}$ are polynomials in $n$ variables $b_0,\ldots,b_{n-1}$
with coefficients from $k$.
In particular, the first column is the vector of monomials 
$(b_0,\ldots,b_{n-1})$, i.e., $h_{i,0}=b_i$ for all $i=0\ldots n-1$.
The second column consists of the coefficient of the presentation
of  $cb^3$ in the basis $\Bb(c)$ and so on.
Solving this linear system we get 
$$
v_{n-1}=\det(A_{(n-1)})/\det(A),
$$
where $A$ is the matrix of the system and the matrix 
$A_{(n-1)}$ is got by replacing
the last column of $A$ by the vector $x$ (of free terms).
Observe that both $\det(A_{(n-1)})$ and $\det(A)$ are homogeneous polynomials
(in $n$ variables $b_0,b_1,\ldots,b_{n-1}$)
of degrees $1+3+\ldots+(2n-3)=(n-1)^2$ and $1+3+\ldots+(2n-1)=n^2$ 
respectively.
Hence, the map $V_c\ra k$ given by $b=(b_0,\ldots,b_{n-1})\mapsto v_{n-1}$
is the regular map (observe that the determinant $\det(A)$ is non-zero
for all $b\in V_c$, since $cb^2$ is primitive). 
Thus, the subset $W_c$ is open in $S^*$.
The fact that $W_c$ is non-empty follows from Sublemma \ref{vanpol} below.
\end{proof}


\begin{sublemma}\label{vanpol}
If $x\neq 0$, then the polynomial $\det(A_{n-1})$ is non-trivial.
\end{sublemma}

\begin{proof} We may assume the field $k$ is algebraically closed.
We have the presentation of the determinant 
$$
\det(A_{(n-1)})= \Delta_{0}x_{n-1}-
\Delta_{1}x_{n-2}+\ldots +(-1)^{n-1}\Delta_{n-1}x_0,
$$
where $\Delta_i$, $i=0,\ldots,n-1$, is the determinant of the 
$(n-1-i,n-1)$-minor of $A_{(n-1)}$.
For each monomial $b_0^{m_0}b_1^{m_1}\ldots b_{n-1}^{m_{n-1}}$
we define the weight to be the sum $\sum_{i=0}^{n-1} i\cdot m_i$.
Observe that each polynomial $\Delta_i$ consists of monomials of 
weight $\geq i$. Indeed, it is enough to consider
the simple extension $S=k[t]/t^n$ over $k$ (see \ref{algcl}).
In this case the element $h_{i,j}(b_0,\ldots,b_{n-1})$ 
of the respective matrix $A$ 
is the zero polynomial if $i<j$ 
and consists of monomials
of the weight $i-j$ otherwise, i.e., we have the following 
matrix of weights
$$\left(.
\begin{matrix}
0 & 0 & 0 & \ldots & 0 \\
1 & 0 & 0 & \ldots & 0 \\
2 & 1 & 0 & \ldots & 0 \\
\vdots & \vdots & \vdots &  & \vdots \\
n-1 & n-2 & n-3 & \ldots & 0
\end{matrix}\right)
$$
where the first column corresponds to the vector of monomials 
$(b_0,\ldots,b_{n-1})$.

Now it is easy to see that the polynomial $\Delta_i$
consists of monomials of the weight that is greater or equal than the weight
of the product of diagonal elements of the $(n-1-i,n-1)$-minor of $A_{(n-1)}$,
i.e., precisely $i$.


Now each minor can be viewed
as the sum $\Delta_i= h_i+ h_{>i}$ of two homogeneous polynomials 
(in $n$-variables $b_0,b_1,\ldots,b_{n-1}$) of degree $(n-1)^2$,
where $h_i$ is the sum of monomials of weight $i$ and $h_{>i}$
is the sum of monomials of weight strictly bigger than $i$.


We claim that $h_i\neq 0$.
Indeed, for $i=0$ we have $h_0=b_0^{(n-1)^2}$, i.e.,
$h_0\neq 0$.
For $i=1$ we have $h_1= -C^1_{2n-3} b_0^{n(n-2)}b_1=-(2n-3)b_0^{n(n-2)}b_1$
which is non-zero if the characteristic of $k$ doesn't divide $2n-3$.
For $i>1$ the polynomial $h_i$ contains the unique monomial (the monomial
with the maximal power of $b_0$) 
$(-1)^i(2(n-i)-1)b_0^{n(n-2)}b_i$ which is non-zero if the characteristic
of $k$ doesn't divide $2(n-i)-1$.

Now it follows immediately that the polynomials $\Delta_i$, $i=0,\ldots,n-1$,
are linearly independent. And we are done.
\end{proof}
 
We will need one more fact concerning the polynomials $\Delta_i$:
\begin{sublemma}\label{vanprod}.
For any integers $0\leq i,j \leq n-1$
The polynomials $\Delta_i\Delta_j$, where $i\leq j$,
are linearly independent.
\end{sublemma}

\begin{proof}
The product $\Delta_i\Delta_j$ is the homogeneous polynomial of
degree $2(n-1)^2$ and can be represented as follows
$\Delta_i\Delta_j=h_ih_j+g_{>i+j}$, where $h_ih_j$ is the sum
of monomials of weight $i+j$ and $g_{>i+j}$ consists of monomials
of weight strictly bigger than $i+j$.
Observe now that $h_ih_j$ contains the unique monomial (the monomial
with the maximal power of $b_0$) 
$(-1)^{i+j}(2(n-i)-1)(2(n-j)-1)b_0^{2n(n-2)}b_ib_j$ 
(see the proof of the previous Sublemma). 
\end{proof}

Now we are ready to prove the main result of this section:
\begin{thm}\label{genpos}
Let $S$ be a simple extension of degree $n$ of a field $k$ of characteristic 
$0$. Let $c$ be a primitive element of $S/k$.
Let $q$ be a regular quadratic form over $k$ of rank $m$.
and 
$x=(x^{(1)},\ldots,x^{(m)})$ be a vector in $S^m$ such that
$q(x)\neq 0$.
By $\{x,c\}=(x_{n-1}^{(1)},\ldots,x_{n-1}^{(m)})$ 
we denote the vector of the $(n-1)$-th  coordinates
of $x$ in the basis $\Bb(c)$, i.e., $\{x,c\}_j=\{x^{(j)},c\}$.
Then the subset 
$$
U_{c,x,q}=\{b\in V_c \mid q(\{xb^{-1},cb^2\})\neq 0\}
$$
is non-empty and open in $S^*$.
\end{thm}

\begin{rem} According to \ref{charrem} the Theorem is not true
if the characteristic of the residue field $k$ is non-zero.
Take $k=\bar{{\mathbb F}}_3$, $S=k[t]/(t^3-1)$, $q(x)=x^2$ and $c=x=t$.
\end{rem}

\begin{proof}
The proof is a little modification of the proof of \ref{cxprim}.
We use the notation introduced in the proof of Lemma \ref{cxprim}.

Clearly, $U_{c,x,q}$ is open (by the same arguments as in \ref{cxprim}).
The main problem is to show that $U_{c,x,q}$ is non-empty.
Hence, we have to prove that the polynomial 
$q(\det(A_{(n-1)}^{(1)}),\ldots,\det(A_{(n-1)}^{(m)}))$ is non-trivial,
where $A_{(n-1)}^{(j)}$ denotes the matrix corresponding to the element
$x^{(j)}$. 
Let 
$$\det(A_{(n-1)}^{(j)})= \Delta_{0}x_{n-1}^{(j)}-
\Delta_{1}x_{n-2}^{(j)}+\ldots +(-1)^{n-1}\Delta_{n-1}x_0^{(j)}
$$
be the representation as in the proof of \ref{vanpol}, where
$x^{(j)}=\sum_{i=0}^{n-1} x_i^{(j)}c^i$ is the presentation of $x^{(j)}$
in the basis $\Bb(c)$.
Let $q(x)=\sum_j a_j(x^{(j)})^2$ be our quadratic form.

Then, we have
$$
(\det A)^2 \cdot q(\{xb^{-1},cb^2\})=\sum_j a_j(\det A_{(n-1)}^{(j)})^2=
$$
$$
=\sum_j a_j(\sum_i (-1)^ix_{n-1-i}^{(j)}\Delta_i)^2
$$
Now if we replace $\Delta_i$ by $(-1)^it^{n-1-i}$ we get precisely
\begin{equation}
\tag{*}
\sum_j a_j(\sum_i (-1)^ix_{n-1-i}^{(j)}\Delta_i)^2=
q(x^{(1)}(t),\ldots,x^{(m)}(t))
\end{equation}
as the polynomial in $R[t]$, where $x^{(j)}(t)=\sum_i x_i^{(j)}t^i$.

Assume that the polynomial 
$q(\{xb^{-1},cb^2\})$ is trivial.
Since the polynomials $\Delta_i\Delta_j$ are linearly independent
(by Sublemma \ref{vanprod}), this implies that
the polynomials $\Delta_i\Delta_j$ in the sum (*) have zero coefficients.
In particular, the $t^i$ have trivial coefficients as well,
i.e., the polynomial $q(x^{(1)}(t),\ldots,x^{(m)}(t))$ is trivial.
This contradicts with the hypothesis of the Theorem that the image
of $q(x^{(1)}(t),\ldots,x^{(m)}(t))$ by means of the canonical map
$R[t]\ra R[t]/p_c(t)=S$, i.e., precisely $q(x)$, is non-trivial.
\end{proof}


\section{The Knebusch's Norm Principle}


\begin{ntt}
Let $q_S$ be a quadratic form  of rank $m$ over a ring $S$.
By $D^0(q_S)$ ($D^1(q_S)$) we denote the set of all even (odd) products
of invertible elements of $S$ represented by $q_S$, i.e.,
$$
D^i(q_S)=\{\prod_{i=0}^lq(x_i) \mid x_i\in S^m,\,q(x_i)\in S^*,\, l\equiv i\, {\rm mod}\, 2 \},\quad i=0,1.
$$
Observe that $D^0(q_S)$ forms the subgroup of the group $D(q_S)$
generated by all invertible elements represented by $q_S$.
And $D^1(q_S)$ is just a subset of $D(q_S)$.
Clearly, if $c\in D^i(q_S)$, $i=0,1$, and $b\in D^0(q_S)$,
then $cb\in D^i(q_S)$.

Let $n$ be a positive integer, then we set $D^n(q_S)=D^0(q_S)$ if $n$ is even
and $D^n(q_S)=D^1(q_S)$ if $n$ is odd.

\end{ntt}

The goal of the present section is to prove the following

\begin{thm}\label{normpr} Let $R$ be a semi-local domain
with residue fields of characteristic 0.  
Let $S/R$ be a simple extension
of degree $n$. Let $q$ be a regular quadratic form over $R$ of rank $m$.
Let $c\in S^*$ be an element represented by the form $q_S$, i.e.,
$c=q_S(x)$ for some vector $x\in S^m$.  
Then $N^S_R(c)\in D^n(q_R)$, where $N^S_R$ is the norm map.
In particular, there is an inclusion between the subgroups of $R^*$
$$
N^S_R(D^0(q_S))\subset D^0(q_R),
$$
where $D^0(q_S)$ is the subgroup generated by all products of
two elements of $S^*$ represented by $q_S$.
\end{thm}

\begin{lemma}\label{squares}
In the hypothesis of Theorem \ref{normpr} we have
$(S^*)^2\subset D^0(q_S)$.
\end{lemma}

\begin{proof}
Let $r\in R^*$ be a value of the quadratic form $q$, i.e., $r=q(y)$
for some $y\in R^m$.
Then for any $b\in S^*$ we have $b^2=q(yb)q(y/r)\in D^0(q_S)$.
\end{proof}

\begin{proof}[The Proof of Theorem \ref{normpr}]
We prove by induction on the degree $n$ of the simple extension $S/R$.
The case $n=1$ is trivial.
Assume $n>1$.

In order to make the induction step we use the following idea:
Since any square can be viewed
as the product of two elements represented by the quadratic  form $q$,
the multiplication by a square doesn't change a lot, hence,
it gives some freedom in the choice of elements. 
As a consequence, by applying general position arguments (Theorem \ref{genpos})
we may control the leading coefficient of the polynomial $h$
(see the equation (**)) and make it invertible. 
Since it is invertible, we may apply 
the induction hypothesis and we are done.


More precisely, let $c=q_S(x)$ for some $x\in S^m$. 
We want to show $N^S_R(c)\in D^n(q)$.
By \ref{primelts} and Lemma \ref{cprim}
the element $c$ can be written as the product $c=cb^2\cdot (1/b)^2$, 
where $cb^2=q_S(xb)\in D^n(q_S)$ is primitive.
Hence, by multiplicativity of the norm map and Lemma \ref{squares}
we may assume $c$ (replaced by $cb^2$) is primitive.

Now we mimic the proof of \cite{KN} (see also \cite[VII.5.1]{La}).
Replace $c$ by it's inverse $c^{-1}$. We get the equation $1=cq_S(x)$,
where $c$ is primitive.
More precisely, we have
\begin{equation}
\tag{*}
1=cq_S(x^{(1)}(c),\ldots,x^{(m)}(c)),
\end{equation}
where $x^{(j)}(c)\in S$ is the $j$-th coordinate of the vector $x$ written
in the basis $\Bb(c)$. 
According to Theorem \ref{genpos}
we may assume that the value of the quadratic form $q_S$
on the last coefficients of the vectors $x^{(j)}$, i.e., 
$q(x^{(1)}_{n-1},\ldots,x^{(m)}_{n-1})$, is invertible in $R$.
In fact, it is enough to consider the quotient modulo the maximal
ideal of $R$ (see \ref{primelts}), i.e., the simple extension $\bar{S}/k$.
The open subset $U_{c,x,q}$ from \ref{genpos} is non-empty
and open. Take any element $b$ from $U_{c,x,q}$
and replace $c$ by $cb^2$ and $x$ by $x/b$.  

Consider the pull-back of the equation (*) by means of the canonical map
$R[t]\ra R[t]/p_c(t)=S$.
Since $tq(x(t))-1$ lies in the principal ideal $(p_c(t))$ of 
the polynomial ring $R[t]$ there is
a polynomial $h(t)$ such that 
\begin{equation}
\tag{**}
1+p(t)h(t)=tq_S(x^{(1)}(t),\ldots,x^{(m)}(t)).
\end{equation}
Since $R$ is a domain
the leading coefficient of the left hand side of (**)
coincides with the leading coefficient of $h$, denoted by $r$, 
and coincides with the leading
coefficient of the right hand side that is
$r=q(x^{(1)}_{n-1},\ldots,x^{(m)}_{n-1})$,
where $r$ is invertible in $R$.
Clearly, we have $n+\deg h=2(n-1)+1$. So that $\deg h=n-1$.

As in \cite{KN} we have 
$N^S_R(c)=(-1)^np_c(0)$ and $1+p_c(0)h(0)=0$.
Hence, we have 
$$
N^S_R(c)^{-1}=(-1)^{n-1}g(0)r,
$$
where  $g(t)=h(t)/r$ 
is the monic polynomial of degree $\deg h=n-1$.
Observe that the norm of the primitive element $u=t$
of the respective simple extension $T=R[t]/g(t)$ over $R$
is precisely $N^T_R(u)=(-1)^{n-1}g(0)$.
Hence, in order to show that $N^S_R(c)\in D^n(q)$ it is enough
to show that $N^T_R(u)\in D^{n-1}(q)$. But this follows by
the induction hypothesis, since $u$ is represented by the quadratic form $q_T$.
 Indeed,
taking (**) modulo the principal ideal (g(t))
we get similar to (*) the equation
$1=uq_T(x(u))$, i.e., $u=q_T(x(u)/u)$.
\end{proof}

\section{Grothendieck-Serre's conjecture for the case of the spinor group}

\begin{ntt}
Let $R$ be a semi-local domain with the residue fields
of characteristic different from 2 
and $q$ be a regular quadratic form over $R$.
Following \cite[IV.6]{Kn} we define the spinor group (scheme) $Spin_q$
to be $Spin_q(R)=\{x\in S\Gamma_q(R)\mid x\sigma(x)=1\}$, 
where $\sigma$ is the canonical involution. Recall that
$S\Gamma_q(R)=\{c\in C_0(V,q)^* \mid cVc^{-1}\subset V\}$, 
where $C_0(V,q)$ is the even part of the Clifford algebra of
the respective quadratic space $(V,q)$ over $R$.
\end{ntt}

The goal of the present section is:
\begin{thm}\label{groth}
Let $R$ be a semi-local regular ring containing a field
of characteristic 0. Let $K$ be it's quotient field.
Let $q$ be a regular quadratic form over $R$.
Then the induced map on the sets of principal homogeneous spaces
$$
H^1_{et}(R,Spin_q)\ra H^1_{et}(K,Spin_q)
$$
has trivial kernel, 
where $Spin_q$ is the spinor group for the quadratic form $q$.
\end{thm}

\begin{rem}
Observe that the theorem is the particular case $G=Spin_q$ 
of Grothendieck-Serre's conjecture on principal homogeneous spaces 
\cite{Gr},
which states for a flat reductive group scheme $G$ over $R$
the induced map $H^1_{et}(R,G)\ra H^1_{et}(K,G)$ has trivial kernel.
\end{rem}

\begin{proof}
The proof is based on the results of papers \cite{Za} and \cite{Za1}.

Assume $R$ is a semi-local regular ring of geometric type 
over a field of characteristic $0$. Let $K$ be it's quotient field.
We have the following commutative diagram (see \cite[IV.8.2.7]{Kn}):
$$
\xymatrix{
SO_q(R) \ar[r]^{SN} \ar[d] & R^*/(R^*)^2 \ar[r] \ar[d] &
H^1_{et}(R, Spin_q) \ar[r] \ar[d] &  H^1_{et}(R,SO_q) \ar[d] \\
SO_q(K) \ar[r]^{SN} & K^*/(K^*)^2 \ar[r] &
H^1_{et}(K, Spin_q) \ar[r] &  H^1_{et}(K,SO_q),
}
$$
where $SN: SO_q(R) \ra H^1_{et}(R,\mu_2)=R^*/(R^*)^2$ is the spinor norm.
The main result of paper \cite{OP} says that the vertical arrow on the right
hand side has trivial kernel (see also \cite[3.4]{Za}). 
Thus, in order to show that the middle one
has trivial kernel it is enough to check that the induced map on the
cokernels $coker(SN)(R)\ra coker(SN)(K)$ is injective. 

Consider the group scheme
$F:S\mapsto coker(SN)(S)$.
According to the theorem of section 2 of \cite{Za} 
to prove the mentioned injectivity
we have to show that the functor $F$ satisfies all the axioms 
C,TE,TA,TB, E of sections 1 and 2 of \cite{Za}. 
In fact, all the axioms, excluding the existence
of transfer map, i.e., TE, holds by the same arguments as in 
sections 3.2 and 3.4 of \cite{Za}.
Hence, in order to prove the injectivity 
we have to produce  
a well-defined transfer map $Tr_R^S: F(S) \ra F(R)$ 
for any finite surjective extension of semi-local rings $S/R$ 
(see the axiom TE of \cite{Za}).

Observe that for any finite surjective extension $S/R$ of semi-local
rings there is already a well-defined norm map 
$N_R^S: S^*/(S^*)^2 \ra R^*/(R^*)^2$.
Hence, if we show that
the norm map $N^S_R$ is compatible with the spinor norm, i.e., 
$$
N^S_R(SN(SO_q(S)))\subset SN(SO_q(R)),
$$ 
then taking $Tr_R^S=N_R^S$ on the quotients modulo the images of $SN$
we get the desired transfer map.

In fact, instead of finite surjective extensions we may consider
only finite etale extensions $S/R$ of semi-local rings. Indeed,
if we replace
the Geometric Presentation Lemma \cite[10.1]{OP} used in the section 1.1
of \cite{Za} 
by it's stronger (etale) version from \cite[6.1]{Za1},
then nothing will be changed in the proof of the injectivity theorem
of section 2 of \cite{Za}.

According to the definition of the spinor norm \cite[IV.6]{Kn}, 
\cite[III.3.21]{Ba} to show
the norm map commutes with the spinor norm is equivalent
to show the norm map commutes with the
functor $D: S \mapsto D^0(q_S)$ (that sends any $R$-algebra $S$ to the
subgroup $D^0(q_S)$), i.e.,
$N^S_R(D^0(q_S)) \subset D^0(q_R)$.
Hence, we have to prove the analog of Knebusch's Norm Principle
for quadratic forms in the case of finite etale extensions
of semi-local rings. 
But this is done by Theorem \ref{normpr}.

Finally, to extend our result 
to the case of a semi-local regular
ring containing a field of characteristic $0$ 
we use Popesky's approximation theorem
\cite[7.5]{Za1}. We refer to the item 1 of section 5 of \cite{Za} 
for the precise arguments.
\end{proof}

Now we are ready to prove the main theorem of this paper

\begin{proof}[The Proof of Theorem \ref{mainthm}] 
We use the notations of the proof of Theorem \ref{groth}.
Let $K$ and $L$ be the quotient fields of the semi-local rings $R$ and $S$
respectively. Observe that $L/K$ is a finite field extension.
We have the commutative diagram of abelian groups
$$
\xymatrix{
S^*/(S^*)^2 \ar[r]^{can} \ar[d]_{N^S_R} & coker(SN)(S) \ar[r]^{i_S} & coker(SN)(L) \ar[d]^{N^L_K} \\
R^*/(R^*)^2 \ar[r]^{can} & coker(SN)(R) \ar[r]^{i_R} & coker(SN)(K) \\
}
$$
where $can$ is the quotient map, the maps $i_R$ and $i_S$
are injective according to the proof of \ref{groth} and
$N^L_K$ is the norm map for the finite field extension $L/K$ taken
modulo the images of the spinor norms.
Observe that $N^L_K$ is well-defined 
since the Norm Principle holds in the case of finite field extensions 
\cite{KN}.

The diagram immediately implies that the norm map $N^S_R$
is well-defined on the quotients, i.e., 
$N^S_R(SN(SO_q(S)))\subset SN(SO_q(R))$ or, equivalently,
$$
N^S_R(D^0(q_S))\subset D^0(q_R).
$$
\end{proof}

\begin{rem}
Observe that under some restrictions
the proofs of \ref{groth} and \ref{mainthm}
imply that 
the Grothendieck-Serre's conjecture for the spinor group 
is equivalent to the Knebusch's Norm Principle.
\end{rem}

\begin{thebibliography}{999}

\bibitem{Ba} R.~Baeza, Quadratic forms over semilocal rings,
Lect. Notes in Math. 655, 1978.
\bibitem{Gr} A.~Grothendieck, Le groupe de Brauer. II. 
Th\'eorie cohomologique. (French)
Dix Expos\'es sur la Cohomologie des Sch\'emas, North-Holland, Amsterdam; 
Masson, Paris, 1968, pp.67-87.
\bibitem{La} T.Y.~Lam, The algebraic theory of quadratic forms,
Math. Lect. Notes Ser., W.A.Benjamin, Inc., 1973.
\bibitem{KN} M.~Knebusch, Ein Satz \"uber die Werte von quadratischen Formen
\"uber K\"orpern. (German) {\it Invent. Math.} 12: 300-303, 1971.
\bibitem{Kn} M.-A.~Knus, Quadratic and Hermitian Forms over Rings,
Grund. der Math Wiss. 294, Springer-Verlag, Berlin, 1991.
\bibitem{OP} M.~Ojanguren, I.~Panin,
Rationally trivial hermitian spaces are locally trivial,
{\it Math. Z.} 237: 181-198, 2001.
\bibitem{Za} K.~Zainoulline, On Grothendieck's conjecture on principal
homogeneous spaces for some classical algebraic groups, {\it Algebra i Analiz}
12(1): 150-184, 2000 (Russian); transl. in 
{\it St.Petersburg Math. J.} 12(1): 117-143, 2001.
(the draft version is available on http://www.math.uiuc.edu/K-theory/0332/)
\bibitem{Za1} K.~Zainoulline,
The Purity Problem for Functors with Transfers,
{\it K-Theory J.} 22(4): 303-333, 2001.
\end{thebibliography}

\paragraph{Current address (till 31/08/2003):} 
Fakult\"at f\"ur Mathematik, Universit\"at Bielefeld,
Postfach 100131, D-33501 Bielefeld, Germany


\paragraph{Permanent address:}
Laboratory of Algebra and Number Theory,
Steklov Mathematical Institute at St.Petersburg,
Fontanka 27,
191011 St.Petersburg,
Russia
\end{document}

