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\begin{document}

 \title{A deterministic  polynomial--time algorithm for the first
 Bertini theorem}
\author{Alexander L.~Chistov%
 \\[2ex]
St.~Petersburg Institute for Informatics and
Automation of the\\ Academy of Sciences of Russia\\
14th line 39, St.~Petersburg 199178, Russia,\\
e-mail: labta@iias.spb.su  or sliss@iias.spb.su }
\date{\Large 2004}

\newtheorem{thms}{THEOREM}
\newtheorem{lems}{LEMMA}
\newtheorem{rems}{REMARK}
\newtheorem{defns}{DEFINITION}
\newtheorem{props}{PROPOSITION}
\newtheorem{coros}{COROLLARY}

\maketitle

\begin{abstract}
Consider a projective algebraic variety $W$ which is an irreducible
component of a set of all common zeroes of a family of homogeneous
polynomials of degrees less than $d$ in $n+1$ variables in
zero--characteristic.  Consider a linear system on $W$ given by homogeneous
polynomials of degree $d'$.  Under the conditions of the first Bertini
theorem for $W$ and this linear system we show how to construct an
irreducible divisor in general position from the statement of this theorem.
This algorithm is deterministic and polynomial in $(dd')^n$ and the size
of input.
\end{abstract}




\newpage
 \section*{Introduction}

The first Bertini theorem and its different variants
were analyzed approximately 50--60 years ago, see the survey in \cite{1}.
In the present paper we suggest the
deterministic polynomial time algorithm for
a general version of this theorem
in the case of zero--characteristic.
The algorithmic point of view requires to pay
more attention to details here than
one paid previously. As a result we obtain a simple explicit description
of an open in the Zariski topology subset
from the statement of the first Bertini theorem.
So this theorem ceased to be purely existential (at least in
zero--characteristic). More precisely, consider an irreducible hypersurface
which is a set of all common zeroes of
a homogeneous polynomial $\Phi$ in $X_0,\ldots , X_n$ in an
$n$-dimensional projective space in
zero--characteristic, $n\ge 2$. Let the degree $\deg_{X_0}\Phi\ge 1$.
Consider a linear system on this hypersurface given by
$X_1,\ldots ,X_n$
(the general case, as we shall see, can be reduced to this one).
Let $R=\mbox{\rm Res}_{X_0}(\Phi,\partial\Phi/\partial X_0)$ be
the discriminant of $\Phi$ with respect to
$X_0$, see below the for more details.
Denote by $R_0$ the square free part of $R$ with
respect to
$X_0$ (i.e., the separable polynomial of the maximal degree dividing the
discriminant,
which is here and below in this paper
by definition the resultant of the polynomial and
its partial derivative with respect to the considered variable).
Then a linear form $L$ in $X_1,\ldots ,X_n$ satisfies the
assertion of the first Bertini
theorem in the considered case whenever the polynomial in $n-1$ variable
$R_0\bmod L$
is separable.


This paper applies the deterministic polynomial--time algorithms from
\cite{11} (the correction of the statement of Lemma~2 \cite{11}
see in \cite{12}), \cite{12}, \cite{13}. All
the cited papers in their turn
can be considered as a continuation of our previous ones.
They are based on
real algebraic geometry or, more precisely, use
real structures, see \cite{6}, for obtaining
polynomial--time algorithms for algebraic varieties
over algebraically closed fields of zero--characteristic.
Especially we would like to underline here the importance of the
fundamental algorithm
from \cite{9} for \cite{11}, \cite{12}, \cite{13} and the present paper.
The results of \cite{11} allow to decide
whether two points belong to the same irreducible component of an algebraic
variety. Besides that, in \cite{11}
we suggest an algorithm for
the first Bertini theorem in the case of finite
morphisms, and show
how to construct linear sections (from the
linear system of all linear forms)
in general position of an algebraic
variety. In \cite{12} we
investigate rational morphisms from the algorithmic point of view in the
case when
the dimension of the variety coincides with the one of its image. In
\cite{13}
we construct the objects in general position
related to a rational morphism in the general case.
To obtain the algorithms in \cite{13} we
suggest a new method for solving
systems of polynomial equations and inequalities
with squares of absolute values, see \cite{6},
in the case when these systems contain
generic linear forms. In the present paper we need to strengthen
this method for the case of generic homogeneous polynomials (in place of
generic linear forms) of the given degree.
This requires to use the result of \cite{14}, see Section~2.



In \cite{11} and the present paper
we attract the transcendental
methods of theory of analytical functions, namely, the group of monodromy
of an algebraic function to substantiate the algorithms.
Theorem~3 Section~4 uses these methods and generalizes
Theorem~3 \cite{11} for the case of not necessarily finite morphisms.
Basing on this result we get an irreducibility
criterion for multivariable polynomials,
see below, which generalizes the one from \cite{11}.


Now we proceed to more detailed statements. Let $k$ be a field of
zero--characteristic with algebraic closure $\overline{k}$.
Let $X_0,X_1,\ldots $ be independent
variables over $k$.
Denote by ${\Bbb P}^n(\overline{k})$ the projective space over
the field $\overline{k}$ with coordinates $X_0,\ldots , X_n$. We
shall suppose that ${\Bbb P}^n(\overline{k})$ is defined over
$k$.  For arbitrary homogeneous polynomials $g_1,\ldots
,g_m\in\overline{k}[X_0,\ldots , X_n]$ we shall denote by
${\cal Z}(g_1,\ldots ,g_m)$ the set of all common zeroes of
polynomials $g_1,\ldots ,g_m$ in ${\Bbb P}^n(\overline{k})$.
The similar notations will be used for the sets of zeroes of
ideals and polynomials with other fields of coefficients in affine
and projective spaces (this will be seen from the context). For a
field $k$ and a family of elements $a'=\{a'_i\}_{i\in I}$
from an extension of $k$ we
shall denote by $k(a')$ (respectively $k[a']$) the field
(respectively the ring) generated over $k$
by all the elements $a'_i$, $i\in I$.
For two (or more) families $a'$, $a''$ the
notations $k(a',a'')$, $k[a',a'']$ have the similar sense.
Now suppose that $a'$ is a family of independent variables over $k$ and
$f\in k[a']$. Suppose that
$b'=\{b'_i\}_{i\in I}$ is a family of elements from $k$ with the same
indices as $a'$. Then we shall denote by $f(b')$ the value of $f$ at
$a'_i=b'_i$,
$i\in I$, and also we use other similar notations.



In the algorithmic part of the paper we shall
suppose that the field $k$ is finitely generated
over the field of rational numbers ${\Bbb Q}$.
Let $V\subset{\Bbb P}^n(\overline{k})$ be a
projective algebraic variety  which is a
set of all common zeroes of a family of
homogeneous polynomials in $X_0,\ldots , X_n$ of
degrees less than $d$
with coefficients from a field $k$.
Let $W$ be a defined over $k$ and irreducible over
$k$ component of $V$ and $W$ is given
by its point which belongs to only one
irreducible over $k$ component of $V$, see  Theorem~1 \cite{11} and also
below for details.
Let $e_0,\ldots ,e_r\in k[X_0,\ldots , X_n]$,
be homogeneous polynomials such that $\deg e_i=d'\ge 1$
for every nonzero $e_i$, $0\le i\le r$.
We shall suppose in what follows that $\dim W\cap{\cal Z}(e_0,\ldots ,
e_r)<\dim W$. Let
$$
p\, :\,
W\setminus{\cal Z}(e_0,\ldots , e_r)
\rightarrow{\Bbb P}^r,\quad (X_0:\ldots : X_n)\mapsto(e_0:\ldots : e_r).
\eqno (1)
$$
be the regular morphism.
Set $W'=\overline{p(W\setminus{\cal Z}(e_0,\ldots ,e_r))}$
(here and below in the similar situations the bar denotes closure
with respect to the Zariski topology). Hence $W'$ is irreducible over $k$
closed subvariety of ${\Bbb P}^r(\overline{k})$.
We shall suppose in this paper that $\dim W'>0$.
Note that one can consider the rational morphism $W\rightarrow{\Bbb
P}^r(\overline{k})$ of projective algebraic varieties given by the same
rule but to be more accurate in what follows we deal
with regular morphisms of quasiprojective algebraic varieties.
Denote by
$$
p'\, :\, W\setminus{\cal Z}(e_0,\ldots , e_r)\rightarrow W'
$$
the dominant morphism induced by $p$.
Suppose that $\dim W=n-s$ and $\dim W'=n-\sigma>0$ (one can compute
$s$ and $\sigma$ using the algorithm from \cite{7}, see \cite{13}).
So $0\le s\le\sigma<n$.
Let $v_j=\sum_{0\le i\le n}v_{j,i}X_i$, $s+1\le j\le\sigma$,
be generic linear forms in $X_0,\ldots ,
X_n$, i.e.,
the family of their
coefficients $v_{j,i}$ has transcendency degree $(\sigma-s)(n+1)$ over $k$.
Denote by $k_v$ the field generated by all the coefficients of
linear forms $v_j$, $s+1\le j\le\sigma$, over $k$.
Denote by $W_v$ the intersection of $W$ with the linear subspace
${\cal Z}(v_{s+1},\ldots , v_\sigma)$
of codimension $\sigma-s$ (hence $W_v$ is defined over $k_v$).
Denote by $k_v$ the field generated by all the coefficients of
linear forms $v_j$, $s+1\le j\le\sigma$, over $k$.
Set $v=(v_{s+1},\ldots , v_\sigma)$ and $\upsilon=\widetilde{v}=
\{v_{j,i},\,s+1\le
j\le\sigma,
0\le i\le n\}$ to be the family of all the coefficients of linear forms
$v_j$, $s+1\le j\le\sigma$
(one should not confuse $v$ and $\upsilon$).



Let $n-\sigma\ge 2$. Then our aim is to construct within the time polynomial
in $(dd')^n$
and the size of input
linear forms $L'_{\sigma+1},\ldots , L'_{n-1}\in k[X_0,\ldots ,X_r]$ with
integer coefficients of length
$O(n\log d+(n-\sigma)\log d')$
satisfying for all $\sigma\le i\le n$ the following conditions.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item The intersections  $(W\setminus{\cal Z}(e_0,\ldots ,
e_r))\cap{\cal
Z}(e'_{\sigma+1},\ldots ,
e'_i)$ and $W'\cap{\cal
Z}(L'_{\sigma+1},$ $\ldots , L'_i)$
are transversal (see the definition below); if $\sigma+1\le i\le n-1$ then
these intersections are the irreducible
over $k$ varieties.
\item The morphism $p'_{v,i}\, :\,(W_v\setminus{\cal Z}(e_0,\ldots ,
e_r))\cap{\cal Z}(e'_{\sigma+1},\ldots ,
e'_i)\rightarrow W'(\overline{k_v})
\cap{\cal Z}(L'_{\sigma+1},\ldots ,$ $L'_i)$ induced by $p'$ is
dominant. Denote $p'_{v,\sigma}=p'_v$.
\item Then the degree of $p'_{v,i}$
coincides with the degree of $p'_v$.
\item The degree of the closure
$\overline{(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap{\cal
Z}(e'_{\sigma+1},\ldots ,
e'_i)}$ is maximal possible (when one varies linear forms $L'_j$,
$\sigma+1\le j\le n-1$).
\end{enumerate}
Even more conditions of general position hold, e.g.,
$(L'_{\sigma+1},\ldots , L'_{n-1})\in{\cal U}'''_{n-1}$,
where  ${\cal U}'''_{n-1}\subset{\Bbb A}^{(r+1)(n-\sigma-1)}$ is
a nonempty open in the Zariski topology subset introduced in \cite{13},
see below.
For every $\sigma+1\le j\le n-1$ we shall
define an open in the Zariski
topology subset ${\cal U}_j\subset{\Bbb A}^{(r+1)(n-j)}$
such that if $(L'_{\sigma+1},\ldots , L'_{n-1})\in{\cal U}_j$ then
(i)--(iv) hold for all $\sigma+1\le i\le j$.



In particular,
under conditions of the first Bertini theorem
for the morphism $p'$, see below, we show how to find
an element from the open set
${\cal U}$ from the statement of this theorem
within the time
polynomial in $(dd')^n$ and the size of input.
Notice also once more that so far the first Bertini theorem
has been mainly existential.
There has not been explicit description of an
open in the Zariski topology subset ${\cal U}$
from the statement of this theorem.

\par\medskip\noindent{\bf REMARK~1}\hspace{0.1em} {\it  In Remark~1 from
\cite{13} we give a sufficient condition when (i)
and (ii) and (iii) imply (iv).
It is interesting to find other such conditions.
}\par\medskip




Denote by $k_u$ the field generated over $k$ by the
algebraically independent over $k$ elements
$u_{j,w}$, $j\in\{0,\sigma+1,\ldots , n\}$, $0\le
w\le r$.
Let $K$ be a field containing $k_u$
and such that $K$ and $k_v$ are
linearly disjoint over $k$.
Let $l'_{j,w}\in \overline{k}$,
$j\in\{0,\sigma+1,\ldots , n\}$, $0\le
w\le r$.  Let $L'_j=l'_{j,0}X_0+\ldots + l'_{j,r}X_r$
(respectively $U_j=u_{j,0}X_0+\ldots + u_{j,r}X_r$),
$j\in\{0,\sigma+1,\ldots ,
n\}$, be linear forms from $\overline{k}[X_0,\ldots ,X_r]$
(respectively $K[X_0,\ldots , X_n]$).
Set $e'_j=L'_j(e_0,\ldots , e_r)$ (respectively $u_j=U_j(e_0,\ldots ,
e_r)$), $j\in\{0,\sigma+1,\ldots ,n\}$. Put $L'=(L'_0,L'_{\sigma+1},\ldots ,
L'_n)$
(respectively $U=(U_0,U_{\sigma+1},\ldots , U_n)$).
Set $u$ (respectively $l'$) to be the family of all $u_{j,i}$,
(respectively $l'_{j,i}$) $j\in\{0,\sigma+1,\ldots , n\}$,
$0\le i\le r$.
In what follows we shall define some objects corresponding to $L'$. The
similar objects corresponding to
$U$ are defined by replacing the field $k$ by $K$.
Below we shall consider also the case when linear forms $L'_j$ have
coefficients from
$\overline{K}$. In this case the objects corresponding to $L'$ are defined
by the extension of the field $k$ till $\overline{K}$.


Recall some notations from \cite{13}.
Let us identify the affine space \\
${\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$
(respectively ${\Bbb A}^{(r+1)(n-i)}(\overline{k})$)
with the set of all
$(n-\sigma+1)$--tuples $L'=(L'_0,L'_{\sigma+1},\ldots , L'_n)$
(respectively $(n-i)$--tuples
$(L'_{i+1},\ldots , L'_n)$) of linear forms
$L'_j\in\overline{k}[X_0,\ldots , X_r]$.
For convenience of notation for $i=\sigma+1$
we shall identify $1$--tuples $(L'_{\sigma+1})$ with the linear form
$L'_{\sigma+1}$.
Recall that in Theorem~1 \cite{13} we define the open in the Zariski topology
subset ${\cal U}'_0\subset{\Bbb
A}^{(r+1)(n-\sigma+1)}(\overline{k})$ in the case when $s=\sigma$.

In \cite{13} the set ${\cal U}'_0$ is defined in the general case.
Namely, let ${\cal V}$ be the open in the Zariski topology subset from
the statement of Theorem~1 \cite{12} for $p'_v$, $k_v$
in place of $p'$, $k$ respectively.
Then ${\cal U}'_0=
{\cal V}\cap{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$.
In brief,
$L'\in{\cal U}'_0$ if and only if the numbers of elements of
$\#W_v\cap{\cal Z}(e'_{\sigma+1},\ldots , e'_n)\setminus({\cal
Z}(e'_0)\cup\overline{(V\setminus W)})$ (this is
a subset of ${\Bbb P}^n(\overline{k_v})$)
is $\deg p'_v\deg W'$ and
$\#(W'\cap{\cal Z}(L'_{\sigma+1},\ldots ,L'_n))=\deg W'$,
i.e., the last two intersections are finite and have the maximal possible
number of elements, see \cite{12}. Hence, if $L'\in{\cal U}'_0$ then
$W'\cap{\cal Z}(L'_0,\ldots , L'_r)=\emptyset$ in ${\Bbb
P}^r(\overline{k_v})$ and $W\cap{\cal Z}(e'_0,e'_{\sigma+1},\ldots ,
e'_n)=W\cap{\cal Z}(e_0,\ldots , e_r)$ in ${\Bbb P}^n(\overline{k})$.


\par\medskip\noindent{\bf REMARK~2}\hspace{0.1em} {\it  Obviously
if $L'=(L'_0,L'_{\sigma+1},\ldots , L'_n)\in{\cal U}'_0$, and
$\gamma$ is a nondegenerate linear transformation of $L'_{\sigma+1},\ldots ,
L'_n$ over $\overline{k}$ then
$(L'_0,\gamma(L'_{\sigma+1}),$ $\ldots , \gamma(L'_n))$ $\in{\cal U}'_0$.
}\par\medskip


Suppose that $L'\in{\cal U}'_0$. Consider the morphism
\begin{eqnarray*}
&&p_{L'}\, :\, W
\setminus{\cal Z}(e'_0,e'_{\sigma+1}\ldots , e'_n)\rightarrow
{\Bbb P}^{n-\sigma}(\overline{k}),\\
&&(X_0:\ldots :
X_n)\mapsto(e'_0:e'_{\sigma+1}:\ldots : e'_n)\nonumber.
\end{eqnarray*}
In \cite{13} for an arbitrary $L'$ we define algebraic varieties $W_i(L')$,
$W^{(\beta)}_i$,
$\sigma\le i\le n+1$ (they are subvarieties of $W$).
In the present paper we shall need only the case when
$\sigma\le i\le n$ and
$L'\in{\cal U}'_0$. In what follows unless we state otherwise we assume
that $L'\in{\cal U}'_0$. Then
\begin{itemize}
\item $W_i(L')$, $\sigma\le
i\le n$, is the union of all the irreducible components $E$ of
$W\cap{\cal Z}(e_{\sigma+1},\ldots , e_i)$ such that $\dim
E=\dim\overline{p(E\setminus{\cal Z}(e_0,\ldots , e_r)}=n-i+\sigma-s$.
\end{itemize}
Hence $W_\sigma(L')=W$. Set
$W^{(\beta)}_\sigma(L')=\emptyset$, $\beta=1,2,3$. Further, see
\cite{13}, for $\sigma+1\le i\le n$ the dimension $\dim
W_{i-1}(L')\cap{\cal Z}(e'_i)\le n+\sigma-s-i$
and $\dim W_n(L')\cap{\cal Z}(e'_0)\le \sigma-s-1$.
The dimension of each irreducible component of
$W^{(\beta)}_i(L')$, $\beta=1,2,3$, is $n-\sigma+s-i$.
Namely
$W_{i-1}(L')\cap{\cal Z}(e'_i)=W_i(L')\cup W^{(1)}_i(L')\cup
W^{(2)}_i(L')
\cup W^{(3)}_i(L')$,
where:
\begin{itemize}
\item $W^{(1)}_i(L')$ is the union of all
irreducible components $E$ of $W_{i-1}(L')\cap{\cal Z}(e'_i)$ such that
$E$ is an irreducible component of $W\cap
{\cal Z}(e_0,\ldots , e_r)$;
\item $W^{(2)}_i(L')$ is the union of all
irreducible components $E$ of $W_{i-1}(L')\cap{\cal Z}(e'_i)$ such that
$E\subset{\cal Z}(e_0,\ldots , e_r)$ but $E$ is not
an irreducible component of $W\cap{\cal Z}(e_0,\ldots , e_r)$;
\item $W^{(3)}_i(L')$ is the union of all
irreducible components $E$ of $W_{i-1}(L')\cap{\cal Z}(e'_i)$ such that
$E\not\subset{\cal Z}(e_0,\ldots , e_r)$ and
$\dim\overline{p_{L'}
(E\setminus{\cal Z}(e'_0,e'_{\sigma+1}\ldots , e'_n))}<n-i+\sigma-s$,
see \cite{13}.
\end{itemize}
Since $L'\in{\cal U}'_0$ we have, see \cite{13},
\begin{itemize}
\item $W^{(3)}_i(L')$ is the union of all
irreducible components $E$ of $W_{i-1}(L')\cap{\cal Z}(e'_i)$ such that
 $E\not\subset{\cal Z}(e_0,\ldots , e_r)$ and
$\dim\overline{p(E\setminus{\cal Z}(e_0,\ldots , e_r))}<n-i+\sigma-s$.
\end{itemize}
Hence if $W^{(3)}_i(L')\ne\emptyset$ then $\dim W^{(3)}_i(L')\ge 1$
(in particular, $W^{(3)}_n(L')=\emptyset$). Also we have
$W^{(2)}_{\sigma+1}(L')=\emptyset$.
Set $D_i(L')=\deg W_i(L')$ and $D^{(\beta)}_i(L')=
\deg W^{(\beta)}_i(L')$, $\beta=1,2,3$.

Let $W_j$, $j\in J_i(L')$ (respectively $j\in J^{(\beta)}_i(L')$,
$\beta=1,2,3$),
be the family of all the irreducible over $\overline{k}$
components  of $W_i(L')$ (respectively $W^{(\beta)}_i(L')$), and
$\iota_j(L')=i(W_{i-1}(L'),{\cal
Z}(e'_i); W_j)$
denote the index of intersection of $W_{i-1}(L')$
and ${\cal Z}(e'_i)$ at
$W_j$, see \cite{13}.
Let $W^{(1)}_i=\bigcup_{j\in J^{(1)}_i}W_j$ be the union of all the
irreducible components of dimension
$n-\sigma+s-i$ of $W\cap{\cal Z}(e_0,\ldots , e_r)$.
We suppose in what follows without loss of generality that
$J^{(1)}_i(L')\subset J^{(1)}_i$ for every $L'\in{\cal U}'_0$.


Let $\sigma+1\le\alpha\le n$ be an
integer. By Lemma~11 \cite{12}, condition
$$
W_i(L')=\overline{W\cap{\cal Z}(e'_{\sigma+1},\ldots , e'_i)
\setminus{\cal Z}(e_0,\ldots , e_r)}
\eqno (2)
$$
holds for all $\sigma+1\le i\le\alpha$
if and only if $W^{(3)}_i(L')=\emptyset$
for all $\sigma+1\le i\le\alpha$.
If (2) holds for every $\sigma+1\le i\le\alpha$ then, see \cite{13},
$W^{(1)}_i(L')=W^{(1)}_i$
for every $\sigma+1\le i\le\alpha$ does not depend on the choice of $L'$.
Hence in this case that
$J^{(1)}_i(L')=J^{(1)}_i$ for every $\sigma+1\le i\le\alpha$ does not depend
on the choice of $L'$.

We set $D_i=D_i(U)$,
$D^{(\beta)}_i=D^{(\beta)}_i(U)$, $\beta=1,2,3$,
and $\iota_j=\iota_j(U)$ for every $j\in J^{(1)}$.

In \cite{13} we introduce also the notation $W_i(L')=W(L_{\sigma+1},\ldots ,
L_i)$ and
similarly for $W^{(1)}_i(L')$, $W^{(2)}_i(L')$, $D_i(L')$ (in place of
$W_i(L')$) and so on; the considered
objects depend actually only on $L_{\sigma+1},\ldots , L_i$.

Let $\sigma\le\alpha\le n$ be an integer.
{\it Denote by ${\cal U}'_\alpha$,
the image of ${\cal U}'_0$
under the linear mapping
$\pi_\alpha\, :\, {\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})
\rightarrow{\Bbb A}^{(r+1)(\alpha-\sigma)}(\overline{k})$,
$L'\mapsto(L'_{\sigma+1},\ldots , L'_\alpha)$.}
Then ${\cal U}'_\alpha$ is an open with respect to the
Zariski topology set
since the image under a linear projection of an open
set is again open.

{\it
Denote by ${\cal U}''_\alpha\subset{\cal U}'_\alpha$ the subset of
all $(L'_{\sigma+1},\ldots ,
L'_\alpha)$ such that
$$
D(L'_{\sigma+1},\ldots , L'_i)=D_i
\eqno (3)
$$
for all $\sigma+1\le i\le \alpha$. Set also ${\cal U}''_0$
to be the set of all $L'\in{\cal U}'_0$ such that
$(L'_{\sigma+1},\ldots , L'_n)\in{\cal U}''_n$

Denote by  ${\cal U}'''_\alpha\subset{\cal U}''_\alpha$ the subset
of all $(L'_{\sigma+1},\ldots ,
L'_\alpha)$ such that
$$
D^{(2)}(L'_{\sigma+1},\ldots , L'_i)=D^{(2)}_i
\eqno (4)
$$
for all $\sigma+1\le i\le \alpha$. Set also ${\cal U}'''_0$
to be the set of all $L'\in{\cal U}''_0$ such that
$(L'_{\sigma+1},\ldots , L'_n)\in{\cal U}'''_n$.}


If $(L'_{\sigma+1},\ldots , L'_\alpha)\in{\cal U}'_\alpha$ then,
by Lemma~11 \cite{12}, the indices of intersection
$\iota_j(L'_{\sigma+1},\ldots , L'_i)=1$ for all
$j\in J(L'_{\sigma+1},\ldots , L'_i)$, $\sigma+1\le i\le \alpha$.
If $(L'_{\sigma+1},\ldots , L'_\alpha)\in{\cal U}''_\alpha$ then,
see \cite{13}, the indices of intersection
$\iota_j(L'_{\sigma+1},\ldots , L'_i)=\iota_j$ for all
$j\in J^{(1)}_i$, $\sigma+1\le i\le \alpha$.
Further, ${\cal U}''_\alpha$, $U'''_\alpha$ are
nonempty open in the Zariski topology subsets of ${\Bbb
A}^{(r+1)(\alpha-\sigma)}(\overline{k})$, see \cite{13}.
By Lemma~14 \cite{12} applied to
$W_v$, $p'_v$, $\alpha$
in place of $W$, $p'$, $i+1$ if $D_i(L')=D_i$ for all
$\sigma+1\le i\le\alpha$ then
$W^{(3)}_i(L')=\emptyset$ for all $\sigma+1\le i\le\alpha$.
Therefore, if $L'\in{\cal U}''_0$ then
condition (2) holds for all $\sigma\le i\le n-1$.


{\it Denote for brevity ${\cal U}'={\cal U}'_{\sigma+1}$
(respectively ${\cal U}''={\cal U}''_{\sigma+1}$, ${\cal
U}'''={\cal U}'''_{\sigma+1}$).} Notice that
${\cal U}''={\cal U}'''$.

Notice, see \cite{13}, that $D_i$ is the maximal possible value of $D_i(L')$;
if $D_i(L')=D_i$ then
all the indices of intersection
$\iota_j(L')=\iota_j$, $j\in J^{(1)}$ are minimal possible;
$D^{(2)}_i$ is the maximal possible value
of $D^{(2)}_i(L')$ under conditions $D_j=D_j(L')$ for all $\sigma+1\le j\le
n$, $D^{(2)}_j(L')=D^{(2)}_j$ for all $\sigma+2\le j<i$.



\medskip Let $W_1,W_2\subset{\Bbb P}^n(\overline{k})$
be two quasiprojective algebraic varieties
defined over $k$. Let $W_3$ be an arbitrary defined over $k$ and irreducible
over $k$
component of $W_1\cap W_2$. We
shall say that the intersection
$W_1\cap W_2$ is transversal at $W_3$ if and
only if
there is a point $x\in W_3$
such that $x$ is a smooth point of
every $W_i$, $i=1,2$, and the
intersection of the tangent spaces
$T_{x,W_1}\cap T_{x,W_2}$ of $W_1$ and $W_2$ at the point $x$ is transversal,
i.e., $\dim(T_{x,W_1}\cap T_{x,W_2})=n-\dim(T_{x,W_1})-\dim(T_{x,W_2})$.
We shall say that the intersection
$W_1\cap W_2$ is transversal if and only if it is transversal at every its
defined over $k$ and irreducible over $k$ component.  In the similar way one
defines transversality for intersection of a finite number of
algebraic varieties.

By, e.g., Lemma~21 \cite{13}, if the intersection $W_1\cap W_2$ is
transversal
and $y$ is an arbitrary
smooth point of $W_1\cap W_2$ then $y$ is a smooth point of every $W_i$,
$i=1,2$ and the intersection of the tangent spaces
$T_{y,W_1}\cap T_{y,W_2}$ of $W_1$ and $W_2$ at the point $y$ is transversal.
Also the following simple result may be useful here (we leave its proof
to the reader).

\par\medskip\noindent{\bf PROPOSITION~1}\hspace{0.1em} {\it  Let $W_i$,
$i\in I$, be a finite family of
quasiprojective subvarieties of ${\Bbb P}^n(\overline{k})$.
Suppose that the
intersection $\bigcap_{i\in I}W_i$ is transversal. Assume that for an
arbitrary
family $E_i$, $i\in I$, such that $E_i$ is an irreducible component of
$W_i$, $i\in I$, the sum
$\sum_{i\in I}(n-\dim E_i)=\sum_{i\in I}\mbox{\rm codim} E_i\le
n$.
Let $I_1\subset I$. Suppose that for every irreducible component $E$ of
$\bigcap_{i\in I_1}W_i$ the intersection
$E\cap\bigcap_{i\in I\setminus I_1}W_i\ne\emptyset$ (in particular, this
condition  holds
if all the algebraic varieties $W_i$, $i\in I$, are projective,
i.e., closed in ${\Bbb P}^n(\overline{k})$). Then the intersection
$\bigcap_{i\in I_1}W_i$ is also transversal.
}\par\medskip




Now we are going to formulate a variant of the first
Bertini theorem (mainly for references in what follows).
Notice that all the
objects introduced above for the
varieties $V$, $W$ and the morphism $p$
can be defined also in the same way for the case when $k$ is a perfect field
of nonzero characteristic and the extension $k(W)\supset k(W')$
is separable. In this case similarly to zero--characteristic
${\cal U}'_\alpha$, ${\cal U}''_\alpha$, ${\cal
U}'''_\alpha$ are open in the Zariski topology subsets of the corresponding
affine spaces
(we leave the details to the reader; it is not our aim to consider
the case of nonzero characteristic in this paper,
except, may be, the Appendix).
Recall that if $L'\in{\cal U}''$ then
$W^{(3)}(L'_{\sigma+1})=\emptyset$ and
$W(L'_{\sigma+1})=\overline{W\cap{\cal
Z}(e'_{\sigma+1})\setminus{\cal Z}(e_0,\ldots , e_r)}$.
(it is proved for nonzero characteristic similarly to case
zero--characteristic, see Lemma~14 \cite{12}).
If $V=W$ then the set ${\cal U}'$, ${\cal
U}''$, ${\cal U}'''$ does not
depend on the choice of $f_1,\ldots ,f_m$.
From the given here variant of the first
Bertini theorem one can
deduce the usual one, see e.g., \cite{17}, on the language of linear systems.
One needs only to put $V=W$ and $k=\overline{k}$.

\medskip\noindent
{\bf
THEOREM (a version of the Bertini theorem)}\quad
{\it Let $k$ be a perfect field (now it is not necessarily of
zero--characteristic)
and $W\subset{\Bbb P}^n(\overline{k})$
be a defined over $k$ and irreducible over $k$
projective algebraic variety with
$\dim W=n-s$. Let the varieties $V$, $W$,
the polynomials $e_0,\ldots , e_r$, $r\ge 2$,
the bound for the degree $d'$ and the morphisms $p$, $p'$ be as above.
Suppose that $\dim W'>1$ (hence $s\le n-2$)
and the extension of fields of rational
functions $k(W)\supset k(W')$ is separable (may be of the
nonzero transcendency degree).
Denote by ${\cal U}$ the subset of
all linear forms $L'_{\sigma+1}\in{\cal U}''$
such that the irreducible over $\overline{k}$ components of the algebraic
varieties $W$
and $W(L'_{\sigma+1})$ are in the one--to--one
correspondence by the rule $W_1\mapsto W_2$ where $W_2$ is the unique
(the uniqueness is claimed)
irreducible over $\overline{k}$ component of
$W_1\cap{\cal Z}(e'_{\sigma+1})$ such that $W_2\subset W(L'_{\sigma+1})$
(hence the intersection of $W$ and ${\cal Z}(e'_{\sigma+1})$
is transversal at
each irreducible component $W_2$ of $W(L'_{\sigma+1})$).
Then ${\cal U}$ is a nonempty open in the Zariski topology defined over
$k$ subset of ${\Bbb A}^{n+1}(\overline{k})$.
For every $L'_{\sigma+1}\in{\cal U}(k)$
the algebraic variety $W(L'_{\sigma+1})$ is defined over $k$
and irreducible over $k$.
}

\medskip
One can deduce the formulated theorem
from, e.g., \cite{17}.
For convenience of the reader we give an independent
proof of this fact in the Appendix.
The fact that ${\cal U}$ is nonempty and open in
the Zariski topology follows from Lemma~24 which is based on Theorem~4.
Here it is rather
difficult to show (in nonzero characteristic) that ${\cal
U}\ne\emptyset$.

Let ${\cal U}_\alpha$, $\sigma+1\le\alpha\le n-1$,
be the subset of all $(\alpha-\sigma)$--tuples
$(L'_{\sigma+1},\ldots , L'_\alpha)\in{\cal U}'''_\alpha$ such that
for every $\sigma+1\le i\le\alpha$ (or, which is equivalent, for $i=\alpha$,
see Lemma~11 from \cite{12})
the irreducible over $\overline{k}$ components of the algebraic
varieties $W$
and $W(L'_{\sigma+1},\ldots , L'_i)$ are in the one--to--one
correspondence by the rule $W_1\mapsto W_2$ where $W_2$ is the unique
(the uniqueness is claimed)
irreducible over $\overline{k}$ component of
$W_1\cap{\cal Z}(e'_{\sigma+1},\ldots , e'_i)$
such that $W_2\subset W(L'_{\sigma+1},\ldots , L'_i)$.
Hence the intersection of $W$ and ${\cal Z}(e'_{\sigma+1},\ldots ,e'_i)$
is transversal at
each irreducible component $W_2$ of $W(L'_{\sigma+1},\ldots , L'_i)$,
$\sigma+1\le i\le\alpha$.
Applying the formulated variant of
the first Bertini theorem $\alpha-\sigma$ times one
gets that
${\cal U}_\alpha$, $\sigma+1\le\alpha\le n-1$, is  nonempty.
By Lemma~24 the set ${\cal U}_\alpha$ is a nonempty open in the
Zariski topology in
${\Bbb A}^{(r+1)(\alpha-\sigma)}(\overline{k})$ provided that
the extension
$k(W)\supset k(W')$ is separable.

{\it
Denote by ${\cal U}_n$ (respectively ${\cal U}_0$) the subset of all
$(L'_{\sigma+1},\ldots , L'_n)\in{\cal U}'''_n$ (respectively
$L'\in{\cal
U}'''_0)$ such that
$(L'_{\sigma+1},\ldots , L'_{n-1})\in{\cal U}_{n-1}$.}

\par\medskip\noindent{\bf REMARK~3}\hspace{0.1em} {\it  Recall that Remark~2
\cite{13} related  to the extensions
of ground field holds for all ${\cal U}'_\alpha$,
${\cal U}''_\alpha$, ${\cal U}'''_\alpha$. The similar remark is
true now for all ${\cal U}_\alpha$
and ${\cal U}^{({\rm v})}_\alpha$ defined below.
}\par\medskip


Now let $\lambda=(L_{s+1},\ldots , L_\sigma)$ and all
$L_j=\sum_{0\le i\le n}l_{j,i}X_i\in \overline{k}[X_0,\ldots ,
X_n]$, $s+1\le j\le\sigma$.
Denote by $\widetilde{\lambda}$ the family of coefficients $l_{j,i}$,
$s+1\le i\le \sigma$,
$0\le i\le n$ of linear forms $L_j$.
Unless we state otherwise
we shall suppose in what follows (the other assumptions,
see at the beginning of Section~1)
that the following properties holds for $\lambda$.
\begin{itemize}
\item[(C)] All the coefficients $l_{j,i}\in k$, $s+1\le j\le\sigma$,
$0\le i\le n$. The dimension
$\dim W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)=n-\sigma>0$;
the last intersection is transversal.
The algebraic variety $W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$ is
irreducible over $k$.
Denote $W_\lambda=W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$,
$V_\lambda=V\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$.
The algebraic variety
$W_\lambda\not\subset\overline{V_\lambda\setminus W_\lambda}$.
\end{itemize}
Recall that if condition (A) from \cite{13} holds
then (C) is fulfilled.
Denote by
\setcounter{equation}{4} \begin{eqnarray}
&&p_\lambda\, :\, W_\lambda\setminus{\cal
Z}(e_0,\ldots , e_r)\rightarrow{\Bbb P}^r(\overline{k}),\label{5}\\
&&p'_\lambda\, :\, W_\lambda\setminus{\cal
Z}(e_0,\ldots , e_r)\rightarrow W'\label{6}
\end{eqnarray}
the morphisms induced by $p$ and $p'$.
Suppose that the morphism $p'_\lambda$ is dominant.
Let us replace $(V,W,p,p')$ by $(V_\lambda,W_\lambda,p_\lambda,p'_\lambda)$
in the definitions of all $W_i(L')$,
$W(L'_{\sigma+1},\ldots , L'_n)$,
${\cal U}'_\alpha$, ${\cal U}''_\alpha$, ${\cal U}'''_\alpha$
and so on. We shall denote the
obtained objects by
$W_{\lambda,i}(L')$, $W_\lambda(L'_{\sigma+1},\ldots , L'_n)$,
${\cal U}'_{\lambda,\alpha}$, ${\cal U}''_{\lambda,\alpha}$,
${\cal
U}'''_{\lambda,\alpha}$ and so on.
Notice that the field
$k(v_\lambda)=k$
since $s_\lambda=\sigma_\lambda=
\sigma$.
It will be convenient for us to consider the morphism $p'_\lambda$ together
with $p'_v$, see Theorem~2 below. Let us define $V_v$, $W_v$ and so on
extending the ground field $k$ till $k_v$.


Consider the basis over $k$ of
the linear space of all homogeneous polynomials
of degree $d'$
consisting of all different monomials
$X_0^{i_0}\cdots X_n^{i_n}$,
$i_0+\ldots + i_n=d'$, $i_0,\ldots , i_n\ge 0$. Denote by
$I$ the set of the all considered multiindices $(i_0,\ldots , i_n)$. So the
number of elements
$\#I={n+d'\choose n}$.
Let $\widetilde{e}_{n+1}$ be a generic homogeneous polynomial of degree $d'$
in $X_0,\ldots ,X_n$,
i.e., the family $c=\{c_i\}_{i\in I}$
of its coefficients
in the considered basis has transcendency degree ${n+d'\choose n}$
over $K_v$. Set $K'=K(v,c)\supset k_v(c,u)=k(c,\upsilon,u)$.

Let $e'_{n+1}$ be a homogeneous polynomial with coefficients from $k$.
Denote by $c'=\{c'_i\}_{i\in I}$
the family of coefficients
of $e'_{n+1}$ in the considered basis.

Let $\lambda$ be as above and $L'\in{\cal U}'_{\lambda,0}$
(unless we state otherwise we shall suppose
that these assumptions hold when we consider the morphisms
(7)--(11); but sometimes the assumptions will be other).
Let us define the morphisms
\setcounter{equation}{6} \begin{eqnarray}
&&p_{\lambda,L'}\, :\, W_\lambda\setminus
{\cal Z}(e'_0,e'_{\sigma+1},\ldots ,
e'_n)\rightarrow{\Bbb
P}^{n-\sigma}(\overline{k}),\label{7}\\
&&(X_0:\ldots :
X_n)\mapsto(e'_0:e'_{\sigma+1}:\ldots : e'_n),\nonumber\\
&&p_{c',\lambda}\, :\, W_\lambda\setminus{\cal Z}(e_0,\ldots ,
e_r,e'_{n+1})\rightarrow
{\Bbb P}^{r+1}(\overline{k}),\label{8}\\
&&(X_0:\ldots : X_n)\mapsto(e_0:\ldots : e_r:e'_{n+1});\nonumber\\
&&p_{c',\lambda,L'}\, :\, W_\lambda\setminus
{\cal Z}(e'_0,e'_{\sigma+1}\ldots ,e'_{n+1})\rightarrow
{\Bbb P}^{n-\sigma+1}(\overline{k}),\label{9}\\
&&(X_0:\ldots : X_n)\mapsto(e'_0:e'_{\sigma+1}:\ldots : e'_{n+1});\nonumber\\
&&p_{\lambda,L',*}\, :\, W_\lambda\cap{\cal Z}(e'_{\sigma+1},\ldots ,
e'_{n-1})\setminus
{\cal Z}(e'_0,e'_n)\rightarrow{\Bbb
P}^1(\overline{k}); \label{10}\\
&&(X_0:\ldots :
X_n)\mapsto(e'_0:e'_n),\nonumber\\
&&p_{c',\lambda,L',*}\, :\, W_\lambda\cap{\cal Z}(e'_{\sigma+1},\ldots ,
e'_{n-1})\setminus
{\cal Z}(e'_0,e'_n,e'_{n+1})\rightarrow{\Bbb P}^2(\overline{k})
,\label{11}\\
&&(X_0:\ldots : X_n)\mapsto(e'_0:e'_n:e'_{n+1}).\nonumber
\end{eqnarray}
Put
\setcounter{equation}{11} \begin{eqnarray}
&&W'(c',\lambda)=
\overline{p_{c',\lambda}(W_\lambda\setminus{\cal Z}(e_0,\ldots ,
e_r,e'_{n+1}))}\subset{\Bbb P}^{r+1}(\overline{k}),\label{12}\\
&&W'(c',\lambda,L')=
\overline{p_{c',\lambda,L'}(W_\lambda\setminus{\cal Z}
(e'_0,e'_{\sigma+1}\ldots ,e'_{n+1}))}
\subset{\Bbb P}^{n-\sigma+1}(\overline{k}),\label{13}\\
&&W'(c',\lambda,L',*)=\nonumber\\
&&\overline{p_{c',\lambda,L',*}(W_\lambda\cap{\cal
Z}(e'_{\sigma+1},\ldots , e'_{n-1})
\setminus{\cal Z}
(e'_0,e'_n,e'_{n+1}))}. \label{14}
\end{eqnarray}
Hence $W'(c',\lambda,L',*)\subset{\Bbb P}^2(\overline{k})$.
Since $L'\in{\cal U}'_{\lambda,0}$ the morphism $p_{\lambda,L'}$ is
dominant. If the morphism $p_{\lambda,L'}$ is dominant and $\dim
W_\lambda=n-\sigma$
(without any additional assumptions for $\lambda\in{\Bbb
A}^{(\sigma-s)(n+1)}(\overline{k})$
and $L'\in{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$) then
$\dim W'(c',\lambda,L')=$ $\dim W'(c',\lambda)
=n-\sigma$. By \cite{13} if assertions (c) and (d) of
Theorem~2 \cite{13} hold for $\lambda$ and $L'$
then the dimension of every irreducible
component of $W_\lambda\cap{\cal
Z}(e'_{\sigma+1},\ldots , e'_{n-1})\setminus{\cal Z}
(e'_0,e'_n,e'_{n+1})$ is $1$,
and the morphism $p_{\lambda,L',*}$ is dominant.
If the morphism $p_{\lambda,L',*}$ is dominant and $\dim
W_\lambda\cap{\cal
Z}(e'_{\sigma+1},\ldots , e'_{n-1})\setminus{\cal Z}
(e'_0,e'_n,e'_{n+1})=1$
(without any additional assumptions for $\lambda\in{\Bbb
A}^{(\sigma-s)(n+1)}(\overline{k})$
and $L'\in{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$) then
$\dim W'(c',\lambda,L',*)=\dim W'(c',\lambda,*)=1$.



For an arbitrary polynomial $\psi$ we shall denote by $\psi_0$ its square
free part, i.e., the separable (that is  without multiple factors)
polynomial of the maximal degree dividing $\psi$. This
polynomial $\psi_0$ is defined up to a factor from
$k$; we choose and fix it.
Now let $\psi\in \overline{k}[a',Z_0,Z_{\sigma+1},\ldots ,
Z_{n+1}]$ where $a'$ are some other
variables (or variables and families of variables). Then we put
$\psi_*=\psi(a',Z_0,0,\ldots , 0,Z_n,Z_{n+1})\in
\overline{k}[a',Z_0,Z_n,Z_{n+1}]$.
If $\psi\in \overline{k}[a',Z_0,Z_{\sigma+1},\ldots ,
Z_n]$ then we shall assume that $\psi_*\in \overline{k}[a',Z_0,Z_n]$.


Let $c'$, $\lambda=(L_{s+1},\ldots , L_\sigma)$, $L'=(L'_0,L'_{\sigma+1},
\ldots , L'_n)\in{\cal U}'_{\lambda,0}$ be as above. Recall that
we suppose that all linear forms $L_j\in k[X_0,\ldots , X_n]$,
$s+1\le j\le\sigma$;
$L'_w\in\overline{k}[X_0,\ldots ,X_r]$, $\sigma+1\le w\le n$.
Denote for brevity by $A_{\lambda,L'}$ the set of isolated
points of the intersection
$$
(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(L_{s+1},\ldots , L_\sigma)\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_n).
\eqno (15)
$$
Hence $\#A_{\lambda,L'}=\deg p'_\lambda\deg W'
\le\deg p'_v\deg W'$, see \cite{13}.

{\it Denote by $\Phi\in \overline{k}[Z_0,Z_{\sigma+1},\ldots ,
Z_{n+1}]$ the
separable polynomial of the minimal
degree with respect to all the variables such that
$\Phi(e'_0,e'_{\sigma+1},$ $\ldots , e'_{n+1})$ is
vanishing on $W_\lambda$.} We choose and fix $\Phi$ (it is uniquely defined
up to a
nonzero factor from $\overline{k}$ even under less restrictive assumptions,
namely,
that the morphism $p_{\lambda,L'}$ is dominant,
$\dim W_\lambda=n-\sigma$, $\lambda\in{\Bbb
A}^{(\sigma-s)(n+1)}(\overline{k})$
and $L'\in{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$).
{\it Denote by
$$
R=\mbox{\rm Res}_{Z_{n+1}}(\Phi,\partial\Phi/\partial
Z_{n+1})\in
\overline{k}[Z_0,Z_{\sigma+1},\ldots , Z_n]
$$
the discriminant of the polynomial $\Phi$ with respect to $Z_{n+1}$} (this
definition of the
discriminant as the resultant
of the polynomial and its partial derivative is different
from the generally accepted by a factor
$\mbox{\rm lc}_{Z_{n+1}}\Phi$, but it is
more convenient for us).
{\it Denote by $R_0\in \overline{k}[Z_0,Z_{\sigma+1},\ldots , Z_n]$
the square free part of $R$.}
If all linear forms $L'_j$, $\sigma+1\le j\le n$,
have coefficients from $k$ then we choose
$\Phi$ to be a polynomial with coefficients from $k$.
Hence $R$, $R_0$ also have coefficients from $k$ in the considered case.
We shall denote also
$\Phi=\Phi_{c',\lambda,L'}$,
$R=R_{c',\lambda,L'}$, $R_0=R_{c',\lambda,L',0}$  when the
dependence on $e'_{n+1}$,
$\lambda$, $L'$ will be important.
Obviously $W'(c',\lambda,L')={\cal Z}(\Phi_{c',\lambda,L'}(X_0,
\ldots , X_{n-\sigma+1}))$, the degrees
$\deg_{Z_0,Z_{\sigma+1},\ldots , Z_{n+1}}\Phi_{c',\lambda,L'}=\deg
W'(c',\lambda,L')$ and
$\deg_{Z_{n+1}}\Phi_{c',\lambda,L'}=\deg r_{c',\lambda,L'}$,
see (23) below.
{\it
Hence if all linear forms $L'_j\in k[X_0,\ldots ,X_r]$,
$j\in\{0,\sigma+1,\ldots , n\}$ then one can
compute $\deg_{Z_{n+1}}\Phi_{c',\lambda,L'}$ and
$\deg_{Z_0,Z_{\sigma+1},\ldots , Z_{n+1}}\Phi_{c',\lambda,L'}$ using
Theorem~1 \cite{12}.}
We have $W'(c',\lambda)\cap{\cal Z}(U_0,U_{\sigma+1},\ldots ,
U_n,X_{n+1})=\emptyset$ in ${\Bbb P}^{n-\sigma+1}(\overline{k_u})$.
Hence $\deg_{Z_0,Z_{\sigma+1},\ldots , Z_{n+1}}\Phi_{c',\lambda}=\deg
W'(c',\lambda)$.
If 1), 2) and 3) holds, see below Definition~1, then
by Lemma~2
the polynomials $\Phi_{c,\lambda,L',*}$,
$R_{c,\lambda,L',*}$ and $R_{c,\lambda,L',*,0}$
are nonzero.
Set
$$
a(c',\lambda,L')=\deg_{Z_0,Z_n}R_{c',\lambda,L',*,0},\quad
b(c',\lambda,L')=\deg_{Z_0,Z_n}R_{c',\lambda,L',0,*}.
$$
Obviously  $a(c',\lambda,L')\le b(c',\lambda,L')$. In Theorem~3 Section~4
the case is considered when this inequality becomes the equality, $d'=1$,
and condition 4), see below, holds.


\par\medskip\noindent{\bf DEFINITION~1}\hspace{0.1em} {\it
We shall say that $(c',\lambda,L')$ satisfies (*)
(or (*) holds for $(c',\lambda,L')$)
if and only if the following conditions are fulfilled.
\begin{enumerate} \renewcommand{\labelenumi}{\arabic{enumi})}
\item All the linear forms $L_j\in k[X_0,\ldots , X_n]$,
$s+1\le j\le\sigma$;
$L'_w\in\overline{k}[X_0,\ldots ,X_r]$, $w\in\{0,\sigma+1,\ldots , n\}$.
\item Further, $L'$ and $\lambda$ satisfy assertions (c) and (d)
of Theorem~2 \cite{13}
(now we do not suppose that $L_j$ and $L'_j$ have integer
coefficients as in assertions (a) and (b) of this theorem).
In the other words, $\lambda$ satisfies condition (A) from \cite{13};
the morphism $p'_\lambda$,
is dominant; the degree $\deg p'_\lambda=\deg p'_v$;
$D_{\lambda,i}=D_i$, $D^{(1)}_{\lambda,i}=D^{(1)}_i$,
$D^{(2)}_{\lambda,i}=D^{(2)}_i$ for all
$\sigma \le i\le n$; and $L'\in{\cal U}'''_{\lambda,0}\subset
{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$.
\item The polynomial $e'_{n+1}\in k[X_0,\ldots , X_n]$ is
homogeneous of degree $d'$,
and number of elements
$$
\#(e'_{n+1}/e'_0)(W_\lambda\cap{\cal Z}(e'_{\sigma+1},\ldots , e'_n)
\setminus{\cal Z}(e_0,\ldots , e_r))=
\deg p'_v\deg W',
\eqno (16)
$$
\item Finally, the polynomial $\Phi_{c',\lambda,L',*}$ does not
have irreducible factors
from $\overline{k}[Z_0,Z_n]$,
i.e., for every irreducible factor $\Psi$ of $\Phi_{c',\lambda,L',*}$ the
degree \\ $\deg_{Z_{n+1}}\Psi\ge 1$.
\end{enumerate}
Let $c_0,\lambda_0,L^{(0)}$ be similar to $c',\lambda,L'$. We shall say that
$(c_0,\lambda_0,L^{(0)})$ satisfies (*) (respectively 1), 2), 3), 4)) if and
only if it satisfies (*) (respectively 1), 2), 3), 4))
in place of $(c',\lambda,L')$.
Let us extend the ground field $k$ till $k_u$.
We shall say that $(c',\lambda)$ satisfies property (*)
(respectively 1), 2), 3), 4)) if and only if
$(c,\lambda,U)$ satisfies (*)
(respectively 1), 2), 3), 4)).
}\par\medskip

By Lemma~10 $(c',\lambda)$ satisfies (*) if and only if
$(c',\lambda,U)$ satisfies 1)--3) from (*)
(for the ground field $k_u$).

\par\medskip\noindent{\bf REMARK~4}\hspace{0.1em} {\it  One can modify 2)
replacing (d) by $L'\in{\cal U}'_0$, and
suggest new versions of the results given below.
Then (iv), see above,
will not necessarily hold.
One can also weaken (c). The reader can suggest other such
modifications.
}\par\medskip


Under conditions 1), 2) and 3) the number of elements
$\#(e'_{n+1}/e'_0)(W_\lambda\cap{\cal Z}(e'_{\sigma+1},\ldots , e'_n))$
is maximal possible (when it is finite, see \cite{13}),
and \\ $\deg_{Z_{n+1}}\Phi_{c',\lambda,L'}=\deg p'_v\deg W'$.
By assertion (e) of
Theorem~2 \cite{13} (c) and (d) imply $L'\in{\cal U}'''_0$.
If 1), 2), and 3) hold then $\deg_{Z_{n+1}}\Phi=\deg p'_v\deg W'$.
Below we shall see how to construct at least one element $(c',\lambda,L')$
satisfying (*).

Recall that under conditions 1), 2) and 3) the polynomial
$\Phi_{c',\lambda,L',*}\ne 0$.
Then $\deg_{Z_0,Z_n,Z_{n+1}}
\Phi_{c',\lambda,L',*}=
\deg_{Z_0,Z_{\sigma+1},\ldots , Z_{n+1}}\Phi_{c',\lambda,L'}=\deg
W'(c',$ $\lambda,L')$.
{\it Assume that 1), 2) and 3) hold. Then by Lemma~3 property
4) is satisfied for $(c',\lambda,L')$ if and only if
$\deg W'(c',\lambda,L')=
\deg W'(c',\lambda,L',*)$.}





\par\medskip\noindent{\bf REMARK~5}\hspace{0.1em} {\it  Suppose that
$(c',\lambda,L')$ satisfies 1) and, more than that,
all the linear forms
$L'_j$ have coefficients from $k$. Then one can decide whether
$(c',\lambda,L')$
satisfies (*) by the algorithms from \cite{12} and \cite{13} with the time
bounds
similar to the ones from Theorem~1 \cite{12} and Theorem~2 \cite{13}.
More precisely, under conditions 1) and 2)
to decide whether $(c',\lambda,L')$
satisfies 4) it is sufficient to compute
$\deg W'(c',\lambda,L')$ and
$\deg W'(c',\lambda,L',*)$ by Theorem~1 \cite{12}.
}\par\medskip






We shall consider $(c',\lambda,L')$ as a
point of
${\Bbb A}^\nu(\overline{k})$ where $\nu={n+d'\choose n}+
(\sigma-s)(n+1)+(n-\sigma)(r+1)$. The coordinates of $(c',\lambda,L')$ are
$c'_i$, $i\in I$; $l_{j,i}$, $s+1\le j\le\sigma$, $0\le i\le n$;
$l'_{j,w}$, $\sigma\le j\le n$, $0\le w\le r$. Similarly
$(c',\lambda)$ is a point of
${\Bbb A}^{\nu_1}(\overline{k})$ where $\nu_1=
{n+d'\choose n}+(\sigma-s)(n+1)$.












Let $(c_1,\lambda_1,L^{(1)})$ (respectively $(c_1,\lambda_1)$) be a generic
point of an
affine linear subspace ${\cal L}_{c_1,\lambda_1,L^{(1)}}$ (respectively
${\cal L}_{c_1,\lambda_1}$)
of ${\Bbb A}^\nu(\overline{k})$ (respectively ${\Bbb
A}^{\nu_1}(\overline{k})$)
such that $c_1$ and $\lambda_1$ are defined over $k$.


More explicitly, the generic point $(c_1,\lambda_1)$
is given as follows. A point
$(c_0,\lambda_0,$ $L^{(0)})\in{\cal L}_{c_1,\lambda_1,L^{(1)}}$ is given.
Let $c_{0,i}$, $i\in I$; $l_{0,j,i}$, $s+1\le j\le\sigma$, $0\le i\le n$;
$l^{(0)}_{j,i}$, $j\in\{0,\sigma+1,\ldots , n\}$, $0\le i\le r$
for $(c_0,\lambda_0,L^{(0)})$ be similar to the corresponding
coefficients $c'_i$; $l_{j,i}$;
$l'_{j,i}$ defined for $(c',\lambda,L')$.
Further, suppose that
$c_1=\{c_{1,i}\}_{i\in I}$; $\lambda_1=(L_{1,s+1},\ldots , L_{1,\sigma})$,
$L_{1,j}=\sum_{0\le i\le n}l_{1,j,i}X_i$, $s+1\le j\le\sigma$;
$L^{(1)}=(L^{(1)}_{\sigma+1},\ldots , L^{(1)}_{n})$, $L^{(1)}_j=\sum_{0\le
i\le
r}l^{(1)}_{j,i}X_i$,
$j\in\{0,\sigma+1,\ldots , n\}$;
\setcounter{equation}{16} \begin{eqnarray}
&&c_{1,i}=c_{0,i}+\sum_{1\le w\le n_1}c_{1,i,w}\tau_w,
\quad i\in I,\label{17}\\
&&l_{1,j,i}=l_{0,j,i}+\sum_{1\le w\le n_1}l_{1,j,i,w}\tau_w,
\quad s+1\le j\le\sigma,\,0\le i\le n,\label{18}\\
&&l^{(1)}_{j,i}=l^{(0)}_{j,i}+\sum_{1\le w\le n_1}l^{(1)}_{j,i,w}\tau_w,
\quad j\in\{0,\sigma+1,\ldots , n\},\,0\le i\le r, \label{19}
\end{eqnarray}
where $\nu_1\ge 0$ is bounded from above by a polynomial in $(d')^n$ and $r$;
$\tau_1,\ldots , \tau_{n_1}$
is a family of algebraically independent elements over $k$;
and all $c_{1,i,w}\in k$, $l_{1,j,i,w}\in k$,
$l^{(1)}_{j,i,w}\in\overline{k}$. The point
$(c_1,\lambda_1)$ is given in the similar
way, one should only
delete (19). Set
$$
e_{j,n+1}=\sum_{i=(i_0,\ldots , i_n)\in I}c_{1,j,i}X_0^{i_0}\cdots X_n^{i_n},
\quad j=0,1.
\eqno (20)
$$
We do not assume that $n_1$ is minimal possible, but when it is necessary
one can always construct
solving a linear system new representations
(17), (18), (19) giving a
generic point of the same
linear space and such that $n_1$ is minimal possible.
In this case $n_1\le\nu$ (respectively $n_1\le\nu_1$).

If $(0,\ldots , 0)\in
{\cal L}_{c_1,\lambda_1,L^{(1)}}$ (respectively
${\cal L}_{c_1,\lambda_1}$) we shall omit
sometimes $c_{0,i}$, $l'_{j,i}$, $l^{(0)}_{j,i}$
from (17)--(19)
but remember that they exist.

Let us extend the field $k$ till $k(\tau_1,\ldots , \tau_{n_1})$.
We shall say that $(c_1,\lambda_1,L^{(1)})$ (respectively $(c_1,\lambda_1)$)
satisfies (*) (or this property holds for $(c_1,\lambda_1,L^{(1)})$) if and
only if
(*) holds for $(c_1,\lambda_1,L^{(1)})$ (respectively $(c_1,\lambda_1)$) and
the field
$k(\tau_1,\ldots , \tau_{n_1})$ in place of $(c',\lambda,L')$
(respectively $(c',\lambda)$) and $k$.
{\it Consider the following property of $(c_0,\lambda_0,L^{(0)})$.
\begin{enumerate} \renewcommand{\labelenumi}{\arabic{enumi})}
\setcounter{enumi}{4}
\item $\deg W'(c_0,\lambda_0,L^{(0)})=\deg W'(c_1,\lambda_1,L^{(1)})$.
\end{enumerate}
We shall say that $(c_4,\lambda_4,L^{(4)})$ (which is similar to
$(c_0,\lambda_0,L^{(0)})$) satisfies 5) for $(c_1,\lambda_1,L^{(1)})$
if and only if this property holds with
$(c_4,\lambda_4,L^{(4)})$ in place of $(c_0,\lambda_0,L^{(0)})$.}
We shall omit ``for $(c_1,\lambda_1,L^{(1)})$'' when
this will not lead to an ambiguity.

Suppose that $(c_0,\lambda_0,L^{(0)})$ satisfies conditions 1) and 2)
from (*).
Let $\overline{\lambda_0}=(L_{0,0},\ldots , L_{0,n+1})$,
see condition (A) from the
Introduction of \cite{13}.
Set $\overline{\lambda_1}=(L_{0,0},\ldots , L_{0,s},L_{1,s+1},$ $\ldots ,$
$L_{1,\sigma},L_{0,\sigma+1},\ldots , L_{0,n+1})$.
Then condition (A) \cite{13} holds for
$\lambda_1$, see \cite{13} Section~1 for details.
Suppose that $(c_0,\lambda_0,L^{(0)})$ satisfies
(*) and 5).
Then by Lemma~9 the element $L^{(1)}\in{\cal U}'''_0$ and (*) holds
for $(c_1,\lambda_1,L^{(1)})$.
We shall always choose the
polynomial $\Phi_{c_1,\lambda_1,L^{(1)}}$ to be a separable polynomial
from the ring
$\overline{k}[\tau_1,\ldots ,\tau_{n_1},Z_0,Z_{\sigma+1},\ldots , Z_{n+1}]$
of the minimal degree with respect to all the variables.
If all $l^{(1)}_{j,i,w}\in k$ we choose $\Phi_{c_1,\lambda_1,L^{(1)}}$ to be
a polynomial with coefficients from $k$.
The polynomials
$R_{c_1,\lambda_1,L^{(1)}},R_{c_1,\lambda_1,L^{(1)},0}\in
\overline{k}[\tau_1,\ldots
,\tau_{n_1},Z_0,
Z_{\sigma+1},\ldots , Z_n]$;
$R_{c_1,\lambda_1,L^{(1)},*},$ $R_{c_1,\lambda_1,L^{(1)},0,*},$
$R_{c_1,\lambda_1,L^{(1)},*,0}\in \overline{k}[\tau_1,$
$\ldots ,\tau_{n_1},Z_0,Z_n]$.

One can consider particular cases when some of $\tau_j$ coincide with
elements of the families
$c$, $\upsilon$, $u$. More precisely,
it is convenient to suppose in the case, when one
(or more)  element from $c_1,\lambda_1,L^{(1)}$ coincides with $c,
\upsilon,u$ respectively, that
in (17)--(19)
there are $\tau_1,\ldots ,\tau_{n_1}$ together with
algebraically independent of them elements
from families $c,\upsilon,u$ in place of
$\tau_1,\ldots ,\tau_{n_1}$ from the initial case.
Further, if we consider, e.g., the points of the
type $(c_1,v,L^{(1)})$ (respectively $(c_1,v)$)
we suppose that all the coefficients of $v_{j,i}$ in
(17), (19) are zeroes. Similarly for the point $(c_1,v,U)$
(respectively $(c_1,\lambda_1,U)$)
all the coefficients of $v_{j,i}$ and $u_{j_1,i_1}$
(respectively of $u_{j_1,i_1}$)
in (17) (respectively (17), (18)) are zeroes.

For example, we
may consider one of the eight cases $(c',\lambda,L')$,
$(c',v,L')$, $(c,\lambda,U)$, $(c',v,U)$, $(c,\lambda,L')$,
$(c,v,L')$, $(c,\lambda,U)$, $(c,v,U)$
as $(c_1,\lambda_1,L^{(1)})$.
Set $\Phi_{c,v,U}=F$, $R_{c,v,U}=\Delta$, $b(c,v,U)=b$;
$a(c,v,U)=a$,
$R_{c,v,U,0}=\Delta_0$. In the other cases for convenience
sometimes we omit generic elements $c$, $v$, $U$ from the
notations. For example, $R_{c_1,\lambda_1,U}=R_{c_1,\lambda_1}$,
$\Phi_{c_1,v,L'}=\Phi_{c_1,L'}$,
$\Phi_{c,v,L'}=\Phi_{L'}$, $R_{c,\lambda,U}=R_\lambda$,
$b(c,\lambda_1,U)=b(\lambda_1)$, $W'(c,\lambda_1)=W'(\lambda_1)$ and so on
(in the case of $(c,v,U)$ this operation may lead to an ambiguity,
so we have introduced the new notations).


Obviously for $(c,v,U)$ property (*) holds.
For $(c,\lambda,L')$ one can omit 3)
from (*) (and obtain the equivalent property).
{\it Consider the following property of $(c_1,\lambda_1,L^{(1)})$.}
\begin{itemize}
\item[(**)]
{\it All $l^{(0)}_{j,i}$ and $l^{(1)}_{j,i,w}$ from (19)
are elements from $k$. Further, a point
$(c_3,\lambda_3,$ $L^{(3)})$ $\in{\cal L}_{c_1,\lambda_1,L^{(1)}}$ is
given such that $L^{(3)}$ and $\lambda_3$ satisfy 1) and 2) from (*)
(in place of $L'$ and $\lambda$), and
all the coefficients $l^{(3)}_{j,i}$ corresponding to $L^{(3)}$
are elements from $k$.}
\end{itemize}
In Section~3 we shall construct
{\it an algorithm to decide whether (*) holds for
$(c_1,\lambda_1,L^{(1)})$ satisfying (**).
If (*) is fulfilled for $(c_1,\lambda_1,L^{(1)})$
we have at the output of this decision algorithm a point
$(c_4,\lambda_4,L^{(4)})\in{\cal L}_{c_1,\lambda_1,L^{(1)}}$
such that (*) holds for $(c_4,\lambda_4,L^{(4)})$, the degree
$\deg W'(c_4,\lambda_4,L^{(4)})=\deg W'(c_1,\lambda_1,L^{(1)})$, and
all the coefficients $c_{4,i}$, $l_{4,j,i}$,
$l^{(4)}_{j,i}$
corresponding to $(c_4,\lambda_4,L^{(4)})$
are obtained from (17)--(19) by
the substitution $\tau'_i$ for
$\tau_i$, $1\le i\le n_1$, where all $\tau'_i$ are integers of length
$O(n\log d+(n-\sigma)\log d')$. The working time of this
algorithm is the same as for the one from assertion (e) of Theorem~2.}


Suppose that we have constructed by this decision algorithm a point
$(c_4,\lambda_4,$ $L^{(4)})$.
Then replacing in (17)--(19) $(c_0,\lambda_0,L^{(0)})$ by
$(c_4,\lambda_4,L^{(4)})$ we can suppose without loss of generality that
$(c_0,\lambda_0,L^{(0)})$ satisfies (*) and 5).
{In what follows, unless it is not state otherwise,
we shall suppose that if (*) holds for $(c_1,\lambda_1,L^{(1)})$ then
(*) and 5) holds for $(c_0,\lambda_0,L^{(0)})$.}



Actually we need this decision algorithm only for partial cases such as,
e.g., $(c_1,\lambda_1,L^{(1)})=
(c,v,L')$. In the last case one can always construct the required
$(c_3,\lambda_3,L^{(3)})$
using \cite{13}.

The partial case of this decision algorithm is
$(c_1,\lambda_1,L^{(1)})=(c',\lambda,U)$.
In this case one can also construct
$(c_3,\lambda_3,L^{(3)})=(c',\lambda,L^{(3)})$
using \cite{13}. {\it Therefore, one can decide whether
$(c',\lambda)$ satisfies (*).}

Note also that extending the field $k$ till $k^{(1)}$ (where the degree
$[k^{(1)}:k]$ is finite)
one can generalize the property (**) and this decision
algorithm for the case when $l^{(0)}_{j,i},l^{(1)}_{j,i,w},l^{(3)}_{j,i},
l^{(4)}_{j,i}\in k^{(1)}$.

A decision algorithm without knowing a point $(c_3,\lambda_3,L^{(3)})$
(if it is possible to describe it) requires
probably a revision of the results from \cite{12} and \cite{13}.




By Lemma~10 the polynomial $R_{c_1,\lambda_1,0,*}$ is nonzero separable,
and $a(c_1,\lambda_1)$ $=b(c_1,\lambda_1)$.
In particular, $\Delta_{0,*}$ is nonzero, do not have
multiple factors with respect to any variables, and $a=b$.




The definition of the open set ${\cal U}$
(and more generally, ${\cal U}_\alpha$, $\alpha\in\{0,\sigma+1,\ldots ,
n\}$)
is an existential one. So far there has not been a
simple description of any open subset of the set ${\cal U}$.
Now we are going to define explicitly a nonempty defined over $k$
open in the Zariski topology subset
${\cal U}^{({\rm v})}
\subset{\cal U}$ (and more generally, ${\cal U}^{({\rm v})}_\alpha
\subset{\cal U}_\alpha$,
$\alpha\in\{0,\sigma+1,\ldots ,n\}$) in
the case of zero--characteristic. This subset has a
simple structure. Actually we shall define more other open in
the Zariski topology subsets of ${\cal U}$.




\par\medskip\noindent{\bf DEFINITION~2}\hspace{0.1em} {\it  Let
$(c_1,\lambda_1)$
be as above and satisfy (*).
Then ${\cal U}_{c_1,\lambda_1,0}^{({\rm v})}$
is the set of all $L'\in{\cal U}'''_{\lambda_1,0}(\overline{k})=
{\cal U}'''_{\lambda_1,0}\cap{\Bbb A}^{(r+1)(n-
\sigma+1)}(\overline{k})$
(recall here that ${\cal U}'''_{\lambda_1,0}$ is defined for
the ground field
$k(\tau_1,\ldots ,\tau_{n_1})$, see above and \cite{13}; also $L'\in
{\cal U}'''_{\lambda_1,0}$ implies $L'\in
{\cal U}'''_0$) such that
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item $(c_1,\lambda_1,L')$ satisfies property (*),
\item the degree
$\deg W'(c_1,\lambda_1)=
\deg_{Z_0,Z_{\sigma+1},\ldots , Z_n}\Phi_{c_1,\lambda_1,L'}$,
\item $b(c_1,\lambda_1)=a(c_1,\lambda_1,L')$.
\end{enumerate}
For every $\sigma+1\le\alpha\le n$ denote by ${\cal
U}_{c_1,\lambda_1,\alpha}^{({\rm v})}$
the image of ${\cal
U}_{c_1,\lambda_1,0}^{({\rm v})}$
under the linear projection
$\pi_{0,\alpha}\, :\,{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})
\rightarrow{\Bbb A}^{(r+1)(\alpha-\sigma)}(\overline{k})$,
$L'\mapsto(L'_{\sigma+1},\ldots ,$ $L'_\alpha)$. Set ${\cal
U}_{c_1,\lambda_1}^{({\rm v})}$ to be ${\cal
U}_{c_1,\lambda_1,\sigma+1}^{({\rm v})}$.
}\par\medskip




\par\medskip\noindent{\bf REMARK~6}\hspace{0.1em} {\it  Obviously
$\pi_{0,\alpha}({\cal U}''_0)\subset{\cal
U}''_\alpha$,
$\pi_{0,\alpha}({\cal U}'''_0)\subset{\cal U}'''_\alpha$,
$\pi_{0,\alpha}({\cal U}_0)\subset{\cal
U}_\alpha$ for all
$\alpha$ but
we do not define ${\cal U}''_\alpha$, ${\cal U}'''_\alpha$ and
${\cal
U}_\alpha$ just
as the projections of
${\cal U}''_0$, ${\cal U}'''_0$ and ${\cal U}_0$; our
definitions are more fine.
In the case of ${\cal
U}_{c_1,\lambda_1,\alpha}^{({\rm v})}$ one can also give the similar more
fine
definitions separately for every $\alpha$
(it requires more other definitions above).
In this way one can obtain the objects
which might be different from $\pi_{0,\alpha}({\cal
U}_{c_1,\lambda_1,0}^{({\rm v})})$.
But here to simplify the presentation we bound
ourselves by the definitions given above.
}\par\medskip



Definition~2 is especially simple for
the particular case $(c_1,\lambda_1)=(c,v)$.
Then one can replace
${\cal U}'''_{\lambda_1,0}$ by ${\cal U}'''_0$ and omit (i).



By Lemma~11 the set
${\cal U}_{c_1,\lambda_1,0}^{({\rm v})}$
is nonempty open with respect to the Zariski topology
in ${\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$.
Further, see the proof of the same lemma,
$L'\in {\cal U}_{c_1,\lambda_1,0}^{({\rm v})}$
if and only if
$L'\in{\cal U}'''_{\lambda_1,0}(\overline{k})$,
the polynomials $\Phi_{c_1,\lambda_1,L'}$ and
$\Phi_{c_1,\lambda_1}|_{u=l'}$
coincide up to a nonzero factor from $\overline{k}[\tau_1,\ldots
,\tau_{n_1}]$, the polynomial $\Phi_{c_1,\lambda_1,*}|_{u=l'}$
is nonzero and does not have irreducible factors
from $\overline{k}[\tau_1,\ldots ,\tau_{n_1},\,Z_0,Z_n]$,
and $R_{c_1,\lambda_1,L',*,0}$ coincides with
$R_{c_1,\lambda_1,0,*}|_{u=l'}$
up to a nonzero factor from $\overline{k}[\tau_1,\ldots ,$ $\tau_{n_1}]$.


\par\medskip\noindent{\bf REMARK~7}\hspace{0.1em} {\it  Suppose that (i) is
fulfilled, and all
linear forms $L'_j\in k[X_0,\ldots ,$ $X_r]$,
$j\in\{0,\sigma+1,\ldots , n\}$. Then one
can decide whether (ii) (respectively (iii))
holds by Theorem~2, see below.
}\par\medskip



For convenience sometimes
we omit generic elements $c$, $v$ from the notations, cf. above.
For example ${\cal U}^{({\rm v})}_{c,\lambda,0}={\cal
U}^{({\rm v})}_{\lambda,0}$, ${\cal U}^{({\rm v})}={\cal
U}^{({\rm v})}_{c,v}$, ${\cal U}^{({\rm v})}_\alpha={\cal
U}^{({\rm v})}_{c,v,\alpha}$ for $\alpha\in\{0,\sigma+1,\ldots , n\}$.
We get four nonempty
sets ${\cal U}_{c',\lambda,0}^{({\rm v})}$,
${\cal U}_{c',0}^{({\rm v})}$, ${\cal U}_{\lambda,0}^{({\rm v})}$,
${\cal U}_0^{({\rm v})}$.


Let $(c_1,\lambda_1,L^{(1)})$, $n_1$
be as above, and $(c_2,\lambda_2,L^{(2)})$,
$n_2$ be similar to
$(c_1,\lambda_1,$ $L^{(1)})$, $n_1$.
{\it Set $(c_2,\lambda_2,L^{(2)})\le (c_1,\lambda_1,L^{(1)})$ if and only if
${\cal L}_{c_2,\lambda_2,L^{(2)}}\subset{\cal
L}_{c_1,\lambda_1,L^{(1)}}$.}
For example $(c_1,\lambda_1,$ $L^{(1)})\le(c,v,U)$ for any
$(c_1,\lambda_1,L^{(1)})$.
If $(c_2,\lambda_2,L^{(2)})\le (c_1,\lambda_1,L^{(1)})$
 we shall suppose in what follows
without loss of generality for convenience of notation that the family
$\tau_1,\ldots ,\tau_{n_2}$ defined
for $(c_2,\lambda_2,L^{(2)})$ is a
part of the family $\tau_1,\ldots ,\tau_{n_1}$ defined
for $(c_1,\lambda_1,L^{(1)})$, and
(17)--(19) for $(c_2,\lambda_2,L^{(2)})$ are obtained from
(17)--(19) for $(c_1,\lambda_1,L^{(1)})$
 by the substitution of $\tau'_{n_2+1},\ldots ,$
$\tau'_{n_1}\in k$ for $\tau_{n_2+1},\ldots ,
\tau_{n_1}$. So we do not suppose that $(c_0,\lambda_0,L^{(0)})$ is the same
for $(c_1,\lambda_1,L^{(1)})$ and $(c_2,\lambda_2,L^{(2)})$.
In the similar way we define the relation
$(c_1,\lambda_1)\le(c_2,\lambda_2)$ (with the analogous assumptions on
the families $\tau_1,\ldots,\tau_{n_i}$, $i=1,2$);
one need only to omit above everywhere $L^{(1)}$, $L^{(2)}$ and (19).








\par\medskip\noindent{\bf THEOREM~1}\hspace{0.1em} {\it  Let $\lambda$, $L'$
and a
homogeneous polynomial $e'_{n+1}$ of degree $d'$
with the corresponding family of coefficients $c'$
be as above, and $(c',\lambda)$ satisfy (*). Let
$(c_2,\lambda_2,L^{(2)})\le
(c_1,\lambda_1,L^{(1)})$ (respectively $(c_2,\lambda_2)\le
(c_1,\lambda_1)$) be as above, and the both
satisfy (*). Then the following assertions hold.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item ${\cal U}_{c',\lambda,0}^{({\rm v})}$,
${\cal U}_{c',0}^{({\rm v})}$,
${\cal U}_{\lambda,0}^{({\rm v})}$, and
${\cal U}_0^{({\rm v})}$
are nonempty open in the Zariski
topology subsets of ${\cal U}_0$. More generally, the sets
${\cal U}^{({\rm v})}_{c_1,\lambda_1,\alpha}$
defined above are nonempty open
in the Zariski topology subsets of ${\cal U}_\alpha$
for all $\alpha\in\{0,\sigma+1,\ldots , n\}$. In particular,
${\cal U}^{({\rm v})}\subset{\cal U}$.
\item For every $(c_1,\lambda_1)$ for an arbitrary
$L'\in{\cal U}^{({\rm v})}_{c_1,\lambda_1,0}$
there exists $(c',\lambda)$ such that $(c',\lambda)$ satisfies (*),
$(c',\lambda)\le (c_1,\lambda_1)$,
$\deg W'(c',\lambda,L')=\deg W'(c',\lambda)=\deg W'(c_1,\lambda_1)$,
$a(c',\lambda,L')=b(c',\lambda)=b(c_1,\lambda_1)$, and, therefore,
$L'\in{\cal U}^{({\rm v})}_{c',\lambda,0}\subset
{\cal U}^{({\rm v})}_{c_1,\lambda_1,0}$.
\item Besides that, in assertion (b) the pair  $(c',\lambda)$ always can be
obtained from
$(c_1,\lambda_1)$ by the substitution in (17), (18) values
$\tau'_i$ for
$\tau_i$, $1\le i\le n_1$, where all $\tau'_i$ are integers of length
$O(n\log d+(n-\sigma)\log d')$. In particular,
if $c_1=c$
(respectively $\lambda_1=\upsilon$) then $c'$ is
a family of integers
of lengths $O(n\,\log d+(n-\sigma)\log d')$ (respectively all linear forms
$L_j$, $s\le j\le \sigma$,
have integer coefficients of length $O(n\,\log d+(n-\sigma)\log d')$).
\item In the notation from (c) let
${\mathfrak l}\subset{\Bbb A}^{\nu_1}(\overline{k})$,
$\nu_1={n+d'\choose n}+(\sigma-s)(n+1)$ see above, be a line
defined over $k$ and such
that $(c',\lambda)\in{\mathfrak l}\subset{\cal L}_{c_1,\lambda_1}$,
where ${\cal
L}_{c_1,\lambda_1}$ is a
affine linear subspace corresponding to $(c_1,\lambda_1)$, see above. Then
for all
$(c_3,\lambda_3)\in{\mathfrak l}(k)$, apart of at most a
polynomial in $d^n(d')^{n-\sigma}$ number, assertion (b) holds for
$(c_3,\lambda_3)$ in place of $(c',\lambda)$.
\item The integers $a(c_2,\lambda_2,L^{(2)})\le
a(c_1,\lambda_1,L^{(1)})$, $b(c_2,\lambda_2,L^{(2)})\le
b(c_1,\lambda_1,L^{(1)})$, $\deg_{Z_0,Z_{\sigma+1},\ldots , Z_{n+1}}
\Phi_{c_2,\lambda_2,L^{(2)}}\le
\deg_{Z_0,Z_{\sigma+1},\ldots , Z_{n+1}}\Phi_{c_1,\lambda_1,L^{(1)}}$.
\item
The integers $b(c_2,\lambda_2)\le
b(c_1,\lambda_1)$, the degrees
$\deg W'(c_2,\lambda_2)\le\deg W'(c_1,$ $\lambda_1)$.
If $\deg W'(c_2,\lambda_2)=\deg W'(c_1,$ $\lambda_1)$ and
$b(c_2,\lambda_2)=b(c_1,\lambda_1)$ then
${\cal U}^{({\rm v})}_{c_2,\lambda_2,0}\subset
{\cal U}^{({\rm v})}_{c_1,\lambda_1,0}$.
\item For any $(c_1,\lambda_1)$  there is $L'=(L'_0,
L'_{\sigma+1},\ldots , L'_n)
\in{\cal U}^{({\rm v})}_{c_1,\lambda_1,0}$ such that all
linear forms $L'_j$,  $j\in\{0,\sigma+1,\ldots ,n\}$,
have integer coefficients of length $O(n\log d+(n-\sigma)\log d')$.
\end{enumerate}
}\par\medskip


\medskip
Theorem~2 uses in its proof Theorem~3, Section~4 which
generalizes Theorem~3 from \cite{11} for the case of rational morphisms.
From Theorem~3 and its Corollary~1
we get immediately the following result.

\medskip\noindent
{\bf CRITERION FOR IRREDUCIBILITY}\quad
{\em Let
$\Phi=\sum_{0\le i\le D}\Phi_iX_0^i\in k[X_0,$ $\ldots , X_n]$, $n\ge1$,
be a homogeneous polynomial
over a field of zero--characteristic, and all
$\Phi_i\in k[X_1,$ $\ldots ,$ $X_n]$.
Let $\deg_{X_0}\Phi=D\ge 1$.
Suppose that the discriminant $R$ of $\Phi$
with respect to $X_0$ is nonzero. Let $R_0$ be
square free part of
$R$ (i.e., $R_0$ is separable and ${\cal Z}(R_0)={\cal Z}(R)$).
Let $L$ be a nonzero linear form in $X_1,\ldots, X_n$ with
coefficients from $k$, and
$A=k[X_0,\ldots , X_n]/(L)$
(hence $A$ is isomorphic to the ring of polynomials in
$n$ variables and $\deg_{X_0}a$ is defined for every $a\in A$).
\begin{enumerate} \renewcommand{\labelenumi}{\arabic{enumi})}
\item
Suppose that the polynomial $R_0\bmod L\in A$ is separable.
Then the irreducible over $k$ factors of degree with respect to $X_0$ at
least $1$
of the polynomials $\Phi$ and $\Phi\bmod L\in A$ are in
the one--to--one correspondence
by the rule $\Psi\mapsto\Psi\bmod L$.
\item Suppose additionally that
the greatest common divisor $\mbox{\rm
G\,C\,D\,}\{\Phi_i\bmod L\,
:\, 0\le i\le D\}=1$ in $A$.
Then every irreducible factor of $\Phi$ (respectively $\Phi\bmod L$) is of
degree with respect to $X_0$ at least $1$.
\end{enumerate}
Besides that, for every linear subspace ${\cal L}$
of linear forms in $X_1,\ldots ,X_n$ with $\dim{\cal L}\ge 3$
there is $L\in{\cal L}$
satisfying all the formulated conditions from 1)
(respectively 1) and 2) whenever
$\mbox{\rm G\,C\,D\,}\{\Phi_i\,
:\, 0\le i\le D\}=1$ in $k[X_0,\ldots , X_n]$).
In particular, under conditions of 1) and 2) if $\Phi$ is
irreducible over $k$ then
$\Phi\bmod L$ is irreducible over $k$.}










\medskip
Now we proceed to the precise statements required for Theorem~2.
Let  homogeneous polynomials
$f_1,\,\ldots\,$, $f_m\in k[X_0,\,\ldots\, ,X_n]$ be given, $m\ge
1$.
Consider the
 closed algebraic set or algebraic variety (in this paper we
 do not distinguish these two concepts)
$$
 V=\{(x_0: \ldots: x_n): \: f_i(x_0, \ldots, x_n)=0, \; x_i\in
\overline{k},\;\forall 1 \le i \le m\} \subset {\Bbb{P}}
^n(\overline{k})
 \: .
$$
This is a set of all common zeros of polynomials $f_1, \ldots, f_m$ in
${\Bbb{P}}  ^n(\overline{k})$, where $\overline{k}$ is an algebraic
closure of $k$.  Hence $V={\cal Z}(f_1,\ldots ,f_m)$.
Let $e_0,\ldots ,e_r\in k[X_0,\ldots , X_n]$
be homogeneous polynomials such that the degree
$\deg_{X_0,\ldots , X_n}e_j=d'$ if $e_j\ne 0$ for every
$0\le j\le r$.

Now let  the field $k={\Bbb Q}(t_1,\,\ldots\, ,t_l,\theta )$
where $t_1,\,\ldots\, ,t_l$ are
 algebraically independent over the field ${\Bbb Q}$ and $\theta$ is
 algebraic over ${\Bbb Q}  (t_1,\,\ldots\, ,t_l)$ with the minimal
 polynomial $\psi_1\in{\Bbb Q}  [t_1,\,\ldots\, ,t_l,Z]$ and leading
 coefficient $ {\mathrm  lc}_Z\psi_1$ of $\psi_1$ is equal to 1.
 We shall represent each polynomial $f=f_i$ (and similarly $e_j$) in the form
$$
 f=\frac{1}{a_0} \sum_{i_0, \ldots, i_n} \sum_{0 \leq j <
 \deg_Z \psi_1} a_{i_0, \ldots, i_n,j} \theta^j X_0^{i_0} \cdots X_n^{i_n}
 \, ,
$$
 where $a_0,a_{i_0, \ldots, i_n,j} \in {\Bbb Z}  [t_1, \ldots, t_l]$,
 $\mbox{\rm G\,C\,D\,}_{i_0, \ldots, i_n,j}
 (a_0,a_{i_0, \ldots, i_n,j})=1$.
 Define the length ${\rm l}(a)$ of an integer $a$ by the formula
 ${\rm l}(a)=\min\{s \in {\Bbb Z}  : \: |a|<2^{s-1}\}$.
 The length of coefficients ${\rm l}(f)$ of the polynomial $f$ is defined to
 be the maximum of lengths of coefficients from ${\Bbb Z}$ of polynomials
 $a_0,a_{i_0, \ldots, i_n,j}$ and the degree
$$
 \deg_{t_\gamma} (f)=\max_{i_0, \ldots, i_n,j} \{\deg_{t_\gamma} (a_0),
 \deg_{t_\gamma} (a_{i_0, \ldots, i_n,j})\} \, ,
$$
 where $1 \leq \gamma \leq l$.
 In the similar way we shall define degrees and lengths
of integer coefficients of other polynomials,
in particular $\deg_{t_\gamma} F$ and ${\rm l}(F)$ are defined.

 \noindent
 We shall suppose that we have the following bounds
 \begin{eqnarray*}
&& \deg_{X_0, \ldots, X_n} (f_i)<d, \; \deg_{t_\gamma}(f_i)<d_2, \;
 {\rm l}(f_i)<M, \\
&&\deg_{X_0, \ldots, X_n} (e_j)=d'\;\mbox{or}\; e_j=0, \;
\deg_{t_\gamma}(e_j)<d'_2, \;
 {\rm l}(e_j)<M', \\
&& \deg_Z (\psi_1)<d_1, \; \deg_{t_\gamma} (\psi_1)<d_1, \;
{\rm l}(\psi_1)<M_1 \: .
 \end{eqnarray*}
for all $1\le i\le m$, $0\le j\le r$, $1\le\gamma\le l$.
 The size ${\rm L}(f)$ of the polynomial $f$ is defined to be the product of
 ${\rm l}(f)$ to the number of all the coefficients from ${\Bbb Z}$ of
$f$ in
 the dense representation.
 We have
$$
 {\rm L}(f_i)<({d+n \choose n}d_1+1)d_2^l M
$$
 Similarly ${\rm L}(\psi_1)<d_1^{l+1} M_1$.

\par\medskip\noindent{\bf REMARK~8}\hspace{0.1em} {\it  Unless
we state otherwise, in what follows we suppose $l$ to be fixed.
The working time of the algorithm from Theorem~2, see below, is
essentially the same as for solving systems of polynomial equations with a
finite set of solutions in the projective space.
 So this theorem can be formulated also in the case when $l$ is not fixed,
see \cite{5}. Notice that the constants $O(\ldots)$,
see Theorem~2 and lemmas below,
in the estimate of the lengths of integer coefficients
of linear forms $L'_j$, $L_j$ are absolute; they does not depend on $l$.
}\par\medskip


We shall represent a point $z\in V$ with coordinates from a
finite extension of $k$ as follows. An index $0\le i_0\le n$ is
known such that $X_{i_0}(z)\ne 0$  and an isomorphism of fields
$$
k(z)=k((X_1/X_{i_0})(z),\ldots , (X_n/X_{i_0})(z))=
k[\eta]\simeq k[Z]/(\psi_2)
\eqno (21)
$$
is given where $\eta=\sum_{0\le i\le n}\nu_i(X_i/X_{i_0})(z)$,  the
coefficients $\nu_i\in {\Bbb Z}$ are given and $\psi_2\in k[Z]$ is
minimal polynomial of $\eta$ over $k$ with leading coefficient
$\mbox{\rm lc}_Z\psi_2=1$ (so the point $z$ is defined up
to a conjugation over $k$).

Let $g\in k[\eta]$ be an arbitrary element. Then $g=G(\eta)$ for the
uniquely defined polynomial
$A\in k[Z]$ such that $\deg_ZG<\deg_Z\Phi$. The length of integer
coefficients ${\rm l}(g)$,
the size ${\rm L}(g)$ and the degrees $\deg_{t_\alpha}g$,
$1\le \alpha\le l$, of $g$
are defined by the formulas
$$
{\rm l}(g)={\rm l}(G),\quad {\rm L}(g)={\rm L}(G), \quad
\deg_{t_\alpha}g=\deg_{t_\alpha}G.
$$
Let us define the size of the point $z$ to be
${\rm L}(\psi_2)+\sum_{0\le i\le n}{\rm L}(X_i/X_{i_0})$.



\par\medskip\noindent{\bf THEOREM~2}\hspace{0.1em} {\it
Let the field $k$ of zero--characteristic, the algebraic variety $V$,
the polynomials $F$, $f_1,\ldots ,f_m$ and $e_0,\ldots ,e_r$, $r\ge 2$,
be given as above. Let a point $x^{(0)}\in V$ be given and $x^{(0)}$
belongs to only one irreducible over $k$ component of $V$.
Denote this component by $W$.
Let $W'$ be as above, see (1).
Then one can compute the dimension $\dim W'$. Suppose that
$\dim W'=n-\sigma>1$ (so $s\le\sigma\le n-2$).
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item Let $(c_1,\lambda_1)$
be given by (17), (18).
Suppose that $(c_0,\lambda_0)$ from (17), (18)
satisfies (*) and 5) (and hence $(c_1,\lambda_1)$ also satisfies (*)
by Lemma~9).
Then one can compute the integer $b(c_1,\lambda_1)$ and the degree
$\deg W'(c_1,\lambda_1)$. Further, one can
construct $L'=(L'_0,L'_{\sigma+1},$ $\ldots , L'_n)
\in{\cal U}^{({\rm v})}_{c_1,\lambda_1,0}$ satisfying assertion (b) of
Theorem~1,
i.e., all $L'_j$ are linear from $k[X_0,\ldots ,X_r]$
with integer coefficients of length $O(n\log d+(n-\sigma)\log d')$.
\item In particular, one can construct an element $L'$ from an open in the
Zariski topology
subset ${\cal U}^{({\rm v})}_0$ of ${\cal U}_0$ such that all $L'_j$
have integer coefficients satisfying the estimate from (a). Hence one can
compute a linear form
$L'_{\sigma+1}\in{\cal U}^{({\rm v})}$
with integer coefficients satisfying the estimate from (a).
\item Under conditions from (a)
let $L'=(L'_0,L'_{\sigma+1},\ldots , L'_n)$ be given,
and all $L'_j\in k[X_0,\ldots ,X_r]$, $j\in\{0,\sigma+1,\ldots ,n\}$.
Let $(c_1,\lambda_1,L')$ satisfy (**), see above.
Then one can decide whether
$L'\in{\cal U}^{({\rm v})}_{c_1,\lambda_1,0}$.
If $L'\in{\cal U}^{({\rm v})}_{c_1,\lambda_1,0}$
one can construct $(c',\lambda)$
such that assertions from (b) and (c) of
Theorem~1 hold.
\item Let $L'=(L'_0,L'_{\sigma+1},\ldots , L'_n)$ be given,
and all $L'_j\in k[X_0,\ldots ,X_r]$, $j\in\{0,\sigma+1,\ldots ,n\}$.
Then, in particular, one can decide
whether $L'\in{\cal U}^{({\rm v})}_0$.
If $L'\in{\cal U}^{({\rm v})}_0$
one can construct $(c',\lambda,L')$
such that $\deg W'(c',\lambda,$ $L')=\deg W'(c,v)$,
$a(c',\lambda,L')=b$, assertion (c) of Theorem~1 holds for
$(c',\lambda,L')$ and
$(c_1,\lambda_1)=(c,v)$.
\item Let $(c_1,\lambda_1,L^{(1)})$
be given by (17)--(19) and satisfy (**).
Then one can  decide whether $(c_1,\lambda_1,L^{(1)})$ satisfies (*).
If (*) holds for $(c_1,\lambda_1,L^{(1)})$ then one can
compute $\deg W'(c_1,\lambda_1,L^{(1)})$, $\deg
W'(c_1,\lambda_1)$,
$a(c_1,\lambda_1,L^{(1)})$, $b(c_1,\lambda_1)$, and construct
$(c_4,\lambda_4,L^{(4)})\in
{\cal L}_{c_1,\lambda_1,L^{(1)}}$ satisfying (*), condition 5) and
such that $a(c_4,\lambda_4,L^{(4)})=a(c_1,\lambda_1,L^{(1)})$.
Finally, $(c_4,\lambda_4,L^{(4)})$ is obtained
from $(c_1,\lambda_1,L^{(1)})$
by the substitution in (17)--(19) values $\tau'_i$ for $\tau_i$,
$1\le i\le n_1$, where all $\tau'_i$ are integers of
lengths $O(n\log d+(n-\sigma)\log d')$.
\end{enumerate}
The working time of the algorithm from (b) is polynomial in
$d^n$, $(d')^n$, $d_1$, $d_2$, $d'_2$,
$M$, $M'$, $M_1$, $m$, $r$,
the size ${\rm L}(x^{(0)})$ of the point $x^{(0)}$.
The working time of the algorithm from (a),
(respectively (c), (d), (e))
is polynomial in
the same values and the size of input, i.e., for (a) it is additionally
${\rm L}(c_1)$, ${\rm L}(\lambda_1)$; for (c) it is
${\rm L}(c_1)$, ${\rm L}(\lambda_1)$, ${\rm L}(L')$; for
(d) it is  ${\rm L}(L')$; and for (e) it is ${\rm L}(c_1)$, ${\rm
L}(\lambda_1)$,
${\rm L}(L^{(1)})$.
}\par\medskip


\par\medskip\noindent{\bf REMARK~9}\hspace{0.1em} {\it   One can
compute also $b(c_1,\lambda_1,L^{(1)})$ within the
time for the algorithm from
assertion (e). But we do not need it in the present
paper.
}\par\medskip



\par\medskip\noindent{\bf REMARK~10}\hspace{0.1em} {\it
In what follows in many cases, when one says that $a_1,\ldots , a_\mu$
(here may be any other family of variables or coefficients of
some polynomials, may be only one variable)
are infinitesimals, par abus de langage it is convenient
to understand that every $a_1,\ldots , a_\mu$ is an infinitesimals or zero.
Usually we state explicitly that $a_1,\ldots , a_\mu$ can be equal to zero.
But even if we forgot to state this explicitly somewhere,
the reader will see
it easily from the context. This will not lead to an ambiguity.
For example, par abus de langage it is more convenient
to say that all the coefficients of a
linear form $\widetilde{L}_0$ are
infinitesimals, but still we usually say that
all the nonzero coefficients of a linear form $\widetilde{L}_0$ are
infinitesimals in this case.
}\par\medskip



\section{Several lemmas}\label{s1}


\medskip In the next seven lemmas the ground field is $\overline{K}$, and
we consider the points
$(c_j,\lambda_j,L^{(j)})\in{\Bbb A}^\nu(\overline{K})$ for integer $j$
(actually for $j=0,3,4,5$). Below, see Definition~3, we formulate property
(D) for
$(\lambda_j,L^{(j)})$.
For $(c_j,\lambda_j,L^{(j)})$ the family
$c_j=\{c_{j,i}\}_{i\in I}$ similarly to $j=0$, see the Introduction, the
polynomial $e_{j,n+1}$ corresponds to
$c_j$ similarly to (20), $\lambda_j=(L_{j,s+1},\ldots , L_{j,\sigma})$,
$L^{(j)}=(L^{(j)}_{\sigma+1},\ldots ,
L^{(j)}_n)$, all $L_{j,i}\in\overline{K}[X_0,\ldots , X_n]$,
$s+1\le i\le\sigma$,
and $L^{(j)}_w\in\overline{K}[X_0,\ldots ,X_r]$ are linear forms,
$e^{(j)}_w=L^{(j)}_w(e_0,\ldots , e_r)$,
$w\in\{0,\sigma+1,\ldots ,n\}$, see the Introduction. The morphisms
$p_{\lambda_j,L^{(j)}}$, $p_{\lambda_j,L^{(j)},*}$ are defined similarly to
(7), (10).
The varieties $W'(c_j,\lambda_j,L^{(j)})$, $W'(c_j,\lambda_j,L^{(j)},*)$
are defined similarly to (13), (14).
Put $A_{\lambda_j,L^{(j)}}$ to be the set of all the
isolated points from ${\Bbb P}^n(\overline{K})$ of the intersection
$$
(W\setminus{\cal Z}(e_0,\ldots ,
e_r))\cap{\cal Z}(L_{j,s+1},\ldots , L_{j,\sigma})\cap
{\cal Z}(e^{(j)}_{\sigma+1},\ldots , e^{(j)}_n),
\eqno (22)
$$
cf. (15). The polynomial $\Phi_{c_j,\lambda_j,L^{(j)}}$ for
$(c_j,\lambda_j,L^{(j)})$ is also defined similarly to the case $j=0$,
see the Introduction. The integers $b(c_j,\lambda_j,L^{(j)})$,
$a(c_j,\lambda_j,L^{(j)})$ are defined similarly to the case of the
point $(c',\lambda,L')$ from the Introduction.
These definitions are always possible for the introduced objects
corresponding to $(c_j,\lambda_j,L^{(j)})$
under the formulated below assumptions,
see also the remarks about our assumptions and their variants
in the Introduction.

Set $W_{\lambda_j}=W\cap{\cal Z}(L_{j,s+1},\ldots , L_{j,\sigma})$ in
${\Bbb P}^n(\overline{K})$, cf. the Introduction.
For every $\sigma\le j\le n$ put $\widetilde{W}_{\lambda_j,i}(L^{(j)})$ to
be the
closure with respect to the Zariski topology
of the algebraic variety $(W_{\lambda_j}\setminus{\cal Z}(e_0,\ldots ,
e_r))\cap
{\cal Z}(e^{(j)}_{\sigma+1},\ldots , e^{(j)}_i)$ in ${\Bbb
P}^n(\overline{K})$.

\par\medskip\noindent{\bf DEFINITION~3}\hspace{0.1em} {\it  We shall say
that $(\lambda_j,L^{(j)})$ (respectively
$(c_j,\lambda_j,L^{(j)})$) satisfies property (D) if and only if the
following conditions holds.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item The number of elements
$N_{\lambda_j,L^{(j)}}=\#A_{\lambda_j,L^{(j)}}=\deg q_v\deg W'$.
\item $A_{\lambda_j,L^{(j)}}\cap({\cal
Z}(e^{(j)}_0)\cup\overline{(V\setminus W)})=\emptyset$.
\item For every $\sigma\le i\le n$ every irreducible over $\overline{K}$
component of $\widetilde{W}_{\lambda_j,i}(L^{(j)})
\setminus{\cal Z}(e_0,\ldots , e_r)$ contains a point from
$A_{\lambda_j,L^{(j)}}$. Hence for every $\sigma\le i\le n$ the dimension
$\dim\widetilde{W}_{\lambda_j,i}(L^{(j)})=n-i$.
\item For every $\sigma\le i\le n$ the
degree $\deg \widetilde{W}_{\lambda_j,i}(L^{(j)})=D_i$, see the Introduction.
\end{enumerate}
}\par\medskip

(the notation $(c_j,\lambda_j,L^{(j)})$ is not essential here; one can take
in this definition
any other point similar to $(c_j,\lambda_j,L^{(j)})$, say,
$(c_*,\lambda_*,L^*)$ and
any other field similar to $K$; this will be seen from the context).
Now we would like to make some observations
related  to property (D) which are useful in what follows.




By Lemma~17 (ii) \cite{13}
if $\#A_{\lambda_j,L^{(j)}}=\deg p_v\deg W'$ then for every $x\in
A_{\lambda_j,L^{(j)}}$
the intersection of $n-s+1$ algebraic varieties $W$, ${\cal
Z}(L_{j,s+1}),\ldots ,{\cal Z}(L_{j,\sigma})$,
${\cal Z}(e^{(j)}_{\sigma+1}),$ $\ldots ,{\cal Z}(e^{(j)}_n)$ at the
point $x$ is transversal
(in particular $x$ is a smooth point of these $n-s+1$ algebraic varieties).
Therefore, if property (D) holds for $(\lambda_j,L^{(j)})$ then intersection
(22) coincides with $A_{\lambda_j,L^{(j)}}$ (one should take $i=n$
in (iii) to see it).

If (D) holds then the morphism
$p_{\lambda_j,L^{(j)},i}\, :\,
\widetilde{W}_{\lambda_j,i}(L^{(j)})\setminus{\cal
Z}(e^{(j)}_0,e^{(j)}_{\sigma+1}\ldots, e^{(j)}_n)$ $\rightarrow{\Bbb
P}^{n-i}(\overline{K})$ induced by $p_{\lambda_j,L^{(j)}}$ is dominant
(recall that this means by definition that the restriction of
$p_{\lambda_j,L^{(j)},i}$
to every irreducible component of
$\widetilde{W}_{\lambda_j,i}(L^{(j)})\setminus{\cal
Z}(e^{(j)}_0,e^{(j)}_{\sigma+1}\ldots, e^{(j)}_n)$ is dominant)
by Lemma~9 \cite{13}.
In particular,
the morphisms $p_{\lambda_j,L^{(j)}}$ and $p_{\lambda_j,L^{(j)},*}$ are
dominant.

Suppose that $\dim W_{\lambda_j}=n-\sigma$ and the morphism
$p_{\lambda_j,L^{(j)}}$  is
dominant (now we do not assume that (D) holds).
Then by Lemma~17 (ii) \cite{13} the degree $\deg
p_{\lambda_j,L^{(j)}}\le
\deg q_v\deg W'$. Further if for every $\sigma\le i\le n$
the dimension
$\dim\widetilde{W}_{\lambda_j,i}(L^{(j)})=n-i$ and the morphism
$p_{\lambda_j,L^{(j)},i}$ is dominant
then obviously, cf. also the proof of Lemma~11 \cite{12}, the degree
$\deg p_{\lambda_j,L^{(j)},i}\le\deg p_{\lambda_j,L^{(j)}}\le
\deg q_v\deg W'$. Hence in the considered case if (i) holds then
by Lemma~2 \cite{12}
the degree $\deg p_{\lambda_j,L^{(j)},i}=\deg q_v\deg W'$, and (iii) holds.
Consider the following condition.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\setcounter{enumi}{4}
{\it\item For every $\sigma\le i\le n$
the dimension $\dim\widetilde{W}_{\lambda_j,i}(L^{(j)})=n-i$ and the
morphism $\deg p_{\lambda_j,L^{(j)},i}$ is dominant.}
\end{enumerate}
Therefore, the conjugation of (i), (ii), (iv) and (v) is equivalent to
property (D), i.e.,
if one replaces in (D) condition (iii) by (v) then the obtained property is
equivalent to (D).

Let $k'$ be a purely transcendental extension of $k$ of finite transcendency
degree such that all the coefficients of
all linear forms $L_{j,i}$, $L^{(j)}_w$
belong to the algebraic closure of $k'$
(hence $W$ is irreducible over $k'$).
Let ${\cal U}'_{k',0}$ (respectively ${\cal U}''_{k',0}$)
be the set ${\cal U}'_0$ (respectively ${\cal U}''_0$) defined for
the ground field $k'$,
see Remark~2 \cite{13} (and Remark~3).
The sets ${\cal U}'_{k',0}$ and ${\cal U}''_{k',0}$
depend only on $\overline{k'}$, see
Remark~2 \cite{13}.
Let $W_i(L^{(j)})$, see the Introduction, be
defined for the ground field $k'$.
Obviously if
$(c_j,\lambda_j,L^{(j)})$ satisfies conditions (i) and (ii) from (D) then
$L^{(j)}\in{\cal U}'_{k',0}$ (to prove this one can assume without loss
of generality that the
coefficients of the linear forms from $v$ are infinitely close to the
corresponding coefficients of the linear forms from
$\lambda_j$, and apply the implicit function theorem).
Further, if
$(c_j,\lambda_j,L^{(j)})$ satisfies property (D) then
$L^{(j)}\in{\cal U}''_{k',0}$, and by the B\'ezout theorem
$\widetilde{W}_{\lambda_j,i}(L^{(j)})=W_i(L^{(j)})\cap
{\cal Z}(L_{j,s+1},\ldots , L_{j,\sigma})$ for every $\sigma\le i\le n$.
Consider the following condition.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\setcounter{enumi}{5}
{\it\item $L^{(j)}\in{\cal U}''_{k',0}$.}
\end{enumerate}
(the truth of this condition does not depend on the choice of $k'$).
Then, assuming as above that the
coefficients of the linear forms from $v$ are infinitely close to the
corresponding coefficients of the linear forms from
$\lambda_j$ and applying
Lemma~17 \cite{13} and the implicit function theorem, one proves that
the conjugation of (i), (ii) and (vi) is equivalent to property (D).

Now assume that (v) holds for $(\lambda_j,L^{(j)})$ (we do not suppose that
(D) is fulfilled). Let $k'$ and $W_i(L^{(j)})$ be as above.
Denote by $E$ the union of all the irreducible over $\overline{K}$
components
$E_1$ of $W\cap{\cal Z}(e^{(j)}_{\sigma+1},\ldots ,
e^{(j)}_i)$ such that $E_1\cap{\cal Z}(L_{j,s+1},\ldots , L_{j,\sigma})$
contains an irreducible component of $\widetilde{W}_{\lambda_j,i}(L^{(j)})$.
Then $E\cap{\cal Z}(L_{j,s+1},\ldots , L_{j,\sigma})\supset
\widetilde{W}_{\lambda_j,i}(L^{(j)})$.
Further, obviously the algebraic variety
$E\subset W_i(L^{(j)})$.
Then $\deg W_i(L^{(j)})\le D_i$
according to Lemma~12 \cite{13} with the ground field $k'$ in place
of $k$
(actually the proof of this inequality does not depend on the ground field;
we leave the details to the reader). Hence by the B\'ezout theorem
$\deg \widetilde{W}_{\lambda_j,i}(L^{(j)})\le D_i$.


If $\lambda_j$ satisfies (C) (in place of $\lambda$) and
$L^{(j)}\in{\cal U}''_{\lambda_j,0}$
(so now all the linear forms $L_{j,w}$ have coefficients from $k$,
and all the linear forms $L^{(j)}_w$ have the coefficients from
$\overline{k}$) then
by Lemma~11 \cite{12} and Lemma~14 \cite{12} the pair  $(\lambda_j,L^{(j)})$
satisfies property (D).

Suppose that the all the coefficients of all
linear forms $L_{j,\nu}$, $L^{(j)}_w$ belong to a subfield
$k''$ of $\overline{K}$ such that $k''\supset k$. Let
$\lambda_*=(L_{*,s+1},\ldots , L_{*,\sigma})$,
$L^*=(L^*_0,L^*_{\sigma+1},\ldots ,
L^*_n)$ where all $L_{*,\nu}\in\overline{K}[X_0,\ldots , X_n]$,
$L^*_w\in\overline{K}[X_0,\ldots ,X_r]$ are linear forms
such that all the nonzero coefficients of linear forms
$L_{j,\nu}-L_{*,\nu}$, $s+1\le\nu\le\sigma$,
$L^{(j)}_w-L^*_w$, $w\in\{0,\sigma+1,\ldots , n\}$, are
infinitesimals with respect to the field $k'$.
In this case by the implicit function theorem, Lemma~13 \cite{13}
and condition (2) (for $L^*$ in place of $L'$) if
$(\lambda_j,L^{(j)})$ satisfies (D) then
$(\lambda_*,L^*)$ also satisfies (i), (ii) and (vi). Therefore, (D)
holds for $(\lambda_*,L^*)$.





\par\medskip\noindent{\bf LEMMA~1}\hspace{0.1em} {\it  Let us extend the
ground field $k$ till $\overline{K}$.
Let $(c_0,\lambda_0,L^{(0)})\in{\Bbb A}^\nu(\overline{K})$
(so all the elements $c_{0,i}$,
the coefficients of all linear
forms $L_{0,j}$, $L^{(0)}_w$ are elements from
$\overline{K}$). Suppose that $(\lambda_0,L^{(0)})$ satisfies property (D).
Then\\
$(\mbox{\rm lc}_{Z_{n+1}}
\Phi_{c_0,\lambda_0,L^{(0)}})(1,0,\ldots , 0)\ne 0$.
Therefore,
$(\mbox{\rm lc}_{Z_{n+1}}
\Phi_{c_0,\lambda_0,L^{(0)}})_*\ne 0$, i.e., the degree
$\deg_{Z_{n+1}}\Phi_{c_0,\lambda_0,L^{(0)},*}=
\deg_{Z_{n+1}}\Phi_{c_0,\lambda_0,L^{(0)}}$.
In particular, these assertions hold for any point
$(c_0,\lambda_0,L^{(0)})\in{\Bbb A}^\nu(\overline{k})$ satisfying
conditions 1), 2) from (*).
}\par\medskip

\noindent {\bf PROOF} \quad By property (D) and our observations
the dimensions
$\dim W_{\lambda_0}=n-\sigma$,
$\dim (W_{\lambda_0}\setminus{\cal
Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e^{(0)}_{\sigma+1},\ldots , e^{(0)}_{n-1})=1$, and
the morphisms $p_{\lambda_0,L^{(0)}}$, $p_{\lambda_0,L^{(0)},*}$
are dominant.
Consider the regular morphisms
\setcounter{equation}{22} \begin{eqnarray}
&&r_{c_0,\lambda_0,L^{(0)}}\, :\,W'(c_0,\lambda_0,L^{(0)})
\setminus\{(0:\ldots : 0:1)\}\rightarrow{\Bbb
P}^{n-\sigma}(\overline{k}),\label{23}\\
&&(X_0:\ldots :
X_{n-\sigma+1})\mapsto(X_0:\ldots : X_{n-\sigma});\nonumber\\
&&q_{c_0,\lambda_0,L^{(0)}}\, :\,
W_\lambda\setminus{\cal Z}(e'_0,e'_{\sigma+1},\ldots ,
e'_n)\rightarrow W'(c_0,\lambda_0,L^{(0)}).\label{24}\\
&&(X_0:\ldots : X_n)\mapsto(e'_0:e'_{\sigma+1}:\ldots : e'_{n+1}).\nonumber
\end{eqnarray}
Then $p_{\lambda_0,L^{(0)}}=r_{c_0,\lambda_0,L^{(0)}}\circ
q_{c_0,\lambda_0,L^{(0)}}$, the morphisms
$r_{c_0,\lambda_0,L^{(0)}}$ and $q_{c_0,\lambda_0,L^{(0)}}$ are dominant.
Recall that
$W'(c_0,\lambda_0,L^{(0)})={\cal Z}(\Phi_{c_0,\lambda_0,L^{(0)}}
(X_0,\ldots , X_{n-\sigma+1}))$. The number of elements
$\#A_{\lambda_0,L^{(0)}}=\deg p'_v\deg W'$ is maximal
possible (when it is finite), see Lemma~17 (i) \cite{13}. Hence
by Lemma~2 \cite{12}  there is an affine neighborhood
$U'$ of the point $z=(1:0:\ldots : 0)\in{\Bbb
P}^{n-\sigma}(\overline{k})$ such that
$p_{\lambda_0,L^{(0)}}^{-1}(U')\rightarrow U'$
is a finite dominant morphism. The composite of two dominant morphisms is
finite if and only if
the both of them are finite.
Hence the morphism $r^{-1}_{c_0,\lambda_0,L^{(0)}}(U')\rightarrow U'$
induced by
$r_{c_0,\lambda_0,L^{(0)}}$ is finite dominant.
The variety $U'$ is normal. It is known that in this case the leading
coefficient $\mbox{\rm
lc}_{Z_{n+1}}\Phi_{c_0,\lambda_0,L^{(0)}}$ does not vanish at any
point of $r^{-1}_{c_0,\lambda_0,L^{(0)}}(U')$.
Hence $(\mbox{\rm lc}_{Z_{n+1}}
\Phi_{c_0,\lambda_0,L^{(0)}})(1,0,\ldots , 0)\ne 0$.
The lemma is proved.



\par\medskip\noindent{\bf LEMMA~2}\hspace{0.1em} {\it  Under conditions of
Lemma~1
suppose additionally that the number of elements
$\#(e_{0,n+1}/e^{(0)}_0)(A_{\lambda_0,L^{(0)}})=\deg p'_v\deg W'$.
Then the polynomial $\Phi_{c_0,\lambda_0,L^{(0)}}(1,$
$0,\ldots , 0,Z_{n+1})$ is separable. Hence
$\Phi_{c_0,\lambda_0,L^{(0)},*}$ is separable
with respect to $Z_{n+1}$.
Therefore, $R_{c_0,\lambda_0,L^{(0)},*}$
and $R_{c_0,\lambda_0,L^{(0)},*,0}$ are nonzero.
In particular, these assertions hold for any point
$(c_0,\lambda_0,L^{(0)})\in{\Bbb A}^\nu(\overline{k})$ satisfying
conditions 1)--3) from (*).
}\par\medskip

\noindent {\bf PROOF} \quad Indeed, the polynomial
$\Phi_{c_0,\lambda_0,L^{(0)}}(1,0,\ldots , 0,Z_{n+1})$ is nonzero
by Lemma~1.
Hence under our conditions the polynomial
$\Phi_{c_0,\lambda_0,L^{(0)}}(1,0,\ldots , 0,$ $Z_{n+1})$ is separable.
The lemma is proved.

\par\medskip\noindent{\bf DEFINITION~4}\hspace{0.1em} {\it  Suppose that the
polynomial $\Phi_{c_0,\lambda_0,L^{(0)}}
\in\overline{K}[Z_0,Z_{\sigma+1},\ldots , Z_n]$ is defined for
$(c_0,\lambda_0,L^{(0)})$, see the Introduction.
Suppose that $\Phi_{c_0,\lambda_0,L^{(0)},*}=
\Phi_{c_0,\lambda_0,L^{(0)}}(Z_0,0,\ldots ,$ $0,Z_n,Z_{n+1})\ne 0$.
Let us represent $\Phi_{c_0,\lambda_0,L^{(0)},*}=g_1g_2$ where
$g_1\in\overline{K}[Z_0,Z_n]$,
$g_2\in\overline{K}[Z_0,Z_n,Z_{n+1}]$ and $\deg g_1$ is maximal possible.
Let us choose and fix $g_2$ (it is uniquely defined up to a nonzero factor
from $\overline{K}$).
Denote $\Phi_{c_0,\lambda_0,L^{(0)},+}=g_2$.
}\par\medskip



\par\medskip\noindent{\bf LEMMA~3}\hspace{0.1em} {\it  Under conditions of
Lemma~2 the algebraic variety
$W'(c',\lambda,L',*)={\cal Z}(\Phi_{c,\lambda,L,+}(X_0,X_1,X_2))$,
and the polynomial $\Phi_{c,\lambda,L,+}$ is separable.
}\par\medskip

\noindent {\bf PROOF} \quad Let
\setcounter{equation}{24} \begin{eqnarray}
&&q_{c_0,\lambda_0,L^{(0)},*}\, :\, W_\lambda\cap{\cal
Z}(e'_{\sigma+1},\ldots ,
e'_{n-1})\setminus
{\cal Z}(e'_0,e'_n)\nonumber\\
&&\rightarrow W'(c_0,\lambda_0,L^{(0)},*),\label{25}\\
&&r_{c_0,\lambda_0,L^{(0)},*}\,
:\,W'(c_0,\lambda_0,L^{(0)},*)\setminus\{(0:0:1)\}
\rightarrow{\Bbb
P}^1(\overline{k}),\label{26}
\end{eqnarray}
be the morphisms induced by $q_{c_0,\lambda_0,L^{(0)}}$,
$r_{c_0,\lambda_0,L^{(0)}}$, see the proof of
Lemma~2. By Lemma~2 the morphism
$p_{\lambda_0,L^{(0)},*}$
is dominant. Hence $q_{c_0,\lambda_0,L^{(0)},*}$,
$r_{c_0,\lambda_0,L^{(0)},*}$ are also dominant. Obviously
$W'(c_0,\lambda_0,L^{(0)},*)\subset
{\cal Z}(\Phi_{c_0,\lambda_0,L^{(0)},*}(X_0,X_1,X_2))$.
Since $\#(e_{0,n+1}/e^{(0)}_0)(A_{\lambda_0,L^{(0)}})=\deg p'_v\deg W'$
the variety $W'(c_0,\lambda_0,L^{(0)},*)=\\ {\cal
Z}(\Phi_{c_0,\lambda_0,L^{(0)},+}(X_0,X_1,X_2))$.
The lemma is proved.



\par\medskip\noindent{\bf LEMMA~4}\hspace{0.1em} {\it  Under conditions of
Lemma~1
let ${\mathfrak l}\subset{\Bbb A}^{\nu}(\overline{K})$ be
a line and the point $y_0=(c_0,\lambda_0,L^{(0)})\in{\mathfrak l}$.
Let $(c_4,\lambda_4,L^{(4)})$ be a
generic point of the line ${\mathfrak l}$.
Then for all $y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
$y$ satisfies property (D),
the number of elements
$\#(e_{3,n+1}/e^{(3)}_0)(A_{\lambda_3,L^{(3)}})=
\#(e_{4,n+1}/e^{(4)}_0)(A_{\lambda_4,L^{(4)}})\ge
\#(e_{1,n+1}/e^{(1)}_0)(A_{\lambda_0,L^{(0)}})$,
the degree
$\deg W'(c_3,\lambda_3,$ $L^{(3)})=\deg W'(c_4,\lambda_4,L^{(4)})\ge
\deg W'(c_0,$ $\lambda_0,L^{(0)})$.
}\par\medskip

\noindent {\bf PROOF} \quad  Let $\varepsilon$ be
the coordinate function on ${\Bbb
A}^1(\overline{K})$.
Let
$(c_5,\lambda_5,L^{(5)})\in {\mathfrak l}$ and
$(c_0,\lambda_0,L^{(0)})\ne(c_5,\lambda_5,L^{(5)})$.
We have $e_{3,n+1}=e_{0,n+1}+\varepsilon^{(3)} e_{5,n+1}$,
$L_{3,i}=L_{0,i}+\varepsilon^{(3)} L_{5,i}$,
$L^{(3)}_w=L^{(0)}_w+\varepsilon^{(3)} L^{(5)}_w$, where
$\varepsilon^{(3)}\in\overline{K}$ for all $i,w$.
We shall assume without loss of generality that
$e_{4,n+1}=e_{0,n+1}+\varepsilon e_{5,n+1}$,
$L_{4,\nu}=L_{0,\nu}+\varepsilon L_{5,\nu}$,
$L^{(4)}_w=L^{(0)}_w+\varepsilon L^{(5)}_w$ for
the generic point
$(c_4,\lambda_4,L^{(4)})$ of the line ${\mathfrak l}$.
Consider the subvariety of ${\Bbb P}^n(\overline{K})\times{\Bbb
A}^1(\overline{K})$
\setcounter{equation}{26} \begin{eqnarray}
&&{\cal V}_i=(V\times{\Bbb A}^1(\overline{K}))\cap{\cal
Z}(L_{0,s+1}+\varepsilon L_{5,s+1},\ldots ,
L_{0,\sigma}+\varepsilon L_{5,\sigma})\nonumber\\
&&\cap{\cal Z}(e^{(0)}_{\sigma+1}+\varepsilon e^{(5)}_{\sigma+1},\ldots ,
e^{(0)}_i+\varepsilon e^{(5)}_i), \quad i=s,n-1,n  \label{27}
\end{eqnarray}
(so for $i=s$ one can omit here the variety
${\cal Z}(e^{(0)}_{\sigma+1}+\varepsilon e^{(5)}_{\sigma+1},\ldots ,
e^{(0)}_i+\varepsilon e^{(5)}_i)$). Denote by ${\cal W}_i$
the union of all the irreducible over $\overline{K}$
components $E$ of ${\cal V}_i$ such that
$E\cap(A_{\lambda_0,L^{(0)}}\times\{0\})\ne\emptyset$.
Hence ${\cal W}_i\subset W\times{\Bbb A}^1(\overline{K})$.
The dimension of every irreducible component of ${\cal W}_i$ is
$n-i+1$. Further, ${\cal W}_i\cap{\cal
Z}(\varepsilon)=\widetilde{W}_{\lambda_0,i}(L^{(0)})\times\{0\}$.
For every $\sigma\le i\le n$
by the B\'ezout theorem there is at most a polynomial in $d^n(d')^{n-\sigma}$
number of point $y\in{\mathfrak l}$ such that ${\cal W}_i\cap{\cal
Z}(\varepsilon-\varepsilon^{(3)})$ has an irreducible component $E$
contained in ${\cal Z}(e_0,\ldots , e_r)$ (since in this case $E$ is
necessarily the irreducible component of
${\cal W}_i\cap{\cal Z}(e_0,\ldots , e_r)$).
Therefore, for all $y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
${\cal W}_i\cap{\cal
Z}(\varepsilon-\varepsilon^{(3)})$ is a union of some
irreducible components of $\widetilde{W}_{\lambda_3,i}(L^{(3)})\times
\{\varepsilon^{(3)}\}$.
We shall show that $\deg {\cal W}_i\cap{\cal
Z}(\varepsilon-\varepsilon^{(3)})\ge D_i$ for all
$y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number
for every $\sigma\le i\le n$. On the other hand, we know, see the
observations above, that
always $\deg \widetilde{W}_{\lambda_3,i}(L^{(3)})\le D_i$.
Hence this will imply that ${\cal W}_i\cap{\cal
Z}(\varepsilon-\varepsilon^{(3)})=\widetilde{W}_{\lambda_3,i}(L^{(3)})\times
\{\varepsilon^{(3)}\}$ for all
$y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number
for every $\sigma\le i\le n$.


There are
homogeneous polynomials $h_1,\ldots , h_s\in k[X_0,\ldots , X_n]$ such that
$W$ is an irreducible component of ${\cal Z}(h_1,\ldots , h_s)$ in
${\Bbb P}^n(\overline{k})$, and the intersection
of $s$ hypersurfaces ${\cal Z}(h_1),\ldots ,{\cal Z}(h_s)$ at
each point from $A_{\lambda_0,L^{(0)}}$ is
transversal.
We choose $h_1,\ldots , h_s$ using linear projections ${\Bbb
P}^n(\overline{k})\rightarrow{\Bbb P}^{n-s+1}(\overline{k})$ in general
position such that $\deg h_i\le\deg W\le d^s$ (we leave the details to the
reader).
There are linearly independent linear forms $\widetilde{L}_{\sigma+1},\ldots
,\widetilde{L}_n\in k[X_0,\ldots , X_n]$ such that
for every $\sigma\le i\le n$
the intersection of $n$ hypersurfaces
${\cal Z}(h_1),\ldots ,{\cal Z}(h_s)$,
${\cal Z}(L_{0,s+1}),\ldots ,{\cal Z}(L_{0,\sigma})$,
${\cal Z}(e^{(0)}_{\sigma+1}),\ldots , {\cal Z}(e^{(0)}_i)$,
${\cal
Z}(\widetilde{L}_{i+1}),\ldots ,$ ${\cal Z}(\widetilde{L}_n)$ is
transversal at every point $\xi\in
\widetilde{W}_{\lambda_0,i}(L^{(0)})\cap{\cal
Z}(\widetilde{L}_{i+1},\ldots ,\widetilde{L}_n)=B_i$
(hence by (iv) the number of elements $\#B_i=D_i$),
for every point $\xi\in B_i$
the differential $d_{\xi}p_{\lambda_0,L^{(0)},i}$ is an isomorphism,
and $B_i\cap({\cal Z}(e^{(0)}_0)\cup\overline{(V\setminus
W)})=\emptyset$.
These are conditions of general position for $\widetilde{L}_{\sigma+1},\ldots
,\widetilde{L}_{n}$ due to property
(D) for $(\lambda_0,L^{(0)})$ and our first observation related  to (D), see
above.  Denote by $\Delta_i$ the Jacobian of
$n$ polynomials $h_1,\ldots , h_s$,
$L_{0,s+1}+\varepsilon L_{5,s+1},\ldots ,L_{0,\sigma}+\varepsilon L_{5,n}$,
$e^{(0)}_{\sigma+1}+\varepsilon e^{(5)}_{\sigma+1},\ldots ,
e^{(0)}_i+\varepsilon e^{(5)}_i$,
$\widetilde{L}_{i+1},\ldots ,\widetilde{L}_n$ with respect to $X_1,\ldots
,X_n$.
Obviously $\deg_{X_0,\ldots , X_n,Z}\Delta_i$ is bounded from above by a
polynomial in $d^n(d')^{n-\sigma}$.
Performing a nondegenerate linear transformation of coordinates
in ${\Bbb P}^n(\overline{K})$ we shall assume without loss of generality
that $X_0(\xi)\ne 0$ for every $\xi\in B_i$
and $(B_i\times\{0\})\cap{\cal Z}(\Delta_i)=\emptyset$
for all $\sigma\le i\le n$.



Set $C_i$ to be the union of all the irreducible components $C$ of
${\cal W}_i\cap{\cal Z}(\widetilde{L}_{i+1},\ldots ,$
$\widetilde{L}_n)$
such that $C\cap{\cal Z}(\varepsilon)\cap(B_i\times\{0\})\ne\emptyset$.
Hence $C_i\cap{\cal Z}(\varepsilon)=B_i\times\{0\}$ and
$C_i$ is a curve.
By the B\'ezout theorem the degree $\deg C_i\le 2^{n-s}d^s(d')^{n-\sigma}$
($\deg C_i$
is by definition the degree of $C\setminus{\cal Z}(L_0)$ in the affine
space ${\Bbb A}^n(\overline{K})\times{\Bbb A}^1(\overline{K})$
where $L_0\in k[X_0,\ldots , X_n]$ is an arbitrary
linear form which does not vanish on any irreducible component of $C_i$).
The linear projection $C_i\rightarrow{\Bbb A}^1(\overline{K})$,
$((X_0:\ldots : X_n),\varepsilon)\mapsto \varepsilon$, is dominant.
Hence for all $\varepsilon^{(3)}\in\overline{K}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
the number of elements $+\infty>\#(C_i\cap{\cal
Z}(\varepsilon-\varepsilon^{(3)}))\ge
\#(C_i\cap{\cal Z}(\varepsilon))=D_i$
for every $\sigma\le i\le n$.
Recall that $C_i$ is a union of irreducible components of
${\cal W}_i\cap{\cal Z}(\widetilde{L}_{i+1},\ldots ,$
$\widetilde{L}_n)$.
Hence by the B\'ezout theorem for all $\varepsilon^{(3)}\in\overline{K}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
every isolated point of  $C_i\cap{\cal
Z}(\varepsilon-\varepsilon^{(3)})$ is an isolated point of
${\cal W}_i\cap{\cal Z}(\widetilde{L}_{i+1},\ldots ,$
$\widetilde{L}_n)\cap{\cal Z}(\varepsilon-\varepsilon^{(3)})$.
Therefore for all $\varepsilon^{(3)}\in\overline{K}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
$\deg{\cal W}_i\cap{\cal Z}(\varepsilon-\varepsilon^{(3)})\ge D_i$.
For these $y$
the algebraic variety ${\cal W}_i\cap{\cal
Z}(\varepsilon-\varepsilon^{(3)})=
\widetilde{W}_{\lambda_3,i}(L^{(3)})\times
\{\varepsilon^{(3)}\}$ and
$\deg\widetilde{W}_{\lambda_3,i}(L^{(3)})=D_i$, see above.



By the B\'ezout theorem the number of points from $C_i\cap{\cal
Z}(\Delta_i)$ is bounded from above by a
polynomial in $d^n(d')^{n-\sigma}$. Hence for all $y\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
for every point $(\xi,\varepsilon^{(3)})\in
(\widetilde{W}_{\lambda_3,i}(L^{(3)})\times\{\varepsilon^{(3)}\})\cap
{\cal Z}(\widetilde{L}_{i+1},\ldots ,\widetilde{L}_n)$ the differential
$d_\xi p_{\lambda_3,L^{(3)},i}$ is an isomorphism for all $\sigma\le i\le n$.
Thus, we have proved that for all $y\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
conditions (i), (iv) and (v), see the beginning of the section, hold
with $j=3$.

Recall that $B_i\cap({\cal
Z}(e^{(0)}_0)\cup\overline{(V\setminus W)})=\emptyset$, see above.
Hence the number of elements $\#(C_i\cap({\cal
Z}(e^{(0)}_0+\varepsilon e^{(5)}_0)\cup\overline{(V\setminus W)}))$ is finite
and it is bounded from above by a polynomial in $d^n(d')^{n-\sigma}$ by the
B\'ezout theorem. Taking $i=n$ we get that (ii) with $j=3$ is fulfilled for
all $y\in{\mathfrak l}$ except at most a
polynomial in $d^n(d')^{n-\sigma}$ number.
For such $y$ conditions (i), (ii), (iv), (v) hold. Hence
also (i)--(iv) are satisfied,
see observations at the beginning of the section. Thus, property (D) holds
for all
$y\in{\mathfrak l}$ except at most a polynomial in $d^n(d')^{n-\sigma}$
number.
In particular, (D) is fulfilled for the generic point
$(c_4,\lambda_4,L^{(4)})$
since there are infinitely many isomorphic over $\overline{K}$ generic
points of ${\mathfrak l}$ in ${\mathfrak l}(\overline{K(\varepsilon)})$.

The polynomial
$\Phi_{c_4,\lambda_4,L^{(4)}}\in
\overline{k}[\varepsilon,Z_0,Z_{\sigma+1},\ldots , Z_{n+1}]$
is separable and $\varepsilon$ does not divide
$\Phi_{c_4,\lambda_4,L^{(4)}}$. Hence
$\Phi_{c_4,\lambda_4,L^{(4)}}|_{\varepsilon=0}\ne 0$ and
$\Phi_{c_0,\lambda_0,L^{(0)}}$
divides $\Phi_{c_4,\lambda_4,L^{(4)}}|_{\varepsilon=0}$. Hence
$\deg W'(c_4,\lambda_4,L^{(4)})\ge\deg W'(c_0,\lambda_0,L^{(0)})$,\\
$\#(e_{4,n+1}/e^{(4)}_0)(A_{\lambda_4,L^{(4)}})=
\deg_{Z_{n+1}}\Phi_{c_4,\lambda_4,L^{(4)}}\ge\deg_{Z_{n+1}}\Phi_{c_0,\lambda_0,L^{(0)}}=$\\
$\#(e_{0,n+1}/e^{(0)}_0)(A_{\lambda_0,L^{(0)}})$.
Let us represent $\Phi_{c_4,\lambda_4,L^{(4)}}=\sum_{0\le
i\le\delta}\Phi_iZ_{n+1}^i$,
where all $\Phi_i\in\overline{k}[\varepsilon,Z_0,Z_{\sigma+1},\ldots ,
Z_n]$, $\delta=\deg_{Z_{n+1}}\Phi_{c_4,\lambda_4,L^{(4)}}$.
Then $\mbox{\rm G\,C\,D\,}_{0\le i\le\delta}\{\Phi_i\}=1$
according to the definition of
$\Phi_{c_4,\lambda_4,L^{(4)}}$.


On the other hand, for all $y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
the polynomial
$\Phi_{c_4,\lambda_4,L^{(4)}}|_{\varepsilon=\varepsilon^{(3)}}$
is separable, the degree $\deg_{Z_{n+1}}\Phi_{c_3,\lambda_3,L^{(3)}}=
\#(e_{3,n+1}/e^{(3)}_0)(A_{\lambda_3,L^{(3)}})=
\#(e_{4,n+1}/e^{(4)}_0)(A_{\lambda_4,L^{(4)}})=$ \\
$\deg_{Z_{n+1}}\Phi_{c_4,\lambda_4,L^{(4)}}$, and, cf.
the proof of Lemma~8 below,\\
$\mbox{\rm G\,C\,D\,}_{0\le i\le\delta}
\{\Phi_i|_{\varepsilon=\varepsilon^{(3)}}\}=1$.
Hence for these $(c_3,\lambda_3,L^{(3)})$ the polynomials\\
$\Phi_{c_3,\lambda_3,L^{(3)}}$ and
$\Phi_{c_4,\lambda_4,L^{(4)}}|_{\varepsilon=\varepsilon^{(3)}}$
coincide up to a nonzero factor from $\overline{K}$.
Thus, $\deg W'(c_3,\lambda_3,L^{(3)})=
\deg W'(c_4,\lambda_4,L^{(4)})\ge\deg
W'(c_0,$  $\lambda_0,L^{(0)})$, and\\
$\#(e_{3,n+1}/e^{(3)}_0)(A_{\lambda_3,L^{(3)}})=
\#(e_{4,n+1}/e^{(4)}_0)(A_{\lambda_4,L^{(4)}})$ $\ge
\#(e_{0,n+1}/e^{(0)}_0)(A_{\lambda_0,L^{(0)}})$.  The lemma is proved.








\par\medskip\noindent{\bf LEMMA~5}\hspace{0.1em} {\it  Under conditions of
Lemma~4 suppose
that the number of elements $\#(e_{0,n+1}/e^{(0)}_0)(A_{\lambda_0,L^{(0)}})=
\#(e_{4,n+1}/e^{(4)}_0)(A_{\lambda_4,L^{(4)}})=\deg p'_v\deg W'$
and $\deg W'(c_0,\lambda_0,L^{(0)})$ $=
\deg W'(c_4,\lambda_4,L^{(4)})$. Then
for all $y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
$b(c_3,\lambda_3,L^{(3)})=b(c_4,\lambda_4,L^{(4)})$ $\ge
b(c_0,\lambda_0,L^{(0)})$.
}\par\medskip

\noindent {\bf PROOF} \quad Let $\varepsilon^{(0)}=0$. For $j=0,3$ for all
$y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
the polynomials $\Phi_{c_j,\lambda_j,L^{(j)}}$ coincide with
$\Phi_{c_4,\lambda_4,L^{(4)}}|_{\varepsilon=\varepsilon^{(j)}}$.
Hence $(R_{c_4,\lambda_4,L^{(4)},0}|_{\varepsilon=0})_0=
R_{c_0,\lambda_0,L^{(0)},0}$ and
$R_{c_4,\lambda_4,L^{(4)},0}|_{\varepsilon=\varepsilon^{(3)}}$ $=
R_{c_3,\lambda_3,L^{(3)},0}$ for all
$y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number.
This implies the assertion of the lemma. The lemma is proved.

\par\medskip\noindent{\bf LEMMA~6}\hspace{0.1em} {\it  Under conditions of
Lemma~4 suppose additionally that
 the number of elements
$\#(e_{0,n+1}/e^{(0)}_0)(A_{\lambda_0,L^{(0)}})=\deg p'_v\deg W'$.
Then for all $y=(c_3,\lambda_3,$ $L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
$\deg W'(c_3,\lambda_3,L^{(3)},*)$ $=
\deg W'(c_4,\lambda_4,L^{(4)},*)\ge\deg
W'(c_0,\lambda_0,L^{(0)},*)$.
}\par\medskip

\noindent {\bf PROOF} \quad Let us choose in the Definition~4
(with the field $\overline{K}$ in place of
$\overline{k}$) for $\Phi_{c_4,\lambda_4,L^{(4)},+}$ the
polynomials
$g_1\in\overline{k}[\varepsilon,Z_0,Z_n]$ of maximal possible degree
with respect to $\varepsilon,Z_0,Z_n$ and $g_2\in
\overline{k}[\varepsilon,Z_0,$ $Z_n,Z_{n+1}]$.
Notice that $\Phi_{c_4,\lambda_4,L^{(4)},+}$ is separable,
$\varepsilon$ does not divide $\Phi_{c_4,\lambda_4,L^{(4)},+}$.
Hence $\Phi_{c_4,\lambda_4,L^{(4)},+}|_{\varepsilon=0}\ne 0$,
and $\Phi_{c_0,\lambda_0,L^{(0)},+}$
divides $\Phi_{c_4,\lambda_4,L^{(4)},+}|_{\varepsilon=0}$. Therefore,
the degree
$\deg W'(c_4,\lambda_4,L^{(4)},$ $*)\ge\deg W'(c_0,\lambda_0,L^{(0)},*)$
by Lemma~3.
Let $\Phi_{c_4,\lambda_4,L^{(4)},+}=\sum_{0\le
i\le\delta}\varphi_iZ_{n+1}^i$,
where all $\varphi_i\in\overline{k}[\varepsilon,Z_0,Z_{\sigma+1},\ldots ,
Z_n]$, $\delta=\deg p'_v\deg W'$.
Then $\mbox{\rm G\,C\,D\,}_{0\le i\le\delta}\{\varphi_i\}$ $=1$
according to the definition of
$\Phi_{c_4,\lambda_4,L^{(4)},+}$.
On the other hand, for all $y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number,
the polynomial
$\Phi_{c_4,\lambda_4,L^{(4)},+}|_{\varepsilon=\varepsilon^{(3)}}$
is separable,  and, cf.
the proof of Lemma~8, the greatest common divisor
$\mbox{\rm G\,C\,D\,}_{0\le i\le\delta}
\{\varphi_i|_{\varepsilon=\varepsilon^{(3)}}\}=1$.
Hence for these $(c_3,\lambda_3,L^{(3)})$ the polynomials
$\Phi_{c_3,\lambda_3,L^{(3)},+}$ and
$\Phi_{c_4,\lambda_4,L^{(4)},+}|_{\varepsilon=\varepsilon^{(3)}}$
coincides up to a nonzero factor from $\overline{K}$,
and, therefore, $\deg W'(c_3,\lambda_3,L^{(3)},*)=
\deg W'(c_4,\lambda_4,L^{(4)},*)\ge\deg
W'(c_0,\lambda_0,L^{(0)},*)$.
The lemma is proved.


\par\medskip\noindent{\bf LEMMA~7}\hspace{0.1em} {\it  Under conditions of
Lemma~4 suppose
the that the number of elements
$\#(e_{0,n+1}/e^{(0)}_0)(A_{\lambda_0,L^{(0)}})=
\#(e_{4,n+1}/e^{(4)}_0)(A_{\lambda_4,L^{(4)}})$ and the degrees $\deg
W'(c_0,\lambda_0,L^{(0)})$
$=\deg W'(c_4,\lambda_4,L^{(4)})=\deg W'(c_0,\lambda_0,L^{(0)},*)=
\deg W'(c_4,$ $\lambda_4,L^{(4)},*)$. Then
for all $y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number
$a(c_3,\lambda_3,L^{(3)})=
a(c_4,\lambda_4,L^{(4)})\ge a(c_0,\lambda_0,L^{(0)})$.
}\par\medskip

\noindent {\bf PROOF} \quad Let $\varepsilon^{(0)}=0$ and
$\varepsilon^{(3)}$ be as above.
By Lemma~3 and Lemma~6
for all $y=(c_3,\lambda_3,L^{(3)})\in{\mathfrak l}$,
except at most a polynomial in $d^n(d')^{n-\sigma}$ number
$\Phi_{c_j,\lambda_j,L^{(j)},+}=\Phi_{c_j,\lambda_j,L^{(j)},*}=$
$\Phi_{c_4,\lambda_4,L^{(4)},*}|_{\varepsilon=\varepsilon^{(j)}}$, $j=0,3$.
Now the required assertion follows from the definitions of
$a(c_j,\lambda_j,L^{(j)})$,
$j=0,3,4$ immediately. The lemma is proved.

\medskip
Recall that in the notation from the Introduction we have the natural
projection
${\Bbb A}^\nu(\overline{k})\rightarrow{\Bbb A}^{\nu_1}(\overline{k})$
such that $(c,\lambda,L')\mapsto (c',\lambda)$.



\par\medskip\noindent{\bf LEMMA~8}\hspace{0.1em} {\it  Let ${\Bbb
A}^1(\overline{k})$ be defined over $k$ and
have a fixed coordinate function $\varepsilon$ defined over $k$. In the
notation
from the Introduction let  $\iota\, :\,{\Bbb
A}^1(\overline{k})\rightarrow {\Bbb A}^\nu(\overline{k})$
(respectively ${\Bbb A}^1(\overline{k})\rightarrow{\Bbb
A}^{\nu_1}(\overline{k})$) be an affine linear embedding inducing
the defined over $k$ morphism $\iota\, :\,{\Bbb
A}^1(\overline{k})\rightarrow {\Bbb A}^{\nu_1}(\overline{k})$
(respectively defined over $k$ itself).
Let $y_\varepsilon=\iota(\varepsilon)=
(c_4,\lambda_4,L^{(4)})$ be the generic
point of the line ${\mathfrak l}=\iota({\Bbb A}^1(\overline{k}))$.
Suppose that the point
$\iota(0)=(c_0,\lambda_0,L^{(0)})$
(respectively $\iota(0)=(c_0,\lambda_0)$)
satisfies property (*) from the Introduction
and $\deg W'(c_0,\lambda_0,L^{(0)})=
\deg W'(c_4,\lambda_4, L^{(4)})$ (respectively satisfies property (*) and
$\deg W'(c_0,\lambda_0,U)=\deg W'(c_4,\lambda_4, U)$).
Then for all $y\in k$, except at most a
polynomial in $d^n(d')^{n-\sigma}$ number,
property (*) holds for $\iota(y)= (c_3,\lambda_3,L^{(3)})$
(respectively for $\iota(y)= (c_3,\lambda_3)$) and the degree
$\deg W'(c_3,\lambda_3,L^{(3)})=
\deg W'(c_4,\lambda_4, L^{(4)})$ (respectively
$\deg W'(c_3,\lambda_3,U)$ $=\deg W'(c_4,\lambda_4,U)$).
More than that, the similar statement holds if one replaces (*)
by the conjugation of conditions 1) and 2)
(or 1), 2) and 3)) from (*).
}\par\medskip

\noindent {\bf PROOF} \quad If $(c_0,\lambda_0)$ satisfies property (*) and
$\deg W'(c_0,\lambda_0,U)=\deg W'(c_4,$ $\lambda_4,U)$ then
there exists $L^{(0)}\in{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$ such
that $\deg W'(c_0,\lambda_0,L^{(0)})=\deg W'(c_4,\lambda_4,U)$.
For any point $(c_3,\lambda_3)$ from the statement of the lemma
$\deg W'(c_3,\lambda_3,L^{(0)})\le\deg W'(c_3,\lambda_3,U)\le
\deg W'(c_4,\lambda_4,U)$.
Hence the case of $(c_0,\lambda_0)$ is reduced to
$(c_0,\lambda_0,L^{(0)})$.
The last assertion of the lemma is proved similarly to the first one.

So we shall consider the first assertion for the case of
$(c_0,\lambda_0,L^{(0)})$. The intersection (15) with $j=0$
is finite and
is transversal, see \cite{13}. Considering the corresponding
Jacobian we get that it remains transversal for all points
$(c_3,\lambda_3,L^{(3)})\in\iota({\Bbb A}^1(\overline{k}))$ in place of
$(c_0,\lambda_0,L^{(0)})$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number.
Hence assertion (d) of Theorem~1
\cite{12}, assertion (f) of Theorem~2 \cite{13} and the B\'ezout
theorem imply that properties 1),  2) and 3)
from (*) hold for all
$(c_3,\lambda_3,L^{(3)})\in\iota({\Bbb A}^1(\overline{k}))$ in place of,
except at most a polynomial in
$d^n(d')^{n-\sigma}$ number.

Consider in details condition 4) from (*). Let us choose
$\Phi_{c_4,\lambda_4,L^{(4)}}\in\overline{k}[\varepsilon,Z_0,$
$Z_{\sigma+1},\ldots , Z_{n+1}]$ where $\varepsilon$ does not divide the last
polynomial. Denote for brevity
$\delta=\deg_{Z_{n+1}}\Phi_{c_4,\lambda_4,L^{(4)}}$. Let
$\Phi_{c_4,\lambda_4,L^{(4)},*}=\sum_{0\le i\le\delta}
\varphi_i Z_{n+1}^i$ where all
$\varphi_i\in\overline{k}[\varepsilon,Z_0,Z_n]$. By the B\'ezout
theorem all the degrees $\deg_{\varepsilon,Z_0,Z_n}\varphi_i$ are bounded
from
above by a polynomial in $d^n(d')^{n-\sigma}$, cf. \cite{12}
Lemma~7. The greatest common divisor $\mbox{\rm G\,C\,D\,}_{0\le
i\le\delta}\{\varphi_i(0,$ $Z_0,Z_n)\}=1$ in
$\overline{k}[Z_0,Z_n]$ by condition 4) from (*) for $\iota(0)$
and since $\deg W'(c_0,\lambda_0,L^{(0)})=
\deg W'(c_4,\lambda_4, L^{(4)})$.
Hence there are polynomials $A_{j,i}
\in\overline{k}[\varepsilon,Z_0,Z_n]$, $j=0,n$, $0\le i\le\delta$,
and $0\ne A_j\in\overline{k}[\varepsilon]$, $j=0,n$, such that the
degrees with respect to $\varepsilon,Z_0,Z_n$ of all $A_{j,i},A_j$
are bounded from above by a polynomial in $d^n(d')^{n-\sigma}$,
and $\sum_{0\le i\le\delta}A_{j,i}\varphi_i=A_jZ_j^\mu$, $j=0,n$.
Now if $\varepsilon^{(3)}\in k$,
$\iota(\varepsilon^{(3)})=
(c_3,\lambda_3,L^{(3)})$, properties 1), 2), 3) hold for
$(c_3,\lambda_3,L^{(3)})$ and $(A_0A_n)(\varepsilon^{(3)})\ne 0$
then the polynomial $\Phi_{c_3,\lambda_3,L^{(3)},*}=
\Phi_{c_4,\lambda_4,L^{(4)},*}|_{\{\varepsilon=\varepsilon^{(3)}\}}$
does not have irreducible factors from $\overline{k}[Z_0,Z_n]$,
i.e., property 4) holds and $\deg W'(c_3,\lambda_3,L^{(3)})=
\deg W'(c_4,\lambda_4, L^{(4)})$. The lemma is proved.








\par\medskip\noindent{\bf LEMMA~9}\hspace{0.1em} {\it  Let
$(c_1,\lambda_1,L^{(1)})$ (respectively $(c_1,\lambda_1)$) be
given by
(17)--(19) (respectively (17), (18)), see the
Introduction.
In the notation from the Introduction suppose that
$(c_0,\lambda_0,L^{(0)})$ (respectively $(c_0,\lambda_0)$)
satisfies property (*) and 5) (respectively satisfies (*) and
$\deg W'(c_0,\lambda_0,U)=\deg W'(c_1,\lambda_1, U)$).
Then $(c_1,\lambda_1,L^{(1)})$
(respectively $(c_1,\lambda_1)$) satisfy (*).
More than that, the similar statement holds if one replaces (*)
by the conjunction of conditions 1) and 2)
(or 1), 2) and 3)) from (*).
}\par\medskip

\noindent {\bf PROOF} \quad The generic point $(c_4,\lambda_4,L^{(4)})$
(respectively $(c_4,\lambda_4)$) of the line
$\iota({\Bbb A}^1(\overline{k}))$ from Lemma~8 satisfies (*) and 5)
(respectively (*) and $\deg W'(c_0,\lambda_0,U)$ $=\deg W'(c_4,\lambda_4,
U)$)
if one extends ground field. Hence extending subsequently $k$ till
$k(\tau_1)$, $k(\tau_1,\tau_2),\ldots , k(\tau_1,\ldots
,\tau_{n_1})$ and applying each time Lemma~8 we get the
assertion of the lemma. The lemma is proved.






\par\medskip\noindent{\bf LEMMA~10}\hspace{0.1em} {\it  Let
$(c_1,\lambda_1)$ be
given by
(17), (18) and satisfy 1)--3) from (*), see the
Introduction. Then the following assertions hold.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item $R_{c_1,\lambda_1,0,*}=
R_{c_1,\lambda_1,0}(\tau_1,\ldots ,
\tau_{n_1},u,Z_0,0,\ldots , 0,Z_n)$ is a nonzero separable element
of the ring of polynomials $\overline{k}[\tau_1,\ldots
,\tau_{n_1},u,Z_0,Z_n]$ (with respect to all the variables
$\tau_1,\ldots ,\tau_{n_1}$, $u,Z_0, Z_n$). In the other words,
$a(c_1,\lambda_1)=b(c_1,\lambda_1)$, see the Introduction.
\item The polynomial $\Phi_{c_1,\lambda_1,*}$ does not have irreducible
factors from $k[\tau_1,\ldots ,\tau_{n_1},\,$ $u,Z_0,Z_n]$.
\end{enumerate}
In particular (see the notation in the Introduction), $\Delta_{0,*}$ is a
nonzero separable element of the ring of polynomials
$\overline{k}[c,\upsilon,u,Z_0,Z_n]$, and $F_*$ does not have irreducible
factors from $\overline{k}[c,\upsilon,u,Z_0,Z_n]$.
}\par\medskip

\noindent {\bf PROOF} \quad Let
$\widetilde{Z}_{\sigma+1},\ldots , \widetilde{Z}_{n-1}$ be linear
forms in $Z_0,Z_{\sigma+1},\ldots , Z_n$ with coefficients
from $k$ in general position. Let us identify
$k[Z_0,Z_{\sigma+1},\ldots , Z_n]/(\widetilde{Z}_{\sigma+1},\ldots ,$
$\widetilde{Z}_{n-1})=k[Z_0,Z_n]$.

Let $\delta=\deg p'_v\deg W'$.
By Lemma~2 $\Phi_{c_1,\lambda_1,*}\ne 0$.
Let us represent $\Phi_{c_1,\lambda_1}=\sum_{0\le
i\le\delta}\Phi_iZ_{n+1}^i$ where
all $\Phi_i\in\overline{k}[\tau_1,\ldots ,\tau_{n_1},u,Z_0,Z_n]$.
By its definition the polynomial $\Phi_{c_1,\lambda_1}$ does not have
irreducible
factor from $\overline{k}[\tau_1,\ldots ,\tau_{n_1},u,Z_0,\ldots ,$ $Z_n]$.
Hence $\mbox{\rm G\,C\,D\,}_{0\le i\le\delta}\{\Phi_i\}=1$ in the
last ring, i.e., $\mbox{\rm codim}{\cal Z}(\Phi_0,\ldots
,\Phi_\delta)\ge 2$ in the corresponding affine space.
Recall that $\widetilde{Z}_{\sigma+1},\ldots , \widetilde{Z}_{n-1}$ are in
general position. Hence considering the codimension we get $\mbox{\rm
G\,C\,D\,}_{0\le
i\le\delta}\{\Phi_i\bmod(\widetilde{Z}_{\sigma+1},
\ldots , \widetilde{Z}_{n-1})\}=1$ in the ring
$\overline{k}[\tau_1,\ldots ,$ $\tau_{n_1},u,$ $Z_0,\ldots ,
Z_n]/(\widetilde{Z}_{\sigma+1},
\ldots , \widetilde{Z}_{n-1})$ $=\overline{k}[\tau_1,\ldots ,
\tau_{n_1},u,Z_0,Z_n]$.



Consider the affine space ${\Bbb A}^N(\overline{k})$ with the coordinate
functions from the families
$\tau_1,\ldots ,\tau_{n_1}$, $u$, $Z_0,Z_{\sigma+1},\ldots ,
Z_n$. Let ${\Bbb A}^N(\overline{k})\subset{\Bbb P}^N(\overline{k})$ in
the natural way. Let ${\cal W}$ be an irreducible component of the
closure in
${\Bbb P}^N(\overline{k})$ of the algebraic variety
${\cal Z}(R_{c_1,\lambda_1,0})$.
Let us apply Theorem~1 \cite{13} to every morphism ${\cal
W}\rightarrow{\Bbb
P}^{n-\sigma+1}(\overline{k})$ induced by the linear projection to given by
$Z_0,Z_{\sigma+1},\ldots,
Z_n$. Then since $\widetilde{Z}_{\sigma+1},\ldots , \widetilde{Z}_{n-1}$
are in general position, we obtain that
$0\ne R_{c_1,\lambda_1,0}\bmod(\widetilde{Z}_{\sigma+1},\ldots ,
\widetilde{Z}_{n-1})\in \overline{k}[\tau_1,\ldots
,\tau_{n_1},u,\,Z_0,Z_{\sigma+1},\ldots ,
Z_n]/(\widetilde{Z}_{\sigma+1},\ldots ,$ $\widetilde{Z}_{n-1})$ is a
separable element of the last ring.
By their definition $U_j$ are
generic linear forms. Hence $U_0$,
$\widetilde{Z}_j(U_0,U_{\sigma+1},\ldots , U_{n-1})$, $\sigma+1\le
j\le n-1$, $U_n$ is also a family of generic linear forms in
$X_0,\ldots , X_n$ with respect to the field $k$. Therefore, the
polynomials $\Phi_{i,*}$, $R_{c_1,\lambda_1,0,*}$ are obtained from
$\Phi_i\bmod(\widetilde{Z}_{\sigma+1},
\ldots , \widetilde{Z}_{n-1})$,
$R_{c_1,\lambda_1,0}\bmod(\widetilde{Z}_{\sigma+1},\ldots ,
\widetilde{Z}_{n-1})$ by an automorphism induced by a linear
transformation over $k$ of the variables from the family
$u$. Therefore, the greatest common divisor
$\mbox{\rm G\,C\,D\,}_{0\le
i\le\delta}\{\Phi_{i,*}\}=1$ and
$R_{c_1,\lambda_1,0,*}$ is nonzero separable.
The lemma is proved.


\par\medskip\noindent{\bf LEMMA~11}\hspace{0.1em} {\it  Let
$(c_1,\lambda_1)$ be from the Introduction and satisfies (*).
Then the set of all $L'$ such that $L'\in{\cal
U}'''_{\lambda_1,0}(\overline{k})$ and
conditions (i), (ii) (respectively (i)--(iii)) from Definition~2 hold for
$(c_1,\lambda_1,L')$ is a nonempty open in the Zariski topology subset of
${\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$.
Hence ${\cal U}^{({\rm v})}_{c_1,\lambda_1,0}$
is a nonempty open in the Zariski topology subset of ${\Bbb
A}^{(r+1)(n-\sigma+1)}(\overline{k})$.
}\par\medskip

\noindent {\bf PROOF} \quad By conditions 3) and 2) from (*) the polynomial
$e_{1,n+1}$ does not vanish on any irreducible component of
$W_{\lambda_1}$.
Hence $X_{r+1}$ does not vanish on any irreducible component of
$W'(c_1,\lambda_1)$.
Therefore,
$W'(c_1,\lambda_1)\cap{\cal Z}(U_0,U_{\sigma+1},\ldots ,
U_n,$ $X_{r+1})=\emptyset$, the degree
$\deg W'(c_1,\lambda_1)=\\
\deg_{Z_0,Z_{\sigma+1},\ldots , Z_n}\Phi_{c_1,\lambda_1}$,
cf. Remark~5 from \cite{12}.
Hence, see Lemma~13 \cite{8}
and Remark~6 \cite{13} if
$\deg_{Z_0,Z_{\sigma+1},\ldots , Z_n}\Phi_{c_1,\lambda_1,L'}=\deg
W'(c_1,\lambda_1)$ then the polynomials $\Phi_{c_1,\lambda_1,L'}$ and
$\Phi_{c_1,\lambda_1}|_{u=l'}$ (the last polynomial
is obtained by the substitution
of $l'_{i,j}$ for $u_{i,j}$ for all $i,j$)
coincide up to a nonzero factor from $\overline{k}[\tau_1,\ldots
,\tau_{n_1}]$.

Further, let $L'\in
{\cal U}'''_{\lambda_1,0}(\overline{k})$ and $(c_1,\lambda_1,L')$
satisfies
2), 3) from (*) (the set of all such $L'$ is a nonempty open in the Zariski
topology subset of
${\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$). Then
condition (ii) of Definition~2
holds for $L'$ if and only if $W'(c_1,\lambda_1)\cap
{\cal Z}(L'_0,L'_{\sigma+1},\ldots ,
L'_n,$ $X_{r+1})=\emptyset$. Therefore the set ${\cal W}$ of all
$L'\in{\cal U}'''_{\lambda_1,0}(\overline{k})$  such that conditions
2), 3) from (*) and (ii) of Definition~2 are fulfilled for
$(c_1,\lambda_1,L')$
is a nonempty open in the Zariski topology subset of ${\Bbb
A}^{(r+1)(n-\sigma+1)}(\overline{k})$.

Let $\delta=\deg p'_v\deg W'$.
Let us represent $\Phi_{c_1,\lambda_1,*}=\sum_{0\le
i\le\delta}\varphi_iZ_{n+1}^i$ where
all $\Phi_i\in\overline{k}[\tau_1,\ldots ,\tau_{n_1},u,Z_0,Z_n]$. By Lemma~10
there are polynomials $A_{j,i}
\in\overline{k}[\tau_1,\ldots ,\tau_{n_1},u,Z_0,Z_n]$, $j=0,n$, $0\le
i\le\delta$,
and $0\ne A_j\in\overline{k}[\tau_1,\ldots ,\tau_{n_1},u]$, $j=0,n$,
such that  $\sum_{0\le i\le\delta}A_{j,i}\varphi_i=A_jZ_j^\mu$, $j=0,n$.
Now let $L'\in{\cal W}$. Then $(A_0A_n)|_{\{u=l'\}}\ne 0$
implies condition 4) from (*) for $(c_1,\lambda_1,L')$. Hence the set of
$L'\in{\cal W}$
such that condition 4) from (*) holds for $(c_1,\lambda_1,L')$ is nonempty.
On the other hand, obviously if $L'\in{\cal W}$ and $\mbox{\rm
G\,C\,D\,}\{\varphi_i|_{\{u=l'\}}\}=1$ in
$\overline{k}(\tau_1,\ldots ,\tau_{n_1})[Z_0,Z_n]$ then there is a
neighborhood ${\cal W}'\subset{\cal W}$
of $L'$ with respect to Zariski topology such that for every
$L''\in{\cal
W}'$ we have
$\mbox{\rm G\,C\,D\,}\{\varphi_i|_{\{u=l''\}}\}=1$
(here $l''$ for $L''$ is similar to $l'$ defined for $L'$) in
$\overline{k}(\tau_1,\ldots ,\tau_{n_1})[Z_0,Z_n]$.
This implies the assertion of the lemma related  to conditions (i) and (ii).

Denote by ${\cal W}''$ the set of all $L'\in{\cal
U}'''_{\lambda_1,0}(\overline{k})$ such that condition (i), (ii) of
Definition~2
hold.
Then $L'\in{\cal U}^{(5)}_{c_1,\lambda_1}$ if and only if
$L'\in{\cal W}''$ and
$R_{c_1,\lambda_1,0,*}|_{\{u=l'\}}$
coincides with $R_{c_1,\lambda_1,L',*,0}$
up to a nonzero factor from $\overline{k}[\tau_1,\ldots ,\tau_{n_1}]$,
i.e., if and only if $R_{c_1,\lambda_1,0,*}|_{\{u=l'\}}$ is separable with
respect to the variables
$Z_0,Z_n$.
The polynomial $R_{c_1,\lambda_1,0,*}$ is nonzero and separable by Lemma~10.
Hence, considering the discriminants of $R_{c_1,\lambda_1,0,*}$ with respect
to $Z_0$ and $Z_n$ we
deduce that the set of all $L'\in{\cal W}'$ such that
$R_{c_1,\lambda_1,0,*}|_{\{u=l'\}}$ coincides with $R_{c_1,\lambda_1,L',*,0}$
is a nonempty open in the Zariski topology subset of ${\cal W}''$.
The lemma is proved.




\par\medskip\noindent{\bf LEMMA~12}\hspace{0.1em} {\it  In the notation from
the Introduction let
$(c_2,\lambda_2,L^{(2)})\le(c_1,\lambda_1,$ $L^{(1)})$.
Assume that $(c_i,\lambda_i,L^{(i)})$ satisfies
conditions 1)--3) from (*) if one extends the field $k$ till
$k(\tau_1,\ldots ,\tau_{n_i})$, $i=1,2$. Suppose that
$n_1=n_2+1$, and (17)--(19) for
$(c_2,\lambda_2,L^{(2)})$ are obtained from (17)--(19)
for $(c_1,\lambda_1,L^{(1)})$ by the substitution of
$\tau'_{n_1}\in k$ for $\tau_{n_1}$, see the Introduction. Then the
polynomial $\Phi_{c_2,\lambda_2,L^{(2)}}$ coincides with
$\Phi_{c_1,\lambda_1,L^{(1)}}|_{\{\tau_{n_1}=\tau'_{n_1}\}}$ up to
a nonzero factor from $\overline{k}[\tau_1,\ldots
,\tau_{n_2},Z_0,$ $Z_{\sigma+1},\ldots , Z_n]$. In particular, if
$\deg_{Z_0,Z_{\sigma+1},\ldots ,
Z_{n+1}}\Phi_{c_2,\lambda_2,L^{(2)}}=\deg_{Z_0,Z_{\sigma+1},\ldots ,
Z_{n+1}}$ $\Phi_{c_1,\lambda_1,L^{(1)}}$ then
$\Phi_{c_2,\lambda_2,L^{(2)}}$ coincides with
$\Phi_{c_1,\lambda_1,L^{(1)}}|_{\{\tau_{n_1}=\tau'_{n_1}\}}$ up to
a nonzero factor from $\overline{k}[\tau_1,$ $\ldots
,\tau_{n_2}]$.
}\par\medskip

\noindent {\bf PROOF} \quad Extending the ground field $k$ till
$k(\tau_1,\ldots ,\tau_{n_2})$
we shall suppose without loss of generality that $n_1=1$ and $n_2=0$. Let
${\Bbb A}^1(\overline{k})$ has the coordinate
function $\tau_1$.
Let ${\cal W}$ be the union of all the
irreducible over $\overline{k}$ components $E$ of
$(W\times{\Bbb A}^1(\overline{k}))\cap{\cal Z}(L_{1,s+1},\ldots ,
L_{1,\sigma})$ such that $E$ is not contained in any
hyperplane ${\cal Z}(\tau_1-z)$, $z\in\overline{k}$.
Then an arbitrary polynomial $h\in\overline{k}[X_0,\ldots , X_n,\tau_1]$
homogeneous with respect to $X_0,\ldots , X_n$ vanishes on
$W_{\lambda_1}(\overline{k(\tau_1)})$
if and only if $h$ vanishes on ${\cal W}$.

Besides that, $\dim
E=n-\sigma+1$ for every component $E$ of ${\cal W}$. Hence
$\dim{\cal W}\cap{\cal Z}(\tau_1-\tau'_1)
=n-\sigma$. The dimension of every irreducible component $E_1$ of
$(W\times{\Bbb A}^1(\overline{k}))\cap{\cal Z}(L_{1,s+1},\ldots ,
L_{1,\sigma})$ is at least $n-\sigma+1$.
In particular this is true for every irreducible component
$E_1\subset{\cal Z}(\tau_1-\tau'_1)$.
Obviously $(W\times{\Bbb A}^1(\overline{k}))\cap{\cal
Z}(L_{1,s+1},\ldots ,
L_{1,\sigma})\cap{\cal Z}(\tau_1-\tau'_1)=W_{\lambda_2}\times\{\tau'_1\}$
and $\dim W_{\lambda_2}=n-\sigma$.
Hence ${\cal W}\cap{\cal
Z}(\tau_1-\tau'_1)=W_{\lambda_2}\times\{\tau'_1\}$.

The element $\tau_1-\tau'_1$
does not divide $\Phi_{c_1,\lambda_1,L^{(1)}}$. Hence the
polynomial
$\Phi_{c_1,\lambda_1,L^{(1)}}|_{\{\tau_1=\tau'_1\}}$ $\ne
0$. Let us substitute in
$\Phi_{c_1,\lambda_1,L^{(1)}}$ the
elements $L^{(1)}_j(e_0,\ldots ,$ $e_r)$ for $Z_j$,
$j\in\{0,\sigma+1,\ldots , n\}$, and $e_{1,n+1}$ for $Z_{n+1}$. The
obtained polynomial vanishes on
$W_{\lambda_1}(\overline{k(\tau_1)})$ and, therefore, on ${\cal W}$.
Hence, if one substitutes in
$\Phi_{c_1,\lambda_1,L^{(1)}}|_{\{\tau_1=\tau'_1\}}$ the
elements $L^{(2)}_j(e_0,\ldots , e_r)$ for $Z_j$,
$j\in\{0,\sigma+1,\ldots , n\}$, and $e_{2,n+1}$ for $Z_{n+1}$, then the
obtained polynomial vanishes on
$W_{\lambda_2}$.
Therefore, the degree
$\deg_{Z_{n+1}}\Phi_{c_1,\lambda_1,L^{(1)}}|_{\{\tau_1=
\tau'_1\}}=\deg p'_v\deg W'$. This implies the required assertions
immediately. The lemma is proved.

\par\medskip\noindent{\bf LEMMA~13}\hspace{0.1em} {\it  Under conditions of
Lemma~12
let $\gamma\ge 0$ be the most
integer such that $(\tau_{n_1}-\tau'_{n_1})^\gamma$ divides
$\Phi_{c_1,\lambda_1,L^{(1)},*}$ (it exists by Lemma~2). Then
the polynomial $(\,\Phi_{c_1,\lambda_1,L^{(1)},*}/(\tau_{n_1}-
\tau'_{n_1})^\gamma\,)|_{\{\tau_{n_1}=\tau'_{n_1}\}}$ coincides
with  $\Phi_{c_2,\lambda_2,L^{(2)},*}$ up to a nonzero factor from
$k(\tau_1,\ldots ,\tau_{n_2},Z_0,Z_n)$.
Suppose additionally that
$\deg_{Z_0,Z_{\sigma+1},\ldots ,
Z_{n+1}}\Phi_{c_2,\lambda_2,L^{(2)}}=\deg_{Z_0,Z_{\sigma+1},\ldots ,
Z_{n+1}}$ $\Phi_{c_1,\lambda_1,L^{(1)}}$.
Then $\gamma=0$ and
$\Phi_{c_1,\lambda_1,L^{(1)},*}|_{\{\tau_{n_1}=\tau'_{n_1}\}}$ coincides
with  $\Phi_{c_2,\lambda_2,L^{(2)},*}$ up to a nonzero factor from
$k(\tau_1,\ldots ,\tau_{n_2})$.
}\par\medskip

\noindent {\bf PROOF} \quad  Similarly to the proof of Lemma~12 we shall
assume
without loss of generality that $n_2=0$ and $n_1=1$. Let ${\cal W}_1$ be
the union of all the
irreducible over $\overline{k}$ components $E$ of
$(W\times{\Bbb A}^1(\overline{k}))\cap{\cal Z}(L_{1,s+1},\ldots ,
L_{1,\sigma})\cap{\cal Z}(e^{(1)}_{\sigma+1},\ldots , e^{(1)}_{n-1})$
such that $E$ is not contained in any
hyperplane ${\cal Z}(\tau_1-z)$, $z\in\overline{k}$ and
$E\not\subset{\cal Z}(e_0,\ldots , e_r)$.
Then an arbitrary polynomial $h\in\overline{k}[X_0,\ldots , X_n,\tau_1]$
homogeneous with respect to $X_0,\ldots , X_n$ vanishes on
$W_{\lambda_1}(\overline{k(\tau_1)})\cap
{\cal Z}(e^{(1)}_{\sigma+1},\ldots , e^{(1)}_{n-1})\setminus{\cal
Z}(e_0,\ldots , e_r)$
if and only if $h$ vanishes on ${\cal W}_1$.

By property 2) from (*) the dimension
$\dim(W_{\lambda_j}(\overline{k(\tau_1,\ldots ,\tau_{n_j})}\cap
{\cal Z}(e^{(1)}_{\sigma+1},\ldots,$ $e^{(1)}_{n-1})\setminus{\cal
Z}(e_0,\ldots, e_r))=1$, $j=1,2$. Hence $\dim{\cal W}_1=2$.
The dimension of every irreducible component of
$(W\times{\Bbb A}^1(\overline{k}))\cap{\cal Z}(L_{1,s+1},\ldots ,
L_{1,\sigma})\cap{\cal Z}(e^{(1)}_{\sigma+1},\ldots , e^{(1)}_{n-1})$
is at least $2$. Therefore, similarly to the proof of
Lemma~12
we get $({\cal W}_1\setminus{\cal Z}(e_0,\ldots ,
e_r))\cap{\cal Z}(\tau_1-\tau'_1)=
(W_{\lambda_2}\cap
{\cal Z}(e^{(1)}_{\sigma+1},\ldots, e^{(1)}_{n-1})\setminus{\cal
Z}(e_0,\ldots, e_r))\times\{\tau'_1\}$.

By Lemma~2 the polynomial
$\Phi_{c_1,\lambda_1,L^{(1)},*}/(\tau_1-\tau'_1)^\gamma$
is nonzero.
Let us substitute $Z_0=L^{(1)}_0(e_0,\ldots , e_r)$,
$Z_n$ $=L^{(1)}_n(e_0,\ldots , e_r)$, $Z_{n+1}=e_{1,n+1}$ in \\
$\Phi_{c_1,\lambda_1,L^{(1)},*}/(\tau_1-\tau'_1)^\gamma$.
By Lemma~3 the obtained polynomial vanishes on
$W_{\lambda_1}(\overline{k(\tau_1)})\cap
{\cal Z}(e^{(1)}_{\sigma+1},\ldots , e^{(1)}_{n-1})\setminus{\cal
Z}(e_0,\ldots , e_r)$ and, therefore, on ${\cal W}_1$.
Hence if one substitutes
$Z_0=L^{(2)}_0(e_0,\ldots , e_r)$, $Z_n=L^{(2)}_n(e_0,\ldots ,e_r)$ in
$(\Phi_{c_1,\lambda_1,L^{(1)},*}/(\tau_1-\tau'_1)^\gamma)|_{\{\tau_1=\tau'_1\}}$
then the obtained polynomial vanishes on
$W_{\lambda_2}\cap
{\cal Z}(e^{(1)}_{\sigma+1},\ldots, e^{(1)}_{n-1})\setminus{\cal
Z}(e_0,\ldots, e_r)$. Therefore, by Lemma~3 the polynomial
$\Phi_{c_2,\lambda_2,L^{(2)},+}$ divides \\
$(\Phi_{c_1,\lambda_1,L^{(1)},*}/(\tau_1-\tau'_1)^\gamma)|_{\{\tau_1=\tau'_1\}}$.
By Lemma~2 the degree
$\deg_{Z_{n+1}}\Phi_{c_1,\lambda_1,L^{(1)},*}/(\tau_{n_1}$ $-
\tau'_{n_1})^\gamma=\deg_{Z_{n+1}}
\Phi_{c_2,\lambda_2,L^{(2)},*}=\deg p'_v\deg W'$.
This implies the first assertion of the lemma.
The second assertion follows from the second one of Lemma~12
immediately. The lemma is proved.





\medskip Suppose that the field $k$ is supplied with a real structure
(it induces the real structure on $\overline{k}$), see \cite{6}.
Let $\varepsilon_i>0$ be an infinitesimal with respect to the
field $k_i=k(\varepsilon_1,\ldots , \varepsilon_{i-1})$ for every
integer $i\ge 1$. The field $\overline{k_i}$ is supplied with the
real structure, see \cite{6}. The mapping of the standard part
$\mbox{\rm st}_i\, :\,{\Bbb
P}^n(\overline{k_i})\rightarrow{\Bbb P}^n(\overline{k})$ is
defined (we shall use the same notation for all $n$, it will not
lead to an ambiguity), see \cite{6}. Namely, for every
$z\in{\Bbb P}^n(\overline{k_i})$ the element $\mbox{\rm
st}_i(z)\in{\Bbb P}^n(\overline{k})$ is infinitely
close to $z$ ($\mbox{\rm st}_i(z)$ is uniquely defined
by this property). For every $z\in{\Bbb A}^n(\overline{k_i})$
such that every the coordinate of $z$ is not infinitely large with
respect to the field $k$ the element $\mbox{\rm
st}_i(z)\in{\Bbb A}^n(\overline{k})$. In particular,
if $z\in\overline{k_i}$ is not infinitely large with respect to
the field $k$ then $\mbox{\rm st}_i(z)\in\overline{k}$.

\par\medskip\noindent{\bf LEMMA~14}\hspace{0.1em} {\it  In the notation from
the Introduction suppose that the
point $(c',\lambda,$ $L')\in{\Bbb A}^\nu(\overline{k})$ satisfies
properties 1) and 2) from (*). Let $i\ge 1$ be an integer. Suppose
that $(c'',\lambda',L'')\in{\Bbb A}^\nu(\overline{k_i})$, and
$\mbox{\rm st}_i((c'',\lambda',L''))=(c',\lambda,L')$.
Let us replace in (8)--(14), (7),
(10) the elements $c'$,
$\lambda$, $L'$, and the field $k$ by $c''$, $\lambda'$, $L''$,
and the field $\overline{k_i}$ respectively
(we do not assume now that (C) is fulfilled for
$(c'',\lambda',L'')$ over any field). Then  $\dim
W'(c'',\lambda')=\dim W'(c'',\lambda',L'')=n-\sigma$, $\dim
W'(c'',\lambda',*)=\dim W'(c'',\lambda',L'',*)=1$,
the morphisms $p_{\lambda',L''}$,
$p_{\lambda',L'',*}$ are dominant, and $\mbox{\rm
st}_i(W'(c'',\lambda'))\supset W'(c',\lambda)$,
$\mbox{\rm st}_i(W'(c'',\lambda',L''))\supset
W'(c',\lambda,L')$,
$\mbox{\rm st}_i(W'(c'',\lambda',*))\supset
W'(c',\lambda,*)$,
$\mbox{\rm
st}_i(W'(c'',\lambda',$ $L'',*))\supset
W'(c',\lambda,L',*)$.
}\par\medskip

\noindent {\bf PROOF} \quad Let $\lambda'=(L'_{s+1},\ldots ,
L'_\sigma)$, $L''=(L''_0,L''_{\sigma+1},\ldots
, L''_n)$, and $e''_j=L''_j(e_0,$ $\ldots , e_r)$
for all $j$. Properties 1)
and 2) for $(c',\lambda,L')$ imply
that $(c',\lambda,L')$ satisfies (D), see the beginning of the section.
Therefore, see the last observation before Lemma~1, the point
$(c'',\lambda',L'')$
also satisfies (D). Therefore, the morphisms
$p_{\lambda',L''}$ and $p_{\lambda',L'',*}$ are dominant.


Therefore, $\dim W'(c'',\lambda')=\dim
W'(c'',\lambda',L'')=n-\sigma$, $\dim W'(c'',\lambda',*)=
\dim W'(c'',\lambda',L'',*)=1$. For
every $y\in W_{\lambda'}$ which is not infinitely close to
${\cal Z}(e'_0,e'_{\sigma+1},\ldots , e'_n)$ with respect to
the field $k$ the equality $p_{c',\lambda,L'}(\mbox{\rm
st}_i(y))=\mbox{\rm
st}_i(p_{c',\lambda,L'}(y))$ holds. Hence using the
transfer principle or directly one obtains that $\mbox{\rm
st}_i(W'(c'',\lambda',L''))$ contains a dense with
respect to the Zariski topology subset of $W'(c',\lambda,L')$. By
Lemma~5 \cite{13} every irreducible component of $\mbox{\rm
st}_i(W'(c'',\lambda',L''))$ has dimension $n-\sigma$.
Therefore, $\mbox{\rm
st}_i(W'(c'',\lambda',L''))\supset W'(c',\lambda,L')$.
The inclusions $\mbox{\rm
st}_i(W'(c'',\lambda'))\supset W'(c',\lambda)$,
$\mbox{\rm st}_i(W'(c'',\lambda',*))\supset
W'(c',\lambda,*)$ and
$\mbox{\rm st}_i(W'(c'',\lambda',L'',$ $*))\supset
W'(c',\lambda,L',*)$ are proved in the similar way. The lemma is
proved.







\section{Some facts from linear algebra}\label{s2}

By definition in any characteristic
of the ground field there exists only one
$(0\times 0)$--ma\-t\-rix, and its determinant is $1$.
Further, for every integer $m_1\ge 0$ there is only one $(m_1\times
0)$--ma\-t\-rix
(respectively $(0\times m_1)$--ma\-t\-rix). Let $m_2\ge 0$ be an integer.
If $A_1$ is an $(m_1\times m_2)$--ma\-t\-rix and
$A_2$ is the $(m_2\times 0)$--ma\-t\-rix then $A_1A_2$ is the
$(m_1\times 0)$--ma\-t\-rix.  If $A_1$ is the $(0\times m_1)$--ma\-t\-rix and
$A_2$ is an $(m_1\times m_2)$--ma\-t\-rix then $A_1A_2$ is the
$(0\times m_2)$--ma\-t\-rix.

According to \cite{14} for every integers
$m_1\ge m_2\ge 0$ one can construct within the time polynomial in $m_1$ a
family of
$((m_1-m_2)\times m_1)$--ma\-t\-ri\-ces $A_i$, $1\le i\le m_2(m_1-m_2)+1$
(they depend on $m_1$ and $m_2$)
satisfying the following conditions.
\begin{itemize}
\item All the ma\-t\-ri\-ces $A_i$ have integer coefficients of length
$O(m_1)$ (here and below
the constant of $O(\ldots)$ is absolute) and the rank
$\mbox{\rm rank}(A_i)=m_1-m_2$.
\item For every $(m_2\times m_1)$--ma\-t\-rix $A$ with coefficients from a
field of zero--characteristic (actually in \cite{13} the case of
arbitrary characteristic is considered)
with $\mbox{\rm rank}(A)=m_2$ there exists $1\le i\le
m_2(m_1-m_2)+1$ such that the determinant
$$
\mbox{\rm det}\left(
\begin{array}{l}
A\\
A_i
\end{array}
\right)\ne 0.
$$
\end{itemize}
Hence one can construct within the time polynomial in $m_1$ an
$(m_1\times m_1)$--ma\-t\-rix $B_i$ with integer coefficients
of length $O(m_1)$
such that $A_iB_i=(0,\Delta_i)$ where $\Delta_i$ is a diagonal
$(m_1-m_2)\times(m_1-m_2)$--ma\-t\-rix, all the nonzero
coefficients of $\Delta_i$
coincide, and the determinant
$\mbox{\rm det} \Delta_i\ne 0$. Let us represent
$B_i=(C_i,C'_i)$ where
$C_i$ (respectively $C'_i$) is the ma\-t\-rix with $m_2$
(respectively $m_1-m_2$) columns.
We shall denote also $A_i$, $B_i$, $C_i$ $C'_i$ by
$A_{m_1,m_2,i}$, $B_{m_1,m_2,i}$, $C_{m_1,m_2,i}$ $C'_{m_1,m_2,i}$
respectively if the dependence on $m_1$, $m_2$ is important.



\par\medskip\noindent{\bf LEMMA~15}\hspace{0.1em} {\it  Let $m_1$, $m_2$,
$m_3$ be arbitrary integers such that
$m_1\ge m_3$, $m_2\ge m_3$, $m_3\ge 0$. Let $A=(a_{i,j})$
be a $(m_2\times m_1)$-ma\-t\-rix
with coefficients from $\overline{k_\alpha}$ where the field $k_\alpha=
k(\varepsilon_1,\ldots ,\varepsilon_{\alpha})$, see Section~1, $\alpha\ge 0$.
Suppose that $\mbox{\rm rank}(A)=m_3$. Let $\Delta$ be an
$(m_3\times m_3)$--minor of
$A$ (here it is up to a sign the
determinant of the ma\-t\-rix $(a_{i_v,j_w})$ for some integers
$1\le i_1<\ldots < i_{m_3}\le m_2$, $1\le j_1<\ldots < j_{m_3}\le m_3$)
such that $\Delta\ne 0$ and for any other $(m_3\times m_3)$--minor
$\Delta_1$ of
$A$ the element $\Delta_1/\Delta$ is not infinitely large with respect to
the field $k$.
Then the following assertions hold
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item There are integers $i,j$ such that
$1\le j\le m_3(m_1-m_3)+1$, $1\le i\le m_3(m_2-m_3)+1$,
the determinant $\Delta_{i,j}=\mbox{\rm det}
(C_{m_2,m_3,i}^TAC_{m_1,m_3,j})\ne 0$
(here $T$ denotes transposition of ma\-t\-ri\-ces), and
$\Delta/\Delta_{i,j}$ is not infinitely large with respect to the field $k$
(hence the standard part $\mbox{\rm
st}_\alpha(\Delta/\Delta_{i,j})$ is a nonzero element of
$\overline{k}$).
\item Suppose that $m_2=m_3$. Then there is $1\le j\le m_3(m_1-m_3)+1$
such that the determinant $\Delta_j=\mbox{\rm det}
(AC_{m_1,m_3,j})\ne 0$ and
$\Delta/\Delta_j$ is not infinitely large with respect to the field $k$.
Hence the standard part $\mbox{\rm
st}_\alpha(\Delta/\Delta_j)$ is a nonzero element of
$\overline{k}$.
\item Suppose that $m_2=m_3$. Then there exists an $(m_1\times
m_1)$--ma\-t\-rix $G=(g_{i,j})$ such that every the coefficient
$g_{i,j}\in\overline{k_\alpha}$ is not infinitely large with respect to the
field $k$;
$\mbox{\rm
det}(G)\ne 0$ and $\mbox{\rm
det}(G)$ is not an infinitesimal with respect to $k$;
the ma\-t\-rix $AG=(A',0)$, where $A'$
has $m_3$ columns.
The determinant $\mbox{\rm det}(A')\ne 0$,
and $\Delta/\mbox{\rm det}(A')$ is not infinitely large
with respect to the field $k$.
\end{enumerate}
}\par\medskip

\noindent {\bf PROOF} \quad We shall assume without loss of generality
that $m_3\ge 1$.
Let us show that (ii) implies (i). Indeed denote
by $A'''$ the subma\-t\-rix of $A$
constituted from the rows with induces $i_1,\ldots , i_{m_3}$.
Let us apply (ii) to $A'''$. Then $A'''C_{m_1,m_3,j}$
is a subma\-t\-rix of $AC_{m_1,m_3,j}$, the determinant $\mbox{\rm
det}(A'''C_{m_1,m_3,j})\ne 0$,
and $\Delta/\mbox{\rm det}(A'''C_{m_1,m_3,j})$ is not infinitely
large with respect to
the field $k$.
So to prove (i) it is sufficient to apply (ii) for the second time
to the ma\-t\-rix $(AC_{m_1,m_3,j})^T$. The required assertion is  proved.

Let us prove (iii). Notice that, see \cite{6},
for all $z_1,z_2\in\overline{k_\alpha}$ such that $z_1\ne 0$ and
$z_2/z_1$ is not infinitely large with respect to the field $k$
there is $c\in \overline{k}$ such that $(z_2-cz_1)/z_1$ is an infinitesimal
with respect to the field $k$.
Set $Q=A$
as the base of the recursion.
We shall describe a recursion by an
$(m_3\times m_1)$--ma\-t\-rix $Q=(q_{i,j})$ with
coefficients from $\overline{k_\alpha}$. At each step
of this recursion we use the
elementary transformations of columns of $Q$ over the field
$\overline{k_\alpha}$
and permutations of rows (the permutations of rows are used only for
convenience of notation)
such that all the coefficients of the ma\-t\-ri\-ces corresponding to these
transformations of columns are not infinitely large with respect to the
field $k$.
At the end of the recursion we shall obtain a ma\-t\-rix
$Q$ which has the form $(A',0)$. This is sufficient for the proof of
assertion (iii).

Let $0\le i<m_3$ be an integer.
The recursive assumption is that at the beginning of $(i+1)$--th step
$$
Q=Q_i=\left(
\begin{array}{ll}
Q'_i,&0\\
Q''_i&Q'''_i
\end{array}
\right),
\eqno (28)
$$
where $Q'_i=(q'_{i,j})$ is a
$(i\times i)$--ma\-t\-rix, coefficients
$q'_{1,1},\ldots , q'_{i,i}$ are nonzero, $q'_{i,j}=0$ if $j>i$, and
for all
$j,j_1,i_1$ satisfying the inequalities
$0\le j\le i$, $j\le j_1\le m_1$, $1\le i_1\le m_2$ the quotient
$q_{i_1,j_1}/q_{j,j}$ is not infinitely large with respect to the field $k$.
The ma\-t\-rix $Q'''_i\ne 0$.

Let us describe the general step of the recursion. Let us choose a nonzero
coefficient $z$ of the
ma\-t\-rix
$Q'''_i$ such that for any other coefficient $z'$ of $Q'''_i$ the element
$z'/z$ is not infinitely large with respect to the field $k$.
Using the permutations of rows and columns we shall suppose without loss of
generality that $z=q_{i+1,i+1}$.
Now by an appropriate elementary transformation of columns we transfer
$Q'''_i$
to $\widetilde{Q}_i$
where the first row of $\widetilde{Q}_i$ is $(q_{i+1,i+1},0,\ldots , 0)$.
By definition the image the ma\-t\-rix $Q_i$
after all these transformations is $Q_{i+1}$
(to define $Q'_{i+1}$, $Q''_{i+1}$, $Q'''_{i+1}$
one need to replace $i$ by $i+1$ in (28)).
Since rank $\mbox{\rm rank}(A)=m_3$ the ma\-t\-rix $Q'''_{i+1}\ne
0$ or $i+1=m_3$. Thus, if $i+1<m_3$ we proceed to the next step of the
recursion.
If $i+1=m_3$ we set $A'=Q'_{m_3}$. The existence of $G$ and $A'$ is proved.
We have $A=(A',0)G^{-1}$. Hence the minor $\Delta=\mbox{\rm
det}(A')\mbox{\rm det}(G')$ for an
$(m_3\times m_3)$--subma\-t\-rix $G'$ of $G^{-1}$. This implies the last
assertion. Thus, (iii) is proved.

Let us prove (ii). We shall use the notation from the proof of (iii).
Let $G_0=\mbox{\rm st}_\alpha(G)=(\mbox{\rm
st}_\alpha(g_{i,j}))$ be
the standard part of the ma\-t\-rix $G$
(it is ma\-t\-rix with coefficients from $\overline{k}$).
Hence by (iii) the determinant $\mbox{\rm det}(\mbox{\rm
st}_\alpha(G))\ne 0$.
By (iii) and the formula for the adjoint ma\-t\-rix $(E+H)G=G_0$
where $E$ is identity
ma\-t\-rix, and all the nonzero coefficients of $H$ are elements from
$\overline{k_\alpha}$
which are infinitely small with respect to the field $k$.
There is $1\le i\le m_3(m_1-m_3)+1$ such that
$$
\mbox{\rm det}
\left(\begin{array}{l}
(A',0)G_0^{-1}\\
A_{m_1,m_3,i}
\end{array}
\right)=c\mbox{\rm det}(A'),
$$
where $0\ne c\in\overline{k}$.
Then
$$
\mbox{\rm det}
\left(\begin{array}{l}
A\\
A_{m_1,m_3,i}
\end{array}
\right)=
\mbox{\rm det}
\left(\begin{array}{l}
(A',0)G_0^{-1}(E+H)\\
A_{m_1,m_3,i}
\end{array}
\right)=c_1\mbox{\rm det}(A'),
$$
where $0\ne c_1\in\overline{k_\alpha}$, and $c_1$, $c_1^{-1}$ are not
infinitely large with respect to the
field $k$. Therefore, $\mbox{\rm
det}(AC_{m_1,m_3,i})=c_2\mbox{\rm det}(A')$,
where $0\ne c_2\in\overline{k_\alpha}$, and $c_2$, $c_2^{-1}$ are not
infinitely large with respect to the
field $k$. By (iii) $\Delta/\mbox{\rm det}(A')$ is not infinitely
large with respect to the field $k$.
This implies (ii). The lemma is proved.

\medskip
Let $(c_1,\lambda_1,L^{(1)})$ be given by (17)--(19),
see the Introduction. Now
we shall suppose without loss of generality that $n_1$
is minimal possible and hence $n_1\le\nu$, see the Introduction,
is bounded from above by a polynomial in
$rd(d'+1)^n$, see the Introduction.
Consider systems
of polynomial equations and inequalities
with squares of absolute values
containing generic homogeneous polynomials.
Recall that in \cite{13} a method for solving such systems is suggested
(only generic linear forms occur in \cite{13}).
Now we are going to strengthen this method
for the case of a system satisfying the following conditions.
\begin{enumerate}\renewcommand{\labelenumi}{\arabic{enumi})}
\item The coefficients of all the equalities and inequalities of this system
are elements from a finite extension
of some $k_\alpha=k(\varepsilon_1,\ldots ,
\varepsilon_{\alpha})$, see Section~1, where $\alpha=O(1)$.
\item The indeterminates of this system are $\tau_1,\ldots ,\tau_{n_1}$;
$X_{i,j}$, $0\le j\le n$,
$1\le i\le\nu_2$, where $\nu_2=O(1)$ (is considered as a constant,
$\nu_2<n$); and some other $O(n)$ variables.  Denote them, say,
by $Y_\gamma$, $\gamma\in\Gamma$.
\item This system contains the inequality $\sum_{1\le i\le
n_1}|\tau_i|^2\le\varepsilon_1$.
All other equations and inequalities of this system have coefficients from a
finite extension $k'$ of
$k(\varepsilon_2,\ldots ,\varepsilon_\alpha)$.
\item Every equation or inequality of this system, apart of
$\sum_{1\le i\le n_1}|\tau_i|^2\le\varepsilon_1$, is equivalent to
$A\ge 0$ or $A=0$ where $A$ belongs to the least class of functions
satisfying the following properties. This class is
closed under the operations of the addition of two functions,
multiplication of two functions, multiplication of a function to a scalar
from $\overline{k(\varepsilon_2,\ldots ,\varepsilon_\alpha)}$.
The function $B\mapsto|B|^2$ belongs to this class. Finally, this class
contains all the
scalars from $\overline{k(\varepsilon_2,\ldots ,\varepsilon_\alpha)}$;
the functions $X_{i,j}$, $0\le j\le n$,
$1\le i\le\nu_2$; $e_{1,n+1}(X_{i,0},\ldots , X_{i,n})$, see (20);
$L_{1,j}(X_{i,0},\ldots , X_{i,n})$, $s+1\le j\le\sigma$;
$e^{(1)}_w(X_{i,0},\ldots , X_{i,n})$, $w\in\{0,\sigma+1,\ldots ,n\}$,
where $1\le i\le\nu_2$; and $Y_\gamma$, $\gamma\in\Gamma$.
\end{enumerate}
Denote the considered system by $(\star)$ for brevity.

Now suppose that for every $1\le i\le \nu_2$
the algebraic varieties $V^{(i)}$, $W^{(i)}$
are similar (and given in the analogous way) to $V$, $W$ from the
Introduction.
We are interested whether system $(\star)$ has a solution $\tau_w=\tau^*_w$,
$X_{i,j}=x^*_{i,j}$, $Y_\gamma=y^*_\gamma$ such that all $\tau^*_w$,
$x^*_{i,j}$,
$y^*_\gamma$ are elements from $\overline{k_\alpha}$, and
$$
(x^*_{i,0}:\ldots : x^*_{i,n})\in W^{(i)}(\overline{k_\alpha}), \quad 1\le
i\le\nu_2.
\eqno (29)
$$

The total number of variables of system $(\star)$
may be $\ge{\cal P}(rd(d'+1)^n)$ for a polynomial ${\cal
P}$. Hence applying the general algorithms
for constructing the described
solution of such systems, see \cite{6} and Theorem~2 \cite{12},
we get an estimate polynomial in
$(d')^{{\cal P}(rd(d'+1)^n)}$ for the working time
of the decision algorithm and the algorithm
for constructing a solution of system $(\star)$.
But the suggested method allows to reduce the working time.


Namely, let $1\le n_3\le\min\{n_1,\nu_2(n-s+2)\}$ be an integer.
Let $B_{n_1,n_3,j}$,
$C_{n_1,n_3,j}$, $1\le j\le n_3(n_1-n_3)+1$, be the ma\-t\-ri\-ces
defined at the beginning of the section.
Put
$$
\left(\begin{array}{l}
T_{1,n_3,j}\\
\vdots\\
T_{n_1,n_3,j}
\end{array}
\right)=B^{-1}_{n_1,n_3,j}
\left(\begin{array}{l}
\tau_1\\
\vdots\\
\tau_{n_1}
\end{array}
\right).
$$
For an arbitrary linear form $H=\sum_{1\le i\le n_1}h_i\tau_i$ with respect
to $\tau_1,\ldots ,\tau_{n_1}$
represent $H=\sum_{1\le i\le n_1}h_{i,n_3,j}\times$ $T_{i,n_3,j}$
(hence all $h_{i,n_3,j}$ are linear combinations of $h_i$ with
integer coefficients).
By definition set
$$
H_{n_3,j}=\sum_{1\le i\le n_3}h_{i,n_3,j}T_{i,n_3,j}.
$$
Thus, $e_{1,n+1,n_3,j}$; $L_{1,w_1,n_3,j}$, $s+1\le w_1\le \sigma$;
$L^{(1)}_{w_2,n_3,j}$, $w_2\in\{0,\sigma+1,\ldots ,n\}$, are given.

Let us replace in system $(\star)$ in all the equations and inequalities
$e_{1,n+1}$; $L_{1,w_1}$;
$e^{(1)}_{w_2}$ by
$e_{1,n+1,n_3,j}$; $L_{1,w_1,n_3,j}$, $s+1\le w_1\le \sigma$;
$e^{(1)}_{w_2,n_3,j}$, $w_2\in\{0,\sigma+1,\ldots ,n\}$ respectively.
Let us replace the inequality $\sum_{1\le i\le
n_1}|\tau_i|^2\le\varepsilon_1$ by
$\sum_{1\le i\le n_3}|T_{i,n_3,j}|^2\le\varepsilon_1$. Let us replace the
indeterminates
$\tau_1,\ldots ,\tau_{n_1}$ of system $(\star)$ by
$T_{1,n_3,j},\ldots , T_{n_3,n_3,j}$.
Denote the obtained system by $(\star,n_3,j)$.
Hence system $(\star,n_3,j)$ has $O(n)$ indeterminates $T_{1,n_3,j},\ldots ,
T_{n_3,n_3,j}$; $X_{i,j}$, $0\le j\le n$,
$1\le i\le\nu_2$; $Y_\gamma$, $\gamma\in\Gamma$.



Similarly to system $(\star)$ we shall consider a solution
$T_{i,n_3,j}=T'_{i,n_3,j}$,
$X_{i,j}=x^*_{i,j}$, $Y_\gamma=y^*_\gamma$ such that all $T'_{i,n_3,j}$,
$x^*_{i,j}$,
$y^*_\gamma$ are elements from $\overline{k_\alpha}$ and (29) hold.



\par\medskip\noindent{\bf LEMMA~16}\hspace{0.1em} {\it  System $(\star)$ has
a solution
over the field $\overline{k_\alpha}$ satisfying
(29) if and only if
there exist $0\le n_3\le\min\{n_1,\nu_2(n-s+2)\}$
and $1\le j\le n_3(n_1-n_3)+1$ such that system $(\star,n_3,j)$
has a solution over the field $\overline{k_\alpha}$ satisfying (29).
Besides that, using the proof of the lemma for a given solution of system
$(\star,n_3,j)$
over the field $\overline{k_\alpha}$ satisfying (29)
one can construct explicitly the solution of system $(\star)$
over the field $\overline{k_\alpha}$ satisfying (29).
}\par\medskip

\noindent {\bf PROOF} \quad Suppose that system
$(\star,n_3,j)$ has a solution
over the field $\overline{k_\alpha}$ satisfying (29)
for some $0\le n_3\le\min\{n_1,\nu_2(n-s+2)\}$
and $1\le j\le n_3(n_1-n_3)+1$.
Set $T'_{i,n_3,j}=0$ for all $n_3+1\le i\le n_1$.
Then the linear system $T_{i,n_3,j}=T'_{i,n_3,j}$, $1\le i\le n_1$, with
respect to
$\tau_1,\ldots ,\tau_{n_1}$ has a solution $\tau_i=\tau^*_i\in
\overline{k(\varepsilon_2,\ldots ,\varepsilon_\alpha)}$, $1\le i\le n_1$.
Hence $\sum_{1\le i\le n_1}|\tau_i|^2\le c\varepsilon_1$
for an integer $c\ne 0$ with length $O(n_1)$.
Now to get a solution of system $(\star)$ satisfying (29)
it is sufficient to perform the replacement $\varepsilon_1\mapsto
c^{-1}\varepsilon_1$. Thus, system $(\star)$
has a solution over the field $\overline{k_\alpha}$ satisfying (29).

Conversely, suppose that system $(\star)$
has a solution satisfying (29).
By 3) and the transfer principle , see \cite{3},
one can suppose without loss of generality that
all $\tau^*_w$, $x^*_{i,j}$, $y^*_\gamma$ are elements from
$\overline{k(\varepsilon_2,\ldots ,\varepsilon_\alpha})$.
Set $e_{1,n+1}(x^*_{i,0},\ldots ,  x^*_{i,n})=e^*_{1,n+1,i}$,
$L_{1,j}(x^*_{i,0},\ldots ,  x^*_{i,n})=L^*_{1,j,i}$,
$e^{(1)}_w(x^*_{i,0},\ldots ,  x^*_{i,n})=e^{(1,*)}_{w,i}$
for all $j,w,i$. Hence  $e^*_{1,n+1,i}$, $L^*_{1,j,i}$, $e^{(1,*)}_{w,i}$
are linear forms in $\tau_1,\ldots ,\tau_{n_1}$ with coefficients from
$\overline{k(\varepsilon_2,\ldots ,\varepsilon_n)}$.
Let $A$ be the $((n-s+2)\nu_2\times n_1)$--ma\-t\-rix constituted of the
coefficients from
$\overline{k(\varepsilon_2,\ldots ,\varepsilon_n)}$ of the
linear forms $e^*_{1,n+1,i}$, $1\le i\le\nu_2$; $L^*_{1,j,i}$, $s+1\le
j\le\sigma$,
$1\le i\le\nu_2$; $e^{(1,*)}_{w,i}$, $w\in\{0,\sigma+1,\ldots ,n\}$, $1\le
i\le\nu_2$.
Set $n_3=\mbox{\rm rank}(A)$. Hence $0\le
n_3\le\min\{n_1,\nu_2(n-s+2)\}$.

Let us apply Lemma~15 (i)
to the ma\-t\-rix $A$ with $m_1=n_1$, $m_2=\nu_2(n-s+2)$,
$m_3=n_3$. Thus the ma\-t\-ri\-ces $B_{n_1,n_3,j}$ and $C_{n_1,n_3,j}$
are defined by this lemma. In what follows $j$ is fixed in the proof.

Set $T^*_{i,n_3,j}=T_{i,n_3,j}(\tau^*_1,\ldots ,\tau^*_{n_1})$
for all $1\le i\le n_1$.
Let us define linear forms $\widetilde{e}_{1,n+1,i}$,
$\widetilde{L}_{1,w_1,i}$, $\widetilde{e}^{(1)}_{w_2,i}$
by the equalities  $\widetilde{e}_{1,n+1,i}(T_{1,n_3,j},\ldots ,
T_{n_1,n_3,j})=e^*_{1,n+1,i}$,
$\widetilde{L}_{1,w_1,i}(T_{1,n_3,j},\ldots , T_{n_1,n_3,j})=L^*_{1,w_1,i}$,
$\widetilde{e}^{(1)}_{w_2,i}(T_{1,n_3,j},\ldots ,
T_{n_1,n_3,j})=e^{(1,*)}_{w_2,i}$.
Put $e^*_{1,n+1,i}(\tau^*_1,\ldots ,\tau^*_{n_1})=l^*_{1,n+1,i}$,
$L^*_{1,w_1,i}(\tau^*_1,\ldots ,\tau^*_{n_1})=l^*_{1,w_1,i}$,
$e^{(1,*)}_{w_2,i}(\tau^*_1,\ldots ,\tau^*_{n_1})=l^{(1,*)}_{w_2,i}$
for all $w_1,w_2,i$.

Consider the linear system relatively to the
variables
$T_{1,n_3,j},\ldots ,$ $T_{n_1,n_3,j}$
$$
\left\{
\begin{array}{ll}
\widetilde{e}_{1,n+1,i}(T_{1,n_3,j},\ldots ,
T_{n_1,n_3,j})=l^*_{1,n+1,i}, \\
\widetilde{L}_{1,w_1,i}(T_{1,n_3,j},$ $\ldots ,
T_{n_1,n_3,j})=l^*_{1,w_1,i},& s+1\le w_1\le\sigma,\;1\le i\le\nu_2,\\
\widetilde{e}^{(1)}_{w_2,i}(T_{1,n_3,j},\ldots ,
T_{n_1,n_3,j})=l^{(1,*)}_{w_2,i},& w_2\in\{0,\sigma+1,\ldots ,n\},
\;1\le i\le\nu_2.
\end{array}\right.
\eqno (30)
$$
System (30)
has a solution $T_{i,n_3,j}=T^*_{i,n_3,j}$, $1\le i\le n_1$.
Let us represent the ma\-t\-rix $AB_{n_1,n_3,j}=(A',A'')$
where $A'$ has $n_3$ columns.
By Lemma~15 we have $A''=A'Z$ where
$Z=(z_{i,w})$, and every the coefficient
$z_{i,w}\in
\overline{k(\varepsilon_2,\ldots ,\varepsilon_\alpha)}$
is not infinitely large with respect to the field $k$.
Therefore, there is also a solution of linear system (30)
$T_{i,n_3,j}=T'_{i,n_3,j}$, $1\le i\le n_3$, and $T_{i,n_3,j}=0$,
$n_3+1\le i\le n_1$,
where all $T'_{i,n_3,j}=T^*_{i,n_3,j}+\sum_{n_3+1\le w\le n_1}z_{i,w}
T^*_{w,n_3,j}$. Hence system $(\star,n_3,j)$
has a solution over the field $\overline{k_\alpha}$ satisfying (29).
The lemma is proved.

\par\medskip\noindent{\bf REMARK~11}\hspace{0.1em} {\it
In what follows applying Lemma~16 for the considered $j$ and $n_3$
we shall decide whether system $(\star,n_3,j)$ has a
solution and construct this solution if it exists using Theorem~2 \cite{12}
and Theorem~2 \cite{11}.
}\par\medskip






\section{A decision algorithm for property (*)}\label{s3}

In this section a point $(c_1,\lambda_1,L^{(1)})$ satisfying (**) is given,
see the Introduction.
The following observations will be used in the algorithm
for assertion (e) of
Theorem~2 described in this  section and Section~6.
Suppose that $k_i\subset K$ for some $i$
\begin{itemize}
\item Suppose that $(c_4,\lambda_4,L^{(4)})\in{\cal
L}_{c_1,\lambda_1,L^{(1)}}\otimes\overline{k_i}$
satisfies conditions of Lemma~1 (in place of $(c_0,\lambda_0,L^{(0)})$).
Then using the algorithms from
\cite{5}, \cite{11} one can compute the number of elements
$\#(e_{4,n+1}/e^{(4)}_0)(A_{\lambda_4,L^{(4)}})$. In particular, this can be
done if
conditions 1), 2) from (*) hold for
$(c_4,\lambda_4,L^{(4)})$.

\item Suppose that $(c_4,\lambda_4,L^{(4)})\in{\cal
L}_{c_1,\lambda_1,L^{(1)}}\otimes\overline{k_i}$
satisfies conditions of Lemma~2 (in place of $(c_0,\lambda_0,L^{(0)})$).
Then using the
algorithm from \cite{12} one can compute $\deg W'(c_4,\lambda_4,L^{(4)})$.
In particular, this can be done if
conditions 1)--3) from (*) hold for
$(c_4,\lambda_4,L^{(4)})$.

\item Suppose that $(c_4,\lambda_4,L^{(4)})\in{\cal
L}_{c_1,\lambda_1,L^{(1)}}\otimes\overline{k_i}$
satisfies conditions of Lemma~2 (in particular, this is true if
conditions 1)--3) from (*) hold for
$(c_4,\lambda_4,L^{(4)})$),
and $\deg W'(c_4,\lambda_4,L^{(4)})=
\deg W'(c_1,\lambda_1,L^{(1)})$. Then, see Definition~4 and Lemma~3,
by the algorithms from \cite{5}, \cite{11},
\cite{12} one can compute $W'(c_4,\lambda_4,L^{(4)},*)$ and after that
$\deg W'(c_4,\lambda_4,L^{(4)},*)$.

\item Suppose that $(c_4,\lambda_4,L^{(4)})\in{\cal
L}_{c_1,\lambda_1,L^{(1)}}\otimes\overline{k_i}$
satisfies conditions of Lemma~2 and $\deg W'(c_4,\lambda_4,L^{(4)})=
\deg W'(c_1,\lambda_1,L^{(1)})=\deg W'(c_4,\lambda_4,L^{(4)},*)=
\deg W'(c_1,\lambda_1,L^{(1)},*)$ (in particular, it is true if property (*)
hold for
$(c_4,\lambda_4,L^{(4)})$ and $\deg W'(c_4,\lambda_4,L^{(4)})=
\deg W'(c_1,\lambda_1,L^{(1)})$, $\deg W'(c_4,$ $\lambda_4,L^{(4)},*)=
\deg W'(c_1,\lambda_1,L^{(1)},*)$). Then
$\Phi_{c_4,\lambda_4,L^{(4)},*}=\Phi_{c_4,\lambda_4,L^{(4)},+}$ and,
therefore, by the
algorithms from \cite{5},
\cite{11}, \cite{12} one can compute the polynomials
$\Phi_{c_4,\lambda_4,L^{(4)},*}$,
$R_{c_4,\lambda_4,L^{(4)},*}$, and $R_{c_4,\lambda_4,L^{(4)},*,0}$.
Hence also one can find the integer $a(c_4,\lambda_4,L^{(4)})$.
\end{itemize}
We shall use this observations in the algorithm for assertion (e) of
Theorem~2 described in this and the next sections.


Assume that property (**) is satisfied, see the Introduction.
At first our aim is to decide whether 1)--3) from (*)
hold for $(c_1,\lambda_1,L^{(1)})$.
If 1)--3) are fulfilled for $(c_1,\lambda_1,L^{(1)})$, we shall
construct $(c_5,\lambda_5,L^{(5)})$ similar to
$(c_3,\lambda_3,L^{(3)})$ and
such that conditions 1)--3) from (*) hold for $(c_5,\lambda_5,L^{(5)})$.
In what follows in this and next sections
we shall assume that  $c_5,\lambda_5,L^{(5)}$ are obtained by
substituting in (17)--(19) values $\tau_i=\tau_{5,i}$, $1\le i\le
n_1$.

We describe a recursive algorithm by $(c_5,\lambda_5,L^{(5)})$.
As the base of the recursion
set $(c_5,\lambda_5,L^{(5)})=(c_3,\lambda_3,L^{(3)})$.
Let us describe the general step of this recursion.
Let us compute $\deg p'_v\deg W'$ points from the set
$S=
A_{\lambda_5,L^{(5)}}=(W_{\lambda_5}\setminus{\cal Z}(e_0,\ldots , e_r))
\cap{\cal Z}(e^{(5)}_{\sigma+1},\ldots ,
e^{(5)}_n)$. Properties 1) and 2) from (*) for $(c_5,\lambda_5,L^{(5)})$
imply
the transversality of intersection (22) with $j=5$, see Lemma~17 (ii)
\cite{13}.
Let us compute the number of elements $\#(e_{5,n+1}/e^{(5)}_0)(S)$. If
$\#(e_{5,n+1}/e^{(5)}_0)(S)<
\deg p'_v\deg W'$ then we construct a pair of distinct points $x_1,x_2\in
S$ such that
$(e_{5,n+1}/e^{(5)}_0)(x_1)\ne (e_{5,n+1}/e^{(5)}_0)(x_2)$.
Let us construct a linear form $L_{1,0}\in k[X_0,\ldots , X_n]$ with integer
coefficients of length $O(1)$ such that
$L_{1,0}(x_j)\ne 0$ for $j=1,2$.
Denote for brevity
$X^{(j)}=(X_{j,0},\ldots , X_{j,n})$,
$j=1,2$, and for an arbitrary polynomial $p$ in $X_0,\ldots , X_n$ set
$p(X^{(j)})=
p(X_{j,0},\ldots ,$ $X_{j,n})$.
Consider the system of polynomial equations and inequalities with respect to
the indeterminates $X_{j,0},\ldots , X_{j,n}$, $j=1,2$;
$\tau_1,\ldots ,\tau_{n_1}$ with coefficients from the field $\overline{k_3}$
$$
\left\{\begin{array}{ll}
\sum_{1\le i\le n_1}|\tau_i|^2\le\varepsilon_1,\\
\sum_{0\le i\le n}|X_{j,i}-(X_i/L_{1,0})(x_j)L_{1,0}|^2\le\varepsilon_2,
& j=1,2,\\
\sum_{0\le i\le n}
|(e_{5,n+1}-e_{0,n+1}+e_{1,n+1})(X^{(1)})-\\
(e_{5,n+1}-e_{0,n+1}+e_{1,n+1})(X^{(2)})|^2\ge
\varepsilon_3,\\
(L_{5,i}-L_{0,i}+L_{1,i})(X^{(j)})=0,& s+1\le i\le\sigma,\;j=1,2,\\
(e^{(5)}_w-e^{(0)}_w+e^{(1)}_w)(X^{(j)})=0,& \sigma+1\le w\le n,\;j=1,2,\\
(e^{(5)}_0-e^{(0)}_0+e^{(1)}_0)(X^{(j)})=1, & j=1,2.
\end{array}\right.
\eqno (31)
$$
\par\medskip\noindent{\bf LEMMA~17}\hspace{0.1em} {\it  If
$(c_1,\lambda_1,L^{(1)})$ satisfies conditions 1)--3) from (*)
then
system (31) has a solution $X_{j,w}=x^*_{j,w}$, $\tau_i=\tau^*_i$ from
$\overline{k_3}$ such that
$$
(x^*_{j,0}:\ldots : x^*_{j,n})\in W(\overline{k_3}),\quad j=1,2.
\eqno (32)
$$
}\par\medskip

\noindent {\bf PROOF} \quad There is an element $\varepsilon'>0$
such that $\varepsilon'$ is an infinitesimal with respect to the field $k_2$
and
$\varepsilon_3$ is an infinitesimal with respect to the field
$k_2(\varepsilon')$. By the transfer principle, see \cite{3},
it is sufficient to construct a solution of system (31) with
$\overline{k(\varepsilon')}$ in place of
$\overline{k_3}$ in (32).
Set in Lemma~4 the element $(c_0,\lambda_0,L^{(0)})$ to be the generic
point
$(c_1,\lambda_1,L^{(1)})$ of
${\cal L}_{c_1,\lambda_1,L^{(1)}}$ (one need to extend here the ground
field $k$
till $k(\tau_1,\ldots ,\tau_{n_1})$). Subsequently applying $n_1$ times
Lemma~4 we get a point $(c_3,\lambda_3,L^{(3)})\in{\cal
L}_{c_1,\lambda_1,L^{(1)}}$
satisfying conditions 1)--3) from (*) and such that it is obtained from
$(c_1,\lambda_1,L^{(1)})$
by the substitution integers $\tau'_i$ for $\tau_i$ in (17)--(19).
Consider the line ${\mathfrak l}$ containing $(c_5,\lambda_5,L^{(5)})$ and
$(c_3,\lambda_3,L^{(3)})$.
Let us choose a point $(c_*,\lambda_*,L^*)\in{\mathfrak
l}(\overline{k(\varepsilon')})$ which is
infinitely close to $(c_5,\lambda_5,L^{(5)})$.
Let $\tau_i=\tau^*_i$, $1\le i\le n_1$, corresponds to $(c_*,\lambda_*,L^*)$.
Then obviously for the considered $\tau^*_i$, $1\le i\le n_1$, there is a
solution
(32) of system (31) with $\overline{k(\varepsilon')}$ in place of
$\overline{k_3}$. The lemma is proved.

\medskip Notice that system (31)
satisfies conditions 1)--4) from Section~2.
Hence using Lemma~16 and Remark~11 we decide whether system (31) has a
solution
satisfying (32), and if it exists we construct such a solution.
Suppose that such a solution does not exist. Then by Lemma~17 the point
$(c_1,\lambda_1,L^{(1)})$ does not satisfy 3), and hence (*) does not hold.



Suppose that we have constructed a solution
of system (31) satisfying (32).
Substituting in (17)--(19) and (20)
$\tau_{5,i}+\tau^*_i$ for $\tau_i$, $1\le i\le n_1$,
we get a point $(c_*,\lambda_*,L^*)$ and the polynomial $e^*_{n+1}$
with coefficients from
a subfield $k'$ of $\overline{k_3}$, where the extension
$k'\supset k$ is finitely generated. We can suppose without loss of
generality that $k'\subset\overline{K}$. We have $L^{(5)}\in{\cal
U}''_0$,
$A_{\lambda_5,L^{(5)}}\cap{\cal Z}(e^{(5)}_0)=\emptyset$  and
$A_{\lambda_5,L^{(5)}}\cap\overline{V\setminus W}=\emptyset$
by 1) and 2).
Hence $L^*\in{\cal U}''_0$ (where ${\cal U}''_0$
is defined for the ground field
$\overline{k'}$, see Remark~2 \cite{13}), $A_{\lambda_*,L^*}\cap{\cal Z}(
e^*_0)=\emptyset$
and $A_{\lambda_*,L^*}\cap\overline{V\setminus W}=\emptyset$.
Since intersection (22) is transversal for $j=5$, the implicit function
theorem implies that
intersection (22) is transversal with $\lambda_*$, $L^*$ in place of
$\lambda_j$, $L^{(j)}$,
and $\#A_{\lambda_*,L^*}=\deg p'_v\deg W'$.
We have
$\#(e^*_{n+1}/e^*_0)(A_{\lambda_*,L^*})>
\#(e_{5,n+1}/e^{(5)}_0)(A_{\lambda_5,L^{(5)}})$.
Now conditions of Lemma~4 hold for $(c_*,\lambda_*,L^*)$
in place of $(c_0,\lambda_0,L^{(0)})$.

By Lemma~4 and the observations at the beginning of the section
one can apply the algorithm of
reduction to integer coefficients,
see \cite{12}, Proposition~2, and replace subsequently all values $\tau^*_i$
by
integers $\tau'_i$ of length $O(n\log d+(n-\sigma)\log(d'))$.
Denote by $(c',\lambda,L')$ the obtained point. Then
$L'\in{\cal U}''_0$, the number of
elements $\#A_{\lambda,L'}=
\deg p'_v\deg W'$, $A_{\lambda,L'}\cap{\cal Z}(e'_0)=\emptyset$, and
$A_{\lambda,L'}\cap\overline{V\setminus W}=\emptyset$,
the dimensions $\dim W_{\lambda}=n-\sigma$,
$\dim (W_{\lambda}\setminus{\cal
Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_{n-1})=1$,
the number of elements
$\#(e'_{n+1}/e'_0)(A_{\lambda,L'})\ge\#(e^*_{n+1}/e^*_0)(A_{\lambda_*,L^*})
>\\ \#(e_{5,n+1}/e^{(5)}_0)(A_{\lambda_5,L^{(5)}})$,
the morphisms $p_{\lambda,L'}$,
$p_{\lambda,L',*}$ are dominant.
Let us apply Lemma~8 to the case when $\iota({\Bbb A}^1(\overline{k}))$
is the line containing
$(c',\lambda,L')$ and $(c_5,\lambda_5,L^{(5)})$, and $\iota(0)=
(c_5,\lambda_5,L^{(5)})$. Then using Theorem~2 (f) \cite{13},
Lemma~8 and Lemma~4 we
construct a point $(c'',\lambda',L'')$
similar to $(c_5,\lambda_5,L^{(5)})$ and such that the number of elements
$\#(e''_{n+1}/e''_0)(A_{\lambda',L''})\ge\#(e'_{n+1}/e'_0)(A_{\lambda,L'})>
\#(e_{5,n+1}/e^{(5)}_0)(A_{\lambda_5,L^{(5)}})$. After that we replace
$(c_5,\lambda_5,L^{(5)})$
by $(c'',\lambda',L'')$ and proceed to the next step of the recursion.
Thus, finally we shall come to the case when the number of elements
$\#(e_{5,n+1}/e^{(5)}_0)(S)=\deg p'_v\deg W'$.
Then $(c_1,\lambda_1,L^{(1)})$ and $(c_5,\lambda_5,L^{(5)})$
satisfies conditions 1)--3).
This step of the described recursion is final.


\medskip
In what follows we shall assume without loss of generality that
$(c_1,\lambda_1,L^{(1)})$ satisfies 3).
Now we shall describe a new recursion. As the base of the recursion
we take the constructed point $(c_5,\lambda_5,L^{(5)})$. At the end of this
recursion we shall obtain $(c_5,\lambda_5,L^{(5)})$ such that conditions
1)--3) from (*) hold for $(c_5,\lambda_5,L^{(5)})$ and
$\deg W'(c_5,\lambda_5,L^{(5)})=\deg W'(c_1,\lambda_1,L^{(1)})$.

Let us describe the general step of this recursion. At the beginning of the
step a
point $(c_5,\lambda_5,L^{(5)})$ satisfying conditions 1)--3) from (*) is
given. Let us define $r_{c_5,\lambda_5,L^{(5)}}$,
$p_{c_5,\lambda_5,L^{(5)}}$, $q_{c_5,\lambda_5,L^{(5)}}$,
$p_{\lambda_5,L^{(5)}}$ similarly to $r_{c_0,\lambda_0,L^{(0)}}$,
$p_{c_0,\lambda_0,L^{(0)}}$, $q_{c_0,\lambda_0,L^{(0)}}$,
$p_{\lambda_0,L^{(0)}}$, see the proof of Lemma~1 and the
Introduction.
At first our aim is to construct linearly independent linear forms
$\widetilde{L}_j\in
k[X_0,\ldots ,X_{n-\sigma}]$,
$j\in\{0,\sigma+1,\ldots ,n\}$ with integer coefficients
of length $O(n^2\log(dd')$ (additionally
applying the algorithm of reduction to integer
coefficients, see Proposition~2 \cite{12} one can obtain the estimate
$O(n\log d+(n-\sigma)\log(d'))$,
but it is not principal now) and an integer
$t$ of the same length satisfying the following conditions.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item The $(n-\sigma+1)$--tuples of linear forms $(\widetilde{L}_0,
\widetilde{L}_{\sigma+1},\ldots ,\widetilde{L}_{n-1},\widetilde{L}_n+
tX_{n-\sigma+1})$ satisfies the assertion of Theorem~1 \cite{12} for the
morphism
$q_{c_5,\lambda_5,L^{(5)}}$ (in place of the $(n-\sigma+1)$--tuples
$L'$ for $p'$). Hence
$\#W'(c_5,\lambda_5,$ $L^{(5)})\cap{\cal
Z}(\widetilde{L}_{\sigma+1},\ldots ,$ $\widetilde{L}_{n-1},\widetilde{L}_n+
tX_{n-\sigma+1})=\deg W'(c_5,\lambda_5,L^{(5)})$,
$W'(c_5,\lambda_5,L^{(5)})
\cap{\cal Z}(\widetilde{L}_{\sigma+1},\ldots ,$
$\widetilde{L}_{n-1},\widetilde{L}_n+tX_{n-\sigma+1})\cap{\cal
Z}(\widetilde{L}_0)=\emptyset$
(here we assume that $X_0,\ldots ,
X_{n-\sigma+1}$ are the coordinate
functions on ${\Bbb P}^{n-\sigma+1}(\overline{k})$),
for every $z\in W'(c_5,\lambda_5,$ $L^{(5)})\cap{\cal
Z}(\widetilde{L}_{\sigma+1},\ldots ,\widetilde{L}_{n-1},\widetilde{L}_n+
tX_{n-\sigma+1})$ the inverse image
$q_{c_5,\lambda_5,L^{(5)}}^{-1}(z)$ consists of one point
$z_1$, and $z_1$ is a smooth point of $W_{\lambda_5}$, see \cite{12}.

\item Put $L_j=\widetilde{L}_j(L^{(5)}_0,L^{(5)}_{\sigma+1},\ldots ,$
$L^{(5)}_n)$, and
$L=(L_0,L_{\sigma+1},\ldots ,L_n)$,
$j\in \{0,\sigma+1,\ldots ,n\}$. Then
the element $(c_5,\lambda_5,L)$ satisfies conditions 1)--3)
from (*).

\item For every $z\in W'(c_5,\lambda_5,L^{(5)})
\cap{\cal Z}(\widetilde{L}_{\sigma+1},\ldots ,$
$\widetilde{L}_{n-1},\widetilde{L}_n+tX_{n-\sigma+1})$ put
$L_z=(L_0,L_{\sigma+1},\ldots
,L_{n-1},L_n-(\widetilde{L}_n/\widetilde{L}_0)(z)L_0)$.
Then $(c_5,\lambda_5,L_z)$ (in place of $(c_0,\lambda_0,L^{(0)})$)
satisfies conditions of Lemma~2.

\item Denote
$\eta^{(5)}_j=\widetilde{L}_j(e^{(5)}_0,e^{(5)}_{\sigma+1},\ldots ,
e^{(5)}_n)=L_j(e_0,\ldots , e_r)$ for
all
$j\in\{0,\sigma+1,\ldots ,n\}$.
For every $z\in W_{\lambda_5}\cap{\cal
Z}(\eta^{(5)}_{\sigma+1},\ldots
,\eta^{(5)}_{n-1},\eta^{(5)}_n+te_{5,n+1})\setminus{\cal
Z}(e_0,\ldots ,$ $e_r)$ the number of elements
$\#p_{\lambda_5,L}^{-1} p_{\lambda_5,L}(z)=\deg
p'_v\deg W'$, and the intersection
$p_{\lambda_5,L}^{-1} p_{\lambda_5,L}(z)\cap
(\overline{V\setminus W})=\emptyset$.
Hence by Lemma~2 \cite{12} the differential
$d_zp_{\lambda_5,L}$ is an isomorphism. Therefore,
again by this lemma
and (i) the differentials $d_zq_{c_5,\lambda_5,L}$ are isomorphisms
for all considered $z$. This implies, see below, that
the differentials $d_zq_{c_5,\lambda_5,L^{(5)}}$ are isomorphisms
for all considered $z$.
\end{enumerate}

Applying Theorem~1 \cite{12} and Remark~2
to the morphism $q_{c_5,\lambda_5,L^{(5)}}$ we
construct linear forms $\widetilde{L}_j\in k[X_0,\ldots
, X_{n-\sigma}]$,
$j\in\{0,\sigma+1,\ldots ,n\}$ and an integer $t'$
satisfying (i) with $t'$ in place of $t$.
By assertion (d) Theorem~1 \cite{12} we shall assume without loss of
generality that
$\widetilde{L}_j$,
$j\in\{0,\sigma+1,\ldots ,n\}$ are linearly independent.
Consider a replacement $\widetilde{L}_0$ by $\widetilde{L}_0+t''X_0$ and
$\widetilde{L}_j$ by $\widetilde{L}_j+t''X_{j-\sigma}$, $j\in
\{\sigma+1,\ldots ,n\}$,
with an integer $t''$ of length $O(n\log d+(n-\sigma)\log d')$.
By Theorem~2 (f) \cite{13} and Lemma~8 using such a replacement with an
appropriate $t''$ (it is constructed by the enumeration) we shall
assume without loss of generality that additionally
the element $(c_5,\lambda_5,L)$ satisfies (ii).
Notice that the morphism $p_{\lambda_5,L}$ is distinct from
$p_{\lambda_5,L^{(5)}}$ only
by an automorphism of ${\Bbb P}^{n-\sigma}(\overline{k})$.
Similarly $p_{c_5,\lambda_5,L}$ is distinct from
$p_{c_5,\lambda_5,L^{(5)}}$ only
by an automorphism of ${\Bbb P}^{n-\sigma+1}(\overline{k})$.
Hence the algebraic variety $W'(c_5,\lambda_5,L)$ is distinct from
$W'(c_5,\lambda_5,L^{(5)})$ only
by an automorphism of ${\Bbb P}^{n-\sigma+1}(\overline{k})$.
Therefore, $\deg W'(c_5,\lambda_5,L^{(5)})=\deg W'(c_5,\lambda_5,L)$.
Note also that $W'(c_5,$ $\lambda_5,L^{(5)})={\cal
Z}(\Phi_{c_5,\lambda_5,L^{(5)}}(X_0,\ldots , X_{n-\sigma+1}))$,
$W'(c_5,\lambda_5,L)={\cal
Z}(\Phi_{c_5,\lambda_5,L}(X_0,\ldots , X_{n-\sigma+1})))$ and
$$
\Phi_{c_5,\lambda_5,L^{(5)}}(X_0,\ldots ,
X_{n-\sigma+1})=\mu\Phi_{c_5,\lambda_5,L}
(\widetilde{L}_0,\widetilde{L}_{\sigma+1},\ldots ,
\widetilde{L}_n,X_{n-\sigma+1}),
\eqno (33)
$$
where $0\ne\mu\in\overline{k}$.



Now by (i) every irreducible component of the variety
$C_{\lambda_5,L}=W_{\lambda_5}\cap{\cal Z}(\eta^{(5)}_{\sigma+1},\ldots
,$ $\eta^{(5)}_{n-1})
\setminus{\cal Z}(e_0,\ldots , e_r)$ is a
curve.
By conditions 1)--3) from (*) for $(c_5,\lambda_5,L)$
the morphism $p_{\lambda_5,L,*}$ is dominant,
and for all points $z\in C_{\lambda_5,L}$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number,
the differential $d_zp_{\lambda_5,L}$ is an isomorphism,
$\#p_{\lambda_5,L}^{-1} p_{\lambda_5,L}(z)=\deg
p'_v\deg W'$, the intersection
$p_{\lambda_5,L}^{-1} p_{\lambda_5,L}(z)\cap
(\overline{V\setminus W})=\emptyset$.
Let $z\in C_{\lambda_5,L}$ and $z'=q_{c_5,\lambda_5,L^{(5)}}(z)$.
Then by Lemma~2 for all points $z\in C_{\lambda_5,L}$, except at most a
polynomial in
$d^n(d')^{n-\sigma}$ number, the point
$(c_5,\lambda_5,L_{z'})$ (in place of $(c_0,\lambda_0,L^{(0)})$)
satisfies conditions of Lemma~2.
Notice that by \cite{5} for every integer $t$ one can compute the algebraic
variety $C_{\lambda_5,L}\cap{\cal Z}(\widetilde{e}_n+te_{5,n+1})$.
Hence one can construct an integer $t$ of length
$O(n^2\log (dd'))$ such that (i) and (ii) hold, the
variety $C_{\lambda_5,L}\cap{\cal Z}(\widetilde{e}_n+te_{5,n+1})$ is
zero--dimensional, and
conditions (iii) and (iv) are fulfilled.
Thus, the required linear forms $\widetilde{L}_j$ and the integer $t$ are
constructed.

Denote for brevity $\eta_j=\widetilde{L}_j(e^{(5)}_0-e^{(0)}_0+e^{(1)}_0,
e^{(5)}_{\sigma+1}-e^{(0)}_{\sigma+1}+e^{(1)}_{\sigma+1},\ldots ,
e^{(5)}_n-e^{(0)}_j+e^{(1)}_j)$, $j\in\{0,\sigma+1,\ldots ,n\}$.
Let $\tau^*_i\in\overline{K}$ be arbitrary (in what follows we assume that
$K\supset k_3$).
Let us define $(c_*,\lambda_*,L^*)\in{\cal
L}_{c_1,\lambda_1,L^{(1)}}\otimes\overline{K}$
substituting in (17)--(19) the
values $\tau_i=\tau_{5,i}+\tau^*_i$, $1\le i\le n_1$.
Let $\lambda_*=(L_{*,s+1},\ldots , L_{*,\sigma})$,
$L^*=(L^*_0,L^*_{\sigma+1},\ldots , L^*_n)$,
$e^*_j=L^*_j(e_0,\ldots , e_r)$,
$j\in\{0,\sigma+1,\ldots ,n\}$.
Set  $\eta^*_j=\widetilde{L}_j(e^*_0,e^*_{\sigma+1},\ldots ,e^*_n)$,
$j\in\{0,\sigma+1,\ldots ,n\}$.
Thus, $\eta_j^*=\eta_j|_{\tau_i=\tau^*_i,\,1\le i\le n_1}$ and
$\eta_j^{(5)}=\eta_j|_{\tau_i=0,\,1\le i\le n_1}$.
Put $\overline{L}_j=\widetilde{L}_j(L^*_0,L^*_{\sigma+1},\ldots , L^*_n)$,
$j\in \{0,\sigma+1,
\ldots ,n\}$, and
$\overline{L}=(\overline{L}_0,\widetilde{L}_{\sigma+1},\ldots ,
\overline{L}_n)$.
Hence $\eta^*_j=\overline{L}_j(e_0,\ldots , e_r)$ for
all $j\in\{0,\sigma+1,\ldots ,n\}$.

Consider the system of equations and inequalities with
respect to
the indeterminates $X_0,\ldots , X_n$, $Y_{\sigma+1},\ldots ,Y_n$,
$\tau_1,\ldots ,\tau_{n_1}$ with coefficients from the field $k_2$
$$
\left\{
\begin{array}{ll}
\sum_{1\le i\le n_1}|\tau_i|^2\le\varepsilon_1,     \\
\eta_j-Y_j\eta_0=0,& \sigma+1\le j\le n-1,\\
\eta_n-Y_n\eta_0+\\
+t(e_{5,n+1}-e_{0,n+1}+e_{1,n+1})=0,\\
\eta_0=1,\\
L_{5,j}-L_{0,j}+L_{1,j}=0,& s+1\le j\le \sigma,\\
\sum_{\sigma+1\le i\le n}|Y_i|^2\le\varepsilon_2,\\
\sum_{0\le i\le n}|e_i|^2\ge\varepsilon_2^{-1}.
\end{array}
\right.
\eqno (34)
$$

\par\medskip\noindent{\bf LEMMA~18}\hspace{0.1em} {\it  Suppose that
$(c_1,\lambda_1,L^{(1)})$ satisfies
conditions 1)--3) from (*),
conditions (i)--(iv) hold,
$\deg W'(c_1,$ $\lambda_1,L^{(1)})>\deg W'(c_5,\lambda_5,L^{(5)})$.
Then system (34) has a solution
$X_w=x^*_w$, $0\le w\le n$; $\tau_i=\tau^*_i$, $1\le i\le n_1$; $Y_j=y^*_j$,
$\sigma+1\le j\le n$, from
$\overline{k_2}$ such that
$$
(x^*_0:\ldots : x^*_n)\in W(\overline{k_2}).
\eqno (35)
$$
For arbitrary solution (35) the point
$(c_*,\lambda_*,L^*)$ satisfies conditions of Lemma~2 (in place of
$(c_0,\lambda_0,L^{(0)})$)
and $\deg W'(c_*,\lambda_*,L^*)>\deg W'(c_5,\lambda_5,L^{(5)})$.
}\par\medskip

\noindent {\bf PROOF} Obviously it is sufficient to prove this lemma with
$k_3$ in place of $k_2$.
Put $S'=W'(c_5,\lambda_5,L^{(5)})\cap{\cal Z}(\widetilde{L}_{\sigma+1},
\ldots ,\widetilde{L}_{n-1},\widetilde{L}_n+tX_{n+1})$.
Let $\tau_i^*,y^*_j\in\overline{k_3}$, $1\le i\le
n_1$, $\sigma+1\le j\le n$
be arbitrary (not necessarily related  to a solution of system (34))
such that
$\sum_{1\le i\le n_1}|\tau^*_i|^2\le\varepsilon_1$ and
$\sum_{\sigma+1\le j\le n}|y^*_j|^2\le\varepsilon_2$.
Notice that according to the observations from the beginning of Section~1
the point $(c_*,\lambda_*,L^*)$ satisfies (D) and
conditions of Lemma~2. Therefore,
the objects $q_{c_*,\lambda_*,L^*}$, $W'(c_*,\lambda_*,L^*)$ and
so on are defined. Let $z\in S'$ and
$q_{c_5,\lambda_5,L^{(5)}}(z_1)=z$, see (i).

Let us show that for every $z\in S'$ there is the unique point $z^*\in
S^*=W'(c_*,\lambda_*,L^*)\cap{\cal
Z}(\widetilde{L}_{\sigma+1}-y^*_{\sigma+1}\widetilde{L}_0,\ldots ,
\widetilde{L}_{n-1}-y^*_{n-1}\widetilde{L}_0,
\widetilde{L}_n-y^*_n\widetilde{L}_0+tX_{n+1})$ such that
$q_{c_*,\lambda_*,L^*}^{-1}(z^*)$
contains a point $z^*_1$ which is infinitely close to $z_1$
(hence also $z^*$ is infinitely close to $z$).
Note that  by (i) $\widetilde{L}_{\sigma+1}/\widetilde{L}_0,\ldots ,
\widetilde{L}_{n-1}/\widetilde{L}_0,
(\widetilde{L}_n+tX_{n+1})/\widetilde{L}_0$ generate
the linear space ${\mathfrak
m}_z/{\mathfrak m}_z^2$ where ${\mathfrak m}_z$ is the maximal ideal of
the
local ring ${\cal O}_{z,W'(c_5,\lambda_5,L^{(5)})}$ of the algebraic
variety
$W'(c_5,\lambda_5,L^{(5)})$ at $z$.
By (iv) the differential $d_{z_1}q_{c_5,\lambda_5,L^{(5)}}$ is an
isomorphism. Therefore,
the dual mapping to $d_{z_1}q_{c_5,\lambda_5,L^{(5)}}$ is also an
isomorphism.
Hence $\eta^{(5)}_{\sigma+1}/\eta^{(5)}_0,\ldots ,
\eta^{(5)}_{n-1}/\eta^{(5)}_0,
(\eta^{(5)}_n+te_{5,n+1})/\eta^{(5)}_0$ generate the linear space ${\mathfrak
m}_{z_1}/{\mathfrak m}_{z_1}^2$ where ${\mathfrak m}_{z_1}$ is the maximal
ideal of the local ring ${\cal O}_{z_1,W_{\lambda_5}}$ of the algebraic
variety
$W_{\lambda_5}$ at $z_1$.
Hence $z_1$ is a smooth point of hypersurfaces
${\cal Z}(\eta^{(5)}_j)$, $\sigma+1\le j\le n-1$;
${\cal Z}(\eta^{(5)}_n+te_{5,n+1})$,
and the intersection of the tangent spaces of the algebraic varieties
$W_{\lambda_5}$, ${\cal Z}(\eta^{(5)}_{\sigma+1}),\ldots ,$
${\cal Z}(\eta^{(5)}_{n-1}),$ ${\cal Z}(\eta^{(5)}_n+te_{5,n+1})$ at
the point $z_1$ is transversal.

On the other hand, by Lemma~17 (ii) \cite{13} and (iii)
the intersection $W\cap{\cal Z}(L_{5,s+1},$ $\ldots , L_{5,\sigma})\cap
{\cal Z}(\eta^{(5)}_{\sigma+1},\ldots , \eta^{(5)}_{n-1},\eta^{(5)}_n-
(\widetilde{L}_n/\widetilde{L}_0)(z)\eta^{(5)}_0)$ is transversal at
$z_1$. Hence the intersection
$W\cap{\cal Z}(L_{5,s+1},\ldots , L_{5,\sigma})$ is also transversal
at $z_1$.
Therefore, the intersection $W\cap{\cal Z}(L_{5,s+1},\ldots ,
L_{5,\sigma})\cap
{\cal Z}(\eta^{(5)}_{\sigma+1})\cap\ldots
\cap{\cal Z}(\eta^{(5)}_{n-1})\cap{\cal Z}(\eta^{(5)}_n+te_{5,n+1})$
is
transversal at $z_1$. Finally, by the implicit function theorem there is
one and only one
point $z_1^*\in q_{c_*,\lambda_*,L^*}^{-1}(S^*)$ such that $z_1^*$ is
infinitely close to $z_1$. The required assertion is proved.



Now suppose that $y\in S^*$, $q_{c_*,\lambda_*,L^*}(y_1)=y$, and
$y_1$ is not infinitely close to ${\cal Z}(e_0,\ldots , e_r)$.
Then obviously $y_1$ is infinitely close to
$q_{c_5,\lambda_5,L^{(5)}}^{-1}(S')$ and, hence, $y$ is infinitely close to
$S'$. Conversely suppose that
$y_1$ is  infinitely close to ${\cal Z}(e_0,\ldots , e_r)$.
Let us show that $y$ is not infinitely close to
$S'$. Indeed,
suppose contrary, that $y$ is infinitely close to $z\in S'$.
Therefore, $\widetilde{L}_0(y)\ne 0$.
Set
$\overline{L}_y=(\overline{L}_0,\overline{L}_{\sigma+1},\ldots ,
\overline{L}_{n-1},\overline{L}_n-
(\widetilde{L}_n/\widetilde{L}_0)(y)\overline{L}_0)$.
Notice that the coefficients of the linear forms from
$\overline{L}_{y}$ are infinitely
close to the coefficients of the
corresponding linear forms from $L_z$, see condition (iii).
By property (D) for $(c_5,\lambda_5,L_z)$, see (iii), the intersection
$(W\setminus{\cal Z}(e_0,\ldots , e_r))
\cap{\cal Z}(L_{5,s+1},\ldots , L_{5,\sigma})\cap
{\cal Z}(\eta^{(5)}_{\sigma+1})\cap\ldots\cap{\cal
Z}(\eta^{(5)}_{n-1})\cap{\cal
Z}(\eta^{(5)}_n-(\widetilde{L}_n/\widetilde{L}_0)(z)\eta^{(5)}_0)=
A_{\lambda_5,L_z}$
is transversal and consists of $\deg p'_v\deg W'$ points. By Lemma~17
\cite{13} and the implicit function theorem the intersection
$(W\setminus{\cal Z}(e_0,\ldots , e_r))
\cap{\cal Z}(L_{*,s+1},\ldots , L_{*,\sigma})\cap
{\cal Z}(\eta^*_{\sigma+1})\cap\ldots\cap{\cal
Z}(\eta^*_{n-1})\cap{\cal
Z}(\eta^*_n-(\widetilde{L}_n/\widetilde{L}_0)(y)\eta^*_0)=
A_{\lambda_*,\overline{L}_{y}}$
is transversal and consists of $\deg p'_v\deg W'$ points. By the implicit
function theorem the points
from $A_{\lambda_*,\overline{L}_y}$ are infinitely close to the
corresponding
points from $A_{\lambda_5,L_z}$.
But by (i) $q_{c_5,\lambda_5,L^{(5)}}^{-1}(z)$ consist of one point, and
$q_{c_5,\lambda_5,L^{(5)}}^{-1}(z)\subset A_{\lambda_5,L_z}$,
$q_{c_*,\lambda_*,L^*}^{-1}(y)\subset A_{\lambda_*,\overline{L}_{y}}$.
Hence $q_{c_*,\lambda_*,L^*}^{-1}(y)$ also consists of one point
which is infinitely close to the one from
$q_{c_5,\lambda_5,L^{(5)}}^{-1}(z)$.
Thus, $y_1$ is not infinitely close to ${\cal
Z}(e_0,\ldots , e_r)$. This contradiction proves the required assertion.


Let us show that $\widetilde{L}_0$ does not vanish on any irreducible
component of $W'(c_*,$ $\lambda_*,L^*)$.
Indeed by (ii) every irreducible component of $W_{\lambda_5}$ contains a
point from $A_{\lambda_5,L}$,
and $\eta^{(5)}_0$ does not vanish at any points from $A_{\lambda_5,L}$.
Notice that the coefficients of the linear forms $\overline{L}_j$ are
infinitely close to the corresponding coefficients of the linear forms
$L_j$, $j\in\{0,\sigma+1,\ldots , n\}$.
Hence by Lemma~17 \cite{13} and the implicit function theorem,
see the last observation related to property (D) at the beginning of
Section~1, every irreducible component of
$W_{\lambda_*}$ contains a point from $A_{\lambda_5,\overline{L}}$, and
$\eta^*_0=\widetilde{L}_0(e^*_0,e^*_{\sigma+1},\ldots , e^*_n)$ does not
vanish at any points from $A_{\lambda_5,\overline{L}}$.
Therefore, every irreducible component of $W'(c_*,\lambda_*,L^*)$ contains a
point from $q_{c_*,\lambda_*,L^*}(A_{\lambda_5,\overline{L}})$, and
$\widetilde{L}_0$ does not vanish at any points from
$q_{c_*,\lambda_*,L^*}(A_{\lambda_5,\overline{L}})$.
This implies the required assertion.



Let us prove the existence of the required solution of system (34).
By Lemma~4 and the algorithm of reduction to integer
coefficients, cf. the
proof of the existence of $(c_*,\lambda_*,L^*)$ in Lemma~17,
there are $\tau^*_i\in k(\varepsilon_3)$ which are
infinitesimal with respect to the field $k$
such that $(c_*,\lambda_*,L^*)$ satisfies conditions of
Lemma~4 (for any line ${\mathfrak l}$ containing
$(c_*,\lambda_*,L^*)$) in place of $(c_0,\lambda_0,L^{(0)})$, and
$\deg W'(c_*,\lambda_*,L^*)=\deg W'(c_1,\lambda_1,L^{(1)})$.
We have $\deg W'(c_*,\lambda_*,L^*)>\deg W'(c_5,\lambda_5,L^{(5)})$.
Hence as we have proved the B\'ezout theorem implies the existence of
a point $z\in W'(c_*,\lambda_*,L^*)\cap
{\cal Z}(\widetilde{L}_{\sigma+1},$ $\ldots
,\widetilde{L}_{n-1},\widetilde{L}_n+tX_{n-\sigma+1})$, which
is not infinitely close to $S'$. Further, there exists a point $y\in
W'(c_*,\lambda_*,L^*)$ with the coordinates from
$\overline{k(\varepsilon_3)}$
such that $y$ is infinitely close to $z$ with respect to the field $k$,
the inverse image
$q_{c_*,\lambda_*,L^*}^{-1}(y)\ne\emptyset$ and $\widetilde{L}_0(y)\ne 0$.
Put
$y_j^*=(\widetilde{L}_j/\widetilde{L}_0)(y)$ for $\sigma+1\le j\le
n-1$ and
$y_n^*=((\widetilde{L}_n+tX_{n-\sigma+1})/\widetilde{L}_0)(y)$. Then every
nonzero
$y_j^*\in\overline{k(\varepsilon_3)}$, $j\in \{\sigma+1,\ldots ,n\}$, is
infinitesimal with respect to
the field $k$. Let $y_1\in q^{-1}_{c_*,\lambda_*,L^*}(y)$.
By previous considerations
the point $y_1$ is infinitely close to ${\cal Z}(e_0,\ldots ,
e_r)$. Thus, system (34) has a
solution satisfying (35) (with $k_3$ in place of $k_2$).

Conversely, if (34) has a
solution satisfying (35) then by the previous considerations and the
B\'ezout theorem
$\deg W'(c_*,\lambda_*,L^*)>\deg W'(c_5,\lambda_5,L^{(5)})$. The lemma is
proved.

\medskip Let us return to the description of the general step of the
considered recursion.  Notice that system (34)
satisfies conditions 1)--4) from Section~2.
Hence using Lemma~16 and Remark~11 we decide whether system (34) has a
solution
satisfying (35), and if such a solution exists we construct it.
Suppose that such a solution does not exist. Then by Lemma~18 the degree
$\deg W'(c_1,\lambda_1,L^{(1)})=\deg W'(c_5,\lambda_5,L^{(5)})$ and the
considered step is
final.

Suppose that we have constructed a solution of system (34)
satisfying (35), see the statement of Lemma~18.
Then $\deg W'(c_*,\lambda_*,L^*)>\deg W'(c_5,\lambda_5,L^{(5)})$ by this
lemma. Similarly to the previous recursion conditions of Lemma~4 hold for
$(c_*,\lambda_*,L^*)$ in place of $(c_0,\lambda_0,L^{(0)})$.
By Lemma~4 and the observations from the beginning of the section one can
apply the algorithm of reduction to integer coefficients,
see \cite{12}, Proposition~2, and replace subsequently all values $\tau^*_i$
by
integers $\tau'_i$ of length $O(n\log d+(n-\sigma)\log(d'))$.
Denote by $(c',\lambda,L')$ the obtained point.
Then
$L'\in{\cal U}''_0$, the number of
elements $\#A_{\lambda,L'}=
\deg p'_v\deg W'$, $A_{\lambda,L'}\cap{\cal Z}(e'_0)=\emptyset$, and
$A_{\lambda,L'}\cap\overline{V\setminus W}=\emptyset$,
the dimensions $\dim W_{\lambda}=n-\sigma$,
$\dim (W_{\lambda}\setminus{\cal
Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_{n-1})=1$,
the number of elements
$\#(e'_{n+1}/e'_0)(A_{\lambda,L'})=\#(e_{5,n+1}/e^{(5)}_0)(A_{\lambda_5,L^{(5)}})$,
the morphisms $p_{\lambda,L'}$,
$p_{\lambda,L',*}$ are dominant, and
$\deg W'(c',\lambda,L')\ge\deg W'(c_*,\lambda_*,L^*)$.
Let us apply Lemma~8 to the case when $\iota({\Bbb A}^1(\overline{k}))$
is the line containing
$(c',\lambda,L')$ and $(c_5,\lambda_5,L^{(5)})$, and $\iota(0)=
(c_5,\lambda_5,L^{(5)})$. Then using Theorem~2 (f) \cite{13}, Lemma~8 and
Lemma~4 we
construct a point $(c'',\lambda',L'')$ which
is similar to $(c_5,\lambda_5,L^{(5)})$, and
$\deg W'(c'',\lambda',L'')\ge\deg W'(c',\lambda,L')>\deg
W'(c_5,\lambda_5,L^{(5)})$. After that we replace
$(c_5,\lambda_5,L^{(5)})$
by $(c'',\lambda',L'')$ and proceed to the next step of the recursion.
The general step and all the recursion are described.


\medskip
Now we shall describe one more recursion. As the base of the recursion
we take the constructed point $(c_5,\lambda_5,L^{(5)})$. At the end of this
recursion we shall obtain $(c_5,\lambda_5,L^{(5)})$ such that conditions
1)--3) from (*) hold for $(c_5,\lambda_5,L^{(5)})$ and
$\deg W'(c_5,\lambda_5,L^{(5)})=\deg W'(c_1,\lambda_1,L^{(1)})$,
$\deg W'(c_5,\lambda_5,L^{(5)},*)=\deg W'(c_1,$ $\lambda_1,L^{(1)},*)$.

Let us describe the general step of this recursion. At the beginning of the
step a
point $(c_5,\lambda_5,L^{(5)})$ satisfying conditions 1)--3) from (*)
is given, and
$\deg W'(c_5,\lambda_5,L^{(5)})=\deg W'(c_1,\lambda_1,L^{(1)})$.
Applying the algorithm from \cite{5} let us compute,
see the observations at the beginning of the section,
the variety $W'(c_5,\lambda_5,L^{(5)},*)$
and the polynomial $\Phi_{c_5,\lambda_5,L^{(5)},+}$, see Lemma~3.
Hence $W'(c_5,\lambda_5,L^{(5)},*)={\cal
Z}(\Phi_{c_5,\lambda_5,L^{(5)},+}(X_0,X_1,X_2))$ by
Lemma~3. By properties 1) and 2) from (*) for $(c_5,\lambda_5,L^{(5)})$
the intersection $(W_{\lambda_5}\setminus{\cal Z}(e_0,\ldots ,
e_r))\cap{\cal Z}(e^{(5)}_{\sigma+1})\cap\ldots\cap{\cal
Z}(e^{(5)}_n)=
A_{\lambda_5,L^{(5)}}$
is transversal at each its point, consists of $\deg p'_v\deg W'$ isolated
points,
and $A_{\lambda_5,L^{(5)}}\cap{\cal Z}(e^{(5)}_0)=\emptyset$.
By property 3) from (*) and Lemma~2 \cite{12} the intersection
$W'(c_5,\lambda_5,L^{(5)})\cap{\cal Z}(X_1,\ldots , X_{n-\sigma})$ is
also transversal
at every point from $q_{c_5,\lambda_5,L^{(5)}}(A_{\lambda_5,L^{(5)}})$, the
number of elements
$\#q_{c_5,\lambda_5,L^{(5)}}(A_{\lambda_5,L^{(5)}})=\deg p'_v\deg W'$,
and for every $z\in A_{\lambda_5,L^{(5)}}$ the differential
$d_zq_{c_5,\lambda_5,L^{(5)}}$
is an isomorphism. Let us define $r_{c_5,\lambda_5,L^{(5)},*}$,
$p_{\lambda_5,L^{(5)},*}$, $q_{c_5,\lambda_5,L^{(5)},*}$ similarly to
$r_{c_0,\lambda_0,L^{(0)},*}$,
$p_{\lambda_0,L^{(0)},*}$, $q_{c_0,\lambda_0,L^{(0)},*}$,
see the proof of Lemma~3 and the Introduction.
By Lemma~11 \cite{12} every irreducible component of
$(W_{\lambda_5}\setminus{\cal Z}(e_0,\ldots , e_r))\cap{\cal
Z}(e^{(5)}_{\sigma+1},$ $\ldots , e^{(5)}_{n-1})=C_{\lambda_5,L^{(5)}}$
contains a point from $A_{\lambda_5,L^{(5)}}$.
Hence the intersection
$$
(W_{\lambda_5}\setminus{\cal Z}(e_0,\ldots ,e_r))\cap{\cal
Z}(e^{(5)}_{\sigma+1})\cap\ldots\cap{\cal Z}(e^{(5)}_{n-1})
\eqno (36)
$$
 is transversal,
each irreducible component of $C_{\lambda_5,L^{(5)}}$ is a curve, and the
variety
$W'(c_5,\lambda_5,L^{(5)},*)$ is the union of all the irreducible components
$E$ of
$W'(c_5,\lambda_5,L^{(5)})\cap{\cal Z}(X_1,\ldots , X_{n-\sigma-1})$
such that $E\cap
q_{c_5,\lambda_5,L^{(5)}}(A_{\lambda_5,L^{(5)}})\ne\emptyset$.
Therefore, the intersection $W'(c_5,\lambda_5,L^{(5)})\cap{\cal
Z}(X_1,\ldots , X_{n-\sigma-1})$ is
transversal at every irreducible component $E$ of
$W'(c_5,\lambda_5,L^{(5)},*)$.
We have $\deg p_{\lambda_5,L^{(5)},*}=\deg r_{c_5,\lambda_5,L^{(5)},*}=
\deg p_{\lambda_5,L^{(5)}}=\deg r_{c_5,\lambda_5,L^{(5)}}=\deg p'_v\deg W'$.
Hence
the morphism $q_{c_5,\lambda_5,L^{(5)},*}$ induces the isomorphism of
algebras of rational
functions on $C_{\lambda_5,L^{(5)}}$ and $W'(c_5,\lambda_5,L^{(5)},*)$
respectively (by definition these
algebras are isomorphic to the direct products
of fields of rational functions of all irreducible components of
$C_{\lambda_5,L^{(5)}}$ and $W'(c_5,\lambda_5,L^{(5)},*)$
respectively). Similarly $q_{c_5,\lambda_5,L^{(5)}}$ induces the
birational isomorphism.
Further,  $q_{c_5,\lambda_5,L^{(5)},*}$  (respectively
$q_{c_5,\lambda_5,L^{(5)}}$)
is an isomorphism in a neighborhood with respect to the Zariski topology of
the finite set $A_{\lambda_5,L^{(5)}}$.
Let $z\in W'(c_5,\lambda_5,L^{(5)},*)$ and $X_0(z)\ne 0$. Put
$L^{(5)}_z=(L^{(5)}_0,L^{(5)}_{\sigma+1},\ldots ,
L^{(5)}_{n-1},L^{(5)}_n-(X_{n-\sigma}/X_0)(z)L^{(5)}_0)$.

Let $z\in C_{\lambda_5,L^{(5)}}$ and
$z'=q_{c_5,\lambda_5,L^{(5)}}(z)$.
Then by the B\'ezout theorem
(we leave the details to the reader here) for all
$z\in C_{\lambda_5,L^{(5)}}$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number,
the morphisms $q_{c_5,\lambda_5,L^{(5)},*}$
and $q_{c_5,\lambda_5,L^{(5)}}$ are isomorphisms in some neighborhoods of
$z$ with respect to the Zariski topology, $z\not\in\overline{V\setminus W}$,
intersection (36) is transversal at $z$
(in particular $z$ is a smooth point of $C_{\lambda_5,L^{(5)}}$),  the
number of elements $p_{\lambda_5,L^{(5)}}^{-1}
p_{\lambda_5,L^{(5)}}(z)=\deg p'_v\deg W'$,
$e^{(5)}_0(z)\ne 0$, the element $(c_5,\lambda_5,L^{(5)}_{z'})$ satisfies
conditions of Lemma~2 (in place of $(c_0,\lambda_0,L^{(0)})$).

Let $t',t''$ be integers. Set
$B_{\lambda_5,L^{(5)}}=C_{\lambda_5,L^{(5)}}\cap{\cal
Z}(e^{(5)}_0+t'e^{(5)}_n+t''e_{5,n+1})$
in ${\Bbb P}^n(\overline{k})$.
Recall that we know explicitly the polynomial
$\Phi_{c_5,\lambda_5,L^{(5)},+}$.
Hence one can construct integers $t',t''$ with length bounded from above
$O(n\log d+(n-\sigma)\log(d'))$ such that:
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\setcounter{enumi}{4}
\item for
every $z\in B_{\lambda_5,L^{(5)}}$ the
morphism $q_{c_5,\lambda_5,L^{(5)},*}$ (respectively
$q_{c_5,\lambda_5,L^{(5)}}$) is an isomorphism
in a neighborhood with respect to the Zariski topology of the point $z$,
\item for
every $z\in B_{\lambda_5,L^{(5)}}$ the number of elements
$p_{\lambda_5,L^{(5)}}^{-1}
p_{\lambda_5,L^{(5)}}(z)=\\ \deg p'_v\deg W'$.
\item for
every $z\in B_{\lambda_5,L^{(5)}}$
intersection (36) is transversal at $z$,
\item $q_{c_5,\lambda_5,L^{(5)}}(B_{\lambda_5,L^{(5)}})=
W'(c_5,\lambda_5,L^{(5)},*)\cap
{\cal Z}(X_0+t'X_1+t''X_2)$,\\
$\#q_{c_5,\lambda_5,L^{(5)}}(B_{\lambda_5,L^{(5)}})=\deg
W'(c_5,\lambda_5,L^{(5)},*)$,
and $q_{c_5,\lambda_5,L^{(5)}}(B_{\lambda_5,L^{(5)}})\cap{\cal Z}(X_0)=
\emptyset$,
\item for
every $z\in q_{c_5,\lambda_5,L^{(5)}}(B_{\lambda_5,L^{(5)}})$
the element $(c_5,\lambda_5,L^{(5)}_z)$ satisfies
conditions of Lemma~2 (in place of $(c_0,\lambda_0,L^{(0)})$).
\end{enumerate}
Consider the system of equations and inequalities with
respect to
the indeterminates $X_0,\ldots , X_n$, $Y_0$,
$\tau_1,\ldots ,\tau_{n_1}$ with coefficients from the field $k_2$
$$
\left\{
\begin{array}{ll}
\sum_{1\le i\le n_1}|\tau_i|^2\le\varepsilon_1,     \\
e^{(5)}_j-e^{(0)}_j+e^{(1)}_j=0,& \sigma+1\le j\le n-1,\\
e^{(5)}_0-e^{(0)}_0+e^{(1)}_0+\\
+t'(e^{(5)}_n-e^{(0)}_n+e^{(1)}_n)+\\
+t''(e_{5,n+1}-e_{0,n+1}+e_{1,n+1})=Y_0,\\
e^{(5)}_0-e^{(0)}_0+e^{(1)}_0=1,\\
L_{5,j}-L_{0,j}+L_{1,j}=0,& s+1\le j\le \sigma,\\
|Y_0|^2\le\varepsilon_2,\\
\sum_{0\le i\le n}|e_i|^2\ge\varepsilon_2^{-1}.
\end{array}
\right.
\eqno (37)
$$

\par\medskip\noindent{\bf LEMMA~19}\hspace{0.1em} {\it  Suppose that
$(c_1,\lambda_1,L^{(1)})$ satisfies
conditions 1)--4) from (*),
conditions (v)--(ix) hold, $\deg W'(c_1,
\lambda_1,L^{(1)})=\deg W'(c_5,$ $\lambda_5,L^{(5)})$,
$\deg W'(c_1,$ $\lambda_1,L^{(1)},*)>\deg W'(c_5,\lambda_5,L^{(5)},*)$.
Then system (34) has a solution
$X_w=x^*_w$, $0\le w\le n$; $\tau_i=\tau^*_i$, $1\le i\le n_1$; $Y_0=y^*_0$,
from $\overline{k_2}$ such that
$$
(x^*_0:\ldots : x^*_n)\in W(\overline{k_2}).
\eqno (38)
$$
For arbitrary solution (38) the point
$(c_*,\lambda_*,L^*)$ satisfies conditions of Lemma~2 (in place of
$(c_0,\lambda_0,L^{(0)})$), $\deg W'(c_1,$ $\lambda_1,L^{(1)})=
\deg W'(c_5,\lambda_5,L^{(5)})$
and \\ $\deg W'(c_*,$ $\lambda_*,L^*)>\deg W'(c_5,\lambda_5,L^{(5)})$.
}\par\medskip

\noindent {\bf PROOF} Obviously it is sufficient to prove this lemma with
$k_3$ in place of $k_2$.
Denote for brevity $S'=q_{c_5,\lambda_5,L^{(5)}}(B_{\lambda_5,L^{(5)}})$
(in the proof of this lemma the notations $S'$ and $S^*$ are distinct
from the ones introduced in the proof of Lemma~18, but they
have the analogous sense).
Let $\tau_i^*,y^*_0\in\overline{k_3}$, $1\le i\le
n_1$,
be arbitrary (not necessarily related  to a solution of system (34))
such that
$\sum_{1\le i\le n_1}|\tau^*_i|^2\le\varepsilon_1$ and
$|y^*_0|^2\le\varepsilon_2$.
Notice that according to the observations from the beginning of Section~1
the point $(c_*,\lambda_*,L^*)$ satisfies (D) and
conditions of Lemma~2. Therefore,
the objects $q_{c_*,\lambda_*,L^*}$, $W'(c_*,\lambda_*,L^*)$ and
so on are defined. Let $z\in S'$ and $q_{c_5,\lambda_5,L^{(5)}}(z_1)=z$.

Let us show that for every $z\in S'$ there is the unique point $z^*\in
S^*=W'(c_*,\lambda_*,L^*,*)\cap{\cal
Z}(X_0+t'X_1+t''X_2-y^*_0X_0)$ such that
$q_{c_*,\lambda_*,L^*}^{-1}(z^*)$
contains a point $z^*_1$ which is infinitely close to $z_1$
(hence also $z^*$ is infinitely close to $z$).
Note that  by (v) and (viii) the intersection
$C_{\lambda_5,L^{(5)}}\cap{\cal
Z}(e^{(5)}_0+t'e^{(5)}_n+t''e_{5,n+1})$ is transversal at $z_1$.
Hence by
(vii) the intersection $(W_{\lambda_5}\setminus{\cal Z}(e_0,\ldots
,e_r))\cap{\cal
Z}(e^{(5)}_{\sigma+1})\cap\ldots\cap{\cal Z}(e^{(5)}_{n-1})\cap{\cal
Z}(e^{(5)}_0+t'e^{(5)}_n+t''e_{5,n+1})$ is transversal at $z_1$
(in particular $z_1$ is a smooth point of all the considered varieties).

On the other hand, by Lemma~17 (ii) \cite{13} and (ix)
the intersection $W\cap{\cal Z}(L_{5,s+1},$ $\ldots , L_{5,\sigma})\cap
{\cal Z}(e^{(5)}_{\sigma+1},\ldots , e^{(5)}_{n-1},e^{(5)}_n-
(X_{n-\sigma}/X_0)(z)e^{(5)}_0)$ is transversal at
$z_1$. Hence the intersection
$W\cap{\cal Z}(L_{5,s+1},\ldots , L_{5,\sigma})$ is also transversal
at $z_1$.
Therefore, the intersection $W\cap{\cal Z}(L_{5,s+1},\ldots ,
L_{5,\sigma})\cap
{\cal Z}(e^{(5)}_{\sigma+1})\cap\ldots
\cap{\cal Z}(e^{(5)}_{n-1})\cap{\cal Z}(e^{(5)}_0+
t'e^{(5)}_n+t''e_{5,n+1})$
is transversal at $z_1$. Finally, by the implicit function theorem there is
one and only one
point $z_1^*\in q_{c_*,\lambda_*,L^*,*}^{-1}(S^*)$ such that $z_1^*$ is
infinitely close to $z_1$. The required assertion is proved.



Now suppose that $y\in S^*$, $q_{c_*,\lambda_*,L^*,*}(y_1)=y$, and
$y_1$ is not infinitely close to ${\cal Z}(e_0,\ldots , e_r)$.
Then obviously $y_1$ is infinitely close to
$q_{c_5,\lambda_5,L^{(5)},*}^{-1}(S')$ and, hence, $y$ is infinitely close
to $S'$. Conversely, suppose that
$y_1$ is  infinitely close to ${\cal Z}(e_0,\ldots , e_r)$.
Let us show that $y$ is not infinitely close to
$S'$. Indeed,
suppose contrary, that $y$ is infinitely close to $z\in S'$.
Therefore, $X_0(y)\ne 0$.
Put $L^*_{y}=(L^*_0,\ldots , L^*_{n-1},L^*_n-(X_{n-\sigma}/X_0)(y)L^*_0)$.
Then the nonzero coefficients of the linear forms from $L^*_{y}$ are
infinitely
close to the coefficients of the
corresponding linear forms from $L^{(5)}_z$.

By property (D) for $(c_5,\lambda_5,L^{(5)}_z)$, see (ix),
the intersection
$(W\setminus{\cal Z}(e_0,\ldots ,$ $e_r))
\cap{\cal Z}(L_{5,s+1},\ldots , L_{5,\sigma})\cap
{\cal Z}(e^{(5)}_{\sigma+1})\cap\ldots\cap{\cal
Z}(e^{(5)}_{n-1})\cap{\cal Z}(e^{(5)}_n-(X_{n-\sigma}/X_0)(z)e^{(5)}_0)=
A_{\lambda_5,L^{(5)}_z}$
is transversal and consists of $\deg p'_v\deg W'$ points. By Lemma~17
\cite{13} and
the implicit function theorem the intersection
$(W\setminus{\cal Z}(e_0,\ldots , e_r))
\cap{\cal Z}(L_{*,s+1},\ldots , L_{*,\sigma})\cap
{\cal Z}(e^*_{\sigma+1})\cap\ldots\cap{\cal
Z}(e^*_{n-1})\cap{\cal Z}(e^*_n-(X_{n-\sigma}/X_0)(y)e^*_0)=
A_{\lambda_*,L^*_{y}}$
is transversal and consists of $\deg p'_v\deg W'$ points. By the implicit
function theorem the points
from $A_{\lambda_*,L^*_{y}}$ are infinitely close to the corresponding
points from $A_{\lambda_5,L_z}$.
But by (v) $q_{c_5,\lambda_5,L^{(5)}}^{-1}(z)$ consist of one point.
Further, $q_{c_5,\lambda_5,L^{(5)}}^{-1}(z)\subset A_{\lambda_5,L^{(5)}_z}$,
$q_{c_*,\lambda_*,L^*}^{-1}(y)\subset A_{\lambda_*,L^*_{y}}$.
Hence $q_{c_*,\lambda_*,L^*}^{-1}(y)$ also consists of one point
which is infinitely close to the one from
$q_{c_5,\lambda_5,L^{(5)}}^{-1}(z)$.
Thus, $y_1$ is not infinitely close to ${\cal
Z}(e_0,\ldots , e_r)$. This contradiction proves the required assertion.

Let us show that $X_0$ does not vanish on any irreducible
component of $W'(c_*,\lambda_*,L^*,*)$.
Indeed as we have seen every irreducible component of
$C_{\lambda_5,L^{(5)}}$ contains a
point from $A_{\lambda_5,L^{(5)}}$,
and $e^{(5)}_0$ does not vanish at any points from $A_{\lambda_5,L^{(5)}}$.
Notice that the coefficients of the linear forms $L^*_j$ are
infinitely close to the corresponding coefficients of the linear forms
$L^{(5)}_j$, $j\in\{0,\sigma+1,\ldots , n\}$.
Hence by Lemma~17 \cite{13} and the implicit function theorem,
see the last observation related to property (D) at the beginning of
Section~1, every irreducible component of
$(W_{\lambda_*}\setminus{\cal Z}(e_0,\ldots , e_r))\cap{\cal
Z}(e^*_{s+1},\ldots , e^*_\sigma)$
contains a point from $A_{\lambda_*,L^*}$, and
$e^*_0=\widetilde{L}_0(e^*_0,e^*_{\sigma+1},\ldots , e^*_n)$ does not
vanish at any points from $A_{\lambda_*,L^*}$.
Therefore, every irreducible component of $W'(c_*,\lambda_*,L^*,*)$
contains a point from
$q_{c_*,\lambda_*,L^*,*}(A_{\lambda_*,L^*})$, and
$X_0$ does not vanish at any points from
$q_{c_*,\lambda_*,L^*,*}(A_{\lambda_*,L^*})$.
This implies the required assertion.



Let us prove the existence of the required solution of system (37).
By  Lemma~4, Lemma~6 and the algorithm of reduction to integer
coefficients, cf. the
proof of the existence of $(c_*,\lambda_*,L^*)$ in Lemma~17,
there are $\tau^*_i\in k(\varepsilon_3)$, which are
infinitesimals with respect to the field $k$ or zeroes,
such that $(c_*,\lambda_*,L^*)$ satisfies conditions of
Lemma~4, Lemma~6 (for any line ${\mathfrak l}$ containing
$(c_*,\lambda_*,L^*)$) in place of $(c_0,\lambda_0,L^{(0)})$, and
$\deg W'(c_*,\lambda_*,L^*)=\deg W'(c_1,\lambda_1,L^{(1)})$,
$\deg W'(c_*,\lambda_*,$ $L^*,*)=\deg W'(c_1,\lambda_1,L^{(1)},*)$.
We have $\deg W'(c_*,\lambda_*,L^*,*)>\deg W'(c_5,\lambda_5,$ $L^{(5)},*)$.
Hence as we have proved the B\'ezout theorem implies the existence of
a point $z\in W'(c_*,\lambda_*,$ $L^*,*)\cap
{\cal Z}(X_0+t'X_1+t''X_2)$, which
is not infinitely close to $S'$. Further, there exists a point $y\in
W'(c_*,\lambda_*,L^*,*)$ with the coordinates from
$\overline{k(\varepsilon_3)}$
such that $y$ is infinitely close to $z$ with respect to the field $k$,
the inverse image
$q_{c_*,\lambda_*,L^*,*}^{-1}(y)\ne\emptyset$ and $X_0(y)\ne 0$.
Put
$y_0^*=((X_0+t'X_1+t''X_2)/X_0)(y)$. Then
$y_0^*\in\overline{k(\varepsilon_3)}$ is
infinitesimal with respect to
the field $k$.
Let $y_1\in q^{-1}_{c_*,\lambda_*,L^*,*}(y)$.
By previous considerations
the point $y_1$ is infinitely close to ${\cal Z}(e_0,\ldots ,
e_r)$. Thus, system (37) has a
solution satisfying (38) (with $k_3$ in place of $k_2$).

Conversely, if (37) has a
solution satisfying (38) then by the previous considerations and the
B\'ezout theorem
$\deg W'(c_*,\lambda_*,L^*,*)>\deg W'(c_5,\lambda_5,L^{(5)},*)$. The lemma is
proved.

\medskip Let us return to the description of the general step of the
considered recursion.  Notice that system (37)
satisfies conditions 1)--4) from Section~2.
Hence using Lemma~16 and Remark~11 we decide whether system (37) has a
solution
satisfying (38), and if such a solution exists
we construct it.
Suppose that such a solution does not exist. Then by Lemma~19 the degree
$\deg W'(c_1,\lambda_1,L^{(1)},*)=\deg W'(c_5,\lambda_5,L^{(5)},*)$ and the
considered step is final. In this case we can decide whether
$\deg W'(c_1,\lambda_1,L^{(1)})=\deg W'(c_1,\lambda_1,L^{(1)},*)$.
If $\deg W'(c_1,\lambda_1,L^{(1)})\ne\deg W'(c_1,\lambda_1,L^{(1)},*)$
then condition 4) from (*) does not hold for $(c_1,\lambda_1,L^{(1)})$,
and hence (*) also is not satisfied for $(c_1,\lambda_1,L^{(1)})$.
If $\deg W'(c_1,\lambda_1,L^{(1)})=\deg W'(c_1,\lambda_1,L^{(1)},*)$
then condition 4) from (*) holds for $(c_1,\lambda_1,L^{(1)})$,
and hence (*) also is fulfilled for $(c_1,\lambda_1,L^{(1)})$.


Suppose that we have constructed a solution of system (37)
satisfying (38), see the statement of Lemma~19.
Then $\deg W'(c_*,\lambda_*,L^*,*)>\deg W'(c_5,\lambda_5,$ $L^{(5)},*)$ by
this
lemma. Similarly to the previous recursions the conditions of Lemma~4,
Lemma~6 hold for
$(c_*,\lambda_*,L^*)$ in place of $(c_0,\lambda_0,L^{(0)})$.
By Lemma~4, Lemma~6 and the observations
from the beginning of the section one can
apply the algorithm of reduction to integer coefficients,
see \cite{12}, Proposition~2, and replace subsequently all $\tau^*_i$ by
integers $\tau'_i$ of length $O(n\log d+(n-\sigma)\log(d'))$.
Denote by $(c',\lambda,L')$ the obtained point.
Then
$L'\in{\cal U}''_0$, the number of
elements $\#A_{\lambda,L'}=
\deg p'_v\deg W'$, $A_{\lambda,L'}\cap{\cal Z}(e'_0)=\emptyset$, and
$A_{\lambda,L'}\cap\overline{V\setminus W}=\emptyset$,
the dimensions $\dim W_{\lambda}=n-\sigma$,
$\dim (W_{\lambda}\setminus{\cal
Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_{n-1})=1$,
the number of elements
$\#(e'_{n+1}/e'_0)(A_{\lambda,L'})=\#(e_{5,n+1}/e^{(5)}_0)(A_{\lambda_5,L^{(5)}})$,
the morphisms $p_{\lambda,L'}$,
$p_{\lambda,L',*}$ are dominant,
$\deg W'(c',\lambda,L')=\deg W'(c_5,\lambda_5,L^{(5)})$, and
$\deg W'(c',\lambda,L',*)\ge\deg W'(c_*,\lambda_*,L^*,*)$.
Let us apply Lemma~8 to the case when $\iota({\Bbb A}^1(\overline{k}))$
is the line containing
$(c',\lambda,L')$ and $(c_5,\lambda_5,L^{(5)})$, and $\iota(0)=
(c_5,\lambda_5,L^{(5)})$. Then using Theorem~2 (f) \cite{13},
Lemma~8 and Lemma~4, Lemma~6 we
construct a point $(c'',\lambda',L'')$ which
is similar to $(c_5,\lambda_5,L^{(5)})$, and such that
$\deg W'(c'',\lambda',L'',*)\ge\deg W'(c',\lambda,L',*)>\deg
W'(c_5,\lambda_5,L^{(5)},*)$. After that we replace
$(c_5,\lambda_5,L^{(5)})$
by $(c'',\lambda',L'')$ and proceed to the next step of the recursion.
The general step and all the recursion are described.
The decision algorithm for property (*) is described completely.
The required estimation for the working time of this algorithm (it is the
same as for assertion (e)
of Theorem~2, see the Introduction) follows immediately from the estimates
for working times of the
applied algorithms.


\section{Monodromy and irreducibility}\label{s4}

The most part of this
section and its notations are independent from the ones of the
Introduction and other sections.
Let $n\ge 1$ and $1\le s\le n$ be integers.
Let $k$ be an arbitrary field of zero--characteristic, $V_s$ be a defined
over $k$
projective algebraic variety in ${\Bbb P}^n(\overline{k})$, $1\le s\le
n-1$, and all the irreducible components of $V_s$ have the same
dimension $n-s$.
Let $T_0,
T_{s+1},T_{s+2},\ldots , T_{n+1}$ be linear forms in $X_0,\ldots ,X_n$ with
coefficients from $k$.
Consider the defined over $k$ morphisms
\setcounter{equation}{38} \begin{eqnarray}
&&p\, :\, V_s\setminus{\cal Z}(T_0,
T_{s+1},T_{s+2},\ldots , T_n)
\rightarrow{\Bbb P}^{n-s}(\overline{k}),\nonumber\\
&&(X_0:\ldots : X_n)
\mapsto(T_0: T_{s+1}:T_{s+2}:\ldots : T_n),\label{39}\\
&&p'\, :\,V_s\setminus{\cal Z}(T_0,
T_{s+1},T_{s+2},\ldots , T_{n+1})
\rightarrow{\Bbb P}^{n-s+1}(\overline{k}),\nonumber\\
&&(X_0:\ldots :
X_n)\mapsto
(T_0: T_{s+1}:T_{s+2}:\ldots : T_{n+1})
\label{40}
\end{eqnarray}
(one should not confuse these morphisms $p$ and $p'$ with the ones from the
Introduction).
In this section
we shall suppose that $T_0, T_{s+1},T_{s+2},\ldots , T_n$, (respectively
$T_0, T_{s+1},T_{s+2},\ldots ,$ $T_{n+1}$) are the coordinate functions on
${\Bbb P}^{n-s}(\overline{k})$,
(respectively ${\Bbb P}^{n-s+1}(\overline{k})$).
We shall assume that the morphism $p$ is dominant,
i.e., by definition for every irreducible component $W$ of $V_s$
the restriction $p|_W$ is a dominant morphism.
Then the closure in the Zariski topology
$$
\overline{p'(V_s\setminus{\cal
Z}(T_0,
T_{s+1},T_{s+2},\ldots , T_{n+1}))}={\cal Z}(\Phi),
$$
where
$\Phi\in k[T_0, T_{s+1},T_{s+2},\ldots , T_{n+1}]$ is a homogeneous nonzero
separable polynomial which is uniquely defined up to a
nonzero factor from $k$. We fix $\Phi$.
Denote by $q\, :\, V_s\setminus{\cal Z}(T_0,
T_{s+1},T_{s+2},\ldots , T_n)\rightarrow
{\cal Z}(\Phi)$ the
morphism induced by (40).
Denote by $k(V_s)$ and $k({\cal Z}(\Phi))$
the algebras of defined over $k$ rational
functions on
$V_s$ and ${\cal Z}(\Phi)$ respectively (by definition these
algebras are isomorphic to the direct products
of fields of rational functions of all defined over $k$ components of $V_s$
and ${\cal Z}(\Phi)$ respectively).
We shall suppose that the morphism $q$ is a birational isomorphism
in the sense that it induces the isomorphism of separable al\-ge\-b\-ras
$k({\cal Z}(\Phi))\simeq k(V_s)$.
Let $r\, :\, {\cal Z}(\Phi)\setminus{\cal Z}(T_0, T_{s+1},$
$T_{s+2},\ldots , T_n)\rightarrow{\Bbb P}^{n-s}$ be the
morphism of the linear projection to the coordinates $T_0, T_{s+1}$,
$T_{s+2},\ldots , T_n$.
Hence $p=r\circ q$. The morphism $r$ is defined over $k$.

By definition the degree $\deg p$ of the morphism $p$ is the degree of the
extension of al\-ge\-b\-ras
of rational functions
$[k(V_s):k({\Bbb P}^{n-s}(\overline{k}))]=\dim_{k({\Bbb
P}^{n-s}(\overline{k}))}k(V_s)$. Similarly one defines the degree
of an arbitrary dominant morphism.
In particular, $\deg r$ is defined.
We have $\deg p=\deg r=\deg_{T_{n+1}}\Phi$
since $q$ is a birational isomorphism.
Conversely, if $\deg p=\deg_{T_{n+1}}\Phi$ then
$q$ is a birational isomorphism. Denote $\deg p=D$.
Hence the integer $D\ge 1$.

Denote by
$R=\mbox{\rm Res}_{T_{n+1}}
(\Phi,\,\partial \Phi/\partial T_{n+1})\in
k[T_0, T_{s+1},T_{s+2},\ldots , T_n]$
the discriminant with respect to $T_{n+1}$ of the polynomial $\Phi$,
see the remark above about the definition of discriminant
at the beginning of the Introduction. Hence
$0\ne R$ is a homogeneous
with respect to $T_0, T_{s+1},T_{s+2},\ldots , T_n$ polynomial.
Let $R_0$ be square free part of $R$, i.e., $R_0$ is separable and
${\cal Z}(R)={\cal Z}(R_0)$.
The polynomial $R_0$ is uniquely defined up to a nonzero factor from $k$.
We choose and fix $R_0$. Let $\Phi_D\in k[T_0,T_{s+1},\ldots , T_n]$
be the coefficient in $T_{n+1}^D$ of the polynomial
$\Phi$. Hence $\Phi_D$ divides $R$.

Denote by $A^{(s)}$ the set of isolated points of $V_s\cap{\cal
Z}(T_{s+1},\ldots , T_n)$
in ${\Bbb P}^n(\overline{k})$.
Consider the condition
$$
A^{(s)}\cap{\cal Z}(T_0)=\emptyset ,\,\&\,
\#(T_{n+1}/T_0)(A^{(s)})=\deg p.
\eqno (41)
$$
Let $T_0,T_{s+1},T_{s+2},\ldots , T_{n+1}$ be arbitrary linear forms with
coefficients from $k$
such that $p$ is dominant and condition (41) holds
(we do not suppose now in advance that $q$ is birational).
Then by Lemma~2 \cite{12} the morphism $q$ is a birational isomorphism. We
shall suppose till the end of the proof of Theorem~3
that $p$ is dominant and condition (41)
holds.

\par\medskip\noindent{\bf LEMMA~20}\hspace{0.1em} {\it  Let $V_s$, $p$, $q$,
$D$ and $\Phi$, $\Phi_D$ be as above.
Suppose that for a point $z\in {\Bbb
P}^{n-s}(\overline{k})\setminus{\cal Z}(T_0,T_{s+1},\ldots , T_n)$,
the number of elements $\#p^{-1}(z)=D$. Then $\Phi_D(z)\ne 0$.
}\par\medskip

\noindent {\bf PROOF} \quad The proof of this lemma is similar
to the one of Lemma~2 with $d'=1$.
It is sufficient to replace in Lemma~2 $W$ $p_{\lambda,L'}$,
$r_{c',\lambda,L'}$, $q_{c',\lambda,L'}$ by $V_s$, $p$, $q$, $r$.
The only difference here is that $V_s$ is not necessarily
irreducible over $k$. The lemma is proved.







\medskip
Suppose that the morphism $p$ is regular dominant and finite. Then
$p'(V_s)=q(V_s)={\cal Z}(\Phi)$, $\deg\Phi=\deg_{T_{n+1}}\Phi=D$,
$\deg p=\deg V_s=\deg q(V_s)=D$, and the
morphisms $q$ and $r$ are also finite. In this case
we choose leading coefficient
$\mbox{\rm lc}_{T_{n+1}}\Phi=1$.
Now
$A^{(s)}=V_s\cap{\cal Z}(T_{s+1},\ldots , T_n)$. Consider the
condition
$$
A^{(s)}=V_s\cap{\cal Z}(T_{s+1},\ldots , T_n),\,\&\,
A^{(s)}\cap{\cal Z}(T_0)=\emptyset ,\,\&\,
\#(T_{n+1}/T_0)(A^{(s)})=\deg
V_s.
\eqno (42)
$$
If $T_0,T_{s+1},T_{s+2},\ldots , T_{n+1}$ are arbitrary linear forms with
coefficients from $k$
satisfying (42) (we do not suppose now that $p$ is dominant and $q$ is
birational)
then the morphism $p$ is regular dominant finite and $q$ is birational
finite, see e.g., \cite{7}, Section~2. Besides that, (42) implies that
$\#A^{(s)}=D$ and the intersection
$V_s\cap{\cal Z}(T_{s+1},\ldots , T_n)$ is transversal at each point
by the B\'ezout theorem. Condition (41) is a consequence of (42).

\medskip
For an arbitrary polynomial $g\in\overline{k}[T_0,T_{s+1},\ldots , T_{n+1}]$
set
$g_*=g(T_0,0,\ldots ,0,$ $T_n,T_{n+1})$.
We shall suppose in what follows that $R_{0,*}$ is nonzero.
We shall assume
without loss of generality that $T_0$ does not divide
$R_{0,*}$
replacing if it is necessary $T_0$ by
$T_0+\lambda T_n$ with an appropriate integer $\lambda$.
Since $R_{0,*}\ne 0$ the polynomial
$\Phi_{D,*}\ne 0$.  Hence $\deg_{T_{n+1}}\Phi_*=\deg_{T_{n+1}}\Phi$.
Let $\varphi\in k[T_0,T_n]$ be the
greatest common divisor
of the coefficients in $T_{n+1}^j$, $0\le j\le D$,
of the polynomial $\Phi_*$. The polynomial $\varphi$
is uniquely defined up to a nonzero factor
from $k$. Let us choose and fix $\varphi$. Set
$\Phi_+=\Phi_*/\varphi$,
$R_+=R_*/\varphi^{2D-1}$.
Hence $\deg_{T_{n+1}}\Phi_+=\deg_{T_{n+1}}\Phi$.

\par\medskip\noindent{\bf REMARK~12}\hspace{0.1em} {\it  Actually by
Lemma~11 \cite{12} the equality
$\deg_{T_{n+1}}\Phi= \deg_{T_{n+1}}\Phi_+$
follows directly from (41)
}\par\medskip


Let us identify ${\cal Z}(T_{s+1},\ldots , T_{n-1})\subset{\Bbb
P}^n(\overline{k})$ (respectively ${\Bbb
P}^{n-s+1}(\overline{k})$, ${\Bbb
P}^{n-s}(\overline{k})$) with ${\Bbb P}^{s+1}(\overline{k})$
(respectively ${\Bbb
P}^2(\overline{k})$, ${\Bbb
P}^1(\overline{k})$).
Denote by
$V_{s,+}$  the union of all
irreducible components $W$ of
$V_s\cap{\cal Z}(T_{s+1},\ldots , T_{n-1})\subset{\Bbb
P}^{s+1}(\overline{k})$ such that
$W\not\subset{\cal Z}(g)$ for any $0\ne g\in k[T_0,T_n]$.
Let us repeat the described construction to $V_{s,+}\subset{\Bbb
P}^{s+1}(\overline{k})$
in place of $V_s\subset{\Bbb P}^n(\overline{k})$. Now $n$ and $s$ are
replaced by $s+1$ and $s$ respectively.  By Lemma~2 \cite{12},
by condition (41), and
since $\deg_{T_{n+1}}\Phi_+=\deg_{T_{n+1}}\Phi$,
one can replace
the linear forms $T_0,T_{s+1},$ $T_{s+2},\ldots , T_{n+1}$ by
$T_0,T_n,T_{n+1}$.
The new set $A^{(s)}_+=A^{(s)}$ is the same as it was initially.
We get the objects
$p_+$, $q_+$, $r_+$, $\Phi_+$, $R_+$, $R_{0,+}$, $A^{(s)}_+$
which are similar to $p$, $q$, $r$, $\Phi$, $R$, $R_0$, $A^{(s)}$
and satisfy the analogous
properties by the similar reasons. In particular, $\deg p=\deg p_+$.
Note that if $p$ is finite then
$\varphi\in k$ and
$V_{s,+}=V_s\cap{\cal Z}(T_{s+1},\ldots , T_{n-1})$.



Now we can formulate our irreducibility criterion.
\par\medskip\noindent{\bf THEOREM~3}\hspace{0.1em} {\it
Let $n\ge 3$ and $1\le s\le n-2$ be integers.
Let $V_s$ be an equidimensional projective algebraic variety in ${\Bbb
P}^n(\overline{k})$ with $\dim V_s=n-s$.
Let $T_0,T_{s+1},$ $T_{s+2},\ldots , T_{n+1}$ be linear forms from
$k[X_0,\ldots , X_n]$ such that the morphism $p$ is dominant and condition
(41) holds. Let the polynomial $\Phi$ and its
discriminant $R$ be as above. Suppose that
the polynomial $R_{0,*}=R_0(T_0,0,\ldots ,0,T_n)$ is nonzero and separable.
Then $\deg p=\deg p_+$ and the irreducible
over $\overline{k}$ (respectively $k$) components of
algebraic varieties $V_s$ and $V_{s,+}$ are
in the one--to--one correspondence by the rule $W\mapsto W_1$
where $W_1$ is the unique irreducible component of $V_{s,+}$
such that $W_1\subset W\cap{\cal
Z}(T_{s+1},\ldots , T_{n-1})$.  In particular,
if $p$ is finite then $V_{s,+}=V_s\cap{\cal
Z}(T_{s+1},\ldots , T_{n-1})$ and $W_1=W\cap{\cal
Z}(T_{s+1},\ldots , T_{n-1})$.
}\par\medskip

\noindent {\bf PROOF} \quad
Since $V_s$ and $V_{s,+}$ are defined over $k$
it is sufficient to prove only
the assertion related to the irreducible over $\overline{k}$ components.
Denote by $N$ (respectively $N_+$) the number of all irreducible over
$\overline{k}$ components of $V_s$
(respectively $V_{s,+}$).
The equality $A^{(s)}=A^{(s)}_+$ entails that for every irreducible component
$W$ of $V_s$ there is an irreducible component $W_1$  of $V_{s,+}$ such that
$W_1$ is an irreducible component of
$W\cap{\cal Z}(T_{s+1},\ldots , T_{n-1})$.
Condition (41), Lemma~2 \cite{12} and the
equality $A^{(s)}=A^{(s)}_+$
imply that the intersection
$V_s\cap{\cal Z}(T_{s+1},\ldots , T_{n-1})$
is transversal at each irreducible component of $V_{s,+}$. Hence the
inequality $N_+\ge N$ holds, and if $N_+=N$ then the required one--to--one
correspondence takes place. So it is sufficient to prove that $N=N_+$.
Note that $N$ (respectively $N_+$) is the number of irreducible over
$\overline{k}$ components of ${\cal Z}(\Phi)$ (respectively
${\cal Z}(\Phi_+)$) since $q$ (respectively $q_+$) is a birational
isomorphism.

We shall assume without loss of generality that $k$ is a subfield
of the field of complex numbers ${\Bbb C}$.
Recall that $R_{0,*}\in k[T_0,T_n]$ is a nonzero homogeneous polynomial
and $T_0$ does not divide
$R_{0,*}$.
Let $y_1,\ldots , y_u$ be all the pairwise different roots of the
polynomial $R_{0,*}(1,T_n)$.
Hence $u=\deg R_0$ by the condition of the lemma.
Let us choose a point $y\in{\Bbb C}$ such that $R_{0,*}(y)\ne 0$ and
for all
$1\le i_1\ne i_2\le u$ the points $y,y_{i_1},y_{i_2}$ do not belong
to the same line in ${\Bbb C}$ (considered as ${\Bbb R}^2$).
Let $0<m_1\in{\Bbb R}$. Denote by $b_i$ the closed disc
with the radius $m_1$ around $y_i$, $1\le i\le u$.
Let us choose $m_1$ so small that any line in ${\Bbb C}$ trough $y$
intersects
at most one disc $b_i$, $1\le i\le u$, and $|y_i-y_j|>3m_1$, $|y-y_i|>m_1$
for all $1\le i\ne j\le u$.
Denote by $c_i$, $1\le i\le u$,
the circumference with the radius $m_1$ around $y_i$.
Denote by $z_i$ the intersection of the interval $[y,y_i]$ with $c_i$.
Denote by $e_i$ the union of the interval $[y,z_i]$ and $c_i$.
Denote by $\gamma_i$, $1\le i\le u$ the path from $y$ to $z_i$ along
$[y,z_i]$, then by $c_i$
around $y_i$ counterclockwise, and back from $z_i$ to $y$ along
the interval $[z_i,y]$.
Set $e=\cup_{1\le i\le u}e_i$ and $E_z=(T_{n+1}/T_0)(r_+^{-1}(1:z))$
for every $z\in{\Bbb C}$.
Note that for every $z\in {\Bbb C}\setminus{\cal Z}(R)$
the number of elements $\#E_z=D$.

Set $E=E_y$. Hence $\#E=D$.
Let $a\in E$. Denote by $\xi_a$ the analytical function in
$Z$ defined in a neighborhood of $y$
such that $\Phi_+(1,Z,\xi_a)=0$ and $\xi_a(y)=a$.
The result of the analytic continuation of $\xi_a$ along $\gamma_i$ is
$\xi_{\sigma_i(a)}$
where $\sigma_i$ is the uniquely defined permutation of $E$. Denote
by $S_D$ the group
of permutations of
$E$ and by $G$ the subgroup of $S_D$ generated by all $\sigma_i$,
$1\le i\le u$. Then it is known, see, e.g., \cite{4}, \cite{16},
that the number $N_+$ of irreducible over
$\overline{k}$ components of
$V_{s,+}$ coincides with the number
of classes of transitivity of $G$ on $E$.

Let $\mu_i,\nu_i\in k$, $s+1\le i\le n-1$ and $0<\delta\in{\Bbb R}$.
Denote by ${\cal A}_\delta$ the condition $\sum_{s+1\le i\le n-1}
(|\mu_i|^2+|\nu_i|^2)<\delta$.
Set
$\widetilde{T}_i=T_i$, $i=0,n,n+1$
and $\widetilde{T}_i=T_i-\mu_i T_0-\nu_iT_n$ for all
$s+1\le i\le n-1$. By Lemma~2 \cite{12}
there is $0<\delta_0\in{\Bbb R}$ such that under
the condition ${\cal A}_{\delta_0}$
the linear forms $\widetilde{T}_i$
satisfy the same properties (considered above) as $T_i$,
$i\in\{0,s+1,s+2,\ldots , n+1\}$ do. In what follows
 we shall suppose that ${\cal A}_{\delta_0}$
holds. Then
for $\widetilde{T}_i$ the objects
$\widetilde{A}_s$, $\widetilde{\Phi}$, $\widetilde{R}$
and so on are defined.
They correspond to $A^{(s)}$, $\Phi$, $R$ and so on.
Note that $\Phi=\widetilde{\Phi}$, $R=\widetilde{R}$ (as polynomials in
$X_0,\ldots , X_n$), $N=\widetilde{N}$,
$u=\widetilde{u}=\deg R_0$ and
\setcounter{equation}{42} \begin{eqnarray}
&&\widetilde{\Phi}_*=\Phi(T_0,\mu_{s+1}T_0+\nu_{s+1}T_n,\ldots ,
\mu_{n-1}T_0+\nu_{n-1}T_n,T_n, T_{n+1}),\label{43}\\
&&\widetilde{R}_*=R(T_0,\mu_{s+1}T_0+\nu_{s+1}T_n,\ldots ,
\mu_{n-1}T_0+\nu_{n-1}T_n,T_n),\label{44} \\
&&\widetilde{\Phi}_*=\widetilde{\varphi}\widetilde{\Phi}_+,
\label{45}
\end{eqnarray}
where $\widetilde{\varphi}\in k[T_0,T_n]$ divides $\widetilde{\Phi}_{D,*}$
and $\widetilde{\Phi}_{D,*}$ divides $\widetilde{R}_{0,*}$.
Hence by Lemma~1 \cite{11} applied to the polynomial $R_*(1,Z)/\mbox{\rm
lc}_Z(R_*(1,Z))$
there is $0<\delta_1<\delta_0$ such that ${\cal A}_{\delta_1}$
entails $|\widetilde{y}_i-y_i|<m_1/2$ for every $1\le i\le u$.
We shall suppose in what follows that
${\cal A}_{\delta_1}$  holds.
Now we choose $\widetilde{y}=y$ and $\widetilde{m}_1=m_1/2$.

Let $m_2=\inf\{|a_1-a_2|\, :\, a_1,a_2\in E_z\,\&\,a_1\ne a_2\,\&\,z\in e\}$.
Then $m_2>0$ due to the compactness of $e$.
By Lemma~20 for every $z\in e$ the polynomial $\Phi_*(1,z,Z)/\mbox{\rm
lc}_Z\Phi_*(1,z,Z)=\Phi_+(1,z,Z)/\mbox{\rm
lc}_Z\Phi_+(1,z,Z)$.
Hence by
Lemma~1 \cite{11} applied to the polynomials $\Phi_*(1,z,Z)/\mbox{\rm
lc}_Z\Phi_*(1,z,Z)$, $z\in e$, by (43),
(45) and
due to the compactness of $e$ there is $0<\delta_2\le \delta_1$
such that ${\cal A}_{\delta_2}$
entails that for every $z\in e$ for every
$a\in E_z$ there is the unique $\widetilde{a}\in\widetilde{E}_z$ such that
$|a-\widetilde{a}|<m_2/3$.  We shall suppose
in what follows that ${\cal A}_{\delta_2}$
holds.

Now there is the bijection $\tau\, :\, E\rightarrow\widetilde{E}$,
$a\mapsto\widetilde{a}$, $|a-\widetilde{a}|<m_2/3$.
Hence $\tau$ induces the isomorphism $S_D\rightarrow\widetilde{S}_D$.
By our construction
the path $\gamma_i$ is homotopic to $\widetilde{\gamma}_i$ in ${\Bbb
C}\setminus{\cal Z}(\widetilde{R}_*)$,
for every $1\le i\le u$. Therefore,
$\widetilde{\sigma}_i=\tau\circ\sigma_i\circ\tau^{-1}$,
for every $1\le i\le u$.
Hence the groups $G\simeq\widetilde{G}$ and $N_+=\widetilde{N}_+$.

Denote by ${\cal L}_{n-s+1}$,
(respectively ${\cal L}_{n-s-1}$) the vector
space of linear forms with coefficients from
$\overline{k}$ with the basis $T_0,T_{s+1}
,\ldots , T_n$ (respectively $T_{s+1},
,\ldots ,$ $T_{n-1}$).
Let ${\cal L}$ be a linear subspace of the space ${\cal L}_{n-s+1}$.
Denote by $I({\cal L})\subset \overline{k}[T_0,T_{s+1},
,\ldots ,T_{n+1}]$ the ideal generated by ${\cal L}$.
For every polynomial $g\in
\overline{k}[T_0,T_{s+1},\ldots , T_{n+1}]$ denote
$$
g\bmod{\cal L}=g\bmod I({\cal L})\in
\bigl(\overline{k}[T_0,T_{s+1},
,\ldots , T_n]/I({\cal L})\bigr)[T_{n+1}]=A_{\cal L}.
$$
The ring $A_{\cal L}$ is
isomorphic to the ring of polynomials in $n-s+2-
\dim{\cal L}$ variables over $\overline{k}$.
For every $\psi\in A_{\cal L}$
the degree $\deg_{T_{n+1}}\psi$
is defined in the natural way.
According to the first Bertini theorem applied subsequently $n-s-1$ times
(one should delete each time fixed components)
in any neighborhood in the classic topology
of the vector subspace ${\cal L}_{n-s-1}\subset{\cal L}_{n-s+1}$
(the neighborhood of ${\cal L}_{n-s-1}$ is considered here in the
corresponding
Grassman variety)
there is an $(n-s-1)$--dimensional vector subspace
${\cal L}\subset{\cal L}_{n-s+1}$
satisfying the following properties.
\begin{itemize}
\item The polynomial $R\bmod{\cal L}\ne 0$. Hence
$\Phi_D\bmod{\cal L}\ne 0$,
$\deg_{T_{n+1}}\Phi=D$, and
the polynomial $\Phi\bmod{\cal L}$ does not have multiple factors
$\psi$ with $\deg_{T_{n+1}}\psi\ge 1$.
\item For every irreducible over $\overline{k}$ factor $\Psi$ of $\Phi$
with $\deg_{T_{n+1}}\Psi\ge 1$ the polynomial
$\Psi\bmod{\cal L}$ has only one irreducible over $\overline{k}$
factor $\psi$ with $\deg_{T_{n+1}}\psi\ge 1$, and hence $\deg_{T_{n+1}}\psi=
\deg_{T_{n+1}}\Psi$.
\end{itemize}
Therefore, one can choose ${\cal L}$ to be
a linear subspace generated by $\widetilde{T}_{s+1},\ldots ,
\widetilde{T}_{n-1}$
for some $\mu_i$, $\nu_i$, $s+1\le i\le n-1$,
satisfying ${\cal A}_{\delta_2}$. In this case $\widetilde{N}_+=
\widetilde{N}$. Thus,
$N=\widetilde{N}=\widetilde{N}_+=N_+$. The theorem is
proved.

\medskip
Let $n\ge 1$ and $1\le s\le n-1$ be integers.
Till the end
of the section we shall not suppose that condition (41) holds.
We shall assume only that $p$ is a
dominant morphism and $q$ is a birational isomorphism.
Now let $L\in
k[T_0,T_{s+1},$ $T_{s+2},\ldots , T_n]$ be a nonzero linear form.
Denote $A_L=k[T_0,T_{s+1},$ $T_{s+2},\ldots , T_n]/(L)$.
So $A_L$ is isomorphic to the ring of polynomials in $n-s$ variables.
The following conditions (i), (ii) and (iii) are pairwise equivalent.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item $R_0\bmod L\in A_L$ is separable,
\item The intersection ${\cal Z}(R)\cap{\cal Z}(L)$
is transversal in
${\Bbb P}^{n-s}(\overline{k})$.
\item The degree $\deg{\cal Z}(R)=
\deg({\cal Z}(R)\cap{\cal Z}(L))$ in ${\Bbb
P}^{n-s}(\overline{k})$.
\end{enumerate}
More than that there is a
nonempty open in the Zariski topology subset ${\cal U}_{s,1}$
of the space of
linear forms in $T_0,T_{s+1},\ldots , T_n$ such that for every $L\in
U_{s,1}$
conditions (i)--(iii) holds.

Denote by $V_{s,L}$ the union of all the irreducible
components $W$ of
$V_s\cap{\cal Z}(L)$ such that
$W\not\subset{\cal Z}(g)$ for any nonzero element
$g\in k[T_0,T_{s+1},\ldots , T_n]/(L)$.

The following conditions (iv) and (v)
are equivalent by Lemma~2 \cite{12}.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\setcounter{enumi}{3}
\item The dimension $\dim V_{s,L}=n-s-1$, further
$p$ induces the rational dominant
morphism $p_L\, :\, V_{s,L}\rightarrow{\Bbb P}^{n-s-1}(\overline{k})$
(i.e., $p|_W$ is a rational dominant morphism
for every irreducible component $W$ of $V_{s,L}$), and $\deg p_L=\deg p$.
\item The dimension $\dim V_{s,L}=n-s-1$ and
there is a point $z\in{\cal Z}(L)
\subset{\Bbb P}^{n-s}(\overline{k})$ such that the number of elements
$\#p^{-1}(z)=\deg p$.
\end{enumerate}
More than that there is a
nonempty open in the Zariski topology subset ${\cal U}_{s,2}$
of the space of
linear forms in $T_0,T_{s+1},\ldots , T_n$ such
that for every $L\in U_{s,2}$
conditions (iv), (v) holds.

Denote by $V_{s,1}$ the union of all irreducible over $\overline{k}$ closed
subvarieties $W$ of $V_s$
such that $\dim W=n-s-1$ and the dimension of the closure
$$
\dim \overline{p(W\setminus{\cal Z}(T_0,T_{s+1},\ldots , T_n))}<n-s-1.
$$
Note that there is only a finite number of such subvarieties $W$, and the
variety $V_{s,1}$
is defined over $k$ (it may be empty). The
conditions (vi) and (vii) are equivalent.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\setcounter{enumi}{5}
\item $V_{s,L}=V_s\cap{\cal Z}(L)$.
\item $\dim V_{s,1}\cap{\cal Z}(L)\le n-s-2$, i.e., for every irreducible
component $W$ of $V_{s,1}$
the algebraic variety $W\cap{\cal Z}(L)\ne W$.
\end{enumerate}
Besides that, there is a nonzero linear form $L$ satisfying (vi) and (vii)
if and only if $\dim(V_s\cap{\cal Z}(T_0,T_{s+1},\ldots , T_n))
\le\dim(V_s)-2$. More than that if the last condition holds then there is a
nonempty open in the Zariski topology subset ${\cal U}_{s,3}$
of the space of
linear forms in $T_0,T_{s+1},\ldots , T_n$ such that for every
$L\in U_{s,3}$
conditions (vi) and (vii) are satisfied.

\par\medskip\noindent{\bf COROLLARY~1}\hspace{0.1em} {\it  Let $n\ge 1$ and
$1\le s\le n-1$ be integers.  Let $V_s$ be an
equidimensional projective algebraic variety in ${\Bbb
P}^n(\overline{k})$ with $\dim V_s=n-s$.  Let $T_0,T_{s+1},$ $T_{s+2},\ldots
, T_{n+1}$ be linear forms from $k[X_0,\ldots , X_n]$ such that the
morphism $p$ is dominant and $q$ is birational, see above.
Let $L\in k[T_0,T_{s+1},$ $T_{s+2},\ldots , T_n]$ be a nonzero linear form.
Suppose that the equivalent conditions (i)--(iii) hold and the equivalent
conditions (iv) and (v) are satisfied, see above. Then the irreducible over
$\overline{k}$ (respectively irreducible over $k$) components of $V$ and
$V_{s,L}$ are in the one--to--one correspondence by the rule $W\mapsto W_1$
where $W_1$ is the uniquely defined component $V_{s,L}$ such that
$W_1\subset W\cap{\cal Z}(L)$. Besides that, if
the equivalent conditions
(vi) and (vii) hold then $\dim(V_s\cap{\cal
Z}(T_0,T_{s+1},\ldots , T_n)) \le\dim(V_s)-2$.  The set of nonzero linear
forms $L$ satisfying (i)--(iii) (respectively (iv)--(vi)) is a
nonempty open in the Zariski topology subset
${\cal U}_{s,1}$ (respectively ${\cal U}_{s,2}$) of the space of all
linear forms in $T_0,T_{s+1},\ldots , T_n$. If $\dim(V_s\cap{\cal
Z}(T_0,T_{s+1},\ldots , T_n)) \le\dim(V_s)-2$ then the set of nonzero linear
forms $L$ satisfying (vi) and (vii) is a nonempty open in the
Zariski topology subset ${\cal U}_{s,3}$ of the
space of all linear forms in $T_0,T_{s+1},\ldots , T_n$.
}\par\medskip

\noindent {\bf PROOF} \quad Let ${\Bbb P}^{n+1}(\overline{k})$ has
homogeneous coordinates $X_0,\ldots ,X_{n+1}$.
Let us identify ${\Bbb P}^n(\overline{k})={\cal
Z}(X_{n+1})\subset{\Bbb P}^{n+1}(\overline{k})$.
Let $I(V_s)$ be homogeneous ideal of $V_s$.
Denote by $\widetilde{V}_s\subset{\Bbb P}^{n+1}(\overline{k})$ the
projective algebraic variety which is the set of all common zeroes
of $I(V_s)$. Hence $V_s=\widetilde{V}_s\cap{\cal Z}(X_{n+1})$ in
${\Bbb P}^{n+1}(\overline{k})$.
Set $\widetilde{T}_{s+1}=L$ (one should
not confuse the tilde here and below
with the one from the proof of the theorem).
By (v) and Lemma~20 there are
linear forms
$\widetilde{T}_0,\widetilde{T}_{s+2},\ldots , \widetilde{T}_n$ in
$T_0,T_{s+1},$ $T_{s+2},\ldots , T_n$
with coefficients from $k$ such that the linear forms
$\widetilde{T}_0,\widetilde{T}_{s+1},\ldots , \widetilde{T}_n$
are linearly independent,
the number of elements $\#r^{-1}({\cal
Z}(\widetilde{T}_{s+1},$ $\ldots , \widetilde{T}_n))=\deg\Phi$ and
$r^{-1}({\cal Z}(\widetilde{T}_{s+1},\ldots ,
\widetilde{T}_n))\cap{\cal Z}(\widetilde{T}_0)=\emptyset$.
Let us replace in (41)
$n$, $V_s$, $T_0,$ $T_{s+1},\ldots , T_n$ and $T_{n+1}$ by
$n+1$, $\widetilde{V}_s$, $\widetilde{T}_0,$ $\widetilde{T}_{s+1},\ldots ,
\widetilde{T}_n,$ $X_{n+1}$ and
$T_{n+1}$ respectively.
Then the obtained condition holds. We get the polynomials
$\widetilde{\Phi}$, $\widetilde{R}$ corresponding
$\widetilde{V}_s$ and the new linear forms
according to the construction described before the Theorem~3.
Obviously $\widetilde{\Phi}=\Phi$ and $\widetilde{R}=R$
as polynomials in $X_0,\ldots , X_{n+1}$. Now applying
Theorem~3 in the considered situation
we get all the assertions of the corollary. The corollary is proved.


\section{Proof of Theorem~1}\label{s5}



By Lemma~11
to prove (a) it is sufficient to show that ${\cal U}^{({\rm
v})}_{c_1,\lambda_1,0}$ is a subset of
${\cal U}_0$.
Replacing the field $k$ by $k(\tau_1,\ldots
,\tau_{n_1})$
one can suppose without loss of generality that $n_1=0$, and
$(c_1,\lambda_1)=(c',\lambda)$, where
$(c',\lambda)$ is from the Introduction and satisfies (*).


Let $L'\in{\cal U}_{c',\lambda,0}^{({\rm v})}$.
Using the Veronese isomorphism of degree $d'$, see \cite{15}, we can suppose
without loss of generality that $d'=1$, i.e., that
$e'_0,e'_{\sigma+1},\ldots , e'_{n+1}$ are linear forms in
$X_0,\ldots , X_n$ with coefficients from $k$. Let us replace in Theorem~3
$V_s,s$ and all $T_j$
by $W_\lambda,\sigma$ and $e'_j$ respectively and apply this theorem.


Then the irreducible over $\overline{k}$ components of $W_\lambda$ and
$W_{\lambda,n-1}(L')$ are in the one--to--one
correspondence by the rule
$$
E\mapsto\overline{E\cap{\cal
Z}(e'_{\sigma+1},\ldots , e'_{n-1})\setminus{\cal Z}(e_0,\ldots , e_r)}.
\eqno (46)
$$
The irreducible over $\overline{k}$ components of $W$ and $W\cap{\cal
Z}(L_{s+1},\ldots , L_\sigma)$ are in the one--to--one
correspondence by the rule
$E\mapsto E\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)$,
see \cite{13}.
The intersections $(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_{n-1})$ and
$(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_{n-1})$ are transversal, see \cite{13}.
Therefore, the irreducible over $\overline{k}$ components of $W$ and
$W_{\lambda,n-1}(L')$ are in the one--to--one
correspondence by the rule
$$
E\mapsto\overline{E\cap{\cal
Z}(L_{s+1},\ldots , L_\sigma)
\cap{\cal Z}(e'_{\sigma+1},\ldots , e'_{n-1})\setminus{\cal
Z}(e_0,\ldots , e_r)}.
$$
Thus, the irreducible over $\overline{k}$ components of $W$ and
$W_{n-1}(L')$ are in the one--to--one
correspondence by (46).
Hence $L'\in{\cal U}_0$. Assertion (a) is proved.

Let us prove (e). We can suppose without loss of generality that $n_1=n_2+1$,
and conditions of Lemma~12 hold.
Denote for brevity $\Phi_i=\Phi_{c_i,\lambda_i,L^{(i)}}$,
$R_i=R_{c_i,\lambda_i,L^{(i)}}$, $i=1,2$; $\overline{\Phi_1}=
\Phi_1|_{\{\tau_{n_1}=\tau'_{n_1}\}}$, $\overline{R_1}=
R_1|_{\{\tau_{n_1}=\tau'_{n_1}\}}$, $\overline{R_{1,0}}=
R_{1,0}|_{\{\tau_{n_1}=\tau'_{n_1}\}}$, $\overline{R_{1,*}}=
R_{1,*}|_{\{\tau_{n_1}=\tau'_{n_1}\}}$
(here the index $0$ denotes taking square free part, the  index $*$ is also
defined in the Introduction).
Set $\delta=\deg p'_v\deg W'$.
Then by Lemma~12 we have
$\overline{\Phi_1}=\varphi\Phi_2$, where $0\ne
\varphi\in\overline{k}[\tau_1,\ldots ,\tau_{n_2},\,Z_0,Z_{\sigma+1},$
$\ldots ,
Z_n]$.
Hence $\deg_{Z_0,Z_{\sigma+1},\ldots ,
Z_{n+1}}\Phi_{c_2,\lambda_2,L^{(2)}}\le
\deg_{Z_0,Z_{\sigma+1},\ldots , Z_{n+1}}\Phi_{c_1,\lambda_1,L^{(1)}}$.

Further, $\overline{R_1}=\varphi^{2\delta-1}R_2$. Therefore,
$\overline{R_{1,0}}\ne 0$ and
$R_{2,0}$ divides $\overline{R_{1,0}}$. Hence $b(c_2,\lambda_2, L^{(2)})\le
b(c_1,\lambda_1, L^{(2)})$.

We have $R_{1,*}\ne 0$ by Lemma~2.
By Lemma~13 the polynomial $(\,\Phi_{1,*}/(\tau_{n_1}-
\tau'_{n_1})^\gamma\,)|_{\{\tau_{n_1}=\tau'_{n_1}\}}=
\varphi_1 \Phi_{2,*}$, $(\,R_{1,*}/(\tau_{n_1}-
\tau'_{n_1})^{\gamma(2\delta-1)}\,)|_{\{\tau_{n_1}=\tau'_{n_1}\}}=
\varphi_1^{2\delta-1} R_{2,*}$ where
$\varphi_1\in\overline{k}(\tau_1,\ldots ,\tau_{n_2},Z_0,Z_n)$. By condition
4) from (*) for $\Phi_2$ the element
$\varphi_1\in\overline{k}[\tau_1,\ldots ,\tau_{n_2},$ $Z_0,Z_n]$. Hence
$\deg R_{1,*,0}\ge\deg R_{2,*,0}$. Thus, $a(c_2,\lambda_2, L^{(2)})\le
a(c_1,\lambda_1, L^{(2)})$. Assertion (e) is proved.

Assertion (d) follows from Lemma~8 and the B\'ezout theorem immediately.

Let us prove (b) and (c) by the induction on $n_1$. Let $n_2+1=n_1$, and the
point
$(c_2,\lambda_2)$ is obtained from $(c_1,\lambda_1)$ by the substitution an
integer $\tau'_{n_1}$ for $\tau_{n_1}$ in (17), (18). By (d)
there is $\tau'_{n_1}$ with length $O(n\log d+(n-\sigma)\log d')$ such that
$(c_2,\lambda_2,L')$ satisfies the same properties as $(c_1,\lambda_1,L')$
does.
Thus, one can replace $(c_1,\lambda_1,L')$ by $(c_2,\lambda_2,L')$, and
assertions (b) and (c) are proved by induction.

Let us prove (g). Consider the point $(c_1,\lambda_1,U)$.
If one extends the ground field then $U\in{\cal U}^{({\rm
v})}_{c_1,\lambda_1,0}$.
Hence by the B\'ezout theorem and Lemma~8 using the algorithm of reduction
to integer coefficients, see Proposition~2 \cite{12},
one can replace subsequently all the coefficients
of the linear forms $U_j$, $j\in\{0,\sigma+1,\ldots , n\}$ by integers
of lengths $O(n\log d+(n-\sigma)\log d')$ and obtain the required $L'$.
Assertion (g) is proved.

Let us prove (f). The inequalities from (f) are
particular cases of the ones from (e),
see the definitions from the Introduction.
Suppose that $\deg W'(c_2,\lambda_2)=\deg W'(c_1,$ $\lambda_1)$ and
$b(c_2,\lambda_2)=b(c_1,\lambda_1)$.
Let $L'\in{\cal U}^{({\rm v})}_{c_2,\lambda_2,0}$
By (c) we can suppose without loss of generality that
$(c_2,\lambda_2)=(c',\lambda)$. Then by Lemma~9 property
(*) holds for $(c_1,\lambda_1,L')$. Further, (ii) and (iii) from
Definition~2 for
$(c',\lambda,L')$ imply (ii) and (iii) for $(c_1,\lambda_1,L')$. Thus,
$L'\in{\cal U}^{({\rm v})}_{c_1,\lambda_1,0}$. Assertion (f) is proved.
The theorem is proved.



\section{Proof of Theorem~2}\label{s6}

Obviously, assertion (a) of this theorem follows from (e)
and Definition~2. Assertion (b) follows from (a). Assertion (c) follows from
(e)
and Definition~2.
Assertion (d) is a partial case of (c). So it is sufficient to prove only
(e).

Notice that $\deg W'(c_1,\lambda_1,U)=\deg W'(c_1,\lambda_1)$ and
$b(c_1,\lambda_1)=b(c_1,\lambda_1,U)$ by definition, and
$b(c_1,\lambda_1,U)=a(c_1,\lambda_1,U)$ by Lemma~10.
So it is sufficient to prove all the
statements from (e)
apart of the ones related to computing $\deg W'(c_1,\lambda_1)$ and
$b(c_1,\lambda_1)$. After that, to compute $\deg W'(c_1,\lambda_1)$ and
$b(c_1,\lambda_1)$, one can take $L^{(1)}=U$.


Under conditions from Section~3 let $\deg W'(c_1,\lambda_1,L^{(1)})=
\deg W'(c_1,\lambda_1,$ $L^{(1)},*)$ and
$(c_5,\lambda_5,L^{(5)})\in{\cal L}_{c_1,\lambda_1,L^{(1)}}$
be the element constructed by the decision algorithm from Section~3.
Hence $\deg W'(c_1,\lambda_1,L^{(1)})=\deg W'(c_5,\lambda_5,L^{(5)})=
\deg W'(c_5,\lambda_5,L^{(5)},*)$ and the both
$(c_1,\lambda_1,L^{(1)})$, $(c_5,\lambda_5,L^{(5)})$
satisfy property (*). Further, condition 5), see the Introduction, holds
for $(c_5,\lambda_5,L^{(5)})$.

We shall describe a recursive algorithm by $(c_5,\lambda_5,L^{(5)})$.
At the end of the recursion we shall obtain an element
$(c_5,\lambda_5,L^{(5)})$
such that $a(c_5,\lambda_5,L^{(5)})=a(c_1,\lambda_1,L^{(1)})$.
As the base of the recursion we take the element $(c_5,\lambda_5,L^{(5)})$
constructed in Section~3.

Let us describe the general step of this recursion.
Let us compute $a(c_5,\lambda_5,$ $L^{(5)})$ and
$R_{c_5,\lambda_5,L^{(5)},*,0}$,
see the observations at the beginning of Section~3.
Replacing if it is necessary $e^{(5)}_0$ by $e^{(5)}_0+\mu e^{(5)}_n$ for an
appropriate
integer $\mu$ with length $O(n\log d+(n-\sigma)\log(d'))$, we shall assume in
what follows without loss of generality that
$Z_0$ does not divide $R_{c_5,\lambda_5,L^{(5)},*,0}$.
Denote for brevity
$\xi_j=e^{(5)}_j-e^{(0)}_j+e^{(1)}_j$, $j\in\{0,\sigma+1,\ldots
,n\}$,
and $e_{n+1}=e_{5,n+1}-e_{0,n+1}+e_{1,n+1}$
(the homogeneous polynomials $e^{(5)}_j,e^{(0)}_j,e^{(1)}_j$,
$e_{5,n+1},e_{0,n+1},e_{1,n+1}$
are defined in the Introduction and Section~3).
Let $X^{(j)}$ be the family of variables $X_{j,0},\ldots , X_{j,n}$, $1\le
j\le 4$.
For an arbitrary polynomial $p$ in $n+1$ variables denote for brevity
$p(X^{(j)})=
p(X_{j,0},\ldots , X_{j,n})$.
Set
\begin{eqnarray*}
&&A_0=\varepsilon_4-|e_{n+1}(X^{(1)})-e_{n+1}(X^{(2)})|^2,\quad
A_1=|e_{n+1}(X^{(1)})|^2-\varepsilon_4^{-1},    \\
&&B_0=\varepsilon_4-|e_{n+1}(X^{(3)})-e_{n+1}(X^{(4)})|^2,\quad
B_1=|e_{n+1}(X^{(3)})|^2-\varepsilon_4^{-1},\\
&&I_0=\{1,2\},\quad I_1=\{1\},\quad J_0=\{3,4\},\quad J_1=\{3\}.
\end{eqnarray*}
Let $(w_1,w_2)\in\{(0,0),(0,1),(1,1)\}$.
Consider the system of polynomial equations and
inequalities with squares of absolute values with respect to
the variables $\tau_1,\ldots ,\tau_{n_1}$,
$X_{j,0},\ldots , X_{j,n}$, $j\in I_{w_1}\cup J_{w_2}$ with coefficients
from the field $k_4$
$$
\left\{
\begin{array}{ll}
\sum_{1\le i\le n_1}|\tau_i|^2\le\varepsilon_1,     \\
\sum_{\sigma+1\le i\le n-1}|\xi_i|^2\le\varepsilon_2,\\
\varepsilon_3\le|\xi_n(X^{(1)})-\xi_n(X^{(3)})|^2\le\varepsilon_2,\\
A_{w_1}=0,\\
B_{w_2}=0,\\
\xi_i(X^{(j)})=0, & j\in I_{w_1}\cup J_{w_2},\;
\sigma+1\le i\le n-1,\\
\xi_n(X^{(j_1)})=\xi_n(X^{(j_2)}), & (j_1,j_2)\in I_{w_1}^2\cup J_{w_2}^2,\\
\xi_0(X^{(j)})=1,& j\in I_{w_1}\cup J_{w_2},\\
L_j(X^{(j)})=0,& j\in I_{w_1}\cup J_{w_2},\;s+1\le j\le\sigma.
\end{array}
\right.
\eqno (47)
$$

\par\medskip\noindent{\bf LEMMA~21}\hspace{0.1em} {\it  The integer
$a(c_5,\lambda_5,L^{(5)})<a(c_1,\lambda_1,L^{(1)})$
if and only if there is a pair $(w_1,w_2)\in\{(0,0),(0,1),(1,1)\}$
such that system (47) has a
solution $X_{j,i}=x_{j,i}^*$, $0\le j\le n$,
$j\in I_{w_1}\cup J_{w_2}$ satisfying the condition
$$
(x_{j,0}^*:\ldots : x_{j,n}^*)\in W(\overline{k_5}),\quad j\in I_{w_1}\cup
J_{w_2}.
\eqno (48)
$$
}\par\medskip

\noindent{\bf PROOF}\quad Suppose that
$a(c_5,\lambda_5,L^{(5)})<a(c_1,\lambda_1,L^{(1)})$. There is an element
$\varepsilon'$
such that $\varepsilon'>0$ be an infinitesimal with respect to the field
$k_2$ and $\varepsilon_3$ be an infinitesimal
with respect to the field $k_2(\varepsilon')$.
By the transfer principle, see \cite{3}, it is sufficient to construct a
solution (48) of system (47)
with the field $\overline{k(\varepsilon',\varepsilon_4)}$ in place of
$\overline{k_4}$.

By Lemma~4, Lemma~6, Lemma~7 and the algorithm of reduction to
integer coefficients, cf. the
proof of the existence of $(c_*,\lambda_*,L^*)$ in Lemma~17,
there are $\tau^*_i\in k(\varepsilon')$ which are
infinitesimal with respect to the field $k$
such that for $(c_*,\lambda_*,L^*)$ satisfies conditions of
Lemma~4, Lemma~6, Lemma~7
(for any line ${\mathfrak l}$ containing
$(c_*,\lambda_*,L^*)$) in place of $(c_0,\lambda_0,L^{(0)})$,
the degrees $\deg W'(c_*,\lambda_*,L^*)=\deg W'(c_1,\lambda_1,L^{(1)})$,
$\deg W'(c_*,\lambda_*,L^*,*)=\deg W'(c_1,\lambda_1,$ $L^{(1)},*)$, and
$a(c_*,\lambda_*,$ $L^*)=a(c_1,\lambda_1,L^{(1)})$
(recall that the point $(c_*,\lambda_*,L^*)$ is defined in Section~3 before
system (34)). Let us choose the polynomial $\Phi_{c_*,\lambda_*,L^*}$
(it is uniquely defined up to a factor from $\overline{k(\varepsilon')}$)
such that the following property holds.
All the coefficients of
$\Phi_{c_*,\lambda_*,L^*}$
are not infinitely large with respect to the field $k$, and
some nonzero coefficient $\varphi$ of $\Phi_{c_*,\lambda_*,L^*}$
is not infinitesimal with respect to the $k$.
Recall that in general situation, if the mapping the standard part
$\mbox{\rm st}\, :\, K\setminus\{\mbox{infinitely large
elements}\}\rightarrow k$ is defined, then for an
arbitrary polynomial $f$ with coefficients from
$K\setminus\{\mbox{infinitely large elements}\}$ the standard part
$\mbox{\rm st}(f)$ is the polynomial with coefficients from $k$
such that all the nonzero coefficients of
the polynomial $f-\mbox{\rm st}(f)$ are infinitesimal with
respect to $k$.
Now by Lemma~14 (with the field $k(\varepsilon')$ and $\mbox{\rm
st}_{\varepsilon'}$
in place of $k_i$ and $\mbox{\rm st}_i$ respectively)
and since $\deg W'(c_*,\lambda_*,L^*)=\deg
W'(c_5,\lambda_5,L^{(5)})$,
the standard part $\mbox{\rm
st}_{\varepsilon'}(\Phi_{c_*,\lambda_*,L^*})$ coincides with
$\Phi_{c_5,\lambda_5,L^{(5)}}$  up to a nonzero factor from $\overline{k}$.
Hence by Lemma~1 applied to $\Phi_{c_5,\lambda_5,L^{(5)}}$ also
$\mbox{\rm st}_{\varepsilon'}(\Phi_{c_*,\lambda_*,L^*,*})$
coincides
with $\Phi_{c_5,\lambda_5,L^{(5)},*}$ up to a nonzero factor from $k$.
We have $\deg_{Z_0,Z_n,Z_{n+1}}\Phi_{c_*,\lambda_*,L^*,*}=
\deg_{Z_0,Z_n,Z_{n+1}}\Phi_{c_5,\lambda_5,L^{(5)},*}=
\deg W'(c_1,\lambda_1,L^{(1)},*)$ and
$\deg_{Z_{n+1}}\Phi_{c_*,\lambda_*,L^*,*}=
\deg_{Z_{n+1}}\Phi_{c_5,\lambda_5,L^{(5)},*}=\deg p'_v\deg W'$ by our
construction.
Hence \\ $\mbox{\rm
st}_{\varepsilon'}(R_{c_*,\lambda_*,L^*,*}))=R_{c_5,\lambda_5,L^{(5)},*}$.
Since $a(c_5,\lambda_5,L^{(5)})<a(c_*,\lambda_*,L^*)$,
the polynomial $R_{c_*,\lambda_*,L^*,*}$ has more pairwise distinct roots
than
$R_{c_5,\lambda_5,L^{(5)},*}$. Therefore, there are two distinct infinitely
close roots
$\xi_{n,0}$, $\xi_{n,1}$ of the polynomial
$R_{c_*,\lambda_*,L^*,*}(1,Z_n)\in \overline{k(\varepsilon')}[Z_n]$.
Hence $\xi_{n,0}-\xi_{n,1}\in \overline{k(\varepsilon')}$ is an
infinitesimal with respect to the field $k$.

Therefore, for every $i=0,1$ one of two cases take place:
\begin{enumerate} \renewcommand{\labelenumi}{\arabic{enumi})}
\item the polynomial $\Phi_{c_*,\lambda_*,L^*,*}(1,\xi_{n,i},Z_{n+1})$
has a multiple root $Z_{n+1}=\xi_{n+1,i}$,
\item $(\mbox{\rm lc}_{Z_{n+1}}
(\Phi_{c_*,\lambda_*,L^*,*}))(1,\xi_{n,i})=0$
(recall that leading coefficient\\ $\mbox{\rm lc}_{Z_{n+1}}
(\Phi_{c_*,\lambda_*,L^*,*})\in \overline{k}[Z_0,Z_n]$).
\end{enumerate}
Permuting $i=0,1$ we can suppose without loss of generality that if 1)
holds for some $i=0,1$ then 1) holds for $i=0$.

For convenience of notation set $I_{1,1}=I_0$, $I_{1,2}=I_1$,
$I_{2,1}=J_0$, $I_{2,2}=J_1$.
%$A_{1,0}=A_0$, $A_{1,1}=A_1$, $A_{2,0}=B_0$, A_{2,1}=B_1$.
If w) (here ${\rm w}=1$ or ${\rm w}=2$)
is fulfilled for $i$ then by the transfer principle
there are $\widetilde{\xi}_{n,i}\in\overline{k(\varepsilon',\varepsilon_4)}$
and
$\widetilde{\xi}_{n+1,j}\in\overline{k(\varepsilon',\varepsilon_4)}$,
$j\in I_{i,w}$, satisfying the following conditions:
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item the element
$\widetilde{\xi}_{n,i}$  is infinitely close to
$\xi_{n,i}$ with respect to the field
$\overline{k(\varepsilon')}$,
\item if $w=1$ then
$|\widetilde{\xi}_{n+1,j_1}-\widetilde{\xi}_{n+1,j_2}|^2=\varepsilon_4$,
where $j_1,j_2\in I_{i,1}$,
$j_1\ne j_2$, and every element $\widetilde{\xi}_{n+1,j}$, $j\in
I_{i,w}$ is infinitely close to $\xi_{n+1,i}$,
\item if $w=2$ then $|\widetilde{\xi}_{n+1,j}|^2=\varepsilon_4^{-1}$, where
$j\in I_{i,2}$.
\end{enumerate}
Notice that $(1:\widetilde{\xi}_{n,i}:\widetilde{\xi}_{n+1,j})$ is a generic
point of
$W'(c_1,\lambda_1,L^{(1)},*)$
for all considered $i,w,j$.
Further, $q_{c_*,\lambda_*,L^*,*}$ is a dominant morphism
of curves and
$\deg q_{c_*,\lambda_*,L^*,*}=1$. Hence
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\setcounter{enumi}{3}
\item
$\#q^{-1}_{c_*,\lambda_*,L^*,*}((1:\widetilde{\xi}_{n,i}:
\widetilde{\xi}_{n+1,j}))=1$
for all $j\in I_{i,w}$.
\end{enumerate}
Now for $i=0,1$ set $w_i=0$ if 1) holds, and $w_i=1$ if 2) holds.
Put $(x^*_{j,0}:\ldots : x^*_{j,n})\in
q^{-1}_{c_*,\lambda_*,L^*,*}((1:\widetilde{\xi}_{n,i}:
\widetilde{\xi}_{n+1,j}))$, $j\in
I_{w_1}\cup
I_{w_2}$. Then according to our construction $X_{j,i}=x_{j,i}^*$, $0\le j\le
n$,
$j\in I_{w_1}\cup J_{w_2}$ is solution of system (47)
satisfying condition (48) (with the field
$k(\varepsilon',\varepsilon_4)$ in place of $k_4$).

Conversely, suppose that there is
solution (48) of system (47).
Then as it follows from the description of the algorithm for the considered
recursion the inequality
$a(c_5,\lambda_5,L^{(5)})<a(c_1,\lambda_1,L^{(1)})$ holds.
The lemma is proved.

\medskip Let us return to the description of the general step of the
recursion. Notice that system (47)
satisfies conditions 1)--4) from Section~2.
Hence using Lemma~16 and Remark~11
we decide whether system (47) has a solution
satisfying (48), and if such a solution exists
we construct it.
Suppose that such a solution does not exist. Then by Lemma~21
the integer
$a(c_5,\lambda_5,L^{(5)})=a(c_1,\lambda_1,L^{(1)})$ and the
considered step is final. Put
$(c_4,\lambda_4,L^{(4)})=(c_5,\lambda_5,L^{(5)})$.
The point $(c_4,\lambda_4,L^{(4)})$ satisfies all the statements from
Theorem~2 (e).

Suppose that we have constructed a solution of system (47)
satisfying (48), see the statement of Lemma~21.
Then from system (47) and (48) we get immediately, cf. the
proof of Lemma~21, that
$a(c_*,\lambda_*,L^*)>a(c_5,\lambda_5,L^{(5)})$ by this
lemma. Similarly to the previous recursions conditions of Lemma~4,
Lemma~6 and Lemma~7 hold for
$(c_*,\lambda_*,L^*)$ in place of $(c_0,\lambda_0,L^{(0)})$.
By Lemma~4, Lemma~6, Lemma~7 and the observations
from the beginning of Section~3 one can
apply the algorithm of reduction to integer coefficients,
see \cite{12}, Proposition~2, and replace subsequently all values $\tau^*_i$
by
integers $\tau'_i$ of length $O(n\log d+(n-\sigma)\log(d'))$.
Denote by $(c',\lambda,L')$ the obtained point.
Then
$L'\in{\cal U}''_0$, the number of
elements $\#A_{\lambda,L'}=
\deg p'_v\deg W'$, $A_{\lambda,L'}\cap{\cal Z}(e'_0)=\emptyset$, and
$A_{\lambda,L'}\cap\overline{V\setminus W}=\emptyset$,
the dimensions $\dim W_{\lambda}=n-\sigma$,
$\dim (W_{\lambda}\setminus{\cal
Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_{n-1})=1$,
the number of elements
$\#(e'_{n+1}/e'_0)(A_{\lambda,L'})=
\#(e_{5,n+1}/e^{(5)}_0)(A_{\lambda_5,L^{(5)}})$,
the morphisms $p_{\lambda,L'}$,
$p_{\lambda,L',*}$ are dominant,
$\deg W'(c',\lambda,L',*)=\deg W'(c_5,\lambda_5,L^{(5)})$,
$\deg W'(c',\lambda,L',*)=\deg W'(c_5,\lambda_5,L^{(5)},*)$,
and $a(c',\lambda,L')\ge a(c_*,\lambda_*,L^*)$.
Let us apply Lemma~8 to the case when $\iota({\Bbb A}^1(\overline{k}))$
is the line containing
$(c',\lambda,L')$ and $(c_5,\lambda_5,L^{(5)})$, and $\iota(0)=
(c_5,\lambda_5,L^{(5)})$. Then using Theorem~2 (f) \cite{13}, Lemma~8 and
Lemma~4, Lemma~6,
Lemma~7 we construct a point $(c'',\lambda',L'')$, which
is similar to $(c_5,\lambda_5,L^{(5)})$ and such that
$a(c'',\lambda',L'')\ge a(c',\lambda,L')>a(c_5,\lambda_5,L^{(5)})$.
After that we replace $(c_5,\lambda_5,L^{(5)})$
by $(c'',\lambda',L'')$ and proceed to the next step of the recursion.
The general step and all the recursion are described.

The required estimation for the working time of the present algorithm
for assertion (e)
of Theorem~2 follows immediately from the estimates
for working times of the
applied algorithms. Assertion (e) of Theorem~2 is proved.
Thus, Theorem~2 is proved, see beginning of this section.









\section*{Appendix: a proof of a version of the first Bertini theorem
in arbitrary characteristic}

To make our presentation in \cite{11}, \cite{12}, \cite{13} and the present
paper self--contained
we give an independent proof of the first Bertini theorem
(and some other related to it results) for the case of
affine algebraic varieties in arbitrary characteristic, see Theorem~4 below.
Notice here that in our presentation we have only two
references to the first Bertini theorem
and only to the partial case of hypersurfaces of affine space in
zero--characteristic.
Namely, they are in the proofs of Theorem~3
of the present paper and Theorem~3 \cite{11}.
Still for convenience of the reader and for possible future applications
we decided to consider in Theorem~4 a more general situation of affine
algebraic varieties over algebraically closed
fields of arbitrary characteristic.
Lemma~24 implies the general version of
the first Bertini theorem formulated in the Introduction.

In this section we use the notations which may be different from the ones of
the Introduction. Let $k$ be a perfect field of arbitrary characteristic
with algebraic closure $\overline{k}$.
Let $B$ be an arbitrary commutative Noetherian integral ring.
We shall denote by $B'$ the integral closure
of $B$ in its field of fractions. For a finitely generated
$B$--module $M$ denote
by $\mbox{\rm Ass}_B(M)$ the set of all
associated prime ideals of $M$
(they are prime ideals of $B$).
Recall that an ideal of $B$ is called radical if it
coincides with its nil--radical. In what follows
the morphisms of algebraic varieties are regular morphisms if it
is not stated otherwise.

Let $n\ge 2$ be an integer.
Let ${\Bbb A}^n(\overline{k})$ be affine
space over the field $\overline{k}$ with the
coordinate functions $X_1,\ldots ,X_n$.
Let $x\in {\Bbb A}^n(\overline{k})$ be the point with the coordinates
$X_i(x)=0$, $1\le i\le n$.

Let $V$ be an affine
algebraic variety defined over $\overline{k}$
and irreducible over $\overline{k}$. Let $p\, :\, V\rightarrow
{\Bbb A}^n(\overline{k})$ be a
dominant separable morphism.
We shall identify the ring $\overline{k}[X_1,\ldots ,X_n]
\subset\overline{k}[V]$.
For an element $g\in\overline{k}[X_1,\ldots ,X_n]$
we shall denote by ${\cal Z}(g)$ the subset of all
zeroes of the polynomial $g$ in
${\Bbb A}^n(\overline{k})$. By $V\cap{\cal Z}(g)$
we shall denote for convenience of notation the subset
$p^{-1}({\cal Z}(g))\subset V$.

We shall suppose additionally that
$$
\dim p^{-1}(x)\le n-2.
\eqno (49)
$$

Denote by $B=\overline{k}[V]$  the ring of
regular functions of the algebraic variety $V$.
Let
$z\in V$ be a point. By
${\mathfrak M}_z$ denote
the maximal ideal of $B$ corresponding to the point $z$.

Let $y\in B$ be a
primitive element of the separable algebraic
extension $\overline{k}(V)\supset \overline{k}(X_1,\ldots , X_n)$, i.e.,
$$
\overline{k}(V)=\overline{k}(X_1,\ldots , X_n)[y].
\eqno (50)
$$
Let $f\in \overline{k}[X_1,\ldots ,X_n,Y]$ be minimal polynomial of $y$
over $\overline{k}(X_1,\ldots , X_n)$ (here $Y$ is a new variable) and
$f$ be irreducible in the ring $\overline{k}[X_1,\ldots
,X_n,Y]$. So the degree $D=\deg_Yf\ge 1$. Let
$\deg_{X_1,\ldots ,X_n,Y}f=m_1$
and $f^{(m_1)}$ be the homogeneous
with respect to $X_1,\ldots ,X_n,Y$
form of degree $m_1$ of the polynomial
$f$.


The element $y$ is integral over
$\overline{k}[X_1,\ldots ,X_n]$ if the morphism $p$ is finite.
Denote by $\mbox{\rm lc}_Yf$ the leading coefficient of the
polynomial $f$ with respect to $Y$.
We choose $f$ such that $\mbox{\rm lc}_Yf=1$
if $p$ is finite.

Besides that, we shall suppose that {\it if $n=2$ then $p$ is finite.}

Denote by
$$
 \Delta_1=\mbox{\rm Res}_Y(f,
\partial f/\partial Y)\in\overline{k}[X_1,\ldots ,X_n]
\eqno (51)
$$
the discriminant of the polynomial $f$ with respect to $Y$.
So $\Delta_1\ne 0$ since $f$ is separable.
\par\medskip\noindent{\bf REMARK~13}\hspace{0.1em} {\it
By definition we assume that the discriminant $\Delta_1$
is the determinant of the $(2D-1)\times(2D-1)$
Sylvester matrix of the resultant. Although
in the case of nonzero characteristic
of the ground field the degree $\deg_Y\partial f/\partial Y$
may be less than $D-1$ the partial derivative
 $\partial f/\partial Y$ is considered formally as a polynomial of
degree $D-1$ when one represents this matrix.
Some higher coefficients of $\partial f/\partial Y$ may be equal to zero
in this representation. Recall once more that this
definition of the
discriminant is different
from the generally accepted by a factor
$\mbox{\rm lc}_Yf$, but it is
more convenient for us
}\par\medskip


Let us identify the set of all
linear forms from $\overline{k}[X_1,\ldots ,X_n]$ with
${\Bbb A}^n(\overline{k})$. Let ${\mathfrak P}$ be a prime ideal of
$\overline{k}[X_1,\ldots ,X_n]$
(respectively $B$, $B'$). Suppose that
$$
{\cal Z}({\mathfrak P})\not\subset{\cal Z}(X_1,\ldots ,X_n),
$$
where ${\cal Z}(X_1,\ldots ,X_n)$ and ${\cal Z}({\mathfrak P})$ are
considered as
subvarieties of ${\Bbb A}^n(\overline{k})$ (respectively $V$, the
normalization of $V$).
Then the set $S_{\mathfrak P}$ of linear forms $L\in\overline{k}[X_1,\ldots
,X_n]$
such that $L\not\in{\mathfrak P}$ is
nonempty and open in the Zariski topology in
${\Bbb A}^n(\overline{k})$ since
$$
S_{\mathfrak P}=\bigcup_{z\in{\cal Z}({\mathfrak
P})}\Bigl\{L\in{\Bbb A}^n(\overline{k})\, :\, L(z)\ne 0\Bigr\}.
$$


Consider
$B'/B$ as a $B$--module.
Set
\begin{eqnarray*}
&&{\cal B}_0=\mbox{\rm Ass}_B
(B'/ B),\qquad
{\cal B}_1=\{{\mathfrak P}\in {\cal B}_0\, :\,
{\cal Z}({\mathfrak P})\not\subset p^{-1}(x)\},\\
&&{\cal B}_2=\{{\mathfrak P}\in {\cal B}_0\, :\,
\dim{\cal Z}({\mathfrak P})=n-1\}.
\end{eqnarray*}
Consider the set ${\cal U}_{1}$ (respectively ${\cal U}_{2}$)
of linear forms
$L$ in $X_1,\ldots ,X_n$ such that $L\not\in{\mathfrak P}$ for every
${\mathfrak P}\in{\cal B}_1$
(respectively ${\cal B}_2$).
Then
${\cal U}_{1}$ (respectively ${\cal U}_{2}$)
is a nonempty open in the Zariski topology subset of
${\Bbb A}^n(\overline{k})$ and ${\cal U}_1\subset{\cal U}_2$.
Note that ${\cal U}_{1}={\cal U}_2={\Bbb A}^n(\overline{k})$ if
${\cal B}_1=\emptyset$,
in particular if $B=B'$ (i.e., if $V$ is normal)
since in the latter case
$\mbox{\rm Ass}_B(B'/B)=\emptyset$.
Note also that if the set of singular points of $V$
has dimension at most $n-2$ then ${\cal B}_2=\emptyset$
and ${\cal U}_2={\Bbb A}^n(\overline{k})$.

Let $L\in\overline{k}[X_1,\ldots ,X_n]$ be a nonzero linear form.
Consider $B/LB$ as a $B$--module. Set
$$
{\cal B}_{1,L}=({\cal B}_0\setminus{\cal B}_1)\cap
\mbox{\rm Ass}_B(B/LB),\quad
{\cal B}_{2,L}=
({\cal B}_0\setminus{\cal B}_2)\cap
\mbox{\rm Ass}_B(B/LB).
$$

Let us show that there is a nonzero element
$\Delta_3\in\overline{k}[X_1,\ldots ,X_n]$ such that the
extensions of rings
$$
B[\Delta_3^{-1}]\supset\overline{k}[X_1,\ldots
,X_n,\,\Delta_3^{-1}][y]\supset\overline{k}[X_1,\ldots
,X_n][\Delta_3^{-1}]
\eqno (52)
$$
are integral. Indeed, by (50)
one can choose $\Delta_3$ to be the product of $\mbox{\rm lc}_Yf$
and all the leading coefficients
of minimal polynomials over $\overline{k}[X_1,\ldots ,X_n]$ of the elements
of a finite system of generators of the
ring $B$ over $\overline{k}[X_1,\ldots ,X_n]$.
In the case when $p$ is finite we choose $\Delta_3=1$.
Fix such an element $\Delta_3$.

Set ${\cal B}_3=\mbox{\rm Ass}_{B'}(B'/\Delta_3 B')$ where
$B'/\Delta_3 B'$ is considered as a $B'$--module.
Hence the elements of ${\cal B}_3$ are prime ideals of $B'$. By
${\cal U}_{3}$ denote the set
of all linear forms $L\in\overline{k}[X_1,\ldots ,X_n]$ such that
$L\not\in{\mathfrak P}$ for every ${\mathfrak P}\in{\cal B}_3$.
Since $B'$ is integrally closed,
the dimension $\dim{\cal Z}({\mathfrak P})=n-1$ for every ${\mathfrak
P}\in{\cal B}_3$, see \cite{2}.
Therefore, ${\cal U}_{3}$ is
a nonempty open in the Zariski topology subset
of ${\Bbb A}^n(\overline{k})$.
Note that if $p$ is finite then ${\cal U}_{3}={\Bbb
A}^n(\overline{k})$.
Note also that if
$L\in{\cal U}_{3}$ then $L$ does not divide
$\Delta_3$ in the ring $\overline{k}[X_1,\ldots ,X_n]$.


\par\medskip\noindent{\bf THEOREM~4}\hspace{0.1em} {\it  Let $n\ge 2$.
Let $V$, $B$, $p\, :\, V\rightarrow {\Bbb A}^n(\overline{k})$, $y$, $f$,
$\Delta_1$, $\Delta_3$, $m_1$, $f^{(m_1)}$, ${\cal U}_{1}$,
${\cal U}_{2}$,
${\cal U}_{3}$, ${\cal B}_0$, ${\cal B}_1$,
${\cal B}_2$, ${\cal B}_3$, ${\cal B}_{1,L}$ and
${\cal B}_{2,L}$
be as above.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item Then there is a nonzero
linear form $L\in\overline{k}[X_1,\ldots , X_n]$ in
$X_1,\ldots , X_n$ such that  the ideal
$(f,L)\subset\overline{k}[X_1,\ldots ,X_n,Y]$
is radical and if $n\ge 3$ it is prime.
The element $\Delta_1\bmod L\in \overline{k}[X_1,\ldots ,X_n]/(L)$
is nonzero (i.e., $L$ does not divide $\Delta_1$) and
$$
f^{(m_1)}\bmod L\in (\,\overline{k}[X_1,\ldots ,X_n]/(L)\,)[Y]
$$
is also nonzero.
\item  The set ${\cal U}_{4}$
of linear forms $L$ satisfying
the properties from (i)
is a nonempty open in the Zariski topology
subset of ${\Bbb A}^n(\overline{k})$.
\item Let
$L\in{\cal U}_{1}\cap{\cal U}_{3}
\cap{\cal U}_{4}$ (respectively $L\in{\cal U}_{2}\cap{\cal U}_{3}
\cap{\cal U}_{4}$).
Then the ideal
$$
LB={\mathfrak p}\cap{\mathfrak q},
$$
where the ideal ${\mathfrak p}$ is the radical of the ideal
$LB$ and  if $n\ge 3$ it is prime.
The ideal
${\mathfrak q}$ is an intersection of ${\mathfrak P}$-primary ideals for all
prime ideals ${\mathfrak P}\in{\cal B}_{1,L}$
(respectively ${\mathfrak P}\in{\cal B}_{2,L}$),
herewith we suppose that the intersection over the empty set
(when ${\cal B}_{1,L}=\emptyset$ or ${\cal B}_{2,L}=\emptyset$)
is $B$.
Hence if $L\in{\cal U}_{1}\cap{\cal U}_{3}
\cap{\cal U}_{4}$ and $p$ is finite then ${\mathfrak q}$ is an intersection
of
${\mathfrak M}_z$-primary ideals for some $z\in p^{-1}(x)$
or ${\mathfrak q}=B$.
Besides that,
for every ${\mathfrak P}\in{\cal B}_{1,L}$
(respectively ${\mathfrak P}\in{\cal B}_{2,L}$)
the dimension
$\dim{\cal Z}({\mathfrak P})\le n-2$ and therefore,
the ${\mathfrak P}$--primary component of the ideal $LB$ is embedded.
\item Let $L\in{\cal U}_{2}\cap{\cal U}_{3}\cap{\cal U}_{4}$.
Then the algebraic variety $V\cap {\cal Z}(L)$
has dimension $\dim V-1$ and if $n\ge 3$ then
$V\cap {\cal Z}(L)$ is irreducible
over $\overline{k}$.
Further, the projection induced by $p$
$$
V\cap{\cal Z}(L)\rightarrow  {\cal Z}(L)
\eqno (53)
$$
is a dominant and separable morphism of
affine algebraic varieties
(here ${\cal Z}(L)\subset {\Bbb A}^n(\overline{k})$).
Besides that if $p$ is a finite morphism then
morphism (53) is also finite.

Denote by $\overline{k}(V\cap{\cal Z}(L))$ the
total quotient ring of the ring of regular functions of the
algebraic variety
$V\cap{\cal Z}(L)$; hence it is the field of rational functions
of this algebraic variety if $n\ge 3$.
Then
\setcounter{equation}{53} \begin{eqnarray}
&&[\overline{k}(V\cap{\cal Z}(L)):\overline{k}({\cal
Z}(L))]=
[\overline{k}(V):\overline{k}(X_1,\ldots ,X_n)],\label{54}
\end{eqnarray}
herewith the left part of this equality is the dimension of the total
quotient ring over the field
$\overline{k}({\cal Z}(L))$ and the right part is the degree of the
extension of fields of rational functions
$\overline{k}(V)\supset\overline{k}({\Bbb A}^n(\overline{k}))$.
Denote by
$y\bmod L$ the image of the element $y$
in $\overline{k}[V\cap{\cal Z}(L)]$.
Then minimal polynomial of the element $y\bmod L$
over $\overline{k}({\cal Z}(L))$ is $f\bmod L$ and
$y\bmod L$ is a primitive element of
the separable algebra $\overline{k}(V\cap{\cal Z}(L))$ over
$\overline{k}({\cal Z}(L))$.
\item Let $n\ge 3$.
Let ${\cal L}$ be a linear subspace
of the space of all linear forms
from $\overline{k}[X_1,\ldots ,X_n]$ with $\dim{\cal L}\ge 3$.
Suppose that for every ${\mathfrak P}\in{\cal B}_3$ such that
$\dim{\cal Z}({\mathfrak P})=n-1$ the dimension of the closure
$$
\dim\overline{p({\cal Z}({\mathfrak P}))}\ge n-2.
\eqno (55)
$$
Then ${\cal L}\cap{\cal U}_2\cap{\cal U}_3\ne\emptyset$.
\item Let $n\ge 3$.
Let ${\cal L}$ be a linear subspace
of the space of all linear forms
from $\overline{k}[X_1,\ldots ,X_n]$ with $\dim{\cal L}\ge 3$. Then
${\cal L}\cap{\cal U}_4\ne\emptyset$.
\end{enumerate}
}\par\medskip

\noindent {\bf PROOF} \quad
Let $n=2$. Then we define ${\cal U}_{4}$ to be the set of all linear
forms $L\in\overline{k}[X_1,X_2]$
such that $L$ does not divide $\Delta_1 f^{(m_1)}$. Hence ${\cal U}_{4}$
is a nonempty open in the Zariski topology subset
of ${\Bbb A}^2(\overline{k})$ in this case.
Recall that by our assumption
the leading coefficient $\mbox{\rm lc}_Yf=1$ if $n=2$.
Hence if $L\in{\cal U}_{4}$ then $\Delta_1\bmod L\ne 0$ and therefore,
the polynomial $f\bmod L$ is separable. Therefore, the ideal
$(f,L)$ is radical. Also $f^{(m_1)}\bmod L\ne 0$. Assertions (i)
and (ii) are proved for $n=2$.

Now our aim is to reduce (iii), (iv) and (v) to
(i) and (ii) for $n\ge 3$ (assertion (i) and (ii) for $n\ge 3$ will
be proved later). So we assume now that (i) and (ii) hold for $n\ge 3$.
Let $\widetilde{V}$ be the normalization of $V$.
Hence $\overline{k}[\widetilde{V}]=B'$.

Denote by $S_1=\{\Delta_3^N\, :\, 0\le N\in{\Bbb Z}\}$ the
multiplicatively closed subset of $\overline{k}[X_1,\ldots ,X_n]$.
Denote for brevity $A=\overline{k}[X_1,\ldots ,X_n]$. By $V_1$
denote the affine algebraic variety with the
ring of regular functions $A[Y]/(f)$.

\par\medskip\noindent{\bf LEMMA~22}\hspace{0.1em} {\it
Let $L\in{\cal U}_{3}\cap{\cal U}_{4}$. Then the
ideal $LB'$ of the ring $B'$ is radical
(respectively prime if $n\ge 3$) and
the natural ho\-mo\-mor\-ph\-isms
\setcounter{equation}{55} \begin{eqnarray}
&&B'/LB'\rightarrow S_1^{-1} (B'/LB'),\quad \label{56}\\
&&A[Y]/(f,L)\rightarrow S_1^{-1} A[Y]/(f,L) \quad \label{57}
\end{eqnarray}
are
monomorphisms. The natural ho\-mo\-mor\-ph\-ism $A[Y]/(f,L)\rightarrow
B'/LB'$ is a
monomorphism.
Hence one can identify  $A[Y]/(f,L)\subset B'/LB'$. Finally under
this identification the algebras $A[Y]/(f,L)$ and $B'/LB'$ have the same
total quotient ring,
i.e., one can identify the total quotient rings
$\overline{k}(\widetilde{V}\cap{\cal Z}(L))$ and
$\overline{k}(V_1)$ of the rings of regular
functions  of the algebraic varieties $\widetilde{V}\cap{\cal
Z}(L)$ and $V_1$.
}\par\medskip

\noindent {\bf PROOF} \quad
Let us prove that (56) is a monomorphism.
It is sufficient to prove that the
multiplication to $\Delta_3\, :\, B'/LB'\rightarrow B'/LB'$
is a monomorphism, see \cite{2}.
Let $z\in B'$ and $\Delta_3z=Lz_1$ for some
$z_1\in B'$. Since $L\in{\cal U}_{3}$, the multiplication to
$L\, :\,  B'/\Delta_3\,B'\rightarrow B'/\Delta_3\,B'$ is a monomorphism.
Therefore, there is $z_2\in B'$
such that $z_1=\Delta_3 z_2$. Since $B'$ is integral, the element $z=Lz_2$.
Hence (56) is a monomorphism.

Since $L\in{\cal U}_{3}$,
the linear form $L$ does not divide $\Delta_3$.
Therefore, the ho\-mo\-mor\-ph\-ism (57)
is a monomorphism.

The extensions of rings (52) are integral, and the element $y$ is
separable
over $\overline{k}(X_1,\ldots ,X_n)$. Hence, see, e.g., \cite{2},
in this case one can identify
$$
S_1^{-1} B'\subset\frac{1}{\Delta_1}S_1^{-1} A[Y]/(f,L).
$$
Let us show that under this identification
$$
L\,S_1^{-1} B'=\Bigl(\,\frac{L}{\Delta_1}S_1^{-1} A[Y]/(f,L)\,\Bigr)\cap
S_1^{-1} B'.
\eqno (58)
$$
Indeed, let $(L/\Delta_1)a_1=a_2$ where $a_1\in S_1^{-1} A[Y]/(f,L)$ and
$a_2\in S_1^{-1} B'$. Set $a=a_2/L$.
Then $La\in S_1^{-1} B'$ and $\Delta_1 a\in S_1^{-1} A[Y]/(f,L)$.
Consider the minimal polynomial $\psi\in A[Z]$
(here $Z$ is a new variable)
of the element $a$ over $\overline{k}(X_1,\ldots ,X_n)$ such that
$\psi$ is an irreducible element of the ring $A[Z]$.
Then the leading coefficient $\mbox{\rm lc}_Z\psi$ divides
$\mbox{\rm G\,C\,D\,}(L^D,\Delta_1^D)$
in the ring $S_1^{-1} A$
Hence $\mbox{\rm lc}_Z\psi=\lambda \Delta_3^N$ where
$0\ne \lambda\in\overline{k}$ and an integer $N\ge 0$. Therefore, $a\in
S_1^{-1} B'$,
since the last ring is integrally closed. The
required assertion is proved.

Let us show that $(\,L S_1^{-1} B'\,)\cap
(\,S_1^{-1}A[Y]/(f,L)\,)=L S_1^{-1}A[Y]/(f,L)$.
Indeed, $L$ does not divide $\Delta_1$ and $S_1^{-1}A[Y]/(f,L)$ is a free
$S_1^{-1}(A/(L))$--module. Hence the required assertion follows from
(58).
Now one can identify
$$
S_1^{-1}A[Y]/(f,L)\subset S_1^{-1} (B'/LB')\subset
\frac{1}{\Delta_1\bmod L}S_1^{-1} A[Y]/(f,L)\subset
\overline{k}(V_1\cap{\cal Z}(L)).
\eqno (59)
$$
Since (56) and (57) are monomorphisms, (59)
implies that $A[Y]/(f,L)\rightarrow B'/LB'$ is also
a monomorphism. Therefore, the first and the
last statements of the lemma again follows from (59).
The lemma is proved.

\medskip Now we are going to prove (iii) and (iv).
By Lemma~22 the ideal
$LB'$ of $B'$ is radical and if $n\ge 3$ it is prime.
Hence the ideal $LB'\cap B$ is radical in $B$ and if $n\ge 3$ it is prime.
Since $B'$ is integral over $B$ the ideal
$LB'\cap B$ is the radical of the ideal $LB$ of $B$.
Consider the morphism $\gamma\, :\, B/LB\rightarrow
B'/LB'$
induced by the inclusion $B\subset B'$. Denote $E_1=\mbox{\rm
Ker}\gamma$ and $E_2=\mbox{\rm Im}\gamma$.
So we have the exact sequence
$$
0\rightarrow E_1\rightarrow B/LB\rightarrow E_2
\rightarrow 0.
$$
Hence $\mbox{\rm Ass}_B(B/LB)\subset\mbox{\rm
Ass}_B(E_1)\cup\mbox{\rm Ass}_B(E_2)$.
We have $\mbox{\rm Ass}_B(E_2)=\\
\mbox{\rm Ass}_B(B/(LB'\cap B))$
since $E_2\simeq B/(LB'\cap B)$.

The inclusion $B\subset B'$ and the multiplication to
$L\, :\, B'\rightarrow B'$  induce the commutative diagram
of ho\-mo\-mor\-ph\-isms of $B$--mo\-du\-les
$$
\begin{CD}
0@>>> B@>>>B'@>>>B'/B@>>> 0   \\
@.  @VV{L}V  @VV{L}V@VV{L}V  @.    \\
0@>>> B@>>> B'@>>>B'/B@>>> 0
\end{CD}
$$
with the exact lines.
Considering Ker--Coker sequence corresponding to this diagram
we get the exact sequence
$$
0\longrightarrow E_1\longrightarrow B'/B\stackrel{L}{\longrightarrow}
B'/B\longrightarrow 0,
$$
i.e., $E_1$ can be identified with the kernel of the ho\-mo\-mor\-ph\-ism of
multiplication to $L\, :\, B'/B
\rightarrow B'/B$. But $L\in{\cal U}_1$ (respectively $L\in{\cal
U}_2$). Hence
$\mbox{\rm Ass}_B(E_1)\subset{\cal B}_{1,L}$ (respectively
$\mbox{\rm Ass}_B(E_1)\subset{\cal B}_{2,L}$).
Further, $\mbox{\rm Ass}_B(B/(LB'\cap B))\cap({\cal
B}_0\setminus{\cal B}_2)=\emptyset$ since
for every ${\mathfrak P}\in \mbox{\rm Ass}_B(B/(LB'\cap B))$ the
dimension $\dim{\cal Z}({\mathfrak P})=n-1$ and for every
${\mathfrak P}\in{\cal B}_0\setminus{\cal B}_2$ the dimension
$\dim{\cal Z}({\mathfrak P})\le n-2$.
Note that ${\cal B}_0\setminus{\cal B}_1
\subset{\cal B}_0\setminus{\cal B}_2$.
Therefore,
$\mbox{\rm Ass}_B(E_1)={\cal B}_{1,L}$ (respectively
$\mbox{\rm Ass}_B(E_1)={\cal B}_{2,L}$) and also
for every ${\mathfrak P}\in\mbox{\rm Ass}_B(E_1)$
the set ${\cal Z}({\mathfrak P})$ is not
an irreducible component of ${\cal Z}(LB'\cap B)$.


Thus,
$$
\overline{k}[V\cap{\cal Z}(L)]\simeq B/(B\cap LB')\subset B'/LB'\simeq
\overline{k}[\widetilde{V}\cap{\cal Z}(L)].
\eqno (60)
$$
Now (60) and Lemma~22 imply that one can identify
$$
A[Y]/(f,L)\subset B/(B\cap LB')\subset B'/LB',
$$
and the rings $A[Y]/(f,L)$, $B/(B\cap LB')$, $B'/LB'$ have the same
total quotient ring.

Finally we need to prove (54). It is sufficient
to show that
$$
[\overline{k}(V_1\cap {\cal Z}(L)):
\overline{k}({\cal Z}(L))]=
[\overline{k}(V_1):\overline{k}(X_1,\ldots ,X_n)].\eqno (61)
$$
But it is equivalent to the equality $\deg_Yf=\deg_Y(f\bmod L)$
which is satisfied since $L$ does not divide $\Delta_1$.

Thus assertions (iii) and (iv) (modulo (i) and (ii) for $n\ge 3$) are
proved. Let us prove (v). If ${\mathfrak P}\in{\cal
B}_2\setminus{\cal B}_3$ then
$\dim\overline{p({\cal Z}({\mathfrak P}))}=\dim{\cal Z}({\mathfrak
P})=n-1$, since extensions of rings
(52) are integral. Therefore, for every ${\mathfrak P}\in{\cal
B}_2\cup{\cal B}_3$ the dimension
$\dim\overline{p({\cal Z}({\mathfrak P}))}\ge n-2$. Hence
$\dim({\mathfrak P}\cap{\cal L})\le 2$
for every ${\mathfrak P}\in{\cal B}_2\cup{\cal B}_3$. Thus
${\cal L}\setminus{\mathfrak P}\ne\emptyset$
for every ${\mathfrak P}\in{\cal B}_2\cup{\cal B}_3$. This implies
(v) (modulo (i) and (ii) for $n\ge 3$).

\medskip
Now our aim is to prove  assertions (i), (ii) and (vi) of Theorem~4
for $n\ge 3$.
Recall that $V_1$ is the affine algebraic
variety with ring of regular functions
$A[Y]/(f)$ and the ring of polynomials
$A=\overline{k}[X_1,\ldots ,X_n]$.

Denote by ${\cal U}'_n$ the set of linear forms
$L=\sum_{1\le i\le n}l_iX_i\in\overline{k}[X_1,\ldots ,X_n]$ such that
all $l_1,\ldots , l_n\in\overline{k}$ and $l_n\ne 0$. For every
$L\in{\cal U}'_n$
the isomorphism
$$
\overline{k}[X_1,\ldots ,X_n]/(L)\simeq \overline{k}[X_1,\ldots ,X_{n-1}]
\eqno (62)
$$
induced by the substitution $X_n=-(l_1X_1+\ldots + l_nX_n)/l_n$ holds.
In what follows for every linear form
$L\in{\cal U}'_n$ we shall identify the rings in the left and right
parts of (62) using this isomorphism.

Recall that the set of linear forms ${\cal U}_4$
is defined in assertion (ii) of Theorem~4.
If $f\bmod~L$ is irreducible over $\overline{k}$ then
the condition $\Delta_1\bmod L\ne 0$
(i.e., that $L$ does not divide $\Delta_1$) is
equivalent
to $(\partial f/\partial Y)\bmod L\ne 0$ and $\deg_Yf\bmod L=\deg_Yf$.
Hence ${\cal U}_4\cap{\cal U}'_n$ is the subset
of all linear forms from ${\cal U}'_n$
such that the element $f\bmod~L$ is irreducible in the ring
$\overline{k}[X_1,\ldots ,X_{n-1},Y]$, the degrees
$$
\deg_{Y,X_1,\ldots ,X_n}(f\bmod~L)=
\deg_{Y,X_1,\ldots ,X_n}f,\quad \deg_Y(f\bmod~L)=\deg_Yf
$$
and $(\partial f/\partial Y)\bmod L\ne 0$.

Due to symmetry of indices $1,\ldots , n$ assertions (i) and (ii)
of Theorem~4
follow immediately from the following lemma.

\par\medskip\noindent{\bf LEMMA~23}\hspace{0.1em} {\it
Let $n\ge 3$ and $f\in \overline{k}[X_1,\ldots , X_n,Y]$
be an irreducible polynomial such that
$\partial f/\partial Y\ne 0$.
Then ${\cal U}_4\cap{\cal U}'_n$ is a
nonempty open in the Zariski topology subset of ${\cal U}'_n$.
}\par\medskip

\noindent {\bf PROOF} \quad
At first our aim is to show that
${\cal U}_4\cap{\cal U}'_n\ne\emptyset$,
i.e., to construct some linear form $L$ from
${\cal U}_4\cap{\cal U}'_n$.
Denote $\deg_Yf=D$ and $\deg_{Y,X_1,\ldots ,X_n}f=m_1$.


Let us represent $f=\sum_{i,j\ge 0}f_{i,j}Y^j$ where
$f_{i,j}\in\overline{k}[X_1,\ldots ,X_n]$ are
homogeneous polynomials with $\deg f_{i,j}=i$
for all $i,j$.
Let $\alpha,\beta\in \overline{k}$ and
$$
(\Delta_1 \sum_{0\le i\le m_1}f_{i,m_1-i}
)(X_1,X_2,\ldots ,
X_{n-2},\alpha X_1,\beta  X_1)\ne 0.
\eqno (63)
$$
Then $\Delta_1
\sum_{0\le i\le m_1}f_{i,m_1-i}\not\in
(X_{n-1}-\alpha X_1,X_n-\beta X_1)$ where
$(X_{n-1}-\alpha X_1,X_n-\beta X_1)$ is the ideal of the ring
$\overline{k}[X_1,\ldots ,X_n]$ generated by the
elements $X_{n-1}-\alpha X_1$, $X_n-\beta X_1$. Hence
$$
 f\not\in (X_{n-1}-\alpha X_1,X_n-\beta X_1).\eqno (64)
$$

 Denote by
$\widetilde{f}\in\overline{k}[X_1,\ldots , X_n,Y]$ the polynomial such
that
$$
\widetilde{f}(X_1,X_2,\ldots ,
X_{n-2},X_{n-1}-\alpha X_1,X_n-\beta  X_1,Y)=f.
$$
Then $\widetilde{f}$ is an irreducible element of the ring
$\overline{k}[X_1,\ldots , X_n,Y]$.
Since $\widetilde{f}$ is irreducible
the polynomial $\widetilde{f}(X_1,X_2,\ldots , X_{n-1},X_{n-1}X_n,Y)$ is
reducible if and only if $X_{n-1}$ divides it, i.e., if and only
if $\widetilde{f}\in (X_{n-1},X_n)$.
This is impossible by (64). Hence
$\widetilde{f}(X_1,X_2,\ldots , X_{n-1},X_{n-1}X_n,Y)$ is
also an irreducible
element of the ring $\overline{k}[X_1,\ldots , X_n,Y]$.
Suppose that
$$
\widetilde{f}(X_1,X_2,\ldots , X_{n-1},X_{n-1}X_n,Y)\in
\overline{k}[X_1,\ldots , X_{n-1},Y]. \eqno (65)
$$
Then we can take $L=X_n-\alpha X_1$.
So we shall suppose in what follows in the proof that
(65) does not hold.

Suppose that there is $\lambda\in\overline{k}$ such that the polynomial
$\widetilde{f}(X_1,\ldots , X_{n-1},\lambda X_{n-1},$ $Y)$
is irreducible in the
ring
$\overline{k}[X_1,\ldots , X_{n-1},Y]$. Then obviously the linear form
$L=\lambda(X_1-\beta X_n)-(X_1-\alpha X_{n-1})$ satisfies the
statement of the theorem.

Consider the lexicographic order on the $n$--tuples $(i_1,\ldots , i_n)$,
where $0\le i_1,\ldots , i_n\in{\Bbb Z}$. Namely, set $(i_1,\ldots ,
i_n)>(j_1,\ldots , j_n)$ if and only if
there is $1\le r\le n$ such that $i_1=j_1,\ldots , i_{r-1}=j_{r-1}$ but
$i_r>j_r$.
Let  a field $K\supset\overline{k}$.
Let
$$
F=\sum_{i_1,\ldots , i_n\ge 0}F_{i_1,\ldots , i_n}X_1^{i_1}\ldots
X_{n-1}^{i_{n-1}}Y^{i_n}\in K[X_1,\ldots , X_{n-1},Y]
$$
be an arbitrary
nonzero polynomial where all $F_{i_1,\ldots , i_n}\in K$. Set
\begin{eqnarray*}
&&(j_1,\ldots , j_n)=\max\{(i_1,\ldots , i_n)\, :\, F_{i_1,\ldots , i_n}\ne
0\},\\
&&\mbox{\rm lex}(F)=X_1^{j_1}\ldots X_{n-1}^{j_{n-1}}Y^{j_n},\\
&&\mbox{\rm lexc}(F)=F_{j_1,\ldots , j_n}.
\end{eqnarray*}
Note that the condition $\mbox{\rm lex}(F)\ne 1$
is equivalent to $F\not\in\overline{k}$.
For every polynomials $F_1, F_2$ similar to $F$ the equalities $\mbox{\rm
lex}(F_1)\mbox{\rm lex}(F_2)=
\mbox{\rm lex}(F_1F_2)$,
$\mbox{\rm lexc}(F_1)\mbox{\rm
lexc}(F_2)=\mbox{\rm lexc}(F_1F_2)$ hold.

Suppose that
\begin{enumerate} \renewcommand{\labelenumi}{(\Alph{enumi})}
\setcounter{enumi}{1}
\item for every $\lambda\in\overline{k}$  the polynomial
$\widetilde{f}_\lambda=\widetilde{f}(X_1,\ldots , X_{n-1},\lambda X_{n-1},Y)$
is not irreducible in the ring $\overline{k}[X_1,\ldots , X_{n-1},Y]$.
\end{enumerate}


Let $\widetilde{f}_\lambda=g_\lambda h_\lambda$
where $g_\lambda, h_\lambda\in
\overline{k}[X_1,\ldots , X_{n-1},Y]\setminus\overline{k}$.
Hence there is $\lambda_0\in\overline{k}$
such that $\mbox{\rm lex}(g_\lambda)=
\mbox{\rm lex}(g_{\lambda_0})$,
$\mbox{\rm lex}(h_\lambda)=
\mbox{\rm lex}(h_{\lambda_0})$ for
infinitely many
$\lambda\in\overline{k}$. Note that $\mbox{\rm
lex}(g_{\lambda_0})\ne 1$ and
$\mbox{\rm lex}(h_{\lambda_0})\ne 1$ since
$g_{\lambda_0},h_{\lambda_0}\not\in\overline{k}$.
Now consider the polynomials $G,H$ in $X_1,\ldots , X_{n-1},Y$
with indeterminate coefficients such that
$\mbox{\rm lex}(G)=
\mbox{\rm lex}(g_{\lambda_0})$,
$\mbox{\rm lex}(H)=
\mbox{\rm lex}(h_{\lambda_0})$ and
\begin{eqnarray*}
&&\deg_{Y,X_1,\ldots ,X_n}G,
\deg_{Y,X_1,\ldots ,X_n}H\le\\
&&\deg_{Y,X_1,\ldots ,X_n}
\widetilde{f}(X_1,X_2,\ldots , X_{n-1},X_{n-1}X_n,Y).
\end{eqnarray*}
In particular $\mbox{\rm lexc}(G),
\mbox{\rm lexc}(H)$ are independent variables.
Denote for brevity $\widetilde{f}_{X_n}=
\widetilde{f}(X_1,X_2,\ldots , X_{n-1},X_{n-1}X_n,Y)$.
The equality $\widetilde{f}_{X_n}=GH$ and the
conditions $\mbox{\rm lexc}(G)\ne 0,
\mbox{\rm lexc}(H)\ne 0$
are equivalent to a system of polynomial equations
$$
\Psi_1=\ldots =\Psi_a=0,\; Z\mbox{\rm lexc}(G)
\mbox{\rm lexc}(H)=1\eqno (66)
$$
in the indeterminate coefficients of
$G$ and $H$, $X_n$ and $Z$ (here $Z$ is a new variable)
with coefficients in the field $\overline{k}$.
Now the condition that $f_\lambda=g_\lambda h_\lambda$,
$\mbox{\rm lex}(g_\lambda)=
\mbox{\rm lex}(g_{\lambda_0})$ and
$\mbox{\rm lex}(h_\lambda)=
\mbox{\rm lex}(h_{\lambda_0})$
is equivalent to the existence of a solution of system (66) with
$X_n=\lambda$.
Hence there are infinitely many $\lambda\in\overline{k}$ such that system
(66) has a solution
with $X_n=\lambda$. Therefore there is an irreducible component of variety of
solutions of system (66)
which is not contained in any hyperplane ${\cal Z}(X_n-c)$,
$c\in\overline{k}$.
Hence there is an irreducible curve $C$ which is contained in the variety of
solutions of system
(66) and is not contained in any hyperplane ${\cal Z}(X_n-c)$,
$c\in\overline{k}$.
Then the field of rational functions $\overline{k}(C)$ is algebraic over the
field $\overline{k}(X_1)$.
Thus the polynomial $\widetilde{f}_{X_n}$ is reducible in
the ring $\overline{k(X_n)}[X_1,\ldots , X_{n-1},Y]$ where
$\overline{k(X_n)}$ is algebraic closure of the field $\overline{k}(X_n)$.

Let $\widetilde{f}_{X_n}=\widetilde{g}\,\widetilde{h}$ where
$$
\widetilde{g},\,\widetilde{h}\in \overline{k(X_1)}[X_1,\ldots , X_{n-1},Y]
\setminus
\overline{k(X_n)},
$$
$\deg_Y\widetilde{g}=n_1$, $\deg_Y\widetilde{h}=n_2$.
Let us show that  $n_1\ge 1$ and $n_2\ge 1$.
Indeed, suppose that, say, $n_2=0$. Then
$\widetilde{h}\in \overline{k(X_n)}[X_1,\ldots , X_{n-1}]\setminus
\overline{k(X_n)}$ and $n_1=\deg_Yf>0$.
Represent
\begin{eqnarray*}
&&\widetilde{f}_{X_n}=\sum_{0\le i\le n_1}\widetilde{f}_iY^i,
\qquad\widetilde{g}=\sum_{0\le i\le n_1}\widetilde{g}_iY^i,\qquad
\mbox{where}\\
&&\widetilde{f}_i\in\overline{k}[X_1,\ldots ,X_n],\qquad
\widetilde{g}_i\in\overline{k(X_n)}[X_1,\ldots , X_{n-1}],
\qquad 0\le i\le n_1.
\end{eqnarray*}
The subrings
\setcounter{equation}{66} \begin{eqnarray}
&&\overline{k}[X_1,\ldots ,X_n,\widetilde{h}], \label{67}\\
&&\overline{k}[X_1,\ldots ,X_n,\widetilde{g}_0,\ldots ,\widetilde{g}_{n_1},
\widetilde{h}] \label{68}
\end{eqnarray}
of the ring $\overline{k(X_n)}[X_1,\ldots , X_{n-1}]$
are of dimension $n$ and integral.
Denote by $S'\subset {\Bbb A}^{n+1}(\overline{k})$ and
$S''\subset{\Bbb A}^{n+n_1+2}(\overline{k})$ the affine algebraic
varieties
with the rings of regular functions (67) and (68) respectively.
There is a polynomial
$0\ne Q\in\overline{k}[X_n]$ such
that all the coefficients of the polynomials
$Q\widetilde{g}_0,\ldots ,Q\widetilde{g}_{n_1}$,
$Q\widetilde{h}$ (considered as
polynomials in $X_1,\ldots , X_{n-1}$
with coefficients from $\overline{k(X_n)}$) are integral over
$\overline{k}[X_n]$.

Since $\dim S'=n$ and ring (67) is integral
there is an  irreducible polynomial (uniquely defined up to a nonzero
constant factor from $\overline{k}$)
$$
H=\sum_{0\le i\le N}H_iZ^i\in \overline{k}[X_1,\ldots ,X_n,Z]
$$
which is a minimal polynomial of $\widetilde{h}$ over the field
$\overline{k}(X_1,\ldots ,X_n)$, herewith all
$H_i\in\overline{k}[X_1,\ldots ,X_n]$, $H_N\ne 0$
and the ring
$$
\overline{k}[S']=
\overline{k}[X_1,\ldots ,X_n,\widetilde{h}]\simeq\overline{k}[X_1,\ldots
,X_n,Z]/(H).
\eqno (69)
$$
The polynomial $H_0\not\in\overline{k}[X_n]$ since
$\deg_{X_1,\ldots , X_{n-1}}H_0\ge\deg_{X_1,\ldots , X_{n-1}}
\widetilde{h}\ge 1$.

The embeddings of the rings of regular functions
$\overline{k}[X_1,\ldots ,X_n]\subset
\overline{k}[S']\subset\overline{k}[S'']$
induce the finite dominant morphisms
$$
S''\setminus{\cal Z}(Q)\stackrel{\rho_2}{\longrightarrow}S'
\setminus{\cal Z}(Q)\stackrel{\rho_1}{\longrightarrow}
{\Bbb A}^n(\overline{k})\setminus {\cal Z}(Q).
$$

The polynomial $Q$ does not belong to the radical of the ideal $(H_0)$ in
the ring
$\overline{k}[X_1,\ldots ,X_n]$ since $Q\in\overline{k}[X_n]$ and
$\deg_{X_1,\ldots , X_{n-1}}H_0\ge 1$.
Hence $Q$ does not belong to the radical of the ideal
$(H_0,Z)$ in the ring
$\overline{k}[X_1,\ldots ,X_n,Z]$. But the ideal $(H_0,Z)=(H,Z)$
in this ring. Therefore
$\emptyset\ne{\cal Z}(\widetilde{h})\setminus{\cal Z}(Q)$
in $S'\setminus{\cal Z}(Q)$.
Since the morphism $\rho_2$ is finite and dominant
$\emptyset\ne{\cal Z}(\widetilde{h})\setminus{\cal Z}(Q)$
in $S''\setminus{\cal Z}(Q)$.
Further, $\widetilde{f}_i=\widetilde{g}_i\widetilde{h}$ for all $1\le i\le
n_1$. Hence
$$
\emptyset\ne{\cal Z}(\widetilde{h})\setminus{\cal
Z}(Q)\subset{\cal Z}(f_0,\ldots , f_{n_1})\setminus{\cal Z}(Q)
$$
in $S''\setminus{\cal Z}(Q)$.
Hence $\dim {\cal Z}(f_0,\ldots , f_{n_1})\setminus{\cal Z}(Q)\ge
n-1$
in $S''\setminus{\cal Z}(Q)$. Therefore
$\dim {\cal Z}(f_0,\ldots , f_{n_1})\setminus{\cal Z}(Q)\ge n-1$
in ${\Bbb A}^n(\overline{k})\setminus {\cal Z}(Q)$ since the
morphism $\rho_1\circ\rho_2$
is finite and dominant. This means that $\mbox{\rm
G\,C\,D\,}\{f_0,\ldots , f_{n_1}\}\ne 1$
in the ring $\overline{k}[X_1,\ldots ,X_n]$, i.e., $\widetilde{f}_{X_n}$ is
reducible in the ring
$k[X_1,\ldots , X_n,Y]$.  This contradiction proves the required
assertion that
$n_1\ge 1$ and $n_2\ge 1$.

Now
$$
f=\widetilde{f}\left(X_1,\ldots , X_{n-2},X_{n-1}-\alpha X_1,
\left(\frac{X_n-\beta X_1}{X_{n-1}-\alpha X_1}\right)
(X_{n-1}-\alpha X_1),Y\right)
$$
is a reducible element of the ring
$$
\overline{k\left(\frac{X_n-\beta X_1}{X_{n-1}-\alpha X_1}\right)}
\left(X_1,\ldots , X_{n-1}\right)\left[Y\right]\eqno (70)
$$
Namely, represent
$f/\mbox{\rm lc}_Yf=gh$ where
$g$ and $h$ are polynomials from the ring
(70) with leading coefficients $\mbox{\rm lc}_Yg=1$,
$\mbox{\rm lc}_Yh=1$ and
$\deg_Yg\ge 1$, $\deg_Yh\ge 1$.

Denote for brevity
$k_{\alpha,\beta}=\overline{k}((X_n-\beta X_1)/(X_{n-1}-\alpha X_1))$.
Denote by $k'$ the field generated
by coefficients of the polynomials $g$, $h$ from ring (70)
over the field
$$
k_{\alpha,\beta}(X_1,\ldots ,X_{n-1})=\overline{k}(X_1,\ldots ,X_n).
$$
Hence
$k'\subset \overline{k}_{\alpha,\beta}(X_1,\ldots ,X_{n-1})$ and the
extension $k'\supset k_{\alpha,\beta}(X_1,\ldots ,X_{n-1})$
is finite separable. Set $k'_{\alpha,\beta}=k'\cap
\overline{k}_{\alpha,\beta}$.
Hence the extension $k'_{\alpha,\beta}\supset k_{\alpha,\beta}$
is finite separable.

Let us show that $k'=k'_{\alpha,\beta}(X_1,\ldots ,X_{n-1})$.
The definition of $k'$ implies that
$k'\subset k''_{\alpha,\beta}(X_1,\ldots ,X_{n-1})$,
where $k''_{\alpha,\beta}\supset k_{\alpha,\beta}$ is a finite (and
separable) Galois extension.
Now the required assertion follows from the Galois theory.

The degree of
the extension $k'_{\alpha,\beta}\supset k_{\alpha,\beta}$
is greater than~$1$ since, otherwise, the polynomial
$f\in\overline{k}[X_1,\ldots ,X_n,Y]$ is also
reducible by the Gauss lemma.
It is known that in this case there is a discrete valuation $v\,
:\, k_{\alpha,\beta}
\rightarrow {\Bbb Z}\cup\{+\infty\}$
of the field $k_{\alpha,\beta}$ over $\overline{k}$ (or which is the same $v$
is zero on
$\overline{k}\setminus\{0\}$)
and a discrete valuation $w$ which is an extension of $v$ to
$k'_{\alpha,\beta}$
such that $w$ is ramified over $v$. The valuation $v$ is defined by a
uniformizing element
$$
\pi=\frac{\mu_1(X_n-\beta X_1)+\mu_2(X_{n-1}-\alpha X_1)}
{\mu_3(X_n-\beta X_1)+\mu_4(X_{n-1}-\alpha X_1)}
$$
such that $\mu_1,\mu_2,\mu_3,\mu_4\in\overline{k}$,
$\mu_1\mu_4-\mu_3\mu_2\ne 0$
and $v(\pi)=1$.

Let $y_1,\ldots , y_D$ be all the roots of the polynomial $f$
in the algebraic closure of the field $\overline{k}(X_1,\ldots ,X_n)$.
Then the field
$\overline{k}(X_1,\ldots ,X_n)[y_1,\ldots ,$ $y_D]$ contains
$k'_{\alpha,\beta}$
since the coefficients of $g$ and $h$
are polynomials in $y_1,\ldots ,y_D$.
Let us continue  $w$ till a discrete valuation
$w_1$ of the field $k'_{\alpha,\beta}(X_1,\ldots , X_{n-1})$ such that
$w_1(z)\ge 0$ for every $z\in\overline{k}[X_1,\ldots ,X_n]$ and
$$
w_1(\mu_1(X_n-\beta X_1)+\mu_2(X_{n-1}-\alpha X_1))=1.
$$
Denote the field
$k''=\overline{k}(X_1,\ldots ,X_n)[y_1,$ $\ldots ,y_D]$. Hence $k''$
is a finite separable
extension of $\overline{k}(X_1,\ldots ,X_n)$ and of
$k'_{\alpha,\beta}(X_1,\ldots
,X_{n-1})$.
Further, let us continue $w_1$ till a discrete valuation $w_2$ of the field
$k''$.
By our construction there is an element
$z_1\in k'_{\alpha,\beta}(X_2,\ldots ,X_n)$ such that
$w_2(z_1)=1/e$ where $e>1$ is an integer.

By $w_3$ (respectively $w_4$) denote the restriction
of $w_2$ to the field $\overline{k}(X_1,\ldots ,$ $X_n)[y]$,
(respectively $\overline{k}(X_1,\ldots ,X_n)$)
herewith we suppose without loss of generality that $y=y_1$.
Since $k''$ is a minimal Galois extension of
$\overline{k}(X_1,\ldots ,X_n)$ containing
$\overline{k}(X_1,\ldots ,X_n)[y]$ and $w_2$ is ramified over $w_4$,
then, see, e.g., \cite{2},
the discrete valuation $w_3$ is also ramified over $w_4$.
Therefore  $\mu_1(X_n-\beta X_1)+\mu_2(X_{n-1}-\alpha X_1)$ divides
the discriminant $\Delta_1$ of the polynomial $f$ with respect to $Y$.

Now for all $\beta,\alpha\in \overline{k}$
satisfying (63) there are
$\mu_1,\mu_2\in\overline{k}$ such that
$(\mu_1,\mu_2)\ne (0,0)$ and
$$
(\mu_1(X_n-\beta X_1)+\mu_2(X_{n-1}-\alpha X_1))
\quad\mbox{divides}\quad \Delta_1
\eqno (71)
$$
in the ring
of polynomials $\overline{k}[X_1,\ldots ,X_n]$.
This is possible only in the case when $\Delta_1=0$
which contradicts to the separability of the
polynomial $f$. Thus the assumption (B) leads to a contradiction.
The existence of the required linear form $L$ is proved.

Now we are going to prove that the set
${\cal U}_4\cap{\cal U}'_n$
is open in the Zariski topology.
By definition $L\in{\cal U}_4$ if and only if the following two
conditions
hold:
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item for every $L\in {\cal U}_{4}$ the ideal $(f,L)$ is prime
in $\overline{k}[X_1,\ldots ,X_n,Y]$,
\item for every $L\in {\cal U}_{4}$ the polynomial
$\Delta_1
\sum_{0\le i\le m_1}f_{i,m_1-i}\not\in (L)$.
\end{enumerate}
Obviously the set of linear  forms satisfying (b)
is a open in the Zariski topology subset of ${\Bbb
A}^n(\overline{k})$.
Hence it is sufficient to prove that the set of all $L\in{\cal U}'_n$
satisfying (a)
is a open in the Zariski topology subset of ${\Bbb
A}^n(\overline{k})$. Let $X_0$ be a
new variable.
Then the homogeneous polynomial in $X_0,\ldots ,X_{n-1}$, $Y$ with
coefficients in the field $\overline{k}$
$$
\overline{f}=
X_0^{\deg_{X_1,\ldots ,X_n,Y}f}f(X_1/X_0,\ldots , X_n/X_0, Y/X_0)
$$
is irreducible over $\overline{k}$.
Let $L=\sum_{1\le i\le n}l_iX_i$  satisfy (a),
where all $l_i\in\overline{k}$, $l_n\ne 0$.
Let $\sum_{1\le i\le n}u_iX_i$ be a linear form with
coefficients $u_i\in\overline{k}$. Suppose that $l_n+u_n\ne 0$.
Then the linear form $L+\sum_{1\le i\le n}u_iX_i$ satisfies (a)
if and only if the polynomial
$$
\overline{f}\left(X_0,\ldots , X_{n-1},
-\sum_{1\le i\le n-1}(l_n+u_n)^{-1}(l_i+u_i)X_i,Y\right)
\eqno (72)
$$
is irreducible over the field $\overline{k}$. Consider the set $P'$ of all
homogeneous polynomials in $X_0,\ldots , X_{n-1},Y$ of degree
$\deg_{X_1,\ldots ,X_{n-1},Y}f$. Let
$n_0$ be the number of coefficients
of such polynomials in dense representation.
Let us identify the set $P'$ with the affine space ${\Bbb
A}^{n_0}(\overline{k})$.
Then it is known, see e.g., \cite{15},
that the set $P''$ of all irreducible over $\overline{k}$ polynomials is a
nonempty (recall that $n\ge 3$) open
in the Zariski topology subset of ${\Bbb A}^{n_0}(\overline{k})$.
Hence there is a polynomial $\Phi\in\overline{k}[X_1,\ldots , X_n]$ such
that $\Phi(0,\ldots , 0)\ne 0$ and
if $\Phi(u_1,\ldots , u_n)\ne 0$ then
polynomial (72) is irreducible
over $\overline{k}$. Thus, ${\cal U}_4\cap{\cal U}'_n$ is open in
the Zariski topology subset of ${\Bbb A}^n(\overline{k})$.
The lemma is proved.

Let us prove (vi). We can suppose without loss of generality
performing a nondegenerate linear transformation of $X_1,\ldots ,X_n$
that $X_1,X_{n-1},X_n\in{\cal L}$.
We have proved that in this case ${\cal L}\cap{\cal U}_{4}\ne
\emptyset$. This implies (vi).
The theorem is proved.

\par\medskip\noindent{\bf REMARK~14}\hspace{0.1em} {\it  The sets ${\cal
U}_i$, $1\le i\le 4$, from the
statement of Theorem~4 satisfy the following property.
For all $0\ne\lambda\in\overline{k}$ and $L\in{\cal U}_i$
the linear form $\lambda L\in{\cal U}_i$.
}\par\medskip



\par\medskip\noindent{\bf COROLLARY~2}\hspace{0.1em} {\it
Under conditions of Theorem~4 let $L\in{\cal U}_{2}\cap{\cal U}_{3}
\cap{\cal U}_{4}$ be a linear form. Let $W$ be the set of
all smooth point of the algebraic variety $V\cap{\cal Z}(L)
\setminus p^{-1}(x)$.
Then $W$ is a dense open in the Zariski topology
subset of the algebraic variety
$V\cap{\cal Z}(L)$, every point $z\in W$ is a
smooth point of $V$, the local ring ${\cal
O}_{z,V\cap{\cal Z}(L)}$
of the algebraic variety $V\cap{\cal Z}(L)$ at the point $z$
is isomorphic to ${\cal O}_{z,V}/L{\cal O}_{z,V}$.
In particular if $V\subset{\Bbb A}^N(\overline{k})$ for some
integer $N$ and $p$ is the linear
projection to the first $n$ coordinates then
the intersection of the tangent spaces
$$
T_{z,V}\cap{\cal Z}(L)
\eqno (73)
$$
of $V$ and ${\cal Z}(L)$ (we identify $T_{z,{\cal Z}(L)}$ with
${\cal Z}(L)$)
at the point $z$ is transversal, i.e.,
$\dim T_{z,V}\cap{\cal Z}(L)=n-1$ and there is a system of local
parameters
$h_1,\ldots , h_{N-n}$ of $V$ at the point $z$ such that $L,h_1,\ldots ,
h_{N-n}$ is a system
of local parameters of $V\cap{\cal Z}(L)$ at the point $z$.
}\par\medskip

\noindent {\bf PROOF} \quad
We can suppose without loss of generality
that $V\subset{\Bbb A}^N(\overline{k})$
and $p$ is the linear
projection to the first $n$ coordinates.
Obviously $W$ is a dense open in
the Zariski topology subset of $V\cap{\cal Z}(L)$.
Let $z\in W$ be a smooth point of $V\cap{\cal Z}(L)$.
Denote by ${\cal O}_z$ the local ring at the point $z$ of the
algebraic variety ${\Bbb A}^N(\overline{k})$.
Denote by $I(V)$ the ideal of the affine algebraic variety $V\subset{\Bbb
A}^N(\overline{k})$. So $\overline{k}[V]=A_1/I(V)$
where $A_1=\overline{k}[X_1,\ldots ,X_N]$.
Then by assertion (iii) the equality
$$
{\cal O}_{z,V\cap{\cal Z}(L)}={\cal O}_z/(I(V){\cal
O}_z+L{\cal O}_z)
$$
holds. Hence $L$ is one of local parameters at the point $z$ of the
algebraic variety $V\cap{\cal Z}(L)$
(considered as a subvariety of ${\Bbb A}^N(\overline{k})$)
and other such local parameters can be chosen from $I(V)$. Hence
there is a system of local parameters $L,h_1,\ldots , h_{N-n}$ in the
point $z$ of $V\cap{\cal Z}(L)$ such that
$h_1,\ldots , h_{N-n}\in I(V)$.

This implies that the point $z$ is a smooth point of $V$, the
tangent space $T_{z,V}={\cal Z}(d_zh_1,\ldots , d_zh_{N-n})$ and the
intersection of tangent spaces (73) is transversal.
The corollary is proved.



\par\medskip\noindent{\bf COROLLARY~3}\hspace{0.1em} {\it  Let $n\ge 3$.
Let $f\in\overline{k}[X_1,\ldots ,X_n]$ be an irreducible
polynomial with $\deg f\ge 1$. Suppose that $f$
is not a homogeneous polynomial relative to
$X_1,\ldots ,X_n$ or $n\ge 4$.  Let ${\cal L}$ be
a linear subspace of the space of
all linear forms in $X_1,\ldots ,X_n$ with coefficients in $\overline{k}$.
Suppose that the dimension $\dim{\cal L}\ge 3$.
Then there is a nonempty
open in the Zariski topology subset ${\cal U}'''$ of
${\cal L}$ such that for every $L\in {\cal U}'''$
the element $f\bmod L\in\overline{k}[X_1,\ldots ,X_n]/(L)$
is irreducible.
}\par\medskip

\noindent {\bf PROOF} \quad
Suppose that $f$ is not a homogeneous polynomial relative to
$X_1,\ldots ,$ $X_n$. Consider the polynomial
$$
F=Y^{\deg_{X_1,\ldots
,X_n}f}f(X_1/Y,\ldots , X_n/Y).
$$
Then $F\in\overline{k}[X_1,\ldots ,X_n,Y]$ is irreducible.
The polynomial $F\bmod L$ is irreducible
if and only if $f\bmod L$ is irreducible.
If characteristic $\mbox{\rm char}(\overline{k})=0$ then set
$F_1=F$.
If $\mbox{\rm char}(\overline{k})=p>0$ then set
$s\ge 0$ to be maximal integer such that $F\in \overline{k}[X_1,\ldots
,X_n,Y^{p^s}]$,
and denote $F_1=F(X_1,\ldots ,X_n,Y^{p^{-s}})$. Then
$F_1\in\overline{k}[X_1,\ldots ,X_n,Y]$ is irreducible and
$\partial F/\partial Y\ne 0$.
If $\mbox{\rm char}(\overline{k})>0$ and $s\ge 1$ then there is
$1\le i\le n$
such that $\partial F/\partial X_i\ne 0$ since $f$ is irreducible and,
therefore, separable.
We can suppose without loss of generality
performing a linear transformation of
$X_1,\ldots ,X_n$ that $i=n$.

Applying
Theorem~4 (vi)
to the algebraic variety ${\cal Z}(F_1)$ and the polynomial $F_1$
(in place of $V$ and $f$) we conclude that
there is a nonempty open in the Zariski topology subset ${\cal
U}'\subset{\cal L}$
such that for every $L\in{\cal U}'$ the polynomial $F_1\bmod L$ is
irreducible and
$\partial (F_1\bmod L)/\partial Y\ne 0$.
Let $L=l_1X_1+\ldots + l_nX_n$ where all
$l_i\in\overline{k}$ and
$l_n\ne 0$. Then, see above, $F_1\bmod L\in\overline{k}[X_1,\ldots
,X_{n-1}]$.
It is known, see, e.g., \cite{5}, that in the considered case the polynomial
$F\bmod L$ is irreducible
if and only if $F\bmod L$ is separable
with respect to $X_1,\ldots ,X_{n-1},Y$.
If $\mbox{\rm char}(\overline{k})=0$ or $s=0$ the last
condition holds. If $\mbox{\rm char}(\overline{k})>0$ and $s\ge
1$ then $\partial F/\partial X_n\ne 0$.
Hence in this case the condition $\partial (F\bmod L)/\partial X_{n-1}\ne 0$
defines
an open in the Zariski topology subset ${\cal U}''\subset{\cal
L}$. Set ${\cal U}'''={\cal U}'\cap{\cal U}''$.
Thus ${\cal U}'''$ is the required open in the Zariski topology subset
of ${\cal L}$.


In the case when $n\ge 4$ one can perform a nondegenerate linear
transformation of $X_1,\ldots ,X_n$ and suppose without loss of generality
that leading coefficient $\mbox{\rm lc}_{X_n}f=1$
and $\partial F/\partial X_1\ne 0$.
Besides that, we can assume without loss of generality that
$X_n\not\in{\cal L}$ if $\dim{\cal L}<n$.
After that we
set $X_n=Y$ and applying Lemma~23
obtain a linear form $L$ satisfying the required conditions. Now
similarly to the proof of Lemma~23
we get the existence of the required open in the Zariski
topology subset ${\cal U}'''$ of ${\cal L}$ in the
considered case. The corollary is proved.




\par\medskip\noindent{\bf LEMMA~24}\hspace{0.1em} {\it  Let $k$ be a perfect
field of arbitrary characteristic.
In the notations from  the Introduction for every
$\sigma+1\le\alpha\le n-1$
the set ${\cal U}_\alpha$ is nonempty and open in the
Zariski topology in
${\Bbb A}^{(r+1)(\alpha-\sigma)}(\overline{k})$.
}\par\medskip

\noindent{\bf PROOF}\quad
Using the Veronese embedding we shall suppose without loss of generality that
$d'=1$.
By  Theorem~4
there exists $\lambda=
(L_{s+1},\ldots , L_\sigma)$, see the Introduction, such that
the irreducible over $\overline{k}$ components of the algebraic
varieties $W$ and
$W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$ are in the one--to--one
correspond by the rule $E\mapsto
E\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$,
and the intersection $W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$ is
transversal
(all these are conditions of general position for $\lambda$).
Replacing $W$ by $W_\lambda$ we shall assume in what follows without loss of
generality that $s=\sigma$.
In the case of the nonzero characteristic replacing $k$ by its purely
transcendental
extension $k(t)$ we shall suppose without loss of generality that the field
$k$ is infinite.

Let us consider at first the case of zero--characteristic (at the end of the
proof we shall return
to the case of arbitrary characteristic).
Let us show that ${\cal U}_\alpha\ne\emptyset$.  Let $L''\in{\cal
U}''_0$.
Denote $p_{L''}=p_{\lambda,L''}$ (now $\lambda=()$ since $s=\sigma$).
Suppose that for an integer $\sigma\le i<\alpha$
\begin{itemize}
\item[(*)] linear forms $L'_{\sigma+1},\ldots , L'_i$, $\sigma+1\le
i<\alpha$ exist such that all $L'_j$,
$s+1\le j\le i$ are
linear combinations of $L''_0,L''_{\sigma+1},\ldots , L''_n$ with
coefficients from $k$, and
$(L'_{\sigma+1},\ldots , L'_i)\in{\cal U}_i$.
\end{itemize}
Then by Lemma~11 \cite{12} and Lemma~14 \cite{12}
the morphism $W\cap{\cal Z}(L'_{\sigma+1},\ldots ,
L'_i)\rightarrow{\cal Z}(X_1,\ldots ,$ $X_{i-\sigma})=
{\Bbb P}^{n-i}(\overline{k})$ induced by $p_{L''}$ is dominant.
It induces the morphism of affine algebraic varieties $\mbox{\rm
con}(W)\cap{\cal Z}(L'_{\sigma+1},\ldots ,
L'_i)\rightarrow{\Bbb A}^{n-i+1}(\overline{k})$. Let us represent
$$
\mbox{\rm con}(W)\cap{\cal Z}(L'_{\sigma+1},\ldots ,
L'_i)\setminus{\cal Z}(L''_0,L''_{\sigma+1},\ldots ,
L''_n)=\bigcup_{j\in J}E_j,
$$
where $E_j$, $J\in J$, is a finite family of affine nonempty open in the
Zariski topology subsets of $\mbox{\rm con}(W)\cap{\cal
Z}(L'_{\sigma+1},\ldots ,
L'_i)\setminus{\cal Z}(L''_0,L''_{\sigma+1},\ldots , L''_n)$.
Hence by Theorem~4 applied to every morphism $E_j\rightarrow{\Bbb
A}^{n-i+1}(\overline{k})$, $j\in J$,
there is a linear form $L''_{i+1}$ (from the intersection of open in the
Zariski topology
subsets from the statement of Theorem~4
over all $j\in J$)
such that (*) holds with $i+1$ in place of $i$.
Thus, we have proved the existence of
$(L'_{\sigma+1},\ldots , L'_\alpha)\in{\cal U}_\alpha$ by the induction
on $i$.

Let us prove that ${\cal U}_\alpha$ is open in the Zariski topology
subset of ${\Bbb A}^{(n-\alpha)(n+1)}(\overline{k})$.
At first we shall show that ${\cal U}_\alpha$ is a constructive subset
of
${\Bbb A}^{(n-\alpha)(n+1)}(\overline{k})$.
Probably this can be proved using the first order theory of algebraically
closed fields. But we give the direct proof.

Recall that by definition a
subset of an algebraic variety
$F$ is constructive
if and only if it is a union $\cup_{\gamma\in\Gamma}(F'_\gamma\cap
F''_{\gamma})$, where the number of elements $\#\Gamma<+\infty$,
all $F'_\gamma$ are open in the Zariski topology subsets of $F$, and all
$F''_\gamma$ are closed in the Zariski topology subsets of $F$.
We shall use the following criterion.
Suppose $E\subset F$, and for every closed subvariety $E'\subset F$
such that $E'\cap E$ is dense with respect to the Zariski topology in $E'$
there exists a nonempty open in the Zariski topology subset $E''\subset E'$
such that $E''\subset E$. Then $E$ is a constructive
subset of $F$. The proof of this criterion is by the induction on
$\dim\overline{E}$
(here $\overline{E}$ is the closure of $E$ with respect to Zariski topology).
Indeed, there is a dense open in the Zariski topology subset $E'''\subset
\overline{E}$ such that $E'''\subset E$.
One can replace $E$ by $E\setminus E'''$. Obviously
$\dim\overline{E\setminus E'''}<\dim\overline{E}$.
In the given criterion one can assume additionally that $E'$ is irreducible
over the algebraic closure of the ground field.

Let us show that ${\cal U}_\alpha$ satisfies the formulated criterion.
Let $E'$ be a closed irreducible over $\overline{k}$
subvariety of ${\Bbb A}^{(n-\alpha)(n+1)}(\overline{k})$
and $(L'_{\sigma+1},\ldots , L'_\alpha)\in E'\cap{\cal U}_\alpha$. Let
us extend the ground field
$k$ till $k_u$, see the Introduction. We shall assume without loss of
generality that the all the
coefficients of the linear forms $U_i$, $\sigma+1\le i\le\alpha$ are
infinitely close to the corresponding
coefficients of the linear forms $L'_i$.
Suppose that $(L''_{\sigma+1},\ldots , L''_\alpha)\in E'\cap{\cal
U}_\alpha$, all
linear forms $L''_i-L'_i$, $\sigma+1\le i\le\alpha$, have coefficients from
$\overline{k_u}$,
which are infinitely small with respect to $k$ (or zeroes). Then by
Lemma~13 \cite{13}
$(L''_{\sigma+1},\ldots , L''_\alpha)\in{\cal U}''_\alpha$.
By Lemma~15 \cite{13} $\mbox{\rm st} (W(L''_{\sigma+1},\ldots ,
L''_\alpha))=
W(L'_{\sigma+1},\ldots , L'_\alpha)$. Hence by Lemma~5 \cite{13} the degree
of every irreducible over $\overline{k_u}$ component of
$W(L''_{\sigma+1},\ldots , L''_\alpha)$
is at least $\deg W(L'_{\sigma+1},\ldots , L'_\alpha)$.
Therefore, $W(L''_{\sigma+1},\ldots , L''_\alpha)$ is irreducible over
$\overline{k_u}$.

Further, one can choose  $(L''_{\sigma+1},\ldots ,L''_\alpha)$ such that
additionally
$\xi=(L''_{\sigma+1},\ldots ,$ $L''_\alpha)$ is a generic point of $E'$.
Denote by $\overline{k}(\xi)$ the field generated over $\overline{k}$ by the
coordinates of the
point $\xi$.
We shall identify it with $\overline{k}(E')$.
There exists a linear projection $\pi\, :\, W(L''_{\sigma+1},\ldots ,
L''_\alpha)\rightarrow{\Bbb P}^{n-\alpha+1}(\overline{k})$,
$(X_0:\ldots : X_n)\mapsto(Z_0:\ldots : Z_{n-\alpha+1})$ such that all $Z_j$,
$0\le j\le n-\alpha+1$
are linear forms
with coefficients from $k$, the image $\pi(W(L''_{\sigma+1},\ldots ,
L''_\alpha))={\cal Z}(H)$, $H\in \overline{k}(\xi)[X_0,\ldots ,
X_{n-\alpha+1}]$ is separable,
the morphism $W(L''_{\sigma+1},\ldots ,
L''_\alpha)\rightarrow{\cal Z}(H)$ induced by $\pi$ is finite and
$\deg W(L''_{\sigma+1},\ldots ,
L''_\alpha)$ $=\deg H$ (these are conditions of general position for the
linear
projection).
Hence the number of irreducible over $\overline{k(\xi)}$ factors
of $H$ is the
number of irreducible over $\overline{k}$ components of $W$.
Replacing $H$ and all $L''_j$ by
$cH$ and $cL''_j$ with an appropriate $0\ne c\in \overline{k}[E']$ we shall
assume without loss of generality that
$H\in\overline{k}[E'][X_0,\ldots , X_{n-\alpha+1}]$ and all $L''_j\in
\overline{k}[E'][X_0,\ldots , X_n]$.
Obviously the polynomial $H(Z_0,\ldots , Z_{n-\alpha+1})$ vanishes on
$W(L''_{\sigma+1},\ldots ,
L''_\alpha)$. Now consider the ho\-mo\-mor\-ph\-ism
$\overline{k}[E']\rightarrow\overline{k}$
giving a point $\widetilde{\xi}\in E'$.
This ho\-mo\-mor\-ph\-ism is extended in the natural way till the
ho\-mo\-mor\-ph\-isms of the rings of polynomials with coefficients from
$\overline{k}[E']$.
Denote by $\widetilde{H}$ and $\widetilde{L}_j$ the images of $H$ and $L'_j$
under these ho\-mo\-mor\-ph\-isms.
There is a nonempty open in the Zariski topology subset $E^{(1)}\subset E'$
such that if
$\widetilde{\xi}\in E^{(1)}$ then $(\widetilde{L}_{\sigma+1},\ldots
,\widetilde{L}_\alpha)\in{\cal U}''_\alpha$,
the image $\pi(W(\widetilde{L}_{\sigma+1},\ldots ,
\widetilde{L}_\alpha))={\cal Z}(\widetilde{H})$, the morphism
$W(\widetilde{L}_{\sigma+1},\ldots ,
\widetilde{L}_\alpha)\rightarrow{\cal Z}(\widetilde{H})$ induced by
$\pi$ is finite, and
$\deg W(\widetilde{L}_{\sigma+1},\ldots ,
\widetilde{L}_\alpha)=\deg\widetilde{H}$, $\widetilde{H}\in
\overline{k}[X_0,\ldots ,
X_{n-\alpha+1}]$ is separable.
This assertion is straightforward (we leave the details to the reader).
Finally there exists a nonempty open in the Zariski topology subset
$E^{(2)}\subset
E^{(1)}$ such that for every
$\widetilde{\xi}\in E^{(2)}$
the number of the irreducible over $\overline{k}$ factors of the polynomial
$\widetilde{H}$ is equal to the number
of irreducible over $\overline{k(\xi)}$ factors of $H$. Indeed,
in Lemma~23 we have proved that ${\cal U}_4\cap{\cal U}'_n$ is open
in the Zariski topology, and
the last assertion is proved in the similar way.

Thus, the set ${\cal
U}_\alpha$ is constructive. Finally, as we have seen in the proof
for every point of
$L'\in{\cal U}_\alpha$ there exists
an infinitely small neighborhood ${\cal W}$ of $L'$ in ${\cal
U}''_\alpha$
such that ${\cal
W}\subset{\cal U}_\alpha$.
Obviously it is possible only in the case if ${\cal U}_\alpha$ is an
open in the Zariski topology.
The lemma is proved for zero--characteristic.

In the case of nonzero characteristic it is sufficient to generalize all the
lemmas from
\cite{12}, \cite{13} which are used in the proof to the case of
infinite ground field of nonzero characteristic
under condition that the extension $k(W)\supset k(W')$ is separable.
It is possible without difficulties. We leaves the details to the reader.
One can define infinitesimals in the case of nonzero characteristic using
valuations, see \cite{2}.
For example, in the most simple case for a transcendental over $k$ element
$\varepsilon$
we have the valuation $v\, :\, k(\varepsilon)\rightarrow{\Bbb
Z}\cup\{+\infty\}$ over $k$
such that $v(\varepsilon)=1$. It can be extended till
the valuation $\overline{v}\,
:\,\overline{k(\varepsilon)}\rightarrow{\Bbb Q}\cup
\{+\infty\}$. By definition a nonzero element
$a\in\overline{k(\varepsilon)}$ is
infinitesimal if and only if
$\overline{v}(a)>0$. The lemma is proved.












 \newpage

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