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\begin{document}

 \title{Polynomial--time computation of the degree of a
 dominant morphism in zero--characteristic II}
\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 dominant rational morphism from $W$ to
$W'$ given by homogeneous
polynomials of degree $d'$. We suggest algorithms to construct objects
in general position related  to the morphism
$W\rightarrow W'$ generalizing the
algorithms from the first part
of the paper to the case $\dim W>\dim W'$.
These algorithms are
deterministic and polynomial in $(dd')^n$ and the size
of input.
\end{abstract}




\newpage

\section*{Introduction}

This paper continues \cite{9}. We investigate general rational morphisms
in zero--characteristic from
algorithmic point of view now.  Everything is prepared for the next paper,
where an algorithm for the
first Bertini theorem is constructed. The last algorithm will use
Theorem~2, its statement see at the end of the Introduction. The second
application of the results of the present
paper is the substantiation of
the method of representing an algebraic variety using
representative systems of points of its irreducible components
suggested in \cite{7}. Namely, in one of the next papers we shall
consider this substantiation as well as an algorithm for constructing the
smooth stratification described in \cite{7} in details. Theorem~1, see below,
will be the main tool there. Besides that
we shall see in the same next, that Theorem~1 allows to compute
the intersections  and the indices of
intersection of algebraic varieties given by method from \cite{7}
(although the irreducible components of the intersections
not always can be given using the method from \cite{7} with a good
upper bound for degrees of corresponding polynomials).

In the present paper we need to
elaborate a new method
for solving
systems of polynomial equations and inequalities
with squares of absolute values, see \cite{3},
in the case when these systems contain
generic linear forms (or even generic homogeneous
polynomials of given degrees; but in this
paper we need only the case
of linear forms). Namely, let it is a system in $(n+1)\nu$ variables
$X_{i,j}$, $0\le j\le n$,
$1\le i\le\nu$, where $\nu=O(1)$ (is considered as a constant,
$\nu<n$), and also in
linear forms $U_\alpha=\sum_{0\le j\le n}u_{\alpha,j}X_{i,j}$,
$1\le\alpha\le \mu$, where
all $u_{\alpha,j}$ are variables and $\mu=O(n)$.
Suppose also that this
system depends on other $O(n)$ variables. So the total number of variables
may be $\ge Cn^2$ for a constant $C$. Hence applying the general method
for solving such systems, see \cite{3}, we get an estimate polynomial in
$d^{n^2}$ for the working time of the algorithm for constructing a solution,
where $d$ is a bound for degrees of all equations and inequalities.
In brief, the suggested method allows to
replace the generic linear forms by
the ones in $\nu$ variables $X_{i,j_1},\ldots ,
X_{i,j_\nu}$, where
$0\le j_1<\ldots < j_\nu\le n$ are chosen, and obtain a new system.
So there are only $\mu\nu=O(n)$
variables $u_{\alpha,j_\beta}$ in the new systems corresponding to the
different choices of
$j_1,\ldots , j_\nu$. This method
reduces constructing a solution of the initial system to constructing the
ones of ${n+1\choose\nu}$ new systems in $O(n)$ variables.

The results of the present paper are based actually on
this method (although we do not describe it formally separately
from the considered problems).
It is applied in Section~1 to strengthen Theorem~1 \cite{8},
Section~4 and Section~6.
Also we would like to underline
here the importance of \cite{5} and its
consequence Theorem~2 \cite{9} in the present paper.
Besides that, now we use
\cite{8} (the correction of the statement of Lemma~2 \cite{8}
see in \cite{9}), \cite{9} and other our previous papers.



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
${\Bbb P}^n(\overline{k})$, $n\ge 0$, 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 ${\Bbb Q}$.
The field $k$ is supplied with a real structure if the embedding
$k\rightarrow R[\sqrt{-1}]$ is given,
where $R$ is a real ordered field.

Consider a system of polynomial
equations in $n$ variables and a fixed number of inequalities with
squares of absolute values
with coefficients from the field $k$.
Let the degrees of equations and inequalities be less than $d$,
see the details in \cite{3} and also in Section~1 of \cite{9},
Then, see Proposition~1 \cite{3},
one can construct a real structure on $k$ and decide within the time
polynomial in $d^n$ and the size of input whether this system has a solution
in ${\Bbb A}^n(\overline{k})$ relative to this real structure.
When such a solution exists one can construct it
within the same time. The coordinates of the solution belong to a finite
extension of the field $k$. This finite extension is constructed and
is given by a primitive element and its minimal polynomial.
We shall use this proposition many times without special
reference to it.

Since $k$ is an arbitrary finitely generated field over ${\Bbb Q}$ this
result is valid
also in the case if one replace the field of coefficients $k$ by
a finite extension of
the field $k_i=k(\varepsilon_1,\ldots ,\varepsilon_i)$ for fixed $i$,
where $\varepsilon_j>0$ for $j\ge 1$
is an infinitesimal with respect to the field $k_{j-1}$, see below.
Theorem~2 \cite{9}, which is based on \cite{5}, strengthens
this result. Actually in this paper the most part of systems
of polynomial equations and
inequalities is solved by Theorem~2 \cite{9}.


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 from \cite{8}
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 ${\Bbb P}^r(\overline{k})$.
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 $\dim W'=n-\sigma$ (one can compute
$s$ and $\sigma$ using the algorithm from \cite{4}, see below).
So $0\le s\le\sigma\le n$.
Let $v_j=\sum_{0\le i\le n}v_{j,i}X_i$, $0\le j\le n+1$, be generic
linear forms in
$X_0,\ldots , X_n$, i.e., the transcendency degree of the family of all
coefficients $v_{j,i}$ over $k$ is
$(n+2)(n+1)$. By definition ,
$v=(v_{s+1},\ldots , v_\sigma)$, $\overline{v}=(v_0,\ldots , v_{n+1})$ and
$k_v$ (respectively $k_{\overline{v}}$) is
the field generated over $k$ by all
the $v_{j,i}$, $s+1\le j\le\sigma$,
$0\le i\le n$ (respectively $0\le j\le n+1$,
$0\le i\le n$). Denote by $\widetilde{v}$ the family of coefficients
$v_{j,i}$, $0\le j\le n+1$, $0\le i\le n$. So $k_{\overline{v}}=
k(\widetilde{v})$.
Now $p$ and $p'$ induce the morphisms
\begin{eqnarray*}
&&p_v\, :\, (W\cap{\cal Z}(v_{s+1},\ldots , v_\sigma))
\setminus{\cal Z}(e_0,\ldots , e_r)\rightarrow
{\Bbb P}^r(\overline{k_v}),\\
&&p'_v\, :\, (W\cap{\cal Z}(v_{s+1},\ldots , v_\sigma))
\setminus{\cal Z}(e_0,\ldots , e_r)\rightarrow W'(k_v)
\end{eqnarray*}
defined over the field $k_v$. For convenience of notation set $q_v=p'_v$.
So $q_v$ is a dominant morphism.
The dimension $\dim W\cap{\cal Z}(v_{s+1},\ldots ,
v_\sigma)\setminus
{\cal Z}(e_0,\ldots , e_r)=n-\sigma$.
We compute, see Theorem~2 below, the degree $\deg p'_v$.




Let us apply Theorem~1 \cite{9} to the morphism $p'_v$ (in place of $p'$)
replacing the field $k$ by $k_v$ in the
statement of this theorem. Denote by ${\cal V}\subset{\Bbb
A}^{(r+1)(n-\sigma+1)}(\overline{k_v})$
the nonempty open in the Zariski topology
subset of $(n-\sigma+1)$-tuples of linear forms from assertion (d) of
Theorem~1 \cite{9} applied to $p'_v$.
{\it Set ${\cal U}'_0={\cal V}\cap{\Bbb
A}^{(r+1)(n-\sigma+1)}(\overline{k})$.} Then obviously ${\cal U}'_0$
is a nonempty open in the Zariski topology subset of ${\Bbb
A}^{(r+1)(n-\sigma+1)}(\overline{k})$.
Hence $L'\in{\cal U}'_0$ if and only if (30)--(33), see
Section~4, holds with
$v$ and ground field $k_v$ in place of $\lambda$ and $k$ respectively.
The set ${\cal U}'_0$ is invariant
with respect to the action of the Galois group $\mbox{\rm
Gal}(\overline{k}/k)$ and, therefore, is defined over $k$.
We shall construct an element from ${\cal U}'_0$. More
precisely, we shall define below
nonempty open in the Zariski topology sets
${\cal U}'''_{n+1}\subset{\cal
U}'''_0\subset{\cal U}''_0\subset{\cal U}'_0$
related  to the morphism $p$, and
construct an element from ${\cal U}'''_0$ within the polynomial
time, see Theorem~2.

Let $K$ be a field containing
algebraically independent over $k$ elements
$u_{j,w}$, $j\in\{0,\sigma+1,\ldots , n\}$, $0\le
w\le r$, and such that $K$ and $k_{\overline{v}}$ are
linearly disjoint over $k$. Denote by $u$ the family of elements
$u_{j,w}$, $j\in\{0,\sigma+1,\ldots , n\}$, $0\le
w\le r$.
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 $U=(U_0,U_{\sigma+1},\ldots ,$ $U_n)$).
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 $K$.

Let $L'$ be an arbitrary as above
(but may be $L'\not\in{\cal U}'_0$).
In what follows unless we state otherwise we shall suppose that the linear
forms
linear forms $L'_0,L'_{\sigma+1},\ldots , L'_n$ are linearly independent.
We shall consider the morphisms
\setcounter{equation}{1} \begin{eqnarray}
&&p_{L'}\, :\, W
\setminus{\cal Z}(e'_0,e'_{\sigma+1}\ldots , e'_n)\rightarrow
{\Bbb P}^{n-\sigma}(\overline{k}),\label{2}\\
&&p_{v,L'}\, :\, W\cap{\cal Z}(v_{s+1},\ldots , v_\sigma)
\setminus{\cal Z}(e'_0,e'_{\sigma+1}\ldots , e'_n)\rightarrow
{\Bbb P}^{n-\sigma}(\overline{k_v}),\label{3}
\end{eqnarray}
defined over the fields $k$ and $k_v$ respectively and given by the same
formula
$$
(X_0:\ldots :
X_n)\mapsto(e'_0:e'_{\sigma+1}:\ldots : e'_n).
$$
For every $\sigma\le i\le n$ denote by
\begin{itemize}
\item $W_i(L')$ the union
of all irreducible over $\overline{k}$ components $E$ of $W\cap{\cal
Z}(e'_{\sigma+1},\ldots ,$ $e'_i)$ such that $\dim E=n-i+\sigma-s$ and
 the restriction
$p_{L'}|_E\, :\, E\setminus{\cal Z}(e'_0,e'_{\sigma+1}\ldots , e'_n)
\rightarrow{\cal Z}(X_1,\ldots , X_{i-\sigma})$
is dominant (here ${\cal Z}(X_1,\ldots , X_{i-\sigma})$ $\subset
{\Bbb P}^{n-\sigma}(\overline{k})$ is a linear subspace).
\end{itemize}
Hence $E\not\subset{\cal Z}(e'_0,t'_{\sigma+1}\ldots , e'_n)$. The
morphism $p_{L'}|_{W_i(L')}\, :\, W_i(L')
\setminus{\cal Z}(e'_0,e'_{\sigma+1},$ $\ldots , e'_n)
\rightarrow{\cal Z}(X_1,\ldots , X_{i-\sigma})$ is dominant if
$W_i(L')\ne\emptyset$ (one can not exclude here that $W_i(L')$
is empty for some $i$). In particular $W_\sigma(L')=W$ if and only if the
morphism $p_{L'}$ is dominant.
Note that $W_i(L')$ depends only on $L'_{\sigma+1},\ldots , L'_i$.
So we shall denote also $W_i(L')=W(L'_{\sigma+1},\ldots , L'_i)$.
The morphism $p_{L'}|_{W_i(L')}$ is dominant or $W_i(L')=\emptyset$,
for every $\sigma+1\le i\le n$. Therefore,
$$
\dim W(L'_{\sigma+1},\ldots , L'_{i-1})\cap{\cal Z}(e'_i)\le
n-i+\sigma-s,
\eqno (4)
$$
and
$$
\dim W(L'_{\sigma+1},\ldots , L'_n))\cap{\cal Z}(e'_0)\le \sigma-s-1.
\eqno (5)
$$
Denote for brevity $L'_{n+1}=L'_0$, $e'_{n+1}=e'_0$,
$W_{n+1}(L')=W(L'_{\sigma+1},\ldots , L'_{n+1})=\emptyset$.
%Put ${\cal U}'_{n+1}=\{(L'_{\sigma+1},\ldots , L'_{n+1})\, :\,
%(L'_{n+1},L'_{\sigma+1},\ldots , L'_n)\in{\cal U}'_0\}$.
Therefore,
$$
W_{i-1}(L')\cap{\cal Z}(e'_i)=W_i(L')\cup W^{(1)}_i(L')
\cup W^{(2)}_i(L')\cup W^{(3)}_i(L'),
\quad \sigma+1\le i\le n+1,
$$
where:
\begin{itemize}
\item $W^{(1)}_i(L')$ is the union of all irreducible over $\overline{k}$
components $E$
of $W_{i-1}(L')\cap{\cal Z}(e'_i)$
such that $E$ is an irreducible component of
$W\cap{\cal Z}(e'_0,e'_{\sigma+1}\ldots , e'_n)$;
\item $W^{(2)}_i(L')$ is the union of all the irreducible over $\overline{k}$
components $E$
of $W_{i-1}(L')$ $\cap{\cal Z}(e'_i)$
such that $E$ is contained in $W^{(1)}_j(L')$ for some $s\le j<i$;
\item $W^{(3)}_i(L')$ is the union of all irreducible over $\overline{k}$
components $E$ of $W_{i-1}(L')\cap{\cal Z}(e'_i)$
such that $E\not\subset{\cal Z}(e'_0,e'_{\sigma+1}\ldots , e'_n)$ and\\
$\dim\overline{p_{L'}
(E\setminus{\cal Z}(e'_0,e'_{\sigma+1}\ldots , e'_n))}<n-i+\sigma-s$.
Hence $W^{(3)}_i(L')=\emptyset$ for $i=n,n+1$.
\end{itemize}
Set also $W^{(\beta)}(L')=W^{(\beta)}_\sigma(L')=\emptyset$,
$W^{(\beta)}_0(L')=W^{(\beta)}_{n+1}(L')$,
$\beta=1,2,3$.

Notice that if $L'\in{\cal U}'_0$ then $W'\cap{\cal
Z}(L'_0,L'_{\sigma+1},\ldots , L'_n)=\emptyset$ in ${\Bbb
P}^n(\overline{k})$.
By Lemma~8
if $W'\cap{\cal
Z}(L'_0,L'_{\sigma+1},\ldots , L'_n)=\emptyset$ then
$W\cap{\cal Z}(e_0,\ldots , e_r)=
W\cap{\cal Z}(e'_0,e'_{\sigma+1},\ldots , e'_n)$
(actually in this lemma we have a more strong assertion).
Therefore, the dimension of every irreducible component of
$W\cap{\cal Z}(e_0,\ldots , e_r)$ is at least $\sigma-s-1$.
Hence for arbitrary $L'$ (may be, $W'\cap{\cal
Z}(L'_0,L'_{\sigma+1},\ldots , L'_n)\ne\emptyset$) every
irreducible component of
$W^{(1)}_{n+1}(L')$ is an irreducible component of dimension $\sigma-s-1$ of
$W\cap{\cal Z}(e_0,\ldots , e_r)$.
If $L'\in{\cal U}'_0$ then for every $\sigma+1\le i\le n$ every
irreducible component of
$W^{(1)}_i(L')$ is an irreducible component of dimension $n-\sigma-s-i$ of
$W\cap{\cal Z}(e_0,\ldots , e_r)$.
If $L'\in{\cal U}'_0$ and
$\bigcup_{\sigma+1\le j\le i-1}W^{(3)}_j(L')$ does not contain any
irreducible component $E$
of $W\cap{\cal Z}(e_0,\ldots , e_r)$ such that $\dim E=n-\sigma+s-i$
(in particular, if $\bigcup_{\sigma+1\le j\le i-1}W^{(3)}_j(L')=\emptyset$,
or if $L'\in{\cal U}''_0$,
see below) then, conversely, every irreducible component
of dimension $n-\sigma-s-i$ of the variety $W\cap{\cal Z}(e_0,\ldots ,
e_r)$ is an irreducible component $W^{(1)}_i(L')$.

Further, for every $z\in W'$ each irreducible component $E$ of $p^{-1}(z)$
is an irreducible component of
$p_{L'}^{-1} p_{L'}p^{-1}(z)$ if $L'\in{\cal U}'_0$. Hence $\dim
E\ge\sigma-s$.
The last inequality is
the theorem about dimensions of fibers for the morphism
$p'$, cf. \cite{12}.
Besides that, for every $\sigma+1\le i\le n+1$
\begin{itemize}
\item if $W'\cap{\cal
Z}(L'_0,L'_{\sigma+1},\ldots , L'_n)=\emptyset$ then
$W^{(1)}_i(L')$ is the union of all irreducible over $\overline{k}$
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 if $W'\cap{\cal
Z}(L'_0,L'_{\sigma+1},\ldots , L'_n)=\emptyset$ then $W^{(3)}_i(L')$
is the union of all irreducible over $\overline{k}$
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}


{\it In what follows in the Introduction we assume that $W'\cap{\cal
Z}(L'_0,L'_{\sigma+1},\ldots , L'_n)$ $=\emptyset$.}
Let us represent $W_i(L')$ and $W^{(\beta)}_i(L')$, $\beta=1,2,3$
as the unions of pairwise distinct irreducible
over $\overline{k}$ algebraic varieties:
\begin{eqnarray*}
&&W_i(L')=\bigcup_{j\in J_i(L')}W_j,\quad \sigma\le i\le n,\\
&&W^{(\beta)}_i(L')=\bigcup_{j\in J^{(\beta)}_i(L')}W_j
,\quad \sigma\le i\le n+1,\; \beta=1,2,3,
\end{eqnarray*}
(actually $J^{(2)}_{\sigma+1}(L')=\emptyset$ and
$W^{(2)}_{\sigma+1}(L')=\emptyset$).
%Set $J_0(L')=J_{n+1}(L')$,
%$J^{(\beta)}_0(L')=J^{(\beta)}_{n+1}(L')$, $\beta=1,2,3$.
Denote by $W^{(1)}_i$ the union of all the irreducible
components of dimension
$n-i+\sigma-s$ of $W\cap{\cal Z}(e_0,\ldots , e_r)$.
%Put also $W^{(1)}_0=W^{(1)}_{n+1}$.
Let
$W_j$, $j\in J^{(1)}_i$, be the family of all the irreducible
components of $W^{(1)}_i$. We shall assume in
what follows without loss of generality that $J^{(1)}_i(L')\subset
J^{(1)}_i$.

Denote by $\iota_j(L')$, $j\in J_{i+1}(L')$
(respectively $\iota_j(L')$, $j\in J^{(\beta)}_{i+1}(L')$, $\beta=1,2,3$),
$\sigma\le i\le n$,
the index of intersection $i(W_i(L'),{\cal Z}(e'_{i+1});W_j)$ of the
algebraic varieties
$W_i(L')$ and ${\cal Z}(e'_{i+1})$ at $W_j$, see \cite{10}.
Hence by the B\'ezout theorem, see \cite{10},
$$
d'\deg W_i(L')=
\sum_{j\in J_{i+1}(L')}\iota_j(L')\deg W_j+\sum_{1\le\beta\le 3}
\sum_{j\in J^{(\beta)}_i(L')}
\iota_j(L')\deg W_j.
\eqno (6)
$$
Set
\begin{eqnarray*}
&&D_i(L')=\deg W_i(L')=\sum_{j\in J_i(L')}\deg W_j,
\quad \sigma\le i\le n+1,\\
&&D^{(\beta)}_i(L')=\deg W^{(\beta)}_i(L')=
\sum_{j\in J^{(\beta)}_i(L')}
\deg W_j,\quad \sigma\le i\le n+1,\,\beta=1,2,3.
\end{eqnarray*}
%and $D_0(L')=D_{n+1}(L')$, $D^{(\beta)}_0(L')=D^{(\beta)}_{n+1}(L')$,
%$\beta=1,2,3$.
We shall denote also $D_\alpha(L')=D(L'_{\sigma+1},\ldots ,
L'_\alpha)$, $J_\alpha(L')=J(L'_{\sigma+1},\ldots , L'_\alpha)$,
$W_\alpha(L')=W(L'_{\sigma+1},\ldots ,
L'_\alpha)$ for
$\sigma\le \alpha\le n+1$; and
$W^{(\beta)}_\alpha(L')=W^{(\beta)}(L'_{\sigma+1},\ldots ,$
$L'_\alpha)$,
$D^{(\beta)}_\alpha(L')=D^{(\beta)}(L'_{\sigma+1},\ldots ,
L'_\alpha)$,
$J^{(\beta)}_\alpha(L')=J^{(\beta)}(L'_{\sigma+1},\ldots ,$ $L'_\alpha)$;
for $\sigma+1\le \alpha\le n+1$, $\beta=1,2,3$.



Set
\setcounter{equation}{6} \begin{eqnarray}
&&D_i=D_i(U),\; J_i=J_i(U),\quad \sigma\le i\le n+1, \nonumber\\
&&D^{(\beta)}_i=D^{(\beta)}_i(U),\; J^{(\beta)}_i=J^{(\beta)}_i(U), \quad
\sigma\le i\le n+1,\;\beta=1,2,3,
\nonumber\\
&&\iota_j=\iota_j(U),\quad j\in J^{(\beta)}_i,  \, \sigma\le i\le n+1,
\;\beta=1,2,3, \label{7}
\end{eqnarray}
and $D_0=D_{n+1}$, $D^{(\beta)}_0=D^{(\beta)}_{n+1}$,
$J^{(\beta)}_0=J^{(\beta)}_{n+1}$,
$\beta=1,2,3$. The inequality $D_i(L')\le D_i$ holds for every $\sigma\le
i\le n$, see Lemma~12.
{\it If $L'\in{\cal U}'_0$ then by Lemma~11 from
\cite{9} the index of intersection
$\iota_j(L')=1$ for every $j\in J_i(L')$, $\sigma+1\le i\le n$.}
In particular, $\iota_j=1$ for every $j\in J_i$, $\sigma+1\le i\le n$.
Further, if $L'\in{\cal U}'_0$ and $J^{(1)}_i(L')=J^{(1)}_i$ then
$\iota_j(L')\ge \iota_j$ for
every $j\in J^{(1)}_i$, see Lemma~11 of the present paper
(actually the last inequality is fulfilled under the weaker conditions from
this lemma).
Besides that, if additionally $i=n+1$
then $\iota_j(L')=\iota_j$, see Lemma~12.


Denote by $W'''$
the closure in the Zariski topology of all points $z\in W'$
such that $\dim p^{-1}(z)>s-\sigma$.
Denote by $W^{(4)}$ the closure in the Zariski topology of the
union of all irreducible components $E$ of $p^{-1}(W''')$ such
that the dimension of closure
$\dim\overline{p(E)}<\sigma-s+\dim E$.
By the theorem about dimension of fibers, see \cite{12} or above,
for every $z\in W'''$ the dimension
of every irreducible component (if it exists,
$p^{-1}(z)$ may be empty)
of $p^{-1}(z)$ is at least $\sigma-s+1$.
Hence the quasiprojective algebraic variety $W^{(4)}\setminus{\cal
Z}(e_0,\ldots , e_r)$
is exactly the set of all points
$x\in W\setminus{\cal Z}(e_0,\ldots , e_r)$ such
that for every irreducible component $E$ of the fiber
$p^{-1}p(x)$ if $x\in E$ then $\dim E>\sigma-s$.
Hence if $L'\in{\cal U}'_0$ then
$W^{(3)}_i(L')\subset W^{(4)}$, $\sigma+1\le i\le n-1$.


Let $\sigma\le i\le n$ be an integer.
We shall need also conditions
(if the dimension here is negative then the variety is empty)
\setcounter{equation}{7} \begin{eqnarray}
&&\dim (\,W^{(4)}\setminus{\cal Z}(e_0,\ldots , e_r)\,)\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_i)\nonumber\\
&&\le n-i+\sigma-s-1, \label{8}\\
&&\dim W(L'_{\sigma+1},\ldots , L'_i)\cap(\overline{V\setminus W})\nonumber\\
&&\le n-i+\sigma-s-1.\label{9}
\end{eqnarray}


By Theorem~1 \cite{9} if $L'=(L'_0,L'_{\sigma+1},\ldots ,
L'_n)\in{\cal U}'_0$
then
(9) holds (recall that (4), (5) are always fulfilled),
$W_n(L')$ is the closure with respect to the Zariski topology of
$W\cap{\cal Z}(e'_{\sigma+1},\ldots ,
e'_n)\setminus{\cal Z}(e_0,\ldots , e_r)$
and $D_n(L')=D_n=\deg p'_v\deg W'$.
By Lemma~11 \cite{9} if $L'\in{\cal U}'_0$ then $W_i(L')$
is exactly the union of all the irreducible over $\overline{k}$
components $E$ of $W\cap{\cal Z}(e'_{\sigma+1},\ldots , e'_i)$ such that
$E(\overline{k_v})$ contains at least one of $\deg p'_v\deg W'$ points from
$W_v\cap{\cal Z}(e'_{\sigma+1},$ $\ldots , e'_n)\setminus{\cal
Z}(e_0,\ldots , e_r)$ (or, which is equivalent, $E$ contains an irreducible
component of $W_n(L')$).

Further, let $\sigma\le\alpha\le n$ be an integer
and $L'\in{\cal U}'_0$.
By Lemma~11 \cite{9} condition (8)
holds for all $\sigma\le i\le\alpha$ if and only if
$$
W_i(L')=\overline{W\cap{\cal Z}(e'_{\sigma+1},\ldots , e'_i)
\setminus{\cal Z}(e_0,\ldots , e_r)}
\eqno (10)
$$
for all $\sigma\le i\le\alpha$.
In this case
$W^{(3)}_i(L')=\emptyset$ for all $\sigma+1\le i\le\alpha$, and
therefore,
by (4) and (5) the algebraic variety $W^{(1)}_i(L')$,
$\sigma+1\le i\le\alpha$, coincides with
the union of all the irreducible components of dimension $n-i+\sigma-s$ of
$W\cap{\cal Z}(e_0,\ldots , e_r)$. Hence $W^{(1)}_i(L')$ and
$J^{(1)}_i(L')=J^{(1)}_i$
do not depend on $L'$ under the considered conditions.


One can decide whether $L'\in{\cal U}'_0$ by Lemma~25 Section~4
within the time polynomial in the size of input and $(dd')^n$,
i.e., polynomial in the size of $L'$ and in the values
from the estimate for the working time of the
algorithm of Theorem~2.

Let $\sigma\le\alpha\le n+1$ be an integer.
Let us identify the set of all $(\alpha-\sigma)$--tuples
$(L'_{\sigma+1},\ldots ,
L'_\alpha)$ of linear forms $L'_j\in \overline{k}[X_0,\ldots ,X_r]$
(respectively $L'_j\in\overline{k_v}[X_0,\ldots ,X_r]$) with
${\Bbb A}^{(r+1)(\alpha-\sigma)}(\overline{k})$
(respectively ${\Bbb A}^{(r+1)(\alpha-\sigma)}(\overline{k_v})$).
For convenience of notation for $\alpha=\sigma+1$
we shall identify $1$--tuples $(L'_{\sigma+1})$ with the linear form
$L'_{\sigma+1}$. Recall that $L'_0=L'_{n+1}$.
{\it Denote by ${\cal U}'_\alpha$, $\sigma\le\alpha\le n+1$,
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. Besides that, $\pi_{n+1}$ is an isomorphism.





\medskip{\it Let $\sigma\le\alpha\le n+1$ be an integer.
Denote by ${\cal U}''_\alpha\subset{\cal U}'_\alpha$ the subset of
all $(L'_{\sigma+1},\ldots ,
L'_\alpha)$ such that
$$
D(L'_{\sigma+1},\ldots , L'_i)=D_i
\eqno (11)
$$
for all $\sigma+1\le i\le \alpha$. Set also ${\cal
U}''_0={\cal
U}'_0\cap\pi_n^{-1}({\cal U}''_n)=\pi_{n+1}^{-1}({\cal
U}''_{n+1})$.}
If $(L'_{\sigma+1},\ldots , L'_\alpha)\in{\cal U}''_\alpha$ then, see
Lemma~12 Section~3, the indices of intersection
$\iota_j(L'_{\sigma+1},\ldots , L'_i)=\iota_j$ for all
$j\in J^{(1)}_i$, $\sigma+1\le i\le \alpha$, and as we have noticed
$\iota_j=1$ for all $j\in J_i(L'_{\sigma+1},\ldots ,
L'_i)$, $\sigma+1\le i\le \alpha$.
Further, ${\cal U}''_\alpha$ is a
nonempty open in the Zariski topology subset of ${\Bbb
A}^{(r+1)(\alpha-\sigma)}(\overline{k})$, see Lemma~13 Section~3.
By Lemma~14 \cite{9} 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 (8) and (10) hold for all $\sigma\le i\le n-1$.

If one extends the ground field then $U\in{\cal U}''$. Hence (6)
implies
\setcounter{equation}{11} \begin{eqnarray}
&&(d')^{n-\sigma+1}\deg W=\nonumber\\
&&\sum_{\sigma+1\le i\le n+1}\Bigl(\,
\sum_{j\in J^{(1)}_i}
(d')^{n-i+1}\iota_j\deg W_j+
\sum_{j\in J^{(2)}_i}
(d')^{n-i+1}\iota_j\deg W_j\,\Bigr). \label{12}
\end{eqnarray}


{\it Let $\sigma\le\alpha\le n+1$ be an integer.
Denote by  ${\cal U}'''_\alpha\subset{\cal U}''_\alpha$ the subset
of all $(L'_{\sigma+1},\ldots ,
L'_\alpha)$ such that
$$
D^{(2)}(L'_{\sigma+1},\ldots , L'_i)=D^{(2)}_i
\eqno (13)
$$
for all $\sigma+1\le i\le \alpha$. Set also ${\cal U}'''_0={\cal
U}''_0\cap\pi_n^{-1}({\cal U}'''_n)$, i.e., ${\cal U}'''_0$
is a set of all $L'\in{\cal U}''_0$ such that
$(L'_{\sigma+1},\ldots , L'_n)\in{\cal U}'''_n$.}
Then ${\cal U}'''_\alpha$ is a
nonempty open in the Zariski topology subset of ${\Bbb
A}^{(r+1)(\alpha-\sigma)}(\overline{k})$, see Lemma~16 Section~3.
Further, $\pi_{n+1}(U)\in{\cal U}'''_{n+1}(\overline{K})$.
Obviously $\pi_{n+1}({\cal U}'''_0)\supset{\cal U}'''_{n+1}$,
and if $s=\sigma$ then the last inclusion becomes the equality.


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

To formulate Theorem~2 without any restrictions on $s$ and $\sigma$
we need to consider also the case when
$W$ is a union of some defined over $k$ and irreducible over $k$
components of $V$ (i.e., $W$ is
not necessarily irreducible over $k$)
and $\dim W=n-s=0$ (actually for $n=s$ Theorem~2 is trivial).
In the considered case $p$, $p'$, $W'$, are
defined as above. We have $W\cap{\cal Z}(e_0,\ldots , e_r)=\emptyset$,
$\dim W'=n-\sigma=0$.
Let $L'=(L'_0)$ where $L'_0$ is a linear form from $\overline{k}[X_0,\ldots
,X_r]$.
The element $L'=(L'_0)\in{\cal U}'_0$,
if and only if $W'\cap{\cal Z}(L'_0)=\emptyset$ in ${\Bbb
P}^r(\overline{k})$.
The morphism $p'$ is dominant if and only if $L'\in{\cal U}'_0$.
For arbitrary $L'\in{\cal U}'_0$
put $W_n(L')=W$; $W_{n+1}(L')=\emptyset$, $W^{(j)}_n(L')=\emptyset$,
$W^{(j)}_{n+1}(L')=\emptyset$ for $j=1,2,3$.
Set ${\cal U}'''_0={\cal U}''_0={\cal U}'_0={\cal
U}'''_{n+1}={\cal U}''_{n+1}={\cal U}'_{n+1}$.
All the other objects defined above for irreducible $W$ are defined in
the considered case in the similar way.

\par\medskip\noindent{\bf REMARK~1}\hspace{0.1em} {\it  The B\'ezout theorem
implies that if $L'\in{\cal U}'_0$,
$J^{(2)}_i(L')=\emptyset$ for all $s+2\le i\le n$ and
condition (10) holds for all $\sigma+1\le i\le n-1$ then
for every $\sigma+1\le i\le n+1$ the degree
$D_i(L')=D_i$ and
$\iota_j(L')=\iota_j$ for every $j\in J^{(1)}_i$.
}\par\medskip


\par\medskip\noindent{\bf REMARK~2}\hspace{0.1em} {\it  Let $K_1$ be an
algebraically closed field and
$K_1\supset k$ linearly disjoint with $k(W)$ over $k$.
Let us choose an extension
$K_2\supset k$ such that $W$ is irreducible over $K_2$ and
the algebraic closure $\overline{K_2}=K_1$
(e.g., $K_2$ may be an arbitrary purely transcendental an extension of $k$
such that $\overline{K}=K_1$). Let us define the sets ${\cal
U}'_\alpha$, ${\cal U}''_\alpha$, ${\cal U}'''_\alpha$ for
the ground field $K_1$ (and also $K_2$) replacing in the
definitions of these sets the ground field $k$ by $K_2$. Denote
for brevity  the obtained sets by ${\cal U}'_{K_1,\alpha}$,
${\cal U}''_{K_1,\alpha}$, ${\cal U}'''_{K_1,\alpha}$
(we use this more detailed notation only in this remark).
Then obviously ${\cal U}'_{K_1,\alpha}$, ${\cal
U}''_{K_1,\alpha}$, ${\cal U}'''_{K_1,\alpha}$ do not depend
on the choice of the field $K_2$ such as above.
In particular, the initial ${\cal U}'_\alpha$, ${\cal
U}''_\alpha$, ${\cal U}'''_\alpha$ defined for the ground
field $k$ depend only on $\overline{k}$.
Further, obviously ${\cal
U}'_\alpha= {\cal U}'_{K_1,\alpha}(\overline{k})$, ${\cal
U}''_\alpha= {\cal U}''_{K_1,\alpha}(\overline{k})$,
${\cal U}'''_\alpha= {\cal
U}'''_{K_1,\alpha}(\overline{k})$ for all $\alpha$.
On the other hand, probably also ${\cal
U}'_\alpha(K_1)={\cal U}'_{K_1,\alpha}$, ${\cal
U}''_\alpha(K_1)={\cal U}''_{K_1,\alpha}$, ${\cal
U}'''_\alpha(K_1)={\cal U}'''_{K_1,\alpha}$. We leave to prove
the last three equalities to the reader. We shall not need them in this
paper.
Similarly the reader may prove that if $V$ and $W$ are defined over a
subfield
$k'$ of $k$ (the extension $k\supset k'$ may be transcendental)
then ${\cal U}'_\alpha$, ${\cal U}''_\alpha$, ${\cal
U}'''_\alpha$ are also defined over $k'$.
}\par\medskip





Recall some details from \cite{8}.
Let
$\overline{\lambda}=(L_0,\ldots ,L_{n+1})$, and all $L_j\in k[X_0,\ldots ,$
$X_n]$, $0\le j\le n+1$, are linear forms. Suppose that $\overline{\lambda}$
(i.e., linear
forms
$L_0,\ldots ,L_{n+1}$) satisfy assertion of Theorem~1 \cite{8} for the
algebraic variety $V$. Then, see Section~2 \cite{8}, for every $1\le j\le
n-2$ such that $V$ has an
irreducible component of dimension $n-j$
the equality
$v_j(\overline{\lambda})=u_j(\overline{\lambda})$ holds, where
$v_j(\overline{\lambda}),u_j(\overline{\lambda})\le d^j(d^j-1)$ are integers
defined in \cite{8}. But now the notations $v_j$ and $u_j$ are already
used in the other sense. So we shall denote in this paper these integers by
$b_j(\overline{\lambda}),
a_j(\overline{\lambda})$ in place of
$v_j(\overline{\lambda}),u_j(\overline{\lambda})$ from \cite{8} respectively.
More precisely, in Section~1
for arbitrary $\overline{\lambda}$ and $V$ satisfying property (B), see the
beginning of Section~1,
we define the integers $b_j(\overline{\lambda})=b_j(\overline{\lambda},V)$,
$a_j(\overline{\lambda})=a_j(\overline{\lambda},V)$
for all $0\le j\le n$.

In \cite{8} we did not construct $\overline{\lambda}$ in general
position with maximal possible
$b_j(\overline{\lambda})$, see Remark~3 \cite{8}.
We applied there Remark~2 \cite{8} as a weak reduction to general position.
In this paper it is also possible but it makes
the presentation of results slightly awkward.
So we manage to avoid the application of Remark~2 \cite{8}.
Instead of it we strengthen
the result of \cite{8} and construct
$\overline{\lambda}$ in general position
with maximal possible $b_j(\overline{\lambda})$ in see Section~1
using the elaborated method, see the beginning of the Introduction.
So it is sufficient
now to apply the simple algorithm of reduction to integer coefficients, see
Proposition~2, \cite{9}.






Let $\lambda=(L_{s+1},\ldots , L_\sigma)$ where all $L_j\in k[X_0,\ldots ,
X_n]$, $s+1\le j\le\sigma$,
are linear forms.
Consider the following property
\begin{enumerate} \renewcommand{\labelenumi}{(\Alph{enumi})}
\item For
$\lambda$ an element $\overline{\lambda}$ is given
(so linear forms $L_w$, $s+1\le w\le\sigma$, of $\lambda$ and
$\overline{\lambda}$ coincide);
Theorem~1
\cite{8} holds for the algebraic variety $V$ and the linear forms
$L_0,\ldots ,L_{n+1}$; and for every $0\le j\le
n$ such that $V$ has an
irreducible component of dimension $n-j$ the
integer $b_j(\overline{\lambda})=a_j(\overline{\lambda})$ is maximal
possible (if one varies $\overline{\lambda}$, see Section~1 for details).
\end{enumerate}
In this case we shall say for brevity that $\lambda$ satisfies condition (A)
for $V$, or that
(A) holds for $\lambda$ and $V$.



We shall suppose in what follows unless we state otherwise that
the dimension
$\dim W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)=n-\sigma$;
the last intersection is transversal; if $n-\sigma>0$ then
the algebraic variety $W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$ is
irreducible over $k$.
Denote $W_\lambda=W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$,
$V_\lambda=V\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$.
We shall assume in what follows that
$\dim(W_\lambda\cap(\overline{V_\lambda\setminus W_\lambda})<\dim W_\lambda$.
Notice that if (A) holds for $\lambda$
then all these properties are fulfilled.
Denote by
\setcounter{equation}{13} \begin{eqnarray}
&&p_\lambda\, :\, W_\lambda\setminus{\cal
Z}(e_0,\ldots , e_r)\rightarrow{\Bbb P}^r(\overline{k}),\label{14}\\
&&p'_\lambda\, :\, W_\lambda\setminus{\cal
Z}(e_0,\ldots , e_r)\rightarrow W'\label{15}
\end{eqnarray}
the morphisms induced by $p$ and $p'$.
Suppose that
$$
\dim W_{\lambda}\cap{\cal Z}(e_0,\ldots , e_r)<\dim W_{\lambda}
\eqno (16)
$$
and the morphism $p'_\lambda$ is dominant.
Notice that if $n-\sigma>0$ then $W_{\lambda}$ is irreducible over $k$
by our assumption.
Hence if $n-\sigma>0$ and
the morphism $p'_\lambda$ is dominant then condition (16) holds.
Let us replace $(V,W,p,p')$ by $(V_\lambda,W_\lambda,p_\lambda,p'_\lambda)$
in the definitions of all $W_i(L')$,
$W(L'_{\sigma+1},\ldots , L_n)$,
${\cal U}'_\alpha$, ${\cal U}''_\alpha$, ${\cal U}'''_\alpha$,
$p_{L'}$ and so on. We shall denote the
obtained objects by
$W_{\lambda,i}(L')$, $W_\lambda(L'_{\sigma+1},\ldots , L_n)$,
${\cal U}'_{\lambda,\alpha}$, ${\cal U}''_{\lambda,\alpha}$,
${\cal U}'''_{\lambda,\alpha}$, $p_{\lambda,L'}$ and so on.
Notice that the field
$k_{v_\lambda}=k$ and $p_{\lambda,v_\lambda,L'}=p_{\lambda,L'}$
since $s_\lambda=\sigma_\lambda=
\sigma$.
It will be convenient for us to consider the morphism $p'_\lambda$
(set $q_\lambda=p'_\lambda$)
together
with $p'_v$, see Theorem~2 below. Let us define $V_v$, $W_v$ and so on
extending the ground field $k$ till $k_v$. Notice that condition (A) holds
for $v$
and $V$ if one extend $k$ till $k_{\overline{v}}$.


\medskip
Giving these definitions we proceed to the precise statements.
Let  homogeneous polynomials
$f_1,\,\ldots\,$, $f_m\in k[X_0,\,\ldots\, ,X_n]$ be given, $n\ge 0$, $m\ge
1$.
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 $F\in{\Bbb Q}  [t_1,\,\ldots\, ,t_l,Z]$ and leading
 coefficient $ {\mathrm  lc}_ZF$ of $F$ is equal to 1.
 We shall represent each polynomial $f=f_i$ (and similarly $e_j$) in the form
$$
 f=\frac{1}{a_0} \sum_{i_0, \ldots, i_n} \sum_{0 \leq j <
 \deg_Z F} a_{i_0, \ldots, i_n,j} \theta^j X_0^{i_0} \cdots X_n^{i_n}
 \, ,
$$
 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 (F)<d_1, \; \deg_{t_\gamma} (F)<d_1, \; {\rm l}(F)<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 ${\rm L}(F)<d_1^{l+1} M_1$.



\par\medskip\noindent{\bf REMARK~3}\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{2}. 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
$$
k(z)=k((X_1/X_{i_0})(z),\ldots , (X_n/X_{i_0})(z))=
k[\eta]\simeq k[Z]/(\Phi)
\eqno (17)
$$
is given where $\eta=\sum_{0\le i\le n}c_i(X_i/X_{i_0})(z)$,  the
coefficients $c_i\in {\Bbb Z}$ are given and $\Phi\in k[Z]$ is
minimal polynomial of $\eta$ over $k$ with leading coefficient
$\mbox{\rm lc}_Z\Phi=1$ (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
${\rm L}(\Phi)+\sum_{0\le i\le n}{\rm L}(X_i/X_{i_0})$.




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 Lemma~21 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.


\par\medskip\noindent{\bf THEOREM~1}\hspace{0.1em} {\it  Let the field $k$,
the algebraic varieties $V$, $W$, $W'$,
and polynomials $F$, $f_1,\ldots ,f_m$, $e_0,\ldots , e_r$, and the
point $x^{(0)}\in W$ be given as above.
One can compute the dimensions $\dim W=n-s$, $\dim W'=n-\sigma$
(hence $n\ge\sigma\ge s\ge 0$).
Let $L'=(L'_0,L'_{\sigma+1},\ldots ,L'_n)\in{\cal U}'_0$ be given.
Suppose that all the linear forms $L'_j$, $j\in\{0,\sigma+1,\ldots ,n\}$ have
coefficients from a finitely generated extension $k'$ of $k$ which is given
by the minimal polynomial of a primitive element of $k'$ over a purely
trancendental extension of $k$ of fixed transcendency degree.
In the statement of the theorem we assume that all the points
$y',y'',x$, see below, are given
similarly to (17) but with the field $k'$ in place of $k$.
Set
$L'_{n+1}=L'_0$, $e'_j=L'_j(e_0,\ldots , e_r)$ for all $j$.
Put $\widetilde{V}_i=(V\cup{\cal Z}(e'_{i+1}))\cap{\cal
Z}(e'_{\sigma+1},\ldots , e'_i)$
and $\iota=i-\sigma+s$ for all $\sigma\le i\le n$.
Then one can construct an element
$\overline{\lambda}=(L_0,L_1,\ldots , L_{n+1})$,
where $L_0,L_1,\ldots , L_{n+1}$ are linear forms in $X_0,\ldots ,
X_n$ with integer coefficients of length $O(n\,\log d+(n-\sigma)\log d')$
such that the following assertions (a) and (b) hold.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}

\item For every $\sigma\le i\le n$ the intersection of the algebraic
varieties $\widetilde{V}_i$ and
${\cal Z}(L_{\iota+1},\ldots , L_n)$ is transversal,
$W_i(L')\cap{\cal Z}(L_0,L_{\iota+1},\ldots , L_n)=\emptyset$, and the
number of elements
$\#(L_{n+1}/L_0)(W_i(L')\cap{\cal Z}(L_{\iota+1},\ldots , L_n))
=D_i(L')$ in ${\Bbb P}^n(\overline{k})$.
Further, for every $\sigma\le i\le n$
the assertion of Theorem~3, see Section~1, holds for $\widetilde{V}_i$
in place of $V$. In particular, one can construct all the points from
$W_i(L')\cap{\cal Z}(L_{\iota+1},\ldots, L_n)$
for every $\sigma\le i\le n$.

\item For every $\sigma+1\le i\le n+1$ the intersection of the algebraic
varieties $W_{i-1}(L')\cap{\cal Z}(e'_i)$
and ${\cal Z}(L_{\iota+1},\ldots , L_n)$ is transversal,
$W_{i-1}(L')\cap{\cal Z}(e'_i)\cap
{\cal Z}(L_0,L_{\iota+1},\ldots , L_n)=\emptyset$, and
$\#(L_{n+1}/L_0)(W_{i-1}(L')\cap{\cal Z}(e'_i)\cap
{\cal Z}(L_{\iota+1},\ldots ,
L_n))=D_i(L')+D^{(1)}_i(L')+
D^{(2)}_i(L')+D^{(3)}_i(L')$ in ${\Bbb P}^n(\overline{k})$.
For every $\sigma+1\le i\le n+1$
the assertion of Theorem~3, see Section~1,
holds for $W_{i-1}(L')\cap{\cal Z}(e'_i)$
in place of $V$ (notice that here $W_{i-1}(L')\cap{\cal Z}(e'_i)$ is
not given explicitly by a system of polynomial equations).
In particular, one can construct the finite set $W_{i-1}(L')\cap{\cal
Z}(e'_i)\cap
{\cal Z}(L_{\iota+1},\ldots , L_n)$, and for every irreducible over
$\overline{k}$ component $E$ of
$W_{i-1}(L')\cap{\cal Z}(e'_i)$ the intersection
$E\cap{\cal Z}(L_{\iota+1},\ldots ,
L_{n-1})$ is irreducible over $\overline{k}$.
Further, for any two points $y',y''\in W_{i-1}(L')\cap{\cal Z}(e'_i)\cap
{\cal Z}(L_{\iota+1},\ldots , L_n)$ one can decide whether
$y'$ and $y''$ belong to the same defined over $k'$ and irreducible over
$k'$
(respectively defined over $\overline{k}$ and irreducible over
$\overline{k}$) component of
$W_{i-1}(L')\cap{\cal Z}(e'_i)$ (in the case of
irreducible over $\overline{k}$ components we
assume that the composite $k'(y',y'')$ of the fields
$k'(y')$ and $k'(y'')$ over $k'$
is given explicitly similarly to Theorem~2 \cite{8},
and the bound for the size of the composite holds,
which polynomial in the size of input, cf. \cite{8}).
Finally, for every $y'\in W_{i-1}(L')\cap{\cal Z}(e'_i)\cap
{\cal Z}(L_{\iota+1},\ldots , L_n)$ one can compute the number of points
$y''\in W_{i-1}(L')\cap{\cal Z}(e'_i)\cap
{\cal Z}(L_{\iota+1},\ldots , L_n)$ such that $y'$ and $y''$ belong to
the same defined over $k'$ and irreducible over $k'$
(respectively defined over $\overline{k}$ and irreducible over
$\overline{k}$) component of
$W_{i-1}(L')\cap{\cal Z}(e'_i)$.
\end{enumerate}
Let $\overline{\lambda}$ be an
arbitrary element
satisfying (a) and (b) (not necessarily constructed by the algorithm)
and $\sigma+1\le i\le n+1$ be an integer.
Then the following assertions holds.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\setcounter{enumi}{2}
\item Let a point
$y'\in W_{i-1}(L')\cap{\cal Z}(e'_i)\cap
{\cal Z}(L_{\iota+1},\ldots , L_n)$ be given. Denote by $E$ the unique
defined over $k'$ and irreducible over $k'$ (respectively
defined over $\overline{k}$ and irreducible over
$\overline{k}$) component of $W_{i-1}(L')\cap{\cal Z}(e'_i)$
such that $y'\in W_{i-1}(L')\cap{\cal Z}(e'_i)$.
Then for an arbitrary point $x\in{\Bbb P}^n(\overline{k})$ one can decide
whether $x\in E$, and, more generally one can compute the multiplicity
$\mu(x,E)$ of the point $x$
at $E$
(again if $E$ is irreducible over $\overline{k}$ then we assume that
the composite $k'(x,y')$ of the fields
$k'(x)$ and $k'(y')$ over $k$ is given explicitly similarly to (b)).

\item For a given point
$y'\in W_{i-1}(L')\cap{\cal Z}(e'_i)\cap
{\cal Z}(L_{\iota+1},\ldots , L_n)$ one can
decide whether $y'$ is an element of $W_i(L')$ (respectively
$W^{(1)}_i(L')$,
$W^{(2)}_i(L')$, $W^{(3)}_i(L')$). Hence by (b) and (c)
one can compute the degrees
$\deg W_i(L')$, $\deg W^{(1)}_i(L')$, $\deg W^{(2)}_i(L')$, $\deg
W^{(3)}_i(L')$. Besides that, for every irreducible over $\overline{k}$
component $E$ of $W_{i-1}(L')\cap{\cal Z}(e'_i)$ we have
$E\subset{\cal Z}(e_0,\ldots , e_r)$
if and only if $E\cap{\cal Z}(L_{\iota+1},\ldots , L_n)\subset{\cal
Z}(e_0,\ldots , e_r)$.

\item By (b), (c), (d)
for an arbitrary point $x\in{\Bbb P}^n(\overline{k})$ one can
compute the multiplicities $\mu(x,W_i(L'))$ (respectively
$\mu(x,W^{(1)}_i(L'))$,
$\mu(x,W^{(2)}_i(L'))$,\\ $\mu(x,W^{(3)}_i(L'))$) of the point $x$ at
$W_i(L')$ (respectively
$W^{(1)}_i(L')$,
$W^{(2)}_i(L')$, $W^{(3)}_i(L')$).

\item Under conditions of (c) suppose that $E$ is an defined over
$\overline{k'}$ and irreducible over $\overline{k'}$ component. Then
one can compute the index of intersection
$i(W_{i-1}(L'),{\cal Z}(e'_i);E)$ of the algebraic varieties
$W_{i-1}(L')$ and  ${\cal Z}(e'_i)$
at the component $E$ in ${\Bbb P}^n(\overline{k})$.



\item Let us identify the set of all $(n+2)$--tuples of linear forms from
$\overline{k}[X_0,\ldots ,$ $X_n]$ with the affine space
${\Bbb A}^{(n+1)(n+2)}(\overline{k})$.
Let ${\mathfrak l}\in{\Bbb A}^{(n+1)(n+2)}(\overline{k})$ be a line
defined over $k'$ such that $\overline{\lambda}\in{\mathfrak l}$. Then for
all $\lambda^*\in{\mathfrak l}(k')$
(here ${\mathfrak l}(k')$ is the set of $k'$--points of ${\mathfrak l}$;
we assume that $\lambda^*=(L^*_0,\ldots , L^*_{n+1})$, where all
$L^*_j\in\overline{k}[X_0,\ldots , X_n]$ are linear forms),
except at most a polynomial in $d^n(d')^{n-\sigma}$ number, assertions
(a) and (b) hold with $\lambda^*$
in place of $\overline{\lambda}$. For every element $\lambda^*\in{\mathfrak
l}(k')$ one can decide whether
assertions (a) and (b) hold with $\lambda^*$ in place of
$\overline{\lambda}$.
\end{enumerate}
The working time of the algorithm for computing $s$, $\sigma$ and
constructing $\overline{\lambda}$
(respectively of the algorithm from (a), (b), (d), (f)) 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 size ${\rm L}(L')$ of the element $L'$ (including the size of the minimal
polynomial of the primitive element of $k'$, see above).
The working times of the algorithms from (c) and (e) (respectively (g)) are
polynomial in the same values and  in the size
${\rm L}(x)$ of the point $x$ (respectively
in the same values and  in the size ${\rm L}(\lambda^*)$  of the element
$\lambda^*$).
}\par\medskip




Notice here that the difficulty in the proof of assertion (b) of Theorem~1
in comparison with its assertion (a) (and Theorem~3)
is that some irreducible components of
$W_{i-1}(L')\cap{\cal Z}(e'_i)$ may be
contained in $\overline{V\setminus W}$.




\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 dimensions $\dim W$, $\dim W'$. Suppose that
$\dim W=n-s$ and $\dim W'=n-\sigma$.
Hence the open in the Zariski topology subset ${\cal U}'''_0\subset
{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$ is defined, see
above.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item
Then one can
construct an element
$L'=(L'_0,L'_{\sigma+1},\ldots , L'_n)\in{\cal U}'''_0$
such that linear forms
$L'_j$ have integer coefficients with lengths $O(n\log d+
(n-\sigma)\log d')$ for
all $j\in\{0,\sigma+1,\ldots ,
n\}$. Hence
conditions (4), (5),(8)--(13)
hold for $\sigma+1\le i\le n$.
One can compute the degrees $D_i$, $D^{(1)}_i$, $D^{(2)}_i$,
$\sigma\le i\le n$.
\item For a given
$L'\in{\cal U}'''_0$
satisfying the assertion from (a) (but not necessarily computed in (a);
one can verify that $L'\in{\cal U}'''_0$ by remarks
given at the end of the Introduction)
one can construct $\lambda=
(L_{s+1},\ldots , L_\sigma)$, where all $L_j\in
k[X_0,\ldots , X_n]$
are linear forms with integer coefficients of
length $O(n\log d+(n-\sigma)\log d')$
such that  the following statements (c) and (d) hold.
\item $\lambda$ satisfies conditions (A) and (16), see above;
the morphism $p'_\lambda$,
see (15), 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
$\sigma \le i\le n$.
\item $L'\in{\cal U}'''_{\lambda,0}\subset
{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$.
\item Let $\lambda$ and $L'$ satisfy assertions (c) and (d).
Then $L'\in{\cal U}'''_0$. Further,
$W_{\lambda,i}(L')=W_i(L')\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$,
$W^{(1)}_{\lambda,i}=W^{(1)}_i\cap{\cal Z}(L_{s+1},\ldots ,$
$L_\sigma)$,
$W^{(2)}_{\lambda,i}(L')=W^{(2)}_i(L')\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)$ for all $\sigma\le i\le n$ and all these intersections are
transversal.
Therefore, $D_{\lambda,i}=D_{\lambda,i}(L')=D_i(L')=D_i$,
$D^{(1)}_{\lambda,i}=D^{(1)}_i$,
$D^{(2)}_{\lambda,i}=D^{(2)}_{\lambda,i}(L')=D^{(2)}_i(L')=D^{(2)}_i$ for
all $\sigma\le i\le n$.
\item Let $L'=
(L'_0,L'_{\sigma+1},\ldots , L'_n)\in
{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$ be an arbitrary
element
(i.e., not necessarily the one constructed by the algorithm from this
theorem), where all linear forms $L'_j\in \overline{k}[X_0,\ldots ,X_r]$.
Let $\lambda=
(L_{s+1},\ldots , L_\sigma)\in
{\Bbb A}^{(n+1)(\sigma-s)}(k)$ be an arbitrary element
(not necessarily constructed in (b)), where all $L_j\in
k[X_0,\ldots , X_n]$.
Set $\nu_0=(r+1)(n-\sigma+1)+(n+1)(\sigma-s)$.
Consider the  pair $(\lambda,L')$ as an element of
${\Bbb A}^{\nu_0}(\overline{k})$.
Suppose that assertions (c) and (d) hold for $(\lambda,L')$.
Let us identify ${\Bbb A}^1(\overline{k})=\overline{k}$ using a
coordinate function on ${\Bbb A}^1(\overline{k})$
defined over $k$.
Let  $\iota_0\, :\,{\Bbb
A}^1(\overline{k})\rightarrow
{\Bbb A}^{\nu_0}(\overline{k})$ be an
affine linear embedding.
For every $y\in\overline{k}$ let us represent
$\iota_0(y)=(\lambda_0,L^{(0)})$ where $\lambda_0=
(L_{0,s+1},\ldots , L_{0,\sigma})$,
$L^{(0)}=(L^{(0)}_0,L^{(0)}_{\sigma+1},\ldots , L^{(0)}_n)$,
all $L_{0,j}\in\overline{k}[X_0,\ldots , X_n]$,
$L^{(0)}_w\in\overline{k}[X_0,\ldots ,X_r]$ are linear forms.
Assume that for every $y\in k$ all $L_{0,j}$ have coefficients
from $k$.
Suppose that $\iota_0(0)=(\lambda,L')$.
Then for all $y\in k$, except at most a polynomial
in $d^n(d')^{n-\sigma}$ number, $\iota_0(y)=(\lambda_0,L^{(0)})$
satisfies assertions (c) and (d) in place of
$(\lambda,L')$. Besides that, see condition (A),
for $\alpha\in\{0,\ldots , s,\sigma+1,\ldots , n+1\}$
the linear forms $L_\alpha$ of $\overline{\lambda}$
and $L_{0,\alpha}$ of $\overline{\lambda_0}$ coincide.
\end{enumerate}
The working time of the algorithm from (a) is polynomial in
$d^n$, $(d')^n$, $d_1$, $d_2$, $d'_2$,
$M$, $M'$, $M_1$, $m$, $r$, and
the size ${\rm L}(x^{(0)})$ of the point $x^{(0)}$.
The working time of the algorithm from (b) is polynomial in
the same values and the size
${\rm L}(L')$ of $L'$.
}\par\medskip




\medskip This theorem will be proved in Section~7.
Notice that if one replaces $k$ by $k_v$ then $v$
(in place of $\lambda$) satisfies assertions (c) and (d).
For more detailed information see the algorithms from Section~4 and
Section~5.
Namely, for a given element $L'=(L'_0,L'_{\sigma+1},
\ldots , L'_n)\in{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$,
where all $L'_j\in k[X_0,\ldots , X_n]$
one can decide whether $L'\in{\cal U}'_0$ using Lemma~25.
Using the algorithms from Theorem~1 one can check whether $D_i(L')=D_i$,
$D^{(2)}_i(L')=D^{(2)}_i$, $\sigma\le i\le n$, and hence to decide
whether $L'\in{\cal U}''_0$ and $L'\in{\cal U}'''_0$.
For more details related to assertion (f) see also
Lemma~19, Lemma~18, Lemma~20.

\par\medskip\noindent{\bf REMARK~4}\hspace{0.1em} {\it  Actually in
Section~6 we construct an element
$L'\in{\cal U}'''_0$ from assertion (a) of Theorem~2
such that $\pi_{n+1}(L')\in{\cal U}'''_{n+1}$.
}\par\medskip



\par\medskip\noindent{\bf REMARK~5}\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{Constructing linear sections in general position of an algebraic
variety }\label{s1}

Recall some notations from \cite{8}.
Let $V\subset{\Bbb P}^n(\overline{k})$ be an arbitrary projective
algebraic variety (not necessarily the one from the
Introduction).
For every $0\le j\le n$ denote by $V_j$ the union of all the irreducible
components of $V$ of dimension $j$.
Let
$L_0,L_1,\ldots , L_{n+1}\in k[X_0,\ldots , X_n]$ be linear
forms, $L_w=
\sum_{0\le i\le n}l_{w,i}X_i$, where all $l_{w,i}\in k$, $0\le
w\le n+1$,
$0\le i\le n$.
Put
$\overline{\lambda}=(L_0,L_1,\ldots , L_{n+1})$. Denote by
$\widetilde{\lambda}$ the family of coefficients
$l_{w,i}$, $0\le w\le n+1$, $0\le i\le n$.
Consider the following condition.
\begin{itemize}
\item[(B)] For every $0\le j\le n$ the set
$A_j=V_j\cap{\cal Z}(L_{j+1},\ldots , L_n)$ is finite
(or empty if $V_j=\emptyset$) in ${\Bbb P}^n(\overline{k})$, consists
of smooth points of $V$, and the intersections of tangent
spaces of $V_j$ and ${\cal Z}(L_{j+1},\ldots , L_n)$ at
all these points are transversal. The linear form $L_0$ does not vanish at
any point of
$A_j$ (i.e., $A_j\cap{\cal Z}(
L_0)=\emptyset$), and
the number of elements $\#(L_{n+1}/L_0)(A_j)=\#A_j=\deg V_j$.
\end{itemize}
If $V$ is from the Introduction then
using the algorithms from \cite{4} and \cite{2} for given linear
forms $L_0,\ldots , L_n$
(or which is the same for $\overline{\lambda}$)
one can decide  whether condition (B) holds
within the time polynomial in $d^n$, $d_1$, $d_2$,
$M$, $M_1$, $m$, and
the sum of sizes $\sum_{0\le j\le n+1}{\rm L}(L_j)$ of linear forms $L_j$.


In what follows we shall consider also property (B) of linear forms
with coefficients from an extension of $k$
replacing the ground field $k$ by its
extension. In particular, the integers $a_j(\overline{\lambda})$,
$b_j(\overline{\lambda})$, see below,
are also defined if the coefficients of the linear forms $L_j$ are from an
extension of $k$ and property
(B) holds over the algebraic closure of this
extension.

Suppose that $\overline{\lambda}$ satisfies condition (B).
Let $Z_0,\ldots ,Z_{n+1}$ be independent variables.
For every $0\le j\le n$ denote by $\Phi_j\in
k[Z_0,Z_{j+1},Z_{j+2},\ldots , Z_{n+1}]=B_j$
the nonzero separable polynomial of minimal degree
such that $\Phi_j(L_0,L_{j+1},L_{j+2},\ldots ,$ $L_{n+1})$
is vanishing on $V_j$. Then
 $\deg_{Z_{n+1}}\Phi_j=\deg\Phi_j=\deg V_j$.
Unless we state otherwise, we choose and fix $\Phi_j$ such that the
leading coefficient of $\Phi_j$ with respect to $Z_{n+1}$
is equal to $1$.
Denote by $R_j\in k[Z_0,Z_{j+1},Z_{j+2},$ $\ldots , Z_n]$ the
discriminant
of $\Phi_j$  with respect to of $Z_{n+1}$
and by $R_{j,0}$ square free part of $R_j$ (i.e., $R_{j,0}$ is a
separable polynomial of the most degree dividing $R_j$).
Put $\Phi_{j,*}=\Phi_j(Z_0,0,\ldots , 0,Z_n,Z_{n+1})\in
k[Z_0,Z_n,Z_{n+1}]$,
$R_{j,*}=R_j(Z_0,0,\ldots , 0,Z_n)\in k[Z_0,$ $Z_n]$, $R_{j,0,*}=
R_{j,0}(Z_0,0,\ldots , 0,Z_n)\in k[Z_0,$ $Z_n]$.
If $V_j=\emptyset$ then $\Phi_j=R_j=1$. If $j=n$ then $\Phi_j$, $R_j$,
$\Phi_{j,*}$, $R_{j,*}$ do not depend on $Z_n$.
Denote by $R_{j,*,0}$ square free part of $R_{j,*}$.
If the dependence on $\overline{\lambda}$ is essential then we shall denote
the introduced objects by
$\Phi_{\overline{\lambda},j}$, $R_{\overline{\lambda},j}$,
$R_{\overline{\lambda},j,0}$,
$\Phi_{\overline{\lambda},j,*}$, $R_{\overline{\lambda},j,*}$,
$R_{\overline{\lambda},j,0,*}$ and $R_{\overline{\lambda},j,*,0}$.
In particular extending the field $k$ till
$k_{\overline{v}}$, see the
Introduction, we get
$\Phi_{\overline{v},j}$, $R_{\overline{v},j}$, $R_{\overline{v},j,0}$,
$\Phi_{\overline{v},j,*}$,
$R_{\overline{v},j,*}$,
$R_{\overline{v},j,0,*}$ and $R_{\overline{v},j,*,0}$. Besides that,
we choose the last elements to be the separable polynomials
from the ring $k[\widetilde{v},Z_0,\ldots ,Z_{n+1}]$.



\par\medskip\noindent{\bf LEMMA~1}\hspace{0.1em} {\it  Under previous
conditions for every $0\le j\le n$ the polynomial
$\Phi_{\overline{v},j}(\widetilde{\lambda},Z_0,Z_{j+1},Z_{j+2},\ldots ,
Z_{n+1})$ is nonzero and
coincides with $\Phi_{\lambda,j}$ up to a nonzero from $k$.
}\par\medskip

\noindent {\bf PROOF} \quad Let $\nu\in k$. Then
$v_0,v_{s+1},\ldots , v_n,v_{n+1}+\nu v_0$ are generic linear forms in
$X_0,\ldots , X_n$ over $k$.
More generally, each linear automorphism $\gamma$ of
$k[v_0,v_{j+1},\ldots , v_{n+1}]$ over $k$ induces
the uniquely defined linear automorphism $\gamma'\, :\,
k[\widetilde{v}]\rightarrow k[\widetilde{v}]$ such
that (denote by
$\gamma'(\widetilde{v})$
the family of elements obtained under the action of $\gamma'$ to
the elements of $\widetilde{v}$) the polynomial
$\Phi_{\overline{v},j}(\gamma'(\widetilde{v}),\gamma(v_0),
\gamma(v_{j+1}),$ $\ldots ,$
$\gamma(v_{n+1}))=
\Phi_{\overline{v},j}$.

Hence
it is sufficient to prove the lemma for
$L_0,L_{j+1},\ldots , L_n,L_{n+1}+\nu L_0$ in place of $L_0,L_{j+1},\ldots ,
L_n,L_{n+1}$.
Let us choose $\nu\in k$ such that
$V_j\cap{\cal Z}(L_{j+1},\ldots ,$ $L_n,L_{n+1}+\nu L_0)=\emptyset$.
Then the conditions of Lemma~13 \cite{6} are satisfied for $j$,
$V_j$ and linear forms $L_0,L_{j+1},\ldots , L_n,L_{n+1}+\nu L_0$
in place of $s$, ${\cal Z}({\mathfrak P})$ and  $L_0,L_s,\ldots , L_n$.
The required assertion follows immediately from Lemma~13 \cite{6}
since $\deg\Phi_{\overline{\lambda},j}=\deg\Phi_{\overline{v},j}$.
The lemma is proved.

\par\medskip\noindent{\bf REMARK~6}\hspace{0.1em} {\it  In the similar way
one can strengthen Lemma~13 from \cite{6}
replacing there conditions (63), (64) by the one
${\cal Z}({\mathfrak P})\cap
{\cal Z}(L_0,L_s,\ldots , L_n)=\emptyset$.
}\par\medskip



Now set $b_j(\overline{\lambda})=b_j(\overline{\lambda},V)
=\deg_{Z_0,Z_{j+1},\ldots ,Z_n} R_{j,0}$;
$a_j(\overline{\lambda})=a_j(\overline{\lambda},V)=\\ \deg_{Z_0,Z_n}
R_{j,*,0}$. If condition (B) holds for $\overline{\lambda}$,
and $V$ is from the Introduction,
then, see \cite{8},
using \cite{2} one can construct all $\Phi_{j,*}$, $R_{j,*}$, $0\le j\le
n$ within the time polynomial in $d^n$ and the size of the input.
After that one can compute $R_{j,*,0}$ and $a_j(\overline{\lambda},V)$
within the same time. Let $\lambda^*=(L^*_0,\ldots , L^*_{n+1})$ where
all $L^*_j$ are linear forms with coefficients from a finitely generated
extension $k^*$ of $k$.
If condition (B) holds for $\lambda^*$ then extending the ground field in
one can compute $a_j(\lambda^*,V)$ in the similar way.


Notice that
$b_j(\overline{\lambda})=\deg_{Z_0,Z_n} R_{j,0,*}$. We have
$a_j(\overline{\lambda})
\le b_j(\overline{\lambda})\le
\deg V_j(-1+\deg V_j)$ for all $j$, see \cite{8}.
Obviously $a_j(L')=b_j$ for $j=n-1,n$.
Further, see \cite{8} Theorems~1 and 3 if $a_j(\overline{\lambda})=
b_j(\overline{\lambda})$
(or which is the same $\deg_{Z_0,Z_n} R_{j,*,0}=
\deg_{Z_0,Z_n} R_{j,0,*}$) for all $1\le j\le n-2$ then assertions
(a), (b), (c)
of Theorem~1 \cite{8} holds.
For a given $\overline{\lambda}$ satisfying
assertions (a), (b), (c)
Theorem~1 \cite{8} one can construct by \cite{8} all the objects from
assertions (d) and (e)
of this theorem and prove (d) and (e) (provided that all $L_w$ have integer
coefficients
of length $O(n\log d)$).

Put $b_j=b_j(\overline{v})=b_j(V)=b_j(\overline{v},V)=
\deg_{Z_0,Z_{j+1},\ldots , Z_n}
R_{\overline{v},j,0}$;
 $a_j=a_j(\overline{v})=a_j(V)=a_j(\overline{v},V)=
\deg_{Z_0,Z_n} R_{\overline{v},j,*,0}$. Then
$b_j=a_j$, see \cite{8} for all $0\le j\le n-1$.
We shall identify the set of all
$(n+2)$--tuples of linear forms from $\overline{k}[X_0,\ldots , X_n]$ with
${\Bbb A}^{(n+2)(n+1)}(\overline{k})$. The following lemma is
straightforward, cf. \cite{4}.

\par\medskip\noindent{\bf LEMMA~2}\hspace{0.1em} {\it
There is a defined over $k$ open in the Zariski topology
subset ${\cal V}'\subset {\Bbb A}^{(n+2)(n+1)}(\overline{k})$ such
that for every
$\overline{\lambda}\in{\cal V}'$ the property
(B) holds and $b_j(\overline{\lambda})=
a_j(\overline{\lambda})=b_j$ for all $0\le j\le n$.
}\par\medskip


\par\medskip\noindent{\bf LEMMA~3}\hspace{0.1em} {\it  Let $K$ be an
arbitrary extension of $k$.
Assume that all the linear forms $L_w\in\overline{K}[X_0,\ldots , X_n]$,
$0\le w\le n+1$.
Let $\overline{\lambda}$ satisfies property (B) (over $\overline{K}$), see
above. Let ${\mathfrak l}\subset{\Bbb
A}^{(n+2)(n+1)}(\overline{K})$ be a line containing $\overline{\lambda}$.
Then for all $\lambda^*\in{\mathfrak l}$,
apart of at most a polynomial in $2^n\sum_{0\le j\le n}\deg V_j$
number of exceptions, the
element $\lambda^*$ satisfies (B),
and the inequalities
$b_j(\overline{\lambda})\le b_j(\lambda^*)$,
$a_j(\overline{\lambda})\le a_j(\lambda^*)$ hold for all $0\le j\le n$.
}\par\medskip

\noindent {\bf PROOF} \quad The proof is straightforward
by the B\'ezout theorem, cf. \cite{4}. The lemma is proved.


\par\medskip\noindent{\bf THEOREM~3}\hspace{0.1em} {\it  Let $V$ be from the
Introduction.
Under conditions of Theorem~1 \cite{8} one can construct
 all the objects from the statement of this theorem such that
additionally
$b_j=b_j(\overline{\lambda})=a_j(\overline{\lambda})$ for all $0\le j\le n$
(and hence all $L_w$, $0\le w\le n+1$,
are linear forms with integer coefficients of length $O(n\log d)$, and
the integers
$b_j=b_j(\overline{\lambda})=a_j(\overline{\lambda})=a_j$ are computed).
}\par\medskip


\noindent {\bf PROOF} \quad We shall suppose without loss of generality that
$\dim V<n$, and hence it is sufficient to prove
the assertion of the theorem for all $1\le j\le n-1$. By the previous remarks
it is sufficient to construct $\overline{\lambda}$ such that all $L_w$,
$0\le w\le n+1$,
are linear forms with integer coefficients of length $O(n\log d)$, and
$b_j=b_j(\overline{\lambda})$. The construction of the required
$\overline{\lambda}$ is
recursive. We take as the base of the recursion an arbitrary element
$\overline{\lambda}$
satisfying (B),
i.e., all the required conditions but not necessarily to the condition
$b_j=b_j(\overline{\lambda})$.
(for example one can take as $\overline{\lambda}$ the element constructed
in Theorem~1 \cite{8}).

Let us describe the general step of the recursion.
Let $1\le j\le n-2$ be maximal such that for all $1\le w<j$ the integers
$b_w$ are computed,
$b_w=a_w(\overline{\lambda})$, and $V_j\ne\emptyset$
(the last condition can be verified
by the algorithm from \cite{4}).
Let us compute using the algorithms from \cite{4} and \cite{2} the
polynomials
$\Phi_{\overline{\lambda},j,*}$, $R_{\overline{\lambda},j,*}$ and
$R_{\overline{\lambda},j,*,0}$.
We shall suppose without loss of generality in what follows that $Z_0$ does
not divide
$R_{\overline{\lambda},j,*,0}$
replacing if it is necessary $Z_0$ by $Z_0-\mu Z_n$ for an appropriate
integer $\mu$
bounded from above by $d^{2n}$.

Let $1\le\tau\le 4$ be an integer.
Let $0\le j_1<\ldots < j_\tau\le n$ be integers.
Assume that the field $k$ is supplied with a real structure, see \cite{3}.
Let the field
$k_j=k(\varepsilon_1,\ldots ,\varepsilon_j)$, $0\le j\in{\Bbb Z}$,
where $\varepsilon_j>0$ for $j\ge 1$ is an infinitesimal
with respect to the field $k_{j-1}$.
Let $X_{w,i}$, $1\le w\le 4$, $0\le i\le n$;
$Z_{w,i}$, $i\in\{0,j+1,\ldots , n+1\}=I_j$, $1\le w\le \tau$,
be variables. Denote for brevity $L'_i=L_i+\sum_{1\le w\le
\tau}Z_{w,i}X_{j_w}$, $i\in I_j$
(so all $L'_i$ depend on the choice of $j_1,\ldots , j_\tau$).
Denote $X^{(i)}=(X_{i,0},\ldots , X_{i,n})$, and for an
arbitrary polynomial $c$ in
$n+1$ variables $c(X^{(i)})=c(X_{i,0},\ldots , X_{i,n})$, $1\le i\le 4$.
Assume that $j\le n-2$.
Consider the following system of polynomial equations and inequalities with
squares of absolute values with coefficients from $k_3$
with respect to the variables $X_{w,i}$, $1\le w\le 4$, $0\le i\le n$; and
$Z_{w,i}$, $i\in\{0,j+1,\ldots , n+1\}=I_j$, $1\le w\le \tau$.
$$
\left\{\begin{array}{ll}
\sum_{i\in I_j}\sum_{1\le w\le \tau}|Z_{i,w}|^2\le\varepsilon_1,\\
\varepsilon_2\le |L'_n(X^{(1)})-L'_n(X^{(3)})|^2
\le \varepsilon_1,&\\
|L'_{n+1}(X^{(i)})-L'_{n+1}(X^{(i+1)})|^2=
\varepsilon_3,& i=1,3, \\
L'_\nu(X^{(i_1)})=L'_\nu(X^{(i_2)})=0,&
1\le i_1<i_2\le 4,\quad j+1\le\nu\le n-1,\\
L'_n(X^{(i)})-L'_n(X^{(i+1)})=0,& i=1,3,\\
L'_0(X^{(i)})-1=0,& 1\le i\le 4,
\end{array}\right.
\eqno (18)
$$

\par\medskip\noindent{\bf LEMMA~4}\hspace{0.1em} {\it  Let $1\le j\le n-2$
be an integer.
The inequality $a_j(\overline{\lambda})<b_j$ holds if and only if
there exist $1\le\tau\le 4$ and $0\le j_1<\ldots < j_\tau\le n$ satisfying
the following property.
System (18) has a solution
$X_{w,i}=x^*_{w,i}\in\overline{k_3}$,
$1\le w\le 4$, $0\le i\le n$; $Z_{w,i}=z^*_{w,i}\in\overline{k_3}$,
$i\in\{0,j+1,\ldots , n+1\}$, $1\le w\le \tau$
such that
$$
(x^*_{w,0}:\ldots : x^*_{w,n})\in V_j(\overline{k_3}),\quad
1\le w\le 4.
\eqno (19)
$$
}\par\medskip

\noindent {\bf PROOF} \quad There exists
$\varepsilon'>0$
such that $\varepsilon'$ is an infinitesimal
with respect to the field $k_1$ and
$\varepsilon_2$ is an infinitesimal with respect to the field
$k_1(\varepsilon')$.


Suppose that $a_j(\overline{\lambda})<b_j$.
By Lemma~2 there are linear forms
$L''_w\in\overline{k(\varepsilon')}[X_0,$ $\ldots ,
X_n]$, $0\le w\le n+1$, such that
all the nonzero coefficients of $L''_w-L_w$, $0\le w\le n+1$, are
infinitesimals with respect to the
field $k$ and $a_j(\lambda')=b_j$, where $\lambda'=(L''_0,\ldots ,
L''_{n+1})$.
Let us replace in system (18) all $L'_i$ by $L''_i$ and delete the first
inequality.
Denote the obtained system by (**).

By Lemma~1 all the nonzero coefficients of the polynomial
$R_{\lambda',j,0,*}$ are
infinitely close to
the corresponding coefficients of $R_{\overline{\lambda},j,0,*}$, and
$R_{\lambda',j,0,*}(1,Z_n)$ has more
different roots than $R_{\overline{\lambda},j,0,*}(1,Z_n)$.
Hence by the transfer principle, see \cite{1},
there are roots $Z_n=z_{\alpha,n}\in k(\varepsilon')$, $\alpha=1,3$,
of $R_{\lambda',j,0,*}(1,Z_n)$
such that $z_{1,n}-z_{3,n}$ is an infinitesimal from
$\overline{k(\varepsilon')}$.
Recall that
$\mbox{\rm lc}_{Z_{n+1}}\Phi_{\lambda',j}=1$.
Hence for $\alpha=1,3$ by the transfer principle, applied to
the polynomial $\Phi_{\lambda',j,*}(1,z_{\alpha,n},Z)$, there are
$\widetilde{z}_{\alpha,n}\in \overline{k(\varepsilon')}$,
$\widetilde{z}_{\beta,n+1}
\in\overline{k(\varepsilon',\varepsilon_3)}$,
$\beta=\alpha,\alpha+1$ such that
$\widetilde{z}_{\alpha,n}-z_{\alpha,n}\in
\overline{k(\varepsilon',\varepsilon_3)}$, $\alpha=1,3$,
is infinitesimal with respect to the field
$k_2(\varepsilon')$,
$|\widetilde{z}_{\alpha,n+1}-\widetilde{z}_{\alpha+1,n+1}|^2=\varepsilon_3$,
and $\Phi_{\lambda',j,*}(1,
\widetilde{z}_{\alpha,n},\widetilde{z}_{\beta,n+1})=0$
for $\beta=\alpha,\alpha+1$. Set
$\widetilde{z}_{\beta,n}=\widetilde{z}_{\alpha,n}$ for $\alpha=1,3$,
$\beta=\alpha,\alpha+1$.
The morphism $V_j\rightarrow{\cal Z}(\Phi_{\lambda',j})$,
$(X_0:\ldots : X_n)\mapsto(L''_0:L''_{j+1}:\ldots : L''_n)$, is
finite dominant. Therefore, there is a solution of system (**) giving a
point
from $V_j(\overline{k(\varepsilon',\varepsilon_3)})^4$
and satisfying
additional equations $L''_\nu(X^{(\beta)})=0$, $j+1\le\nu\le n-1$,
$L''_j(X^{(\beta)})=\widetilde{z}_{\beta,j}$, $j=n,n+1$, $\beta=1,2,3,4$.
Let this solution be $X_{w,i}=\widetilde{x}_{w,i}$, $1\le w\le 4$, $0\le
i\le n$.

Now we would like to replace all $L''_i$ by
$L_i+\sum_{1\le w\le \tau}\widetilde{z}_{w,i}X_{j_w}$, $i\in\{0,j+1,\ldots ,
n\}$ for appropriate $1\le\tau\le 4$,
$0\le j_1<\ldots < j_\tau\le n$ and $\widetilde{z}_{w,i}\in
\overline{k(\varepsilon',\varepsilon_3)}$ that are
infinitesimals with respect to the field $k$
(if they are nonzero). Let $\tau\ge 1$ be the
rank of the ma\-t\-rix
$(\widetilde{x}_{w,i})$.
Let us choose a $(\tau\times\tau)$--minor $\Delta_0\ne 0$
of the considered ma\-t\-rix such that for any other
$(\tau\times\tau)$--minor $\Delta_1$ of this ma\-t\-rix $\Delta_1/\Delta_0$
is not infinitely large
with respect to the field $k$. Let $0\le j_1<\ldots < j_\tau\le n$ be the
indices of the
columns of $\Delta_0$.
Then by Cramer's rule one can represent
$$
\widetilde{x}_{w,q}=\sum_{1\le
\nu\le\tau}\mu_{w,q,\nu}\widetilde{x}_{w,j_\nu},
\eqno (20)
$$
$1\le w \le 4$, where every
$\mu_{w,q,\nu}\in\overline{k(\varepsilon',
\varepsilon_3)}$ is not
infinitely large with respect to the field $k$.
By (20) one can find
$\widetilde{z}_{w,i}$ for all $1\le w\le\tau$, $i\in I_j$, such that
$X_{w,i}=\widetilde{x}_{w,i}$, $1\le w\le 4$, $0\le i\le n$,  and
$Z_{w,i}=\widetilde{z}_{w,i}$, $1\le w\le \tau$, $i\in I_j$, is a solution of
system
(18) giving a point from
$V_j^4(\overline{k(\varepsilon',\varepsilon_3)})$.
Thus, we have constructed the required solution for some $1\le\tau\le 4$ and
$0\le j_1<\ldots < j_\tau\le n$ but over the field
$\overline{k(\varepsilon',\varepsilon_3)}$.
Now by transfer principle, see \cite{1}, or directly one sees that there is
also
the required solution over the field $\overline{k_3}$ for the considered
$1\le\tau\le 4$ and
$0\le j_1<\ldots < j_\tau\le n$.

Conversely, suppose that system (18) has a solution for some
 $1\le\tau\le 4$ and
$0\le j_1<\ldots < j_\tau\le n$. Then, as it follows from the description of
the
algorithm below, $a_j(\overline{\lambda})<b_j$. The lemma is proved.

\medskip Let us return to the description of the algorithm for Theorem~3.
Let us enumerate integers $\tau$ and $j_1,\ldots , j_\tau$ such that
$1\le\tau\le 4$ and $0\le
j_1<\ldots < j_\tau\le n$.
Applying Theorem~2 \cite{9} to the algebraic variety $\mbox{\rm
con}(V)\subset{\Bbb A}^{n+1}(\overline{k})$
(which is the set of all zeroes of the homogeneous ideal of $V$ in
${\Bbb A}^{n+1}(\overline{k})$) and using \cite{4} we
decide whether system (18) has a
solution satisfying (19).
If it is true, we construct such a solution. If system (18)
does not have a solution satisfying (19) then we proceed to the next
$\tau$ and $j_1,\ldots , j_\tau$.

Suppose that system (18) does not have a solution satisfying (19)
for all $1\le\tau\le 4$ and
$0\le j_1<\ldots < j_\tau\le n$. Then $b_j=a_j(\overline{\lambda})$
by Lemma~4. If $j+1=n-1$ the element $\overline{\lambda}$ satisfying
assertion of Theorem~3
is constructed. Suppose that $j+1\le n-2$. Then we replace $j$ by $j+1$ and
proceed to
the next step of the described recursion.

Suppose that (18) has a solution satisfying (19)
for some $1\le\tau\le 4$ and $0\le j_1<\ldots < j_\tau\le n$.
Let us compute, see \cite{3},
$z_{w,i}=\mbox{\rm st}_{\varepsilon_3}(z^*_{w,i})$,
$1\le w\le \tau$, $i\in I_j$, where all $z^*_{w,i}$
are from the statement of Lemma~4.
Set $L^*_i=L_i$ for $i\in\{1,\ldots , j\}$, and $L^*_i=L_i+
\sum_{1\le w\le \tau}z_{w,i}X_{j_w}$ for $i\in I_j$. So $L^*_i$ are linear
forms with
coefficients from $\overline{k_3}$. Put $\lambda^*=(L^*_0,\ldots , L^*_n)$.
By Lemma~2 the element $\lambda^*$ satisfies property (B).
From system (18) we get that the polynomial
$R_{\lambda^*,j,*}(1,Z_n)$ has more different roots than
$R_{\overline{\lambda},j,*}(1,Z_n)$.
So $a_j(\overline{\lambda})<a_j(\lambda^*)$ and
$a_w(\overline{\lambda})=a_w(\lambda^*)$
for $1\le w<j$. Applying Lemma~3 and the algorithm of reduction to
integer coefficients, see \cite{9}
Proposition~2, we replace subsequently all the coefficients of linear forms
$L^*_i$ by integers of length
$O(n\,\log d)$ and obtain new linear forms $L^{**}_i$, $0\le i\le n+1$, with
integer coefficients of length $O(n\,\log d)$ satisfying condition (B) and
such that
(denote $\lambda^{**}=(L^{**}_0,\ldots , L^{**}_{n+1})$)
$a_w(\lambda^{**})\ge a_w(\lambda^*)$
for all $1\le w\le j$. In particular
$a_j(\lambda^{**})>a_j(\overline{\lambda})$.
Now we replace $\overline{\lambda}$ by $\lambda^{**}$ and proceed to the
next step of the described
recursion.

Thus, finally we shall construct the element $\overline{\lambda}$ satisfying
the statement of Theorem~3.
Notice that by the B\'ezout theorem the total number of steps of the
described recursion
is bounded from above by a polynomial in $d^n$. The estimation for the
working time of the algorithm for Theorem~3
follows immediately from the ones of the applied algorithms. The theorem is
proved.




\section{Standard parts of algebraic varieties and their
intersections}\label{s2}

Let $K'$ be an arbitrary extension of the field
$\overline{k}$ by infinitesimals, and the algebraic closure
$\overline{K'}$ is supplied with
the real structure, see \cite{3}.
Denote by $\mbox{\rm st}\, :\,{\Bbb P}^n(\overline{K'})
\rightarrow{\Bbb P}^n(\overline{k})$ the mapping of the standard part,
see, e.g.,
\cite{3} (the standard part of an element $x'\in{\Bbb P}^n(\overline{K'})$
is the element $x\in{\Bbb P}^n(\overline{k})$ which is infinitely
close to $x'$).
Let us identify ${\Bbb A}^n(\overline{K'})\subset{\Bbb
P}^n(\overline{K'})\setminus{\cal Z}(X_0)$. Then for every
$x'\in{\Bbb A}^n(\overline{K'})$ the standard part $\mathop{\rm
st}(x')\in
{\Bbb P}^n(\overline{K'})$. Let us define for an arbitrary subset
$S\subset{\Bbb
A}^n(\overline{K'})$ the affine standard part $\mbox{\rm
sta}(E)=\mbox{\rm
st}(E)\cap{\Bbb A}^n(\overline{K'})=\mbox{\rm
st}(E)\setminus{\cal Z}(X_0)$.
In the next three general lemmas
the variety $W$ is not necessarily the one from the
Introduction.

\par\medskip\noindent{\bf LEMMA~5}\hspace{0.1em} {\it  Let
$\widetilde{W}\subset{\Bbb P}^n(\overline{K'})$
(respectively $\widetilde{W}\subset{\Bbb A}^n(\overline{K'})$) be an
arbitrary
irreducible over $\overline{K'}$ projective  algebraic
variety (respectively affine algebraic variety such that $\mbox{\rm
sta}(\widetilde{W})\ne\emptyset$). Let
$W=\mbox{\rm st}(\widetilde{W})$ (respectively $W=\mbox{\rm
sta}(\widetilde{W})$). Then $W$ is a nonempty
projective (respectively affine) algebraic variety in
${\Bbb P}^n(\overline{k})$ (respectively ${\Bbb A}^n(\overline{k})$),
for every irreducible component $W_1$ of $W$ the dimension $\dim W_1=\dim
\widetilde{W}$, and the degree $\deg\widetilde{W}\ge\deg W$.
}\par\medskip

\noindent {\bf PROOF} \quad Let
$\widetilde{W}\subset{\Bbb A}^n(\overline{K'})$ be an affine algebraic
variety.
Denote by $\widetilde{W}_1$ the closure with respect to the Zariski topology
of $\widetilde{W}$ in ${\Bbb P}^n(\overline{K'})$.
Then obviously $\mbox{\rm st}(\widetilde{W})=\mbox{\rm
st}(\widetilde{W}_1)$, and, therefore, the assertions related to
the affine varieties follow from the corresponding ones about projective
varieties. Hence we shall assume in what follows
in the proof that $\widetilde{W}\subset{\Bbb P}^n(\overline{K})$ is a
projective algebraic variety.

For $n=0$ the assertion of the lemma is trivial. Using the induction on
$n$ we shall suppose that this assertion is proved for
projective algebraic varieties in ${\Bbb P}^{n-1}(\overline{K'})$.

We can suppose without loss of generality that
$\widetilde{W}\ne{\Bbb P}^n(\overline{K'})$. Then
there is an irreducible polynomial $\Phi\in \overline{K'}[X_0,\ldots , X_n]$
such that $\widetilde{W}\subset{\cal Z}(\Phi)$.
Multiplying $\Phi$ by a nonzero factor from $K'$
we shall suppose without loss of generality that each coefficient of $\Phi$
is not infinitely large
with respect to $k$ and there is a coefficient of $\Phi$ that is not
infinitesimal with respect to the field $k$.
Let $\mbox{\rm st}(\Phi)\in \overline{k}[X_0,\ldots , X_n]$ be
the polynomial such that all the nonzero coefficients of
$\Phi-\mbox{\rm st}(\Phi)$ are infinitesimals with respect to
$k$. Then $\mbox{\rm st}(\Phi)\ne 0$ and
$W\subset{\cal Z}(\mbox{\rm st}(\Phi))$.

Suppose that $\dim\widetilde{W}=n-1$.
Then $\deg\Phi=\deg\mbox{\rm st}(\Phi)$,
$\widetilde{W}={\cal Z}(\Phi)$ and
$W\subset{\cal Z}(\mbox{\rm st}(\Phi))$.
Let $x\in{\cal Z}(\mbox{\rm st}(\Phi))$.
Let us show that $x\in W$ and hence
$W=\mbox{\rm st}({\cal Z}(\Phi))$. If $n\ge 2$ then,
intersecting $\widetilde{W}$ and $W$ with a hyperplane in general
position trough $x$, and applying the inductive
assumption we get $x\in W$.
If $n=1$ then one can prove that $x\in W$ directly or using the
Veronese embedding, see \cite{10},
${\Bbb P}^1\rightarrow{\Bbb P}^{\deg\Phi}$
of degree $\deg\Phi$
and applying Proposition~6 \cite{3}. Hence
$\deg\widetilde{W}=\deg\Phi=\deg\mbox{\rm st}(\Phi)\ge\deg W$.
The lemma is proved for
$\dim\widetilde{W}=n-1$.

Suppose that $\dim\widetilde{W}\le n-2$ (hence $n\ge 2$).
The set $W$ is nonempty since
for every
point $z\in\widetilde{W}$
the point $\mbox{\rm st}(z)\in W$.
Denote by $\overline{W}$ the closure in the Zariski topology of $W$.

Let $x\in{\Bbb
P}^n(\overline{k})$ be a point. Denote by $\pi\, :\, {\Bbb P}^n
\rightarrow{\Bbb P}^{n-1}$ the linear projection with the center
$x$. Let us choose a point $x$
such that $\mbox{\rm st}(\Phi)(x)\ne 0$ and
$\deg\widetilde{W}=\deg\pi(\widetilde{W})$ (these are conditions of general
position for $x$). Then $x\not\in\overline{W}$ and hence
$\pi(\widetilde{W})$ and $\pi(\overline{W})$
are closed with respect to the Zariski topology
in ${\Bbb P}^{n-1}(\overline{K'})$ and ${\Bbb P}^{n-1}(\overline{k})$
respectively.

Further, for every $z\in\widetilde{W}$ the equality $\mbox{\rm
st}(\pi(z))=\pi(\mbox{\rm st}(z))$ holds
since the morphism $\widetilde{W}\rightarrow\pi(\widetilde{W})$
is finite dominant due to the choice of the point $x$.
Therefore, $\mbox{\rm st}(\pi(\widetilde{W}))=\pi(W)$.
Hence by induction $\pi(W)$ is a projective algebraic variety and
$\dim\pi(W)=\dim\pi(\widetilde{W})$.
Considering the finite morphism $\overline{W}\rightarrow\pi(\overline{W})$
induced by $\pi$
we get $\pi(\overline{W})=\pi(W)$.
But $\dim\pi(\overline{W})=\dim \overline{W}$ and
$\dim\pi(\widetilde{W})=\dim\widetilde{W}$
by the choice of the point $x$. Hence $\dim\overline{W}\le n-2$.

Let $y\in\overline{W}$. We fixe this point for some time.
Since $\dim\overline{W}\le n-2$ one can choose
in advance the point $x$ such that additionally
for every irreducible component
$W_1$ of $\overline{W}$ the morphism $W_1\rightarrow\pi(W_1)$ induced by
$\pi$ is
finite dominant; $\deg\pi(W_1)=\deg W_1$; the irreducible components of
$\overline{W}$ and $\pi(\overline{W})$
are in the one--to--one correspondence by the rule $W_1\mapsto\pi(W_1)$;
the number of elements $\#\pi^{-1}\pi(y)\cap\overline{W}=1$ (these are
conditions of general position for $x$).
As it is proved
there is a point $\widetilde{y}\in\widetilde{W}$ such that $\mbox{\rm
st}(\pi(\widetilde{y}))=\pi(y)$. Now
$\mbox{\rm st}(\widetilde{y})\in W$ and $\pi(\mbox{\rm
st}(\widetilde{y}))=\pi(y)$. Hence $\mbox{\rm
st}(\widetilde{y})=y\in W$.
Since $y\in\overline{W}$ is arbitrary the algebraic variety $\overline{W}=W$.

Finally, by the inductive assumption
$\dim\pi(\widetilde{W})=\dim\pi(W_1)$ and
$\deg\pi(\widetilde{W})$ $\ge \deg\pi(W)$. On the other hand,
$\dim W_1=\dim\pi(W_1)$.
The lemma is proved.

\par\medskip\noindent{\bf LEMMA~6}\hspace{0.1em} {\it  Under conditions of
Lemma~5 let $g\in \overline{k}[X_0,\ldots ,
X_n]$ be a polynomial and
$E$ be an irreducible over $\overline{k}$ component of
$W$ such that $E\cap{\cal Z}(g)$ is a union of some irreducible
components of $W\cap{\cal Z}(g)$ and
$\dim (E\cap{\cal Z}(g))=\dim (W)-1$. Then $\dim
(\widetilde{W}\cap{\cal Z}(g))
=\dim(\widetilde{W})-1$, and $\mbox{\rm
st}(\widetilde{W}\cap{\cal Z}(g))\supset E$ in
${\Bbb P}^n(\overline{k})$ (respectively $\mbox{\rm
sta}(\widetilde{W}\cap{\cal Z}(g))\supset E$ in
${\Bbb A}^n(\overline{k})$). In particular, if
$\dim (W\cap{\cal Z}(g))=\dim (W)-1$ then $\mbox{\rm
st}(\widetilde{W}\cap{\cal Z}(g))=W\cap{\cal Z}(g)$ in
${\Bbb P}^n(\overline{k})$ (respectively $\mbox{\rm
sta}(\widetilde{W}\cap{\cal Z}(g))=W\cap{\cal Z}(g)$ in
${\Bbb A}^n(\overline{k})$).
}\par\medskip

\noindent {\bf PROOF} \quad Similarly to the proof of Lemma~5 the case of
affine
algebraic varieties is reduced to the projective one. So we shall assume in
what follows in the proof that
$\widetilde{W}$ and $W$ are projective varieties.

The polynomial $g$ does not vanish on $\widetilde{W}$ since, otherwise, by
Lemma~5 it vanishes on every irreducible component of $W$. Therefore, $\dim
(\widetilde{W}\cap{\cal Z}(g))
=\dim(\widetilde{W})-1$. We shall assume without loss of
generality that $\widetilde{W}\ne{\Bbb P}^n(\overline{K'})$.
Using the Veronese isomorphism, see \cite{10}, we can suppose without loss of
generality that $g$ is a linear form.

Suppose at first that $\dim W\cap{\cal Z}(g)=\dim W-1$. In this case
we shall prove that $\mbox{\rm
st}(\widetilde{W}\cap{\cal Z}(g))=W\cap{\cal Z}(g)$ in
${\Bbb P}^n(\overline{k})$ using the induction on
$n-\dim\widetilde{W}$.
The base is $\dim\widetilde{W}=n-1$. For the base $\widetilde{W}={\cal
Z}(\Phi)$
and $W={\cal Z}(\mbox{\rm st}(\Phi))$, see the proof of
Lemma~5. Further, $\dim{\cal Z}(\mbox{\rm st}(\Phi),g)=n-2$,
since ${\cal Z}(g)$ is not an irreducible component of $\mbox{\rm
st}(\Phi)$.
One sees that
$\mbox{\rm st}({\cal Z}(\Phi,g))={\cal Z}(\mbox{\rm
st}(\Phi),g)$, and the base of the required assertion is proved.

Let $\dim W\cap{\cal Z}(g)=\dim W-1$ and
$\dim W<n-1$. Let $x\in{\Bbb P}^n(\overline{k})$ be an arbitrary
point, and $\pi\, :\,{\Bbb P}^n(\overline{k})\rightarrow{\Bbb
P}^{n-1}(\overline{k})$ be the linear projection with the center $x$.
Let $y\in W\cap{\cal Z}(g)$.
By Lemma~5 it is sufficient to prove that $y\in\mbox{\rm
st}(\widetilde{W}\cap{\cal Z}(g))$.
Let ${\mathfrak l}\subset{\Bbb P}^n(\overline{k})$ be a line such that
$y\in{\mathfrak l}$.
If we choose ${\mathfrak l}$ in general position then ${\mathfrak l}\cap
W=\{y\}$. Hence
one can find $x$ such that $x\in{\cal
Z}(g)\setminus(W\cup\widetilde{W})$ and the number of elements
$\#\pi^{-1}\pi(y)\cap W=1$. In what follows we assume that $x$ satisfies
these properties.
 Then the morphisms $W\rightarrow\pi(W)$ and
$\widetilde{W}\rightarrow\pi(\widetilde{W})$ induced by $\pi$ are finite
dominant. The standard part $\mbox{\rm
st}(\pi(\widetilde{y}))=\pi(\mbox{\rm
st}(\widetilde{y}))$ for arbitrary $\widetilde{y}\in\widetilde{W}$
since the morphism $\widetilde{W}\rightarrow\pi(\widetilde{W})$ induced by
$\pi$ is finite.
By the inductive assumption
applied to $\pi(\widetilde{W})\subset{\Bbb P}^{n-1}(\overline{K'})$
we have
$\mbox{\rm st}(\pi(\widetilde{W})\cap{\cal Z}(g))=
\pi(W)\cap{\cal Z}(g)$.
But $\pi(W)\cap{\cal Z}(g)=\pi(W\cap{\cal Z}(g))$,
$\pi(\widetilde{W})\cap{\cal Z}(g)=
\pi(\widetilde{W}\cap{\cal Z}(g))$ since $x\in{\cal Z}(g)$.
Hence there is $\widetilde{y}\in\widetilde{W}\cap{\cal Z}(g)$ such that
$\mbox{\rm st}(\pi(\widetilde{y}))=\pi(y)$.
Therefore, $\pi(\mbox{\rm st}(\widetilde{y}))=\pi(y)$.
Obviously $\mbox{\rm st}(\widetilde{y})\in W$. Finally,
$\widetilde{y}=y$ since $\#(\pi^{-1}\pi(y)\cap W)=1$.
The required assertions are proved for the case $\dim W\cap{\cal
Z}(g)=\dim W-1$.

Consider the general case. We need to prove that $\mbox{\rm
st}(\widetilde{W}\cap{\cal Z}(g))\supset E$ in
${\Bbb P}^n(\overline{k})$. Let $y\in W\cap{\cal
Z}(g)\setminus\overline{W\setminus E}$.
By Lemma~5 it is sufficient to prove that $y\in\mbox{\rm
st}(\widetilde{W}\cap{\cal Z}(g))$.
There is a linear subspace ${\cal L}\subset{\Bbb P}^n(\overline{k})$
such that $y\in{\cal L}$, $\dim{\cal L}+\dim W-1=n$,
$y$ is an isolated point of ${\cal W}\cap{\cal L}\cap{\cal
Z}(g)$, $\dim\widetilde{W}\cap{\cal L}=1$,
$\dim W\cap{\cal L}=1$. Denote $\widetilde{C}=\widetilde{W}\cap{\cal
L}$, $C=W\cap{\cal L}$.
By the proved $\mbox{\rm st}(\widetilde{C})=C$. So it is
sufficient to show that
$y\in\mbox{\rm st}(\widetilde{C}\cap{\cal Z}(g))$.

Let us represent $C=C_1\cup C_2$ where $C_1$ (respectively $C_2$) is a union
of all the irreducible components
$C'$ of $C$ such that $C'\not\subset{\cal Z}(g)$ (respectively
$C'\subset{\cal Z}(g)$).
Performing a linear transformation of the coordinates we shall assume
without loss of generality that $(\widetilde{C}\cup
C_1)\cap{\cal Z}(X_0,g)=\emptyset$ in ${\Bbb P}^n(\overline{K'})$.
Consider the finite dominant morphisms $\widetilde{\tau}\,
:\,\widetilde{C}\rightarrow{\Bbb P}^1(\overline{K'})$,
and $\tau_1\, :\, C_1(\overline{K'})\rightarrow{\Bbb P}^1(\overline{K'})$
given by the same formula $(X_0:\ldots : X_n)\mapsto(X_0:g)$. Hence for all
$a\in\overline{K'}$, except at most a finite number,
$\#\widetilde{\tau}^{-1}((1:a))=\deg\widetilde{\tau}$ and
$\#\tau_1^{-1}((1:a))=\deg\tau_1$.
Let $\overline{K'}=R+R\sqrt{-1}$,
where $R$ is the real closed and real ordered field.
For arbitrary two points $z',z''\in{\Bbb
P}^n(\overline{K'})\setminus{\cal Z}(X_0)$ put the distance
$0\le d(z',z'')=
(\sum_{1\le i\le n}|(X_i/X_0)(z')-(X_i/X_0)(z'')|^2)^{1/2}\in R$.
Hence if $z',z''\in{\Bbb P}^n(\overline{k})\setminus{\cal Z}(X_0)$
then $d(z',z'')\in R\cap\overline{k}$.

Since $y$ is an isolated point of $C\cap{\cal Z}(g)$ there is $0<m_2\in
R\cap\overline{k}$
such that for every $y\ne z'\in C\cap{\cal Z}(g)$ the distance
$d(y,z')\ge
m_2$.
There is $0<m_1\in R\cap\overline{k}$ so small that
for all $a\in R$ satisfying the inequalities $-m_1\le a\le m_1$, $a\ne 0$,
\begin{itemize}
\item $\#\widetilde{\tau}^{-1}((1:a))=\deg\widetilde{\tau}$ and
$\#\tau_1^{-1}((1:a))=\deg\tau_1$,
\item for every $z'\in \tau_1^{-1}((1:a))$ there is
$z''\in\tau_1^{-1}((1:0))$
such that the distance $d(z',z'')\le m_2/8$, and, conversely,
for every $z''\in \tau_1^{-1}((1:0))$ there is
$z'\in\tau_1^{-1}((1:a))$
such that the distance $d(z',z'')\le m_2/8$
\end{itemize}
Denote by $c_1$ the semialgebraically connected component, see \cite{1},
of $C\cap\{x\in{\Bbb
P}^n(\overline{K'})\setminus{\cal Z}(X_0)\, :\, |(g/X_0)(x)|^2\le m_1\}=
S_{m_1}$ such that $y\in c_1$. Denote by $c_2$ the union of all the
semialgebraically connected components $c'$ of $S_{m_1}$ such that $c'\ne
c_1$. Then obviously
$c_1\subset C_1$, for all $z'\in c_1$, $z''\in c_2$ the distance
$d(z',z'')\ge (3/4)m_2$, and $c_1\cap{\cal Z}(g)=\{y\}$.

For every semialgebraic set $A\subset{\Bbb P}^n(\overline{K'})\setminus
{\cal Z}(X_0)$ and $z'\in{\Bbb
P}^n(\overline{K'})\setminus{\cal Z}(X_0)$
the distance $d(z',A)=\inf\{d(z',z'')\, :\, z''\in A\}$ is a continuous
function of $z'$
with respect to the topology of the real closed field $R$, see \cite{1}.
Set for every $0\ne a\in R$, $|a|\le m_1$,
$$
\psi(a)=\sum_{z\in\widetilde{\tau}^{-1}(a)}d(z,c_2)\in R.
\eqno (21)
$$
Then by the implicit function theorem $\psi$ is a continuous function of $a$
with respect to
the topology of the real closed field $R$ (we leave the details to the
reader here).

Let $\varepsilon>0$ be an infinitesimal with respect to the field
$\overline{K'}$, and $\widetilde{R(\varepsilon)}$ be
the real closure of $R(\varepsilon)$. Hence $\widetilde{R(\varepsilon)}$ is
a real ordered field, and
the mapping of the standard part $\mbox{\rm st}_\varepsilon\,
:\,{\Bbb P}^n(\overline{K'(\varepsilon)})\rightarrow{\Bbb
P}^n(\overline{K'})$ is defined. Obviously $\mbox{\rm
st}_\varepsilon(\widetilde{C}(\overline{K'(\varepsilon)})=
\widetilde{C}$. Hence $\mbox{\rm st}(\mbox{\rm
st}_\varepsilon(\widetilde{C}(\overline{K'(\varepsilon)}))=C$.
We extend the ground field $K'$ till $K'(\varepsilon)$. Hence now
$\psi\, :\,\{a\in\widetilde{R(\varepsilon)}\, :\, |a|\le m_1\;\&\; a\ne
0\}\rightarrow \widetilde{R(\varepsilon)}$
is a continuous function.


Let us show that there is a point
$\widetilde{y}_\varepsilon\in\widetilde{C}(\overline{K'(\varepsilon)})
\cap{\cal Z}(g-\varepsilon X_0)$
such that $\widetilde{y}_\varepsilon$ is infinitely close to $y$ with
respect to the field $k$, i.e., such that
$\mbox{\rm st}(\mbox{\rm
st}_\varepsilon(\widetilde{y}_\varepsilon))=y$. Indeed, suppose
contrary.
Then $\mbox{\rm st}(\mbox{\rm st}_\varepsilon(z))\in
c_2$ for every
$z\in\widetilde{\tau}^{-1}(\varepsilon)$, and, hence $|\psi(\varepsilon)|\le
m_2/3$.
On the other hand, $C\cap{\cal Z}(g-m_1 X_0)$ is a finite set. Hence as
we have proved
$\mbox{\rm st}(\widetilde{C}\cap{\cal Z}(g-m_1
X_0))=C\cap{\cal Z}(g-m_1 X_0)$.
We have $c_1\cap{\cal Z}(g-m_1 X_0)\setminus{\cal
Z}(X_0)\ne\emptyset$ by the choice of $m_1$ and $m_2$.
Therefore, $\psi(m_1)\ge 2m_2/3$. Now the image of the interval
$\psi([\varepsilon,m_1])$ is a semialgebraically connected set in
$\widetilde{R(\varepsilon)}$, see \cite{1}.
Hence $\psi([\varepsilon,m_1])$ contains the interval
$[\psi(\varepsilon),\psi(m_1)]\supset[m_2/3,2m_2/3]$,
see \cite{1}.
Therefore, there is $a\in[\varepsilon,m_1]$ such that
$m_2/3<\psi(a)<2m_2/3$.
But $\psi(a)\ge 2m_2/3$ if $\mbox{\rm st}(\mbox{\rm
st}_\varepsilon(z))\in c_1$ for at least
one $z\in\widetilde{\tau}^{-1}(a)$ and, otherwise, $\psi(a)\le m_2/3$.
This contradiction proves the existence of $\widetilde{y_\varepsilon}$.
Now put $\widetilde{y}=\mbox{\rm
st}_\varepsilon(\widetilde{y}_\varepsilon)$. Then obviously
$\widetilde{y}\in\widetilde{C}\cap{\cal Z}(g)$
and $\mbox{\rm st}(\widetilde{y})=y$. The lemma is proved.


\par\medskip\noindent{\bf LEMMA~7}\hspace{0.1em} {\it
Let the field $K'$ be as above.
Let $\widetilde{W}_i\subset{\Bbb P}^n(\overline{K'})$
(respectively $\widetilde{W}_i\subset{\Bbb A}^n(\overline{K'})$),
$i=1,2$,  be
projective (respectively affine) algebraic varieties. Suppose that all the
irreducible component of $\widetilde{W}_i$
have the same dimension $n-s_i$, $i=1,2$. Put $W_i=\mbox{\rm
st}(\widetilde{W}_i)\subset{\Bbb P}^n(\overline{k})$
(respectively $W_i=\mbox{\rm
sta}(\widetilde{W}_i)\subset{\Bbb A}^n(\overline{k})$), $i=1,2$
(hence the dimension of every irreducible component of $W_i$ is $n-s_i$ by
Lemma~5). Then the following assertions hold.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item Let $E$ be an irreducible over $\overline{k}$ component of
$W_1\cap W_2$.
Assume that $\dim E=n-s_1-s_2\ge 0$. Then there is an irreducible
over $\overline{K'}$ component
$\widetilde{E}$ of
$\widetilde{W}_1\cap\widetilde{W}_2$ such that $\dim\widetilde{E}=n-s_1-s_2$
and
$\mbox{\rm st}(\widetilde{E})\supset E$ (respectively
$\mbox{\rm sta}(\widetilde{E})\supset E$).
\item Suppose that $\dim(W_1\cap W_2)=n-s_1-s_2$ (if $n-s_1-s_2<0$
then the intersection is empty). Then $\mbox{\rm
st}(\widetilde{W}_1\cap\widetilde{W}_2)=W_1\cap W_2$
(respectively $\mbox{\rm
sta}(\widetilde{W}_1\cap\widetilde{W}_2)=W_1\cap W_2$) and
therefore, $\dim(\widetilde{W}_1\cap\widetilde{W}_2)=n-s_1-s_2$
(respectively the dimension of every irreducible component $\widetilde{E}$
of $\widetilde{W}_1\cap\widetilde{W}_2$
such that $\mbox{\rm sta}(\widetilde{E})\ne\emptyset$ is
$n-s_1-s_2$) by Lemma~5.
\end{enumerate}
}\par\medskip

\noindent{\bf PROOF}\quad Let us prove (a).
By Lemma~5 it is sufficient to consider the case of affine algebraic
varieties.
Set $\Delta=\{(x,x)\in{\Bbb A}^{2n}(\overline{k})\, :\, x\in{\Bbb
A}^n(\overline{k})\}$. Let $X_1,\ldots ,X_n$, $Y_1,\ldots , Y_n$ be the
coordinate functions on
${\Bbb A}^{2n}(\overline{k})$ such that $\Delta={\cal
Z}(X_1-Y_1,\ldots , X_n-Y_n)$.
Put $\delta_i=X_i-Y_i$, $1\le i\le n$.
One identifies $W_1\cap W_2=(W_1\times W_2)\cap{\cal Z}(\delta_1,\ldots
,\delta_n)$
in ${\Bbb A}^{2n}(\overline{k})$ and
$\widetilde{W}_1\cap\widetilde{W}_2=(\widetilde{W}_1\times\widetilde{W}_2)
\cap{\cal Z}(\delta_1,\ldots ,\delta_n)$ in ${\Bbb
A}^{2n}(\overline{K'})$.
There are irreducible over $\overline{k}$ algebraic varieties $E_0\supset
E_1\supset\ldots\supset
E_n=E$ such that
$E_i$ is an irreducible component of $(W_1\times W_2)\cap{\cal
Z}(\delta_1,\ldots ,\delta_i)$ and
$\dim E_i=(n-s_1)+(n-s_2)-i$ for every $0\le i\le n$.



Let us show that that there are irreducible
over $\overline{K'}$ algebraic varieties $\widetilde{E}_0\supset
\widetilde{E}_1\supset\ldots\supset\widetilde{E}_n=\widetilde{E}$ such that
$\widetilde{E}_i$ is an irreducible component of $(\widetilde{W}_1\times
\widetilde{W}_2)\cap{\cal
Z}(\delta_1,\ldots ,\delta_i)$,
$\dim\widetilde{E}_i=(n-s_1)+(n-s_2)-i$ and $\mbox{\rm
sta}(\widetilde{E}_i)\supset E_i$
for every $1\le i\le n$. We shall use the induction on $i$.
The base of the induction $i=0$ follows from the equality $\mbox{\rm
sta}(\widetilde{W}_1\times\widetilde{W}_2)=
W_1\times W_2$ and Lemma~5. Suppose that we have proved that there are
$\widetilde{E}_0,\ldots ,
\widetilde{E}_i$ for some $0\le i\le n-1$.
Then obviously
$$
E_{i+1}\subset\mbox{\rm sta}(\widetilde{E}_i)\cap{\cal
Z}(\delta_{i+1})\subset
(W_1\times W_2)\cap{\cal Z}(\delta_1,\ldots ,\delta_{i+1}).
$$
Therefore, $E_{i+1}$ is an irreducible component of $\mbox{\rm
sta}(\widetilde{E}_i)\cap{\cal Z}(\delta_{i+1})$.
Hence $E_{i+1}$ is an irreducible component of $\mbox{\rm
sta}(\widetilde{E}_i\cap{\cal Z}(\delta_{i+1}))$
by Lemma~6.  By Lemma~5 there is an irreducible over $\overline{K'}$
component $E'$ of
$\widetilde{E}_i\cap{\cal Z}(\delta_{i+1})$
such that $\mbox{\rm sta}(E')\supset E_{i+1}$.
Then $\dim E'=
\dim\widetilde{E}_i\cap{\cal Z}(\delta_{i+1})=(n-s_1)+(n-s_2)-i-1$
by Lemma~5 and
since $\widetilde{E}_i$ is irreducible over $\overline{K'}$.
Now
\begin{eqnarray*}
&&E_{i+1}\subset\mbox{\rm sta}(E')\subset\mbox{\rm
sta}((\widetilde{W}_1\times\widetilde{W}_2)
\cap{\cal Z}(\delta_1,\ldots ,\delta_{i+1}))\subset\\
&&(W_1\times W_2)\cap{\cal Z}(\delta_1,\ldots ,\delta_{i+1}).
\end{eqnarray*}
Therefore, $E_{i+1}$ is an irreducible component of $\mbox{\rm
sta}((\widetilde{W}_1\times\widetilde{W}_2)
\cap{\cal Z}(\delta_1,\ldots ,\delta_{i+1}))$. We have $E'\subset
(\widetilde{W}_1\times\widetilde{W}_2)
\cap{\cal Z}(\delta_1,\ldots ,\delta_{i+1})$ and $\mbox{\rm
sta}(E')\supset E_{i+1}$.
Hence $E'$ is an irreducible component of
$(\widetilde{W}_1\times\widetilde{W}_2)
\cap{\cal Z}(\delta_1,\ldots ,\delta_{i+1})$ by Lemma~5.
Now one can put $\widetilde{E}_{i+1}=E'$. The existence of
$\widetilde{E}_0,\ldots ,\widetilde{E}_n$ is proved by the induction.
Assertion (a) is proved.

Let us prove (b).
Obviously $\mbox{\rm
st}(\widetilde{W}_1\cap\widetilde{W}_2)\subset W_1\cap W_2$
(respectively $\mbox{\rm
sta}(\widetilde{W}_1\cap\widetilde{W}_2)\subset W_1\cap W_2$).
So one can assume without loss of generality that $W_1\cap W_2\ne\emptyset$.
Now (b) follows from (a) immediately. The lemma is proved.







\section{Several lemmas}\label{s3}


In assertion (d) of Theorem~1 \cite{9} we claim that ${\cal
U}'_0$ is a nonempty open in the Zariski topology subset of
${\Bbb A}^{(r+1)(n-s+1)}(\overline{k})$
(recall that there $s=\sigma$).
In the next lemma we shall consider a similar but more general
situation. Below we shall not use Lemma~8 and Lemma~10.
But still we decided to present them here for the completeness.


\par\medskip\noindent{\bf LEMMA~8}\hspace{0.1em} {\it  Let $E\subset W$ be
an arbitrary closed in the Zariski topology
subvariety and
$E'$ be the closure in $W'$ of the image $p(E\setminus{\cal
Z}(e_0,\ldots , e_r))$.
Suppose that $\dim E'\le n-i$, where $\sigma\le i\le n$.
Let $L'_0,L'_{i+1},\ldots , L'_n\in\overline{k}[X_0,\ldots ,X_r]$ be linear
forms and $e'_w=L'_w(e_0,\ldots , e_r)$ for all $w$.
Denote by $p_{E,L'}\, :\, E\setminus{\cal Z}(e'_0,e'_{i+1},\ldots
, e'_n)\rightarrow{\Bbb P}^{n-i}(\overline{k})$, $(X_0:\ldots :
X_n)\mapsto(e'_0:e'_{i+1},\ldots , e'_n)$,
the regular morphism. Let $E_1$ be an arbitrary irreducible over
$\overline{k}$ component of $E$.
Denote by $E'_1$ (respectively $E''_1$)
the closure with respect to Zariski topology of the image
$p(E_1\setminus{\cal Z}(e_0,\ldots , e_r))$
(respectively $p_{E,L'}(E_1\setminus
{\cal Z}(e'_0,e'_{i+1},\ldots , e'_n))$).
Then the following assertions are
equivalent.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item $E\cap{\cal Z}(e'_0,e'_{i+1},\ldots
, e'_n)=E\cap{\cal Z}(e_0,\ldots , e_r)$ in ${\Bbb
P}^n(\overline{k})$, and for every irreducible component $E_1$ of
$E$ the dimension $\dim E''_1=\dim E'_1$.
\item $(E\setminus{\cal Z}(e_0,\ldots , e_r))\cap{\cal
Z}(e'_0,e'_{i+1},\ldots, e'_n)=\emptyset$ in ${\Bbb
P}^n(\overline{k})$, and for every irreducible component $E_1$ of $E$
the dimension $\dim E''_1=\dim E'_1$.
\item $E'\cap
{\cal Z}(L'_0,L'_{i+1},\ldots , L'_n)=\emptyset$
in ${\Bbb P}^r(\overline{k})$.
\end{enumerate}
Further, denote by  ${\cal U}_{E,i}$ the subset of all
$(n-i+1)$--tuples $(L'_0,L'_{i+1},\ldots , L'_n)\in \\
{\Bbb A}^{(r+1)(n-i+1)}(\overline{k})$ satisfying (a), (b) and (c).
Then ${\cal U}_{E,i}$ is a nonempty open in the Zariski topology subset
of ${\Bbb
A}^{(r+1)(n-i+1)}(\overline{k})$.
}\par\medskip

\noindent{\bf PROOF}\quad Obviously (a) and (b) are equivalent, and (c)
implies (a). Let us show that (a) implies (c).
We shall use the induction on $j=n-i$.
If $j=0$ the assertions of the lemma are trivial. The case $j=1$
will be considered
separately.
If $j>1$ then at the inductive step we reduce the case of $j$ to
the one of $j-1$.
For an arbitrary $i$ we shall suppose without loss of generality that $E$ is
irreducible over $\overline{k}$.
Intersecting $E$ with an appropriate linear subspace of codimension $\dim
E-\dim E'$ in ${\Bbb P}^n(\overline{k})$
we shall assume without loss of generality that $\dim E=\dim E'=n-i_1$
(here we leave the details to the reader).
Replacing $L'_0,L'_{i+1},\ldots , L'_n$
by $n-i_1+1$ appropriate linear combinations of these linear forms, we shall
assume without loss of
generality that (a) holds with $i_1$ in place of $i$.
So we shall suppose without loss of generality at the beginning of the
inductive step (and also if $j=1$) that $i_1=i$
and $E$ is irreducible over $\overline{k}$.
Hence the morphism $p_{E,L'}$ is dominant.
At the end of the inductive step
one can obtain again
the general case when $E$ is not necessarily irreducible over $\overline{k}$.

Suppose that $j=1$. Let $\xi'\in
E'\cap{\cal Z}(L'_0,L'_n)$. We shall assume without loss of generality
that
$X_0(\xi)\ne 0$. Let $\widetilde{E}$ be the normalization of $E$. Then, see,
e.g., \cite{10},
we have the dominant regular morphism $\widetilde{p}\,
:\,\widetilde{E}\rightarrow E'$ which is a composite of the
natural projection $\pi\, :\,\widetilde{E}\rightarrow E$ and the morphism
$E\setminus{\cal Z}(e_0,\ldots , e_r)\rightarrow E'$
induced by $p$. Since $\widetilde{E}$ and $E$ are complete varieties
$\widetilde{p}(\widetilde{E})=E'$.
Therefore, there is $\widetilde{\xi}\in\widetilde{E}$ such that
$\widetilde{p}(\widetilde{\xi})=\xi'$. Set
$\xi=\pi(\widetilde{\xi})$. Then obviously $\xi\in E\cap{\cal
Z}(e'_0,e'_n)\setminus{\cal Z}(e_0)$.
This contradiction proves the lemma for $j=n-i=1$.

Suppose that $j>1$,
and $E'\cap{\cal Z}(L'_0,L'_{i+1},\ldots , L'_n)=F\ne\emptyset$.
Suppose that $\dim F<n-i-1$ (respectively $\dim F=n-i-1$).
Let $L''_{i+1}$ be a linear combination of $L'_0,L'_{i+1},\ldots , L'_n$
(respectively $X_0,\ldots ,X_r$) with coefficients from $\overline{k}$.
Denote $e''_{i+1}=L''_{i+1}(e_0,\ldots ,
e_r)$. Let us choose  $L''_{i+1}$
such that $\dim E'\cap{\cal Z}(L''_{i+1})=
\dim E\cap{\cal Z}(e''_{i+1})=
n-i-1$; the morphism $E\cap{\cal Z}(e''_{i+1})\rightarrow
E'\cap{\cal Z}(L''_{i+1})$ induced by $p$ is dominant;
the morphism  $E\cap{\cal Z}(e''_{i+1})\rightarrow{\cal
Z}(L''_{i+1})={\Bbb P}^{n-i-1}(\overline{k})$ induced by $p_{E,L'}$
is dominant;
and if $\dim F=n-i-1$ then $\dim F\cap{\cal Z}(L''_{i+1})=n-i-2$
(these are conditions of general position
for $L''_{s+1}$ in the both considered cases).
Notice that $E\cap{\cal Z}(e''_{i+1})\cap{\cal Z}(e_0,\ldots , e_r)=
E\cap{\cal Z}(e''_{i+1})\cap{\cal Z}(e'_0,e'_{i+1},\ldots
, e'_n)$ and $E'\cap{\cal Z}(L''_{i+1})\cap
{\cal Z}(L'_0,L'_{i+1},\ldots , L'_n)\ne\emptyset$.
There are appropriate linear combinations $L''_0,L''_{i+2},\ldots , L''_n$
of $L'_0,L'_{i+1},\ldots , L'_n$
such that (a) holds with $i+1$, $E\cap{\cal Z}(e''_{i+1})$ and
 $L''_0,L''_{i+2},\ldots , L''_n$ in place of $i$, $E$
and $L'_0,L'_{i+1},\ldots , L'_n$. Further, $E'\cap{\cal
Z}(L''_{i+1})\cap
{\cal Z}(L''_0,L''_{i+2},\ldots , L''_n)\ne\emptyset$.
Thus, by induction we get a contradiction.
Therefore, (a) implies (c). The equivalence of (a), (b) and (c) is proved.

Now the last assertion of the lemma follows from \cite{11}
(and also \cite{6} Section~7).
The lemma is proved.


\par\medskip\noindent{\bf LEMMA~9}\hspace{0.1em} {\it  {\bf (A CRITERION FOR
DOMINANTNESS OF A MORPHISM)}
Let $\pi\, :\, V_1\rightarrow V_2$ be a defined over $k$ morphism of defined
over $k$ algebraic varieties.
Suppose that all the irreducible components of $V_1$
have the same dimension $n_1$, and $V_2$ is an irreducible over $k$
algebraic variety of dimension $n_2$.
Let $n_1\ge n_2$. Let $z\in V_2$. Suppose that for every irreducible over
$k$ component $E$ of $V_1$
there is an irreducible over $k$ (or which is the equivalent irreducible
over $\overline{k}$) component $E_1$ of the fiber
$\pi^{-1}(z)$ such that
$E_1\subset E$ and $\dim E_1=n_1-n_2$. Then the morphism $\pi$ is dominant.
In particular, if $V_1$ is irreducible over $k$ and
$\dim\pi^{-1}(z)=n_1-n_2$
then the morphism $\pi$ is dominant.
}\par\medskip

\noindent {\bf PROOF} \quad This follows from the theorem about dimensions
of fibers, see \cite{12} and also the Introduction.


\par\medskip\noindent{\bf LEMMA~10}\hspace{0.1em} {\it  Let
$\sigma+1\le\alpha\le n+1$. Then the set of all
$(L'_{\sigma+1},\ldots , L'_\alpha)\in{\cal U}'_\alpha$
such that
$J^{(1)}(L'_{\sigma+1},\ldots , L'_i)=J^{(1)}_i$ for all $\sigma+1\le
i\le\alpha$ is a nonempty open in the
Zariski topology subset of
${\cal U}'_\alpha$.
}\par\medskip

\noindent{\bf PROOF}\quad Let $(L'_{\sigma+1},\ldots , L'_\alpha)
\in{\cal U}'_\alpha$.
Let us extend the ground field $k$ till $k_{\overline{v}}$, see the
Introduction.
Let ${\cal L}={\cal Z}(v_{s+1},\ldots ,$
$v_{n-\alpha+\sigma+1})\subset{\Bbb P}^n(\overline{k_{\overline{v}}})$
be a generic  linear subspace.
Hence the intersection of ${\cal L}$ and
$$
W(L'_{\sigma+1},\ldots , L'_{\alpha-1})\cup
(W\cap{\cal Z}(e_0,\ldots , e_r))
$$
is transversal. In particular, by Theorem~1 \cite{9} the number of elements
$\#(W(L'_{\sigma+1},$ $\ldots ,
L'_{\alpha-1})\cap{\cal L})<+\infty$.
Further, $e'_\alpha$ does not vanish at any point from
$W(L'_{\sigma+1},$ $\ldots , L'_{\alpha-1})\cap{\cal L}$.

Set $E=W\cap{\cal L}$.  The dimension of every
irreducible component of
$E$ is $\alpha-\sigma-1$. As an easy consequence of Theorem~1
\cite{9} the
algebraic variety $W(L'_{\sigma+1},\ldots , L'_{\alpha-1})$ has a nonempty
intersection with every
irreducible over $\overline{k}$ component of $W$. Hence the same is true for
the algebraic variety
$W(L'_{\sigma+1},\ldots , L'_{\alpha-1})\cap{\cal L}$.
Therefore,
by Lemma~9 the morphism $E\setminus{\cal
Z}(e'_{\sigma+1},\ldots ,$ $e'_\alpha)\rightarrow
{\Bbb P}^{\alpha-\sigma-1}(\overline{k_{\overline{v}}})$,
$(X_0:\ldots : X_n)\mapsto
(e'_{\sigma+1}:\ldots : e'_\alpha)$, is dominant.
Let us replace in the statement of Lemma~8 element
$(L'_0,L'_{i+1},\ldots ,L'_n)$
by $(L'_{\sigma+1},\ldots , L'_\alpha)$ (hence $n-i+1$ is replaced by
$\alpha-\sigma-1$)
Then condition (a)
of Lemma~8 holds with such a replacement
if and only if for every $\sigma+1\le i\le\alpha$ for every $j\in
J^{(1)}_i$ the algebraic variety $W_j$ is an
irreducible component of $W\cap{\cal Z}(e'_{\sigma+1},\ldots ,
e'_\alpha)$. Now the required assertion
follows from Lemma~8. The lemma is proved.









\par\medskip\noindent{\bf LEMMA~11}\hspace{0.1em} {\it    In the notation
from the Introduction let $L'=(L'_0,L'_{\sigma+1}\ldots , L'_n)$,
where all $L'_j\in\overline{k}[X_0,\ldots , X_n]$ are linear forms.
Let $j\in J^{(1)}_i(L')\cap J^{(1)}_i$ (i.e., $W_j$ is an irreducible
component of
$W^{(1)}_i(L')$ and $W\cap{\cal Z}(e_0,\ldots , e_r)$ simultaneously)
for some  $\sigma+1\le i\le n+1$.
Then the index of intersection $\iota_j(L')\ge
\iota_j$, see (7). Further,
let $\sigma+1\le\alpha\le n+1$ be an integer.
Denote by ${\cal U}^{({\scriptscriptstyle{\rm I}}{\rm v})}_\alpha
\subset{\cal U}'_\alpha$ the subset of all
$(L'_{\sigma+1},\ldots , L'_\alpha)$
such that $J^{(1)}_i(L'_{\sigma+1},\ldots , L'_\alpha)=J^{(1)}_i$
for all  $\sigma+1\le i\le \alpha$ and
$\iota_j(L'_{\sigma+1},\ldots , L'_i)=\iota_j$ for all $j\in J^{(1)}_i$,
$\sigma+1\le i\le\alpha$.
Then ${\cal U}^{({\scriptscriptstyle{\rm I}}{\rm v})}_\alpha$
is a defined over $k$ nonempty open in the Zariski topology subset of
${\Bbb A}^{(r+1)(\alpha-\sigma)}(\overline{k})$.
Finally, for any line ${\mathfrak l}\subset{\Bbb
A}^{(r+1)(\alpha-\sigma)}(\overline{k})$ such that
${\mathfrak l}\cap{\cal U}^{({\scriptscriptstyle{\rm
I}}{\rm v})}_\alpha\ne\emptyset$ the number of elements
$\#(l\setminus{\cal U}^{({\scriptscriptstyle{\rm I}}{\rm v})}_\alpha)$
is bounded from above by a polynomial in
$d^n(d')^{n-\sigma}$.
}\par\medskip

\noindent {\bf PROOF} \quad Set $L'_{n+1}=L'_0$ and the coefficients
$l'_{n+1,\nu}=l'_{0,\nu}$ for all $0\le\nu\le r$, see
the Introduction.

Let ${\Bbb A}^{(r+1)(i-\sigma)}(\overline{k})$ has coordinate functions
$Z_{\beta,\nu}$,
$\sigma+1\le\beta\le i$, $0\le \nu\le r$. Let $z'_i\in{\Bbb
A}^{(r+1)(i-\sigma)}(\overline{k})$
(respectively $z_i\in{\Bbb A}^{(r+1)(i-\sigma)}(\overline{K})$)
be the point with the coordinates $Z_{\beta,\nu}=l'_{\beta,\nu}$
(respectively $Z_{\beta,\nu}=u_{\beta,\nu}$) for all $\beta$, $\nu$, where
$l'_{\beta,\nu}$ (respectively $u_{\beta,\nu}$) are
the coefficients of the linear forms $L'_\nu$ (respectively $U_\nu$), see the
Introduction.
Consider the following closed algebraic subvarieties of ${\Bbb
P}^n(\overline{K})\times {\Bbb A}^{(r+1)(i-\sigma)}(\overline{K})$:
\begin{eqnarray*}
&&E^{(1)}=W\times{\Bbb A}^{(r+1)(i-\sigma)}(\overline{k}),\\
&&E^{(2)}={\cal Z}\Bigl(\sum_{0\le\nu\le r}Z_{\beta,\nu}e_\nu,\quad
\sigma+1\le\beta\le i\,\Bigr),\\
&&E^{(3)}={\cal Z}(Z_{\beta,\nu}-l'_{\beta,\nu},\quad
\sigma+1\le\beta\le i,\; 0\le \nu\le r\,),\\
&&E^{(4)}={\cal Z}(Z_{\beta,\nu}-u_{\beta,\nu},\quad
\sigma+1\le\beta\le i,\; 0\le \nu\le r\,).
\end{eqnarray*}
Notice that $E^{(\gamma)}$ for
$\gamma=1,2,3$ are defined over $\overline{k}$.
The variety
$W_j\times\{z'_i\}$, $j\in J^{(1)}_i(L')$, is an irreducible component
of the intersection
$E^{(1)}\cap E^{(2)}\cap E^{(3)}$ and
the codimension of $W_j\times\{z'_i\}$ is the sum of
the codimensions of
$E^{(1)}$, $E^{(2)}$, $E^{(3)}$ (all the codimensions are considered in
${\Bbb P}^n(\overline{K})\times {\Bbb
A}^{(r+1)(i-\sigma)}(\overline{K})$), i.e.,
the last intersection is proper at $W_j\times\{z'_i\}$.


For arbitrary algebraic varieties $V',V''$
let $i(V',V'';V''')$ be the index of intersection of
algebraic varieties $V'$ and $V''$ at
an irreducible component
$V'''$ of $V'\cap V''$ when this index is defined, see \cite{10}.
Denote by $W_\gamma$, $\gamma\in \Gamma_i$, the
 family of all irreducible over $\overline{k}$ components of the
intersection $E^{(1)}\cap E^{(2)}$. Let $\Gamma_j\subset\Gamma_i$ be the
subset of all
$\gamma$ such that
$W_\gamma\supset W_j\times\{z'_i\}$.
There is $\gamma_j\in\Gamma_j$
such that $W_{\gamma_j}=W_j\times
{\Bbb A}^{(r+1)(i-\sigma)}(\overline{k})$.
Denote by $W_\gamma$, $\gamma\in\Gamma'_j$ the family of all
irreducible component of ${\cal Z}(e'_{\sigma+1},\ldots ,
e'_i)$ such that $W_\gamma\supset W_j$.
Further, since $z_i$ is a
generic point of ${\Bbb A}^{(r+1)(i-\sigma)}(\overline{k})$, the algebraic
variety $W_j\times{\Bbb A}^{(r+1)(i-\sigma)}(\overline{K})$ is
the unique irreducible component $E$ of $E^{(1)}\cap E^{(2)}$ such that $E$
contains $W_j\times\{z_i\}$.

Now, see \cite{10}, by the properties of indices of
intersection (mainly by the one corresponding to the
associativity of intersection) we have
\begin{eqnarray*}
&&i(W,{\cal Z}(e'_{\sigma+1},\ldots , e'_i);W_j)=
i(E^{(1)},{\cal Z}(e'_{\sigma+1},\ldots ,
e'_i)\times\{z'_i\};W_j\times\{z'_i\})= \\
&&\sum_{\gamma\in\Gamma'_j}i(E^{(1)},W_\gamma\times\{z'_i\};
W_j\times\{z'_i\})
i(E^{(2)},E^{(3)};W_\gamma\times\{z'_i\})=\\
&&\sum_{\gamma\in \Gamma_j}i(E^{(1)},E^{(2)};W_\gamma)
i(W_\gamma,E^{(3)};W_j\times\{z'_i\})\ge \\
&&i(E^{(1)},E^{(2)};W_j\times{\Bbb
A}^{(r+1)(i-\sigma)}(\overline{k}))\times\\
&&i(W_j\times{\Bbb
A}^{(r+1)(i-\sigma)}(\overline{k}),E^{(3)};W_j\times\{z'_i\})= \\
&&i(E^{(1)},E^{(2)};W_j\times{\Bbb A}^{(r+1)(i-\sigma)}(\overline{k}))=
i(E^{(1)},E^{(2)};
W_j\times{\Bbb A}^{(r+1)(i-\sigma)}(\overline{K}))\times\\
&&i(W_j\times{\Bbb
A}^{(r+1)(i-\sigma)}(\overline{K}),E^{(4)};W_j\times\{z_i\})=\\
&&i(E^{(1)},{\cal Z}(u_{\sigma+1},\ldots ,
u_i)\times\{z_i\};W_j\times\{z_i\})=
i(W,{\cal Z}(u_{\sigma+1},\ldots , u_i);W_j).
\end{eqnarray*}


Let $(L'_{\sigma+1},\ldots , L'_\alpha)\in
{\cal U}^{({\scriptscriptstyle{\rm I}}{\rm v})}_\alpha$ (respectively
$L'\in{\cal U}^{({\scriptscriptstyle{\rm I}}{\rm v})}_{n+1}$) and
for $\sigma+1\le i\le\alpha$. Then $\Gamma_j=\{\gamma_j\}$ for all
$j\in J^{(1)}_i$ and $\sigma+1\le i\le\alpha$.
Therefore, for every $j\in J^{(1)}_i$
for every $\sigma+1\le i\le\alpha$ there is $w_j\in W_j$ such
that $(w_j,z'_i)\not\in W_\gamma$
for every $\gamma\in\Gamma_i\setminus\{\gamma_j\}$.
Denote by $p_{\alpha,i}\, :\,{\cal U}'_\alpha
\rightarrow{\cal U}'_i$ the natural projection.
The set of $z\in{\cal
U}'_\alpha$ such that
$(w_j,p_{\alpha,i}(z))\not\in W_\gamma$ for all
$\gamma\in\Gamma_i\setminus\{\gamma_j\}$, $j\in J^{(1)}_i$,
$\sigma+1\le i\le\alpha$
is a neighborhood with respect to the Zariski topology of $z'_\alpha$.
This neighborhood is contained in
${\cal U}^{({\scriptscriptstyle{\rm I}}{\rm v})}_\alpha$. If one extends
the ground field $k$ till $K$ then
$(U_{\sigma+1},\ldots , U_\alpha)\in{\cal U}^{({\scriptscriptstyle{\rm
I}}{\rm v})}_\alpha$. Hence ${\cal U}^{({\scriptscriptstyle{\rm
I}}{\rm v})}_\alpha\ne\emptyset$.
Obviously ${\cal U}^{({\scriptscriptstyle{\rm I}}{\rm v})}_\alpha$ is
invariant
with respect to the action of the Galois group $\mbox{\rm
Gal}(\overline{k}/k)$ and, therefore, is
defined over $k$.

To prove the last assertion we shall suppose that $(L'_{\sigma+1},\ldots ,
L'_\alpha)\in{\mathfrak l}\cap
{\cal U}^{({\scriptscriptstyle{\rm I}}{\rm v})}_\alpha$. By the B\'ezout
theorem $\sum_{\gamma\in\Gamma_i}\deg
W_\gamma\le d^s(d')^{i-\sigma}$ for $\sigma+1\le i\le \alpha$.
Therefore, by Theorem~1 (d) \cite{9} there is at most a
polynomial in $d^n(d')^{n-\sigma}$ number of
points $z\in{\mathfrak l}$ such that  $z\not\in{\cal U}'_\alpha$ or
$(w_j,p_{\alpha,i}(z))\in
\bigcup_{\gamma\in\Gamma_i\setminus\{\gamma_j\}}W_\gamma$
for some $j\in J^{(1)}_i$, $\sigma+1\le i\le\alpha$.
The lemma is proved.




\par\medskip\noindent{\bf LEMMA~12}\hspace{0.1em} {\it  In the notation
from the Introduction let $L'=(L'_0,L'_{\sigma+1}\ldots , L'_n)$,
where all $L'_j\in\overline{k}[X_0,\ldots , X_n]$ are linear forms.
Then $D_i(L'_{\sigma+1},\ldots , L'_i)\le D_i$ for $i\le n$, and
$\iota_j(L')=\iota_j$ for every $j\in J^{(1)}_{n+1}$
(recall that $D_{n+1}(L')=D_{n+1}=0$ and
$J^{(1)}_{n+1}(L')\subset J^{(1)}_{n+1}$, see the Introduction).
Let $j\in J^{(1)}_i(L')\cap J^{(1)}_i$ (i.e., $W_j$ is an irreducible
component of
$W^{(1)}_i(L')$ and $W\cap{\cal Z}(e_0,\ldots , e_r)$ simultaneously)
for some  $\sigma+1\le i\le n$ and
$\iota_j(L'_{\sigma+1},\ldots , L'_i)>\iota_j$.
Then $D(L'_{\sigma+1},\ldots , L'_i)<D_i$.
}\par\medskip

\noindent {\bf PROOF} \quad We shall suppose without loss of generality that
the coefficients of the linear
forms $U_{\sigma+1},\ldots , U_\alpha$ belong to a field with a
real structure (i.e., the field $\overline{K}$ is supplied with a
real structure) and are infinitely
close to the corresponding coefficients of
linear forms $L'_{\sigma+1},\ldots , L'_\alpha$.
Obviously condition (8) holds for
$U$ and $K$ (in place of $L'$ and $k$) and hence (10) for
$U$ and $K$ is satisfied.
Now by the definitions and Lemma~7 the inclusion $\mbox{\rm
st}(W_i(U_{\sigma+1},\ldots , U_i))\supset
W_i(L'_{\sigma+1},\ldots , L'_i)$ holds. Hence by Lemma~5 the degree $D_i\ge
D_i(L'_{\sigma+1},\ldots , L'_i)$.

Further, let $j\in J^{(1)}_i$ and $\sigma+1\le i\le n+1$.
Assume that $\iota_j(L'_{\sigma+1},\ldots , L'_i)>\iota_j$ if $i\le n$ and
$\iota_j(L')>\iota_j$ if $i=n+1$.
We shall use
the notation from the proof of Lemma~11.
There is
$\gamma\in\Gamma_j\setminus\{\gamma_j\}$, see the proof of Lemma~11.
Set $\widetilde{W}\times\{z_i\}=W_\gamma\cap
{\cal Z}(Z_{\beta,\nu}-u_{\beta,\nu},\;
0\le \nu\le r, \,\sigma+1\le \beta\le i)$,
where $\widetilde{W}\subset{\Bbb P}^n(\overline{K})$ and
$\dim\widetilde{W}=n-i$.
Then $\widetilde{W}\ne W_j$ since, otherwise, $\gamma=\gamma_j$.
Obviously $\widetilde{W}$ is irreducible over $\overline{k}(u)$
and $\dim\widetilde{W}=n-i+\sigma-s$.

Put $\iota=i-\sigma+s$.
Let us choose linear forms $L_0,L_{\iota+1},\ldots , L_n\in k[X_0,\ldots ,
X_n]$ (this family of linear forms depend on
$i$;
it will not lead to an ambiguity) such that the intersection
$W_\gamma\cap{\cal Z}(Z_{\beta,\nu}-l'_{\beta,\nu},\;0\le\nu\le r,
\;\sigma+1\le\beta\le i)$ and ${\cal
Z}(L_{\iota+1},\ldots , L_n)$ is transversal,
\setcounter{equation}{21} \begin{eqnarray}
&&\mbox{\rm st}(\widetilde{W})\cap{\cal
Z}(L_0,L_{\iota+1},\ldots , L_n)=\emptyset,\label{22}\\
&&W(L'_{\sigma+1},\ldots ,L'_{i-1})\cap{\cal Z}(e'_i)\cap{\cal
Z}(L_0,L_{\iota+1},\ldots , L_n)=\emptyset,\label{23}\\
&&W(L'_{\sigma+1},\ldots ,L'_i)\cap{\cal
Z}(L_{\iota+1},\ldots , L_n)\cap \nonumber\\
&&({\cal Z}(e_0,\ldots , e_r)\cup(\,\overline{V\setminus W}\,))
=\emptyset,\label{24} \\
&&\#(W(L'_{\sigma+1},\ldots ,L'_{i-1})\cap{\cal Z}(e'_i)\cap{\cal
Z}(L_{\iota+1},\ldots , L_n))=\nonumber\\
&&\deg (W(L'_{\sigma+1},\ldots ,L'_{i-1})\cap{\cal Z}(e'_i))
\label{25}
\end{eqnarray}
(these are conditions of general position for linear forms
$L_0,L_{\iota+1},\ldots , L_n$;
some of these conditions will be used later in the
proof of next lemmas).

Let us apply Lemma~7 and Lemma~5 to the affine algebraic varieties
$W_\gamma\setminus{\cal Z}(L_0)$ and
${\cal Z}(Z_{\beta,\nu}-u_{\beta,\nu},\;
0\le \nu\le r, \,\sigma+1\le \beta\le i)\setminus{\cal Z}(L_0)$.
We get $\mbox{\rm
st}(\widetilde{W})\times\{z'_i\}=W_\gamma\cap{\cal
Z}(Z_{\beta,\nu}-
l'_{\beta,\nu},\;0\le \nu\le r, \,\sigma+1\le \beta\le i)\supset
W_j\times\{z'_i\}$. Hence $\mbox{\rm
st}(\widetilde{W})\supset W_j$. Further, $\mbox{\rm
st}(\widetilde{W}\cap{\cal Z}(L_{\iota+1},\ldots , L_n))=
\mbox{\rm st}(\widetilde{W})\cap{\cal Z}(L_{\iota+1},\ldots ,
L_n)\supset W_j\cap{\cal Z}(L_{\iota+1},\ldots , L_n)$
by Lemma~7 and (22). Let us choose points $\xi_j\in W_j\cap
{\cal Z}(L_{\iota+1},\ldots , L_n)$ and $\widetilde{\xi}_j\in
\widetilde{W}\cap{\cal Z}(L_{\iota+1},\ldots ,
L_n)$ such that $\mbox{\rm st}(\widetilde{\xi}_j)=\xi_j$.


Now $\mbox{\rm st}(\widetilde{W})\supset W_j$,
$\widetilde{W}\ne W_j$, and
$\dim\widetilde{W}=\dim W_j$, see above. Therefore,
$\widetilde{W}\not\subset{\cal Z}(e_0,\ldots , e_r)$ in
${\Bbb P}^n(\overline{K})$ since $W_j$ is an irreducible component of
$W\cap{\cal Z}(e_0,\ldots , e_r)$.
Hence if $i=n+1$ we get a contradiction, which implies that
$\iota_j(L')=\iota_j$.
In what follows $i\le n$ in the proof.
Since condition (10) is satisfied for
$U$ and $K$, the variety
$\widetilde{W}$ is a union of some components
of $W(U_{\sigma+1},\ldots ,U_i)$ irreducible over
$\overline{K}$.


We have proved that $\mbox{\rm st}(W(U_{\sigma+1},\ldots ,
U_i))\supset W(L'_{\sigma+1},\ldots , L'_i)$.
Hence \\ $\mbox{\rm st}(W(U_{\sigma+1},\ldots , U_i)\cap{\cal
Z}(L_{\iota+1},\ldots , L_n))\supset W(L'_{\sigma+1},\ldots ,
L'_i)\cap{\cal Z}(L_{\iota+1},\ldots , L_n)$ by Lemma~5 and Lemma~7.
Hence for every point
$\xi\in W(L'_{\sigma+1},\ldots , L'_i)$ $\cap{\cal Z}(L_{\iota+1},\ldots
, L_n)$ there is $\widetilde{\xi}\in
W(U_{\sigma+1},\ldots , U_i)\cap{\cal Z}(L_{\iota+1},\ldots , L_n)$
such that $\widetilde{\xi}$ is infinitely close to $\xi$.
Notice that $\mbox{\rm st}(\widetilde{\xi}_j)=\xi_j\not\in
W(L'_{\sigma+1},\ldots , L'_i)$ by
(24). Hence by the B\'ezout theorem
$D_i>D(L'_{\sigma+1},\ldots , L'_i)$ if $i\le n$.
The lemma is proved.

\par\medskip\noindent{\bf LEMMA~13}\hspace{0.1em} {\it  Let
$\sigma+1\le\alpha\le n+1$ be an integer,
${\cal U}^{({\scriptscriptstyle{\rm I}}{\rm v})}_\alpha$ be the set
from the statement of Lemma~11 and ${\cal U}''_\alpha$ the set defined
in the Introduction.
Then ${\cal U}''_\alpha$ is a defined over $k$ nonempty open in the
Zariski topology subset of
${\cal U}^{({\scriptscriptstyle{\rm I}}{\rm v})}_\alpha$.
Further, let ${\mathfrak l}\subset{\Bbb
A}^{(r+1)(\alpha-s)}(\overline{k})$ be a line such that
${\mathfrak l}\cap{\cal U}''_\alpha\ne\emptyset$. Then the number of
elements $\#({\mathfrak l}\setminus{\cal U}''_\alpha)$ is
bounded from above by a polynomial in $d^n(d')^{n-\sigma}$.
}\par\medskip

\noindent {\bf PROOF} \quad Recall that
$D_n=D_n(L')$ for every $L'\in{\cal U}'_0$ and
$D_{n+1}(L')=D_{n+1}=0$, see the Introduction.
Hence it is sufficient to prove the lemma for $\alpha\le n-1$.
Recall that $V_v$ and $W_v$ are defined, see the
Introduction. Replacing if necessary $(V,W,k)$ by $(V_v,W_v,k_v)$
we shall suppose without loss of generality in what
follows in the proof that $s=\sigma$.

Now there are linear forms $L_0,L_{s+2},\ldots ,
L_n\in k[X_0,\ldots ,X_r]$
such that for every $s+1\le i\le\alpha$ conditions (23)--(25) hold
and additionally
\begin{itemize}
\item the intersection $W\cap{\cal Z}(L_{i+1},\ldots , L_n)$
is transversal and is an
defined over $k$ and irreducible over $k$ algebraic variety.
\end{itemize}

The set $U''_\alpha$ is nonempty since if one extends the
ground field $k$ till $K$ then
$(U_{s+1},\ldots , U_\alpha)\in{\cal U}''_\alpha(\overline{K})$.
Now the last two assertions of
the lemma follows from Theorem~1 (d) \cite{9} applied to the morphisms
\begin{eqnarray*}
&&W\cap{\cal Z}(L_{i+1},\ldots , L_n)
\rightarrow{\Bbb P}^r(\overline{k}),   \\
&&(X_0:\ldots : X_n)\mapsto(e_0:\ldots : e_r),\quad s+1\le i\le\alpha,
\end{eqnarray*}
where $W\cap{\cal Z}(L_{i+1},\ldots , L_n)$ is
considered as an irreducible component
the algebraic variety $V\cap{\cal Z}(L_{i+1},\ldots , L_n)$
(in place of the component $W$ of the variety $V$).
The lemma is proved.

\par\medskip\noindent{\bf LEMMA~14}\hspace{0.1em} {\it  Let
$\sigma+1\le\alpha\le n+1$ be an integer. Let
$(L'_0,L'_{\sigma+1},\ldots ,
L'_n)\in{\cal U}'_0$ and $(L'_{\sigma+1},\ldots ,
L'_{\alpha-1})\in{\cal U}''_{\alpha-1}$. Let
${\mathfrak l}\subset{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$ be a line
such that $(L'_0,L'_{\sigma+1},\ldots ,
L'_n)\in{\mathfrak l}$. Then there is at most a polynomial in
$d^n(d')^{n-\sigma}$ number of elements
$L''=(L''_0,L''_{\sigma+1},\ldots , L''_n)\in{\mathfrak l}$
such that at least one of the
following three conditions fails
\begin{itemize}
\item $(L''_0,L''_{\sigma+1},\ldots ,
L''_n)\in{\cal U}'_0$,
\item $(L''_{\sigma+1},\ldots ,
L''_{\alpha-1})\in{\cal U}''_{\alpha-1}$,
\item $D_\alpha(L'')\ge D_\alpha(L')$.
\end{itemize}
}\par\medskip

\noindent {\bf PROOF} \quad We shall assume without loss of generality that
$\alpha\le n$.
Similarly to the proof of Lemma~13
we shall suppose without loss of generality that $s=\sigma$. Now
the required assertion follows from Lemma~13, assertion (d) Theorem~1
\cite{9}, and Lemma~13 \cite{9} The lemma is proved.


\par\medskip\noindent{\bf LEMMA~15}\hspace{0.1em} {\it  Let
$\sigma+1\le\alpha\le n+1$ be an integer and
$(L'_{\sigma+1},\ldots ,
L'_\alpha)\in{\cal U}''_\alpha$. Let the field
$K'$ and the mapping of the standard part
$\mbox{\rm st}$ be from Section~2.
Let us extend the
ground field $k$ till $K'$.
Let $L''_{\sigma+1},\ldots ,
L''_\alpha\in\overline{K'}[X_0,\ldots ,X_r]$ be linear forms
such that all the nonzero
coefficients of linear forms $L''_j-L'_j$, $\sigma+1\le j\le\alpha$
are infinitesimal with respect to the field $k$. Put $L''_0=L''_{n+1}$
and $L''=(L''_0,L''_{\sigma+1},\ldots , L''_n)$ if
$\alpha=n+1$.
Then for all $\sigma\le i\le\alpha$
\setcounter{equation}{25} \begin{eqnarray}
&&\mbox{\rm st}(W(L''_{\sigma+1},\ldots ,
L''_i))=W(L'_{\sigma+1},\ldots , L'_i),
\label{26}\\
&&\mbox{\rm st}(W^{(2)}(L''_{\sigma+1},\ldots ,
L''_i))=W^{(2)}(L'_{\sigma+1},\ldots , L'_i).
\label{27}
\end{eqnarray}
}\par\medskip

\noindent {\bf PROOF} \quad By Lemma~13 and transfer
principle $(L''_{\sigma+1},\ldots , L''_\alpha)\in{\cal U}''_\alpha$.
Hence $\deg W(L''_{\sigma+1},\ldots ,
L''_i)=\deg W(L'_{\sigma+1},\ldots , L'_i)=D_i$
for all $\sigma\le i\le \alpha$.

By Lemma~5 and Lemma~7 the inclusion
$\mbox{\rm st}(W(L''_{\sigma+1},$ $\ldots ,
L''_i))\supset W(L'_{\sigma+1},$ $\ldots , L'_i)$ holds.
Hence (26) is fulfilled by Lemma~5 since
$\deg W(L''_{\sigma+1},\ldots , L''_i)=\deg W(L'_{\sigma+1},\ldots , L'_i)$.

In particular we have $\mbox{\rm st}(W(L''_{\sigma+1},\ldots ,
L''_{i-1},L'_i))=W(L'_{\sigma+1},\ldots , L'_i)$ and
$\mbox{\rm st}(W(L''_{\sigma+1},$ $\ldots ,
L''_{i-1})=W(L'_{\sigma+1},\ldots , L'_{i-1})$.
Further,  $W^{(1)}(L''_{\sigma+1},\ldots ,
L''_{i-1},L'_i)$ $=W^{(1)}(L'_{\sigma+1},\ldots , L'_i)$, see the
Introduction.
Every irreducible component $E$ of $W^{(2)}(L'_{\sigma+1},\ldots ,
L'_i)$ (respectively of
$W^{(2)}(L''_{\sigma+1},\ldots , L''_i)$) is an irreducible component of
$W(L'_{\sigma+1},\ldots , L'_{i-1})\cap{\cal Z}(e'_i)$  (respectively of
$W(L''_{\sigma+1},\ldots , L''_{i-1})\cap{\cal Z}(e'_i)$) such that
$E\subset W\cap{\cal Z}(e_0,\ldots , e_r)$ and $E$ is not an irreducible
component of $W\cap{\cal Z}(e_0,\ldots , e_r)$.
Now (27) follows from Lemma~6 and the previous remarks.
The lemma is proved.



\par\medskip\noindent{\bf LEMMA~16}\hspace{0.1em} {\it  Let
$\sigma+1\le\alpha\le n+1$ be an integer. Let
$(L'_0,L'_{\sigma+1},\ldots ,
L'_n)\in{\cal U}''_0$ and $(L'_{\sigma+1},\ldots ,
L'_{\alpha-1})\in{\cal U}'''_{\alpha-1}$. Let
${\mathfrak l}\subset{\Bbb A}^{(r+1)(n-\sigma+1)}(\overline{k})$ be a line
such that $(L'_0,L'_{\sigma+1},\ldots ,
L'_n)\in{\mathfrak l}$. Then there are at most a polynomial in
$d^n(d')^{n-\sigma}$ number of elements
$L''=(L''_0,L''_{\sigma+1},\ldots , L''_n)\in{\mathfrak l}$
such that at least one of the
following three conditions fails:
\begin{itemize}
\item $(L''_0,L''_{\sigma+1},\ldots ,
L''_n)\in{\cal U}''_0$,
\item $(L''_{\sigma+1},\ldots ,
L''_{\alpha-1})\in{\cal U}'''_{\alpha-1}$,
\item $D^{(2)}_\alpha(L'')\ge D^{(2)}_\alpha(L')$.
\end{itemize}
In particular, $D^{(2)}_\alpha(L')\le
D^{(2)}_\alpha$. The set ${\cal U}'''_\alpha$ is a defined
over $k$ nonempty open in the
Zariski topology subset of ${\Bbb
A}^{(r+1)(\alpha-\sigma)}(\overline{k})$.
}\par\medskip

\noindent {\bf PROOF} \quad Let the field $K'$
and the mapping $\mbox{\rm st}$ be as above.
Let us extend the ground field $k$ till $K'$. Suppose at first that
$(L''_{\sigma+1},\ldots , L''_\alpha)\in{\cal U}''_\alpha$,
and all the nonzero coefficients of the linear forms $L''_j-L'_j$,
$\sigma+1\le j\le\alpha$
are infinitesimal with respect to the field $k$.
Then by Lemma~15, Lemma~5 and (27)
the degree $D^{(2)}_\alpha(L')\le D^{(2)}_\alpha(L'')$.

Now suppose that $K'=K$ and the family of all the
coefficients of linear forms $L''_j$, $\sigma+1\le j\le\alpha$, is
algebraically independent over $k$. We get
$D^{(2)}_\alpha(L')\le D^{(2)}_\alpha$.
Further, the set ${\cal U}'''_\alpha$ is nonempty since
$(U_{\sigma+1},\ldots ,$ $U_\alpha)\in{\cal U}'''_\alpha(\overline{K})$
if $\alpha\le n$ (respectively $U\in{\cal U}'''_\alpha$ if $\alpha=n+1$).

It is not difficult to prove that one can define ${\cal U}'''_\alpha$
by a formula with quantifiers of the first order theory of algebraically
closed fields (we leave the details to the reader). Hence
${\cal U}'''_\alpha$ is a constructive set, see the
definition in \cite{10}.
As we have proved each point
$(L'_{\sigma+1},\ldots , L'_\alpha)\in{\cal U}'''_\alpha$
has an infinitely
small with respect to $k$ neighborhood ${\cal W}_0$
in ${\Bbb A}^{(r+1)(\alpha-\sigma)}(\overline{K})$ such that ${\cal
W}_0\subset {\cal U}'''_\alpha$.
Obviously that for the ground field ${\Bbb C}$
any constructive set ${\cal W}$ of an affine space is an open in the
Zariski topology subset of this affine space,
provided that ${\cal W}$ contains a neighborhood in the classic topology
of any its point.
Now it follows from the transfer principle, see \cite{1}, that in our
situation ${\cal U}'''_\alpha$
is an open in the Zariski topology subset of ${\Bbb
A}^{(r+1)(\alpha-\sigma)}(\overline{k})$.

It remains to prove the first statement of the lemma.
Let $\varepsilon>0$ be an infinitesimal with respect to the field $k$.
Now suppose that $K'=\overline{k(\varepsilon)}$.
By the induction on $\alpha$ and Lemma~14
it is sufficient to prove that $D^{(2)}_\alpha(L'')<D^{(2)}_\alpha(L')$
for at most a polynomial in $d^n(d')^{n-\sigma}$ number of points
$L''\in{\mathfrak l}$. Let
$L'=(L'_{\sigma+1},\ldots ,L'_n)\in{\mathfrak l}$ and
$(L'_{\sigma+1},\ldots ,L'_{\alpha-1})\in{\cal U}'''_{\alpha-1}$.
Suppose that $\widetilde{L}=
(\widetilde{L}_0,\widetilde{L}_{\sigma+1},\ldots ,
\widetilde{L}_n)\in{\mathfrak
l}(\overline{k(\varepsilon)})$, and $\widetilde{L}_j=L'_j+\varepsilon
L'''_j$,
where $L'''_j\in\overline{k}[X_0,\ldots ,X_r]$ are linear forms
for all $j\in \{0,\sigma+1,\ldots ,n\}$.
Hence $\widetilde{L}$ is a generic
point of ${\mathfrak l}$.

Let linear forms $L_{\iota+1},\ldots , L_n$ be as above, and $i=\alpha$,
$\iota=\alpha-\sigma+s$.
Therefore, (25) holds with $i=\alpha$.
Hence by (25) and Lemma~6 applied $n-\alpha$ times the number of
elements
$$
+\infty>
\#(W^{(2)}_\alpha(\widetilde{L})\cap{\cal
Z}(L_{\iota+1},\ldots , L_n))\ge D^{(2)}_\alpha(L').
$$
Let $\xi$ be an arbitrary point from
$W^{(2)}_\alpha(\widetilde{L})\cap{\cal
Z}(L_{\iota+1},\ldots , L_n)$,
and, say $X_0(\xi)\ne 0$. Using, e.g., the algorithm from
\cite{2} or directly one obtains
that the nonhomogeneous coordinates
$(X_w/X_0)(\xi)=P_w(\varepsilon,\theta)/P(\varepsilon,\theta)$,
$1\le w\le n$, where $\theta$ is an algebraic over
$\overline{k}(\varepsilon)$ element with a
minimal polynomial $0\ne g\in \overline{k}[\varepsilon,Z]$,
$g(\varepsilon,\theta)=0$ and
$P_w,P\in k[\varepsilon,Z]$. Further, the degrees with respect to
$\varepsilon,Z$ of all the polynomials
$P_w$, $P$, $g$ are bounded from above by a polynomial in
$d^s(d')^{i-\sigma}$.
Hence considering the specializations of the generic point we obtain
by Theorem~1 from \cite{9} (or directly) that
the number of elements $L''\in{\mathfrak l}$ such that
$D^{(2)}_\alpha(L'')<D^{(2)}_\alpha(L')$ is
bounded from above by a
polynomial in $d^n(d')^{n-\sigma}$.
The lemma is proved.


\section{Construction of an element from ${\cal U}'_0$}\label{s4}

In this section under conditions of Theorem~2 we shall construct an elements
$L'\in{\cal U}'_0$ and $\lambda$ satisfying
assertion (a)--(d) of this theorem
but with ${\cal U}'_0$ in place of ${\cal U}'''_0$
(obviously if the constructed element
$L'\in{\cal U}'''_0$ then the original
assertions (a)--(d) holds). Let
$e'_j=L'_j(e_0,\ldots , e_r)$, $j\in\{0,\sigma+1,\ldots , n\}$,
for linear forms $L'_j\in\overline{k}[X_0,\ldots , X_r]$
(in Lemma~17 and Lemma~19
 below we shall consider also the case when $L'_j$ have
coefficients from $\overline{K}$; all the corresponding objects are defined
in this case in the similar way with the replacement $k$ by $K$).
Let $\lambda=(L_{s+1},\ldots , L_\sigma)$ where all
$L_j\in \overline{K}[X_0,\ldots , X_n]$ are linear forms.
Set $V_\lambda=V\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$,
$W_\lambda=W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$ in ${\Bbb
P}^n(\overline{K})$.
Denote by $A_{\lambda,L'}$ the set of isolated points
of $W_\lambda\cap{\cal
Z}(e'_{\sigma+1},\ldots , e'_n)\setminus{\cal Z}(e_0,\ldots , e_r)$
in ${\Bbb P}^n(\overline{K})$.
Let $N_{\lambda,L'}=\#A_{\lambda,L'}$ be the number of elements.
Let us define $p_{\lambda,L'}$,
(respectively $p_\lambda$, $p'_\lambda$) by (3),
(respectively (14), (15))
replacing there the field $k_v$ by $K$ (respectively $k$ by $K$).
Recall that $q_v=p'_v$.
Set as previously $q_\lambda=p'_\lambda$.

Let us replace in the definition of the morphism $p_{L'}$ the field
$\overline{k}$ by $\overline{K}$, see
(2). Let $\sigma\le i\le n$ be an integer.
Similarly to the Introduction $W_i(L')$ is a union of all the irreducible
over $\overline{K}$ components $E$ of
$W\cap{\cal Z}(e'_{\sigma+1},\ldots , e'_i)$ such that $\dim
E=n-s+\sigma-i$ and the morphism
$E\setminus{\cal Z}(e'_{i+1},\ldots , e'_n)\rightarrow{\cal
Z}(L'_{\sigma+1},\ldots , L'_i)={\Bbb P}^{n-i}(\overline{K})$ induced by
$p_{L'}$ is dominant. For arbitrary $\lambda$
the variety $W_{\lambda,i}(L')$ is a union of all the irreducible
over the algebraic closure of the ground field components $E$ of
$W_\lambda\cap{\cal Z}(e'_{\sigma+1},\ldots , e'_i)$ such that $\dim
E=n-i$ and the morphism
$E\setminus{\cal Z}(e'_{i+1},\ldots , e'_n)\rightarrow{\cal
Z}(L'_{\sigma+1},\ldots , L'_i)$ induced by
$p_{\lambda,L'}$ is dominant (here ${\cal Z}(L'_{\sigma+1},\ldots ,
L'_i)$ is a linear subspace of dimension $n-i$ in the projective space).


\par\medskip\noindent{\bf LEMMA~17}\hspace{0.1em} {\it  Let $L''\in{\Bbb
A}^{(n-\sigma+1)(r+1)}(\overline{K})$ and $\lambda'=(L'_{s+1},\ldots ,
L_\sigma)\in \\
{\Bbb
A}^{(\sigma-s)(n+1)}(\overline{K})$ where $L''_j\in\overline{K}[X_0,\ldots
,X_r]$,
$j\in\{0,\sigma+1,\ldots , n\}$, and $L'_i\in\overline{K}[X_0,\ldots ,
X_n]$, $s+1\le
i\le\sigma$, are arbitrary
linear forms. Then the following assertions hold.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item $N_{\lambda',L''}\le\deg q_v\deg W'=D_n$.
\item Let $N_{\lambda',L''}=\deg q_v\deg W'=D_n$,
and $x$ be an arbitrary point from $A_{\lambda',L''}$.
Put $e''_j=L''_j(e_0,\ldots , e_r)$, $j\in\{0,\sigma+1,\ldots , n\}$. Then
$x$ is a smooth point of each algebraic variety $W$, ${\cal
Z}(L'_{s+1}),\ldots , {\cal Z}(L'_\sigma)$,
${\cal Z}(e''_{\sigma+1}),\ldots ,{\cal Z}(e''_n)$, and the
intersection of tangent spaces of these
$n-s+1$ algebraic varieties at the point $x$ is transversal, i.e., is
$\{0\}$.
In particular the intersection $W\cap
{\cal Z}(L'_{s+1},\ldots , L'_\sigma)$ is transversal.
\end{enumerate}
}\par\medskip

\noindent{\bf PROOF}\quad We assume without loss
of generality replacing if it is necessary the
linear form $L''_0$ for a new one  that
$A_{\lambda',L''}\cap{\cal Z}(L''_0(e_0,\ldots , e_r))=\emptyset$.

Let us prove (i). We know then $D_n=\deg q_v\deg
W'$, see the Introduction.
Obviously the variety $W_n(L'')$ is the union of all the irreducible over
$\overline{K}$ components
$E$ of $W\cap{\cal Z}(e''_{\sigma+1},\ldots , e''_n)$ such that $\dim
E=\sigma-s$ and
$E\not\subset{\cal Z}(e''_0,e''_{\sigma+1}\ldots , e''_n)$. Then
$A_{\lambda',L''}$ is a
subset of isolated points of the intersection
$W_n(L'')\cap{\cal Z}(L'_{s+1},\ldots , L'_\sigma)$. Hence
$N_{\lambda',L''}\le\deg W_n(L'')$.
Consider the morphism $p_{v,L'',\sigma}\, :\,
W_{v,\sigma}(L'')\setminus{\cal Z}(e''_0,e''_{\sigma+1},\ldots ,
e''_n)\rightarrow{\Bbb
P}^{n-\sigma}(\overline{K_v})$, $(X_0:\ldots :
X_n)\mapsto(e''_0,e''_{\sigma+1}:\ldots : e''_n)$
defined over $\overline{K_v}$, where $W_{v,\sigma}(L'')$ is defined in the
Introduction
(now one should only extend the field $\overline{k}$ till $\overline{K_v}$).
Hence $W_{v,\sigma}(L'')$ is a union of some irreducible
components of $W_v$.
Then the morphism $p_{v,L'',\sigma}$ is dominant, see the Introduction, and
obviously $W_n(L'')\cap{\cal Z}(v_{s+1},\ldots , v_\sigma)$
is contained in the set of isolated points of
$p_{v,L'',\sigma}^{-1}((1:0:\ldots :0))$.
Hence $\deg W_n(L'')\le\deg p_{v,L'',\sigma}$ by Lemma~2 \cite{9}.
Denote by $q_{v,L'',\sigma}$ the restriction of the morphism $q_v$ to
$W_{v,\sigma}(L'')\setminus{\cal Z}(e_0,\ldots , e_r)$.

Consider the morphism of the linear projection
\setcounter{equation}{27} \begin{eqnarray}
q_{L''}\, :\, W'\setminus{\cal Z}(L''_0,L''_{\sigma+1},\ldots , L''_n)
\rightarrow{\Bbb P}^{n-\sigma}(\overline{K}), \label{28} \\
(X_0:\ldots : X_r)\mapsto(L''_0:L''_{\sigma+1}:\ldots : L''_n) \nonumber
\end{eqnarray}
defined over the field $\overline{K}$.
Denote by $\widetilde{W}'$ the union of all the irreducible component $E_2$
of $W'$ such that the
restriction $q_{L''}$ to $E_2\setminus{\cal
Z}(L''_0,L''_{\sigma+1},\ldots , L''_n)$ is dominant. Denote by
$\widetilde{q}$ the
restriction of $q_{L''}$ to
$\widetilde{W}'$. Then the degree of the linear projection
$\deg\widetilde{q}\le\deg\widetilde{W}'\le\deg W'$.
The definitions imply $p_{v,L'',\sigma}=\widetilde{q}\circ q_{v,L'',\sigma}$.
Hence $\deg p_{v,L'',\sigma}=\deg q_{v,L'',\sigma}\deg\widetilde{q}\le\deg
q_v\deg W'$.
Assertion (i) is proved.

To prove (ii)
let us choose a linear form $L'_0\in k[X_0,\ldots , X_n]$
such that $A_{\lambda',L''}\cap{\cal Z}(L'_0)=\emptyset$.
Consider the morphism
\begin{eqnarray*}
&&\overline{p}_{\lambda',L''}\, :\, W\setminus
({\cal Z}(e''_0,e''_{\sigma+1},\ldots , e''_n)\cup{\cal
Z}(L'_0,L'_{s+1},\ldots , L'_\sigma))\\
&&\longrightarrow
({\Bbb P}^{\sigma-s}\times
{\Bbb P}^{n-\sigma})(\overline{k}),\\
&&\overline{p}_{\lambda',L''}(z)=\bigl(\,(L'_0(z):L'_{s+1}(z):\ldots :
L'_\sigma(z)),
(e''_0(z):e''_{\sigma+1}(z):\ldots : e''_n(z))\,\bigr).
\end{eqnarray*}
Hence $\overline{p}_{\lambda',L''}$ depend on the choice of $L'_0$.
According to the proof of (i) if $N_{\lambda',L''}=\deg q_v\deg W'$ then
$N_{\lambda',L''}=\deg W_n(L'')=\deg p_{v,L'',\sigma}=\deg q_v\deg W'$;
the intersection $W_n(L'')\cap{\cal Z}(L'_{s+1},\ldots
, L'_\sigma)$ is transversal
and coincides with $A_{\lambda',L''}$;
$W_v=W_{v,\sigma}(L'')$; every irreducible component of
$W_{v,\sigma}(L'')$ contains
a point from $W_{v,n}(L'')=W_n(L'')\cap{\cal Z}(v_{s+1},\ldots ,
v_\sigma)$
(recall that the last is by Lemma~2 \cite{9}). Hence
every irreducible component of $W_\sigma(L'')$ contains
an irreducible component of $W_n(L'')$.
But $W=W_\sigma(L'')$ since $W_v=W_{v,\sigma}(L'')$.
Therefore, every irreducible
component of $W$ contains
a point from $A_{\lambda',L''}$. Thus, by Lemma~9 the morphism
$\overline{p}_{\lambda',L''}$ is dominant.


Let $\nu_j,\mu_i\in k$, $s+1\le j\le \sigma$, $\sigma+1\le i\le n$.
Set $\lambda''=(L'_{s+1}-\nu_{s+1}L'_0,\ldots , L'_\sigma-\nu_\sigma L'_0)$,
$L'''=(L''_0,L''_{\sigma+1}-\mu_{\sigma+1}L''_0,\ldots , L''_n-\mu_nL''_0)$.
By Lemma~2 \cite{9} there are $\nu_j,\mu_i\in\overline{k}$ such that
$\deg q_v\deg W'=N_{\lambda',L''}\le
N_{\lambda'',L'''}=
\deg\overline{p}_{\lambda',L''}$. On the other hand,
$N_{\lambda'',L'''}\le\deg q_v\deg W'$ by (i).
Hence $\deg\overline{p}_{\lambda',L''}=\\ \deg q_v\deg W'$.
Now (ii) follows from Lemma~2 \cite{9} applied to
$\overline{p}_{\lambda',L''}$.  The lemma is proved.




\par\medskip\noindent{\bf LEMMA~18}\hspace{0.1em} {\it  Let
$\lambda=(L_{s+1},\ldots , L_\sigma)$ where all
$L_j\in k[X_0,\ldots , X_n]$ are linear forms.
Suppose that $\dim W_\lambda=n-\sigma>0$,
and $W_\lambda$ is irreducible over $k$.
Let $L'=(L'_0,L'_{\sigma+1},\ldots , L'_n)$ and all linear forms
$L'_w\in\overline{k}[X_0,\ldots ,X_r]$,
$w\in\{0,\sigma+1,\ldots , n\}$, see above.
Assume that $N_{\lambda,L'}\ge 1$.
Then the following assertions hold.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item If $e'_0$ does not vanish on $A_{\lambda,L'}$ then the morphism
$p_{\lambda,L'}$, see the Introduction, is dominant.
\item The morphism $q_\lambda$, see the Introduction, is dominant.
\item $N_{\lambda,L'}\le\deg p_{\lambda,L'}\le\deg q_\lambda\deg W'$.
Denote by
${\cal V}_\lambda$ the set of all
$L'\in$ ${\Bbb
A}^{(n-\sigma+1)(r+1)}(\overline{k})$ such that $N_{\lambda,L'}=
\deg q_\lambda\deg W'$ and $A_{\lambda,L'}\cap{\cal Z}(e'_0)=\emptyset$.
Then ${\cal V}_\lambda$ is a nonempty defined over $k$
open in the Zariski topology subset of
${\Bbb A}^{(n-\sigma+1)(r+1)}(\overline{k})$.
\item Let $N_{\lambda,L'}=\deg q_\lambda\deg W'$,
and $x$ be an arbitrary point from $A_{\lambda,L'}$.
Then $x$ is a smooth point of each algebraic variety $W_\lambda$,
${\cal Z}(e'_{\sigma+1}),\ldots ,{\cal Z}(e'_n)$, and the
intersection of the tangent spaces of these
$n-\sigma+1$ algebraic varieties at the point $x$
is transversal, i.e., is $\{0\}$.
\item $\deg q_\lambda\le\deg q_v$.
\end{enumerate}
}\par\medskip

\noindent {\bf PROOF} \quad  Let us prove (i). Denote by
$B_{\lambda,L'}$ the subset of all points
$x\in A_{\lambda,L'}$ such that $e'_0(x)\ne 0$.
Suppose that $e'_0$ does not vanish on $A_{\lambda,L'}$.
Then the set $B_{\lambda,L'}$ is nonempty.
The set $B_{\lambda,L'}$  is invariant with respect to the action of the
Galois group $\mbox{\rm Gal}(\overline{k}/k)$.
Therefore,
every irreducible over $k$ component of $W_\lambda$ has a nonempty
intersection with $B_{\lambda,L'}$.
Hence by Lemma~9 the
morphism $p_{\lambda,L'}$, see (3),
is dominant.
Let us define the morphism $q_{L'}\, :\, W'\setminus{\cal
Z}(L'_0,L'_{\sigma+1},\ldots , L'_n)\rightarrow{\Bbb
P}^{n-\sigma}(\overline{k})$ replacing in (28) the element $L''$ by $L'$
and the field $K$ by $k$. Hence $q_{L'}$ is defined over $k$.
Now $p_{\lambda,L'}$ can be represented as a composition
of the morphisms
$W_\lambda\setminus{\cal Z}(e'_0,e'_{\sigma+1},\ldots , e'_n)
\rightarrow W'\stackrel{q_{L'}}{\longrightarrow}{\Bbb
P}^{n-\sigma}(\overline{k})$,
where $W_\lambda\setminus{\cal Z}(e'_0,e'_{\sigma+1},\ldots , e'_n)
\rightarrow W'$ is the restriction of $q_\lambda$.
Hence $q_\lambda$ and $q_{L'}$ are also dominant, and
$\deg p_{\lambda,L'}=\deg q_\lambda\deg q_{L'}\le\deg q_\lambda\deg W'$.
Assertion (i) is proved.

In what follows in the proof of (ii)--(v)
replacing, if it is necessary, the linear form $L'_0$
for a new one we shall assume without loss of generality that
$A_{\lambda,L'}\cap{\cal Z}(e'_0)=\emptyset$.
Hence $A_{\lambda,L'}=B_{\lambda,L'}$.
Now (ii) follows from (i).

Notice that $\deg q_{L'}=\deg W'$
if and only if $W'\cap{\cal Z}(L'_0,L'_{\sigma+1},\ldots ,
L'_n)=\emptyset$ in ${\Bbb P}^n(\overline{k})$.
Hence the set of all $L'$ such that $\deg q_{L'}=\deg W'$ is a nonempty
open in the Zariski topology subset of ${\Bbb
A}^{(n-\sigma+1)(r+1)}(\overline{k})$. Therefore, the set
${\cal V}_\lambda$ is also a nonempty (by Lemma~2 \cite{9})
open in the Zariski topology subset of this affine
space. Assertion (iii) is proved. The set ${\cal V}_\lambda$ is
invariant with respect to the action of the Galois
group $\mbox{\rm Gal}(\overline{k}/k)$, and hence
defined over $k$. Now by Lemma~2 \cite{9} assertion (iv) holds.





Let us define the set ${\cal V}_v$
replacing in the definition of ${\cal V}_\lambda$ the field $k$
by $k_v$ and $\lambda$ by $v$. So ${\cal V}_v$
is a nonempty open in the Zariski topology subset of
${\Bbb A}^{(n-\sigma+1)(r+1)}(\overline{k_v})$.
To prove (v) we shall suppose
without loss of generality that $L'\in{\cal
V}_\lambda\cap{\cal V}_v$. Then (v) follows from assertion (i) of
Lemma~17(vii).
The lemma is proved.



\par\medskip\noindent{\bf LEMMA~19}\hspace{0.1em} {\it  Let the field
$\overline{K}$ be supplied
with a real structure, see \cite{3}, and the mapping
of standard part $\mbox{\rm st}\, :\,
{\Bbb P}^n(\overline{K})\rightarrow{\Bbb P}^n(\overline{k})$ is
defined.
Let $\lambda=(L_{s+1},\ldots ,$ $L_\sigma)$,
$L'=(L'_0,L'_{\sigma+1},\ldots , L'_n)$ where all
$L_j\in \overline{K}[X_0,\ldots , X_n]$, $L'_w\in
\overline{K}[X_0,\ldots ,$ $X_r]$ are linear forms.
Let ${\mathfrak l}\subset{\Bbb A}^{(\sigma-s)(n+1)}(\overline{K})$ be a
line such that $\lambda\in{\mathfrak l}$.
Then the following assertions hold.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item For all $\lambda'\in{\mathfrak l}$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number, the inequality
$N_{\lambda',L'}\ge N_{\lambda,L'}$ is satisfied.
More than that, if $A_{\lambda,L'}\cap({\cal
Z}(e'_0)\cup\overline{V\setminus W})=\emptyset$ then
for all $\lambda'\in{\mathfrak l}$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number,
$\#(A_{\lambda',L'}\setminus({\cal Z}(e'_0)\cup\overline{V\setminus
W}))\ge N_{\lambda,L'}$.

\item If $\lambda'$ is infinitely close to $\lambda$ with respect to the
field $k$ then
$\mbox{\rm st}(A_{\lambda',L'})\supset A_{\lambda,L'}$ and hence
$N_{\lambda',L'}\ge N_{\lambda,L'}$.
\item If $N_{v,L'}=\deg q_v\deg W'$
(i.e., if $L'$ belongs to the nonempty open in the Zariski topology subset
${\cal V}_v$ of the affine space
${\Bbb A}^{(n-\sigma+1)(r+1)}(\overline{k_v})$,
see the proof of Lemma~18)
then the set ${\cal V}_{L'}$ of all
$\lambda\in {\Bbb A}^{(\sigma-s)(n+1)}(\overline{k})$ such
that $N_{\lambda,L'}=\deg q_v\deg W'$
is a nonempty open in the Zariski topology subset of ${\Bbb
A}^{(\sigma-s)(n+1)}(\overline{k})$ defined over $k$.
\end{enumerate}
}\par\medskip

\noindent {\bf PROOF} \quad Let $\lambda''=(L''_{s+1},\ldots ,
L''_\sigma)\in{\mathfrak l}$ and
$\lambda''\ne\lambda$. Let ${\Bbb A}^1(\overline{K})$ has the coordinate
function $Z$.
Consider the subvariety
$$
(W\times{\Bbb A}^1(\overline{K}))\cap{\cal
Z}(L_{s+1}+ZL''_{s+1},\ldots ,
L_\sigma+ZL''_\sigma)\cap{\cal
Z}(e'_{\sigma+1},\ldots , e'_n)
\eqno (29)
$$
of ${\Bbb P}^n(\overline{K})\times{\Bbb
A}^1(\overline{K})$. Let $C$ be the union of all the irreducible components
$C_1$
of (29) such that $C_1\cap (A_{\lambda,L'}\times\{0\})\ne\emptyset$.
Then, since $A_{\lambda,L'}$ is finite
every irreducible component $C_1$ of $C$ is a curve, and $\deg C\le
d^s2^{\sigma-s}(d')^{n-\sigma}$ by the B\'ezout theorem
(the degree $\deg C$
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 $C$).
The natural linear projection
$C\rightarrow{\Bbb A}^1(\overline{K})$, $((X_0:\ldots : X_n),Z)\mapsto
Z$,
is dominant. Hence the number of elements $\#(C\cap{\cal
Z}(Z-a))\ge\#(C\cap{\cal Z}(Z))=N_{\lambda,L'}$
for all $a\in\overline{K}$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number. Denote by $C'$ the union of all the irreducible
components $C_2$
of variety (29) such that $C_2\not\subset C$. Then by the B\'ezout
theorem $\#(C\cap C')$ (respectively under the condition
$A_{\lambda,L'}\cap({\cal
Z}(e'_0)\cup\overline{V\setminus W})=\emptyset$ the number of elements
$\#(C\cap (C'\cup{\cal Z}(e'_0)\cup(\overline{V\setminus
W}\times{\Bbb A}^1(\overline{K})))$)
is bounded from above by a polynomial in $d^n(d')^{n-\sigma}$.
Hence for all $a\in\overline{K}$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number, $C\cap{\cal
Z}(Z-a)\subset A_{\lambda^{(a)},L'}\times\{a\}$ (respectively $C\cap{\cal
Z}(Z-a)\subset (A_{\lambda^{(a)},L'}\setminus({\cal
Z}(e'_0)\cup(\overline{V\setminus
W}\times{\Bbb A}^1(\overline{K})))\times\{a\}$), where
$\lambda^{(a)}=(L_{s+1}+aL''_{s+1},\ldots ,
L_\sigma+aL''_\sigma)$. This implies (a).
Assertion (b) follows from (a) since
$\lambda'$ from (b) is a generic point of ${\mathfrak l}$ (here we leave the
details to the reader).

Using assertion (a) one can replace subsequently all the coefficients
of the linear forms $v_j$, $s+1\le j\le\sigma$,
by the coefficients from $k$ and obtain
linear forms $L_j$, $s+1\le j\le\sigma$, such that
$N_{\lambda,L'}=\deg q_v\deg W'$, cf. the algorithm of reduction to
integer coefficients, Proposition~2 \cite{9}.
Hence ${\cal V}_{L'}$ is nonempty.
It is not difficult to prove (we leave the details to the reader) that
${\cal V}_{L'}$ is a constructive set.
Hence ${\cal V}_{L'}$ is an
open in the Zariski topology subset of
${\Bbb A}^{(\sigma-s)(n+1)}(\overline{k})$ by the implicit function
theorem and assertion
(ii) Lemma~17 (cf. the proof of Lemma~16).
Finally, ${\cal V}_{L'}$ is invariant with respect to
the action of the Galois
group $\mbox{\rm Gal}(\overline{k}/k)$ and hence is defined
over $k$.
The lemma is proved.


\par\medskip\noindent{\bf LEMMA~20}\hspace{0.1em} {\it  Under conditions of
Theorem~2 suppose that
$\lambda=
(L_{s+1},\ldots , L_\sigma)$, where all $L_j\in
k[X_0,\ldots , X_n]$ are linear forms. Suppose that
$\dim W_\lambda=n-\sigma$, the
intersection $W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$ is transversal,
and if $n-\sigma>0$ then $W_\lambda$ is irreducible over $k$. Suppose that
$\dim(W_\lambda\cap(\overline{V_\lambda\setminus W_\lambda})<\dim W_\lambda$,
condition (16) holds,  the morphism $q_\lambda$,
see (15), is dominant, and the degree $\deg q_\lambda=\deg q_v$.
Further, let $L'\in{\cal U}'_{\lambda,0}$. Then
$L'\in{\cal U}'_0$.
}\par\medskip

\noindent {\bf PROOF} \quad By definition, see assertion (d)
of Theorem~1 \cite{9},
$L'\in{\cal U}'_{\lambda,0}$ (respectively $L'\in{\cal U}'_0$) if
and only if
\setcounter{equation}{29} \begin{eqnarray}
&&\#\bigl(\,(W_\lambda\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_n)\bigr)=\deg(q_\lambda)\deg W',
\label{30}\\
&&(W_\lambda\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_n)\cap{\cal
Z}(e'_0)=\emptyset,\label{31}\\
&&(W_\lambda\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_n)\cap(\,\overline{V_\lambda
\setminus W_\lambda}\,)=
\emptyset, \label{32}\\
&&\#\bigl(\,W'\cap{\cal Z}(L'_{\sigma+1},\ldots , L'_n)\,\bigr)=\deg
W'.\label{33}
\end{eqnarray}
(respectively the similar equalities hold
with $W_v$, $V_v$, $q_v$ in place of $W_\lambda$, $V_\lambda$, $q_\lambda$;
and all the varieties are considered
in this case over the field $\overline{k_v}$).
We can assume without loss of generality that the coefficients of all the
linear forms $v_j$ are infinitely close to the corresponding
coefficients of the linear forms $L_j$, $s+1\le j\le \sigma$.
Now (30)--(33) with $V_v$, $W_v$ in place of $V_\lambda$,
$W_\lambda$ follow
from (30)--(33) and Lemma~17 and the implicit
function theorem. Thus,
$L'\in{\cal U}'_0$. The lemma is proved.

\par\medskip\noindent{\bf LEMMA~21}\hspace{0.1em} {\it  Let $W_1,\ldots ,
W_\nu$, $\nu\ge 2$, be
arbitrary quasiprojective algebraic
varieties in ${\Bbb P}^n(\overline{k})$ (or in ${\Bbb
A}^n(\overline{k})$). Let $E$ be an irreducible over $\overline{k}$
component of $W_1\cap\ldots\cap W_\nu$.
Suppose that
the intersection $W_1\cap\ldots\cap W_\nu$ is transversal at $E$,
the definition see in the
Introduction, and $x\in E$ is a smooth point of $W_1\cap\ldots\cap W_\nu$.
Then $x$ is a smooth point of every $W_i$, $i=1,\ldots ,\nu$,
and the intersection $W_1\cap\ldots\cap W_\nu$ is transversal at $x$.
}\par\medskip

\noindent{\bf PROOF}\quad Obviously for every $2\le i\le \nu$ there is only
one irreducible component
$E_i$ of $W_1\cap\ldots\cap W_i$ such that $x\in E_i$, and the intersection
$W_1\cap\ldots\cap W_i$ transversal at $E_i$. Further, $x$ is a
smooth point of the intersection
$W_1\cap\ldots\cap W_i$. Hence using the induction on $i$ it is sufficient
to consider only the case $i=2$.
In what follows $\nu=2$ in the proof.

We shall suppose without loss of generality that
$W_1,W_2$ are subvarieties of ${\Bbb A}^n(\overline{k})$.
The variety $W_1\cap W_2$ is isomorphic to $(W_1\times W_2)\cap\Delta$,
where
$(W_1\times W_2)\cap\Delta\subset{\Bbb A}^n(\overline{k})\times{\Bbb
A}^n(\overline{k})$, and $\Delta=\{(z,z)\, :\, z\in{\Bbb
A}^n(\overline{k})\}$.
The intersection $(W_1\times W_2)\cap\Delta$
is transversal at $(E\times E)\cap\Delta$
if and only if $W_1\cap W_2$ is transversal at $E$.
Hence we shall
assume without loss of generality that
$W_2$ is an affine linear subspace.
Obviously $x$ is a smooth point of $E$.
Let us choose an affine linear subspace ${\cal L}$ such that
$x\in{\cal L}$,
the intersection $E\cap{\cal L}$ is transversal, the intersection
$W_2\cap{\cal L}$ is transversal, and
$x$ is an isolated point of $E\cap{\cal L}$. Now
the index of intersection $i(W_1,W_2\cap{\cal
L};x)=i(W_1,W_2;E)i(E,{\cal L};x)=1$, see \cite{10}.
Hence $x$ is a smooth point of $W_1$,
and the intersection of $W_1$ and the affine linear space
$W_2\cap{\cal L}$ at $x$ is transversal. This implies the assertion of
the lemma. The lemma is proved.

\medskip We shall suppose that the field
$\overline{k}$ is supplied with a real
structure. Let $\varepsilon_j>0$
be an infinitesimal with respect to the field
$k_{j-1}=k(\varepsilon_1,\ldots , \varepsilon_{j-1})$ for every positive
integer $j$.
Then the algebraic closure $\overline{k}_j$ (of the field $k_j$) is supplied
with a real structure in
the natural way, and one can consider and solve systems of polynomial
equations with squares of absolute values with
coefficients from $\overline{k}_j$, see \cite{3}.
The mapping of the standard part $\mbox{\rm st}_j\, :\,{\Bbb
P}^n(\overline{k_j})\rightarrow{\Bbb P}^n(\overline{k})$ is defined in
the natural way, see \cite{3}.

Let $\lambda$ be from the statement of Lemma~18.
Suppose that the linear forms $L'_j\in k[X_0,\ldots ,X_r]$,
$j\in\{0,\sigma+1,\ldots , n\}$.
Let $0\le\beta\le n$ be an integer.
Consider the system of equations and
inequalities with respect to the variables
$X_0,\ldots , X_n$, $Y_{\sigma+1},\ldots , Y_n$, $Z_{s+1},\ldots , Z_\sigma$
with coefficients from $k_1$
$$
\left\{
\begin{array}{ll}
e'_j-Y_je'_0=0,& \sigma+1\le j\le n,\\
L_j-Z_jX_\beta=0,& s+1\le j\le \sigma,\\
\sum_{\sigma+1\le j\le n}|Y_j|^2+\sum_{s+1\le j\le
\sigma}|Z_j|^2\le\varepsilon_1,\\
\sum_{0\le i\le r}|e_j|^2\ge\varepsilon_1^{-1},\\
e'_0=1.
\end{array}
\right.
\eqno (34)
$$


\par\medskip\noindent{\bf LEMMA~22}\hspace{0.1em} {\it  Suppose that $\dim
W_\lambda=n-\sigma>0$, the algebraic variety
$W_\lambda$ is irreducible over $k$,
the intersection $W\cap{\cal Z}(L_{s+1},\ldots ,L_\sigma)$
is transversal, and all the linear forms $L'_j\in k[X_0,\ldots ,X_r]$,
$j\in\{0,\sigma+1,\ldots , n\}$.
Let $\deg q_{L'}=\deg W'$, see the proof of Lemma~18, and
$1\le N_{\lambda,L'}=\deg p_{\lambda,L'}<\deg q_v\deg W'$
 (hence $A_{\lambda,L'}\cap{\cal
Z}(e'_0)=\emptyset$ and the morphisms $q_{\lambda}$ and $p_{\lambda,L'}$ are
dominant by Lemma~18 (ii) and (i)). Then there is $0\le\beta\le
n$ such that system (34) has a solution
$$
(x^*_0,\ldots , x^*_n,y^*_{\sigma+1},\ldots , y^*_n, z^*_{s+1},\ldots ,
z^*_\sigma)\in\overline{k_1}^{2n+1-s}
\eqno (35)
$$
with $x^*=(x^*_0:\ldots : x^*_n)\in W(\overline{k_1})$.
}\par\medskip

\noindent {\bf PROOF} \quad Obviously it is sufficient to prove the lemma
with $k_2$ in place of $k_1$.
Let $\widetilde{L}_j\in\overline{k(\varepsilon_2)}[X_0,\ldots , X_n]$,
$s+1\le j\le \sigma$, be linear forms.
Denote $\widetilde{\lambda}=
(\widetilde{L}_{s+1},\ldots ,\widetilde{L}_\sigma)$ (we shall specify
$\widetilde{\lambda}$
below), $\widetilde{L}=(\widetilde{L}_0,\widetilde{L}_{\sigma+1},\ldots ,
\widetilde{L}_n)$, where $\widetilde{L}_0=L'_0$ and
$\widetilde{L}_j=L'_j-y^*_jL'_0$ for all $\sigma+1\le j\le n$.
Let us replace in system (34)
every equation
$L_j-Z_jX_\beta=0$ by $\widetilde{L}_j=0$, $s+1\le j\le \sigma$.
Denote the obtained system by (*).
We have $\deg p_{v,L'}=\deg q_v\deg W'$ since $\deg q_{L'}=\deg W'$.
Therefore by Lemma~2 \cite{9} (applied over the field
$\overline{k_v(\varepsilon_2}$) there
are $y^*_j\in\overline{k(\varepsilon_2)}$, $s+1\le
j\le \sigma$, such that  all nonzero $y^*_j$ are
infinitesimal with respect to the field $k$; and
 the number of elements
$N_{v,\widetilde{L}}=\deg q_v\deg W'$.
By Lemma~19 (c) there are linear forms
$\widetilde{L}_j\in\overline{k(\varepsilon_2)}[X_0,\ldots , X_n]$,
$s+1\le j\le \sigma$,
such that all the nonzero coefficients of the linear forms
$L_j-\widetilde{L}_j$, $s+1\le j\le \sigma$, are
infinitesimal with respect to the field $k$, and the number of elements
$N_{\widetilde{\lambda},\widetilde{L}}=\deg q_v\deg W'$.
We have $W'\cap{\cal Z}(L'_0,L'_{\sigma+1},\ldots , L'_n)=\emptyset$
since $\deg q_{L'}=\deg W'$.
Now obviously $W'\cap{\cal
Z}(\widetilde{L}_0,\widetilde{L}_{\sigma+1},\ldots ,
\widetilde{L}_n)=\emptyset$.
Hence $A_{\widetilde{\lambda},\widetilde{L}}\cap{\cal
Z}(\widetilde{e}_0)=
\emptyset$, where $\widetilde{e}_0=e'_0$.

By Lemma~18 (iv) and Lemma~21 every point $z\in A_{\lambda,L'}$
is a smooth point of each algebraic variety $W$, ${\cal
Z}(L_{s+1}),\ldots ,{\cal Z}(L_\sigma)$,
${\cal Z}(e'_{\sigma+1}),\ldots ,{\cal Z}(e'_n)$, and the
intersection of tangent spaces of these
$n-s+1$ algebraic varieties at the point $z$ is transversal, i.e., is
$\{0\}$. We have $N_{\lambda,L'}=\deg p_{\lambda,L'}$. Therefore,
$A_{\lambda,L'}=W_{\lambda}\cap{\cal Z}(e'_0,e'_{\sigma+1},\ldots ,
e'_n)$ by Lemma~2 \cite{9}. Hence
for every $z\in A_{\widetilde{\lambda},\widetilde{L}}$ the standard part
$\mbox{\rm st}_2(z)\in A_{\lambda,L'}\cup{\cal
Z}(e_0,\ldots , e_r)$. Since $N_{\widetilde{\lambda},\widetilde{L}}>
N_{\lambda,L'}$ the implicit function theorem implies that
system (*) has a solution
$(x^*_0,\ldots , x^*_n,y^*_{\sigma+1},\ldots , y^*_n)
\in\overline{k_2}^{2n+1-\sigma}$ with $x^*\in W(\overline{k_2})$.
There is $0\le\beta\le n$ such that $X_\beta(x^*)\ne 0$ and
$(X_i/X_\beta)(x^*)$ are not infinitely large with respect to $k$ for all
$0\le i\le n$.
Set $z^*_j=((\widetilde{L}_j-L_j)/X_\beta)(x^*)$. Then (35) with $k_2$
in place of $k_1$
is a solution of system (34) with $x^*\in W(\overline{k_2})$. The
lemma is proved.

\medskip Let us proceed to the describe of the algorithm. Recall
that the point $x^{(0)}\in W$ is given.
We can suppose without loss of generality that
$x^{(0)}\not\in{\cal
Z}(e_0,\ldots , e_r)$, cf. the algorithm from \cite{9}.
Performing  permutations of $X_0,\ldots , X_n$ and $e_0,\ldots , e_r$ we
shall assume without loss of generality that
$(e_0X_0)(x^{(0)})\ne 0$. Using Theorem~5 \cite{4} we construct indices
$1\le i_\sigma<\ldots < i_n\le r$ and
$1\le j_{s+1}<\ldots < j_\sigma\le n$ such
that the family
$X_{j_{s+1}}/X_0,\ldots , X_{j_\sigma}/X_0$, $e_{i_\sigma}/e_0,\ldots ,
e_{i_n}/e_0\in k(W)$
is algebraically independent, and $\sigma$ is minimal possible. For
convenience of notation we shall assume without loss of generality that
$i_w=w$ for $\sigma\le w\le n$ and $j_w=w$ for $s+1\le w\le \sigma$.
Now $\dim W=n-s$ and $\dim W'=n-\sigma$.
Let us apply Theorem~3 to $V\cup{\cal Z}(e_0)$ and construct linear forms
$L_0,\ldots , L_{n+1}$.
At the beginning set $\lambda=(L_{s+1},\ldots , L_\sigma)$,
$\overline{\lambda}=(L_0,\ldots ,L_{n+1})$.
Hence condition (A) holds for $V$ and $\lambda$ and
also condition (16) is satisfied, see the Introduction.

Assume that $n-\sigma=0$. Then put $L'_0=X_0$. Obviously $W'\cap{\cal
Z}(L'_0)=\emptyset$.
Hence $L'=(L'_0)\in{\cal U}'_{\lambda,0}={\cal U}''_{\lambda,0}=
{\cal U}'''_{\lambda,0}=
{\cal U}'_{\lambda,n+1}={\cal U}''_{\lambda,n+1}={\cal
U}'''_{\lambda,n+1}$
in this case. By Lemma~20
the required $\lambda$ and $L'$ are constructed if $n-\sigma=0$.


Suppose that $n-\sigma>0$. The constructed $\lambda$ is the
base of the described below recursion.
Our aim is to construct at the end of this recursion on $\lambda$
\begin{itemize}
\item[$(\dagger)$] an element $(\lambda,L')$ such that
$\lambda=(L_{s+1},\ldots , L_\sigma)$ satisfies
conditions of Lemma~20,
condition (A) for $V$, see the Introduction,
all the linear forms $L_j\in k[X_0,\ldots , X_n]$, $s+1\le j\le\sigma$;
the element $L'$ satisfies conditions (30)--(33)
(with $\deg q_\lambda=\deg q_v$) and all
the linear forms $L'_w\in k[X_0,\ldots , X_n]$, $w\in\{0,\sigma+1,\ldots ,
n\}$. Hence $L'\in{\cal U}'_{\lambda,0}$ and $L'\in{\cal U}'_0$.
\end{itemize}
Besides that all the linear forms $L_j\in k[X_0,\ldots , X_n]$, $s+1\le
j\le\sigma$,
and $L'_w\in k[X_0,\ldots , X_n]$, $w\in\{0,\sigma+1,\ldots ,
n\}$, will have integer coefficients of
length $O(n\log d+(n-\sigma)\log(d'))$.


Consider the morphism
\begin{eqnarray*}
&&\pi\, :\, W\setminus{\cal Z}(X_0e_0)\rightarrow{\Bbb
P}^{n-s}(\overline{k}),\\
&&(X_0:\ldots : X_n)\mapsto(1:X_{s+1}/X_0:\ldots : X_{\sigma}/X_0,\,
e_{\sigma+1}/e_0:\ldots : e_n/e_0).
\end{eqnarray*}
Let us construct using the algorithm from Theorem~1 \cite{9} (b) a point
$z^{(0)}\in{\Bbb P}^{n-s}(\overline{k})$ such that
$$
\#\pi^{-1}(z^{(0)})=\deg\pi\quad\&\quad\pi^{-1}(z^{(0)})\cap
(\overline{V\setminus W})=\emptyset.
\eqno (36)
$$
For convenience of notation, performing a linear transformation of the
coordinates of
${\Bbb P}^{n-s}(\overline{k})$, we shall suppose without loss of
generality that $z^{(0)}=(1:0:\ldots : 0)$. Put
$X^{(0)}=(X_0,X_{\sigma+1},\ldots , X_n)$.



Let us compute using \cite{2} and Theorem~2 \cite{8} the number
$N_{\lambda,X^{(0)}}$.
Suppose that $N_{\lambda,X^{(0)}}=0$. Let $\tau$ be an integer.
Set $L_{0,j}=L_j$ for $j\in\{0,\ldots , s,\sigma+1,\ldots , n+1\}$,
and $L_{0,j}=L_j+\tau X_j$ for $s+1\le j\le\sigma$.
Put $\lambda_0=(L_{0,s+1},\ldots , L_{0,\sigma})$,
$\overline{\lambda_0}=(L_{0,0},\ldots , L_{0,n+1})$.
Let us show that there is at most a polynomial in
$d^n(d')^{n-\sigma}$ number of $\tau$
such that $N_{\lambda_0,X^{(0)}}=0$.
Indeed, let ${\Bbb A}^1(\overline{k})$ has
the coordinate function $Z$.  Consider the subvariety
$$
(W\times{\Bbb A}^1(\overline{k}))\cap{\cal
Z}(ZL_{s+1}+X_{s+1},\ldots ,
ZL_\sigma+X_\sigma)\cap{\cal
Z}(e_{\sigma+1},\ldots , e_n)
\eqno (37)
$$
of ${\Bbb P}^n(\overline{k})\times{\Bbb
A}^1(\overline{k})$. Let $C$ be the union of all the irreducible components
$C_1$
of (37) such that $C_1\cap(\pi^{-1}(z^{(0)})\times\{0\})\ne\emptyset$
(so $C_1\not\subset\overline{V\setminus W}\times{\Bbb
A}^1(\overline{k})$).
Then, cf. \cite{2}, by (36) every irreducible component $C_1$ of $C$ is a
curve.
By the B\'ezout theorem $\deg C\le d^s2^{\sigma-s}(d')^{n-\sigma}$,
cf. the proof of Lemma~19.
Now the required assertion follows from the B\'ezout theorem
and Lemma~9.

Thus, there is an integer
$\tau$ with length $O(n\log d+(n-\sigma)\log(d'))$
such that $N_{\lambda_0,X^{(0)}}>0$.
Applying Lemma~3 to the line ${\mathfrak l}$ containing
$\overline{\lambda}$,
$\overline{\lambda_0}$ and the algorithm of reduction to integer
coefficients, see Proposition~2 \cite{9}, we construct
a new element $\lambda_1$ similar to $\lambda$, satisfying condition (A)
for $V$, and such that $N_{\lambda_1,X^{(0)}}>0$.

Now replacing $\lambda$ by $\lambda_1$
we shall suppose in what follows that $N_{\lambda,X^{(0)}}>0$ and
condition (A) holds for $\lambda$ and $V$.
In particular, the intersection $W\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)$ is transversal and irreducible over $k$ (recall that
$n-\sigma>0$). By Lemma~18 (ii) the morphism $q_\lambda$ is dominant.
We have constructed $\lambda$ for the base of our main recursion.
This is a recursion by $\lambda$.

Let us describe the general step of this recursion.
The recursive assumption is that $\lambda$ satisfies condition (A), and
the morphism $q_{\lambda}$ is dominant
(recall here that if $n-\sigma>0$ and morphism $q_\lambda$ is dominant then
(16) holds, see the Introduction).
Applying Theorem~1 \cite{9} to $W_\lambda$, we construct $L'$ satisfying
(30)--(33). So $N_{\lambda,L'}=\deg q_\lambda\deg W'$.
Note that $W'\cap{\cal Z}(L'_0,L'_{\sigma+1},\ldots , L'_n)=\emptyset$.
For every $0\le\beta\le n$ using Theorem~2 \cite{9}
(recall that it is based on the results of
\cite{5}) we decide whether system
(34) has a solution with $x^*\in W(\overline{k_1})$.
Suppose that such a solution exists for some $\beta$. Set
$L^*_j=L_j-z^*_jX_\beta$,
$s+1\le j\le\sigma$, and $\lambda^*=(L^*_{s+1},\ldots , L^*_\sigma)$.
Put $k^*=k(z^*_{s+1},\ldots , z^*_\sigma)$.
Notice that the algebraic variety $W_{\lambda^*}$ is not necessarily
irreducible over $k^*$. Let us define the morphism
$p_{\lambda^*,L'}$ replacing in the definition of $p_{\lambda,L'}$ the
$(\sigma-s)$--tuples
$\lambda$ by $\lambda^*$ and $k$ by $k^*$.
The morphism $p_{\lambda,L'}$ is dominant by Lemma~18 (i) and (31).

Let $z\in A_{\lambda,L'}$. Let us show that there exists a
point $z^*\in A_{\lambda^*,L^*}$ such that
$\mbox{\rm st}_1(z^*)=z$,
i.e., $\mbox{\rm st}_1(A_{\lambda^*,L^*})\supset A_{\lambda,L'}$.
By Lemma~18 (iv) and Lemma~21 every point $z\in A_{\lambda,L'}$
is a smooth point of each algebraic variety $W$, ${\cal
Z}(L_{s+1}),\ldots ,$ ${\cal Z}(L_\sigma)$,
${\cal Z}(e'_{\sigma+1}),\ldots ,{\cal Z}(e'_n)$, and the
intersection of tangent spaces of these
$n-s+1$ algebraic varieties at the point $z$ is transversal, i.e., is
$\{0\}$. Hence the required assertion follows from the implicit function
theorem.
Similarly $\mbox{\rm st}_1(A_{\lambda^*,L'})\supset
A_{\lambda,L'}$.

Let us show that the morphism $p_{\lambda^*,L'}$ is dominant. By Lemma~5
and
Proposition~6 \cite{3} the equality $\mbox{\rm
st}_1(W_{\lambda^*})=W_{\lambda}$ holds.
We have $\deg W_\lambda=\deg W_{\lambda^*}=\deg W$. Hence again by Lemma~5
for every irreducible over $\overline{k}$ component $E$ of $W_\lambda$ there
is the
unique irreducible over $\overline{k^*}$ component $E^*$ of $W_{\lambda^*}$
such that $\mbox{\rm st}_1(E^*)\supset E$ (since, if $E^*$ is not
unique then
$\deg W_{\lambda^*}>\deg W_\lambda$; we leave the details to the reader).
The morphism $p_{\lambda,L'}$ is dominant and $N_{\lambda,L'}=\deg
p_{\lambda,L'}$.
Hence by Lemma~2 \cite{9} there is a point $z\in E\cap A_{\lambda,L'}$.
As we have proved there is $z^*\in A_{\lambda^*,L'}$ such that
$\mbox{\rm st}_1(z^*)=z$. Then $z^*\in E^*$ by Lemma~5 and
since $z$ is a smooth point of $W_{\lambda}$.
We have $A_{\lambda^*,L'}\cap{\cal Z}(e'_0)=0$ since
$W'\cap{\cal Z}(L'_0,L'_{\sigma+1},\ldots , L'_n)=\emptyset$ by
(31), (33).
Thus, the morphism $p_{\lambda^*,L'}$ is dominant by Lemma~9, and the
required assertion is proved.

We have proved that the algebraic variety
$(W_{\lambda^*}\setminus{\cal Z}(e'_0))\cap
{\cal Z}(e'_{\sigma+1}-y^*_{\sigma+1}e'_0,$ $\ldots , e'_n-y^*_ne'_0)$
contains $N_{\lambda,L'}$ isolated points  such that
the coordinates of each of them satisfy the inequality
$\sum_{0\le i\le r}|e_i|^2<\varepsilon_1^{-1}$
(these points are infinitely close to the corresponding
points from $A_{\lambda,L'}$).
Besides that, there is a solution of system (34)
which is different from all these isolated points.
Therefore, by Lemma~2 \cite{9} the degree
$\deg p_{\lambda^*,L'}>N_{\lambda,L'}=
\deg p_{\lambda,L'}$. Let us construct using  Theorem~1 \cite{8}
and Theorem~2 \cite{8} a representative system of
points of the irreducible over $k^*$
components of the algebraic variety $V_{\lambda^*}$ and choose among them
the points from $W_{\lambda^*}$. Now applying Theorem~1 \cite{9} for the
restrictions of  $p_{\lambda^*,L'}$ to irreducible over $k^*$
components $E^*$ of $W_{\lambda^*}$, we compute for every $E^*$ the degree
$\deg p_{\lambda^*,L'}|_{E^*}$ and
a point $z'$ from the statement of assertion (b) of
Theorem~1 \cite{9} with $p_{\lambda^*,L'}|_{E^*}$ in place of
$\rho$.

Further, by the algorithm of reduction to integer coefficients,
see Proposition~2 \cite{9},
considering the lines through the pairs of
points $z'$ corresponding to different components $E^*$ and
applying assertion (d) of Theorem~1 \cite{9}, we can construct a new point
which is
similar to all $z'$, and is the same for all the components $E^*$. So
we shall suppose without loss of generality that the point $z'$ is the same
for all components $E^*$. Denote $z'=(1:z'_{\sigma+1}:\ldots : z'_n)$.
Hence all $z'_j$ are integers with length $O(n\log d+(n-\sigma)\log(d'))$.
Set $L''_j=L'_j-z'_j L'_0$, $\sigma+1\le j\le n$,
and $L''=(L'_0,L''_{\sigma+1},\ldots , L''_n)$. Then by assertion (b)
of Theorem~1 \cite{9} $N_{\lambda^*,L''}=\deg
p_{\lambda^*,L''}>N_{\lambda,L'}$.
Using Lemma~19 (a)
we replace subsequently the
coefficients of the linear forms
$L^*_j$, $s+1\le j\le\sigma$ by integers with lengths $O(n\log
d+(n-\sigma)\log(d'))$
and obtain linear forms $L'_j$, $s+1\le j\le\sigma$ (denote
$\lambda'=(L'_{s+1},\ldots , L'_\sigma)$)
such that  $N_{\lambda',L''}\ge N_{\lambda^*,L''}>N_{\lambda,L'}$.



Applying Lemma~3, the algorithm of reduction
to integer coefficients, and Lemma~19 (a)
to the line ${\mathfrak l}$ such that $\lambda,\lambda'\in{\mathfrak
l}$
we obtain $\lambda''$
similar to $\lambda'$ such that condition (A) holds for $\lambda''$ and $V$,
and $N_{\lambda'',L''}\ge N_{\lambda',L''}>N_{\lambda,L'}$.
Hence the morphism $q_{\lambda''}$ is dominant
by Lemma~18 (ii).
We have $N_{\lambda,L'}=\deg q_{\lambda}\deg W'$ by our construction, and
$N_{\lambda'',L''}\le\deg q_{\lambda''}\deg W'$ by Lemma~18 (iii).
Therefore, $\deg q_{\lambda''}>\deg q_{\lambda}$.
To avoid the growth of lengths of coefficients by the recursion, again
applying Lemma~3, the algorithm of reduction
to integer coefficients, and Lemma~19 (a), we replace subsequently all the
coefficients
of linear forms $L''_j$, $s+1\le j\le\sigma$
from $\lambda''$ by integers with lengths $O(n\log d+(n-\sigma)\log(d'))$,
obtain new linear forms $L'''_j$, and the element
$\lambda'''=(L'''_{s+1},\ldots , L'''_\sigma)$,
which is similar to $\lambda''$.
Now we replace $\lambda$ by $\lambda'''$ and proceed to the next step of the
recursion. Note that by Lemma~18 (v)
there is at most $\deg q_v$ number of steps
with $N_{\lambda'',L''}>N_{\lambda,L'}$ at the end.

Thus we shall come at the end to the case when for every $0\le\beta\le n$
system (34) does not have a solution with $x^*\in W(\overline{k_1})$.
Then by Lemma~20 and Lemma~22 we have constructed
$(\lambda,L')$, satisfying $(\dagger)$.
The considered step of the recursion is final.
The working time of the described algorithm satisfy the estimate for the
working time from Theorem~2.
This follows from the bounds for the working times of the applied algorithms.



\par\medskip\noindent{\bf REMARK~7}\hspace{0.1em} {\it  Let
$\lambda=(L_{s+1},\ldots , L_\sigma)$ where all
$L_j\in k[X_0,\ldots , X_n]$ are linear forms. Then one can decide whether
$\lambda$ satisfies conditions of Lemma~20, computing
$\deg q_v$ by the described algorithm and
applying the results of \cite{8}, \cite{9}.
If the conditions of Lemma~20 hold then
using the algorithm from \cite{9} one can construct
$L'\in{\cal U}'_{\lambda,0}$ satisfying (10)
for all $\sigma\le i\le n-1$ with
$W_\lambda$, $W_{\lambda,i}(L')$ in place of $W$, $W_i(L')$ respectively.
Suppose additionally that for every irreducible component $E$ of
$(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap{\cal
Z}(e'_{\sigma+1},\ldots , e'_i)$ the intersection
$E\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)\ne\emptyset$
for every $\sigma\le i\le n-1$. Then
(we can assume now without loss of generality that all the coefficients of
all linear forms $U_j-L'_j$,
$j\in\{0,\sigma+1,\ldots , n\}$, are infinitesimal with respect to the
field $k$)
by Lemma~18 (iv),Lemma~5, the implicit function theorem and Lemma~9
condition (10) holds for all $\sigma\le i\le n-1$ for
$W_v$, $W_{v,i}(L')$ in place of $W$, $W_i(L')$.
Therefore, original condition (10)
is satisfied for every $\sigma\le i\le n-1$.
We leave the details to the reader here.
We shall not use this remark in what follows.
}\par\medskip






\par\medskip\noindent{\bf LEMMA~23}\hspace{0.1em} {\it  Let $\nu\ge 0$ and
$m_i\ge 1$, $1\le i\le\nu$, be integers.
Suppose that $f^{(i)}_1,\ldots , f^{(i)}_{m_i}
\in k[X_0,\ldots , X_n]$
are homogeneous polynomials similar to $f_1,\ldots ,f_m$, i.e.,
$f^{(i)}_1,\ldots , f^{(i)}_{m_i}$ satisfy the same
estimations for degrees and lengths of coefficients as $f_1,\ldots ,f_m$ do.
Let $V^{(i)}={\cal Z}(f^{(i)}_1,\ldots , f^{(i)}_{m_i})$
in ${\Bbb P}^n(\overline{k})$. Then one can construct linear forms
$L_0,\ldots , L_{n+1}$ with integer coefficients of length
$O(n\log d+\log\nu)$ such that the assertion of
Theorem~3 holds for $V^{(i)}$ in place of $V$
for all $0\le i\le\nu$ (i.e., linear forms $L_0,\ldots , L_{n+1}$ do not
depend on $i$;
other objects from the statement of this theorem for $V^{(i)}$ depend on
$i$). The working time of this algorithm is
polynomial in $d^n$, $d_1$, $d_2$, $M$, $M_1$, $m$, $\nu$.
}\par\medskip

\noindent {\bf PROOF} \quad
This follows from
Lemma~3 and the algorithm of reduction
to integer coefficients, see Proposition~2 \cite{9}
(we leave the details to the reader). The lemma is proved.

\medskip For an arbitrary
quasiprojective algebraic variety
$E\subset{\Bbb P}^n(\overline{k})$ and a point $x\in{\Bbb
P}^n(\overline{k})$
denote by $\mu(x,E)$ the
multiplicity of the point $x$ at $E$.
For an arbitrary projective algebraic variety $V'\subset{\Bbb
P}^n(\overline{k})$ we shall denote by $\mbox{\rm con}(V')$
the affine algebraic variety
in ${\Bbb A}^{n+1}(\overline{k})$ which is the set of zeroes of the
homogeneous ideal of $V'$.

\par\medskip\noindent{\bf LEMMA~24}\hspace{0.1em} {\it  Let $\nu>\nu_0\ge 0$
be integers.
Suppose that $f^{(i)}_1,\ldots , f^{(i)}_{m_i}
\in k[X_0,$ $\ldots , X_n]$, $V^{(i)}$, $W^{(i)}$, $s_i$, $x^{(0)}_i$ for
$0\le i\le \nu$
are similar to $f_1,\ldots ,f_m$,
$V$, $W$, $s$, $x^{(0)}$ from the Introduction. Suppose that
$f^{(i)}_1,\ldots , f^{(i)}_{m_i}$, $0\le i\le \nu$, satisfy the same
estimations for degrees and lengths of coefficients as $f_1,\ldots ,f_m$ do.
Then one can construct linear forms $L_0,\ldots , L_{n+1}$ from the
statement of Lemma~23 (see also Remark~8 below)
satisfying additionally the following conditions.
For every $0\le j\le\nu_0$, $\nu_0+1\le i\le\nu$
for every $y\in W^{(i)}\cap{\cal
Z}(L_{s_i+1},\ldots , L_n)$ the equality
$\mu(y,W^{(j)})=\min\{\mu(x,W^{(j)})\, :\, x\in W^{(i)}\}$ holds.
The working time of this algorithm is polynomial in $d^n$, $d_1$,
$d_2$, $M$, $M_1$, $m$, $\nu$ and
the sum of sizes $\sum_{0\le i\le\nu}{\rm L}(x^{(0)}_i)$
of the points $x^{(0)}_i$.
}\par\medskip

\noindent {\bf PROOF} \quad We shall describe a recursive algorithm by
linear forms
$L'_0,\ldots , L'_{n+1}$ $\in k[X_0,\ldots , X_n]$ and
a point $z'=(1:z'_1,\ldots ,
z'_n)\in{\Bbb P}^n(\overline{k})$.
Set $\overline{\lambda'}=(L'_0,\ldots ,$ $L'_{n+1})$.

Let us construct using Lemma~23
linear forms $L'_0,\ldots ,$ $L'_{n+1}$ (we change here the notation $L_w$
for $L'_w$).
Denote by $\varphi^{(i)}$ the morphism of the linear projection
${\Bbb P}^n(\overline{k})\setminus{\cal Z}(L'_0,L'_{s_i+1},\ldots ,
L'_n)
\rightarrow{\Bbb P}^{n-s_i}(\overline{k})$, $(X_0:\ldots :
X_n)\mapsto
(L'_0:L'_{s_i+1}:\ldots : L'_n)$, and by
$\varphi_i$ the restriction of $\varphi^{(i)}$ to $W^{(i)}$. So
$\varphi_i$ is a
finite dominant linear projection.
Set $P_i(z')=\varphi_i^{-1}((1:z'_{s_i+1}:\ldots :
z'_n))\subset{\Bbb
P}^n(\overline{k})$ and
$$
\mu(z')=\sum_{0\le j\le\nu_0}\;\sum_{\nu_0+1\le
i\le\nu}\;\sum_{x'\in P_i(z')}\mu(x',W^{(j)})
$$
(we shall use this definition also for other points from
${\Bbb P}^n(\overline{k_2})$ in place of
$z'\in{\Bbb P}^n(\overline{k})$ replacing $k$ by $k_2$).



We take the constructed
$L'_0,\ldots , L'_{n+1}$ and $z'=(1:0:\ldots : 0)$
as the base of the recursion.
The recursive assumption is that
linear forms $L'_0,L'_1-z'_1L'_0,\ldots , L'_n-z'_nL'_0, L'_{n+1}$
(in place of $L_0,\ldots , L_{n+1}$) satisfy assertion of Lem\-ma~23.

Let us describe the general step of this recursion.
For every $\nu_0+1\le i\le\nu$ using Theorem~2 \cite{9} and Theorem~2
\cite{8}
we compute the set $P_i(z')$.

Let us show that using \cite{4} and \cite{8} one can compute the
multiplicity  $\mu(x,W^{(j)})$ for every $x'$ and every $j$. We refer to
Lemma~18 \cite{4}.
Under conditions and in the notation of this lemma $\mu(x,V_{s+1})=\deg
V_{s+1}-\#A$,
where the finite set $A=(V_{s+1}\setminus\{x\})\cap{\cal
Z}(L_{s+2},\ldots , L_n)$, see formula (21) of \cite{4}.
Further, for a defined over $k$ and irreducible over $k$ component $W$ of
$V_{s+1}$ the multiplicity $\mu(x,W)=\deg W-\#A(W)$
where $A(W)=(W\setminus\{x\})\cap{\cal
Z}(L_{s+2},\ldots, L_n)$ (there are two obvious misprints here in \cite{4},
which are immediately corrected from the context;
namely, in formula
(21) of \cite{4} defining $A$ one must replace $V_{s+1}$
by $(V_{s+1}\setminus\{x\})$, and below
in the similar formula for $A(W)$ one must replace
$W$ by $(W\setminus\{x\})$).
If a point $z^{(0)}\in W$ is known then one can compute $\deg W$ and
$\#A(W)$ using Theorem~2 \cite{8}. Thus, one can compute $\mu(x,W)$.
We apply this algorithm for $(V^{(j)},W^{(j)},x^{(0)}_j,x')$ in place of
$(V,W,z^{(0)},x)$ and compute all
the required multiplicities $\mu(x',W^{(j)})$.
After that we construct $\mu(z')$. In what follows we compute $\mu(z^*)$ in
the similar way
for other points $z^*=(1:z^*_1:\ldots : z^*_n)$ in the case when elements
$z^*_j\in\overline{k}_2$.

At the recursive step we
enumerate $i,x',j$ such that
$\nu_0+1\le i\le\nu$, $x'\in P_i(z')$ and $0\le j\le\nu_0$.
For the considered $x'\in P_i(z')$, $j$
we perform the following construction.

Applying Theorem~2 \cite{9}
to the algebraic variety $(\mbox{\rm
con}(V^{(i)})\times\mbox{\rm
con}(V^{(j)}))(\overline{k_2})$
we decide
whether there is a pair $(x'',x''')\in (W^{(i)}\times
W^{(j)})(\overline{k_2})$ such that
$$
\left\{
\begin{array}{ll}
\sum_{0\le w\le n}|(X_w/L'_0)(x'')-(X_w/L'_0)(x')|^2
\le\varepsilon_1,&\\
\varepsilon_2\le\sum_{0\le w\le n}|(X_w/L'_0)(x'')-(X_w/L'_0)(x''')|^2
\le\varepsilon_1,&\\
\varphi^{(j)}(x'')=\varphi^{(j)}(x'''),
\end{array}
\right.
\eqno (38)
$$
cf. system~(2) from \cite{4}.
If there does not exist such a pair $(x'',x''')$, the dimension
$\dim W^{(i)}\le\dim W^{(j)}$ (i.e., $s_i\ge s_j$),
and not all $i,x',j$ are enumerated then
we proceed to the next $i,x',j$.
Suppose that there exists a pair $(x'',x''')\in (W^{(i)}\times
W^{(j)})(\overline{k_2})$ satisfying (38).
Then we compute
$\varphi_i(x'')=(1:z''_{s_i+1}:\ldots : z''_n)$ where all
$z''_w\in\overline{k_2}$. Put $z''_w=z'_w$ for $1\le j\le s_i$ and
$z''=(1:z''_1:\ldots : z''_n)$.
Then $\mu(z'')<\mu(z')$.


Assume that $\dim W^{(i)}>\dim W^{(j)}$,
and there does not exist a pair $(x'',x''')\in (W^{(i)}\times
W^{(j)})(\overline{k_2})$ satisfying (38).
Then we decide whether $x'\in W^{(j)}$ applying Theorem~2 \cite{8}.
If $x'\not\in W^{(j)}$ and not all $i,x',j$ are enumerated
then we proceed to the next $i,x',j$.
Suppose that $x'\in W^{(j)}$.
By the recursive assumption, Theorem~3 and Theorem~1 (a) from \cite{8} the
point
$x'\not\in\overline{V^{(j)}\setminus W^{(j)}}$
in this case. Hence
there is a point $x''\in W^{(i)}(\overline{k_2})$ such that
$$
\left\{
\begin{array}{ll}
\sum_{0\le w\le n}|(X_w/L'_0)(x'')-(X_w/L'_0)(x')|^2
\le\varepsilon_1,&\\
\sum_{1\le w\le m_j}|f^{(j)}_w((X_0/L'_0)(x''),\ldots ,
(X_n/L'_0)(x''))|^2=\varepsilon_2.
\end{array}
\right.
$$
We construct such a point $x''$
applying Theorem~2 \cite{9}
to the algebraic variety $(\mbox{\rm
con}(V^{(i)}))(\overline{k_2})$.
Now $x''\in W^{(i)}(\overline{k_2})\setminus W^{(j)}(\overline{k_2})$
and hence $\mu(x'',W^{(j)})<\mu(x',W^{(j)})$.
Let us compute
$\varphi_i(x'')=(1:z''_{s_i+1}:\ldots : z''_n)$ where all
$z''_w\in\overline{k_2}$. Put $z''_w=z'_w$ for $1\le j\le s_i$ and
$z''=(1:z''_1:\ldots : z''_n)$.
Then $\mu(z'')<\mu(z')$.


Let us return to the general case when $\dim W^{(i)}$, $\dim W^{(j)}$ are
arbitrary.
Suppose that the point $z''$ is constructed in some of the
cases considered above.
It is not difficult to see, by e.g., \cite{4}, that for an arbitrary line
${\mathfrak l}\subset{\Bbb P}^n(\overline{k_2})$ and any point
$x^*\in{\mathfrak l}$ the number of elements
$x^{**}\in{\mathfrak l}$ with $\mu(x^{**},W^{(j)})>\mu(x^*,W^{(j)})$
is bounded from above by
a polynomial in $d^n$, cf. also the proof of Lemma~1 \cite{7}.
Therefore, applying Lemma~3 and
the algorithm of reduction to integer coefficients, see Proposition~2
\cite{9}, to the points $z''$ and $(1:0:\ldots : 0)$
one can construct
a point $\widetilde{z}=(1:\widetilde{z}_1:\ldots : \widetilde{z}_n)
\in{\Bbb P}^n(\overline{k})$
with integer $\widetilde{z}_w$ of length
$O(n\log d+\log\nu)$ such that
$\mu(\widetilde{z})\le\mu(z'')$ and the linear forms
$L'_0,L'_1-\widetilde{z}_1L'_0,\ldots , L'_n-\widetilde{z}_nL'_0,L'_{n+1}$
satisfy assertion Lemma~23 (in place of $L_w$, $0\le w\le n+1$).
Then we replace $z'$ by $\widetilde{z}$
and proceed to the next step of the recursion with the same
$\overline{\lambda'}$.

Finally we shall come to the step when for all
$\nu_0+1\le i\le\nu$
for all $x'\in P_i(z')$, $0\le j\le\nu_0$
the point $z''$ is not constructed by the algorithm.
Then obviously
$\mu(x',W^{(j)})=\min\{\mu(x,W^{(j)})\, :\, x\in W^{(i)}\}$
for all $0\le j\le\nu_0$.
This step is final.
We set $L_0=L'_0$, $L_{n+1}=L'_{n+1}$, and $L_j=L'_j-z'_jL'_0$,
$1\le j\le n$. Now $L_0,\ldots ,L_{n+1}$ satisfy the required properties
by the described construction.
The required bound for the
working time of the described algorithm
follows immediately from the ones for the working times of the applied
algorithms. The lemma is proved.

\par\medskip\noindent{\bf REMARK~8}\hspace{0.1em} {\it  Suppose that under
the conditions of Lemma~23 and Lemma~24
the sum of the
degrees of all the irreducible components of every
$V^{(i)}$ is bounded from above by $D$. Then Lemma~3 and
the algorithm of reduction to
integer coefficients, see Proposition~2 \cite{9}, imply
that one can construct linear forms $L_0,\ldots ,L_{n+1}$ (from the
statements of these lemmas) with integer coefficients
of length $O(n+\log D+\log\nu)$.
}\par\medskip


\par\medskip\noindent{\bf LEMMA~25}\hspace{0.1em} {\it  Let
$L'=(L'_0,L'_{\sigma+1},\ldots ,L'_n)$, where
$L'_j$, $j\in\{0,\sigma+1,\ldots ,n\}$, are linear forms with
coefficients from a finite extension $k'$ of $k$ which is given by
a primitive element and its minimal polynomial.
Then one can decide whether
$L'=(L'_0,L'_{\sigma+1},$
$\ldots , L'_n)\in{\cal U}'_0$. If it is true, one can
compute $\lambda=(L_{s+1},\ldots , L_\sigma)$ satisfying
condition (A), see the Introduction,
conditions of Lemma~20, and
such that $L'\in{\cal U}'_{\lambda,0}$, all $L_j\in k[X_0,\ldots
,X_r]$,
$s+1\le j\le\sigma$, are linear forms with integer coefficients of
length $O(n\log d+(n-\sigma)\log d')$.
The working time of this algorithm 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 $x^{(0)}$, and
the size ${\rm L}(L')$ of the element $L'$ (including the size of the minimal
polynomial of the primitive element of the extension $k'\supset k$).
}\par\medskip

\noindent {\bf PROOF} \quad At first using the described
algorithm one computes  $\deg q_v$ and $\deg W'$.
After that applying Theorem~1 \cite{8} to the algebraic variety
$V\cup{\cal Z}(e'_0)$
(respectively $V\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_n)$) one constructs
linear forms and representative systems
of points of irreducible over $k$ components of this algebraic variety.

Let $W^{(i)}$, $0\le i\le\nu_0$, be all the irreducible over $k$ components
of $V\cup{\cal Z}(e'_0)$.
Let $W^{(i)}$, $\nu_0+1\le i\le\nu$ be all the irreducible over $k$
components of
$V\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_n)$ such that $\dim W_i=\sigma-s$.
Hence $\nu$ is bounded from above by a polynomial in $d^n(d')^{n-\sigma}$
by the B\'ezout theorem.
Set $V^{(i)}=V\cup{\cal Z}(e'_0)$ for $0\le i\le\nu_0$,
and $V^{(i)}=V\cap
{\cal Z}(e'_{\sigma+1},\ldots , e'_n)$ for $\nu_0+1\le i\le\nu$.
By Theorems~1 \cite{8} and 2 \cite{8}
we construct a family of points $x^{(0)}_i\in W^{(i)}$
which are smooth points of $V^{(i)}$. So now one can apply Lemma~24 with
the ground field $k'$ in place of $k$ (one can easily give in the
considered case the new bounds in place of
$d$, $d_1$, $d_2$, $M$, $M_1$ from Lemma~24)
to the constructed family $(V^{(i)},W^{(i)},x^{(0)}_i)$, $0\le i\le\nu$
and the integer $\nu_0$. Thus, we obtain linear forms $L_0,\ldots ,
L_{n+1}$.
Let $0\le j\le\nu_0$ and $\nu_0+1\le i\le \nu$.
Now $W^{(i)}\subset W^{(j)}$ if and only if $W^{(i)}\cap{\cal
Z}(L_{n-\sigma+s+1},\ldots , L_n)\subset W^{(j)}$.
Hence one can decide whether $W^{(i)}\subset W^{(j)}$ using Theorem~2
\cite{8}.
Therefore, one can choose among $W^{(i)}$, $\nu_0+1\le i\le\nu$ all the
irreducible components $E$
such that: $E\subset W$, $E\not\subset{\cal Z}(e'_0)$, and $E\not\subset
W^{(j)}$ whenever $W^{(j)}\ne W$,
$0\le j\le\nu_0$.
We have $\deg E=\#(E\cap{\cal Z}(L_{n-\sigma+s+1},\ldots , L_n))$.
Therefore, one can compute the degree $\delta$ of the union of all such
irreducible components $E$.
By Lemma~24 the element $L'\in{\cal U}'_0$ if and only if $\delta=\deg
q_v\deg W'$.
Set $\lambda=(L_{n-\sigma+s+1},\ldots , L_n)$ for the constructed linear
forms $L_w$.
Finally,
if $\delta=\deg q_v\deg W'$ then $L'\in{\cal U}'_{\lambda,0}$ by the
described construction and Lemma~24. The bound for the length of
integer coefficients of the linear forms $L_j$ follows
from Lemma~23 and Remark~8.
The lemma is proved.


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

The aim of this section is to prove Theorem~1.
The basic construction described here will be used also in the next section.
Let $V$ and $W$ be from the Introduction.
We compute $s$ and $\sigma$ similarly to the algorithm from Section~4.

In what follows $\sigma+1\le i\le n+1$ be arbitrary but fixed
(we effect the construction for all $\sigma+1\le i\le n+1$).
Let us apply Lemma~23 to $n-\sigma+1$ algebraic varieties
$\widetilde{V}_j$, $\sigma\le j\le n$, and construct
linear forms $L_0,\ldots , L_{n+1}$
with integer coefficients of length $O(n\log (dd'))$.
Actually by B\'ezout theorem applying the algorithm of reduction to integer
coefficients, see Proposition~2 \cite{9}, one can get new linear forms
$L_0,\ldots , L_{n+1}$ satisfying the same properties but with
integer coefficients of length $O(n\log d+(n-\sigma)\log d')$.
So we shall suppose in what follows that the last estimate hold.
Now assertion (a) holds for $L_0,\ldots , L_{n+1}$.
In what follows the linear forms $L_0,\ldots , L_{n+1}$ will be changed
recursively in the algorithm.

The further aim is to
construct linear forms $L_0,\ldots , L_{n+1}$ satisfying (a) and (b).
%By Lemma~3 an the algorithm of reduction to integer coefficients for
%this aim it is sufficient to construct $L_0,\ldots , L_{n+1}$
%satisfying (b).
According to Theorem~2 \cite{8} one can decide whether a point $x\in
W_i(L')$.
Hence (replacing $i$ by $i-1$)
one can decide whether $x\in W_{i-1}(L')\cap{\cal Z}(e'_i)=
W_i(L')\cup W^{(1)}_i(L')\cup W^{(2)}_i(L')\cup W^{(3)}_i(L')$, see the
Introduction.

Let us show that the polynomial $e'_i$ does not
vanish on any irreducible component $W_{i-1}(L')\cap{\cal
Z}(L_{\iota+1},\ldots , L_n)$.
Indeed, by Theorem~1 \cite{8} for $\widetilde{V}_{i-1}$
all the points of $W_{i-1}(L')\cap{\cal Z}(L_\iota,\ldots , L_n)$ are
smooth points of
$\widetilde{V}_{i-1}$.
The dimension $\dim W_{i-1}(L')\cap{\cal Z}(e'_i)=n-\iota$,
see the Introduction. Hence every irreducible component of
$W_{i-1}(L')\cap{\cal Z}(L_{\iota+1},\ldots , L_n)$
contains a point from $W_{i-1}(L')\cap{\cal Z}(L_\iota,\ldots ,
L_n)\setminus{\cal Z}(e'_i)$.
The required assertion is proved. Thus, $\dim
W_{i-1}(L')\cap{\cal Z}(L_{\iota+1},\ldots , L_n)\cap{\cal
Z}(e'_i)=0$.



Applying the algorithm from \cite{2} we construct all the defined over $k'$
and
irreducible over $k'$ components
of dimension one of the algebraic variety
$\widetilde{V}_{i-1}\cap{\cal Z}(L_{\iota+1},\ldots , L_n)$, and then
using Theorem~2 \cite{8} choose among them components of
$W_{i-1}(L')\cap{\cal Z}(L_{\iota+1},\ldots , L_n)$.
After that we construct all the points from the set
$S_i=W_{i-1}(L')\cap{\cal Z}(L_{\iota+1},\ldots , L_n)\cap{\cal
Z}(e'_i)$. We shall assume without
loss of generality that $S_i\cap{\cal Z}(L_0)=\emptyset$ and $L_0(x)\ne
0$, since, otherwise,
applying the algorithm of reduction to integer
coefficients one can replace
the linear form $L_0$ for a new one and satisfy these conditions
and all the conditions from (a).

Denote by $\varphi^{(i)}$ the morphism of linear projection
 ${\Bbb P}^n(\overline{k})\setminus{\cal Z}(L_0,L_{\iota+1},\ldots ,
L_n)$ $\rightarrow{\Bbb P}^{n-\iota}(\overline{k})$, $(X_0:\ldots :
X_n)\mapsto
(L_0:L_{\iota+1}:\ldots : L_n)$, and by
 $\varphi_i$ the restriction of $\varphi^{(i)}$ to $W_{i-1}(L')\cap{\cal
Z}(e'_i)$. Hence $\varphi_i$ is a finite dominant linear projection.
Now the projection $\varphi_i\, :\, W_{i-1}(L')\cap{\cal
Z}(e'_i)\rightarrow{\Bbb P}^{n-\iota}(\overline{k'})$,
$(X_0:\ldots : X_n)\mapsto(L_0:L_{\iota+1}:\ldots : L_n)$ is finite dominant.
Let $x'\in{\Bbb P}^n(\overline{k'})$.
Set $\mu_j(x')=\mu(x',W_{j-1}(L')\cap{\cal Z}(e'_j))$,
$\sigma+1\le j\le n+1$.
Let $z'\in{\Bbb P}^{n-\iota}(\overline{k'})$. For every $\sigma+1\le i\le
n$ put
$$
\mu^{(i)}(z')=\Bigl(\#\varphi_i^{-1}(z'),
\,\#(\varphi_i^{-1}(z')\setminus{\cal
Z}(e_0,\ldots , e_r)),\,
\sum_{x'\in\varphi^{-1}_i(z')}\sum_{\sigma+1\le j\le
i-1}\mu_j(x')\Bigr).
$$
Therefore, $\mu^{(i)}(z')\in{\Bbb Z}^3$.
Let $(a_1,b_1,c_1),(a_2,b_2,c_2)\in{\Bbb Z}^3$. Set
$(a_1,b_1,c_1)>(a_2,b_2,c_2)$ if and only if
$(a_1>a_2)\vee((a_1=a_2)\&(b_1>b_2))\vee
((a_1=a_2)\&(b_1=b_2)\&(c_1<c_2))$.
We shall use this definition also with other ground field containing $k$.
This formula defines the linear order on ${\Bbb Z}^3$.
In what follows in the proof
we shall consider this linear order on ${\Bbb Z}^3$.
Let ${\mathfrak l}\subset{\Bbb P}^{n-\iota}(\overline{k'})$ be a line such
that $z'\in{\mathfrak l}$.
The B\'ezout theorem implies that for all $z''\in{\mathfrak l}$, except at
most polynomial in $d^n(d')^{n-\sigma}$ number,
the inequality $\mu^{(i)}(z'')\ge\mu^{(i)}(z')$ holds (here we leave the
details to the reader).
The same assertion is valid for any other ground field $k^*$ containing $k'$.

Let $\nu_{\iota+1},\ldots ,\nu_n\in\overline{k'}$
(or more generally $\nu_{\iota+1},\ldots ,\nu_n$
belong to an extension of $\overline{k'}$) be given.
Then one can construct the set
$\varphi_i^{-1}((1:\nu_{\iota+1}:\ldots :\nu_n))$ as follows.
Denote by ${\cal L}_0$ (respectively ${\cal L}_\nu$)
the subspace of linear forms generated by $L_{\iota+1},\ldots , L_n$
(respectively
$L_{\iota+1}-\nu_{\iota+1}L_0,\ldots , L_n-\nu_nL_0$) and compute the
intersection
${\cal L}'={\cal L}_0\cap{\cal L}_\nu$.
Let us construct applying Theorems~1~\cite{8}~and~2~\cite{8} the set
$S''_{i-1}=W_{i-1}(L')\cap{\cal Z}(L_\iota,\ldots , L_n)$.
If all $\nu_j=0$ then
$\varphi^{-1}_i((1:0:\ldots : 0))=S_i$.
So we shall suppose without loss of generality that not all $\nu_j$ are
zeroes, say, $\nu_{\iota+1}\ne 0$.
Let us construct
using the algorithm from \cite{2} the family $E_a$, $a\in A$, of all the
irreducible components
of dimension $2$ of the algebraic variety $\widetilde{V}_{i-1}\cap{\cal
Z}({\cal L}')$ such that
$E_a\cap S''_{i-1}\ne\emptyset$. Then
again using the algorithm from \cite{2} we construct the intersection
$E_a\cap{\cal Z}(L_{\iota+1}-\nu_{\iota+1}L_0)\cap{\cal Z}(e'_i)$.
The set
$\varphi_i^{-1}((1:\nu_{\iota+1}:\ldots :\nu_n))$ is the union of the last
intersections over all $a\in A$.


Let $\mu^{(i)}(z')=(a(z'),b(z'),c(z'))$.
Our aim is to construct a point $z'\in{\Bbb P}^{n-\iota}(\overline{k'})$
with maximal possible $\mu^{(i)}(z')$.
At first we shall obtain $z'$ such that $a(z')=\deg
W_{i-1}(L')\cap{\cal Z}(e'_i)$.
Let us describe a recursion by $z'$. We
take $z'=(1:0:\ldots : 0)$ as the base of this recursion.
We act similirly to the algorithm
for computing the degree from \cite{4}.
At the general step we
look for a point $z''$ which is infinitely close to
$z'$ and such that $a(z'')>a(z')$ (if $z''$ exists).
Namely, let us construct
all the points $x'\in\varphi^{-1}_i(z')$.
{\it Let $i=j$.} For every $x'$
applying Theorem~2 \cite{9} to the algebraic variety $(\mbox{\rm
con}(\widetilde{V}_{i-1})
\times\mbox{\rm con}(\widetilde{V}_{i-1}))(\overline{k'_2})$ we
decide
whether there is a pair $(x'',x''')\in (W_{i-1}(L')\times
W_{i-1}(L'))(\overline{k'_2})$ such that
$$
\left\{
\begin{array}{ll}
\sum_{0\le w\le n}|(X_w/L_0)(x'')-(X_w/L_0)(x')|^2
\le\varepsilon_1,&\\
\varepsilon_2\le\sum_{0\le w\le n}|(X_w/L_0)(x'')-(X_w/L_0)(x''')|^2
\le\varepsilon_1,&\\
e'_i(x'')=e'_j(x''')=0,&\\
\varphi^{(j)}(x'')=\varphi^{(j)}(x'''),
\end{array}
\right.
\eqno (39)
$$
cf. system~(2) from \cite{4}.
The equality $a(z')=\deg
W_{i-1}(L')\cap{\cal Z}(e'_i)$ does not hold if and only if
there are $x'',x'''$ for some $x'$.
In this case we choose such $x'$, set $z''=\varphi_i(x'')$, apply
the algorithm of reduction to integer coefficients to $z''\in{\Bbb
P}^{n-\iota}(\overline{k'})\setminus{\cal Z}(X_0)$, see Proposition~2
\cite{9},
and obtain a point $\widetilde{z}$ with integer coordinates
such that $a(\widetilde{z})>a(z')$. Hence
$\mu^{(i)}(\widetilde{z})>\mu^{(i)}(z')$.
Then we replace $z'$ by $\widetilde{z}$
and proceed to the next step of the recursion.

Finally we shall come to the case when
for all $x'$ there is no $(x'',x''')$.
Then $a=\deg W_{i-1}(L')\cap{\cal Z}(e'_i)$.
Hence, in particular,
the differential $d_{x'}\varphi_i$ is an isomorphism for every
$x'\in\varphi^{-1}_i(z')$,
cf. \cite{4} (we leave the details to the reader).
According to the B\'ezout theorem and the
algorithm of reduction to integer coefficients
$z'=(1:z'_{i,i-\sigma+s+1}:\ldots :
z'_{i,n})$ where all $z'_{i,j}$ are integers of length
$O(n\log d+(n-\sigma)\log d')$.



Now we are going to maximize $\#(\varphi_i^{-1}(z')\setminus{\cal
Z}(e_0,\ldots , e_r))$.
We enumerate the points $x\in\varphi_i^{-1}(z')\cap{\cal Z}(e_0,\ldots ,
e_r)$.
Consider the following system of polynomial equations and inequalities with
coefficients from $\overline{k'_2}$.
$$
\left\{\begin{array}{ll}
\sum_{0\le j\le r}|e_j|^2=\varepsilon_2,\\
\sum_{0\le i\le n}|X_i-(X_i/L_0)(x)|^2\le\varepsilon_1,\\
e'_i=0,\\
L_0=1.
\end{array}\right.
\eqno (40)
$$
Let us apply Theorem~2 \cite{9} to this system and the variety
$\mbox{\rm con}(V
\cap{\cal Z}(e'_{\sigma+1},\ldots ,$ $e'_{i-1}))$
and construct the set $S''\subset{\Bbb
A}^{n+1}(\overline{k'_2})$
from the statement of this theorem.
Denote by $S'''\subset{\Bbb P}^n(\overline{k'_2})$ the set corresponding
to
$S''$ by the natural projection ${\Bbb
A}^{n+1}\setminus\{0\}\rightarrow{\Bbb P}^n$.
Further, using Theorem~2 \cite{8} we decide whether there is a point $x^*\in
S'''\cap W_{i-1}(L')(\overline{k'_2})$.
Obviously such a point exists if and only if $x\in W_i(L')\cap
W^{(3)}_i(L')$.
Suppose that such a point exists.
Then set $z^*=\varphi_i(x^*)$.
Hence $x^*\in\varphi_i^{-1}(z^*)\setminus{\cal Z}(e_0,\ldots , e_r)$ and
for every $x_1\in \varphi_i^{-1}(z')\setminus{\cal Z}(e_0,\ldots , e_r)$
there
is $x^*_1\in
\varphi_i^{-1}(z^*)\setminus{\cal Z}(e_0,\ldots , e_r)$ which is
infinitely
close to $z_1$
by, e.g., Proposition~6 \cite{3}. Hence the number of elements
$\#(\varphi_i^{-1}(z^*)\setminus{\cal Z}(e_0,\ldots ,
e_r))>\#(\varphi_i^{-1}(z')\setminus{\cal Z}(e_0,\ldots , e_r))$ and,
therefore, $\mu^{(i)}(z^*)>\mu^{(i)}(z')$.
Using the algorithm of reduction to integer coefficients, see Proposition~2
\cite{9}, one can replace subsequently the coordinates of $z^*$
by integers (we find them by the enumeration)
of length $O(n\log d+(n-\sigma)\log(d'))$ and
obtain a new point $z''$ such that
$a(z'')=a(z^*)$, $b(z'')\ge b(z^*)>b(z')$. Hence,
$\mu^{(i)}(z'')\ge\mu^{(i)}(z^*)$. Now we replace $z'$ by $z''$
and proceed to the next step of the recursion.

If there is no point $x^*\in W_{i-1}(L')(\overline{k'_2})$ giving a solution
of (40) does not exists
and not all the points from $\varphi_i^{-1}(z')\setminus{\cal
Z}(e_0,\ldots
, e_r)$ are enumerated then
we proceed to the next point
$x\in\varphi_i^{-1}(z')\setminus{\cal Z}(e_0,\ldots , e_r)$. If all the
points from
$\varphi_i^{-1}(z')\setminus{\cal Z}(e_0,\ldots , e_r)$
are enumerated then we have obtained the required point
$z'$ such that $a(z')=\deg W_{i-1}(L')\cap{\cal Z}(e'_i)$ and $b(z')$ is
maximal possible.


\medskip Applying
the algorithm of reduction to integer coefficients, see
Proposition~2 \cite{9},
one can suppose additionally that the elements
$z'_{i,j}=z'_{i_1,j}$ for all $\sigma+1\le i_1<i\le n$, $i-\sigma+s+1\le
j\le n$,
i.e., $z'_{i,j}=z'_j$, $s+2\le j\le n$, does not depend on $i$.
Then for all $\sigma+1\le i\le n+1$
\setcounter{equation}{40} \begin{eqnarray}
&&W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal Z}(L_0,L_{\iota+1},\ldots ,
L_n)=\emptyset, \label{41}\\
&&\#(W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal
Z}(L_{\iota+1}-z'_{\iota+1}L_0,\ldots , L_n-z'_nL_0))=\nonumber\\
&&\deg (W_{i-1}(L')\cap{\cal Z}(e'_i)). \label{42}
\end{eqnarray}
Further, applying Lemma~3 and
the algorithm of reduction to integer coefficients we shall assume
without loss of generality that the linear forms $L_0,\ldots , L_{s+1},$
$L_{s+2}-z'_{s+2}L_0,\ldots , L_n-z'_nL_0,
L_{n+1}$ satisfies additionally assertion (a) in place of $L_0,\ldots ,
L_{n+1}$ (we leave the details to the reader).
Now replacing $L_j$ by $L_j-z'_jL_0$, $s+2\le j\le n$, we shall suppose in
what follows that $z'_j=0$ for all $s+2\le j\le n$.
Put $\overline{\lambda}=(L_0,\ldots , L_{n+1})$. Hence
$\overline{\lambda}$ satisfies condition (B) from the beginning of Section~1
for the algebraic variety
$W_{i-1}(L')\cap{\cal
Z}(e'_i)$ (in place of $V$) for every $\sigma+1\le i\le n+1$. Besides that,
now $a((1:0:\ldots : 0))$ and $b((1:0:\ldots : 0))$ are maximal
possible for every $\sigma+1\le i\le n+1$.

Recall that the integers $a_\iota(\overline{\lambda},W_{i-1}(L')\cap{\cal
Z}(e'_i))$ and
$b_\iota(W_{i-1}(L')\cap{\cal Z}(e'_i))$ are defined, see Section~1.
Till the end of the proof
we shall denote for brevity $a_\iota(\overline{\lambda})
=a_\iota(\overline{\lambda},W_{i-1}(L')\cap{\cal
Z}(e'_i))$, $b_\iota=b_\iota(W_{i-1}(L')\cap{\cal Z}(e'_i))$,
and use the similar notations for other elements, say,
$\lambda^*$ in place of $\overline{\lambda}$.
Now one can compute the integer
$a_\iota(\overline{\lambda})$ as follows.
Using  the algorithm from \cite{2} for every $\sigma+1\le i\le n+1$
we construct all the irreducible over $k'$
components of dimension $2$ of the intersection
$\widetilde{V}_{i-1}\cap{\cal Z}(L_{\iota+1},\ldots , L_{n-1})$.
By Theorem~1 \cite{8} (one can apply here also Theorem~3 in place of
Theorem~1 \cite{8})
and Theorem~2 \cite{8}
choose among them all the irreducible over $k'$ components
$E$ of $W_{i-1}(L')\cap{\cal Z}(L_{\iota+1},\ldots , L_{n-1})$.
Again applying the algorithm from \cite{2} we construct all the irreducible
over $k'$
components of $E\cap{\cal Z}(e'_i)$ for all $E$. These are all the
irreducible over $k'$ components of
$W_{i-1}(L')\cap{\cal Z}(e'_i)$.
By \cite{2} we construct also the separable
polynomials
$\Phi_{\iota,*}\in k'[Z_0,Z_n,Z_{n+1}]$, see Section~1, corresponding to the
algebraic variety $W_{i-1}(L')\cap{\cal Z}(e'_i)$ in place of $V$ from
Section~1
(notice here that all the irreducible components of
$W_{i-1}(L')\cap{\cal Z}(e'_i)$ have
the dimension $n-\iota$, and, therefore, we associate only one polynomial
$\Phi_{\iota,*}$ to
$W_{i-1}(L')\cap{\cal Z}(e'_i)$). After that we compute the discriminant
$R_{\iota,*}$ of
$\Phi_{\iota,*}$ with respect to $Z_{n+1}$,
its square free part $R_{\iota,*,0}$, and the integer
$a_\iota(\overline{\lambda})=
\deg R_{\iota,*,0}$ for every $\sigma+1\le i\le n+1$.
We shall denote also $\Phi_{\iota,*}=\Phi_{\overline{\lambda},\iota,*}$ and
$R_{\iota,*}=R_{\overline{\lambda},\iota,*}$ if the dependence on
$\overline{\lambda}$ is important, see
Section~1.

Let $1\le\tau\le 4$ be an integer.
Let $0\le i_1<\ldots < i_\tau\le n$ be integers.
Assume that the field $k'$ is supplied with a real structure, see \cite{3}.
Let the field
$k'_j=k'(\varepsilon_1,\ldots ,\varepsilon_j)$, $0\le j\in{\Bbb Z}$,
where $\varepsilon_j>0$ for $j\ge 1$ is an infinitesimal
with respect to the field $k'_{j-1}$.
Let $X_{w,j}$, $1\le w\le 4$, $0\le j\le n$;
$Z_{w,j}$, $j\in\{0,\iota+1,\ldots , n+1\}=I_\iota$, $1\le w\le \tau$,
be variables. Denote for brevity $\overline{L}_j=L_j+\sum_{1\le w\le
\tau}Z_{w,j}X_{i_w}$, $j\in I_\iota$
(so all $\overline{L}_j$ depend on the choice of $j_1,\ldots , j_\tau$).
Denote $X^{(j)}=(X_{j,0},\ldots , X_{j,n})$, and for an
arbitrary polynomial $c$ in
$n+1$ variables $c(X^{(j)})=c(X_{j,0},\ldots , X_{j,n})$, $1\le j\le 4$.
Assume that $\iota\le n-2$.
Consider the following system of polynomial equations and inequalities with
squares of absolute values with coefficients from $k'_3$
with respect to the variables $X_{w,j}$, $1\le w\le 4$, $0\le j\le n$; and
$Z_{w,j}$, $j\in\{0,\iota+1,\ldots , n+1\}=I_\iota$, $1\le w\le \tau$.
$$
\left\{\begin{array}{ll}
\sum_{j\in I_\iota}\sum_{1\le w\le \tau}|Z_{j,w}|^2\le\varepsilon_1,\\
\varepsilon_2\le |\overline{L}_n(X^{(1)})-\overline{L}_n(X^{(3)})|^2
\le \varepsilon_1,&\\
|\overline{L}_{n+1}(X^{(j)})-\overline{L}_{n+1}(X^{(j+1)})|^2=
\varepsilon_3,& j=1,3, \\
\overline{L}_\nu(X^{(j_1)})=\overline{L}_\nu(X^{(j_2)})=0,&
1\le j_1<j_2\le 4,\quad \iota+1\le\nu\le n-1,\\
\overline{L}_n(X^{(j)})-\overline{L}_n(X^{(j+1)})=0,& j=1,3,\\
\overline{L}_0(X^{(j)})-1=0,& 1\le j\le 4,\\
e'_i(X^{(j)})=0,& 1\le j\le 4.
\end{array}\right.
\eqno (43)
$$

\par\medskip\noindent{\bf LEMMA~26}\hspace{0.1em} {\it  Let
$\iota\le\min\{n-2,n+1-\sigma+s\}$.
The inequality
$a_\iota(\overline{\lambda})<
b_\iota$ holds if and only if
there exist $1\le\tau\le 4$ and $0\le i_1<\ldots < i_\tau\le n$ satisfying
the following property. System (43) has a solution
$X_{w,i}=x^*_{w,i}\in\overline{k'_3}$,
$1\le w\le 4$, $0\le i\le n$; $Z_{w,i}=z^*_{w,i}\in\overline{k'_3}$,
$i\in\{0,\iota+1,\ldots , n+1\}$, $1\le w\le \tau$,
such that
$$
(x^*_{w,0}:\ldots : x^*_{w,n})\in W_{i-1}(L')(\overline{k'_3}),\quad
1\le w\le 4.
\eqno (44)
$$
}\par\medskip

\noindent {\bf PROOF} \quad The proof of this lemma is similar to the one
of Lemma~4. The lemma is proved.

\medskip Let us return to the algorithm for assertion (b).
We shall describe a recursion on $\iota$.
The base of the recursion is $\iota=s+1$.
Let us describe the general step of this recursion.
Let $s+1\le\iota\le n-2$ be maximal such that for all $s+1\le w<\iota$ the
integers $b_w(W_{i-1}(L')\cap{\cal Z}(e'_i))$ are computed, and
$b_w(W_{i-1}(L')\cap{\cal Z}(e'_i))=
a_w(\overline{\lambda},W_{i-1}(L')\cap{\cal Z}(e'_i))$.
Let us construct, see above, the polynomials
$\Phi_{\overline{\lambda},\iota,*}$, $R_{\overline{\lambda},\iota,*}$ and
$R_{\overline{\lambda},\iota,*,0}$.
We shall suppose without loss of generality in what follows that $Z_0$ does
not divide
$R_{\overline{\lambda},\iota,*,0}$
replacing if it is necessary $Z_0$ by $Z_0-\mu Z_n$ for an appropriate
integer $\mu$
bounded from above by a polynomial in $d^n(d')^{n-\sigma}$.
The linear forms $L_0-\mu L_n$ satisfies the same properties as $L_0$ does,
see above.


Let us enumerate integers $\tau$ and $i_1,\ldots , i_\tau$ such that
$1\le\tau\le 4$ and $0\le
i_1<\ldots < i_\tau\le n$.
Using Theorem~2 \cite{9} to the algebraic variety $\mbox{\rm
con}(\widetilde{V}_{i-1})\subset{\Bbb A}^{n+1}(\overline{k'})$
and \cite{4} we decide whether system (43) has a
solution satisfying (44).
If it is true, we construct such a solution. If system (43)
does not have a solution satisfying (44) then we proceed to the next
$\tau$ and $i_1,\ldots , i_\tau$.

Suppose that system (43) does not have a solution satisfying (44)
for all $1\le\tau\le 4$ and
$0\le i_1<\ldots < i_\tau\le n$. Then
$a_\iota(\overline{\lambda})=b_\iota$
by Lemma~26.


If $\iota=\min\{n-2,n+1-\sigma+s\}$
the element $\overline{\lambda}$ satisfying
assertion of Theorem~3
is constructed. Suppose that $\iota<\min\{n-2,n+1-\sigma+s\}$.
Then we replace $\iota$ by $\iota+1$ and
proceed to
the next step of the described recursion.

Suppose that (43) has a solution satisfying (44)
for some $1\le\tau\le 4$ and $0\le i_1<\ldots < i_\tau\le n$.
Let us compute, see \cite{3},
$z_{w,j}=\mbox{\rm st}_{\varepsilon_3}(z^*_{w,j})$,
$1\le w\le \tau$, $j\in I_\iota$, where all $z^*_{w,j}$
are from the statement of Lemma~26.
Set $L^*_j=L_j$ for $j\in\{1,\ldots , \iota\}$, and $L^*_j=L_j+
\sum_{1\le w\le \tau}z_{w,j}X_{j_w}$ for $j\in I_\iota$. So $L^*_j$ are
linear forms with
coefficients from $\overline{k'_3}$.
Put $\lambda^*=(L^*_0,\ldots , L^*_n)$.
By Lemma~2 the element $\lambda^*$ satisfies property (B)
for all algebraic varieties $W_{i-1}(L')\cap{\cal Z}(e'_i)$,
$\sigma+1\le i\le n+1$.
From system (43) we get that the polynomial
$R_{\lambda^*,\iota,*}(1,Z_n)$ has more different roots than
$R_{\overline{\lambda},\iota,*}(1,Z_n)$.
So $a_\iota(\overline{\lambda})<
a_\iota(\lambda^*)$ and
$a_w(\overline{\lambda})=
a_w(\lambda^*)$
for $1\le w<\iota$. Applying Lemma~3 and the algorithm of reduction to
integer coefficients, see \cite{9}
Proposition~2, we replace subsequently all the coefficients of linear forms
$L^*_j$ by integers of length
$O(n\,\log d+(n-\sigma)\,\log d')$ and obtain new linear forms $L^{**}_j$,
$0\le j\le n+1$, satisfying the following properties.
All $L^{**}_j$,
$0\le j\le n+1$, have
integer coefficients of length $O(n\,\log d+(n-\sigma)\log d')$ and
satisfy condition (B)
for all algebraic varieties $W_{i-1}(L')\cap{\cal Z}(e'_i)$,
$\sigma+1\le i\le n+1$. Further, $L^{**}_j$,
$0\le j\le n+1$, satisfy assertion (a) of Theorem~1 with $L^*_j$ in
place of $L_j$, $0\le j\le n$, and
(denote $\lambda^{**}=(L^{**}_0,\ldots , L^{**}_{n+1})$)
$a_w(\lambda^{**})\ge a_w(\lambda^*)$
for all $1\le w\le \iota$. In particular,
$a_\iota(\lambda^{**})>a_\iota(\overline{\lambda})$.
Now we replace $\overline{\lambda}$ by $\lambda^{**}$ and proceed to the
next step of the described recursion.

Thus, finally we shall construct the element $\overline{\lambda}$ satisfying
assertions (a) and (b) of Theorem~1.
Notice that by the B\'ezout theorem the total number of steps of the
described recursion
is bounded from above by a polynomial in $d^n(d')^{n-\sigma}$.

\medskip Now any two points $y',y''\in W_{i-1}(L')\cap{\cal
Z}(e'_i)\cap{\cal
Z}(L_{\iota+1},\ldots , L_n)$
belong to the same irreducible over $k'$ (respectively irreducible over
$\overline{k'}$) component
of $W_{i-1}(L')\cap{\cal Z}(e'_i)$ if and only if they belong to the
same irreducible over $k'$ (respectively irreducible over $\overline{k'}$)
component
of the curve $W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal
Z}(L_{\iota+1},\ldots , L_{n-1})$.
We have constructed all the irreducible components of the last algebraic
variety, see above.
Hence one can decide whether $y'$ and $y''$
belong to the same irreducible over $k'$
(respectively irreducible over $\overline{k'}$) component
of the curve $W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal
Z}(L_{\iota+1},\ldots , L_{n-1})$. The details here
are similar to the ones from the proof of Theorem~2 \cite{8}.
Thus, one can compute the number of points $y''\in
W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal
Z}(L_{\iota+1},\ldots , L_n)$ such that $y'$ and $y''$ belong to the same
defined over $k'$ and irreducible over $k'$ component of
$W_{i-1}(L')\cap{\cal Z}(e'_i)$.

Let $E$ be an irreducible over $\overline{k'}$ component of
$W_{i-1}(L')\cap{\cal Z}(e'_i)$ such that $y'\in E$.
To avoid constructing composites $k'(y',y'')$, we compute at first by
\cite{2} the minimal field of definition $k''$ of
$E\cap{\cal Z}(L_{\iota+1},\ldots , L_{n-1})$ containing $k'$.
The field $k''$ coincides with
the minimal field of definition of $E$ containing $k'$, cf. \cite{8}.
After that we implement all the construction over the field $k''$,
i.e., assume that
all the linear forms $L_j\in k''[X_0,\ldots , X_n]$, $0\le j\le n+1$. This
allows to compute the number of points $y''\in
W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal
Z}(L_{\iota+1},\ldots , L_n)$ such that $y'$ and $y''$ belong to $E$.
In the similar way in (c) extending the ground field $k'$ till
$k''$, we reduce the case of an
irreducible over $\overline{k'}$ component $E$ to
the one of a defined over $k''$ and irreducible over $k''$ component,
i.e., for this reduction in (c)
it is sufficient to consider only the case when $E$ is
defined over $k'$ and irreducible over $k'$, and after that
replace $k'$ by $k''$.
In the similar way in (f) extending the ground field $k'$ till
$k''$, we can suppose that $E$ is defined over
the ground field. So in (f) we shall assume without loss of generality that
$E$ is defined over $k'$ and irreducible over $\overline{k'}$.
The algorithms for assertions (a) and (b) are described completely.

\par\medskip\noindent{\bf LEMMA~27}\hspace{0.1em} {\it  Let $\sigma+1\le
i\le n+1$ be an integer.
Let $k^*$ be a finitely generated extension of $k'$ given by the minimal
polynomial of a primitive element
of $k^*$ over a purely trancendental extension of $k'$.
Let $\lambda^*=(L^*_0,\ldots , L^*_{n+1})$, where all
$L^*_j\in k'[X_0,\ldots , X_n]$ are linear forms. Then one can
decide whether the intersection of $\widetilde{V}_{i-1}$ and ${\cal
Z}(L^*_\iota,\ldots , L^*_n)$ is transversal.
Suppose that the intersection $\widetilde{V}_{i-1}$ and ${\cal
Z}(L^*_\iota,\ldots
, L^*_n)$ is transversal. Then the number of elements
$\#(W_{i-1}(L')\cap{\cal Z}(e'_i)
\cap{\cal Z}(L^*_\iota,\ldots , L^*_n))<+\infty$, and one can
construct all the points from $W_{i-1}(L')\cap{\cal Z}(e'_i)\cap
{\cal Z}(L^*_\iota,\ldots , L^*_n)$ in ${\Bbb P}^n(\overline{k'})$.
Further, one can decide
whether the intersection of $W_{i-1}(L')\cap{\cal Z}(e'_i)$ and
${\cal Z}(L^*_{\iota+1},\ldots , L^*_n)$
is transversal.
Hence one can decide whether assertions (a) and (b) hold for $\lambda^*$ in
place of $\overline{\lambda}$.
The working time of the algorithm from this lemma 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 size ${\rm L}(L')$ of the element $L'$,
and the size ${\rm L}(\lambda^*)$ of the element $\lambda^*$
(including the sizes of the minimal
polynomials of the primitive elements of $k'$ and $k^*$).
}\par\medskip

\noindent{\bf PROOF}\quad By (a) and (b) we know the degrees of all the
irreducible
components of the algebraic varieties
$\widetilde{V}_{i-1}$ and $W_{i-1}(L')\cap{\cal Z}(e'_i)$. Hence applying
Theorem~3 and Theorem~2 \cite{8} one can decide whether the intersection of
$\widetilde{V}_{i-1}$
and ${\cal Z}(L^*_\iota,\ldots , L^*_n)$
is transversal (we leave the details
to the reader). Assume that the last intersection is transversal. Then
replacing in the proof of (b) $L_j$ by $L^*_j$, $\iota+1\le j\le n$, we see
that
$W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal Z}(L^*_{\iota+1},\ldots ,
L^*_n)=A^*$ is a finite set
in ${\Bbb P}^n(\overline{k'})$, and construct this set in the similar way.
Now the intersection of $W_{i-1}(L')\cap{\cal Z}(e'_i)$ and
${\cal Z}(L^*_{\iota+1},\ldots , L^*_n)$ is transversal if and only if
the number of elements $\#A^*=\deg
W_{i-1}(L')\cap{\cal Z}(e'_i)$. The lemma is proved.

\medskip To describe the algorithm for (c)
it is sufficient to
consider the case when
$E$ is defined over $k'$ and irreducible over $k'$, see
the end of the proof of (b).
Replacing the linear form $L_0$ for a new one we shall suppose in what
follows in the proof that
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item $L_0(x)\ne 0$,
\item the linear forms $L_0,\ldots , L_{n+1}$ satisfy assertions
(a), (b) of Theorem~1,
\item $L_0,X_1,\ldots , X_n$ are linearly independent over $k'$.
\end{enumerate}
Further, applying Lemma~3 and Lemma~27
one can replace the linear forms $L_1,\ldots ,
L_{n+1}$ for
new ones such that additionally
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\setcounter{enumi}{3}
\item all $L_j$, $\iota+1\le j\le n$, are linear forms in $L_0,X_1,\ldots ,
X_n$ with integer coefficients of length $O(n\log d+(n-\sigma)\log(d'))$.
\end{enumerate}

Let ${\cal L}$ be the subspace of all
linear forms from $k'(x)[X_0,\ldots , X_n]$ vanishing at $x$.
The basis of ${\cal L}$ is $X_i-(X_i/L_0)(x)L_0$, $1\le i\le n$.
Set $L''_j=L_j-(L_j/L_0)(x)L_0\in{\cal L}$,
$\iota+1\le j\le n$. For every $1\le i\le n+1$
let us represent $L_j=\widetilde{L}_j+\mu_j L_0$, where
$\widetilde{L}_j\in k[X_1,\ldots , X_n]$, and $\mu_j$ is an integer.
Then obviously $L''_j=\widetilde{L}_j(X_1-(X_1/L_0)(x)L_0,\ldots ,
X_n-(X_n/L_0)(x)L_0)$.

To compute the multiplicities we describe an
algorithm recursive by $L_0,\ldots ,$ $L_{n+1}$.
We take the constructed $L_0,\ldots , L_{n+1}$
satisfying (i)--(iv) as the base
of this recursion.
At the end of every step which is not final we get
new $L_0,\ldots , L_{n+1}$ such that such that the number of elements
$W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal Z}(L''_{\iota+1},\ldots ,
L''_n)$ is greater than
at the beginning of the step (for old $L_0,\ldots , L_{n+1}$).
At the end of every step (and hence, if the considered step is not final,
at the beginning of the next step) of this recursion conditions (i)--(iv)
hold.


For an arbitrary quasiprojective
algebraic variety $E$ and its point $a$
denote by $\mbox{\rm con}(a,E)$ the tangent cone of
$E$ at the point $x$, see e.g., \cite{4}.
Consider the condition
$$
\mbox{\rm
con}(x,W_{i-1}(L')\cap{\cal Z}(e'_i))
\cap{\cal Z}(L''_{\iota+1},\ldots ,L''_n)=\{x\}.
\eqno (45)
$$
At the end of the recursion
condition (45) will be fulfilled.

Let us describe the general step of the recursion.
We follow the algorithm for computing the multiplicity from \cite{4}.
The only difference here is that the algebraic variety
$W_{i-1}(L')\cap{\cal Z}(e'_i)$
is given in a slightly different form in comparison with \cite{4}.
But still using Theorem~2 \cite{9} we have a certificate to decide
whether a point $x\in W_{i-1}(L')$. This allows to
describe the analog of the algorithm from \cite{4}.
Recall, see the beginning of the section, that for any $\nu_{\iota+1},\ldots
,\nu_n\in\overline{k'}$
we can compute $\varphi_i^{-1}((1:\nu_{\iota+1}:\ldots :\nu_n))$.
Let us compute in this way all the points from
$W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal
Z}(L''_{\iota+1},\ldots , L''_n)=\varphi_i^{-1}(\varphi_i(x))$.
We shall increase recursively the number of elements
$W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal
Z}(L''_{\iota+1},\ldots , L''_n)$.
If it is not possible then $L''_{\iota+1},\ldots , L''_n$ satisfies (45).

Namely, let ${\Bbb A}^{2n+1}(\overline{k'})$ has the coordinates functions
$X_0,\ldots , X_n$, $Y_1,\ldots , Y_n$.
Let us construct the homogeneous polynomial $\widetilde{e}_i\in
k'(x)[X_0,\ldots , X_n]$
such that $\widetilde{e}_i(L_0,$ $X_1,\ldots , X_n)=e'_i$.
Now in place of system (15) from \cite{4}
consider the system of polynomial
equations and inequalities
with squares of absolute values
in $2n+1$ variables with coefficients from $\overline{k'_2}$
$$
\left\{
\begin{array}{ll}
\widetilde{e}_i(1,X_1-(X_1/L_0)(x),\ldots ,X_n-(X_n/L_0)(x))=0,&\\
 \sum_{1\le i\le n}|X_i-(X_i/L_0)(x)|^2= \varepsilon_2^2, &  \\
 \widetilde{L}_j(Y_1-(X_1/L_0)(x),\ldots ,Y_n-(X_n/L_0)(x))=0, & \iota+1\le
j\le n, \\
 \sum_{1\le i\le n}|X_i-Y_i|^2\le\varepsilon_1\varepsilon_2^2,\\
L_0-1=0.
\end{array}
\right.
\eqno (46)
$$
Then, cf. Lemma~8 \cite{4},
this system has a solution $(x^*_0,\ldots , x^*_n,y^*_1,\ldots , y^*_n)\in
{\Bbb A}^{2n+1}(\overline{k'_2})$ such that
$$
(x^*_0:x^*_1:\ldots : x^*_n)\in W_{i-1}(L')
\eqno (47)
$$
if and only if condition (45) does not hold.
One can apply Theorem~2 \cite{9} and Theorem~2 \cite{8} to the algebraic
variety
$(\mbox{\rm
con}(\widetilde{V}_{i-1})
\times{\Bbb A}^n)(\overline{k'_2})\subset{\Bbb
A}^{2n+1}(\overline{k'})$
and construct a solution of system (46) satisfying (47) if it exists.

Suppose that a solution of (46) satisfying (47) is constructed.
Then similarly to \cite{4} one can construct linear forms $L^*_j\in
\overline{k'_2}[X_1,\ldots
,X_n]$, $\iota+1\le j\le n$, such that
$L^*_j(X_1-(X_1/L_0)(x),\ldots ,X_n-(X_n/L_0)(x))=0$,
and the coefficients of the linear forms $L^*_j$ are infinitely close
with respect to the field $k'$ to the
corresponding ones of $\widetilde{L}_j$ for all $\iota+1\le j\le n$.
Notice that the number of elements $\#(W_{i-1}(L')\cap{\cal
Z}(e'_i)\cap{\cal Z}(L^*_{\iota+1},\ldots , L^*_n)>
\#\varphi_i^{-1}\varphi_i(x)$.
Denote by $k^*$ the field generated over $k'$
by the coefficients of linear forms $L^*_j$, $\iota+1\le j\le n$.
%Set the linear forms $L^*_j=\widetilde{L}_j$,
%and the integers $\mu_j=0 for all $j\in\{1,\ldots , \iota,n+1\}$.
Put $\lambda^*=(L_0,L_1\ldots ,L_\iota,
L^*_{\iota+1}+\mu_{\iota+1}L_0,\ldots , L^*_n+\mu_nL_0,L_{n+1})$.
Hence $\lambda^*$ is infinitely close to
$\overline{\lambda}$.
Therefore, by Lemma~2 assertions (a) and (b) of Theorem~1 hold for
$\lambda^*$ and $k^*$ in place of $\overline{\lambda}$ and $k'$. Hence by
Lemma~3,
the B\'ezout theorem and Lemma~27 applying the algorithm of reduction to
integer coefficients, one can replace subsequently
the coefficients of linear forms $L^*_j+\mu_jL_0$, $\iota+1\le j\le n+1$,
in the basis $L_0,X_1,\ldots , X_n$, and obtain new linear forms $L^{**}_j$,
$\iota+1\le j\le n+1$, satisfying the following properties. Put
$L^{**}_j=L_j$
for all $j\in\{1,\ldots , \iota,n+1\}$ and $\lambda^{**}=(L^{**}_0,\ldots ,
L^{**}_{n+1})$.
Then conditions (i)--(iv) are fulfilled for $\lambda^{**}$ in place of
$\overline{\lambda}$ and the number of elements
$W_{i-1}(L')\cap{\cal Z}(e'_i)\cap{\cal Z}(L^{**}_{\iota+1},\ldots ,
L^{**}_n)\ge
\#(W_{i-1}(L')\cap{\cal
Z}(e'_i)\cap{\cal Z}(L^*_{\iota+1},\ldots ,
L^*_n)>\#\varphi_i^{-1}\varphi_i(x)$.
Now we replace $\overline{\lambda}$ by $\lambda^{**}$ and proceed to the
next step of the recursion.

Thus, finally we shall come to the case when
system (46) does not have a solution satisfying (47).
Hence, see above, condition (45) holds.
Then, cf. Lemma~18 from \cite{4}, the multiplicity $\mu(x,E)$
is the number of the points $x^*\in S^*$ such that $x^*\in E$.
Therefore, in this case one can compute the
required multiplicity by (b). The algorithm for (c) is described
completely.

\medskip Now assertion (g) follows from Lemma~3 and
Lemma~27 immediately.


At present we are able to compute $\mu^{(i)}(z')$ for a given point
$z'\in{\Bbb P}^{n-\iota}(\overline{k})$
(and for any point with coordinates from an extension of $\overline{k}$,
when we apply the algorithm of reduction to integer coefficients in the
recursion step, see below).
Let us return to the construction of the
point $z'\in{\Bbb P}^{n-\iota}(\overline{k})$ with maximal possible
$\mu^{(i)}(z')$ (recall that we perform the construction for all
$\sigma+1\le i\le n+1$).
Let us describe a recursion by $z'$. We
take $z'=(1:0:\ldots : 0)$ as the base of this recursion.
At the general step we
look for a point $z''$ which is infinitely close to
$z'$ and such that $\mu^{(i)}(z'')>\mu^{(i)}(z')$ (if $z''$ exists).
Namely, let us construct
all the points $x'\in\varphi^{-1}_i(z')$.
For every $x'$ for every $\sigma+1\le j<i$
applying Theorem~2 \cite{9} to the algebraic variety $(\mbox{\rm
con}(\widetilde{V}_{i-1})
\times\mbox{\rm con}(\widetilde{V}_{j-1}))(\overline{k'_2})$ we
decide
whether there is a pair $(x'',x''')\in (W_{i-1}(L')\times
W_{j-1}(L'))(\overline{k'_2})$ such that
(39) holds,
cf. system~(2) from \cite{4}.
The element $\mu^{(i)}(z')$ is not maximal possible if and only if
there are $x'',x'''$ for some $x'$ and $j$.
In this case we choose such $x'$ and $j$ and  set $z''=\varphi_i(x'')$.
It is not difficult to see using e.g., \cite{4} that for any line ${\mathfrak
l}\subset{\Bbb
P}^{n-\iota}(\overline{k})$ and for any point  $z^*\in{\mathfrak l}$ there
is only a polynomial in $d^n(d')^{n-\sigma}$
number of points $z^{**}\in{\mathfrak l}$ with
$\mu^{(i)}(z^{**})<\mu^{(i)}(z^*)$, cf. also the proof of Lemma~1 \cite{7}.
Hence we apply
the algorithm of reduction to integer coefficients to $z''\in{\Bbb
P}^{n-\iota}(\overline{k'})\setminus{\cal Z}(X_0)$, see Proposition~2
\cite{9},
and obtain a point $\widetilde{z}$ with integer coordinates
such that $\mu^{(i)}(\widetilde{z})>\mu^{(i)}(z')$.
Then we replace $z'$ by $\widetilde{z}$
and proceed to the next step of the recursion.

Finally we shall come to the case when
for all $x'$ and $j$ there is no $(x'',x''')$.
Then $\mu^{(i)}(z')$ is maximal possible.
According to the B\'ezout theorem and the
algorithm of reduction to integer coefficients
$z'=(1:z'_{i,i-\sigma+s+1}:\ldots :
z'_{i,n})$ where all $z'_{i,j}$ are integers of length
$O(n\log d+(n-\sigma)\log d')$.

\medskip Applying
the algorithm of reduction to integer coefficients, see
Proposition~2 \cite{9},
one can suppose additionally that the elements
$z'_{i,j}=z'_{i_1,j}$ for all $\sigma+1\le i_1<i\le n$, $i-\sigma+s+1\le
j\le n$,
i.e., $z'_{i,j}=z'_j$, $s+2\le j\le n$, do not depend on $i$.
Further, applying assertion (g) (proved above) and
the algorithm of reduction to integer coefficients we shall assume
without loss of generality that the linear forms $L_0,\ldots , L_{s+1},
L_{s+2}-z'_{s+2}L_0,\ldots , L_n-z'_nL_0,
L_{n+1}$ satisfies additionally assertions from (a) and (b)
(in place of $L_0,\ldots ,
L_{n+1}$; we leave the details to the reader).
Replacing $L_j$ by $L_j-z'_jL_0$, $s+2\le j\le n$, we shall suppose in
what follows that $z'_j=0$ for all $s+2\le j\le n$.
Put $\overline{\lambda}=(L_0,\ldots , L_{n+1})$. Hence now
$\overline{\lambda}$ satisfies assertions (a) and (b), and
$\mu^{(i)}((1:0:\ldots : 0))$ is maximal possible
for every $\sigma+1\le i\le n+1$.



\medskip Now we are going to prove (d).
Assume that a point $x\in{\Bbb P}^n(\overline{k'})$ is given.
Recall that $z'=(1:0:\ldots : 0)\in{\Bbb
P}^{n-\iota}(\overline{k'})$.
Let us construct a line ${\mathfrak l}_x\subset{\Bbb
P}^{n-\iota}(\overline{k'})$ such that the points
$z',\varphi_i(x)\in{\mathfrak l}_x$. Let us construct a point
$z^*\in{\mathfrak
l}_x(\overline{k'(\varepsilon_1)})$ (here $\varepsilon_1$ is an infinitesimal
with respect to the field $k$) that is infinitely close to
$\varphi_i(x)$. Further
let us construct using the algorithms from \cite{8}, \cite{9} the subset
$S^*\subset
\varphi_i^{-1}(z^*)$ of all the points that are infinitely close to $x$.

The number $\mu^{(i)}(z^*)=
\mu^{(i)}(z')$ is maximal possible. This
immediately implies the following assertions.
The equality $\varphi_i^{-1}(z^*)\setminus{\cal Z}(e_0,\ldots ,$
$e_r)=\varphi_i^{-1}(z^*)\cap(W_i(L')\cup W^{(3)}_i(L'))$ holds.
For an arbitrary irreducible over $\overline{k'}$
component $E$ of $W_{i-1}(L')\cap{\cal
Z}(e'_i)$ the point $x\in E$
if and only if there is $x^*\in S^*$ such that $x^*\in E$.
Besides that, if $E\not\subset{\cal Z}(e_0,\ldots , e_r)$ then $x^*\in
E\setminus{\cal Z}(e_0,\ldots , e_r)$.
Further, every point $x^*\in S^*$ belongs to only one
irreducible component $E$ of $W_{i-1}(L')\cap{\cal Z}(e'_i)$. If $x^*\in
E$ and $x^*\in
E_1$ for some irreducible component
of $W_{j-1}\cap{\cal Z}(e'_j)$
with $\sigma\le j<i$ then $E\subset E_1$.

Now obviously
\begin{itemize}
\item $x^*\in W^{(3)}_i(L')$ if and only if $x^*\not\in W_i(L')$ and
$x^*\not\in{\cal Z}(e_0,\ldots , e_r)$.
\end{itemize}
Therefore, one can decide whether $x\in W^{(3)}_i(L')$. Further,
\begin{itemize}
\item $x^*\in W^{(1)}_i(L')\cup W^{(2)}_i(L')$ if and only if
$x^*\in{\cal Z}(e_0,\ldots , e_r)$.
\end{itemize}
Hence, one can decide whether $x\in W^{(1)}_i(L')\cup W^{(2)}_i(L')$.
Since $i$ is arbitrary one can also decide for every $\sigma\le j<i$ whether
$x\in
W^{(1)}_j(L')\cup W^{(2)}_j(L')$. Extending the ground field and replacing
$x$ by $x^*$
one can decide for every $x^*\in S^*$ for every $\sigma\le
j<i$ whether $x^*\in
W^{(1)}_j(L')\cup W^{(2)}_j(L')$. Now notice that
\begin{itemize}
\item $x^*\in W^{(2)}_i(L')$ if and only if $x^*\in W^{(1)}_i(L')\cup
W^{(2)}_i(L')$ and
$x^*\in W^{(1)}_j(L')\cup W^{(2)}_j(L')$ for some $\sigma\le j<i$,
\item $x^*\in W^{(1)}_i(L')$ if and only if $x^*\in W^{(1)}_i(L')\cup
W^{(2)}_i(L')$ and
$x^*\not\in W^{(1)}_j(L')\cup W^{(2)}_j(L')$ for every $\sigma\le j<i$.
\end{itemize}
Thus, one can decide whether $x$ is an element of $W_i(L')$,
$W^{(1)}_i(L')$, $W^{(2)}_i(L')$, $W^{(3)}_i(L')$, $\sigma\le i\le n+1$.

\par\medskip\noindent{\bf REMARK~9}\hspace{0.1em} {\it  Let $E$ be an
arbitrary irreducible component
of $W^{(3)}_i(L')$ such that $x^*\in E$.
Actually the described construction allows to decide also whether
there is an irreducible component $E_1$ of
$W^{(3)}_j(L')$ with $\sigma+1\le j<i$ such that $E\subset E_1$.
}\par\medskip



Applying the described algorithm to the points from $\varphi_i^{-1}(z')$ we
compute $\deg W_i(L')=\#(\varphi_i^{-1}(z')\cap W_i(L'))$ and
$\deg W^{(j)}_i(L')=\#(\varphi_i^{-1}(z')\cap W^{(j)}_i(L'))$ for $j=1,2,3$.
The algorithm for (d) is described.

Assertion (e) follows from (b), (c) and (d) immediately.

Let us prove (f).
It is sufficient to consider only the case when $E$ is defined over $k'$ and
irreducible over $\overline{k'}$,
see the end of the proof of assertion (b). So we shall assume that $E$ is
defined over $k'$. Let $\overline{\lambda}$ be
the element constructed in (b).
Let us construct homogeneous polynomials $e''_i,e''_0
\in k[X_0,\ldots , X_n$ such that
$\deg e''_i=\deg e''_0=d'$ and the number of elements
\begin{eqnarray*}
&&\#(W_{i-1}(L')\cap{\cal Z}(e''_i)\cap{\cal Z}(L_{\iota+1},\ldots ,
L_n)\setminus({\cal Z}(e'_{i+1}
\cup\overline{V\setminus W})=\\
&&\#(e''_i/e''_0)(W_{i-1}(L')\cap{\cal Z}(e''_i)
\cap{\cal Z}(L_{\iota+1},\ldots , L_n))=D_{i-1}(L')d'.
\end{eqnarray*}
This can be done directly, but we use Theorem~1 \cite{9}.
Namely, let $m_0,\ldots , m_\nu$, $\nu={n+d'\choose n}-1$, be all the
monomials in
$X_0,\ldots , X_n$ of degree $d'$.
Consider the morphism  $\omega\, :\, W_{i-1}(L')\cap{\cal
Z}(L_{\iota+1},\ldots ,
L_n)\rightarrow{\Bbb P}^\nu(\overline{k})$, $(X_0,\ldots ,
X_n)\rightarrow(m_0:\ldots : m_\nu)$,
i.e., $\omega$ is the restriction of the Veronese morphism of degree $d'$ to
$W_{i-1}(L')\cap{\cal Z}(L_{\iota+1},\ldots ,
L_n)$.
We construct $e''_i,e''_0$ applying Theorem~1 \cite{9} to
$\widetilde{V}_{i-1}\cap{\cal Z}(L_{\iota+1},\ldots ,
L_n)$, $W_{i-1}(L')\cap{\cal Z}(L_{\iota+1},\ldots ,
L_n)$ and $\omega$ in place of $V$, $W$ and $p$ respectively.

Set $\widetilde{e}_i=e'_i+\varepsilon e''_i$, where $\varepsilon>0$ is an
infinitesimal with respect to the field $k'$.
Let us extend the ground field $k'$ till $k'(\varepsilon)$.
Now the mapping of the standard part $\mbox{\rm st}_\varepsilon\,
:\,{\Bbb P}^n(\overline{k(\varepsilon)})\rightarrow{\Bbb
P}^n(\overline{k})$ is
defined.
�y Lemma~7 the set $W_{i-1}(L')\cap{\cal
Z}(L_{\iota+1},\ldots , L_n)\cap{\cal Z}(\widetilde{e}_i)$ is finite in
${\Bbb P}^n(\overline{k(\varepsilon)})$.
Applying (b) over the ground field $k'(\varepsilon)$ we construct all the
points (up to conjugation
over $k'(\varepsilon)$; one can compute the number of elements of each class
of conjugated over $k'(\varepsilon)$
points according to (b), see \cite{8} for details) of the finite set
$\widetilde{I}=W_{i-1}(L')\cap{\cal
Z}(L_{\iota+1},\ldots , L_n)\cap{\cal Z}(\widetilde{e}_i)\subset{\Bbb
P}^n(\overline{k(\varepsilon)})$. Denote by $\widetilde{I}_{y'}$ the
subset of all
the points
$y\in \widetilde{I}$ such that $\mbox{\rm
st}_\varepsilon(y)=y'$,
i.e., all the points $y$
which are infinitely close to $y'$ with respect to the field $k'$. Computing
the standard parts of the
points from $\widetilde{I}$ we construct the set $\widetilde{I}_{y'}$
(again the
points from
$\widetilde{I}_{y'}$ are given up to a conjugation over $k'(\varepsilon)$;
recall
also
that the standard parts are computed using Newton--Puiseux expansions, see
\cite{3}), and the number of elements
$\#\widetilde{I}_{y'}$. Now assertion (f) follows from the following lemma.


\par\medskip\noindent{\bf LEMMA~28}\hspace{0.1em} {\it  The index of
intersection $i(W_{i-1}(L'),{\cal
Z}(e'_i);E)=\#\widetilde{I}_{y'}$.
}\par\medskip

\noindent{\bf PROOF}\quad Denote for brevity ${\cal L}={\cal
Z}(L_{\iota+1},\ldots , L_n)\subset{\Bbb P}^n(\overline{k})$.
The intersection $W_{i-1}(L')\cap{\cal Z}(e'_i)$ and ${\cal L}$ is
transversal.
The point $y'\in E\cap{\cal L}$ is a
smooth point of $W_{i-1}(L')\cap{\cal L}$. Hence
there is only one irreducible
over $\overline{k'}$ component $E'$ of $W_{i-1}(L')\cap{\cal L}$ such
that $y'\in E'$. Now by the property
of indices of intersection corresponding to the associativity of intersection
\begin{eqnarray*}
&&i(W_{i-1}(L'),{\cal Z}(e'_i);E)=i(W_{i-1}(L'),{\cal
Z}(e'_i);E)i(E,{\cal L};y')=\\
&&i(E',{\cal Z}(e'_i);y')i(W_{i-1},{\cal L};E')=i(E',{\cal
Z}(e'_i);y')=
i(W_{i-1}(L')\cap{\cal L},{\cal Z}(e'_i);y').
\end{eqnarray*}
But $\widetilde{e}_i$ is infinitely close to to $e'_i$, and the number of
elements
$\#(W_{i-1}(L')\cap{\cal L}\cap{\cal
Z}(\widetilde{e}_i))=D_{i-1}(L')d'$.
It is known that in this case
$i(W_{i-1}(L')\cap{\cal L},{\cal Z}(e'_i);y')=\#\widetilde{I}_{y'}$,
see e.g., \cite{12}. The lemma is proved.

\medskip Assertion (f) is proved.
The required estimations for the working
times of the algorithms constructed in this section
follow immediately from the ones of the applied algorithms.
Theorem~1 is proved completely.









\section{Construction of an element from ${\cal U}'''_{n+1}$}\label{s6}

We shall use the notations from the Introduction. Let
$L'=(L'_0,L'_{\sigma+1},\ldots , L'_n)\in{\cal U}'_0$ be given,
e.g., the element construct in Section~4.
We shall assume that all linear forms $L'_j$ have
coefficients from $k$. The element $L'$
will be changed recursively in the algorithm. Our aim now is to get
at the end of the
recursion an element $L'\in{\cal U}''_0$.


Recall that
$e'_j=L'_j(e_0,\ldots , e_r)$, $j\in\{0,\sigma+1,\ldots , n\}$.
Let an integer $\sigma+1\le i\le n-1$ be given.
This integer $i$ will be changed
recursively (it is a second recursion).
We take $i=\sigma+1$ as the base of this recursion.
The recursive assumption here is that additionally $(L'_{\sigma+1},\ldots ,
L_{i-1})\in{\cal U}''_{i-1}$.
Let the algebraic varieties $\widetilde{V}_j$, $\sigma\le j\le n$,
and linear forms $L_0,\ldots , L_{n+1}$ be
from the statement of Theorem~1. Put
$S_i=W_{i-1}(L')\cap{\cal Z}(L_{\iota+1},\ldots , L_n)\cap{\cal
Z}(e'_i)=\varphi_i^{-1}(z')$, where $z'=(1:0:\ldots : 0)\in{\Bbb
P}^{n-\iota}(\overline{k})$, $\iota=i-\sigma+s$, see Section~5.
Let us construct $L_0,\ldots ,L_{n+1}$ and $S_i$, see Section~5.
Similarly to Section~5 we
suppose without loss of generality that
$L_0(x^{(0)})\ne 0$. We have $S_i\cap{\cal Z}(L_0)=\emptyset$,
 $\deg(W_{i-1}\cap{\cal
Z}(e'_i))=\#\varphi_i^{-1}(z')$.
Set $S^{(0)}_i=\varphi_i^{-1}(z')$ and
$S^{(1)}_i=\varphi_i^{-1}(z')\cap{\cal Z}(e_0,\ldots , e_r)$.
Therefore, by assertion (d) of Theorem~1 the sets
$S^{(1)}_i=\varphi_i^{-1}(z')\cap(W^{(1)}_i(L')\cup
W^{(2)}_i(L'))$
and $\varphi_i^{-1}(z')\setminus S^{(1)}_i=\varphi_i^{-1}(z')\cap(W_i(L')\cup
W^{(3)}_i(L'))$.

As previously we shall assume that the field $\overline{k}$ is supplied with
a real
structure. Let the field $k_j=k(\varepsilon_1,\ldots , \varepsilon_j)$
be as above, see Section~4. Recall that the mapping of the standard part
$\mbox{\rm st}_j\, :\,
{\Bbb P}^n(\overline{k_j})\rightarrow{\Bbb P}^n(\overline{k})$
is defined. We assume that $k_j$ is linearly disjoint with $k_v$ over $k$.
Let $z=(z_0:\ldots : z_n)\in S^{(1)}_i$
where all $z_j\in\overline{k}$. Let
$0\le\alpha\le r$ be an integer. Consider the system of equations and
inequalities with respect to the variables
$X_0,\ldots , X_n$, $Y_{\sigma+1},\ldots , Y_i$
with coefficients from the field $\overline{k_2}$
$$
\left\{
\begin{array}{ll}
e'_j-Y_je_\alpha=0,& \sigma+1\le j\le i,\\
\varepsilon_2\le\sum_{0\le j\le r}|e_j|^2\le\varepsilon_1,\\
\sum_{\sigma+1\le j\le i}|Y_j|^2\le\varepsilon_1\\
\sum_{0\le i\le n}|X_i-(X_i/L_0)(z)|^2\le\varepsilon_1,\\
L_0=1.
\end{array}
\right.
\eqno (48)
$$

\par\medskip\noindent{\bf LEMMA~29}\hspace{0.1em} {\it  Let $L'\in{\cal
U}'_0$ and
$(L'_{\sigma+1},\ldots , L'_{i-1})\in{\cal U}''_{i-1}$.
Suppose that $W^{(3)}_i(L')$ $=\emptyset$. Then
$D_i>D(L'_{\sigma+1},$ $\ldots , L'_i)$ if and only if there are
$z=(z_0:\ldots : z_n)\in S^{(1)}_i$ and
an integer $0\le\alpha\le r$ such that system (48) has a solution
$(x^*_0,\ldots , x^*_n,\,y^*_{\sigma+1},\ldots , y^*_i)
\in{\Bbb A}^{n-\sigma+i+1}(\overline{k}_2)$ with
$x^*=(x^*_0:\ldots : x^*_n)\in W(\overline{k}_2)$.
More precisely, for the considered solution put
\setcounter{equation}{48} \begin{eqnarray}
&&L^*_j=L'_j-y^*_jX_\alpha,\quad
e^*_j=L^*_j(e_0,\ldots , e_r),
\quad \sigma+1\le j\le i, \label{49} \\
&&L^*_j=L'_j,\quad j\in\{0,i+1,\ldots , n\},\quad
L^*=(L^*_0,L^*_{\sigma+1},\ldots , L^*_n).\label{50}
\end{eqnarray}
Then $L^*\in{\cal U}'_0$, $(L^*_{\sigma+1},\ldots , L^*_{i-1})
\in{\cal U}'_{i-1}$,
$x^*\in W_i(L^*)\cup W^{(3)}_i(L^*)$ and $D_i(L^*)>D_i(L')$, see Remark~2.
}\par\medskip

\noindent {\bf PROOF} \quad
Let $D_i>D(L'_{\sigma+1},\ldots , L'_i)$. There exists  $\varepsilon>0$ such
that $\varepsilon$ is an infinitesimal with respect to
the field $k_1$ and $\varepsilon_2$ is an infinitesimal with respect to the
field $k_1(\varepsilon)$. Let us extend the ground field $k$
till $k_2(\varepsilon)$.
Since ${\cal U}''_i$ is open in the
Zariski topology there are linear forms
$\widetilde{L}_{\sigma+1},\ldots ,\widetilde{L}_i\in
\overline{k(\varepsilon)}[X_0,\ldots , X_n]$ such that
$(\widetilde{L}_{\sigma+1},\ldots ,\widetilde{L}_i)\in{\cal
U}''_i$,
and all the nonzero coefficients of
the linear forms $\widetilde{L}_j-L'_j$, $\sigma+1\le j\le i$ are
infinitesimal
with respect to the field $k$.

Set $\widetilde{e}_j=\widetilde{L}_j(e_0,\ldots , e_r)$, $\sigma+1\le j\le
i$. The mapping of the standard part $\mbox{\rm
st}_\varepsilon\, :\,
{\Bbb P}^n(\overline{k(\varepsilon)})
\rightarrow{\Bbb P}^n(\overline{k})$ is defined
similarly to $\mbox{\rm st}_j$.
By Lemma~15 the equality
$\mbox{\rm st}_\varepsilon(W(\widetilde{L}_{\sigma+1}$ $,\ldots
,\widetilde{L}_{i-1}))=W(L'_{\sigma+1},\ldots , L'_{i-1})$ holds.
Hence by Lemma~5 every irreducible component of
$\mbox{\rm st}_\varepsilon(W(\widetilde{L}_{\sigma+1},\ldots
,\widetilde{L}_{i-1})\cap{\cal Z}(\widetilde{e}_i))$ is an irreducible
component of
$W(L'_{\sigma+1},\ldots , L'_{i-1})\cap{\cal Z}(e'_i)$.

Let us show that that every $z\in \varphi_i^{-1}(z')\cap W_i(L')$ is an
smooth
point of the intersection $W\cap{\cal Z}(e'_{\sigma+1},\ldots ,
e'_i)$. Indeed, $z\not\in{\cal Z}(e_0,\ldots , e_r)$ by
assertion (d) of Theorem~1,
and $W_{i-1}(L')\setminus{\cal Z}(e_0,\ldots , e_r)=W\cap{\cal
Z}(e'_{\sigma+1},\ldots , e'_{i-1})\setminus{\cal Z}(e_0,\ldots , e_r)$
since $(L'_{\sigma+1},\ldots , L'_{i-1})\in{\cal U}''_{i-1}$, see the
Introduction.
Hence it is sufficient to prove that $z$ is an smooth point of
$W_{i-1}(L')\cap{\cal Z}(e'_i)$.
But this follows from Theorem~1 (b). The
required assertion is proved.

Recall that the index of intersection
$\iota_j(L')=1$ for every $j\in J_i(L')$,
$\sigma+1\le i\le n$, see the Introduction. Hence the intersection of the
algebraic varieties
$W$, ${\cal Z}(e'_{\sigma+1}),\ldots ,{\cal Z}(e'_i)$ is transversal
at every irreducible component of
$W_i(L')$.
By Theorem~1 every point $z\in\varphi_i^{-1}(z')\cap
W_i(L')$ is a smooth point of $W_i(L')$, and the intersection of $W_i(L')$
and
${\cal Z}(L_{\iota+1},\ldots , L_n)$ is
transversal at $z$.
Now by Lemma~21 {\it the intersection of
$n-s+1$ algebraic varieties
$W$, ${\cal Z}(e'_j)$, $\sigma+1\le j\le i$; ${\cal
Z}(L_w)$, $\iota+1\le w\le n$,
is transversal at each point from $\varphi_i^{-1}(z')\cap W_i(L')$.}

We claim that there is an irreducible component $E'$ of
$W(\widetilde{L}_{\sigma+1},\ldots ,\widetilde{L}_i)$ such that
$\mbox{\rm st}_\varepsilon(E')\not
\subset W(L'_{\sigma+1},\ldots , L'_i)$.
Indeed,
otherwise, $\mbox{\rm st}(W(\widetilde{L}_{\sigma+1},\ldots
,\widetilde{L}_i)\cap{\cal
Z}(L_{\iota+1},\ldots , L_n))\subset
W_i(L')\cap{\cal
Z}(L_{\iota+1},\ldots ,
L_n)=\varphi_i^{-1}(z')\cap W_i(L')$.
Hence
by the implicit function theorem for every $z\in
\varphi_i^{-1}(z')\cap W_i(L')$ there is the
unique $\widetilde{z}$ from
the intersection of
$n-s+1$ algebraic varieties
$W$, ${\cal Z}(\widetilde{e}_j)$, $\sigma+1\le j\le i$; ${\cal
Z}(L_w)$,
$\iota+1\le w\le n$, such that $\mbox{\rm
st}_\varepsilon(\widetilde{z})=z$. Besides that,
the considered intersection is transversal at $\widetilde{z}$.
 Therefore, the degree
$\deg W(\widetilde{L}_{\sigma+1},\ldots ,\widetilde{L}_i)=
\#(W(\widetilde{L}_{\sigma+1},\ldots ,\widetilde{L}_i)\cap
{\cal Z}(L_{\iota+1},\ldots ,
L_n))=D_i(L'_{\sigma+1},\ldots ,$
$L'_i)<D_i$. This contradiction proves the required assertion is proved.

Now $\mbox{\rm st}(E')$ contains an irreducible component $E''$
of $W_i^{(1)}(L')\cup W_i^{(2)}(L')$, and
$E'\ne E''$. There is a point $z\in E''\cap S^{(1)}_i$. Hence there are
$x^*_0,\ldots , x^*_n\in\overline{k(\varepsilon)}$ such that
$x^*=(x^*_0:\ldots : x^*_n)\in
E'\setminus{\cal Z}(e_0,\ldots , e_r)$ is infinitely
close to $z$ with respect to the field $k$, and
$L_0(x^*_0,\ldots , x^*_n)=1$. There is $0\le \alpha\le r$ such
that $e_\alpha(x^*)\ne 0$
and $(e_j/e_\alpha)(x^*)$ is not infinitely large with respect to the field
$k$ for every
$0\le j\le r$. Set $y^*_j=((\widetilde{e}_j-e'_j)/e_\alpha)(x^*)$,
$\sigma+1\le
j\le i$. Then
$y^*_j$ is infinitesimal with respect to the field $k$ for every $\sigma+1\le
j\le i$.
Obviously $(x^*_0,\ldots , x^*_n,\,y^*_1,\ldots , y^*_i)\in
{\Bbb A}^{n-\sigma+i+1}(\overline{k(\varepsilon)})$ is a
solution of system (48) such that $(x^*_0:\ldots : x^*_n)\in
W(\overline{k(\varepsilon)})$.
Hence there is also a similar solution of this system but with the field
$\overline{k}_2$ in place of $\overline{k(\varepsilon)}$.


Conversely, let
$(x^*_0,\ldots , x^*_n,\,y^*_{\sigma+1},\ldots , y^*_i)
\in{\Bbb A}^{n-\sigma+i+1}(\overline{k}_2)$ be a solution of system
(48) such that
$x^*=(x^*_0:\ldots : x^*_n)\in W(\overline{k}_2)$.
Put $\overline{L}_j=L_j-(L_j/L_0)(x^*)L_0$ for $\iota+1\le j\le
n$. We have $L^*\in{\cal U}'_0$, $(L^*_{\sigma+1},\ldots ,L^*_{i-1})\in
{\cal U}''_{i-1}$ by Lemma~15. Hence condition (10) for $i-1$,
$L^*$ and $k_2$ in place of $i$, $L'$ and $k$ holds.
Therefore, see the Introduction,
$x^*\in (W_i(L^*)
\cup W^{(3)}_i(L^*))\cap{\cal
Z}(\overline{L}_{\iota+1},\ldots ,\overline{L}_n)$.
The point $\mbox{\rm st}_2 x^*\not\in
W_i(L')$ since, otherwise,
$\mbox{\rm st}_2 x^*\in
W_i(L')\cap{\cal Z}(L_{\iota+1},
\ldots ,L_n)\cap{\cal Z}(e_0,\ldots ,
e_r)$ which is impossible by the construction from Section~5.
On the other hand by
the implicit function theorem, see above, there
are $D_i(L')$ isolated point from
$W_i(L^*)\cap{\cal
Z}(\overline{L}_{\iota+1},\ldots ,\overline{L}_n)$
that are infinitely close to the corresponding points from
$W_i(L')\cap{\cal Z}(L_{\iota+1},
\ldots ,L_n))$.
Hence $D_i(L^*)\ge D_i(L')$.
Suppose that $x^*\in W_i(L^*)$. Then
$D_i\ge D_i(L^*)>D_i(L')$.
Assume $x^*\in W^{(3)}_i(L^*)$. Denote by $k^*$ the field
generated over $k$ by the
coefficients of all linear forms $L^*_j$,
$j\in\{0,\sigma+1,\ldots , n\}$.
Let $k^*_v=k_v\otimes_k k^*$ be the composite of the fields
$k^*$ and $k_v$ over $k$.
Then by Remark~6 from \cite{9} with the field $k'=
k^*_v$ and $W_v$ in place of $W$ we have
$D_i>D_i(L^*)\ge D_i(L')$.
The lemma is proved.

\medskip
Let us return to the description of the algorithm.
Using the algorithm from Section~5 we decide whether
$W^{(3)}_i(L')=\emptyset$.
Assume that $W^{(3)}_i(L')=\emptyset$.
Let us enumerate $z\in S^{(1)}_i$ and integers $0\le \alpha\le r$. For the
considered $z$ and $\alpha$
let us apply Theorem~2 \cite{9} to the algebraic variety $\mbox{\rm
con}(V)\times
{\Bbb A}^{i-\sigma}(\overline{k})$ (with the coordinate
functions $X_0,\ldots , X_n$, $Y_{\sigma+1},\ldots , Y_i$) and the
semialgebraic
set $U''$
given by system (48).
By this theorem one constructs  the finite set
$S''$ of points of
$\mbox{\rm con}(V)(\overline{k}_2)\times{\Bbb
A}^i(\overline{k}_2)$.
Denote by $\pi$ the natural rational through projection $\mbox{\rm
con}(V)\times{\Bbb A}^i\rightarrow\mbox{\rm
con}(V)\rightarrow V$.
Using Theorem~2 \cite{8} we decide whether $\pi(S'')\cap
W(\overline{k}_2)\ne\emptyset$.
By Theorem~2 \cite{9} and Lemma~29 the intersection $\pi(S'')\cap
W(\overline{k}_2)=\emptyset$ for all $z\in S^{(1)}_i$ and integers $0\le
\alpha\le r$ if and only if $D_i=D_i(L')$.

Suppose that $\pi(S'')\cap W(\overline{k}_2)\ne\emptyset$ for the current
$z\in S^{(1)}_i$ and integer $0\le \alpha\le r$.  Let us construct
$(x^*_0,\ldots , x^*_n,\,y^*_{\sigma+1},\ldots , y^*_i)
\in S''$ such that
$(x^*_0:\ldots : x^*_n)\in W(\overline{k}_2)$.
Let us define $L^*_{\sigma+1},\ldots ,L^*_i$ by (49),
(50).
Let us construct the field $k^*$ defined above by the algorithm from
\cite{2}.
Using the algorithm from \cite{4} we decide whether the point $x^*\in
W_i(L^*)$.

Suppose that $x^*\in W_i(L^*)$.
As we have seen in the proof of Lemma~29 the degree
$D_i(L^*)>D_i(L')$.
By Lemma~14 (with the field $k_2$ in place of $k$) using the
algorithm of reduction to integer coefficients
one can replace subsequently all the coefficients
of linear forms $L^*_0,
L^*_{\sigma+1},\ldots ,L^*_n$ by integers
with lengths $O(n\,\log d+(n-\sigma)\log d')$ and obtain linear forms
$L''_0,L''_{\sigma+1},\ldots , L''_n$ similar to
$L'_0,L'_{\sigma+1},\ldots , L'_n$
such that
$D(L''_{\sigma+1},\ldots , L''_i)\ge D(L^*_{\sigma+1},\ldots
,L^*_i)>D_i(L')$.
Set $L''=(L''_0,L''_{\sigma+1},\ldots , L''_n)$.
Now we replace $L'$ by $L''$ and return proceed to the next step of the
described recursion with the same $i$.
By the B\'ezout theorem for fixed $i$ there might be
at most $d^s(d')^{i-\sigma}$
steps when $x^*\in W_i(L^*)$.

Suppose that $x^*\in W^{(3)}_i(L^*)$ or $W^{(3)}_i(L')\ne\emptyset$.
If $W^{(3)}_i(L')\ne\emptyset$ we set $L^*=L'$, $k^*=k$
(the described below construction is slightly simplified in this case).
Let us define the set $S^{(w)}_i$, $w=0,1$, replacing in the definition of
$S^{(w)}_i$ the linear forms $L'_j$ by $L^*_j$, $j\in\{0,\sigma+1,\ldots ,
n\}$,
and the field $k$ by $k^*$. Let us construct the sets $S^{(w)}_i(L^*)$,
$w=0,1$, replacing in the construction
$S^{(w)}_i$ the linear forms $L'_j$ by $L^*_j$
and the field $k$ by $k^*$. Let us compute a linear form $L'''_i\in
k[X_0,\ldots ,X_r]$
with integer coefficients of length $O(n\log d+(n-\sigma)\log d')$ such that
$(S^{(0)}_i(L^*)\setminus S^{(1)}_i(L^*))\cap{\cal
Z}(L'''_i(e_0,\ldots , e_r))=\emptyset$. Set
$\widetilde{L}_j=L^*_j$ for
$j\in\{0,\sigma+1,\ldots , n\}\setminus\{i\}$ and $\widetilde{L}_{i}=
L^*_i+\varepsilon_3 L'''_i$. Set
$\widetilde{e}_j=\widetilde{L}_j(e_0,\ldots ,e_r)$
for every $j$. Hence all the linear forms
$\widetilde{L}_j$ have coefficients from the
field $k^*(\varepsilon_3)\subset\overline{k_3}$.
We have $W^{(3)}_i(\widetilde{L})\subset W_{i-1}(L^*)\cap W^{(4)}$.
Hence $W^{(3)}_i(\widetilde{L})$ is a union of some irreducible components
of dimension $n-i+\sigma-s$ of the variety $W_{i-1}(L^*)\cap
W^{(4)}$.
In particular, $W^{(3)}_i(L^*)$ is defined over $\overline{k^*}$.
Therefore, $W^{(3)}_i(L^*)\supset W^{(3)}_i(\widetilde{L})$.
Since $\widetilde{L}_i$ does not
vanish on any irreducible component $E$ of
$W^{(3)}_i(L^*)$, the algebraic variety
$W^{(3)}_i(\widetilde{L})=\emptyset$.
Hence $W_i(\widetilde{L})\setminus{\cal Z}(e_0,\ldots ,
e_r)=W_{i-1}(L^*)\cap{\cal Z}(\widetilde{e}_i)
\setminus{\cal Z}(e_0,\ldots , e_r)$ over the field $\overline{k_1}$.

Let us replace in the statement of Lemma~14 \cite{9} $W$, $\widetilde{L}$,
$L'$, $k$, $\varepsilon$,
$i+1$, $s$
by $W_v$, $\widetilde{L}$, $L^*$, $k^*_v$, $\varepsilon_3$, $i$,
$\sigma$ respectively
and apply this lemma.
Then we get $\deg W_i(\widetilde{L})=\deg W_{v,i}(\widetilde{L})>
\deg W_{v,i}(\widetilde{L})=\deg W_i(L^*)$.

By Lemma~14 (with the field $k^*(\varepsilon_3)$ in place of $k$) using the
algorithm of reduction to integer coefficients
one can replace subsequently all the coefficients
of linear forms $\widetilde{L}_0,
\widetilde{L}_{\sigma+1},\ldots ,\widetilde{L}_n$ by integers
with lengths $O(n\,\log d+(n-\sigma)\log d')$ and obtain linear forms
$L''_0,L''_{\sigma+1},\ldots , L''_n$ similar to
$L'_0,L'_{\sigma+1},\ldots , L'_n$ and such that
$D(L''_{\sigma+1},\ldots , L''_i)\ge D(\widetilde{L}_{\sigma+1},\ldots
,\widetilde{L}_i)>D_i(L')$.
Now we replace $L'$ by $L''$ and proceed to the next step of the
described recursion with the same $i$. By the B\'ezout theorem for fixed $i$
there might be at most $d^s(d')^{i-\sigma}$
steps when $x^*\in W^{(3)}_i(L^*)$ or $W^{(3)}_i(L')\ne\emptyset$.

Suppose that $\pi(S'')\cap W(\overline{k}_2)=\emptyset$. Then we proceed to
the next pair $z,\alpha$.
If all these pairs are enumerated then $D_i=D_i(L')$ as we have seen.
In this case if $i<n$ we replace $i$ by $i+1$ and
proceed to the next step of the recursion with the same $L'$.
Finally if in the considered case $i=n-1$ then
$L'\in{\cal U}''_0$.



\medskip We have constructed $L'\in{\cal U}''_0$.
Therefore, see the Introduction, conditions (8)
and (10) hold for all $\sigma+1\le i\le n-1$.
Set this $L'$ as a base of a new recursion.
Our aim is to obtain at the end of this recursion an element
$L'\in{\cal U}'''_0$ such that $\pi_{n+1}(L')\in{\cal U}'''_{n+1}$,
see the Introduction.
Let an integer $\sigma+1\le i\le n+1$. Recall that $L'_{n+1}=L'_0$. This
integer
$i$ will be changed recursively (it is a second recursion).
The recursive assumption here is that additionally $(L'_{\sigma+1},\ldots ,
L_{i-1})\in{\cal U}'''_{i-1}$.
We take $i=\sigma+2$
as the base of this recursion (recall that ${\cal
U}''_{\sigma+1}={\cal U}'''_{\sigma+1}$,
see the Introduction).
Let $L_0,\ldots , L_{n+1}$, the morphism $\varphi_i$, the point $z'$
be as above. Put $S^{(2)}_i=\varphi_i^{-1}(z')\cap W^{(2)}_i(L')$.
Let us compute $S^{(2)}_i$ using the algorithms from Section~5.
Let $0\le\alpha_1<\alpha_2\le r$ be integers and $z\in S^{(2)}_i$.
Denote for brevity $X^{(w)}=(X_{w,0},\ldots , X_{w,n})$, $w=1,2$, and for
an arbitrary polynomial
$c$ in $n+1$ variable set $c(X^{(w)})=c(X_{w,0},\ldots , X_{w,n})$.
Consider the following system of polynomial equations and inequalities with
respect to the variables
$X_{w,0},\ldots , X_{w,n}$, $w=1,2$,
$Y_{\sigma+1},\ldots , Y_i$, $Z_{\sigma+1},\ldots , Z_i$ with coefficients
from the field $\overline{k_4}$
$$
\left\{
\begin{array}{ll}
(e'_j-Y_je_{\alpha_1}-Z_je_{\alpha_2})(X^{(w)})=0,& \sigma+1\le j\le i,\,
w=1,2,\\
\sum_{\sigma+1\le j\le i}(|Y_j|^2+|Z_j|^2)\le\varepsilon_1,&\\
\sum_{0\le j\le n}|X_{w,j}-(X_j/L_0)(z)|^2\le\varepsilon_1,& w=1,2\\
\varepsilon_4\le\sum_{0\le j\le r}|e'_j(X^{(w)})|^2\le\varepsilon_3,&
w=1,2,\\
\sum_{0\le j\le n}|X_{1,j}-X_{2,j}|^2\ge\varepsilon_2,&\\
\sum_{\iota+1\le j\le n}|L_j(X^{(1)})-L_j(X^{(2)})|^2\le\varepsilon_3,&\\
L_0(X^{(w)})=1,& w=1,2.
\end{array}
\right.
\eqno (51)
$$

\par\medskip\noindent{\bf LEMMA~30}\hspace{0.1em} {\it  Let
$L'=(L'_0,L'_{\sigma+1},\ldots ,
L'_n)\in{\cal U}''_0$ and
$(L'_{\sigma+1},\ldots , L'_{i-1})\in{\cal U}'''_{i-1}$. Then
$D^{(2)}_i>D^{(2)}_i(L'_{\sigma+1},$
$\ldots , L'_i)$ if and only if there are
$z\in S^{(2)}_i$ and integers $0\le\alpha_1<\alpha_2\le r$
such that system (51) has a solution
$$
(x^*_{1,0},\ldots , x^*_{1,n},\,x^*_{2,0},\ldots , x^*_{2,n},\,
y^*_{\sigma+1},\ldots , y^*_i,\,z^*_{\sigma+1},\ldots , z^*_i)
\in{\Bbb A}^{2n-2\sigma+2i+2}(\overline{k}_4)
\eqno (52)
$$
with
$x^*_w=(x^*_{w,0}:\ldots : x^*_{w,n})\in
W(\overline{k}_4)$ for $w=1,2$.
}\par\medskip

\noindent {\bf PROOF} \quad
Suppose that $D^{(2)}_i>D^{(2)}_i(L'_{\sigma+1},\ldots , L'_i)$.
There exists $\varepsilon'>0$ such that $\varepsilon'$ is an infinitesimal
with respect to the field $k(\varepsilon_1)$ and
$\varepsilon_2$ is an infinitesimal with respect to the field
$k(\varepsilon_1,\varepsilon')$. There exists $\varepsilon''>0$
such that $\varepsilon''$ is an infinitesimal
with respect to the field $k_3(\varepsilon')$ and
$\varepsilon_4$ is an infinitesimal with respect to the field
$k_3(\varepsilon',\varepsilon'')$. Let us extend the ground field
$k$ till $k_4(\varepsilon',\varepsilon'')$.

Since ${\cal U}'''_i$ is open in the Zariski topology there are linear
forms $\widetilde{L}_{\sigma+1},\ldots ,\widetilde{L}_n\in
\overline{k(\varepsilon')}[X_0,\ldots , X_n]$ such that
$(\widetilde{L}_{\sigma+1},\ldots ,\widetilde{L}_i)\in{\cal U}'''_i$
and all the nonzero coefficients of
the linear forms $\widetilde{L}_j-L'_j$, $\sigma+1\le j\le i$ are
infinitesimal with respect to the field $k$.

Set $\widetilde{e}_j=\widetilde{L}_j(e_0,\ldots , e_r)$, $\sigma+1\le j\le
i$. By Lemma~15 the equality
$\mbox{\rm st}(W^{(2)}(\widetilde{L}_{\sigma+1}$ $,\ldots
,\widetilde{L}_i))=W^{(2)}(L'_{\sigma+1},\ldots , L'_i)$ holds.
By Lemma~6 the intersection $W^{(2)}(\widetilde{L}_{\sigma+1}$ $,\ldots
,\widetilde{L}_i)\cap{\cal Z}(L_0,L_{\iota+1},\ldots , L_n)=\emptyset$ in
${\Bbb P}^n(\overline{k(\varepsilon')})$. Hence, cf. \cite{4},
there exist $z\in S^{(2)}_i$ and $\widetilde{x}_{w,0},\ldots ,
\widetilde{x}_{w,n}
\in\overline{k(\varepsilon')}$ such that
$\widetilde{x}_w=(\widetilde{x}_{w,0}:\ldots : \widetilde{x}_{w,n})\in
W^{(2)}(\widetilde{L}_{\sigma+1}$
$,\ldots,\widetilde{L}_i)(\overline{k(\varepsilon')})$,
$L_0(\widetilde{x}_{w,0},\ldots ,\widetilde{x}_{w,n})=1$,\\
$\sum_{0\le j\le n}|(X_j/L_0)(\widetilde{x}_w)-(X_j/L_0)(z)|^2
<\varepsilon_1$ for
$w=1,2$; $\sum_{0\le j\le
n}|\widetilde{x}_{1,j}-\widetilde{x}_{2,j}|^2
>\varepsilon_2$;
$(L_j/L_0)(\widetilde{x}_1)=(L_j/L_0)(\widetilde{x}_2)$ for
$\sigma+1\le j\le n$.

Further, for $w=1,2$ there exist $x^*_{w,0},\ldots , x^*_{w,n}
\in\overline{k(\varepsilon',\varepsilon'')}$
such that
$x^*_w=(x^*_{1,0}:\ldots : x^*_{1,n})\in W(\widetilde{L}_{\sigma+1}$ $,\ldots
,\widetilde{L}_{i-1})(\overline{k(\varepsilon',\varepsilon'')})$,
$L_0(x^*_{w,0},\ldots , x^*_{w,n})=1$,
$\mbox{\rm st}_{\varepsilon''}(x^*_w)=\widetilde{x}_w$; and
$X_{w,j}=x^*_{w,j}$, $0\le j\le n$, $w=1,2$, is a solution of
the inequalities $\varepsilon_4\le\sum_{0\le j\le r}
|e'_j(X^{(w)})|^2\le\varepsilon_3$.
Therefore, by our construction $X_{w,j}=x^*_{w,j}$
is a solution of all the
equations and inequalities
from system (51) which do not depend on $Y_j$ and $Z_j$.

For all $0\le w_1<w_2\le r$ denote
$\Delta(w_1,w_2)=e_{w_1}(x^*_{1,0},\ldots , x^*_{1,n})
e_{w_2}(x^*_{2,0},\ldots ,$ $x^*_{2,n})-$
$e_{w_2}(x^*_{1,0},\ldots , x^*_{1,n})
e_{w_1}(x^*_{2,0},\ldots , x^*_{2,n})$.
Suppose that not all $\Delta(w_1,w_2)=0$.
Then
there are $0\le\alpha_1<\alpha_2\le r$ such that
$\Delta(\alpha_1,\alpha_2)\ne 0$
and $\Delta(w_1,w_2)/\Delta(\alpha_1,$ $\alpha_2)$ are not infinitely large
with respect to the field
$k$ for all $0\le w_1<w_2\le r$. By Cramer's rule one
can represent
$$
(e_q-\mu_{q,1}e_{\alpha_1}-\mu_{q,2}e_{\alpha_2})(x^*_w)=0,
\eqno (53)
$$
where
$\mu_{q,w}\in\overline{k(\varepsilon',\varepsilon'')}$
is not infinitely large
with respect to the field $k$ for every $0\le q\le r$ and $w=1,2$.

Suppose that
$\Delta(\alpha_1,\alpha_2)=0$ for all $0\le\alpha_1<\alpha_2\le r$.
Let us choose $\alpha_1$ such that $(e_q/e_{\alpha_1})(x^*_w)$
is not infinitely large with respect to $k$ for every $0\le q\le r$.
Represent
$$
e_q(x^*_w)=\mu_q e_{\alpha_1}(x^*_w),
\eqno (54)
$$
for $w=1,2$. Permuting indices we can assume without loss of generality
that $\alpha_1\le r-1$. In this case put $\alpha_2=\alpha_1+1$ and
all $z^*_j=0$,
$\sigma+1\le j\le i$.

Let $\widetilde{e}_j=e'_j+\sum_{0\le q\le r}\nu_{j,q}e_q$ where all
$\nu_{q,j}\in\overline{k(\varepsilon')}$ are infinitesimals with respect to
the field $k$. By (53) or (54)
there is a solution (52) of system (51) with
$\overline{k(\varepsilon',\varepsilon'')}$
in place of $\overline{k}_4$ such that
$x^*_w\in W(\overline{k(\varepsilon',\varepsilon'')})$, $w=1,2$.
Now by the transfer principle, see \cite{1}, or directly we deduce that there
is
also solution (52) with $x^*_w\in W(\overline{k}_4)$, $w=1,2$.

Conversely, suppose that
for some $0\le\alpha_1<\alpha_2\le r$ there is a solution
(52) such that
$x^*_w\in W(\overline{k}_4)$, $w=1,2$. Then set
$\overline{x}_{w,j}=\mbox{\rm
st}_{\varepsilon_3,\varepsilon_4}x^*_{w,j}\in\overline{k}_2$, $0\le
j\le n$, $w=1,2$;
$\overline{x}_w=(\overline{x}_{w,0}:\ldots :\overline{x}_{w,n})
\in{\Bbb P}^n(\overline{k}_2)$, $w=1,2$;
$\overline{y}_j=\mbox{\rm
st}_{\varepsilon_3,\varepsilon_4}y^*_j\in\overline{k}_2$,
$\overline{z}_j=\mbox{\rm
st}_{\varepsilon_3,\varepsilon_4}z^*_j\in\overline{k}_2$,
$\sigma+1\le j\le i$;
$$
\overline{L}_j=L'_j-\overline{y}_jX_{\alpha_1}-
\overline{z}_jX_{\alpha_2},\quad
\overline{e}_j=\overline{L}_j(e_0,\ldots , e_r),\quad \sigma+1\le j\le i;
\eqno (55)
$$
$\widetilde{L}_j=L'_j-y^*_jX_{\alpha_1}-z^*_jX_{\alpha_2}$,
$\widetilde{e}_j=\widetilde{L}_j(e_0,\ldots , e_r)$, $\sigma+1\le j\le i$.
Notice that $\overline{x}_1\ne\overline{x}_2$ and
$(L_j/L_0)(\overline{x}_1)=(L_j/L_0)(\overline{x}_2)$, $\iota+1\le j\le n$,
since $\sum_{0\le j\le
n}|x^*_{1,j}-x^*_{2,j}|^2
\ge\varepsilon_2$ and $\sum_{\iota+1\le j\le n}|(L_j/L_0)(x^*_1)-
(L_j/L_0)(x^*_2)|^2\le\varepsilon_3$.
Further, from system (51) and Lemma~15 we deduce that the point
$\overline{x}_w\in\mbox{\rm
st}_{\varepsilon_3,\varepsilon_4}(W(\widetilde{L}_{\sigma+1},\ldots
,\widetilde{L}_{i-1})\cap
\mbox{\rm st}_{\varepsilon_3,\varepsilon_4}{\cal
Z}(\widetilde{e}_i)\subset W(\overline{L}_{\sigma+1},\ldots
,\overline{L}_{i-1})\cap
{\cal Z}(\overline{e}_i)$, $w=1,2$. The point $\mbox{\rm
st}_{\varepsilon_1,\varepsilon_2}\overline{x}_w=z\not\in
W_i(L')\cup W^{(1)}_i(L')$ by our construction.
Recall that $W^{(1)}_i(L')=W^{(1)}_i(\overline{L})=W^{(1)}_i$ since
$L',\overline{L}\in{\cal U}''_0$,
see the Introduction.
Hence
$\overline{x}_w\not\in W(\overline{L}_{\sigma+1},\ldots
,\overline{L}_i)\cup
W^{(1)}(\overline{L}_{\sigma+1},\ldots ,\overline{L}_i)$ again
by Lemma~15. Therefore,
$\overline{x}_w\in W^{(2)}(\overline{L}_{\sigma+1},\ldots ,
\overline{L}_i)$, $w=1,2$.
On the other hand, we have $\mbox{\rm
st}_2(W^{(2)}(\overline{L}_{\sigma+1},\ldots ,
\overline{L}_i)=W^{(2)}_i(L')$ by Lemma~15 and obviously
$\mbox{\rm st}_2({\cal
Z}(L_{\iota+1}-(L_{\iota+1}/L_0)(\overline{z})L_0,\ldots ,
L_n-(L_n/L_0)(\overline{z})L_0))$ $={\cal
Z}(L_{\iota+1},\ldots , L_n)$. Assume that $\deg
W^{(2)}(\overline{L}_{\sigma+1},\ldots ,
\overline{L}_i)=\deg W^{(2)}_i(L')$. Then by Lemma~7 the
points of $W^{(2)}(\overline{L}_{\sigma+1},\ldots ,
\overline{L}_i)\cap{\cal
Z}(L_{\iota+1}-(L_{\iota+1}/L_0)(\overline{z})L_0,\ldots ,
L_n$ $-(L_n/L_0)(\overline{z})L_0)$ and $S^{(2)}_i$ are in the one--to--one
correspondence by the rule $x\mapsto\mbox{\rm st}_2(x)$. This
contradicts to the existence
of $\overline{x}_1\ne\overline{x}_2$ such that $\mbox{\rm
st}_2(\overline{x}_1)=\mbox{\rm
st}_2)(\overline{x}_2)=z$.
Hence $\deg W^{(2)}(\overline{L}_{\sigma+1},\ldots ,
\overline{L}_i)>\deg W^{(2)}_i(L')$ by Lemma~5. The lemma is proved.





\medskip
Let us return to the description of the algorithm.
Let us enumerate $z\in S^{(2)}_i$ and pairs of integers
$0\le \alpha_1<\alpha_2\le r$. For the
considered $z$ and $\alpha_1,\alpha_2$
let us apply Theorem~2 \cite{9} to the algebraic variety $\mbox{\rm
con}(V)^2\times{\Bbb A}^{2i-2\sigma}(\overline{k})$ and the
semialgebraic set $U''$ given by system (51).
One constructs by this theorem the finite set $S''$ of points of
$\mbox{\rm con}(V)(\overline{k}_4)^2\times{\Bbb
A}^{2i-2\sigma}(\overline{k}_4)$.
Denote by $\pi$ the natural rational through projection $\mbox{\rm
con}(V)^2\times{\Bbb A}^{2i-2\sigma}\rightarrow\mbox{\rm
con}(V)^2\rightarrow V\times V$.
Using Theorem~2 \cite{8} we decide whether $\pi(S'')\cap
W(\overline{k}_4)^2\ne\emptyset$.
By Theorem~2 \cite{9} and Lemma~29 the intersection $\pi(S'')\cap
W(\overline{k}_4)^2=\emptyset$ for all $z\in S^{(2)}_i$ and
pairs of integers $0\le\alpha_1<\alpha_2\le r$
if and only if $D^{(2)}_i=D^{(2)}_i(L')$.

Suppose that $\pi(S'')\cap W(\overline{k}_4)^2\ne\emptyset$ for the current
$z\in S^{(2)}_i$ and integers $\alpha_1,\alpha_2$.  Then we construct a
solution
(52) from $S''$ such that $x^*_w\in W(\overline{k}_2)$
for $w=1,2$. Let us define
$\overline{L}_j$ and $\overline{e}_j$ by (55). Set $\overline{L}_j=L'_j$
for $j=0,i+1,\ldots , n$.
Let us construct all $\overline{L}_j$ and $\overline{e}_j$, see \cite{3}.
As we have seen in the proof of Lemma~30 the degree
$D^{(2)}(\overline{L}_{\sigma+1},\ldots ,\overline{L}_i)>D^{(2)}_i(L')$.
By Lemma~16 (with the field $k_2$ in place of $k$) using the
algorithm of reduction to integer coefficients
one can replace subsequently all the coefficients
of linear forms $\overline{L}_0,\overline{L}_{\sigma+1},\ldots
,\overline{L}_i$ by integers
with lengths $O(n\,\log d+(n-\sigma)\log d')$ and obtain linear forms
$L''_0,L''_{\sigma+1},\ldots , L''_n$ similar to
$L'_0,L'_{\sigma+1},\ldots , L'_n$ and
such that
$D^{(2)}(L''_{\sigma+1},\ldots , L''_i)\ge
D^{(2)}(\widetilde{L}_{\sigma+1},\ldots
,\widetilde{L}_i)>D_i(L')$.
Set $L''=(L''_0,L''_{\sigma+1},\ldots ,$ $L''_n)$.
Now we replace $L'$ by $L''$ and return
to the beginning of the described recursion.

Suppose that $\pi(S'')\cap W(\overline{k}_2)=\emptyset$. Then we proceed to
the next $z$, $\alpha_1,\alpha_2$.
If all $z$, $\alpha_1,\alpha_2$ are enumerated then
$D^{(2)}_i=D^{(2)}_i(L')$
as we have seen it.
In this case
if $i<n$ we replace $i$ by $i+1$ and return to the beginning of the
recursion
with the same $L'$. Finally if in the considered case $i=n$ then
$L'\in{\cal U}'''_0$ and $\pi_{n+1}(L')\in{\cal U}'''_{n+1}$.

The working time of the algorithms from this section 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 $x^{(0)}$, and the sum of
sizes of linear forms $L'_j$, $\sigma\le j\le n$. In particular, if
all the linear forms $L'_j$ have integer coefficients of length
$O(n\log d+(n-\sigma)\log(d'))$ (as in the statement of
Theorem~1 \cite{9}) then this working time is the same as in Theorem~2.
This follows immediately from the estimations for the working times of the
applied algorithms.


\section{Proof of Theorem~2}\label{s7}

Now everything is ready for the proof of Theorem~2.  We have
proved assertion (a) of this theorem and
constructed $L'=(L'_0,L'_{\sigma+1},\ldots , L'_n)\in{\cal U}'''_0$
such that $\pi_{n+1}(L')\in{\cal U}'''_{n+1}$ in
the previous section. Let us prove (b) (and hence also (c) and (d)).
Let us replace the notation $L_0,\ldots , L_{n+1}$ by
$\widetilde{L}_0,\ldots ,\widetilde{L}_{n+1}$ in the statement of Theorem~1
(to avoid an ambiguity in what follows).
Let us construct these
linear forms $\widetilde{L}_0,\ldots ,\widetilde{L}_{n+1}$
from Theorem~1 for the considered $L'$.

Recall some properties of linear forms $\widetilde{L}_0,\ldots
,\widetilde{L}_{n+1}$, which we shall need, see the statement of Theorem~1
and its proof in Section~5.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item $\widetilde{L}_0,\ldots ,\widetilde{L}_{n+1}$ have
integer coefficients of length $O(n\log d+(n-\sigma)\log d')$;
satisfy (in place of $L_0,\ldots ,L_{n+1}$) assertion (a) of Theorem~1.
\item The intersection of the algebraic variety
$W_{i-1}(L')\cap{\cal Z}(e'_i)$ and the linear subspace ${\cal
Z}(\widetilde{L}_{\iota+1},\ldots ,\widetilde{L}_n)$ is
transversal for every $\sigma+1\le i\le n$.
\item Let us replace in the definitions of $\varphi^i$ and $\mu^{(i)}$
in Section~4 linear forms
$L_0,\ldots ,L_{n+1}$ by $\widetilde{L}_0,\ldots ,\widetilde{L}_{n+1}$.
Then $\mu^{(i)}((1:0:\ldots : 0))$ is maximal possible, see the notation in
Section~4 ,for every $\sigma+1\le i\le n$.
\end{enumerate}

Let us apply Lemma~25 to the algebraic varieties
$\widetilde{V}_j$, $\sigma\le j\le n$,
in place of $V^{(i)}$, $0\le i\le\nu$, and construct
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\setcounter{enumi}{3}
\item $\lambda=(L_{s+1},\ldots , L_\sigma)\in
k[X_0,\ldots , X_n]$ satisfying the assertion of
Lemma~25 for the considered $L'\in{\cal U}'''_0$.
\end{enumerate}
Now we identify the affine space ${\Bbb
A}^{(n+2)(n+1)}(\overline{k})$ with the set of all
$(n+2)$--tuples of linear forms from $\overline{k}[X_0,\ldots , X_n]$.
Put $\lambda_0=(0,\ldots , 0,L_{s+1},\ldots , L_\sigma,0)\in
{\Bbb A}^{(n+2)(n+1)}(\overline{k})$ and $\widetilde{\lambda}=
(\widetilde{L}_0,\ldots ,\widetilde{L}_{n+1})$.
Applying Lemma~3 and the
algorithm of reduction
to integer coefficients (it is possible to apply them by
Lemma~19 and Theorem~1 (d) \cite{9})
to the line
${\mathfrak l}$ containing $\widetilde{\lambda}$ and $\lambda_0$
we construct an element $\lambda^*=
(L^*_0,\ldots ,L^*_{n+1})\in{\mathfrak l}$ satisfying
(i)--(iii) in place of $\widetilde{\lambda}$ and such that (iv) holds for
$(L^*_{n-\sigma+s+1},\ldots , L^*_n)$ in place of
$\lambda$.
Then we replace
$\widetilde{\lambda}$ by $\lambda^*$. Thus, we shall assume in what
follows without loss of generality that
$L_j=\widetilde{L}_{j+n-\sigma}$, $s+1\le j\le\sigma$, and (i)--(iv) hold.

For any irreducible component $E$ of $W_{\lambda,i}(L')$ there is an
irreducible component $E_1$ of
$W\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)\cap{\cal
Z}(e'_{\sigma+1},\ldots , e'_i)$ such that $E_1\supset E$.
Obviously $E_1\subset W_i(L')$. Hence
$$
W_i(L')\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)\supset W_{\lambda,i}(L'),\quad \sigma\le i\le n
\eqno (56)
$$
 (and, more than that, these inclusions hold for any
$\lambda,L'$ such that $W_{\lambda,i}(L')$
and $W_i(L')$ are defined).

We have $L'\in{\cal U}'''_0$ and $L'\in{\cal U}'_{\lambda,0}$.
By Lemma~11 \cite{9} an irreducible component $E$ of
$W_i(L')\cap{\cal Z}(L_{s+1},
\ldots , L_\sigma)$ is an irreducible component of $W_{\lambda,i}(L')$ if
and only if
$E$ contains a point from $W_{\lambda,n}(L')$.

The degree $\deg W_n(L')=\deg p_v\deg W'=\deg W_{\lambda,n}(L')$.
Hence by the B\'ezout theorem
$W_n(L')\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)=
W_{\lambda,n}(L')$.
By Lemma~11 \cite{9} for every irreducible component $E_0$ of $W_{v,i}(L')$
the intersection $E_0\cap W_{v,n}(L')\ne\emptyset$. Hence every
irreducible component
$E_1$ of  $W_i(L')$ contains an irreducible component $E_2$ of $W_n(L')$.
By (iv) the variety $E_1\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$
is irreducible for $i<n$.
Therefore, every irreducible component $E$
of $W_i(L')\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)$ contains a point $z$
from $W_{\lambda,n}(L')$. Thus,
$$
W_{i-1}(L')\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)=W_{\lambda,i-1}(L'),\quad\sigma+1\le i\le n+1.
\eqno (57)
$$
Since $L'\in{\cal U}''_0$ condition (10) holds
for all $\sigma+1\le i\le n-1$. Hence by (57)
condition (10) is satisfied for $W_\lambda$,
$p_\lambda$, $L'$
in place of $W$, $p$, $L'$
for all $\sigma+1\le i\le n-1$.
Intersections in the left part of (57) are transversal by (i).
Therefore,
$D_i=D_i(L')=D_{\lambda,i}(L')\le
D_{\lambda,i}$ by (57) and $D_{\lambda,i}\le D_i$ by (56) with $U$ in
place of $L'$.
Hence $D_{\lambda,i}=D_i$.

Intersecting the both sides of (57) with ${\cal Z}(e'_i)$ we get
for every $\sigma+1\le i\le n$
$$
(W_i(L')\cup W^{(1)}_i\cup W^{(2)}_i(L'))\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)=
W_{\lambda,i}(L')\cup W^{(1)}_{\lambda,i}\cup W^{(2)}_{\lambda,i}(L').
\eqno (58)
$$
The codimension of every irreducible component of
the algebraic variety in the right part of (58)
is the sum of codimension
of two intersected varieties from the left part
for every $\sigma+1\le i\le n$.
Hence
\setcounter{equation}{58} \begin{eqnarray}
&&W^{(2)}_i(L')\cap
{\cal Z}(L_{s+1},\ldots , L_\sigma)
\subset W^{(2)}_{\lambda,i}(L'),\label{59}\\
&&W^{(1)}_i\cap{\cal Z}(L_{s+1},\ldots , L_\sigma)\subset
W^{(1)}_{\lambda,i}\cup W^{(2)}_{\lambda,i}(L'), \label{60}
\end{eqnarray}
By (iii) every irreducible component $E_3$ of $W^{(1)}_i\cap{\cal
Z}(L_{s+1},\ldots , L_\sigma)$ is not contained in any irreducible component
of $W^{(1)}_j$ for every $\sigma+1\le j<i$. Hence also $E_3$ is not
contained in any irreducible component
of $W^{(2)}_j(L')$ for every $\sigma+1\le j\le i$. Therefore, $E_3$ is not
contained in any
irreducible component of $W^{(1)}_j\cup W^{(2)}_j(L')$, $\sigma+1\le j\le
i$. Intersecting the last algebraic variety with
${\cal Z}(L_{s+1},\ldots , L_\sigma)$ we get that $E_3$ is not
contained in any irreducible component
of $W^{(1)}_{\lambda,j}\cup W^{(2)}_{\lambda,j}(L')$ for every $\sigma+1\le
j\le i$. Thus, $E_3$ is not an irreducible component of
$W^{(2)}_{\lambda,i}(L')$.
Therefore, $W^{(1)}_i\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)=
W^{(1)}_{\lambda,i}$, $W^{(2)}_i(L')\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)=
W^{(2)}_{\lambda,i}(L')$.
The intersection of $W_{i-1}(L')\cap{\cal
Z}(e'_i)$ and the linear subspace
${\cal Z}(L_{s+1},\ldots , L_\sigma)$ is transversal.
Hence $D^{(1)}_i=D^{(1)}_i(L')=
D^{(1)}_{\lambda,i}(L')=D^{(1)}_{\lambda,i}$ since condition
(10) holds for every $\sigma+1\le i\le n-1$, and also the similar
condition is satisfied for $W_\lambda$, $p_\lambda$, $L'$
in place of $W$, $p$, $L'$ for every $\sigma+1\le i\le n-1$. Finally,
$D^{(2)}_i=D^{(2)}_i(L')=D^{(2)}_{\lambda,i}(L')\le D^{(2)}_{\lambda,i}$.

\par\medskip\noindent{\bf LEMMA~31}\hspace{0.1em} {\it  Let $\lambda$
satisfies condition (A), see the Introduction.
Suppose that $D_i=D_{\lambda,i}$ for every $\sigma\le i\le n$.
Then $D^{(1)}_i\ge D^{(1)}_{\lambda,i}$, $D^{(2)}_i\ge D^{(2)}_{\lambda,i}$,
for all $\sigma+1\le i\le n$
}\par\medskip

\noindent{\bf PROOF}\quad Indeed,
(56) is fulfilled for $U$ in place of $L'$. We have
$U\in{\cal U}''_0$.
Hence in the similar way (57)--(60) hold with $U$ in place of
$L'$. Every irreducible component $E_3$ of
$W^{(1)}_i\cap{\cal
Z}(L_{s+1},\ldots , L_\sigma)$ is not an irreducible component of
$W^{(2)}_{\lambda,i}(U)$ since the last algebraic variety does not have
irreducible components defined over $\overline{k}$. Now replacing $L'$ by
$U$ we deduce as above
$W^{(1)}_i\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)=
W^{(1)}_{\lambda,i}$, $W^{(2)}_i(U)\cap{\cal Z}(L_{s+1},\ldots ,
L_\sigma)=
W^{(2)}_{\lambda,i}(U)$. Hence the required inequalities follows from the
B\'ezout theorem. The lemma is proved.


\medskip Now
$D^{(2)}_i=D^{(2)}_i(L')=D^{(2)}_{\lambda,i}(L')=D^{(2)}_{\lambda,i}$
by Lemma~31.
Using the algorithms from Section~5 we prove that one can
compute $D_i$, $D^{(1)}_i$ and $D^{(2)}_i$, $\sigma\le i\le n$.
Thus, $L'\in{\cal U}'''_{\lambda,0}$ and assertions (b)--(d) hold.


Let us prove (e).
We have $L'\in{\cal U}'''_{\lambda,0}$ and (c), (d) hold. Therefore, see
Section~4, $L'\in{\cal U}'_0$.
As previously the left part of (57) contains its right part.
Since $D_{\lambda,i-1}=D_{i-1}$ the equality (57) holds and
the intersection in the left part of (57) is transversal.
Further, $D_i(L')=D_{\lambda,i}(L')=D_{\lambda,i}=D_i$, $\sigma\le i\le n$.
Hence $L'\in{\cal U}''_0$.
Hence (58) is fulfilled.
Therefore, again (59) and (60) are fulfilled.
Now the equalities $D^{(2)}_{\lambda,i}=D^{(2)}_i$,
$D^{(1)}_{\lambda,i}=D^{(1)}_i$ imply all the assertions of (e).

Let us prove (f), cf. the proof of Lemma~13 \cite{9} and Lemma~19 (a).
Let $(\lambda',L'')\in{\mathfrak l}$, $(\lambda',L'')\ne(\lambda,L')$,
$\lambda'=(L'_{s+1},\ldots , L'_\sigma)$, $L''=(L''_{\sigma+1},\ldots ,
L''_n)$,
where all $L'_j\in k[X_0,\ldots , X_n]$, $s+1\le j\le\sigma$;
$L''_w\in\overline{k}[X_0,\ldots ,X_r]$,
$w\in\{0,\sigma+1,\ldots , n\}$, are linear forms. Let $\varepsilon$ be the
coordinate function of ${\Bbb A}^1(\overline{k})$.
Set $L_{\varepsilon,j}=L_j+\varepsilon L'_j\in k(\varepsilon)[X_0,\ldots ,
X_n]$, $s+1\le j\le\sigma$;
$L^{(\varepsilon)}_w=L'_w+\varepsilon L''_w\in
\overline{k}(\varepsilon)[X_0,\ldots ,X_r]$,
$e^{(\varepsilon)}_w=L^{(\varepsilon)}_w(e_0,\ldots , e_r)$,
$e''_w=L''_w(e_0,\ldots , e_r)$, $w\in\{0,\sigma+1,\ldots , n\}$,
$\lambda_\varepsilon=(L_{\varepsilon,s+1},\ldots , L_{\varepsilon,\sigma})$,
$L^{(\varepsilon)}=(L^{(\varepsilon)}_{\sigma+1},\ldots ,
L^{(\varepsilon)}_n)$.
For every $a\in k$ put $\lambda_a=(L_{a,s+1},\ldots , L_{a,\sigma})$,
$L^{(a)}=(L^{(a)}_{\sigma+1},\ldots , L^{(a)}_n)$, where
$L_{\varepsilon,j}=L_j+a L'_j\in k[X_0,\ldots ,
X_n]$, $s+1\le j\le\sigma$;
$L^{(a)}_w=L'_w+a L''_w\in \overline{k}[X_0,\ldots ,X_r]$,
$e^{(a)}_w=L^{(\varepsilon)}_w(e_0,\ldots , e_r)$,
$w\in\{0,\sigma+1,\ldots , n\}$.

By Lemma~3 for all $a\in k$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number, $\lambda_a$ satisfies condition (A), see
the Introduction. In what follows, we shall assume without loss of
generality condition (A) holds for the element $a$.

For every $\sigma\le i\le n$ consider the subvariety
$$
(W\times{\Bbb A}^1(\overline{k}))\cap{\cal
Z}(L_{\varepsilon,s+1},\ldots ,
L_{\varepsilon,\sigma})\cap{\cal Z}(e^{(\varepsilon)}_{\sigma+1},\ldots
, e^{(\varepsilon)}_i)
\eqno (61)
$$
of ${\Bbb P}^n(\overline{k})\times{\Bbb A}^1(\overline{k})$.
Denote by ${\cal W}_i$ the union of all the irreducible over
$\overline{k}$ components $E$ of variety (61)
such that $E\cap{\cal Z}(\varepsilon)$ contains an irreducible component
of $W_{\lambda,i}(L')\times\{0\}$.
By the definition of ${\cal W}_i$ and ${\cal W}_n$
every irreducible component of ${\cal W}_i$ contains an irreducible
component of the curve
${\cal W}_n$ and ${\cal W}_n\subset{\cal W}_i$.


Generalizing the proof of Lemma~13 \cite{9} (it is straightforward and we
leave the details to the reader) we ascertain that
for all $a\in k$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number, ${\cal W}_i\cap{\cal
Z}(\varepsilon-a)=W_{\lambda_a,j}(L^{(a)})\times\{a\}$,
the degree $\deg W_{\lambda_a,j}(L^{(a)})\ge\deg W_{\lambda,L'}=D_i$
for all $\sigma\le i\le n$, and
$W_{\lambda_a,n}(L^{(a)})\cap\overline{V\setminus W}=\emptyset$. But
obviously $D_{\lambda_a,i}(L^{(a)})\le
D_{\lambda_a,i}$, and
$D_{\lambda_a,i}\le D_i$ by (56) with $U$ in place of $L'$. Hence
$\deg W_{\lambda_a,i}(L^{(a)})=D_{\lambda_a,i}(L^{(a)})=D_{\lambda_a,i}=D_i$
and $L^{(a)}\in{\cal U}''_{\lambda_a,0}\cap{\cal U}''_0$ by (56)
for these $a$ (in what follows in the proof of (f)
we shall always assume that these properties are fulfilled for the
considered $a$). In particular, $\deg W_{\lambda_a,n}(L^{(a)})=D_n=\deg
q_v\deg W'$. Hence $\deg p'_\lambda=\deg p'_v$ in this case.
Besides that, for these $a$ conditions (57)--(60)
hold with $(\lambda_a,L^{(a)})$ in place of $(\lambda,L')$.
Finally, in this case by Lemma~31 the inequalities $D^{(2)}_i\ge
D^{(2)}_{\lambda_a,i}$,
$D^{(1)}_i\ge D^{(1)}_{\lambda,i}$ are satisfied.


Let us choose linear forms $\widetilde{L}_{\sigma+1},\ldots
,\widetilde{L}_n\in k[X_0,\ldots , X_n]$
such that the intersection of the algebraic varieties
$W_{\lambda,i-1}(L')\cap{\cal Z}(e'_i)$ and
${\cal Z}(\widetilde{L}_{i+1},\ldots ,\widetilde{L}_n)$ is transversal
for every $\sigma+1\le i\le n$ (this is a condition of general position for
the considered linear forms).
For every $\sigma+1\le i\le n$ denote by ${\cal W}'_i$
the union of all the irreducible over $\overline{k}$
components $E$ of the intersection ${\cal W}_{i-1}\cap{\cal
Z}(e^{(\varepsilon)}_i)$ such that
$E$ contains an irreducible component of
$(W_{\lambda,i-1}(L')\cap{\cal Z}(e'_i))\times\{0\}$.
Then for all $a\in k$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number, ${\cal W}^{(1)}_i\cap{\cal
Z}(\varepsilon-a)=(W_{\lambda_a,i-1}(L^{(a)})\cap{\cal
Z}(e^{(a)}_i))\times\{a\}$.
Denote by $C''_i$ the union of all the irreducible over $\overline{k}$
components $C''$ of
${\cal W}'_i\cap{\cal Z}(\widetilde{L}_{i+1},\ldots
,\widetilde{L}_n)$ such that $C''$ contains a point from
$(W_{\lambda,i-1}(L')\cap{\cal Z}(e'_i)
\cap{\cal Z}(\widetilde{L}_{i+1},\ldots
,\widetilde{L}_n))\times\{0\}$.
Hence $C''_i$ is a curve.
The natural projection $C''_i\rightarrow{\Bbb
A}^1(\overline{k})$,
$((X_0:\ldots : X_n),\varepsilon)\mapsto \varepsilon$ is dominant.
Hence by the B\'ezout for all $a\in k$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number,
$+\infty>\#(C''_i\cap{\cal Z}(\varepsilon-a))\ge
\#(C''_i\cap{\cal Z}(\varepsilon))=D_i+D^{(1)}_i+D^{(2)}_i$.
Hence for all $a\in k$, except at most a polynomial in
$d^n(d')^{n-\sigma}$ number, the intersection
$W_{\lambda_a,i-1}(L')\cap{\cal Z}(e^{(a)}_i)
\cap{\cal Z}(\widetilde{L}_{i+1},\ldots , \widetilde{L}_n)$
contains at least $D_i+D^{(1)}_i+D^{(2)}_i$ isolated points,
and, therefore, $\deg (W_{\lambda_a,i-1}(L')\cap{\cal Z}(e^{(a)}_i))\ge
D_i+D^{(1)}_i+D^{(2)}_i$.

We have, see above, $D_i(L^{(a)})=
D_{\lambda_a,i}(L^{(a)})=D_{\lambda_a,i}=D^{(j)}_i$ and
$D^{(j)}_{\lambda_a,i}\le D^{(j)}_i$, $j=1,2$.
Hence, see the Introduction, $D^{(j)}_{\lambda_a,i}(L^{(a)})\le
D^{(j)}_{\lambda_a,i}\le D^{(j)}_i$,
$D^{(j)}_i(L^{(a)})\le D^{(j)}_i$ $j=1,2$.
Therefore, $D^{(j)}_{\lambda_a,i}(L^{(a)})=
D^{(j)}_{\lambda_a,i}=D^{(j)}_i$ and $D^{(j)}_i(L^{(a)})=D^{(j)}_i$,
$j=1,2$.
This implies that for the considered $a$ intersection (58) with
$\lambda_a,L^{(a)}$ in place of
$\lambda,L'$ is transversal.
Hence (59)  with
$\lambda_a,L^{(a)}$ in place of
$\lambda,L'$ implies
$W^{(2)}_i(L^{(a)})\cap{\cal Z}(L_{a,s+1},\ldots ,
L_{a,\sigma})=W^{(2)}_{\lambda_a,i}(L^{(a)})$.
Finally from (60) with
$\lambda_a,L^{(a)}$ in place of
$\lambda,L'$ we get
$W^{(1)}_i(L^{(a)})\cap{\cal Z}(L_{a,s+1},\ldots ,
L_{a,\sigma})=W^{(1)}_{\lambda_a,i}(L^{(a)})$.
This implies all the assertions of (f).

The required bounds for the working time of the algorithms from
Theorem~2 follow immediately from the ones for the applied algorithms.
The theorem is proved.

\newpage

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\end{document}
