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\begin{document}

\title
{A purity theorem for linear algebraic groups}

\author{Ivan Panin \footnote{The author thanks very much for the support the
the Ellentuck Fund, the RTN-Network HPRN-CT-2002-00287,
the RFFI-grant 03-01-00633a,
the Swiss National Science Foundation
and INTAS-03-51-3251.} }

\date{01.03.2005}

\maketitle

\begin{abstract}
Given a characteristic zero field $k$ and a dominant morphism of affine algebraic $k$-groups
$\mu: G \to C$ one can
form a functor from $k$-algebras to abelian groups
$R \mapsto {\cal F}(R):=C(R)/\mu(G(R))$.
Assuming  that $C$ is commutative we prove that this functor satisfies
a purity theorem for any regular local $k$-algebra.
Few examples are considered in the very end of the preprint.
\end{abstract}

\section{Introduction}

 Let  ${\cal F}$ be  a covariant functor from the category of
commutative rings to the category
 of abelian groups. For any domain $R$ consider  the sequence
$${\cal F}(R)\to {\cal F}(K)\to \bigoplus_{\mathfrak p}\mathcal F(K)/{\cal F}(R_{\mathfrak p})$$
where $p$ runs over the height $1$ primes of $R$. We say that
{\it $\mathcal F$ satisfies purity for $R$} if this sequence --- which is
clearly a complex --- is exact. The purity for $R$ is equivalent to the following
property
$$
\bigcap_{\textrm{ht} \mathfrak p=1} Im[{\cal F}(R_{\mathfrak p}) \to {\cal F}(K)]=
Im[{\cal F}(R) \to {\cal F}(K)].
$$


Let $k$ be a characteristic zero field and $\bar k$ be its algebraic closure.
By a linear algebraic $k$-group we mean
a reduced affine group $k$-scheme. In particular, a linear algebraic
$k$-group is always $k$-smooth.
The main results of this paper is the following purity theorem




\begin{thm}[A] %%%%% why Theorem (A) instead %%%%%%%%%%%%%of Theorem A as Cartan did?

\label{TheoremA}
Let
$$\mu: G\to C$$
be a dominant morphism of linear algebraic $k$-groups, with $C$
commutative. Let $A$ be a local ring of a
smooth algebraic variety $X$ over $k$.
The functor
$$\mathcal F: R\mapsto C(R)/\mu(G(R))$$
satisfies purity for $A$.
\end{thm}

\begin{thm}[B]
Let $R$ be a regular local
ring containing the field $k$. Let
$$\mu: G\to C$$
be a dominant morphism of linear algebraic $k$-groups, with $C$
commutative. The functor
$$\mathcal F: A\mapsto C(A)/\mu(G(A))$$
satisfies purity for $R$. If $K$ is the fraction field of $R$
this statement can be restated in an explicit way
as follows: given an element $c \in C(K)$
suppose that for each height $1$ prime ideal
$\mathfrak p $ in $ A $ there exist
$a_{\mathfrak p } \in C(R_{\mathfrak p })$,
$g_{\mathfrak p } \in G(K)$
with
$a=a_{\mathfrak p } \cdot \mu(g_{\mathfrak p }) \in C(K)$.
Then there exist
$g_{\mathfrak m } \in G(K)$, $a_{\mathfrak m } \in C(R)$
such that
$$
a=a_{\mathfrak m } \cdot \mu(g_{\mathfrak m }) \in C(K).
$$
\end{thm}
After the pioneering articles
\cite{C-T/P/S}
and
\cite{R}
on purity theorems for algebraic groups,
various versions of purity theorems were proved in
\cite{C-TO},
\cite{PS},
\cite{Z},
\cite{Pa}.
The most general result in the so called constant case was given in
\cite[Exm.3.3]{Z}. This result follows now from our Theorem (A) by
 taking $G$ to be a $k$-rational reductive group, $C=\mathbb G_{m,k}$
and
$\mu: G \to \mathbb G_{m,k}$ a dominant $k$-group morphism. The papers
\cite{PS},
\cite{Z},
\cite{Pa}
contain results for the nonconstant case. However they only consider specific examples
of algebraic scheme morphisms $\mu: G \to C$.

It seems plausible to expect purity theorem in the following context.
Let $R$ be a regular local ring. Let
$\mu: G \to T$ be a smooth dominant morphism of $R$-group schemes
with an $R$-smooth reductive group scheme $G$ and an $R$-torus $T$.
Let  ${\cal F}$ be the covariant functor from the category of
commutative rings to the category of abelian groups
given by
$S \mapsto T(S)/\mu(G(S))$. Then ${\cal F}$ should satisfies purity for $R$
provided that $R$ contains an infinite field.
It might even happen that  ${\cal F}$ satisfies
purity for any regular local ring $R$.

Another functor for which one should expect purity is the following one.
Let $R$ be a regular local ring, $G$ and $G'$  smooth reductive $R$-group schemes,
$\pi: G \to G'$  a smooth dominant $R$-group scheme morphism.
Assume $Z=ker(\pi)$ is  finite
\'etale of multiplicative type over $R$.
The boundary operator
$\delta_{\pi,R}: G'(R) \to H1_{\text{\'et}}(R,Z)$
makes sense and
 is a group homomorphism
\cite[Ch.II, \S 5.6, Cor.2]{Se}.
For an $R$-algebra $S$
set
\begin{equation}
\label{Functor2}
{\cal F}(S)= \textrm{H}^1_{\text{\'et}}(S,Z)/ \textrm{Im}(\delta_{\pi,S}).
\end{equation}
It seems plausible that the functor ${\cal F}$
satisfies purity for $R$
provided that $R$ contains an infinite field.
It may even happen that, for any regular local ring $R$, ${\cal F}$ satisfies
purity. Theorem
\ref{PurityForSubgroup}
states that this is the case if $R$ contains an infinite field $k$
of characteristic zero and $G$, $G'$ and $\pi$
are  already defined  over $k$.




Note that
we use transfers for the functor
$R \mapsto C(R)$, but we do not use at all the norm principle
for the map $\mu: G \to C$.

The preprint is organized as follows. In Section $2$ we construct
 norm maps  following a method from
\cite[Sect.6]{SV}.
In Section $3$ we prove the main geometric Lemmas
\ref{Lemma2}
and
\ref{Lemma1}.
In Section $4$ we discuss
unramified elements.
A key point here is Lemma
\ref{KeyUnramifiedness}.
In Section $5$ we discuss specialization maps.
A key point here is Corollary
\ref{TwoSpecializations}.
In Section $6$ we prove
Theorem (A).
In Section $7$ we prove
Theorem (B).
In Section $8$ we consider the functor
(\ref{Functor2})
and prove in
\ref{PurityForSubgroup}
a purity theorem for it.
Finally in Section $9$ we collect several examples.

\subparagraph{Acknowledgments}
The author thanks very much M.Ojanguren for useful discussions
on the subject of the present article.
%The  author thanks very much for the support the
%RTN-Network HPRN-CT-2002-00287,
%the RFFI-grant 03-01-00633a,
%the Swiss National Science Foundation
%and INTAS-03-51-3251.





\subsection{Norms}
Let $k\subset K\subset L$ be field extensions and assume that $L$
is finite separable over $K$. Let $K^{sep}$ be a separable closure
of $K$ and
$$\sigma_i:K\to K^{sep},\enspace 1\leq i\leq n$$
the different embeddings of $K$ into $L$. As in \S1, let $C$ be a
commutative algebraic group scheme defined over $k$. We can define
a norm map
$${\mathcal N}_{L/K}:C(L)\to C(K)$$
by
$${\mathcal N}_{L/K}(\alpha)=\prod_i C(\sigma_i)(\alpha) \in C(K^{sep})^{{\mathcal
G}(K)} =C(K)\;.
$$
Following Suslin and Voevodsky
\cite[Sect.6]{SV} we
generalize this construction to finite flat ring extensions.
\smallskip
Let $p:X\to Y$ be a finite flat morphism of affine schemes.
Suppose that its rank is constant, equal to $d$. Denote by
$S^d(X/Y)$ the $d$-th symmetric power of $X$ over $Y$.



\begin{lem} There is a canonical section
$$\mathcal N_{X/Y}: Y\to S^d(X/Y)$$
which satisfies the following three properties:
\begin{itemize}
\item
[(1)] Base change: for any map $f:Y'\to Y$ of affine schemes, putting
$X'=X\times_Y Y'$ we have a commutative diagram
$$\xymatrix{Y'\ar[rr]^{\mathcal N_{X'/Y'}}\ar[d]_{f}&&S^d(X'/Y')\ar[d]^{S^d(Id_X\times
f)}\\
             Y\ar[rr]^{\mathcal N_{X/Y}}&&S^d(X/Y)}$$
\item[(2)]
Additivity: If $f_1: X_1\to Y$ and $f_2:X_2\to Y$ are finite
flat morphisms of degree $d_1$ and $d_2$ respectively, then,
putting $X=X_1\coprod X_2$, $f=f_1\coprod f_2$ and $d=d_1+d_2$, we
have a commutative diagram
$$\xymatrix{S^{d_1}(X_1/Y)\times S^{d_2}(X_2/Y)\ar[rr]^(0.63)\sigma&&S^d(X/Y)\\
&&\\
&Y\ar[uul]^{\mathcal N_{X_1/Y}\times \mathcal N_{X_2/Y}}\ar[uur]_{\mathcal N_{X/Y}}& }$$
where $\sigma$ is the canonical imbedding.\\
\item[(3)]
Normalization: If $X=Y$ and $p$ is the identity, then
$\mathcal N_{X/Y}$ is the identity.
\end{itemize}
\end{lem}



\begin{proof} We construct a map $\mathcal N_{X/Y}$ and check that it has the
desired properties. Let $B=k[X]$ and $A=k[Y]$, so that $B$ is a
locally free $A$-module of finite rank $d$. Let $B^{\otimes
d}=B\otimes_A B\otimes_A\cdots \otimes_A B$ be the $d$-fold tensor product
of $B$ over $A$. The permutation group ${\frak S}_d$ acts on
$B^{\otimes d}$ by permuting the factors.  Let $S^d(B)\subseteq
B^{\otimes d}$ be the $A$-algebra of all the ${\frak
S}_d$-invariant elements of $B^{\otimes d}$. We consider
$B^{\otimes d}$ as an $S^d(B)$-module through the inclusion
$S^d(B)\subseteq B^{\otimes d}$ of $A$-algebras. Let $I$ be the
kernel of the canonical  homomorphism $B^{\otimes d}\to
\bigwedge^d(B  )$ mapping $b_1\otimes\cdots\otimes b_d$ to
$b_1\wedge\cdots\wedge b_d$. It is well-known (and easily checked
locally on $A$) that $I$ is generated by all  the elements $x\in
B^{\otimes d}$ such that $\tau(x)=x$ for some transposition
$\tau$. If $s$ is in $S^d(B)$, then $\tau(sx)=\tau(s)\tau(x)=sx$,
hence $sx$ is in $S^d(B)$ too. In other words, $I$ is an
$S^d(B)$-submodule of $B^{\otimes d}$. The induced $S^d(B)$-module
structure on $\bigwedge^d(B)$ defines an $A$-algebra homomorphism
$$\varphi:S^d(B)\to End_A(\bigwedge^d(B))\;.$$
Since $B$ is locally free of rank $d$ over $A$, $\bigwedge^d(B)$
is an invertible $A$-module and we can canonically identify
$End_A(\bigwedge^d(B)$ with $A$. Thus we have a map

$$\varphi:S^d(B)\to A$$
and we define
$$Tr_{X/Y}: Y\to S^d(X)$$
as the morphism of $Y$-schemes induced by $\varphi$. The
verification of properties (1), (2) and (3) is straightforward.
\end{proof}

\smallskip
Let now $C$ be a commutative $k$-group scheme  and $f:X\to C$ any
morphism. We define the norm $N_{X/Y}(f)$ of $f$ as the composite map
\begin{equation}
\label{NormMap}
Y \xra{Tr_{X/Y}}
S^d(X)\xra{S^d(f)} S^d(C)\xra{\times} C
\end{equation}
Here we write $"\times"$ for the group law on $C$.
The norm maps $N_{X/Y}$ satisfy the following conditions
\begin{itemize}
\item[(1)]
Base change: for any map $f:Y'\to Y$ of affine schemes, putting
$X'=X\times_Y Y'$ we have a commutative diagram
$$
\begin{CD}
C(X)                  @>{(id \times f)^{*}}>>            C(X^{\prime})      \\
@V{N_{X/Y}}VV @VV{N_{X^{\prime}/Y^{\prime}}}V    \\
C(Y)                  @>{f^{*}}>>            C(Y^{\prime})
\end{CD}
$$
\item[(2)]
multiplicativity: if $X=X_1 \amalg X_2$ then the diagram commutes
$$
\begin{CD}
C(X)                  @>{(id \times f)^{*}}>>            C(X_1) \times C(X_2)      \\
@V{N_{X/Y}}VV                              @VV{N_{X_1/Y}N_{X_2/Y}}V    \\
C(Y)                  @>{id}>>            C(Y)
\end{CD}
$$
\item[(3)]
normalization: if $X=Y$ and the map $X \to Y$ is the identity then $N_{X/Y}=id_{C(X)}$.
\end{itemize}

\subsection{Geometric lemmas}
In this Section few geometric lemmas will be proved.
Let $k$ be a field of characteristic zero.
The main result of this Section are the lemmas
\ref{Lemma2}
and
\ref{Lemma1}.
Lemma
\ref{Lemma2}
is a refinement of
\cite[Lemma 2]{OP}
and Lemma
\ref{Lemma1}
is a refinement of Quillen's trick.

\begin{lem}
\label{Lemma3}
Let $F$ be a field of characteristic zero and let $S$ be an $F$-smooth algebra
which is a domain of dimension $1$. Let $\mathfrak m_0$ be a maximal ideal with
$S/\mathfrak m_0=F$. Let
$\mathfrak m_1, \mathfrak m_2, \dots, \mathfrak m_n$
be different maximal ideals of $S$
(it might be that $\mathfrak m_0=\mathfrak m_i$ for an index $i$).
Then there exists an element
$\beta \in S$
such that $S$ is finite over $F[\beta]$ and
\begin{itemize}
\item[(1)]
the ideals
$\mathfrak n_i:= \mathfrak m_i \cap F[t]$ ($i=1,2,\dots, n$)
are all different from each other and different  from
$\mathfrak n_0:= \mathfrak m_0 \cap F[t]$
provided that
$\mathfrak m_0$
is different  from
$\mathfrak m_i$'s;
\item[(2)]
the extension $S/F[t]$ is \'etale at each
$\mathfrak m_i$'s and at $\mathfrak m_0$;
\item[(3)]
$F[\beta]/\mathfrak n_i = S/\mathfrak m_i$ for each
$i=1,2,\dots, n$;
\item[(4)]
$\mathfrak n_0= \beta F[\beta]$.
\end{itemize}
\end{lem}

\begin{proof}
Consider  a closed imbedding
$\Spec(S) \hra \Aff^n_F$
and use a generic linear projection to
$\Aff1_F$.

\end{proof}

\begin{lem}
\label{Lemma4}
Under the hypotheses of Lemma
\ref{Lemma3}
let $0 \neq f \in S$
be an element that does not belong to a maximal ideal different of
$\mathfrak m_0, \mathfrak m_1, \dots, \mathfrak m_n$.
Let $N(f)=N_{S/F[\beta]}$ be the norm of $f$. Then for an element
$\beta \in S$ satisfying $(1)$ to $(4)$ of Lemma
\ref{Lemma3}
one has
\begin{itemize}
\item[(a)]
$N(f)=fg$ for an element $g \in S$;
\item[(b)]
$fS + gS= S$;
\item[(c)]
the map $F[\beta]/(N(f)) \to S/(f)$ is an isomorphism.
\end{itemize}
\end{lem}

\begin{proof}
It is straightforward.

\end{proof}


\begin{lem}
\label{Lemma2}
Let $A$ be a local essentially smooth $k$-algebra with a maximal ideal
$\mathfrak m$.
Let $A[t] \subset R$
be an $A[t]$-algebra which is essentially smooth
as a $k$-algebra and is finite over the
polynomial algebra $A[t]$. Assume
$R/\mathfrak m R$
is a domain and assume there exists an element
$h \in R \setminus \mathfrak m R$
such that $R_h$ is $A$-smooth.
Let $\epsilon: R \to A$ be an $A$-augmentation and
$I=ker(\epsilon)$.
Given an $f \in R$ with
$0 \neq \epsilon (f) \in A$
one can find an $s \in R$ satisfying the following conditions
\begin{itemize}
\item[(1)]
$R$ is finite over $A[s]$;
\item[(2)]
$R/sR=R/I \times R/J$ for certain ideal $J$;
\item[(3)]
$J + fR = R$;
\item[(4)]
$(s-1)R + fR = R$;
\item[(5)]
if $N(f)=N_{R/A[s]}(f)$ then $N(f)=fg \in R$ for certain $g \in R$ \\
($R/A[t]$ is a projective module since $R$ and $A[t]$ are regular and $R$ is finite over $A[t]$);
\item[(6)]
$fR + gR = R$;
\item[(7)]
the composition map $A[s]/(N(f)) \to R/(N(f)) \to R/(f)$ is an isomorphism.
\end{itemize}

\end{lem}

\begin{proof}
Replacing $t$ by $t- \epsilon (t)$ we may assume that
$\epsilon (t)=0$.
Since $R$ is finite over $A[t]$ and $R/fR$ is finite over $A$
we conclude that $A[t]/(N(f))$ is finite over $A$, hence
$A/(tN(f))$ is finite over $R$ too. So setting $u=tN(f)$
we get an integral extension $A[t]/A[u]$.
Consider the characteristic polynomial of
the operator
$R \xra{tf} R$ as an $A[u]$-module operator. This polynomial vanishes on
$tf$ and its free coefficient is $N(tf)$ up to a sing. Thus
$N(tf)=tfg$ for an element $g \in R$.
Now replacing
$t$ by $u$ we may assume that $R/A[t]$ is still integral,
$\epsilon (t)=0$ and $t \in fR$.
In fact, $u=N(ft)=ftg$ and whence
$\epsilon (u)=\epsilon (t) \epsilon (fg) =0$
and
$u \in fR$.

We use below ``bar" to denote the reduction modulo the ideal
$\mathfrak mR$. Let
$F=\bar A=A/\mathfrak m$.
By the  assumption of the lemma the $F$-algebra $\bar R$
is an $F$-smooth domain of dimension $1$.
Let
$\mathfrak m_1, \mathfrak m_2, \dots, \mathfrak m_n$
be different maximal ideals of $\bar R$ dividing $\bar f$
and let
$\mathfrak m_0 = ker (\bar {\epsilon})$.
Let $\beta \in \bar R$ be such that the extension
$\bar {R}/F[\beta]$
satisfies the conditions $(1)$ to $(4)$ of Lemma
\ref{Lemma3}.

Let $\alpha \in R$ be a lift of $\beta$, that is
$\bar \alpha=\beta$ in $\bar R$. Replacing $\alpha$ by
$\alpha - \epsilon (\alpha)$
we may assume that
$\epsilon (\alpha) = 0$
and still
$\bar \alpha=\beta$.

Let
$\alpha^n + p_1(t)\alpha^{n-1} + \dots + p_n(t)=0$
be an integral dependence equation for $\alpha$.
Let $N$ be an integer large than the
$max\{2, deg (p_i(t))\}$, where
$i=1, 2, \dots, n$. Then for any
$r \in k^{\times}$
the element
$s= \alpha -rt^N$
is such that $t$ is integral over $A[s]$.
Thus for any
$r \in k^{\times}$
the ring $R$ is integral over $A[s]$.


On the other hand the element
$\bar s = \bar \alpha -r\bar {t}^N$
still satisfies the conditions
$(1)$ to $(4)$ of Lemma
\ref{Lemma3}
because for any
$i=1, 2, \dots, n$ one has
$\bar t \in \mathfrak m_i$.

We claim that the element $s \in R$ is the required element
(for almost all $r \in k^{\times}$).

In fact, for almost all
$r \in k^{\times}$
the element $s$ satisfies the conditions
$(1)$ to $(4)$ of Lemma
\cite[Lemma 2]{OP}.
It remains to show that the conditions
$(5)$ to $(7)$ are satisfied for all
$r \in k^{\times}$.

To prove $(5)$ consider the characteristic polynomial of the
operator $R \xra{f} R$ as an $A[s]$-module operator.
This polynomial vanishes on $f$ and its free coefficient is
$\pm N(f)$ (the norm of $f$). Thus
$f^n - a_1f^{n-1} + \dots \pm N(f)=0$
and $N(f)=fg$ for an element $g \in R$.

To prove $(6)$ one has to check that this $g$ is a unit
modulo the ideal $fR$. It suffices to check that
$\bar g \in \bar R$
is a unit modulo the ideal $\bar f \bar R$. The field
$F=A/\mathfrak m$, the $F$-algebra $S=\bar R$,
its maximal ideals
$\mathfrak m_0, \mathfrak m_1, \dots, \mathfrak m_n$
and the element satisfy the hypotheses of Lemma
\ref{Lemma4}
with $\beta$ replaced by $\bar s$.
Now $\bar g$ is a unit modulo the ideal $\bar f \bar R$
by the item $(b)$ of Lemma
\ref{Lemma4}.

To prove $(7)$ note that $R/fR$ and $A[s]/(N(f))$ are
finite $A$-modules. So it remains to check that the map
$\varphi: A[s]/(N(f)) \to R/fR$
is an isomorphism modulo the maximal ideal
$\mathfrak m$. For that it suffices to verify that the map
$\bar \varphi: F[\bar s]/(N(\bar f)) \to \bar R/\bar f \bar R$
is an isomorphism where
$N(\bar f):=N_{\bar R/F[\bar s]}$.
Now $\bar \varphi$ is an isomorphism by the item $(c)$ of Lemma
\ref{Lemma4}.
The lemma follows.

\end{proof}

\begin{cor}
\label{NormAtZeroAndOne}
Under the hypotheses of Lemma
\ref{Lemma2} let $K$ be the quotient field of $A$,
$R_K=R \otimes_A K$ and $\epsilon_K=\epsilon \otimes_A id: R_K \to K$.
Consider the inclusion
$K[s] \subset R_K$. Then the norm
$N(f) \in K[s]$ does not vanish  at the points $1$ and $0$ of
the affine line $\Aff1_K$.
\end{cor}

\begin{proof}
The condition $(4)$ of
\ref{Lemma2}
implies that $N(f)$ does not vanish at the point $1$.
Since $\epsilon_K (f) \neq 0 \in K$
the conditions $(2)$ and $(3)$ imply that
$N(f)$ does not vanish at $0$ either.

\end{proof}

\begin{lem}
\label{Lemma1}
Let $X$ be a geometrically irreducible $k$-smooth affine variety of
dimension $d$. Let $x \in X$ be a closed point,
$0\neq f \in k[X]$
and $Z$ be the vanishing locus of $f$.
Then there exists a commutative diagram
\begin{equation}
\label{Triangle}
    \xymatrix{
    X \ar[rr]^-{\varrho}\ar[rd]_-{q} && \Aff^d \ar[ld]^-{pr} &   \\
    & \Aff^{d-1}  \\
    }
\end{equation}
with a linear projection $pr$ such that
\begin{itemize}
\item[(1)]
there exists a principle open set $X_{\textrm{h}} \subset X$ containing
the fibre $q^{-1}q(x)$ such that the map
$q: X_{\textrm{h}} \to \Aff^{d-1}$
is smooth;
\item[(2)]
the fibre $q^{-1}q(x)$ is geometrically irreducible;
\item[(3)]
the map $\varrho$ is finite;
\item[(4)]
the map $q|_Z: Z \to \Aff^{d-1}$ is finite;
\item[(5)]
k(x)=k(q(x)).
\end{itemize}
\end{lem}

\begin{proof}
Bertini plus generic projection. Consider a closed imbedding
$X \hra \Aff^n_k$, the closure $\bar X$ of $X$ in the projective
space $\Pro^n_k$ and the complement $X_{\infty}$ of $X$ in $\bar X$.
Let $\bar Z$ be the closure of $Z$ in $\Pro^n_k$ and
$Z_{\infty}$ be the complement of $Z$ in $\bar Z$.
Since $dim X=d$ one has $dim X_{\infty}=d-1$. Set
$\Pro^{n-1}_k=\Pro^n_k - \Aff^n_k$. Now extend the scalars  to an algebraic closure
$\bar k$ of $k$. Let
$x_1, x_2, \dots, x_n \in X(\bar k)$
be the points of
$x \otimes_k \bar k$.
Let
$\Pi \subset \Pro^{n-1}_{\bar k}$
be a  linear subspace of dimension $n-d$ which is general in the sense that
\begin{itemize}
\item[(i)]
$\Pi \cap X_{\infty,\ \bar k}$ consists of finitely many points;
\item[(ii)]
$\Pi \cap Z_{\infty,\ \bar k} = \emptyset$;
\item[(iii)]
if $<\Pi, x_i>$ is the space generated by $\Pi$ and by an $x_i$ then
the $<\Pi, x_i> \cap X_{\bar k}$ is $\bar k$-smooth and irreducible
for each $i=1, 2, \dots, n$;
\item[(iv)]
$<\Pi, x_i> \neq <\Pi, x_j>$ for $i \neq j$.
The requirement $(iii)$ can be satisfied by a Bertini type theorem
for a hyperplane section crossing a given rational point because we are
in characteristic zero.
\end{itemize}
Since the set of linear spaces $\Pi$ satisfying conditions $(i)$ to $(iii)$
is an open set of a Grassmannian we may choose a $\Pi$ which is already defined
over the ground field $k$. We may then choose
a hyperplaneplane $\pi \subset  \Pi$ of dimension $n-d-1$,  defined over $k$
and such that
\begin{equation}
\label{Condition5}
\pi \cap X_{\infty,\ \bar k} \neq \emptyset .
\end{equation}
\end{proof}
Now consider the linear projection
$\Pro^n_k \setminus \pi \to \Pro^n_d$.
Its restriction to $\bar X$ defines a finite morphism
$\bar \rho: \bar X \to \Pro^d_k$.
And its restriction to $X$ defines a finite morphism
$\rho: X \to \Aff^d_k$,
because it is a base change of $\bar \rho$ by means of
the open imbedding
$\Aff^d_k \hra \Pro^d_k$.

The linear projection with the center $\Pi$
defines a morphism
$q: X \to \Aff^{n-1}_k$
which clearly fits in the commutative diagram
(\ref{Triangle})
with a linear projection
$pr: \Aff^d_k \to \Aff^{d-1}_k$.
We claim that these
$\rho$, $q$ and $pr$
are the required morphisms.

To prove this recall that
$\varrho$ is finite. The morphism
$q|_Z: Z \to \Aff^{d-1}_k$
is finite too, because
$\Pi \cap Z_{\infty} = \emptyset$.
Condition $(5)$ is satisfied because
$p \otimes_k \bar k$
separates the closed points $x_i$'s of the scheme
$x \otimes_k \bar k$.
The fibre of the morphism
$p \otimes_k \bar k$
over the point $x_i$ coincides with the
intersection
$<\Pi, x_i> \cap X_{\bar k}$
for each $i=1, 2, \dots, n$.
After the scalar extension $k$ to $\bar k$ the scheme theoretic fibre
of $q$ over $q(x)$ becomes a disjoint union of the fibres of
$q \otimes_k \bar k$
over the points
$(q \otimes_k \bar k)(x_i)$.
Since $k(q(x))=k(x)$ and for each $i$ the intersection
$<\Pi, x_i> \cap X_{\bar k}$
is irreducible we conclude that the fibre $q^{-1}q(x)$
is irreducible too, and thus condition $(2)$ is verified.
Since for each $i$
the intersection
$<\Pi, x_i> \cap X_{\bar k}$
is smooth we conclude that the scheme theoretic fibre
of $q$ over $q(x)$
is smooth too.
Thus condition $(1)$ is satisfied.
The lemma is proved.


\subsection{Unramified elements}
Let $k$ be the characteristic zero field.
We work in this section with {\it the category of commutative $k$-algebras}.
Let  $\mathcal F$ be a covariant functor from the category of commutative
$k$-algebras to the category of abelian groups. Let $K$ be a field containing $k$ and
$R\subset K$ be a $k$-subalgebra  whose field of fractions is $K$. We
define the {\it subgroup of $R$-unramified elements of $K$} as
$$
\mathcal F_{nr,R}(K)=
\bigcap_{\mathfrak p \in Spec(R)^{(1)}} Im [ \mathcal F(R_{\mathfrak p})\to\mathcal F(K) ],
$$
where $Spec(R)^{(1)}$ is the set of hight $1$ prime ideals in $R$.
Obviously the image of $\mathcal F(R)$ in $\mathcal F(K)$ is contained in
$\mathcal F_{nr,R}(K)$. In most cases
$\mathcal F(R_{\mathfrak p})$ injects into $\mathcal F(K)$ and
$\mathcal F_{nr,R}(K)$ is simply the intersection of all $\mathcal F(R_p)$.

A basic functor we are interested in is the following one.
Let
$\mu: G\to C$
be the morphism of linear algebraic $k$-groups from Theorem
\ref{TheoremA}. For a commutative $k$-algebra $R$ set
\begin{equation}
\label{MainFunctor}
{\cal F}(R)=C(R)/\mu(G(R)).
\end{equation}
For an element $\alpha \in C(R)$ we will write $\bar \alpha$ for its
image in ${\cal F}(R)$. In this section we will write  ${\cal F}$
for the functor
(\ref{MainFunctor}),
the only exception being Lemma
\ref{UnramifiednessLemma}.
We will repeatedly use  the following result
\begin{thm}[\cite{C-TO}]
\label{NisnevichThm}
Let $H$ be a linear algebraic group over the field $k$. Let
$R$ be a discrete valuation ring with a fraction field $K$.
Then the map of pointed sets
$$
H^1_{\text{\'et}}(R,H) \to H^1_{\text{\'et}}(K,H)
$$
induced by the inclusion $R \subset K$ has the trivial kernel.
\end{thm}

\begin{cor}[\cite{Ni}] %%%%%% This is due to Nisnevich, the label should be
%%%%%%%%%%%% put as a reference, not as an attribution
\label{NisnevichCor}
Let $R$ be a discrete valuation ring with  fraction field $K$.
The map
${\cal F}(R) \to {\cal F}(K)$
is injective.
\end{cor}

\begin{proof}
Let $H$ be the kernel of $\mu$. Then the boundary map
$\partial: C(R) \to \textrm{H}^1_{\text{\'et}}(R,H)$
fits in a commutative diagram
$$
\CD
C(R)/\mu (G(R)) @>>> C(K)/\mu (G(K)) \\
@VVV @VVV \\
\textrm{H}^1_{\text{\'et}}(R,H) @>>> \textrm{H}^1_{\text{\'et}}(K,H). \\
\endCD
$$
in which the vertical arrows have  trivial kernels.
The bottom arrow has  trivial kernel by Theorem
\ref{NisnevichThm}.
Thus the top arrow has  trivial kernel too.

\end{proof}

\begin{lem}
\label{BoundaryInj}
Let $\mu: G \to C$ be the morphism of linear algebraic groups.
Let $H= \ker (\mu)$.
Then for any field $K$ the boundary map
$\partial: C(K)/{\mu (G(K))} \to \textrm{H}^1_{\text{\'et}}(K,H)$
is injective.
\end{lem}

\begin{proof}
For a $K$-rational point $t \in C$ set
$H_t = \mu^{-1}(t)$. The action by  left multiplication of $H$ on $H_t$
makes $H_t$ into a left principal homogeneous
$H$-space and moreover
$\partial (t) \in \textrm{H}^1_{\text{\'et}}(K,H)$
coincides with the isomorphism class of
$H_t$. Now suppose that $s,t \in C(K)$ are such that
$\partial (s)=\partial (t)$. This means that
$H_t$ and $H_s$ are
isomorphic as  principal homogeneous $H$-spaces.
We must check that for certain $g \in G(K)$ one has
$t=sg$.

Let $K^{sep}$ be a separable closure of $K$. Let
$\psi: H_s \to H_t$
be an isomorphism of  left $H$-spaces. For any
$r \in H_s(K^{sep})$
and
$h \in H_s(K^{sep})$
one has
$$
(hr)^{-1}\psi (hr)= r^{-1}h^{-1}h \psi (r)= r^{-1} \psi (r).
$$
Thus for any
$\sigma \in Gal(K^{sep}/K)$
and any
$r \in H_s(K^{sep})$
one has
$$
r^{-1}\psi(r)=(r^{\sigma})^{-1}\psi (r^{\sigma})=(r^{-1}\psi(r))^{\sigma}
$$
which means that the point
$u=r^{-1}\psi(r)$ is a $Gal(K^{sep}/K)$-invariant point of
$G(K^{sep})$. So $u \in G(K)$.
The following relation shows that the $\psi$ coincides with the right multiplication
by $u$. In fact, for any $r \in H_s(K^{sep})$ one has
$\psi (r)= rr^{-1}\psi (r)= ru$. Since $\psi$ is the right multiplication
by $u$ one has $t=s\mu(u)$, which proves the lemma.

\end{proof}


Let $K$ be a field and
$x: K^* \to \mathbb Z$
be a discrete valuation vanishing on $k$.
Let $A_x$ be the valuation ring of $x$.
Let
$\hat A_x$ and $\hat K_x$
be the completions of $A_x$ and $K$ with respect to $x$.
Let
$i:K \hookrightarrow \hat K_x$
be the inclusion. By Lemma
\ref{NisnevichCor}
the map
${\cal F}(\hat A_x)\to {\cal F}(\hat{K}_x)$
is injective. We will identify
${\cal F}(\hat A_x)$
with its image under this map. Set
$$
{\cal F}_x(K)=i_*^{-1}({\cal F}(\hat A_x)).
$$


The inclusion
$A_x\hookrightarrow K$
induces a map
$
{\cal F}(A_x) \to {\cal F}(K)
$
which is injective by Lemma
\ref{NisnevichCor}.
So both groups
${\cal F}(A)$ and ${\cal F}_x(K)$
are subgroups of
${\cal F}(K)$.
The following lemma shows that
${\cal F}_x(K)$
coincides with the subgroup of $
{\cal F}(K)$
consisting of all elements {\it unramified} at $x$.

\begin{lem}
\label{TwoUnramified}
${\cal F}(A_x)={\cal F}_x(K)$.
\end{lem}

\begin{proof}
We only have to check the inclusion
$
{\cal F}_x(K) \subseteq {\cal F}(A_x)
$.
Let
$
a_x \in {\cal F}_x(K)
$
be an element. It determines the elements
$
a \in {\cal F}(K)
$
and
$
\hat a \in {\cal F}(\hat A_x)
$
which coincide when regarded as elements of
$
{\cal F}(\hat{K}_x)
$.
We denote this common element in
$
{\cal F}(\hat{K}_x)
$
 by
$
\hat a_x
$.
Let $H=\ker (\mu)$ and let
$\partial: C(-) \to \textrm{H}^1_{\text{\'et}}(-,H)$
be the boundary map.

Let
$
\xi=\partial(a)\in \textrm{H}^1_{\text{\'et}}(K,H)
$,
$
\hat \xi=\partial(\hat a)\in \textrm{H}^1_{\text{\'et}}(\hat A_x,H)
$
and
$
\hat \xi_x=\partial(\hat a_x)\in \textrm{H}^1_{\text{\'et}}(\hat K_x,H)
$.
Clearly,
$
\hat \xi
$
and
$
\xi
$
both coincide with
$
\hat \xi_x
$
when regarded as elements of
$
\textrm{H}^1_{\text{\'et}}({\hat K}_x,H)
$.
Thus one can glue
$
\xi
$
and
$
\hat \xi
$
to get a
$
\xi_x \in \textrm{H}^1_{\text{\'et}}(A_x,H)
$
which maps to $
\xi
$
under the map induced by the inclusion
$
A_x \hookrightarrow K
$
and maps to
$
\hat \xi
$
under the map induced by the inclusion
$
A_x \hookrightarrow \hat A_x
$.

We now show that
$
\xi _x
$
has the form
$
\partial (a_x^\prime)
$
for a certain
$
a_x^\prime \in {\cal F}(A_x)
$.
In fact, observe that the image
$
\zeta
$
of
$
\xi
$
in
$
\textrm{H}^1_{\text{\'et}}(K, G)
$
is  trivial.
By Theorem
\ref{NisnevichThm}
the map
$$
\textrm{H}^1_{\text{\'et}}(A_x, G) \to \textrm{H}^1_{\text{\'et}}(K,G)
$$
has  trivial kernel. Therefore the image
$
\zeta_x
$
of
$
\xi_x
$
in
$
\textrm{H}^1_{\text{\'et}}(A_x, G)
$
is trivial. Thus there exists an element
$
a_x^\prime \in {\cal F}(A_x)
$
with
$
\partial(a_x^\prime)=\xi_x \in \textrm{H}^1_{\text{\'et}}(A_x, H).
$

We now prove that
$
a_x^\prime
$
coincides with
$
a_x
$
in
$
{\cal F}_x(K)
$.
Since
$
{\cal F}(A_x)
$
and
$
{\cal F}_x(K)
$
are both subgroups of
$
{\cal F}(K)
$,
it suffices to show that
$
a_x^\prime
$
coincides with the element
$
a
$
in
$
{\cal F}(K)
$.
By Lemma
\ref{BoundaryInj}
the map
$$
{\cal F}(K)\xrightarrow{\partial} \textrm{H}^1_{\text{\'et}}(K, H)
$$
is \underbar {injective}. Thus it suffices to check that
$
\partial(a_x^\prime)=\partial(a)
$
in
$
\textrm{H}^1_{\text{\'et}}(K, H)
$.
This is indeed the case because
$
\partial(a_x^\prime)=\xi_x
$
and
$
\partial(a)=\xi
$,
and
$
\xi_x
$
coincides with
$
\xi
$
when regarded over
$
K
$.
We have proved that
$
a_x^\prime \in {\cal F}(A_x)
$
coincides with
$
a_x
$
in
$
{\cal F}_x(K)
$.
Thus the inclusion
$
{\cal F}_x(K) \subseteq {\cal F}(A_x)
$
is proved, whence the lemma.

\end{proof}



\begin{lem}
\label{KeyUnramifiedness}
Let $A \subset B$ be a finite extension of Dedekind $k$-algebras.
Let $0 \neq f \in B$ be such that $B/fB$ is reduced.

Suppose $N_{B/A}(f)=fg \in B$ for a certain $g \in B$ coprime
with $f$. Suppose the composite map
$A/N(f)A \to B/N(f)B \to B/fB$
is an isomorphism.
Let $F$ and $E$ be the quotient fields of $A$ and $B$ respectively.
Let $\beta \in C(B_f)$ be such that
$\bar \beta \in {\cal F}(E)$
is $B$-unramified. Then, for
$\alpha= N_{E/F}(\beta)$,
the class
$\bar \alpha \in {\cal F}(F)$
is $A$-unramified.
\end{lem}

\begin{proof}
The only primes at which $\bar \alpha$ could be ramified are those which divide
$N(f)$. Let $\mathfrak p$ be one of them. Check that $\bar \alpha$ is
unramified at $\mathfrak p$.

To do this we consider all primes
$\mathfrak q_1, \mathfrak q_2, \dots, \mathfrak q_n$
in $B$ lying over
$\mathfrak p$.
Let
$\mathfrak q_1$
be the unique prime dividing $f$ and lying over
$\mathfrak p$.
Then
$$
\hat B_{\mathfrak p} = \hat B_{\mathfrak q_1} \times \prod_{i \neq 1} \hat B_{\mathfrak q_i}
$$
with
$\hat B_{\mathfrak q_1}=\hat A_{\mathfrak p}$.
If $F$, $E$ are the fields of fractions of $A$ and $B$
then
$$
E \otimes \hat F_{\mathfrak p}=\hat E_{\mathfrak q_1} \times \dots \times \hat E_{\mathfrak q_n}
$$
and
$\hat E_{\mathfrak q_1}=\hat F_{\mathfrak p}$.
We will write
$\hat E_i$ for $\hat E_{\mathfrak q_i}$
and
$\hat B_i$ for $\hat B_{\mathfrak q_i}$.
Let
$\beta \otimes 1= (\beta_1,\dots, \beta_n)
\in C(\hat E_1) \times \dots \times  C(\hat E_n)$.
Clearly for $i \geq 2$
$\beta_i \in C(\hat B_i)$
and
$\beta_1=\mu(\gamma_1)\beta^{\prime}_1$
with
$\beta^{\prime}_1 \in C(\hat B_1)=C(\hat A_{\mathfrak p})$
and
$\gamma_1 \in G(\hat E_1)=G(\hat F_{\mathfrak p})$.
Now
$\alpha \otimes 1 \in C(\hat F_{\mathfrak p})$
coincides with the product
$$
\beta_1N_{\hat L_2/\hat K_{\mathfrak p}}(\beta_2)\cdots N_{\hat L_n/\hat K_{\mathfrak p}}(\beta_n)=
\mu(\gamma_1)
[\beta^{\prime}_1N_{\hat L_2/\hat K_{\mathfrak p}}(\beta_2)\cdots N_{\hat L_n/\hat K_{\mathfrak p}}(\beta_n)].
$$
Thus
$\overline {\alpha \otimes 1}=\bar \beta^{\prime}_1\overline {N_{\hat L_2/\hat K_{\mathfrak p}}(\beta_2)}
\cdots \overline {N_{\hat L_n/\hat K_{\mathfrak p}}(\beta_n)} \in {\cal F}(\hat A_{\mathfrak p})$.
Let
$i: F \hra \hat F_{\mathfrak p}$
be the inclusion and
$i_*:{\cal F}(F) \to {\cal F}(\hat F_{\mathfrak p})$
be the induced map.
Clearly
$i_*(\bar \alpha)=\overline {\alpha \otimes 1}$
in
${\cal F}(\hat F_{\mathfrak p})$.
Now
Lemma
\ref{TwoUnramified}
shows that
the element
$\bar \alpha \in {\cal F}(F)$
belongs to
${\cal F}(A_{\mathfrak p})$.
Hence $\bar \alpha$ is $A$-unramified.

\end{proof}

\begin{lem}[Unramifiedness Lemma]
\label{UnramifiednessLemma}
Let  $\mathcal F$ be a covariant functor from the category of commutative
$k$-algebras to the category of abelian groups.
Let $S$ and $R$ be noetherian domains with  quotient fields $K$ and $L$
respectively.
Let
$
S \xrightarrow{i} R
$
be an injective flat homo\-morph\-ism of finite type and
let
$
j :K \to L
$
be the induced inclusion of the quotient fields.
Then for each localization
$R^{\prime} \supset R$ of $R$
the map
$$
j _*:{\cal F}(K) \to {\cal F}(L)
$$
takes
$S$-unramified elements to
$R^{\prime}$-unramified elements.
\end{lem}

\begin{proof}
Let
$v\in K^*$
and let
$\mathfrak r$
be  height $1$ primes of $R^{\prime}$. Then
$\mathfrak q = R \cap \mathfrak r$
is a height $1$ prime of $R$. Let
$\mathfrak p = S \cap \mathfrak q$.
Since the $S$-algebra $R$ is flat of finite type one has
$\text{ht}(\mathfrak q) \geq \text{ht}(\mathfrak p)$. Thus
$\text{ht}(\mathfrak p)$
is $1$ or $0$. The commutative diagram
$$
\CD
{\cal F}(K) @>>> {\cal F}(L) \\
@AAA @AAA \\
{\cal F}(S_{\mathfrak p}) @>>> {\cal F}(R^{\prime}_{\mathfrak r}) \\
\endCD
$$
shows that the class
$j_* (v)$
is in the image of
${\cal F}(R^{\prime}_{\mathfrak r})$ and
hence the class
$j_* (v) \in {\cal F}(L)$
is $R^{\prime}$-unramified.

\end{proof}

\subsection{Specialization maps}
In this Section we consider the functor $R \mapsto {\cal F}(R)$ defined by
(\ref{MainFunctor}). For a regular ring $R$ containing $k$
and each height one prime $\mathfrak p$ in $R$ we construct specialization maps
$s_{\mathfrak p}: {\cal F}_{nr, R}(K) \to {\cal F}(k(\mathfrak p))$,
where $K$ is the quotient field of $R$ and
$k(\mathfrak p)$ is the residue field of $R$ at the prime $\mathfrak p$.


\begin{defn}
\label{SpecializationDef}
Let
$Ev_{\mathfrak p}: C(R_{\mathfrak p}) \to C(k(\mathfrak p))$
and
$ev_{\mathfrak p}: {\cal F}(R_{\mathfrak p}) \to {\cal F}(k(\mathfrak p))$
be the maps induced by
$R_{\mathfrak p} \to k(\mathfrak p)$.
Define a homomorphism
$s_{\mathfrak p}: {\cal F}_{nr, R}(K) \to {\cal F}(k(\mathfrak p))$
by
$s_{\mathfrak p}(\alpha)=ev_{\mathfrak p}(\tilde \alpha)$
where
$\tilde \alpha$
is a lift of $\alpha$ in
${\cal F}(R_{\mathfrak p})$.
Corollary
\ref{NisnevichCor}
shows that the map $s_{\mathfrak p}$ is well-defined.
It is called the specialization map. The map $ev_{\mathfrak p}$ is called the evaluation
map at the prime $\mathfrak p$.

Obviously for
$\alpha \in C(R_\mathfrak p)$
one has
$s_{\mathfrak p}(\bar \alpha)=\overline {Ev_{\mathfrak p}(\alpha)}
\in {\cal F}(k(\mathfrak p))$.
\end{defn}

\begin{lem}[\cite{H}]
\label{Harder}
Let $H$ be a linear algebraic group over the field $k$.
Let $R$ be a $k$-algebra which is a Dedekind domain with  quotient field $K$.
If $\xi \in \textrm{H}^1_{\text{\'et}}(K,H)$ is an $R$-unramified element for the functor
$\textrm{H}^1_{\text{\'et}}(-,H)$
then $\xi$ can be lifted  to an element of
$\textrm{H}^1_{\text{\'et}}(R,H)$.
\end{lem}

\begin{proof}
Patching.

\end{proof}

\begin{thm}
\label{C-TOthm}
Let $K$ be a field of characteristic zero and let $H$ be a linear algebraic group
over $K$. Then the canonical map
$\textrm{H}^1_{\text{\'et}}(K,H) \to \textrm{H}^1_{\text{\'et}}(K[t],H)$
is bijective.
\end{thm}
\begin{proof}
Let $i: \{0\} \hra \Aff1_K$ be the origin of the affine line.
Let $i^*: \textrm{H}^1_{\text{\'et}}(K[t],H) \to \textrm{H}^1_{\text{\'et}}(K,H)$
be the pull-back map. To prove the theorem we need the following

\begin{clm}
\label{InjectivityAndKernel}
The following two statements are equivalent.
\begin{itemize}
\item[(1)]
For any linear algebraic group $H$ over $K$ the map $i^*$ is injective.
\item[(2)]
For any linear algebraic group $H$ over $K$ the map $i^*$ has the trivial kernel.
\end{itemize}
\end{clm}
Clearly the first statement implies the second one. Now suppose the second statement
holds and prove the first one. For that consider
$\xi, \xi' \in \textrm{H}^1_{\text{\'et}}(K[t],H)$
and set
$\xi_0 = i^*(\xi), \xi'_0= i^*(\xi')$.
Let $H_0$ be the inner form of the group $H$ with respect to the $H$-torsor ${\cal H}_0$.
For any $K$-scheme $S$ there is a well-known bijection
$\phi_S: \textrm{H}^1_{\text{\'et}}(S,H) \to \textrm{H}^1_{\text{\'et}}(S,H_0)$
between non-pointed sets. It takes the $H$-torsor
${\cal H}_0 \times_K S$ to the trivial $H_0$-torsor $H_0 \times_K S$,
where ${\cal H}_0$ is an $H$-torsor over $K$ representing the class $\xi_0$.
These bijections
respect $K$-scheme morphisms.
Suppose
$\xi_0=\xi'_0$, then
$$
i^*(\phi_{\Aff1}(\xi))= \phi_K (\xi_0)= * =\phi_K(\xi'_0)=i^*(\phi_{\Aff1}(\xi'))
\in \textrm{H}^1_{\text{\'et}}(K,H_0).
$$
So
$\phi_{\Aff1}(\xi)= * = \phi_{\Aff1}(\xi') \in \textrm{H}^1_{\text{\'et}}(K[t],H_0)$.
Whence
$\xi = \xi'$ and the injectivity follows.
With this Claim in hand prove the theorem as follows.

The surjectivity of the map $i^*$ is obvious.
To prove its injectivity just use the Claim and
mimic the proof of
\cite[Thm.2.1, Property P2]{C-TO}. Let
$p: \Aff1_K \to \textrm{Spec}(K)$
be the structural morphism.
Since $p \circ i =id$ and $i^*$ is bijective
$p^*$ is  bijective.

\end{proof}


We need the following theorem.
\begin{thm}[Homotopy invariance]
\label{HomInvNonram}
Let $k$ be the field and $\mu: G \to C$ be the algebraic group morphism
from Theorem
\ref{TheoremA}.
Let $k \subset K$ be a field extension and
$K(t)$ be the rational function field in one variable.
Let $R \mapsto {\cal F}(R)$ be the functor defined in
\ref{MainFunctor}. Then one has
$$
{\cal F}(K)={\cal F}_{nr,K[t]}(K(t)).
$$
\end{thm}

\begin{proof}
Injectivity is clear, because the composition
$$
{\cal F}(K) \to {\cal F}_{un}(K(t)) \xrightarrow{s_0} {\cal F}(K)
$$
coincides with the identity (here
$
s_0
$
is the specialization map at the point zero defined in 4.6).

It remains to check the surjectivity. Let
$a\in {\cal F}_{un}(K(t))$
and let
$H=ker(\mu)$.
Then by Lemma
\ref{NisnevichCor}
the element
$
\partial (a)\in H_{et}^1(K(t),H)
$
is a class which for every
$
x\in \Bbb A_K1
$
belongs to the image of
$
H_{et}^1({\cal O}_x,H)
$.
Thus by Lemma
\ref{Harder},
$
\partial (a)
$
can be represented by an element
$
\xi \in H_{et}^1(K[t],H)
$,
where
$
K[t]
$
is the polynomial ring.
By Theorem
\ref{C-TOthm},
the map
$$
H_{et}^1(K,H) \to H_{et}^1(K[t],H)
$$
is an isomorphism. Then
$
\xi = \rho (\xi _0)
$
for an element
$
\xi _0\in H_{et}^1(K,H)
$.
Consider the diagram
$$
\SelectTips{cm}{}
\xymatrix @C=2pc @R=10pt {
{} & a \ar@{|->}[r] & {\xi} \ar@{|->}[r] & {\ast} & {} \\
1 \ar[r] & {{\cal F}(K(t))} \ar[r]^{\partial\qquad} & {H_{et}^1(K(t),H)} \ar[r] & {H_{et}^1(K(t),G)} \\
{} & {} & {} & {} & {} \\
1 \ar[r] & {{\cal F}(K)} \ar[r]^{\partial\qquad} \ar[uu]_{\epsilon} & {H_{et}^1(K,H)} \ar[r] \ar[uu]_{\rho} & {H_{et}^1(K,G)} \ar[uu]_{\eta}\\
{} & {a_0} \ar@{|->}[r] & {\xi_0 \ ,} & {} & {} }
$$
in which all the maps are canonical and all the vertical arrows
have trivial kernels. Since
$
\xi
$
goes to the trivial element in
$
H_{et}^1(K(t),G)
$,
one concludes that
$
\xi_0
$
goes to the trivial element in
$
H_{et}^1(K,G)
$.
Thus there exists an element
$
a_0 \in {\cal F}(K)
$
such that
$
\partial (a_0)=\xi _0.
$
The map
$
{\cal F}(K(t)) \to H_{et}^1(K(t),H)
$
is injective by Lemma
\ref{BoundaryInj}.
Thus
$
\epsilon(a_0)=a.
$

\end{proof}

\begin{cor}
\label{TwoSpecializations}
Let $R \mapsto {\cal F}(R)$ be the functor defined in
(\ref{MainFunctor}).
Let
$$
s_0, s_1: {\cal F}_{un}(K(t)) \to {\cal F}(K)
$$
be the specialization maps at zero and at one
(at the primes (t) and (t-1)). Then $s_0=s_1$.
\end{cor}

\begin{proof}
It is an obvious consequence of Theorem
\ref{HomInvNonram}.
\end{proof}

\subsection{Proof of Theorem (A)}
\begin{proof}
By  assumption there exists a smooth $d$-dimensional
$k$-algebra $S=k[t_1,t_2, \dots, t_n]$ and a prime ideal
$\mathfrak p$ in $R$ such that $A=S_{\mathfrak p}$.
We first reduce the proof to the case in which $\mathfrak p$
is maximal.
To do this,  choose a maximal ideal ${m}$
containing ${p}$. Since $k$ is of characteristic zero it is infinite.
By a standard
general position argument we can find $d$ algebraically
independent elements $X_1,\dots,X_d$ such that $S$ is finite over
$k[X_1,\dots,X_d]$ and \'etale at ${m}$. After a linear
change of coordinates we may assume that $S/{p}$ is finite
over $B=k[X_1,\dots,X_m]$, where $m$ is the dimension of
$S/{p}$. Clearly $S$ is smooth over $B$ at ${m}$ and thus, for
some $h\in S\setminus {m}$, the localization $S_h$ is smooth
over $B$. Let $\cal S$ be the set of nonzero elements of $B$,
$k'={\cal S}^{-1}B$ the field of fractions of $B$ and
$S'={\cal S}^{-1}S_h$. The
prime ideal ${p}'={\cal S}^{-1}{p}_h$ is maximal in $S'$, the
$k'$-algebra $S'$ is smooth and $S=S'_{{p}'}$.

%%% FROM PROBLEM:
%>From now on and till the end of this proof
From now on and till the end of this proof
we assume
that $A={\mathcal O}_{X,x}$ is the local ring of a closed point $x$ of
a smooth $d$-dimensional irreducible affine variety $X$ over $k$. Replacing
$k$ by its algebraic closure in $k[X]$ we may assume furthermore that
this $X$ is a $k$-smooth geometrically irreducible affine variety over $k$.

Let $\xi \in C(K)$ be such that the
$\bar \xi \in {\cal F}(K)$
is $A$-unramified. We may assume that
$\xi \in C(k[X]_\textrm{f})$
for an appropriative non-zero $\textrm{f} \in k[X]$
with a reduced $k$-algebra $k[X]/(\textrm{f})$.
Shrinking $X$ we may assume that
$\bar \xi \in {\cal F}(k[X]_\textrm{f})$
is $k[X]$-unramified.
By Lemma
\ref{Lemma1}
we can find a commutative diagram
\begin{equation}
\label{Diagram1}
    \xymatrix{
    X \ar[rr]^-{\varrho}\ar[rd]_-{q} && \Aff^d \ar[ld]^-{pr} &   \\
    & \Aff^{d-1}  \\
    }
\end{equation}
with a linear projection $pr$ and a function
$\textrm{h} \in k[X]$
such that the conditions
$(1)$ to $(5)$ of are satisfied. Let
$U=\textrm{Spec}(A)$ the point $x \in X$ is the closed point of $U$.
Consider the Cartesian
square of schemes
$$
\xymatrix{
    {} & \mathcal X \ar[rr]^-{p_X}\ar[d]_-{p} && X \ar[d]^-{q} &   \\
    U \ar[ru]^{\delta}\ar[r]^{id} & U \ar[rr]^{r} && \Aff^{d-1}  \\
    }
$$
where
$r=q|_{U}$, ${\cal X}=U \times_{\Aff^{d-1}} X$, $p$
is the first projection and
$\delta$
is the diagonal map.

Applying base change by means of $r$ to
the diagram
(\ref{Diagram1})
we get a commutative triangle
\begin{equation}
    \xymatrix{
    {\cal X} \ar[rr]^-{\lambda}\ar[rd]_-{p} && U \times \Aff1 \ar[ld]^-{pr} &   \\
    & U  \\
    }
\end{equation}
with a finite surjective $\lambda$.
Set
$f=p_X^*(\textrm{f})$
and
$h=p_X^*(\textrm{h})$,
where
$\textrm{h}$ is from the condition $(1)$ of Lemma
\ref{Lemma1}.

Now check that the field $k$, the $k$-algebra $A$ with the maximal ideal
$\mathfrak m:= \mathfrak m_x$,
the ring $R$ of regular functions on the affine scheme
$\cal X$, the inclusion
$\lambda^*: A[t] \hra R$
induced by $\lambda$,
the function $h$, the $A$-augmentation
$\epsilon:= \Delta^* : R \to A$
and the function $f$
satisfy the assumption of Lemma
\ref{Lemma2}.

As a local ring of a $k$-smooth variety $A$ is essentially
$k$-smooth.
Since $r$ is essentially smooth and $X$ is $k$-smooth,
${\cal X}$ is essentially $k$-smooth and so is $R$.
Since $\varrho$ is finite $R$ is finite over $A[t]$ too.
Since $k(q(x))=k(x)$ and the fibre of $q$ over $q(x)$
is geometrically irreducible so is the fibre of $p$ over
$x \in U$. This implies that $R/\mathfrak m$ is a domain.
By base change, the open subscheme ${\cal X}_h$ contains the closed fibre of $p$ and
the restriction of $p$ to ${\cal X}_h$ is smooth.
The assumptions are satisfied.

Now we can find an
$s \in R$ satisfying the conditions $(1)$ to $(7)$ of Lemma
\ref{Lemma2}. The function $s$ defines a finite morphism
$\pi: {\cal X} \to U \times \Aff1$.
Since ${\cal X}$ and
${\cal Y}:=U \times \Aff1$
are regular schemes and $\pi$ is finite surjective
it is finite and flat by a theorem of Grothendieck
\cite[Cor.18.17]{E}.
So for any map $f:{\cal Y'}\to {\cal Y}$ of affine schemes, putting
${\cal X'}={\cal X}\times_{\cal Y} {\cal Y'}$ we have the norm map
given by
(\ref{NormMap})
$$
C({\cal X'}) \to C({\cal Y'}).
$$
Set
${\cal D} = \textrm{Spec}(R/J)$
and
${\cal D}_1 = \textrm{Spec}(R/(s-1)R)$.
Clearly
${\cal D}_1$ is the scheme theoretic pre-image of
$U \times \{1\}$
and the disjoint union
${\cal D}_0:= {\cal D} \coprod \delta (U)$
is the scheme theoretic pre-image of
$U \times \{0\}$
under the morphism $\pi$.
In particular ${\cal D}$ is finite flat over $U$.
We will write $\Delta$ for $\delta (U)$.
Recall that
$\xi \in C(k[X]_\textrm{f})$
and set
$\zeta= p_X^*(\xi) \in C(R_f)$.
Since the function $f \in R$ is co-prime to the ideals
$J$ and $(s-1)R$ one can form the following element
\begin{equation}
\label{RightElement}
\xi' = N_{{\cal D}_1/U}(\zeta|_{{\cal D}_1})N_{{\cal D}/U}(\zeta|_{{\cal D}})^{-1}
\in C(U):=C(A).
\end{equation}

\begin{clm}[Main]
\label{MainClaim}
One has $\bar \xi'_K = \bar \xi_K$ in ${\cal F}(K)$,
where $K$ is the fraction field  of both $k[X]$ and $k[U]=A$.
\end{clm}
To complete the proof, it remains to prove the claim.
To do this it is convenient to fix some notations.
For an $A$-module $M$ set
$M_K= M \otimes_A K$,
for
an ${\cal Y}$-scheme ${\cal Z}$ set
${\cal Z}_K= \textrm{Spec}(K) \times_{{\cal Y}} {\cal Z}$,
for a
${\cal Y}$-morphism
$\varphi: {\cal Z} \to {\cal W}$
set
$\varphi_K= \textrm{Spec}(K) \times_{{\cal Y}} \varphi$.
Clearly one has
$A_K=K$,
$U_K= \textrm{Spec}(K)$,
${\cal X}_K = \textrm{Spec}(R_K)$ and
$(\Aff1 \times U_K)_K$
is just the affine line
$\Aff1_K$.
Its closed subschemes
$U_K \times \{1\}$
and
$U_K \times \{0\}$
coincide with the points
$1$ and $0$ of $\Aff1_K$.
The morphism
$p_K: {\cal X}_K \to \textrm{Spec}(K)$
is smooth by the condition $(1)$ of Lemma
\ref{Lemma2}.
The morphism $\pi_K$ is finite flat and fits in the commutative triangle
\begin{equation}
    \xymatrix{
    {\cal X}_K \ar[rr]^-{\pi_K}\ar[rd]_-{p_K} && \Aff1_K \ar[ld]^-{} &   \\
    & \textrm{Spec}(K)  \\
    }
\end{equation}
One has
${\cal D}_K = \textrm{Spec}(R_K/J_K)$,
${\cal D}_{1,K} = \textrm{Spec}(R_K/(s-1)R_K)$.
Clearly
${\cal D}_{1,K}$ is the scheme theoretic pre-image of
the point $\{1\} \in \Aff1_K$
and the disjoint union
${\cal D}_{0, K}:= {\cal D}_K \coprod \Delta_K$
is the scheme theoretic pre-image of the point
$\{0\} \in \Aff1_K$
under the morphism
$\pi_K$.
Let
$$
\zeta_s:= N_{K({\cal X}_K)/K(\Aff1)}(\zeta)  \in C(K(\Aff1))=C(K(s)).
$$
To prove Claim
\ref{MainClaim}
we need the following one:
\begin{clm}
\label{Claim}
The class
$\bar \zeta_s \in {\cal F}(K(s))$
is $K[s]$-unramified.
\end{clm}
Assuming this claim we complete the proof of Claim
\ref{MainClaim}.
By Claim
\ref{Claim}
we can apply the specialization maps to $\bar \zeta_s$.
By Corollary
\ref{TwoSpecializations}
the specializations at $0$ and $1$ of the element
$\bar \zeta_s$ coincide, that is
\begin{equation}
\label{Rel2}
s_1(\bar \zeta_s)=s_0(\bar \zeta_s) \in {\cal F}(K).
\end{equation}
By Corollary
\ref{NormAtZeroAndOne}
the function
$N_{R_K/K[s]}(f) \in K[s]$
does not vanishes as at $1$, so at $0$ and
by the condition $(5)$ of Lemma
\ref{Lemma2}
one has
$\zeta_s \in C(K[s]_{N(f)})$.
Using the relation between specialization and evaluation maps described in
Definition
\ref{SpecializationDef}
one has a chain of relations
\begin{equation}
\label{Rel3}
\overline {Ev_1(\zeta_s)} = s_1(\bar \zeta_s)= s_0(\bar \zeta_s)=
\overline {Ev_0(\zeta_s)} \in {\cal F}(K).
\end{equation}
Base change and the multiplicativity properties of the norm map
(\ref{NormMap})
imply relations
\begin{equation}
\label{Rel4}
Ev_1(\zeta_s)= N_{{\cal D}_{1,K}/U_K}(\zeta|_{{\cal D}_{1,K}})
\end{equation}
and
\begin{equation}
\label{Rel4}
Ev_0(\zeta_s)=
N_{{\cal D}_K \coprod \Delta_K/U_K}(\zeta|_{{\cal D}_K \coprod \Delta_K})=
N_{{\cal D}_K/U_K}(\zeta|_{{\cal D}_K})N_{\Delta_K/U_K}(\zeta|_{\Delta_K})
\end{equation}
So we have a chain of relations in ${\cal F}(K)$
$$
\overline {N_{{\cal D}_{1,K}/U_K}(\zeta|_{{\cal D}_{1,K}})}=
\overline {Ev_1(\zeta_s)}=\overline {Ev_0(\zeta_s)}=
\overline {N_{{\cal D}_K/U_K}(\zeta|_{{\cal D}_K})}\cdot \overline {N_{\Delta_K/U_K}(\zeta|_{\Delta_K})}
$$
By the normalization property of the norm map
(\ref{NormMap})
one has
$N_{\Delta_K/U_K}(\zeta|_{\Delta_K})=\delta_K^*(\zeta)$.
Since
$\delta_K^*(\zeta)= \xi_K \in C(K)$
we have a chain of relations in ${\cal F}(K)$
$$
\bar \xi_K=
\overline {\delta_K^*(\zeta)}=
\overline {N_{\Delta_K/U_K}(\zeta|_{\Delta_K})}=
\overline {N_{{\cal D}_{1,K}/U_K}(\zeta|_{{\cal D}_{1,K}})} \cdot
(\overline {N_{{\cal D}_K/U_K}(\zeta|_{{\cal D}_K})})^{-1}
$$
By the base change property of the norm map
(\ref{NormMap}) one has
\begin{equation}
\label{Rel1}
\xi'_K=N_{{\cal D}_{1,K}/U_K}(\zeta|_{{\cal D}_{1,K}})N_{{\cal D}_K/U_K}(\zeta|_{{\cal D}_K})^{-1}
\in C(U_K):=C(K).
\end{equation}
So $\bar \xi'_K=\bar \xi_K$.
 Claim
\ref{MainClaim}
follows.

It remains to prove Claim
\ref{Claim}.
Recall that
the class
$\bar \xi \in {\cal F}(k[X]_\textrm{f})$
is $k[X]$-unramified.
Since
$q|_{X_\textrm{h}}:X_\textrm{h} \to \Aff^{d-1}$
is flat (it is smooth)
so is
$r: U \hra X_\textrm{h} \to \Aff^{d-1}$.
Since the
$p_X: {\cal X} \to X$
is the base change of $r$ it coincides
with a composition map
${\cal X} \hra X_\textrm{h} \times_{\Aff^{d-1}} X \xra{q \times id} X$
where $q \times id$ is a flat morphism of finite type.
By Lemma
\ref{UnramifiednessLemma}
the class
$\bar \zeta \in {\cal F}(R_f)$
is $R$-unramified.
Thus the same class
$\bar \zeta$ when regarded in
${\cal F}(K[{\cal X}_K]_f)$
is
$K[{\cal X}_K]$-unramified.

Now check that the inclusion of $K$-algebras
$K[s] \subset R_K=K[{\cal X}_K]$ and
the function $f \in R$ satisfy the hypotheses of Lemma
\ref{KeyUnramifiedness}.
We first check that the $K$-algebra $R_K/fR_K$ is reduced.
Recall that the ring
$k[X]/(\textrm{f})$ is reduced. As we mentioned
above, the morphism
$p_X: {\cal X} \to X$ is essentially smooth,  hence the ring $R/(f)$ is reduced too and thus its localization $R_K/fR_K$ is reduced.
Since the extension
$K[s] \subset R_K=K[{\cal X}_K]$
and the functions $f \in R$
satisfy
the conditions $(5)$ to $(7)$ of Lemma
\ref{Lemma2}
they also satisfy the hypotheses of Lemma
\ref{KeyUnramifiedness}.
Thus by Lemma
\ref{KeyUnramifiedness}
the class $\bar \zeta_s$
is $K[s]$-unramified.
This implies Claim
\ref{Claim}. The proof of the theorem is completed.


\end{proof}

\subsection{Proof of Theorem (B)}
\begin{proof}
To prove Theorem (B)
we now recall a celebrated result of Dorin
Popescu (see \cite{P}
or, for a self-contained proof, \cite{Sw}).

Let $k$ be a field and $R$ a local $k$-algebra. We say that $R$ is
{\it geometrically regular} if $k'\otimes_kR$ is regular for any
finite extension $k'$ of $k$. A ring homomorphism $A\to R$ is
called {\it geometrically regular} if it is flat and for each
prime ideal $\mathfrak q$ of $R$ lying over $\mathfrak p$, $R_\mathfrak q/\mathfrak p R_\mathfrak q =
k(\mathfrak p)\otimes_AR_\mathfrak q$ is geometrically regular over
$k(\mathfrak p)=A_\mathfrak p/\mathfrak p_\mathfrak p$.

Observe that any regular local ring containing a field $k$ is
geometrically regular over the prime field of $k$.

\begin{thm}[Popescu's theorem] A  homomorphism $A\to R$  of
noetherian rings is geometrically regular if and only if $R$ is a
filtered direct limit of smooth $A$-algebras.
\end{thm}

{\it Proof of Theorem B.} Let $R$ be a regular local ring
containing the field $k$. Since $k$ is of characteristic zero
one can apply Popescu's theorem. So $R$ can be presented as a filtered
direct limit of smooth $k$-algebras $A_{\alpha}$,
We
first observe that we may replace the direct system of the
$A_\alpha$'s by a system of  essentially smooth local
$k$-algebras. In fact, if ${\mathfrak  m}$ is the maximal ideal of $R$,
we can replace each $A_\alpha$ by $\left(A_\alpha\right)_{{\mathfrak
p}_\alpha}$, where ${\mathfrak  p}_\alpha={\mathfrak  m}\cap A_\alpha$. Note
that in this case the canonical morphisms
$\varphi_\alpha:A_\alpha\to R$ are local and every $A_\alpha$ is a
regular local ring, in particular a factorial ring.

Let now $L$ be the field of fractions of $R$ and, for each
$\alpha$,  let $K_\alpha$ be the field of fractions of $A_\alpha$.
For each index $\alpha$ let
$\mathfrak a_{\alpha}$
be the kernel of the map
$\varphi_{\alpha}: A_{\alpha} \to R$
and
$B_{\alpha}=(A_{\alpha})_{\mathfrak a_{\alpha}}$.
Clearly, for each
$\alpha$,  $K_{\alpha}$
is the field of fractions of
$B_{\alpha}$.
The composition map
$A_{\alpha} \to R \to L$
factors through $B_{\alpha}$
and hence it also factors through the residue field $k_{\alpha}$
of $B_{\alpha}$.
Since $R$ is a filtering direct limit of the $A_{\alpha}$'s
we see that $L$ is a filtering direct limit of the $B_{\alpha}$'s.

Let $\xi \in C(L)$ be such that the class
$\bar \xi \in {\cal F}(L)$
is $R$-unramified. We need the following two lemmas.
\begin{lem}
\label{WeakGrothendieck}
Let $B={\cal O}_{X,x}$ be the local ring of a $k$-smooth
variety at a point $x$. Let $k(x)$ be its residue field
and $K$ be its field of fractions. Let
$\eta, \rho \in {\cal F}(B)$
be such that
$\eta_K = \rho_K \in {\cal F}(K)$.
Then
$\eta (x)= \rho (x) \in {\cal F}(k(x))$.
\end{lem} %%%%%%% or \eta(x)=\rho(x)?

\begin{lem}
\label{LiftToFiniteLevel}
There exists an index $\alpha$ and an element
$\xi_{\alpha} \in C(B_{\alpha})$ such that the class
$\bar \xi_{\alpha} \in {\cal F}(K_{\alpha})$
is $A_{\alpha}$-unramified.
\end{lem}
Assuming these two claims we complete the proof as follows.
Consider a commutative diagram
$$
\xymatrix{
    A_{\alpha}  \ar[d]\ar[rr]^-{\varphi_{\alpha}}\ar[d]_-{} && R \ar[d]^-{} &   \\
     B_{\alpha} \ar[d]\ar[r]^{} & k_{\alpha} \ar[r]^{} & L \\
     K_{\alpha}.
    }
$$
By Lemma
\ref{LiftToFiniteLevel}
the class
$\bar \xi_{\alpha} \in {\cal F}(K_{\alpha})$
is $A_{\alpha}$-unramified.
Hence by Theorem $A$ there exists an element
$\eta \in C(A_{\alpha})$
such that
$\bar \xi_{\alpha}=\bar \eta \in {\cal F}(K_{\alpha})$.
By Lemma
\ref{WeakGrothendieck}
the elements
$\bar \xi_{\alpha}$
and
$\bar \eta$
have the same image in ${\cal F}(k_{\alpha})$.
Hence $\bar \xi \in {\cal F}(L)$ coincides with
the image of the element $\varphi_{\alpha} (\bar \eta)$
in ${\cal F}(L)$.
It remains to prove the two Lemmas.

{\it Proof of Lemma \ref{WeakGrothendieck}.}
Induction on $dim (B)$. The case of dimension $1$ follows fromTheorem
\ref{NisnevichThm}. To prove the general case choose an $f \in B$
such that $\eta = \rho \in {\cal F}(B_f)$. Let $\pi$ be
a regular parameter in $B$ without common factors with $f$ and let
$\bar B= B/\pi B$. Then for the image
$\overline {\eta - \rho}$ of $\eta - \rho$
in
${\cal F}(\bar B)$
we have
$(\overline {\eta - \rho})_f = \overline {(\eta - \rho)_f} =0
\in {\cal F}(\bar B_{\bar f})$.
By the inductive hypotheses one has
$\overline {\eta - \rho}=0 \in {\cal F}(\bar B)$.
Thus $\eta (x) = \rho (x) \in {\cal F}(k(x))$.

{\it Proof of Lemma \ref{LiftToFiniteLevel}.}
Choose an $f \in R$ such that $\xi$ is defined over $R_f$.
Then $\xi$ is ramified at most at those hight one primes
$\mathfrak p_1, \dots, \mathfrak p_r$
which contains $f$. Since the class
$\bar \xi \in {\cal F}(L)$
is $R$-unramified there exists, for any
$\mathfrak p_i$, an element
$\sigma_i \in G(L)$
and an element
$\xi_i \in C(R_{\mathfrak p_i})$
such that
$\xi=\mu(\sigma_i) \xi_i \in C(L)$.
We may assume that $\xi_i$ is defined over $R_{h_i}$
for some
$h_i \in R - \mathfrak p_i$
and that $\sigma_i$ is defined over $R_{g_i}$
for some $g_i \in R$.

We can find an index $\alpha$ such that $A_{\alpha}$
contains lifts
$f_{\alpha}, h_{1,\alpha}, \dots, h_{r,\alpha}, g_{1\alpha}, \dots, g_{r,\alpha}$
and moreover
\begin{itemize}
\item[(1)]
$C(A_{\alpha, f_{\alpha}})$ contains a lift $\xi_{\alpha}$ of $\xi$,
\item[(2)]
$C(A_{\alpha, h_{i,\alpha}})$ contains a lift of $\xi_{i,\alpha}$ of $\xi_i$,
\item[(3)]
$G(A_{\alpha, g_{i,\alpha}})$ contains a lift of $\sigma_{i,\alpha}$ of $\sigma_i$.
\end{itemize}
Since none of the
$f_{\alpha}, h_{1,\alpha}, \dots, h_{r,\alpha}, g_{1\alpha}, \dots, g_{r,\alpha}$
 vanishes in $R$, the elements
$\xi_{\alpha}, \xi_{1,\alpha}, \dots, \xi_{ir,\alpha}$
and
$\sigma_{1,\alpha}, \dots, \sigma_{r,\alpha}$
may be regarded as elements of
$C(B_{\alpha})$
and
$G(B_{\alpha})$
respectively.

We know that
$\xi_{i,\alpha} \mu (\sigma_{i,\alpha})$
and
$\xi_{\alpha}$
map to the same element in $C(L)$.
Hence replacing $\alpha$ by a larger index, we may assume that
$\xi_{\alpha}= \xi_{i,\alpha} \mu (\sigma_{i,\alpha}) \in C(B_{\alpha})$.
We claim that the class
$\bar \xi_{\alpha} \in {\cal F}(K_{\alpha})$
is $A_{\alpha}$-unramified.
To prove this note that
the only primes at which $\bar \xi_{\alpha}$ could be ramified are those which divide
$f_{\alpha}$. Let
$\mathfrak q_{\alpha}$
be one of them. Check that $\bar \xi_{\alpha}$ is
unramified at $\mathfrak q_{\alpha}$. Let $q_{\alpha} \in A_{\alpha}$ be
a prime element such that
$q_{\alpha}A_{\alpha}=\mathfrak q_{\alpha}$. Then
$q_{\alpha}r_{\alpha} = f_{\alpha}$
for an element
$r_{\alpha}$.
Thus $qr=f \in R$ for the images of $q_{\alpha}$ and $r_{\alpha}$ in $R$.
Since the homomorphism $\varphi_{\alpha}: A_{\alpha} \to R$ is local,
$q \in \mathfrak m_R$. The relation $qr=f$ shows that
$q \in \mathfrak p_i$ for some index $i$. Thus
$q_{\alpha} \in \varphi_{\alpha}^{-1}(\mathfrak p_i)$
and
$\mathfrak q_{\alpha} \subset \varphi_{\alpha}^{-1}(\mathfrak p_i)$.
On the other hand
$h_{i, \alpha} \in A_{\alpha} - \varphi_{\alpha}^{-1}(\mathfrak p_i)$,
because
$h_i \in R - \mathfrak p_i$. Thus
$h_{i, \alpha} \in A_{\alpha} - \mathfrak q_{\alpha}$.
Now the relation
$\xi_{\alpha}= \xi_{i,\alpha} \mu (\sigma_{i,\alpha}) \in C(B_{\alpha})$
with
$\xi_{i,\alpha} \in C(A_{\alpha, h_{i,\alpha}})$
shows that
$\bar \xi_{\alpha}$ is unramified at
$\mathfrak q_{\alpha}$.
Thus
$\bar \xi_{\alpha}$
is unramified at each hight one prime in $A_{\alpha}$ containing $f_{\alpha}$.
Since
$\xi_{\alpha} \in C(A_{\alpha, f_{\alpha}})$
we conclude that
$\bar \xi_{\alpha}$
is $A_{\alpha}$-unramified.
The lemma follows. The theorem is proved.

\end{proof}

\section{Applications}
In this Section we prove a purity theorem for reductive groups.
Let $G$ be a reductive group over the characteristic zero field $k$ and
$Z \xra{i} G$ a closed central subgroup of $G$.
Let $G'=G/Z$ be the factor group,
$\pi: G \to G'$ be the projection. For any $k$-algebra $R$
consider the boundary operator
$\delta_{\pi,R}: G'(R) \to \textrm{H}^1_{\text{\'et}}(R,Z)$.
It is a group homomorphism
\cite[Ch.II, \S 5.6, Cor.2]{Se}.
Set
$$
{\cal F}(R)= \textrm{H}^1_{\text{\'et}}(R,Z)/ Im(\delta_{\pi,R}).
$$
\begin{thm}
\label{PurityForSubgroup}
Let $R$ be a regular local
ring containing the field $k$.
The functor
${\cal F}$
satisfies purity for $R$.
If $K$ is the fraction field of $R$ this statement can be restated in an explicit way
as follows:

given an element $\xi \in \textrm{H}^1_{\text{\'et}}(K,Z)$
suppose that for each height $1$ prime ideal
$\mathfrak p $ in $R$ there exist
$\xi_{\mathfrak p } \in \textrm{H}^1_{\text{\'et}}(R_{\mathfrak p }, Z)$,
$g_{\mathfrak p } \in G'(K)$
with
$\xi=\xi_{\mathfrak p } + \delta_{\pi}(g_{\mathfrak p }) \in \textrm{H}^1_{\text{\'et}}(K,Z)$.
Then there exist
$\xi_{\mathfrak m } \in \textrm{H}^1_{\text{\'et}}(R, Z)$, $g_{\mathfrak m } \in G'(K)$,
such that
$$
\xi=\xi_{\mathfrak m } + \delta_{\pi}(g_{\mathfrak m }) \in \textrm{H}^1_{\text{\'et}}(K,Z).
$$

\end{thm}

\begin{proof}
Since $G$ is reductive the group $Z$ is of  multiplicative type.
So we can find a commutative separable $k$ algebra $l$ and
a closed embedding
$Z \hra R_{l/k}(\mathbb G_{m,\ l})$
into the permutation torus
$T^{+}=R_{l/k}(\mathbb G_{m,\ l})$.
Let
$G^{+}=(G \times T^{+})/Z$
and
$T=T^{+}/Z$,
where
$Z$ is embedded in
$G \times T^{+}$
diagonally.
Clearly
$G^{+}/G=T$.
Consider a commutative diagram
$$
\xymatrix {
{}           & \{1\}                       & \{1\} \\
{}           & {G'} \ar[r]^{id}  \ar[u]          & {G'}   \ar[u]          \\
\{1\} \ar[r] & {G} \ar[r]^{j^+} \ar[u]^{\pi} & {G^+} \ar[r]^{\mu^+} \ar[u]^{\pi^+} & {T} \ar[r] & \{1\} \\
\{1\} \ar[r] & {Z} \ar[r]^{j} \ar[u]^{i} & {T^+} \ar[r]^{\mu} \ar[u]^{i^+} & {T} \ar[u]_{id} \ar[r] &  \{1\} \\
{}           & \{1\} \ar[u]                      & \{1\} \ar[u]\\
}
$$
with exact rows and columns. For a local $k$-algebra $A$ one has
$\textrm{H}^1_{\text{\'et}}(A,T^+)=\{*\}$
by Hilbert $90$
and this diagram gives rise to
a commutative diagram of pointed sets
$$
\xymatrix {
{}           & {}                            & {\textrm{H}^1_{\text{\'et}}(A,G')} \ar[r]^{id}            & {\textrm{H}^1_{\text{\'et}}(A,G')}            \\
{G^+(A)} \ar[r]^{\mu^+_A} & {T(A)} \ar[r]^{\delta^+_A}  & {\textrm{H}^1_{\text{\'et}}(A,G)} \ar[r]^{j^+_*} \ar[u]^{\pi_*} & {\textrm{H}^1_{\text{\'et}}(A,G^+)} \ar[u]^{\pi^+_*}  \\
{T^+(A)} \ar[r]^{\mu_A} \ar[u]^{i^+_*} & {T(A)} \ar[r]^{\delta_A} \ar[u]^{id} & {\textrm{H}^1_{\text{\'et}}(A,Z)} \ar[r]^{\mu} \ar[u]^{i_*} & {\{*\}} \ar[u]^{i^+_*}  \\
{}           & {} & {G'(A)} \ar[u]^{\delta_{\pi}}                      & {} \\
}
$$
with exact rows and columns. It follows that
$\pi^+_*$ has  trivial kernel and one has a chain of group isomorphisms
$$
\textrm{H}^1_{\text{\'et}}(A,Z) / Im (\delta_{\pi,A})= ker (\pi_*)= ker (j^+_*) = T(A)/\mu^+ (G^+(A)).
$$
Clearly these isomorphisms respect $k$-homomorphisms of local $k$-algebras.
The functor
$A \mapsto T(A)/\mu^+ (G^+(A))$
satisfies purity for the regular local $k$-algebra $R$ by Theorem (B).
Hence the functor
$A \mapsto \textrm{H}^1_{\text{\'et}}(A,Z) / Im (\delta_{\pi,A})$
satisfies purity for $R$.
\end{proof}

\section{Examples}
We follow here the notation of The Book of Involutions
\cite{KMRT}.
The field $k$ is a characteristic zero field.
The functors
(\ref{Example1})
to
(\ref{Example12})
satisfy purity for regular local rings containing $k$
as follows either from Theorem
\ref{PurityForSubgroup}
or from
Theorem (B).

\begin{itemize}
\item[(1)]
Let $G$ be a simple algebraic group over the field $k$,
$Z$ a central subgroup, $G'=G/Z$, $\pi: G \to G'$
 the canonical morphism.
For any $k$-algebra $A$ let
$\delta_{\pi,R}: G'(R) \to \textrm{H}^1_{\text{\'et}}(R,Z)$
be the boundary operator. One has a functor
\begin{equation}
\label{Example1}
R \mapsto \textrm{H}^1_{\text{\'et}}(R,Z)/ \textrm{Im}(\delta_{\pi,R}).
\end{equation}

\item[(2)]
Let $(A,\sigma)$ be a finite separable $k$-algebra with an orthogonal involution.
Let
$\pi: \textrm{Spin}(A, \sigma) \to \textrm{PGO}^+(A, \sigma)$
be the canonical morphism of the spinor $k$-group scheme to
the projective orthogonal $k$-group scheme. Let
$Z=ker (\pi)$.
%THE DEFINITION IS GIVEN in
%\cite[just above (30.13)]{KMRT}.\\
For a $k$-algebra $R$ let
$\delta_R: \textrm{PGO}^+(A, \sigma)(R) \to \textrm{H}^1_{\text{\'et}}(R,Z)$
be the boundary operator.
%It coincides with
%the spinor norm transform described in
%\cite[Prop. (13.36)]{KMRT}.
One has a functor
\begin{equation}
\label{Example2}
R \mapsto \textrm{H}^1_{\text{\'et}}(R,Z)/Im(\delta_R).
\end{equation}
In $(2a)$ and $(2b)$ below we describe this functor somewhat more explicitly
following
\cite{KMRT}.

\item[(2a)]
Let $C(A, \sigma)$ be the Clifford algebra. Its center $l$ is an \'etale
quadratic $k$-algebra.
Assume that $deg(A)$ is divisible by $4$.
Let
$\Omega(A,\sigma)$ be the extended Clifford group
\cite[Definition given just below (13.19)]{KMRT}.
Let
$\underline \sigma$
be the canonical involution of
$C(A,\sigma)$
as it is described in
\cite[just above (8.11)]{KMRT}.
Then
$\underline \sigma$
is either orthogonal or symplectic by
\cite[Prop.8.12]{KMRT}.
Let
$
\underline \mu: \Omega(A,\sigma) \to R_{l/k}(\mathbb G_{m,l})
$
be the multiplier map defined in
\cite[just above (13.25)]{KMRT}
by
$\underline \mu (\omega)= \underline \sigma(\omega)\cdot \omega$.
Set
$R_l=R \otimes_k l$.
For a field or a local ring $R$ one has
$
\textrm{H}^1_{\text{\'et}}(R,Z)/Im(\delta_R)=R_l^{\times}/\underline \mu (\Omega(A,\sigma)(R))
$
by
\cite[the diagram in (13.32)]{KMRT}.
Consider the functor
\begin{equation}
\label{Example2a}
R \mapsto R_l^{\times}/\underline \mu (\Omega(A,\sigma)(R)).
\end{equation}
It coincides with the functor
$R \mapsto \textrm{H}^1_{\text{\'et}}(R,Z)/Im(\delta_R)$
on local rings containing $k$.


\item[(2b)]
Now let $deg(A)=2m$ with odd $m$. Let $\tau: l \to l$ be the involution of $l/k$.
The kernel of the morphism
$R_{l/k}(\mathbb G_{m,l}) \xra{id - \tau} R_{l/k}(\mathbb G_{m,l})$
coincides with $\mathbb G_{m,k}$. Thus $id-\tau$ %%%%% this is ugly; one is tempted to write i(d - 1)
induces a $k$-group scheme morphism
which we denote
$\overline {id-\tau}: R_{l/k}(\mathbb G_{m,l})/\mathbb G_{m,k} \hra R_{l/k}(\mathbb G_{m,l})$.
Let
$
\underline \mu: \Omega(A,\sigma) \to R_{l/k}(\mathbb G_{m,l})
$
be the multiplier map defined in
\cite[just above (13.25)]{KMRT}
by
$\underline \mu (\omega)= \underline \sigma(\omega)\cdot \omega$.
Let
$\kappa: \Omega(A,\sigma) \to R_{l/k}(\mathbb G_{m,l})/\mathbb G_{m,k}$
be the $k$-group scheme morphism described in
\cite[Prop.13.21]{KMRT}.
The composition
$\overline {(id-\tau)} \circ \kappa$
lands in
$R_{l/k}(\mathbb G_{m,l})$.
Let
$U \subset \mathbb G_{m,k} \times R_{l/k}(\mathbb G_{m,l})$
be a closed $k$-subgroup consisting of all $(\alpha, z)$ such that
$\alpha^4 = N_{l/k}(z)$.

Set
$\mu_*=(\underline \mu, [\overline {(id-\tau)} \circ \kappa]\cdot\underline \mu2): \Omega(A,\sigma)
\to \mathbb G_{m,k} \times R_{l/k}(\mathbb G_{m,l})$.
This $k$-group scheme morphism lands in $\textrm{U}$.
So we get a $k$-group scheme morphism
$\mu_*: \Omega(A,\sigma) \to \textrm{U}$.
On the level of $k$-rational points it coincides with the one described in
\cite[just above (13.35)]{KMRT}. For a field or a local ring one has
$$
\textrm{H}^1_{\text{\'et}}(R,Z)/Im(\delta_R)=
\textrm{U}(R)/[\{(N_{l/k}(\alpha), \alpha^4) | \alpha \in R_l^{\times}\}\cdot \mu_*(\Omega(A,\sigma)(R))].
$$
Consider the the functor
\begin{equation}
\label{Example2b}
R \mapsto \textrm{U}(R)/[\{(N_{l/k}(\alpha), \alpha^4) | \alpha \in R_l^{\times}\}\cdot \mu_*(\Omega(A,\sigma)(R))].
\end{equation}
It coincides with the functor
$R \mapsto \textrm{H}^1_{\text{\'et}}(R,Z)/Im(\delta_R)$
on local rings containing $k$.

\item[(3)]
Let
$\Gamma(A,\sigma)$
be the Clifford group $k$-scheme of $(A,\sigma)$.
Let
$\textrm{Sn}: \Gamma(A,\sigma) \to \mathbb G_{m,k}$
be the spinor norm map. It is dominant.
Consider the functor
\begin{equation}
\label{Example3}
R \mapsto R^{\times}/\textrm{Sn}(\Gamma(A,\sigma)(R)).
\end{equation}
Purity for this functor was originally  proved in
\cite[Thm.3.1]{Z}.
In fact,
$\Gamma(A,\sigma)$
is $k$-rational.

\item[(4)]
We follow here the Book of Involutions
\cite[\S 23]{KMRT}.
Let $A$ be a separable finite dimensional $k$-algebra with center $l$
and $k$-involution $\sigma$ such that $k$ coincides with all $\sigma$-invariant
elements of $l$, that is $k=l^{\sigma}$.
Consider the $k$-group schemes of
similitudes
of $(A,\sigma)$:
$$
\textrm{Sim}(A,\sigma)(R)=\{a \in A_R^{\times}\ | a\cdot \sigma_R (a) \in l_K^{\times} \}.
$$
We have a $k$-group scheme morphism
$
\mu: \textrm{Sim}(A,\sigma) \to \mathbb G_{m,k}, \ a \mapsto a \cdot \sigma (a).
$
It gives an exact sequence of algebraic $k$-groups
$$
\{1\} \to \textrm{Iso}(A,\sigma) \to \textrm{Sim}(A,\sigma) \to \mathbb G_{m,k} \to \{1\}.
$$
One has a the functor
\begin{equation}
\label{Example4}
R \mapsto R^{\times}/\mu (\textrm{Sim}(A,\sigma)(R)).
\end{equation}
Purity for this functor was originally  proved in
\cite[Thm.1.2]{Pa}.
Various particular cases are obtained
considering unitary, symplectic and orthogonal involutions.
\item[(4a)]
In the case of an orthogonal involution $\sigma$ the connected component
$\textrm{GO}^+(A, \sigma)$
\cite[(12.24)]{KMRT}
of the similitude $k$-group scheme
$\textrm{GO}(A, \sigma):=\textrm{Sim}(A,\sigma)$
has the index two in
$\textrm{GO}(A, \sigma)$.
The restriction of $\mu$ to
$\textrm{GO}^+(A, \sigma)$
is still a dominant morphism to
$\mathbb G_{m,k}$. One has a functor
\begin{equation}
\label{Example4a}
R \mapsto R^{\times}/\mu (\textrm{GO}^+(A,\sigma)(R))
\end{equation}
It seems that its purity
does not follow from
\cite[Thm.1.2]{Pa}. In fact we do not know
whether the norm principle holds for
$\mu: \textrm{GO}^+(A, \sigma) \to \mathbb G_{m,k}$
or not.

\item[(5)]
Let $A$ be a central simple algebra (csa) over $k$ and
$\textrm{Nrd}:GL_{1,A} \to \mathbb G_{m,k}$
 the reduced norm morphism. One has a functor
\begin{equation}
\label{Example5}
R \mapsto R^{\times}/\textrm{Nrd}(\textrm{GL}_{1,A}(R)).
\end{equation}
Purity for this functor was originally  proved in
\cite[Thm.5.2]{C-TO}.
\item[(6)]
Let $(A,\sigma)$ be a finite separable $k$-algebra with
a unitary involution such that its center $l$ is a quadratic extension of $k$.
Let $U(A,\sigma)$ be the unitary $k$-group scheme. Let
$U_l(1)$ be an algebraic tori given by $N_{l/k}=1$.
One has a functor
\begin{equation}
\label{Example6}
R \mapsto \textrm{U}_l(1)(R)/\textrm{Nrd}(\textrm{U}_{A,\sigma}(R)) =
\{\alpha \in R^{\times}_l | N_{l/k}(\alpha)=1 \} /\textrm{Nrd}(\textrm{U}_{A,\sigma}(R))
\end{equation}
where
$\textrm{Nrd}$
is the reduced norm map.
Purity for this functor was originally  proved in
\cite[Thm.3.3]{Z}.

\item[(7)]
With the notation of example $(5)$ choose an integer $d$ and consider
the $k$-group scheme morphism
$\mu: \textrm{GL}_{1,A} \times \mathbb G_{m,k} \to \mathbb G_{m,k}$
given by
$(\alpha,a) \mapsto \textrm{Nrd}(\alpha)\cdot a^d$.
One has a functor
\begin{equation}
\label{Example7}
R \mapsto R^{\times}/\mu [\textrm{GL}_{1,A} \times \mathbb G_{m,k})(R)]=
R^{\times}/\textrm{Nrd}(A^{\times}_R)\cdot {R^{\times}}^d
\end{equation}
Purity for this functor was originally  proved in
\cite[Thm.3.2]{Z}.

\item[(8)]
With the notation of example $(5)$ choose
 an integer $d$ and consider the functor
\begin{equation}
\label{Example8}
R \mapsto
U_l(1)(R)/
[\textrm{Nrd}(U_{A,\sigma}(R))\cdot \{\alpha \in R^{\times}_l | N_{l/k}(\alpha)=1 \}^d]
\end{equation}
Purity for this functor was originally  proved in
\cite[Thm.3.2]{Z}.

\item[(9)]
Let $G_1$,$G_2$,$C$ be affine $k$-group schemes.
  Assume that $C$ is commutative
and let
$\mu_1: G_1 \to C$,
$\mu_2: G_2 \to C$
be $k$-group scheme morphisms and
$\mu: G_1 \times G_2 \to C$ be given by
$\mu(g_1,g_2)= \mu_1(g_1)\cdot \mu_2(g_2)$.
One has a functor
\begin{equation}
\label{Example9}
R \mapsto C(R)/\mu [(G_1 \times G_2)(R)]= C(R)/[\mu_1(G_1(R))\cdot \mu_2(G_2(R))]
\end{equation}
In this style one could get a lot of curious examples of functors, one of which is given here.
\item[(10)]
Let $(A_1,\sigma_1)$ be a finite separable $k$-algebra with an orthogonal involution.
Let $A_2$ be a csa over $k$. Let $\mu_1$ be the multiplier map for
$(A_1,\sigma_1)$
and
$\textrm{Nrd}_2$
be the reduced norm for $A_2$.
One has a functor
\begin{equation}
\label{Example10}
R \mapsto
R^{\times}/[\mu_1(\textrm{GO}^+(A_1,\sigma_1)(R))\cdot {\textrm{Nrd}_2({A^{\times}_{2,R}})} \cdot {R^{\times}}^d].
\end{equation}

\item[(10)]
Let $A$ be a csa of degree $3$ over $k$, $\textrm{Nrd}$  the reduced norm
and $\textrm{Trd}$ be the reduced trace. Consider the cubic form
on the $27$-dimensional $k$-vector space
$A \times A \times A$
given by
$N:= \textrm{Nrd}(x)+\textrm{Nrd}(y)+\textrm{Nrd}(z)-\textrm{Trd}(xyz)$.
Let
$\textrm{Iso}(A,N)$
be the $k$-group scheme of isometries of $N$ and
$\textrm{Sim}(N)$
be the $k$-group scheme of similitudes of $N$.
It is known that
$\textrm{Iso}(N)$ is a normal algebraic subgroup in $\textrm{Sim}(A,N)$
and the factor group coincides with
$\mathbb G_{m,k}$. So we have a canonical $k$-group morphism
(the multiplier)
$\mu: \textrm{Sim}(N) \to \mathbb G_{m,k}$.
Now one has a functor
\begin{equation}
\label{Example11}
R \mapsto R^{\times}/\mu(\textrm{Sim}(N)(R)).
\end{equation}
Note that the connected component of $\textrm{Iso}(N)$ is
a simply connected algebraic $k$-group of the type $E_6$.

\item[(11)]
Let $(A,\sigma)$ be a csa of degree $8$ over $k$
with a symplectic involution. Let $V \subset A$ be the subspace
of all skew-symmetric elements. It is of dimension $28$.
Let
$\textrm{Pfd}$
be the reduced Pfaffian on $V$ and
$\textrm{Trd}$
be the reduced trace on $A$. Consider the degree $4$ form on
the space
$V \times V$
given by
$F:=\textrm{Pfr}(x)+\textrm{Pfr}(y)-1/4\textrm{Trd}((xy)2)-1/16{\textrm{Trd}(xy)}^2$.
Consider the symplectic form on
$V \times V$
given by
$\phi((x_1,y_1),(x_2,y_2))= \textrm{Trd}(x_1y_2-x_2y_1)$.
Let
$\textrm{Iso}(F)$
(resp. $\textrm{Iso}(\phi)$)
be the $k$-group scheme of isometries of the pair $F$
(resp. of $\phi)$).
Let
$\textrm{Sim}(F)$
(resp. $\textrm{Sim}(\phi)$)
be the $k$-group scheme of similitudes of $F$
(resp. of $\phi$).
Set
$G=\textrm{Iso}(F) \cap \textrm{Iso}(\phi)$
and
$G^+=\textrm{Sim}(F) \cap \textrm{Sim}(\phi)$.
It is known that
$G$ is a normal algebraic subgroup in $G^+$
and the factor group coincides with
$\mathbb G_{m,k}$. So we have a canonical $k$-group morphism %%%%%(the multiplier  ???) \\
$\mu: G^+ \to \mathbb G_{m,k}$.
Now one has a functor
\begin{equation}
\label{Example12}
R \mapsto R^{\times}/\mu(G^+(R)).
\end{equation}
Note that $G$ is a simply-connected group of the type $E_7$.

\end{itemize}


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\end{document}


%\bye

%Here is a scketch of our method for the geometric case.
%Given a $k$-smooth variety $X$, a closed point $x \in X$ and an
%${\cal O}_{X,x}$-unramified class
%$\bar \xi \in {\cal F}(k(X))$
%one can find a representative $\xi$
% of $\bar \xi$ such that $\xi \in C(X_\textrm{f})$
%for an appropriate $\textrm{f} \in k[X]$ with $\textrm{f}(x) \neq 0$. Letting
%$U= \textrm{\Spec}({\cal O}_{X,x})$ and $i: U \hra X$
%one can find
% multi-valued morphisms %%%%%%%% what's that? %%%%%%%
%$h_0, h_1: U \to X_{\textrm{f}}$ and
%a multi-valued homotopy
%$H: \Aff^1_U \to X$
%between
%$i + h_0$ and $h_1$. %%%%%%%% what is $i$? %%%%%%%
%Now as a lift in $C(U)$ of  $\bar \xi$ we can choose %%% correct?%%%%%
%$$
%\xi' := h_1^*(\xi)\cdot h_0^*(\xi^{-1}) \in C(U).
%$$
%To check that
%$\overline {\xi'_{k(X)}}=\overline {\xi_{k(X)}} \in {\cal F}(k(X))={\cal F}(k(U))$
%we do the following.
%The restriction of $H$ to the generic point of $\Aff^1_U$
%defines a multi-valued map
%$H_{\eta}: \textrm{Spec}(k(\Aff^1_U)) \to X_{\textrm{f}}$
%and an element
%$\xi_t:= H^*_{\eta}(\xi) \in C(k(\Aff^1_U))$.
%Choosing $H$ carefully we may assume that the class
%$\overline {\xi_t} \in {\cal F}(k(\Aff^1_U))$
%is $\Aff^1_{k(U)}$-unramified and thus it is constant.
%Specializing
%$\overline {\xi_t}$
%at the points $1$ and $0$ we get
%a relation
%$\overline {\xi_{k(U)}} + \overline {h^*_0(\xi)}= \overline {h^*_1(\xi^{-1})}$
%in
%${\cal F}(k(U))$.
%Whence
%$\overline {\xi'_{k(U)}}=\overline {\xi_{k(U)}} \in {\cal F}(k(U))={\cal F}(k(X))$.

