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\begin{document}

 \title{Polynomial--time computation of the degree of a
 dominant morphism in zero--characteristic I}
\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. We show
how to compute the degree of a dominant rational morphism from $W$ to
$W'$. The morphism is given by homogeneous
polynomials of degree $d'$.
This algorithms is deterministic and polynomial in $(dd')^n$ and the size
of input.
\end{abstract}




\newpage
 \section*{Introduction}


Let $k$ be a field of zero--characteristic with algebraic closure
$\overline{k}$. Let $X_0,X_1,\ldots $ be independent
variables over $k$.  In the algorithms we shall suppose that $k$ is
finitely generated over the field of rational numbers ${\Bbb
Q}$. Denote by ${\Bbb P}^n(\overline{k})$ the projective space
over the field $\overline{k}$ with coordinates $X_0,\ldots , X_n$.
We shall suppose that ${\Bbb P}^n(\overline{k})$ is defined
over $k$.  For arbitrary homogeneous polynomials $g_1,\ldots
,g_m\in\overline{k}[X_0,\ldots , X_n]$ 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).

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 $n+1$ variables 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 \cite{8}, Theorem~1 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 $\dim W=\dim W'$ (we can verify this
fact using \cite{6}, see below). Then
we construct an algorithm to compute the degree $\deg p'$
within the time polynomial in $(dd')^n$ and the size of input.
Even more general results are obtained, see Theorem~1.

Notice that the probabilistic algorithm polynomial in $(dd')^n$
and the size of input for computing the degree of a dominant
morphism is easy in arbitrary characteristic of the ground field.
The difficulties arise when one consider the most important model
of deterministic computations with bitwise complexity, cf. the Introduction
of \cite{6}.

In \cite{6}
we described an algorithm to decide whether a morphism $p$ is dominant.
The considered algorithm is polynomial in $(dd')^n$ and the size of  input.
We shall apply this algorithm as a base in the present paper.
But computing the degree of a dominant morphism is a much more difficult
problem than ones considered in \cite{6}.
Now we use essentially the results of
\cite{8} and \cite{7}.

For every $0\le s\le n$ denote by $V_s$ the union of all
the irreducible components of dimension $n-s$ of $V$.
According to  \cite{8}, Theorem~1 one can construct
within the time polynomial in $d^n$ and the size of input
for every $0\le s\le n$ a
zero--dimensional or empty projective algebraic variety $V_{s,0}$ such that
all the points of $V_{s,0}$ are
smooth points of dimension $n-s$ of $V$ and the irreducible over $k$
components of
$V_s$ and $V_{s,0}$ are in the one--to--one correspondence by the rule
$W\mapsto W\cap V_{s,0}$,
see the details in \cite{8}.
So we shall assume that all
$V_{s,0}$, $0\le s\le n$, are constructed.
For every $z\in V_{s,0}$ denote by $W_z$ the uniquely defined irreducible
over $k$ component of $V$ such that $z\in W_z$.
We shall use also Theorem~2 from \cite{8} which allow us to decide
within the time polynomial in $d^n$ and
the size of input whether two (or more)
given points are from the same irreducible over $k$ (or irreducible over
$\overline{k}$) component of $V$,
see the details in the cited theorem.
On the other hand we need to attract \cite{7}.
We discuss the results of this paper and deduce from them some new
corollaries in Section~1.


Notice also one correction in the statement of Lemma~2 from \cite{8}.
One needs to replace there ``a solution from ${\Bbb
A}^{4n+4}(\overline{k}_4)$ ...'' by
``a solution from ${\Bbb A}^{4n+4}(\overline{k}_4)$ giving a point from
$V_s^4(\overline{k}_4)$ ...''.
The proof is without changes. To find such a solution
in the algorithm for Theorem~1 \cite{8}
one needs only to apply the results of \cite{7}. It can be done directly, or
one can use there Theorem~2 Section~1
of the present paper replacing all the inequalities of system (7) \cite{8}
by nonstrict.


Now we proceed to the precise statements.
Let  homogeneous polynomials
$f_1,\,\ldots\,$, $f_m\in k[X_0,\,\ldots\, ,X_n]$ be given, $n\ge 1$, $m\ge
1$.
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 $r\ge 0$ be an integer and $e_0,\ldots ,e_r\in k[X_0,\ldots , X_n]$
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~1}\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~1, 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~1 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
$$
k(z)=k((X_1/X_{i_0})(z),\ldots , (X_n/X_{i_0})(z))=
k[\eta]\simeq k[Z]/(\Phi)
\eqno (2)
$$
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 $a\in k[\eta]$ be an arbitrary element. Then $a=A(\eta)$ for the
uniquely defined polynomial
$A\in k[Z]$ such that $\deg_ZA<\deg_Z\Phi$. The length of integer
coefficients ${\rm l}(a)$,
the the size ${\rm L}(a)$ and the degrees $\deg_{t_\alpha}a$,
$1\le \alpha\le l$, of $a$
are defined by the formulas
$$
{\rm l}(a)={\rm l}(A),\quad {\rm L}(a)={\rm L}(A), \quad
\deg_{t_\alpha}a=\deg_{t_\alpha}A.
$$
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 a 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 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.


Suppose that the dimension $\dim W=\dim W'$.
Denote by $W''\subset W'$ the subset
of all points
$z\in W'$ such that the number of elements $\# p^{-1}(z)\ne \deg p'$
or $z$ is a singular point of $W'$. Then
since the set of all singular points of $W'$ is closed and
by Lemma~2, see below, $W''$ is closed
with respect to the Zariski topology in $W'$
and $W''\ne W'$.
Denote by $W'''$ the closure in the Zariski topology of all points $z\in
W'$ such that
$\dim p^{-1}(z)>0$. So $W'''\subset W''$.

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
$\dim\overline{p(E)}<\dim E$. Therefore, the closure
$$
\overline{p(W^{(4)}\setminus{\cal Z}(e_0,\ldots , e_r))}=W'''.
$$
Further, $p^{-1}(W''')\ne
W\setminus{\cal Z}(e_0,\ldots , e_r)$. Hence
$\dim W^{(4)}=\dim p^{-1}(W''')\le\dim (W)-1$ and $\dim W'''\le\dim
(W)-2$.
Denote by $p^{(4)}\, :\, W^{(4)}\setminus{\cal Z}(e_0,\ldots , e_r)
\rightarrow W'''$ the restriction of $p'$.
By the theorem about dimension of fibers, see \cite{11},
for every $z\in W'''$ the dimension
of every irreducible component (if it exists,
$(p^{(4)})^{-1}(z)$ may be empty)
of $(p^{(4)})^{-1}(z)$ is at least $1$.
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 $\dim E>0$.



\par\medskip\noindent{\bf THEOREM~1}\hspace{0.1em} {\it
Let the field $k$, the algebraic variety $V$
and the polynomials $F$, $f_1,\ldots ,f_m$ and $e_0,\ldots ,e_r$
be given as above. Let a point $x^{(0)}\in V$ be given and $x^{(0)}$
belong to only one irreducible over $k$ component $W$ of $V$.
Let $W'$ be as above, see (1).
Then one can compute $\dim W$ and $\dim W'$.
Suppose that $\dim W=\dim W'=n-s$ where $0\le s\le n$.
Then the following assertions holds.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item One can compute the degree $\deg p'$ of the morphism $p'\,
:\, W\setminus{\cal Z}(e_0,\ldots , e_r)\rightarrow W'$
induced by (1).
\item More precisely,
one can compute indices $0\le i_0<\ldots < i_{n-s}\le r$
such that
$e_{i_0}$ does not vanish on $W$ and the family
$e_{i_1}/e_{i_0},\ldots , e_{i_{n-s}}/e_{i_0}$ is a basis of
transcendency of the field of rational functions
$k(W)$ over $k$. We shall suppose for convenience of notation that $i_j=j$
for all $j$.
Denote by $\rho\, :\, W\setminus{\cal Z}(e_0,\ldots ,
e_{n-s})\rightarrow{\Bbb P}^{n-s}(\overline{k})$,
$(X_0:\ldots : X_n)\mapsto
(e_0:\ldots : e_{n-s})$ the regular dominant morphism. Then one can
construct a point
$z'=(1:z'_1:\ldots : z'_{n-s})
\in{\Bbb P}^{n-s}(\overline{k})$ such that all
$z'_j$ are integers with lengths $O(n\,\log d+(n-s)\log(d'))$;
the number of elements
(here and below $\#$ denotes the number of elements of a set)
$\#\rho^{-1}(z')=\deg\rho$; the number of elements
$\#p(\rho^{-1}(z'))=(\deg\rho)/(\deg p')$;
all the points of $\rho^{-1}(z')$ are smooth points of $V$;
all the points of $p(\rho^{-1}(z'))$ are smooth points of $W'$.
Further, one can construct the sets $\rho^{-1}(z')$ and $p(\rho^{-1}(z'))$.
For an arbitrary $z^{(1)}\in p(\rho^{-1}(z'))$ the number of
elements $\#(p')^{-1}(z^{(1)})=\deg p'$,
cf. Lemma~2, Section~2, and every point from $(p')^{-1}(z^{(1)})$
belongs to the unique irreducible
over $k$ component of $V$, i.e., only to $W$.
\item One can compute the degree $\deg W'$.
More precisely, one can construct linear forms
$L'_0,L'_{s+1},L'_{s+2},\ldots , L'_n\in
k[X_0,\ldots ,X_r]$
satisfying the following properties. Set
$e'_j=L'_j(e_0,\ldots , e_r)$, $j\in\{0,s+1,\ldots , n\}$.
Then
\setcounter{equation}{2} \begin{eqnarray}
&&\#\bigl(\,(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{s+1},\ldots , e'_n)\bigr)=\deg(p')\deg W',
\label{3}\\
&&(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_0,e'_{s+1},\ldots , e'_n)=\emptyset,\label{4}\\
&&(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{s+1},\ldots , e'_n)\cap(\,\overline{V\setminus W}\,)=
\emptyset, \label{5}\\
&&\#\bigl(\,W'\cap{\cal Z}(L'_{s+1},\ldots , L'_n)\,\bigr)=\deg
W'.\label{6}
\end{eqnarray}
Hence, (3), (6) and Lemma~2, see below, imply that
for every point $x'\in W'\cap{\cal Z}(L'_{s+1},\ldots ,
L'_n)$ the number of elements $\#(p')^{-1}(x')=\deg p'$.
Therefore, each point of the intersection in the right part of
(6) is a smooth points of $W'$, and this intersection is
transversal at each its point. Further, again by Lemma~2 each point $x$ of
the intersection
in the right part of (3) is a smooth point of the varieties $W$,
${\cal Z}(e'_{s+1}),\ldots , {\cal Z}(e'_n)$
simultaneously and the intersection of tangent spaces of these $n-s+1$
projective algebraic varieties at $x$ is transversal,
i.e., is $\{0\}$. Besides that, (4),
(3), (6) and Lemma~2 imply
$$
W'\cap{\cal Z}(L'_0,L'_{s+1},\ldots , L'_n)=\emptyset.
\eqno (7)
$$
The constructed linear forms
$L'_0,L'_{s+1},L'_{s+2},\ldots , L'_n$ have
integer coefficients of length
$$
O(n\,\log d+(n-s)\log(d')).
$$
One can construct also
all the points from the intersections
in the right parts of (6), (3).
\item Let us identify the set of $(n-s+1)$--tuples of linear forms from
$\overline{k}[X_0,\ldots ,X_r]$ with ${\Bbb
A}^{(r+1)(n-s+1)}(\overline{k})$. Then by Lemma~2 the set ${\cal U}'_0$
of all $(L''_0,L''_{s+1},L''_{s+2},$ $\ldots , L''_n)\in {\Bbb
A}^{(r+1)(n-s+1)}(\overline{k})$ satisfying
(3), (4), (5), (6) in place of
$(L'_0,$ $L'_{s+1},L'_{s+2},\ldots , L'_n)$ is a nonempty open in the Zariski
topology subset of
${\Bbb A}^{(r+1)(n-s+1)}(\overline{k})$.
Let ${\mathfrak l}'\subset
{\Bbb A}^{(r+1)(n-s+1)}(\overline{k})$ be a line such that
${\mathfrak l}'\cap{\cal U}'_0\ne\emptyset$. Then the number of elements
$\#({\mathfrak l}\setminus{\cal U}'_0)$
is bounded from above by a polynomial
in $d^n(d')^{n-s}$.
\item More than that, one can construct linear forms
$L'_0,L'_{s+1},\ldots , L'_n$ from (c) satisfying additionally
the condition
$$
\dim (\,W^{(4)}\setminus{\cal Z}(e_0,\ldots , e_r)\,)
\cap{\cal Z}(e'_{s+1},\ldots , e'_i)\le n-i-1,\quad
\eqno (8)
$$
in ${\Bbb P}^n(\overline{k})$ for every $s+1\le n-1$.
\item
Let $L'_{s+1},L'_{s+2},\ldots , L'_n\in
\overline{k}[X_0,\ldots ,X_r]$ be arbitrary linear
forms satisfying (3), (6) and (8) for all $s+1\le i\le n-1$.
Then for every $s+1\le i\le n$ the dimensions
\setcounter{equation}{8} \begin{eqnarray}
&&\dim W\cap{\cal Z}(e'_{s+1},\ldots , e'_i)\setminus{\cal
Z}(e_0,\ldots , e_r)=n-i,
\label{9}\\
&&\dim W''\cap{\cal Z}(L'_{s+1},\ldots , L'_i)\le n-i-1 \label{10}
\end{eqnarray}
(here and below in the present paper we suppose that an algebraic variety is
empty if and only if its
dimension is negative; we do not fixe the special negative value for the
dimension of an empty set). Hence
\setcounter{equation}{10} \begin{eqnarray}
&&W''\cap{\cal Z}(L'_{s+1},\ldots , L'_n)=\emptyset,\label{11}\\
&&(\,W^{(4)}\setminus{\cal Z}(e_0,\ldots , e_r)\,)
\cap{\cal Z}(e'_{s+1},\ldots , e'_{n-1})=\emptyset \label{12}
\end{eqnarray}
in ${\Bbb P}^r(\overline{k})$ and ${\Bbb P}^n(\overline{k})$
respectively
Further, each irreducible component of $(\,W\setminus{\cal
Z}(e_0,\ldots ,$ $e_r)\,)\cap{\cal Z}(e'_{s+1})$ is not an irreducible
component
of $W^{(4)}\setminus{\cal Z}(e_0,\ldots , e_r)$, each irreducible
component of $W'\cap{\cal Z}(L'_{s+1})$
is not an irreducible component of $W''$.
Finally, if (5) holds then each irreducible component of
$(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{s+1})$ is not an irreducible
component $(\overline{V\setminus W})\setminus{\cal Z}(e_0,\ldots ,$
$e_r)$.
\end{enumerate}
The working time of the algorithm from (a)--(c) and (e) is polynomial in
$d^n$, $(d')^n$, $d_1$, $d_2$, $d'_2$,
$M$, $M'$, $M_1$, $m$, $r$, $N$ and
the size ${\rm L}(x^{(0)})$ of the point $x^{(0)}$.
}\par\medskip


\par\medskip\noindent{\bf REMARK~2}\hspace{0.1em} {\it  Let us identify
${\Bbb P}^n(\overline{k})={\cal
Z}(X_{n+1})\subset{\Bbb P}^{n+1}(\overline{k})$. Hence $V\subset{\Bbb
P}^{n+1}(\overline{k})$.
Replacing if necessary $n$ by $n+1$ and ${\Bbb P}^n(\overline{k})$ by
${\Bbb P}^{n+1}(\overline{k})$ we shall suppose without loss of
generality that
$s\ge 1$. Besides that, we can suppose that $f_1,\ldots ,f_m$ are linearly
independent over $k$.
}\par\medskip


\par\medskip\noindent{\bf REMARK~3}\hspace{0.1em} {\it  One can combine
conditions (3), (4)
and (5) in the one equivalent
$$
\#\bigl(\,W\cap{\cal Z}(e'_{s+1},\ldots , e'_n)\setminus({\cal
Z}(e'_0)\cup\overline{V\setminus W})\,\bigr)=\deg (p')\deg W'.
$$
}\par\medskip



\section{Preliminaries}\label{s1}

We shall use in this paper the technique from \cite{3},
\cite{5}, \cite{6}, \cite{7}.
Namely, we apply Newton--Puiseux expansions and
the Newton--Puiseux algorithm for constructing roots of a polynomial
in the field of fractional power series in one variable,
we apply infinitesimals, standard parts,
constructing real structures.



Now we need to discuss the results from \cite{7} to deduce some corollaries.
Consider a bounded closed
real algebraic variety $V$ which is a set of all common zeroes of a family
of polynomial $f_1,\ldots ,f_m\in k_0[X_1,\ldots ,X_n]$
of degrees less than $d$ with coefficients from
a real ordered field $k_0$.
Let $R=\widetilde{k_0}$ be a real closure of $k_0$.
The real field $k_0$ is finitely generated
over ${\Bbb Q}$ and is given
is similarly to $k$
from the Introduction.
The nonhomogeneous polynomials $f_i$, $F$ (now $F$ corresponds to $k_0$)
have the same bounds for degrees and lengths of coefficients
as in the Introduction. Additionally
we suppose that for an element $a\in k_0$ one can decide
whether $a>0$ whithin the polynomial time in the size of $a$.

Consider a triangulation of $V$. Consider the cycles
with coefficients from ${\Bbb Z}/2{\Bbb Z}$
of all dimensions of the triangulation of $V$ which are sums of maximal
(by inclusion in the triangulation) simplices.
The main result here is Theorem~1 \cite{7}
which allows to construct within the
time polynomial in $d^n$ and the size of input a
finite subset $S$ of $V$
which has a nonempty intersection with any considered
cycle of any triangulation of $V$.
The number of elements of $S$ is bounded from above by a
polynomial in $d^n$.
The most part of \cite{7} is devoted to the proof of Theorem~1.
The case of an arbitrary real algebraic variety is
reduced to the considered using the
Alexandrov compactification.

Now let $n\ge 1$ and $s\ge 1$ be integers and $g_1,\ldots g_s\in
k_0[X_1,\ldots, X_n]$ be some polynomials
which have the estimations
for degrees and sizes of integer coefficients similar to ones for
$f_1,\ldots,f_m$ from the Introduction. Consider a basic
semialgebraic set (an arbitrary semialgebraic set can be represented as a
union of basic ones)
$$
U=\{x\in R^n\, :\, f_1(x)=\ldots=f_m(x)=0\,\& \,g_1(x)>0\,\&\,\ldots
\,\&\, g_s(x)>0\}.
\eqno (13)
$$
Let $2^{u-1}<n\le 2^u$ for an integer $u$.
Let $s\le s'2^u< s+2^u$ and $s'\in {\Bbb Z}$. If $s<2^u$ set $g_j=1$
for all $s+1\le j\le 2^u$. Set $s_j=s'j$ for
$j=0,\ldots , 2^u-1$ and $s_{2^u}=\max\{s,2^u\}$.
Let $Z_{j,i}$, $1\le j\le 2^{u-i}$, $0\le i\le u$,
be new variables.
Consider the real algebraic variety
$U_1\subset{\Bbb A}^{n+2^{u+1}-1}$
which is the set of all common zeroes of
polynomials
$\prod_{s_{j-1}+1\le i\le s_j}g_i-Z_{j,0}$, $1\le j\le 2^u$;
$f_1,\ldots,f_m$,
and $Z_{j,i}Z_{2j-1,i-1}Z_{2j,i-1}-1$, $1\le j\le 2^{u-i}$, $1\le i\le u$,
(so in the case when $s<2^u$ the
polynomials $1-Z_{j,0}$, $s+1\le j\le 2^u$, belong to the considered family).
Denote by
\begin{eqnarray*}
&&\widetilde{\pi}\, :\, {R}^{n+2^{u+1}-1}\rightarrow {R}^{n},\\
&&(X_1,\ldots ,X_n;\,Z_{j,i},\, 1\le j\le 2^{u-i},\, 0\le i\le u)
\mapsto (X_1,\ldots ,X_n)
\end{eqnarray*}
the linear projection.
Then the  set $U$ coincides with $\widetilde{\pi}(U_2)$ where $U_2$ is a
union of some connected components of $U_1$


{\it Let $W\subset R^n$ be an arbitrary set. By definition the family of all
irreducible
(over $R$) components of $W$ is  $W\cap W'_\alpha$, $\alpha\in A$, where
 $W'_\alpha$, $\alpha\in A$, is the family of all the irreducible (over $R$)
components of the closure of $W$ in $R^n$ with respect to the
Zariski topology.} In this paper and \cite{7} {\it by connected
components of semialgebraic sets we mean
semialgebraic connected components of these sets,} see the
definition in \cite{1}.


Denote by ${\dot U}^{(v)}$ the set of all smooth points of dimension $v$
of $U$ and by
$U^{(v)}$ the closure of ${\dot U}^{(v)}$ in $U$ with respect to the
topology
of the real ordered field $R$. Therefore, ${\dot U}^{(v)}=
\bigcup_{j\in J_v}{\dot U}^{(v)}_j$, where ${\dot U}^{(v)}_j$ are
all the pairwise distinct irreducible over $R$ components of  ${\dot
U}^{(v)}$.
Denote by $U^{(v)}_j$ the closure of ${\dot U}^{(v)}_j$ in $U$ with respect
to the topology of the real field $R$. Hence $U^{(v)}=\bigcup_{j\in
J_v}U^{(v)}_j$, and every $U^{(v)}_j$ is a semialgebraic set.
Now applying Corollary~2 and Remark~2 from \cite{7}
to the real algebraic variety $U_1$ we get the following result.


\par\medskip\noindent{\bf PROPOSITION~1}\hspace{0.1em} {\it  (cf. Theorem~2
\cite{7})
Consider a basic semialgebraic set (13). Then
one can construct a finite  set $S$ of points of $U$  which
has a non--empty intersection with every connected component of every
$U_j^{(v)}$  for every $j\in J_v$, $0\le v\le n$.
The number of points of $S$ is bounded from above by
a polynomial in $((1+s/n)d)^n$. The working time of the
algorithm for constructing $S$ is polynomial
in $((1+s/n)d)^n$, $d_1$, $d_2$, $M$, $M_1$, $m$.
}\par\medskip


So Proposition~1 is an almost immediate consequence of Theorem~1
\cite{7}. In the given here Proposition~1 we remove inaccuracies from
\cite{7}.
Namely, we slightly modify the construction introducing the variables
$Z_{i,j}$ before the statement.
Let $U_i$, $i\in I$, be
the family of all irreducible components of $U$
(or alternatively of all the connected components
of all irreducible components of $U$).
Let $U_i^{(v)}$ be the closure
in the topology of the real ordered
field of the set of all points which are
smooth points of dimension $v$ of $U_i$ and $U$ simultaneously.
Hence $U^{(v)}=\cup_{i\in I}U^{(v)}_i$.
Therefore, each connected component of every $U_i^{(v)}$ is a union
of some connected components of some sets $U^{(v)}_j$, $j\in J_v$
(we leave the details to the reader).
Hence if one adds in the statement
Theorem~2 \cite{7} add ``and $U$ simultaneously'' this theorem
becomes a consequence of Proposition~1.
On the other hand, one needs to
add ``and $U$ simultaneously'' since it may happen that a connected
component $c'$ of an irreducible component $W'$ of $U$ is contained in a
connected component $c''$ of an irreducible component $W''$ of $U$ and $\dim
c'<\dim c''$. One need to exclude all such connected components $c'$ since
they do not correspond to the considered cycles with maximal simlices of the
Alexandrov compactification of $U_1$, see \cite{7} for details.

Now let
$q_1,\ldots, q_{s_1}\in k_0[X_1,\ldots, X_n]$ be some polynomials
which have estimations
for degrees and sizes of integer coefficients similar to ones for
$f_1,\ldots,f_m$, see the Introduction and \cite{7}.
Let $\alpha_i$ denotes $=$ or $\ge$ or $>$ for every $1\le i\le s_1$.
Consider a
semialgebraic set
$$
U'=\{\,x\in R^n\, :\,
q_1(x)\alpha_1 0\quad\&\,\ldots
\,\&\quad q_{s_1}(x)\alpha_{s_1}0\,\}.
\eqno (14)
$$
Let $j\in J_v$, $0\le v\le n$ in the previous notation. Let
${\cal C}'_{j,v}$ be the set of all connected
components of $U_j^{(v)}\cap U'$ in the notation of Proposition~1.
For every $c\in {\cal C}'_{j,v}$ denote by $\overline{c}$ the
closure in $R^n$ with respect to the
topology of the real field $R$ of $c$.
For $c',c''\in {\cal C}'_{j,v}$ set
$c'\sim c''$ if and only if there are $c_1,\ldots , c_\nu\in {\cal
C}_{j,v}$, $\nu\ge 2$,
such that $c_1=c'$, $c_\nu=c''$ and
$\overline{c_i}\cap\overline{c_{i+1}}\ne\emptyset$
for every $1\le i\le\nu-1$. Denote by $\overline{{\cal C}}_{j,v}$ the
set of classes of
equivalence of ${\cal C}'_{j,v}$ by this relation $\sim$.
Set ${\cal C}''_{j,v}=\{\cup c\, :\, c\in \overline{{\cal
C}}_{j,v}\}$ (here $\cup c$ is the union of the sets which are
the elements of $c$), i.e., for every
 $\widetilde{c}\in {\cal C}''_{j,v}$ there is $c'\in{\cal
C}'_{j,v}$ such that $\widetilde{c}=\cup_{c''\sim c'}c''$.
Notice that if $U$ is a closed real algebraic variety in $R^n$ and all
$\alpha_i$ are $=$ or $\ge$
then ${\cal C}''_{j,v}={\cal C}'_{j,v}$.

\par\medskip\noindent{\bf COROLLARY~1}\hspace{0.1em} {\it  In the previous
notation
one can construct a finite  set $S'$ of points of $U\cap U'$  which
has a non--empty intersection with every $c\in {\cal
C}''_{j,v}$ for every $i\in J_v$, $0\le v\le n$.
The number of points of $S'$ is bounded from above by
a polynomial in $((1+(s+s_1)/n)d)^n$. The working time of the
algorithm for constructing $S'$ is polynomial
in $((1+(s+s_1)/n)d)^n$, $d_1$, $d_2$, $M$, $M_1$, $m$.
}\par\medskip

\noindent {\bf PROOF} \quad
Let $\varepsilon>0$ be an infinitesimal with respect to the field $k_0$.
Consider the field $k_0(\varepsilon)$
as a ground field. Let $\widetilde{R(\varepsilon)}$
be a real closure of $R(\varepsilon)$.
Let us replace in (14) each equation $q_i=0$ (respectively inequality
$q_i\ge 0$)
by the conjugation $q_i+\varepsilon>0$ and
$\varepsilon-q_i>0$ (respectively $q_i+\varepsilon>0$). Denote by
$U'_\varepsilon$ the obtained open in the topology
of the real field $\widetilde{R(\varepsilon)}$ semialgebraic set.
Applying Proposition~1 to the set $U\cap U'_\varepsilon$ one constructs the
finite
set of points
$S'_\varepsilon$. Let $c\in C''_{j,v}$.  Then one sees that
$c\subset \cup_{\gamma\in \Gamma}c_\gamma$
where $c_\gamma$, $\gamma\in \Gamma$ are all the
connected
components of the intersection $U^{(v)}\cap U'_\varepsilon$ such that
$c_\gamma\cap c\ne\emptyset$.
For every $z\in \widetilde{R(\varepsilon)}^n$ that is not infinitely large
with respect to $R$ the standard part
$\mbox{\rm st}_\varepsilon(z)\in R^n$ is defined ($\mbox{\rm
st}_\varepsilon(z)$ is
the element which is infinitely close to $z$ with respect to $R$).
Then one sees $\mbox{\rm
st}_\varepsilon(\cup_{\gamma\in\Gamma}c_\gamma)
\supset c$.
On the other hand, each semalgebraic set has only a finite number of
connected components.
Hence according to the given definitions
$\mbox{\rm st}_\varepsilon(c_\gamma)=
\cup_{1\le i\le\mu}c_{\gamma,i}$ where $c_{\gamma,i}$ are some connected
components of
$U^{(v)}\cap U'$ (here $\mu$ depends on $\gamma$).
Besides that, for all $1\le i<j\le\mu$ there are
$1\le i_1,\ldots , i_\omega\le \mu$ such that
$i=i_1$, $j=i_\omega$ and
$\overline{c_{\gamma,i_u}}\,\cap\,\overline{c_{\gamma,i_{u+1}}}\ne\emptyset$
for $1\le u\le\omega-1$.
Thus, $\mbox{\rm
st}_\varepsilon(\cup_{\gamma\in\Gamma}c_\gamma)=c$.
Therefore, $\mbox{\rm st}_\varepsilon(S'_\varepsilon)\cap
c\ne\emptyset$ by Proposition~1.
Now using the algorithm from \cite{3}, \cite{5} we construct
the set $\mbox{\rm st}_\varepsilon(S'_\varepsilon)$.
Put $S'=\mbox{\rm st}_\varepsilon(S'_\varepsilon)$.
The constructed set $S'$ satisfies the required properties.
The estimate for the working time of the algorithm for constructing $S'$
follows immediately from the ones
of the applied algorithms. The corollary is proved.

\medskip The real structure on the field $k$ is an embedding
$k\rightarrow K[\sqrt{-1}]$ where $K$ is a real ordered field, see \cite{4}.
If $k$ is supplied with a real structure then the
absolute value $|a|$ of an element $a\in k$ is defined in the natural way.
Using the algorithm from \cite{4} one can construct
within the polynomial time a real structure on $k$
such that $K$ is a finite extension of $k$, the field  $K$ is given
similarly to  $k_0$, and
\begin{enumerate} \renewcommand{\labelenumi}{(\Alph{enumi})}
\item for every element $a\in K$ one can decide whether $a>0$
within the time polynomial in
$d_1$, $M_1$, and
the size ${\rm L}(a)$ of the element $a$.
\end{enumerate}

We shall use many times Proposition~1 \cite{6},
for solving systems of polynomial equations
and inequalities with squares of absolute values and infinitesimals
within polynomial time
without references to this proposition.

Now we shall change slightly the conditions of Proposition~1 \cite{6}.
Consider inequalities in $X_1,\ldots , X_n$ of the type
$A\ge 0$ where
$A=\sum_{1\le i\le r}a_i|A_i|^2$, $a_i\in K$, and $A_i$ is a superposition of
polynomials with coefficients from $k$
and functions $B\mapsto |B|^2$ for every $i$
The upper bound for the degree of $A$ can be easily
defined if we set $\deg |B|^2=2 \deg B$.

Set
$$
U''=\{\;x\in \overline{k}^n\, :\,
q_1(x)\alpha_1 0\quad\&\,\ldots
\,\&\quad q_{s_1}(x)\alpha_{s_1}0\;\}
\eqno (15)
$$
Suppose that in (15) all $\alpha_i$ are $=$ or $\ge$ and that
all the inequalities
$q_j\ge 0$ have the considered form $A\ge 0$.
For each equation $q_j=0$ from (15)
assume that either it has the form $A=0$ where
$A$ is as above, or $q_j\in k[X_1,\ldots ,X_n]$ (it will be sufficient for
our aims).
Let $f_1,\ldots ,f_m\in k[X_1,\ldots ,X_n]$ be polynomials.
Suppose that $f_1,\ldots ,f_m$, $g_1,\ldots ,g_{s_1}$
satisfy the similar inequalities for degrees and lengths of coefficients as
the polynomials $f_1,\ldots ,f_m$ from
the Introduction do. Let $V={\cal Z}(f_1,\ldots
,f_m)\subset{\Bbb A}^n(\overline{k})$, cf. the Introduction.

\par\medskip\noindent{\bf THEOREM~2}\hspace{0.1em} {\it  Under the
formulated conditions denote by
${\cal C}$ the set of all
connected components of $W\cap U''$, where $W$ runs over all
irreducible over $\overline{k}$
components $W$ of $V$. Then one can
construct a finite set $S''$ of points of $V\cap U''$ which has a
nonempty intersection with every connected component
$c\in {\cal C}$. The
number of points of $S''$ is bounded from above by a polynomial in
$((1+s_1/n)d)^n$. The working time of the algorithm for constructing $S''$ is
polynomial in $((1+s_1/n)d)^n$, $d_1$, $d_2$, $M$, $M_1$, $m$.
}\par\medskip

\noindent {\bf PROOF} \quad
Each system of equations and inequalities with squares of
absolute values over $\overline{k}$ is
equivalent to the system of equations and inequalities
over $\widetilde{K}$ with twice as much of variables.
Hence it is sufficient to apply Corollary~1. The corollary is proved.

\par\medskip\noindent{\bf REMARK~4}\hspace{0.1em} {\it  Since $k$
(respectively $k_0$)
is an arbitrary finitely generated over ${\Bbb Q}$ field
(respectively finitely generated over ${\Bbb Q}$
real ordered field) Corollary~1 and Theorem~2
are valid
also if one replaces the field of coefficients $k$
(respectively $k_0$) by an
extension of $k$ (respectively $k_0$) by an
algebraic element over $k$ (respectively $k_0$)
of degree polynomial in $d^nd_1d_2$
and a fixed number of infinitesimals.
}\par\medskip


In the proof of the following simple proposition
we describe a usefull algorithm of reduction
to general position, cf. also ``the good properties'' from
\cite{5}, \cite{6}. In what follows we shall refer  to it as to
{\it the algorithm of reduction to integer coefficients.}

\par\medskip\noindent{\bf PROPOSITION~2}\hspace{0.1em} {\it  Let ${\cal
U}_i\subset\overline{k}^N$ and $\nu_i\ge
1$, $1\le i\le w$ be integers. Suppose that
for every $z\in{\cal U}_i$ for every line
${\mathfrak l}\subset\overline{k}^N$ through $z$ the number of elements
$\#({\mathfrak l}\setminus{\cal U}_i)\le\nu_i$
for every $1\le i\le w$. Suppose that one can decide for a given point
$z\in\overline{k}$ whether
$z\in{\cal U}_i$, $1\le i\le w$. Suppose that a point
$x_i\in{\cal U}_i$ is given for every
$1\le i\le w$.
Then one can construct a point $x'=(x'_1,\ldots , x'_N)
\in\bigcap_{1\le i\le w}{\cal U}_i$ such that  all $x'_i$, $1\le i\le
N$, are integers of length
$O(\log(1+\sum_{1\le i\le w}\nu_i))$.
}\par\medskip

\noindent {\bf PROOF} \quad
Let $w=1$. Set $x^{(0)}=x_1$. Let $0\le j<N$ be an integer.
Suppose that we have constructed a point
$x^{(j)}=(x^{(j)}_1,\ldots , x^{(j)}_N)\in{\cal U}_1$ such that all
$x^{(j)}_i$, $1\le i\le j$,
are integers of length $O(\log(1+\nu_1))$. Then considering the line
${\mathfrak l}$
parallel to $(j+1)$th coordinate axe such that $x^{(j)}\in{\mathfrak l}$ we
construct a
point $x^{(j+1)}$. Finally we set $x'=x^{(N)}$.
The proposition is proved for $w=1$.

Let $w>1$. Suppose that we have already
constructed a point $z^{(j)}=(z^{(j)}_1,\ldots ,
z^{(j)}_N)\in\bigcap_{1\le i\le j}{\cal U}_i$ for some $1\le
j<w$ such that all $z^{(j)}_i$, $1\le i\le N$, are integers of
length $O(\log(1+\sum_{1\le i\le j}\nu_i))$. Considering a line
through $z^{(j)}$ and $x_{j+1}$ we construct a point
$y^{(j+1)}\in\bigcap_{1\le i\le j+1}{\cal U}_i$. Applying the
algorithm for the case $w=1$ to the set $\bigcap_{1\le i\le
j+1}{\cal U}_i$ and its point $y^{(j+1)}$ we construct a point
$z^{(j+1)}$. Finally we set $x'=z^{(w)}$. The proposition is
proved.




\section{The first auxiliary algorithm for Theorem~1 }\label{s2}

We shall suppose in this section
that $\dim W=\dim W'=n-s$ and the
transcendency degree of the family of rational functions $e_i/e_0\in k(W)$,
$1\le i\le n-s$, is $n-s$. In particular, $e_0$ is not vanishing on $W$.


For an arbitrary projective algebraic variety $V'\subset{\Bbb
P}^n(\overline{k})$ we shall denote by $\mbox{\rm con}(V')$
in what follows the
affine algebraic variety
in ${\Bbb A}^{n+1}(\overline{k})$ which is the set of zeroes of the
homogeneous ideal of $V'$.



Let $\varepsilon_i>0$, $1\le i\in{\Bbb Z}$,
be an infinitesimal with respect to the field
$k_{i-1}=k(\varepsilon_1,\ldots ,\varepsilon_{i-1})$.
One can supply the algebraic closure
$\overline{k_i}$ with a real structure, which extends the one of $k$
$\overline{k_i}$ with a real structure, which extends the one of $k$
see \cite{4}, \cite{5}.

Let $L_0,\ldots ,L_{n+1}$
be linear forms from Theorem~1 \cite{8} for the projective
algebraic variety $V\cup{\cal Z}(e_0)={\cal Z}(f_1e_0,
\ldots , f_me_0)$.
Hence $W\cap{\cal Z}(L_0,L_{s+1},\ldots , L_n)=\emptyset$ in ${\Bbb
P}^n(\overline{k})$, $W\not\subset{\cal Z}(L_0)$ and
$W\cap{\cal Z}(L_{s+1},\ldots , L_n)\cap{\cal Z}(e_0)=\emptyset$.
In what follows we fix $L_1,\ldots , L_{n+1}$. On the other hand, for a
given point $x^*\in{\Bbb P}^n(\overline{k})$
we shall choose $L_0$ such that additionally $L_0(x^*)\ne 0$.


Now we are going to describe the first auxiliary algorithm. In
the input of this algorithm one has a point $x^*\in W$
that is a smooth point of $V$ and $(L_0e_0)(x^*)\ne 0$. In output one has a
point
$x^{**}=x^{(n)}\in W(\overline{k(\varepsilon_2)})$ satisfying
conditions
(a)--(d) with $v=n$, see below.
In this auxiliary algorithm for $s\le v\le n$ one constructs recursively
points $x^{(v)}\in V(\overline{k(\varepsilon_2)})$
and the integers $s+1\le i_{v,v+1}<\ldots < i_{v,n}\le n$ satisfying the
following properties:
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item $(e_0L_0)(x^{(v)})\ne 0$,
\item $\sum_{0\le i\le n}|(X_i/L_0)(x^{(v)})-(X_i/L_0)(x^*)|^2\le
\varepsilon_1$,
\item $L_i(x^{(v)})=L_i(x^*)-\mu_i\varepsilon_2L_0(x^*)$, $s+1\le i\le n$,
where $\mu_i$ are integers (depending on $v$) with lengths $O(n\log d+
(n-s)\log d')$.
\end{enumerate}
Further, denote for brevity $\iota_\alpha=i_{v,\alpha}$, $v+1\le \alpha\le n$
when $v$ is fixed. Set $\widetilde{e}_0=1$ if $v=s$, and
$\widetilde{e}_0=e_0$ if $s+1\le v\le n$.
Denote by
\begin{eqnarray*}
&&\pi_v\, :\, W\setminus{\cal Z}(L_0\widetilde{e}_0,L_0e_1,\ldots ,
L_0e_{v-s},
L_{\iota_{v+1}}\widetilde{e}_0,\ldots
,L_{\iota_n}\widetilde{e}_0)\rightarrow{\Bbb
P}^{n-s}(\overline{k}),\\
&&(X_0:\ldots : X_n)\mapsto(L_0\widetilde{e}_0:L_0e_1:\ldots : L_0e_{v-s}:
L_{\iota_{v+1}}\widetilde{e}_0:\ldots :L_{\iota_n}\widetilde{e}_0)
\end{eqnarray*}
the regular morphism.
Then
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\setcounter{enumi}{3}
\item the differential $d_{x^{(v)}}\pi_v$ of the
morphism $\pi_v$ at the point $x^{(v)}$
is an isomorphism and
hence $\pi_v$ is a dominant morphism.
\end{enumerate}
We shall say that
a point $y\in W$ (which may have coordinates in any
extension of $\overline{k(\varepsilon_2)}$
supplied with the corresponding real structure) satisfies condition (a)
(respectively (b), (c), (d))
if and only if condition (a) (respectively (b), (c), (d)) holds with $y$
in place of $x^{(v)}$.

\medskip
Denote $k'_\beta=k(\varepsilon_2,\ldots , \varepsilon_\beta)\subset k_\beta$
where
$2\le\beta\in{\Bbb Z}$ is fixed (for estimation of the complexity of the
algorithm).
Let a point $\widetilde{x}_v\in V(\overline{k'_\beta})$.
Suppose that $\widetilde{x}_v$ satisfies (a), (b), (d).
Let ${\mathfrak l}\subset{\Bbb P}^{n-s}(\overline{k'_\beta})$ be an
arbitrary line
containing the point $\pi_s(\widetilde{x}_v)$ and $C_1$ be the (unique by
the implicit function theorem)
defined and irreducible over $\overline{k'_\beta}$ component of
$\pi_s^{-1}({\mathfrak l})$ such that
$\widetilde{x}_v\in C_1$. Then there is at most a polynomial in
$d^n(d')^{n-s}$
number of points $x'\in C_1$ such that
at least one of conditions (a), (d) is not satisfied for $x'$.
On the other hand, see, e.g., Proposition~6 \cite{5}, for every point
$z\in{\Bbb P}^{n-s}(\overline{k'_\beta})$ that is infinitely close to
$\pi_s(x^*)$ with respect to the field $k$ there is a point $x''\in
\pi_s^{-1}(z)$ such that
$x''$ is infinitely close to $x^*$ with respect to the field $k$.
Hence, cf. the algorithm of reduction to integer coefficients, Proposition~2,
enumerating
$\mu_i=0,1,\ldots , $ for every $s+1\le i\le n$ and solving
systems of polynomial equations and inequalities with squares of absolute
values
one can replace subsequently the coordinates
$L_i(x^{(v)})$ of $\pi_s(\widetilde{x}_v)$ by
by $L_i(x^*)-\varepsilon_2\mu_iL_0(x^*)$,
where $\mu_i$ are integers with lengths $O(n\log d+(n-s)\log d')$,
and construct
a point $x^{(v)}$ satisfying all the conditions (a)--(d). So it is
sufficient to
construct a point $\widetilde{x}_v$.

The base of the recursion is $v=s$. In this case
obviously $i_{s,\alpha}=\alpha$ for all $s+1\le \alpha\le n$. Let
${\mathfrak l}\subset{\Bbb P}^{n-s}(\overline{k})$ be a line containing
the
points
$\pi_s(x^*)$ and $(1:0:\ldots : 0)$
(this line is unique if these two points do not coincide).
By Theorem~1 \cite{8} for every point $y\in
\pi_s^{-1}((1:0:\ldots : 0))$ the differential $d_y\pi_s$ is an isomorphism.
Let $x^*=(x^*_0:\ldots : x^*_n)$ where all $x^*_i\in\overline{k}$ and
$L_0(x^*_0,\ldots , x^*_n)=1$.
Let us construct solving a system of polynomial equations
and using Theorem~2 \cite{8}
all the points from
$\mbox{\rm con}(W)$
satisfying the system
$$
\left\{\begin{array}{ll}
f_i=0, & 1\le i\le m,\\
L_i=L_i(x^*_0,\ldots , x^*_n),& s+1\le i\le n,\\
L_0=1+\varepsilon_2\\
\end{array}\right.
\eqno (16)
$$
(the set of these points is finite).
After that we choose among the constructed solutions of system (16) a
point
$(\widetilde{x}_{s,0},\ldots , \widetilde{x}_{s,n})\in{\Bbb A}^{n+1}
(\overline{k'_2})$ such that
$\sum_{0\le i\le n}|\widetilde{x}_{s,i}-x^*_i|^2\le\varepsilon_1$
and denote $\widetilde{x}_s=(\widetilde{x}_{s,0}:\ldots :
\widetilde{x}_{s,n})$.
So the point $\widetilde{x}_s\in\pi_s^{-1}({\mathfrak l})$ is infinitely
close to $x^*$. Hence $\widetilde{x}_s$ is a smooth point of $V$ and
satisfies conditions
(a), (b), (d). Therefore, one can construct also $x^{(s)}$.


Now suppose that $s\le v\le n-1$ and the point $x^{(v)}$
is constructed. Our aim is to construct
$x^{(v+1)}$ and all $i_{v+1,\alpha}$, $v+2\le \alpha\le n$.
Let $Z_i$, $s\le i\le v+1$, $T_i$, $v+1\le i\le n$,
be new variables.
Let ${\Bbb A}^{2n-s+3}(\overline{k})$ has the coordinate functions
$X_0,\ldots , X_n$, $Z_s,\ldots , Z_{v+1}$, $T_{v+1},\ldots ,T_n$.
Consider the affine algebraic variety $V_1\subset{\Bbb
A}^{2n-s+3}(\overline{k})$ which
is the set of all common zeroes of the polynomials $f_1,\ldots ,f_m$;
$Z_se_0-1$; $Z_je_0-e_{j-s}$,
$s+1\le j\le v+1$; $T_jL_0-L_{\iota_j}$, $v+1\le j\le n$
(here $\iota_j=i_{v,j}$); $L_0-1$.
Denote for brevity the last family of polynomials
by $g_i\in k[X_0,\ldots , X_n,Z_s ,\ldots , Z_{v+1},T_{v+1},
\ldots , T_n]$,
$1\le i\le m_1=m+n-s+3$.
We shall identify $V_1=V\setminus{\cal Z}(e_0L_0)\subset V$.
So $X_i(x'')$, $0\le i\le n$;
$Z_i(x'')$, $s\le j\le v+1$; $T_i(x'')$, $v+1\le j\le n$,
are defined for every $x''\in V_1$.

In what follows we shall identify for convenience of notations the
tangent space ${\cal Z}(\sum_{1\le i\le n}a_{\gamma,i}dX_i,\,\gamma
\in \Gamma)$ (where $a_{\gamma,i }$ belongs to the coefficients field)
in a point of an affine algebraic variety with the subspace \\
${\cal Z}(\sum_{1\le i\le n}a_{\gamma,i}X_i,\,\gamma\in \Gamma)
\subset{\Bbb A}^n$, herewith ${\Bbb A}^n$
has the coordinates functions $X_1,\ldots ,X_n$.
We shall use the similar identification also for other
affine spaces, in particular, for
$A^{2n-s+3}$ with the coordinates
$X_0,\ldots , X_n$, $Z_{s+1},\ldots , Z_{v+1}$, $T_{v+1},
\ldots , T_n$.





\par\medskip\noindent{\bf LEMMA~1}\hspace{0.1em} {\it  Let $\xi\in V_1$ be a
smooth point (it may have
the coordinates in an arbitrary extension of
$\overline{k}$) and $\xi\in W$.
Let $T_{\xi,V_1}$ be the tangent spaces of $V$ at the
point $\xi$.
Then the intersection $T_{\xi,V_1}\cap{\cal Z}(Z_{s+1},\ldots ,Z_v,
T_{v+1},\ldots , T_n)=\{0\}$ if and only if
the differential $d_\xi\pi_v$ is an isomorphism.
}\par\medskip

\noindent {\bf PROOF} \quad Performing a non--degenerated
linear transformation of $X_0,\ldots , X_n$ we shall suppose without loss of
generality for convenience of notation that $L_0=X_0$.
Let $H_1,\ldots , H_s\in k[X_0,\ldots , X_n]$ be homogeneous polynomials
such that $H_i/X_0^{\deg H_i}$,
$1\le i\le s$, is a system of local parameters in the point $\xi$ of the
projective algebraic variety $V$.
One can choose $H_i$ using linear projections ${\Bbb P}^n(\overline{k})
\rightarrow{\Bbb P}^{n-s+1}(\overline{k})$
in general position
such that $\deg H_i\le\deg W\le d^s$ for all $1\le i\le s$
(we do not use the last fact now in the proof of the lemma
but shall use it later).

Then $H_i(1,X_1,\ldots ,X_n)$, $1\le i\le s$; $Z_se_0-1$;
$(Z_je_0-e_{j-s})(1,X_1,\ldots ,X_n)$,
$s+1\le j\le v+1$; $(T_jL_0-L_{\iota_j})(1,X_1,\ldots ,X_n)$,
$v+1\le j\le n$; $X_0-1$, is a system of local parameters
of $V_1$ in the point $\xi$. Denote the last system of local parameters by
$h_1,\ldots ,h_{n+3}$.
Then $T_{\xi,V_1}\cap{\cal Z}(Z_{s+1},\ldots , Z_v,
T_{v+1},\ldots , T_n)=\{0\}$ if and only if
the differentials of the polynomials $d_\xi h_1,\ldots , d_\xi h_{n+3}$,
$d_\xi Z_{s+1},\ldots , d_\xi Z_v$,
$d_\xi T_{v+1},\ldots , d_\xi T_n$ at the point $\xi$ are linearly
independent.
The differential $d_\xi\pi_v$ is an isomorphism if and only if the
differentials of the rational functions
$d_\xi h_1,\ldots ,$ $d_\xi h_s$, $d_\xi(e_1/e_0),$
$\ldots , d_\xi(e_{v-s}/e_0),$
$d_\xi (L_{\iota_{v+1}}/L_0),\ldots , d_\xi (L_{\iota_n}/L_0)$ are linearly
independent.
Now the lemma is obtained
by the direct comparison of the corresponding Jacobi
matrices taking into account that
$(e_0L_0)(\xi)\ne 0$. The lemma is proved.

\medskip Let
$X'_0,\ldots , X'_n$, $Z'_s,\ldots , Z'_{v+1}$, $T'_{v+1},\ldots , T'_n$,
be new variables.
Consider the system of
equalities and inequalities in $X_0,\ldots , X_n$,
$Z_s,\ldots , Z_{v+1}$, $T_{v+1},\ldots ,$ $T_n$
$X'_0,\ldots , X'_n$, $Z'_s,\ldots , Z'_{v+1}$, $T'_{v+1},\ldots , T'_n$
with coefficients from $\overline{k'_5}$:
$$
\left\{\begin{array}{ll}
 g_i=0, & 1\le i\le m_1,\\
 g_i(X'_0,\ldots , X'_n,Z'_s,\ldots ,
Z'_{v+1}, T'_{v+1},\ldots , T'_n)=0, & 1\le i\le m_1, \\
\sum_{0\le i\le n}|X_i-X_i(x^{(v)})|^2+
\sum_{s+1\le i\le v+1}|Z_i-Z_i(x^{(v)})|^2+\\
\sum_{v+1\le i\le n}|T_i-T_i(x^{(v)})|^2=
\varepsilon_3,\\
\sum_{0\le i\le n}|X_i-X'_i|^2+
\sum_{s+1\le i\le v+1}|Z_i-Z'_i|^2+\\
\sum_{v+1\le i\le n}|T_i-T'_i|^2=\varepsilon_5^2,\\
 Z_i-Z'_i=0, & s+1\le i\le v,\\
 |Z_{v+1}-Z'_{v+1}|^2>\varepsilon_5^2\varepsilon_4.
 \end{array}
 \right. \eqno (17)
$$

By (d), Lemma~1 and Lemma~6 \cite{6} system (17) has a solution
$\xi_1\in\\ {\Bbb A}^{2n-s+2}(\overline{k_5})$.
Let us construct such a solution $\xi_1$.
Put $\widetilde{x}_{v+1,i}=X_i(\xi_1)$, $0\le i\le n$, and
$\widetilde{x}_{v+1}=
(\widetilde{x}_{v+1,0}:\ldots :\widetilde{x}_{v+1,n})$.
Then $\widetilde{x}_{v+1}\in W$ is a smooth point of $V$ satisfying (a) and
(b). Let us construct using Theorem~4
\cite{6} the tangent space
$T_{\widetilde{x}_{v+1},W}$.
By Corollary~2 \cite{6} there is $v+1\le \gamma\le n$
($\gamma=i_0$ in the notation of \cite{6}
Section~3) such that the intersection
$$
T_{\widetilde{x}_{v+1},W}\cap{\cal Z}(Z_{s+1},\ldots ,Z_v,Z_{v+1},
T_{v+1},\ldots , T_{\gamma-1},T_{\gamma+1},\ldots , T_n)=\{0\}.
$$
Let us compute such an integer $\gamma$ constructing the corresponding
intersections.
Now set $i_{v+1,j}=i_{v,j-1}$ for $v+2\le j\le \gamma$, and
$i_{v+1,j}=i_{v,j}$ for $\gamma+1\le j\le n$.
Then by Lemma~1 with $v+1$ in place of $v$ the differential
$d_{\widetilde{x}_{v+1}}\pi_{v+1}$ is an isomorphism.
Hence the point $\widetilde{x}_{v+1}$ satisfies condition (a), (b), (d).
Therefore, by previous remarks one can
construct a point $x^{(v+1)}$ satisfying
(a)--(d). Thus, finally we shall construct $x^{(n)}$.
The first auxiliary algorithm is described.



\section{Several Lemmas}\label{s3}

\par\medskip\noindent{\bf LEMMA~2}\hspace{0.1em} {\it  Let $k$ be a
arbitrary field of zero--characteristic.
Let $W_1$ and $W_2$ be defined over $k$
algebraic varieties and $W_2$ be irreducible over $k$.
Let $\pi\, :\, W_1\rightarrow W_2$ be a dominant (this means that the
restriction of $\pi$ to any
irreducible over $k$ component of $W_1$ is dominant)
regular defined over $k$ morphism.
Let $\dim W_1=\dim W_2$. Denote by $k(W_1)$ the direct product
of the fields of defined over $k$ rational functions
of all irreducible over $k$ components of $W_1$.
Hence the degree $[k(W_1):k(W_2)]$ is defined.
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item Let $z\in W_2$
be a normal point. Denote by $\nu(z)$ the number of connected components of
$\pi^{-1}(z)$. Then $\nu(z)\le [k(W_1):k(W_2)]=\deg\pi$.
In particular, if the number of elements $\#\pi^{-1}(z)<+\infty$
then $\# \pi^{-1}(z)\le [k(W_1):k(W_2)]=\deg\pi$.
\item There is a smooth point $z\in W_2$ such that $\# \pi^{-1}(z)=\deg\pi$.
\item Let $z\in W_2$
be a normal point and $\# \pi^{-1}(z)=\deg\pi$.
Then there is
an open in the Zariski topology subset $U'\subset W_2$ such that $z\in U'$
and the morphism
$\pi^{-1}(U')\rightarrow U'$
induced by $\pi$ is finite.
\item  Let $z\in W_2$
be a normal (respectively smooth)
point and $\# \pi^{-1}(z)=\deg\pi$.  Then
every $x\in \pi^{-1}(z)$ is a normal (respectively smooth) point of
$W_1$ and
the differential $d_x\pi\, :\, T_{x,W_1}\rightarrow T_{z,W_2}$
is an isomorphism of the Zariski tangent spaces  of
$W_1$ and $W_2$ at the points $x$ and $z$ respectively.
One can choose $U'$ such that all the points of $U'$
are normal (respectively smooth) and
the number of elements $\#\pi^{-1}(z_1)=\deg\pi$ for every
$z_1\in U'$.
\end{enumerate}
}\par\medskip

\noindent {\bf PROOF} \quad
We can suppose without loss of generality that $W_2$ is an affine variety,
$W_1$ is
irreducible over $k$, and $k=\overline{k}$.
Denote $[k(W_1):k(W_2)]=\alpha$. Let $z\in W_2$ be an normal point.
Denote by $\widetilde{W}_i$, $i=1,2$, the
normalization of the algebraic variety $W_i$ in the field $k(W_1)$.
So we have the commutative diagram of morphisms of algebraic varieties
$$
\begin{CD}
\widetilde{W}_1@>\widetilde{\pi}>>\widetilde{W}_2  \\
@V\pi_1VV                       @V\pi_2VV\\
W_1            @>\pi>>                  W_2,
\end{CD}
$$
where $\pi_1$ and $\pi_2$ are finite morphisms and
$\widetilde{\pi}$ is a birational isomorphism induced by $\pi$.
The morphism $\pi_2$ is finite and $z$ is a normal point. Therefore,
the coefficients of the
minimal polynomial of each element
of $\overline{k}[\widetilde{W}_2]$ over $\overline{k}(W_2)$ belong to the
the local ring ${\cal O}_{z,W_2}$
of the algebraic variety $W_2$ in the point $z$
provided that leading coefficient of this minimal polynomial is $1$.
Hence
$\#\pi_2^{-1}(z)\le \alpha$.
By the Zariski main theorem, see, e.g., \cite{9},
for every $z'\in \pi_2^{-1}(z)$ if $\widetilde{\pi}^{-1}(z')\ne \emptyset$
then $\widetilde{\pi}^{-1}(z')$ is connected.
For an arbitrary algebraic variety $W$ denote for brevity by
$\nu(W)$ the number of all connected components of $W$.
Hence,
$\#\pi_2^{-1}(z)\ge\nu(\pi_1^{-1}(\pi^{-1}(z)))$. Hence
$\nu(\pi^{-1}(z))\le\alpha$ and (i) is proved.

Assertion (ii) is
known. Under conditions of (iii) the number of elements
$\#\pi_2^{-1}(z)=\#\pi_1^{-1}(\pi^{-1}(z))$ and
$\widetilde{\pi}$ is invertible in some neighborhood in the Zariski topology
$U''$ of the
finite set $\pi_2^{-1}(z)$. So we have the isomorphism of algebraic varieties
$\widetilde{\pi}^{-1}(U'')\simeq U''$
induced by $\widetilde{\pi}$. Set
$U'=W_2\setminus\pi_2(\widetilde{W}_2\setminus U'')$.
Then $z\in U'$ and the morphism
$$
\pi_1^{-1}(\pi^{-1}(U'))=\widetilde{\pi}^{-1}(\pi_2^{-1}(U'))\rightarrow
U'
$$
is finite. Hence, the morphism $\pi^{-1}(U')\rightarrow U'$ is also finite.

To prove (iv)
we shall suppose without loss of generality that all the points of $U'$ are
normal (respectively smooth) points of $W_2$ and $U'$ is affine. Let us
choose a function $y\in k[W_1]$
such that the number of elements $\#y(\pi^{-1}(z))=\#\pi^{-1}(z)$
and $y$ is a primitive element of $k(W_1)$
over $k(W_2)$. Let $\phi\in k(W_2)[Z]$ be the minimal
polynomial of the element $y$ over $k(W_2)$,
and leading coefficient $\mbox{\rm lc}_Z(\phi)=1$.
Since $k[U']$ is integrally closed $\phi\in k[U'][Z]$.
Set the algebraic variety $U'''=\{(\pi(x),y(x))\, :\, x\in \pi^{-1}(U')\}$.
Hence one can identify $U'''={\cal Z}(\phi)\subset  U'\times{\Bbb
A}^1(\overline{k})$.
We have the morphisms $\pi'\, :\, \pi^{-1}(U')\rightarrow U'''$,
$\pi'(x)=(\pi(x),y(x))$,
and $\pi''\, :\, U'''\rightarrow U'$, $\pi''((x_1,x_2))=x_1$. The morphisms
$\pi'$, $\pi''$ are finite dominant and $\pi'$ is a birational isomorphism.

Now $\deg_{Z}\phi=\deg\pi$. Let $\phi=\sum_{i}\phi_iZ^i$, $\phi_i\in k[U']$.
Set $\phi|_z=\sum_{i}\phi_i(z)Z^i\in k[Z]$.
The polynomial $\phi|_z$
has
$\deg\pi$ pairwise distinct roots $Z=y(x)$, $x\in\pi^{-1}(z)$.
Therefore, the derivative $(d (\phi|_z)/d Z)
(y(x))\ne 0$
for every point $x\in\pi^{-1}(z)$. Hence
the differential
$d_{\pi'(x)}\pi''$ is an
isomorphism
for every $x\in\pi^{-1}(z)$. Denote by
$R=\mbox{\rm Res}_Z(\phi,\partial\phi/\partial Z)\in k[U']$ the
discriminant of $\phi$ with respect to $Z$.
Obviously $R(z)\ne 0$.
We choose $U'$ such that the $R(z_1)\ne 0$ for every $z_1\in U'$.
Then $\#\pi^{-1}(z_1)=\deg\pi$ for every $z_1\in U'$.
It is known that the
integral closure of the ring $k[U''']$ is contained in the
mo\-du\-le $R^{-1} k[U''']$.
Hence $k[U''']$ is integrally closed since  $R^{-1}\in k[U''']$. Thus, the
algebraic variety  $U'''$
is normal (respectively obviously $U'''$ is smooth).
In particular every $\pi'(x)$ is a
normal (respectively smooth) point of $U'''$.
By the Zariski main theorem
applied  to the morphism $\pi'$
the point $x$ is a normal (respectively smooth) point of $\pi^{-1}(U')$ and
the
differential $d_x\pi'$ is an isomorphism
for every $x\in\pi^{-1}(z)$. Since $\pi=\pi''\circ\pi'$ on $\pi^{-1}(U')$,
this implies (iv).
The lemma is proved.


\par\medskip\noindent{\bf LEMMA~3}\hspace{0.1em} {\it
Let $W\subset{\Bbb P}^n(\overline{k})$ be a defined over $k$ and
irreducible over $k$ projective algebraic variety and $\dim W=n-s$, $0\le
s\le n$.
Let $g_0,g_{s+1},\ldots , g_n\in k[X_0,\ldots , X_n]$ be homogeneous
polynomials of the same degree and the
morphism
$$
\rho\, :\,
W\setminus{\cal Z}(g_0,g_{s+1},\ldots , g_n)\rightarrow{\Bbb
P}^{n-s}(\overline{k}),\quad(X_0:\ldots : X_n)\mapsto(g_0:g_{s+1}:\ldots :
g_n)
$$
is dominant.
Let ${\mathfrak l}\subset{\Bbb P}^{n-s}(\overline{k})$ be a line. Then the
dimension of every irreducible component of
$\rho^{-1}({\mathfrak l})$ is at least $1$.
}\par\medskip

\noindent {\bf PROOF} \quad
Indeed, after a nondegenerate linear transformation of $g_0,g_{s+1},\ldots ,$
$g_n$ over $k$
we can suppose without loss of generality that $\rho^{-1}({\mathfrak
l})=W\cap{\cal Z}(g_{s+1},\ldots ,$ $g_{n-1})
\setminus{\cal Z}(g_0,g_{s+1},\ldots , g_n)$.
This implies the required assertion. The lemma is proved.

\par\medskip\noindent{\bf LEMMA~4}\hspace{0.1em} {\it  Under conditions of
Lemma~3
let $x\in{\Bbb P}^n(\overline{k})
\setminus{\cal Z}(g_0)$
be a point. Let $x$ be an isolated point of $\rho^{-1}\rho(x)$.
Let ${\mathfrak l}\subset{\Bbb P}^{n-s}(\overline{k})$ be a line
containing
the point $\rho(x)$.
Suppose that there is $\widetilde{z}\in{\mathfrak l}$ such that the
number of elements $\#\rho^{-1}(\widetilde{z})=\deg\rho$.
Let $C$ be the projective
curve such that $C\setminus{\cal
Z}(g_0,g_{s+1},\ldots , g_n)$
is the union of all irreducible
components of $\rho^{-1}({\mathfrak l})$ containing $x$ (such
a curve $C$
exists by Lemma~3). Denote by
$\rho|_C\, :\, C\setminus{\cal Z}(g_0,g_{s+1},\ldots ,
g_n)\rightarrow{\mathfrak l}$ the restriction of $\rho$ to $C$ and
${\mathfrak l}$.
Then the differential $d_x\rho$ is an isomorphism if and only if the
differential $d_x(\rho|_C)$
is an isomorphism.
}\par\medskip

\noindent {\bf PROOF} \quad
Similarly to the proof of
Lemma~3 we shall suppose without loss of generality that
$\rho^{-1}({\mathfrak l})=W\cap{\cal Z}(g_{s+1},\ldots , g_{n-1})
\setminus{\cal Z}(g_0,g_{s+1},\ldots , g_n)$
and $g_n(x)=0$. Denote by $C_\gamma$, $\gamma\in \Gamma$, the
family of all irreducible over $\overline{k}$ components of $C$.
For arbitrary algebraic varieties $E_1$, $E_2$, $E_3$
denote by $i(E_1,E_2;E_3)$ the index of intersection, see \cite{9},
of $E_1$ and
$E_2$ at the variety $E_3$ when this index is defined.
 Then it is known that, see \cite{9},
$$
i(W,{\cal Z}(g_{s+1},\ldots , g_n);x)=
\sum_{\gamma\in\Gamma}i(W,{\cal Z}(g_{s+1},\ldots , g_{n-1});C_\gamma)
i(C_\gamma,{\cal Z}(g_n);x).
$$
Further, if $C$ is irreducible over $\overline{k}$  then
$i(W,{\cal Z}(g_{s+1},\ldots , g_{n-1});C)=1$ since the
corresponding
intersection is transversal by Lemma~2 and due to the
existence of $\widetilde{z}$.
Now $i(W,{\cal Z}(g_{s+1},\ldots , g_n);x)=1$ (respectively
$i(C,{\cal Z}(g_n);x)=1$)
if and only if the differential $d_x\rho$ (respectively $d_x(\rho|_C)$)
is an isomorphism. Hence if $d_x\rho$ is an
isomorphism then $d_x(\rho|_C)$ is an isomorphism.
Conversely, if $d_x(\rho|_C)$ is an isomorphism then  $i(C,{\cal
Z}(g_n);x)=1$.
Hence the curve $C$ is irreducible over $\overline{k}$ and
$i(W,{\cal Z}(g_{s+1},\ldots , g_{n-1});C)=1$. This implies that
$d_x\rho$ is an isomorphism
The lemma is proved.



\medskip Let the infinitesimals $\varepsilon_i>0$, $1\le i\in{\Bbb Z}$,
and
the fields $k_i$, $\overline{k_i}$, $i\ge 0$, supplied with a real structure
be the same as in Section~2.

\par\medskip\noindent{\bf LEMMA~5}\hspace{0.1em} {\it  Under conditions of
Lemma~3 let $W$ be a component of $V={\cal
Z}(f_1,\ldots ,f_m)$.
Suppose that $x=(x_0:\ldots : x_n)\in W$, all $x_i\in\overline{k}$,
$g_0(x)\ne 0$ and $x$ is an isolated point of $\rho^{-1}\rho(x)$.
Set
$\widetilde{x}=(x_0,\ldots , x_n)\in\mbox{\rm con}(W)$.
Consider the system of polynomial equations and inequalities
with respect to the indeterminates
$X_0,\ldots , X_n$, $Y_0,\ldots , Y_n$ with coefficients from
$\overline{k_3}$:
$$
\left\{\begin{array}{ll}
\sum_{s+1\le i\le n}|g_i-g_i(\widetilde{x})|^2=\varepsilon_2,\\
g_i-g_i(Y_0,\ldots , Y_n)=0,& s+1\le i\le n,\\
\sum_{0\le i\le n}|X_i-X_i(\widetilde{x})|^2\le\varepsilon_1,\\
\sum_{0\le i\le n}|Y_i-X_i(\widetilde{x})|^2\le\varepsilon_1,\\
\sum_{0\le i\le n}|X_i-Y_i|^2\ge\varepsilon_3,\\
g_0=g_0(x_0,\ldots , x_n),\\
g_0(Y_0,\ldots , Y_n)=g_0(x_0,\ldots , x_n).
\end{array}\right.
\eqno (18)
$$
Then system (18) has a solution
$$
(x^*_0,\ldots , x^*_n,y^*_0,\ldots , y^*_n)\in
\mbox{\rm con}(W)\times\mbox{\rm con}(W)\subset
{\Bbb A}^{2n+2}(\overline{k})
\eqno (19)
$$
if and only if
the differential $d_x\rho$, see Lemma~3, is not an isomorphism.
}\par\medskip

\noindent {\bf PROOF} \quad System (18) has solution (19)
if and only if
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\item
there are a point
$z^*\in{\Bbb P}^{n-s}(\overline{k})$
and two distinct points $x^*,y^*\in\rho^{-1}(z^*)$ such that $z^*$ is
infinitely close to $\rho(x)$
and $x^*$, $y^*$ are infinitely close to $x$.
\end{enumerate}
Hence by the implicit function theorem it is sufficient to prove that
\begin{enumerate} \renewcommand{\labelenumi}{(\roman{enumi})}
\setcounter{enumi}{1}
\item if the differential
$d_x\rho$ is not an isomorphism then assertion (i) holds.
\end{enumerate}
By Lemma~2 there is a point
$z'\in {\Bbb P}^{n-s}(\overline{k})$ such
that the
number of elements $\#\rho^{-1}(z')=\deg\rho$.
There exists a line
${\mathfrak l}\subset{\Bbb P}^{n-s}(\overline{k})$
such that $\rho(x)\in{\mathfrak l}$, $z'\in{\mathfrak l}$.
Let $C$ and $\rho|_C$ be the same as in Lemma~4.
By Lemma~4 one can replace in (ii) the differential $d_x\rho$ by
$d_x(\rho|_C)$. Further, in (i) one can replace $\rho$ by $\rho|_C$ and
${\Bbb P}^{n-s}(\overline{k})$ by ${\mathfrak l}$.
Using the Veronese isomorphism of degree
$\deg g_0$, see \cite{9},
one can assume without loss of generality
that $\rho$ is a linear projection.
For the linear projection of a curve assertion (ii) is
straightforward, cf. \cite{6}. The lemma is proved.


\par\medskip\noindent{\bf LEMMA~6}\hspace{0.1em} {\it  Under conditions of
Lemma~3 let $z=(1:z_{s+1}:\ldots :
z_n)\in{\Bbb P}^{n-s}(\overline{k})$, the numbers of elements
$\#\rho^{-1}(z)<\deg\rho$ and for every $x\in\rho^{-1}(z)$
the differential $d_x\rho$ is an isomorphism.
Consider the system of polynomial equations and inequalities
with respect to the indeterminates
$X_0,\ldots , X_n$ with coefficients from $\overline{k_2}$:
$$
\left\{\begin{array}{ll}
\sum_{s+1\le i\le n}|g_i-z_i|^2=\varepsilon_2,\\
\sum_{0\le i\le n}|X_i|^2\ge\varepsilon_1^{-1},\\
g_0=1.
\end{array}\right.
\eqno (20)
$$
Then system (20) has a solution
$$
(x^*_0,\ldots , x^*_n)\in\mbox{\rm con}(W)
\subset{\Bbb A}^{n+1}(\overline{k_2}).
\eqno (21)
$$
}\par\medskip

\noindent {\bf PROOF} \quad By Lemma~2 there is a point
$\widetilde{z}=(1:\widetilde{z}_{s+1}:\ldots : \widetilde{z}_n)\in{\Bbb
P}^{n-s}(\overline{k})$ such that the numbers of elements
$\#\rho^{-1}(\widetilde{z})=\deg\rho$. Let ${\mathfrak l}\subset{\Bbb
P}^{n-s}(\overline{k})$ be the line containing the points
$z,\widetilde{z}$.
Then by Lemma~2 and Lemma~5
there is a projective irreducible over $\overline{k}$ curve
$C_1\subset{\Bbb P}^n(\overline{k})$ such that
$C_1\setminus{\cal Z}(g_0,g_{s+1},\ldots , g_n)$
is an irreducible component of $\rho^{-1}({\mathfrak l})$ and the
intersections
$C_1\cap\rho^{-1}(\widetilde{z})\ne\emptyset$, $C_1\cap{\cal
Z}(g_0,g_{s+1},\ldots , g_n)\ne\emptyset$ in
${\Bbb P}^n(\overline{k})$. Considering a branch of the curve $C_1$ at a
point $\xi\in
C_1\cap{\cal Z}(g_0,g_{s+1},\ldots , g_n)$ we get the
required solution of system (20). The lemma is proved.

\par\medskip\noindent{\bf LEMMA~7}\hspace{0.1em} {\it  Let $Q\subset{\Bbb
P}^n(\overline{k})$ be a projective algebraic
variety. Let
$g_0,\ldots , g_r\in \overline{k}[X_0,\ldots , X_n]$ be homogeneous
polynomials with
$\deg g_i= d'$ if $g_i\ne 0$.
Let $\rho_1\, :\, Q\setminus{\cal Z}(g_0,\ldots , g_r)\rightarrow{\Bbb
P}^r(\overline{k})$, $(X_0:\ldots : X_n)\mapsto (g_0:\ldots : g_r)$,
be the regular morphism. Denote by $Q'$ the closure in the Zariski topology
of $\rho_1(Q\setminus{\cal Z}(g_0,\ldots , g_r))$.
Suppose that $\dim Q'\le\nu$.
Then
$\deg Q'\le(d')^\nu\deg Q$ where by definition $\deg Q$ (respectively $\deg
Q'$) is
the sum of the degrees of all the irreducible components of all dimensions
of $Q$ (respectively $Q'$).
Suppose additionally that all the irreducible components of $Q$ are of
dimensions at least $n-s$ and $Q$
is a union of some
irreducible components of the set of all zeroes of homogeneous polynomials
of degrees less than $d$.
Then by the B\'ezout theorem
$\deg Q'\le d^s(d')^\nu$.
}\par\medskip

\noindent {\bf PROOF} \quad
We can suppose without loss of generality that $Q$ is irreducible over
$\overline{k}$ and $\dim Q'=\nu$.
Let ${\cal L}$ be a linear subspace of ${\Bbb
P}^r(\overline{k})$ of codimension $\nu$ in general position.
The degree of the closure
$\deg\overline{\rho_1^{-1}(Q'\cap{\cal L})}\le(d')^\nu\deg Q$
by the B\'ezout theorem. On the other hand, $\deg Q'$ is bounded from above
by the number of
connected components of $\rho_1^{-1}(Q'\cap{\cal L})$ and hence by the
degree $\deg\overline{\rho_1^{-1}(Q'\cap{\cal L})}$.
Therefore, $\deg Q'\le (d')^\nu\deg Q$.
The lemma is proved.

\par\medskip\noindent{\bf LEMMA~8}\hspace{0.1em} {\it
Let $C\subset{\Bbb P}^n(\overline{k})$ be a projective curve of degree
$D$ defined over $k$ and
and $g_0,g_1\in k[X_0,\ldots , X_n]$
be homogeneous polynomials of the same degree $D_1$ such that for every
irreducible over $k$ component $C_1$ of $C$ the
polynomial $g_0$ is not vanishing on $C_1$ and $g_1/g_0\in k(C_1)$ is
transcendental over $k$.
Consider the morphism $\tau\, :\, C\setminus{\cal
Z}(g_0,g_1)\rightarrow{\Bbb P}^1(\overline{k})$,
$\tau(x)=(g_0(x):g_1(x))$.
Then the number of points $z\in{\Bbb P}^1(\overline{k})$ such that
$\#\tau^{-1}(z)<\deg\tau$ is bounded from above
by $DD_1(DD_1-1)$.
}\par\medskip

\noindent {\bf PROOF} \quad Let $k(C)$ be the total quotient ring
of the ring of regular functions $k[C\setminus{\cal Z}(g_0)]$.
Let $C\rightarrow v(C)$, see \cite{9}, be the Veronese isomorphism
of degree $\deg g_0$. Then $\deg v(C)\le DD_1$ by the B\'ezout
theorem. Replacing $C$ by $v(C)$ we shall suppose without loss of
generality that $g_0$ and $g_1$ are linear forms. By the theorem
about primitive element there is a linear form $g_2\in
k[X_0,\ldots , X_n]$ such that the embedding
$k[g_1/g_0,g_2/g_0]\rightarrow k[C\setminus{\cal Z}(g_0)]$
induces the isomorphism of $k(g_1/g_0)[(g_2/g_0)|_C]\simeq k(C)$
and additionally
$C\cap{\cal Z}(e_0,e_1,e_2)=\emptyset$. Consider the regular
morphism $\tau_1\, :\, C\rightarrow{\Bbb P}^2(\overline{k})$,
$\tau_1(x)=(g_0(x):g_1(x):g_2(x))$. Then $\tau_1(C)=
C_2\subset{\Bbb P}^2(\overline{k})$ is a projective curve of
degree at most $DD_1$. Hence $C_2={\cal Z}(\Phi)$ where
$\Phi\in k[X_0,X_1,X_2]$ is a separable polynomial of degree at
most $DD_1$. The morphism $\tau_2\, :\, C\rightarrow C_2$ induced
by $\tau_1$ is finite birational isomorphism. Denote by
$R=\mbox{\rm Res}_{X_2}(\Phi,\partial \Phi/\partial X_2)$ the
discriminant of $\Phi$ with respect to $X_2$ (here and below by
definition the discriminant is the resultant of the polynomial and
its derivative; this definition is different from the generally
accepted by a factor $\mbox{\rm lc}_{X_2}\Phi$, but it
is more convenient for us). Denote by $\tau_3\, :\,
C_2\setminus{\cal Z}(g_0,g_1)\rightarrow{\Bbb
P}^1(\overline{k})$ the morphism induced by $\tau$. Then the
restriction of $\tau_2$ is the isomorphism
$C\setminus\tau^{-1}({\cal Z}(R))\rightarrow
C_2\setminus\tau_3^{-1}({\cal Z}(R))$. Further consider the
projection $\tau_4\, :\, C_2\setminus\tau_3^{-1}({\cal
Z}(R))\rightarrow{\Bbb P}^1(\overline{k})\setminus{\cal
Z}(R)$ which is the restriction of $\tau_3$. Then $\tau_4$ is
\'etale, see the definition in \cite{9}. Note that $\tau_3\circ
\tau_2$ is the restriction of $\tau$. The degree $\deg(R)\le
DD_1(DD_1-1)$. This implies the assertion of the lemma. The lemma
is proved.


\section{Proof of assertions (a)--(d)}\label{s4}

Now we proceed to the description of the algorithm for Theorem~1.
Let us compute $\dim W$ by \cite{6} or
Theorem~2 of \cite{8}.
Let us decide using, e.g., Theorems~1 and 2 from \cite{8}
whether $e_0$ vanishes on $W$.
Performing if necessary a permutation of indices $0,\ldots , r$
we shall suppose without loss of generality in what follows
that $e_0$ does not vanish on $W$. Let $L_1,\ldots ,L_{n+1}$ be linear forms
introduced at the beginning of Section~2.
We have
$W\cap{\cal Z}(L_{s+1},\ldots , L_n)\cap{\cal Z}(e_0)=\emptyset$.
Hence
using Theorems~1 and 2 of \cite{8} we can
construct a point $\widetilde{x}^{(0)}\in W$
that is a smooth point of $V$ and $e_0(\widetilde{x}^{(0)})\ne 0$.
Replacing $x^{(0)}$ by $\widetilde{x}^{(0)}$ we shall suppose in what
follows that
that $e_0(x^{(0)})\ne 0$ and that $x^{(0)}$ is a smooth point of $V$.
Using Theorem~5 of \cite{6} one can decide whether a given family of
rational functions from $k(W)$ is a basis of transcendency of $k(W)$.
As an easy consequently one can compute the transcendency degree over $k$ of
the given family of rational functions from
$k(W)$ (one need only to add to the given
family the known basis of transcendency $k(W)$
and then apply Theorem~5 of \cite{6} to subfamilies of the new
obtaned family of rational functions). So we
compute the transcendency degree of the family of rational functions
$e_1/e_0,\ldots , e_r/e_0\in k(W)$ over $k$, i.e., $\dim W'$.
We shall suppose in what follows without loss of generality in the proof
that $\dim W=\dim W'=n-s$. We compute also applying Theorem~5 of \cite{6}
indices $0=i_0<\ldots < i_{n-s}\le r$ such that the
transcendency degree of the family of rational functions $e_{i_j}/e_{i_0}
\in k(W)$,
$1\le j\le n-s$, is $n-s$. Performing if necessary a permutation of
$0,\ldots , n$
we shall suppose in what follows that $i_j=j$ for all $0\le j\le n-s$.
In what follows applying Lemmas~3--6
we set $W$ to be the considered irreducible component of $V$ and
$g_0=e_0$, $g_i=e_{i-s}$, $s+1\le i\le n$.
Hence the morphism $\pi_n$ from Section~2 is the restriction of $\rho$.

Now we are going to describe a recursion
where $z'\in {\Bbb P}^{n-s}(\overline{k})$ is the current point.
At the end of this recursion we get $z'$ such that the number of
elements
$\#\rho^{-1}(z')=\deg\rho$, all the elements of $\rho^{-1}(z')$ are smooth
points of $V$, and all the elements from $p(\rho^{-1}(z'))$ are smooth
points of $W'$.
As the base set $z'=\rho(x^{(0)})$. At each step $X_0(z')\ne 0$
(we suppose that $X_0,\ldots X_{n-s}$ are homogeneous
coordinates on ${\Bbb P}^{n-s}(\overline{k})$)
and $\rho^{-1}(z')\ne\emptyset$.
For the point $z'$ we shall define the integers $N_i(z')\ge 0$, $0\le i\le
5$.
At the end of each step that is not final we get a point $z''$ similar to
$z'$ such that $N_i(z'')\ge N_i(z')$ for $i=1,3,4,5$
and at least one of these four inequalities is strict.
Further, we shall replace $z'$ by $z''$ and proceed to the next step.
The estimate for the number of steps will be obtained from the B\'ezout
theorem.



Let us describe the general step of this recursion.
Let us represent $z'=(1:z'_1:\ldots : z'_{n-s})$.
Hence all $z'_i\in\overline{k}$, $1\le i\le n-s$.
Applying
Theorem~2 to $\mbox{\rm con}(V)$ and
$$
U''=\bigl\{x\in{\Bbb A}^{n+1}(\overline{k})\, :\, e_i(x)-z'_i=0,
\quad 0\le i\le n-s\bigr\},
$$
we construct a finite set $S_0$
of points of $V$ such that the
intersection of $S_0$ with every connected component of
$\rho^{-1}(z')$ is nonempty.
The number of elements $\#S_0$ is bounded from above
by a polynomial in $(dd')^n$
by Theorem~2. Applying, Theorem~2 \cite{8}
we shall suppose without loss of generality that $S_0=S_0(z')\subset W$.
Hence $S_0(z')\ne\emptyset$ since $\rho^{-1}(z')\ne\emptyset$
by the assumption of the recursion.
Let $x\in S_0$. Then $e_0(x)\ne 0$. Let
$x=(x_0:\ldots : x_n)$ where
all $x_i\in\overline{k}$.
Denote by $U''_x$ the set of points from ${\Bbb A}^{n+1}(\overline{k_1})$
satisfying the following system of polynomial equations and inequalities
with coefficients from $\overline{k_1}$:
$$
\left\{\begin{array}{ll}
\sum_{0\le i\le n}|X_i-x_i|^2=\varepsilon_1,\\
e_i-z'_ie_0=0, & 1\le i\le n-s, \\
e_0=e_0(x).
\end{array}\right.
\eqno (22)
$$
For every $x\in S_0$ using Theorem~2 \cite{8} and
applying Theorem~2 to $\mbox{\rm con}(V)$
and $U''_x$  we decide whether there is a point
$\widetilde{x}\in\rho^{-1}(z')$
that is infinitely close to $x$, i.e., whether $x$ is not an isolated
point of $W$. So deleting some points from $S_0$ we get the subset
{\it $S_1=S_1(z')$ of all isolated points of
$\rho^{-1}(z')$.}
Denote by {\it $S_2=S_2(z')$ the set of all points of $S_1(z')$ which
belong to the unique irreducible over $\overline{k}$ component of $V$} (i.e.,
only to one of the irreducible over $\overline{k}$ components of $W$).
By Theorem~2 \cite{8} using the given point $x^{(0)}\in W$ we construct
the set $S_2(z')$. Put $S_5(z')=p(S_1(z'))$.
The set $S_5(z')$ consists of classes of conjugated over $k$ points of
${\Bbb P}^r(\overline{k})$.
Hence one can easily construct these classes and compute the number of
elements $\#S_5(z')$.





For every $x\in S_1$ using Theorem~2 \cite{8} and
applying Theorem~2 to $\mbox{\rm
con}(V)\times\mbox{\rm con}(V)$
we decide whether system (18) has a solution (19)
and construct this solution if it exists. Set
$x^*=(x^*_0:\ldots : x^*_n)\in W(\overline{k_3})$ and
$z^*=\rho(x^*)\in{\Bbb P}^{n-s}(\overline{k_3})$ for solution (19).
Denote by {\it $S_3=S_3(z')$ (respectively $S_4=S_4(z')$) the set of all
$x\in S_1(z')$ (respectively $x\in S_2(z')$) such that the differential
$d_x\rho$ is an isomorphism.}
Hence by Lemma~5 and Theorem~2
one can construct the sets $S_3(z')$ and $S_4(z')$.






Let $z^{(2)}=(1:z^{(2)}_1:\ldots : z^{(2)}_{n-s})\in{\Bbb
P}^{n-s}(\overline{k_\alpha})$
be an arbitrary point where $0<\alpha\in{\Bbb Z}$ is arbitrary but
fixed
(for estimates of the complexity of algorithms).
Let $X_0(z^{(2)})\ne 0$ and $\rho^{-1}(z^{(2)})\ne\emptyset$.
Let us define the sets $S_i(z^{(2)})$, $1\le i\le 5$,
replacing in the definition of the sets
$S_i$ the point $z'$ by $z^{(2)}$ and the field $k$ by
a finite extension of $k_\alpha$ containing the coordinates of the point
$z^{(2)}$.
We can construct the sets $S_i(z^{(2)})$, $0\le i\le 5$, similarly to
$S_i(z)$ replacing the ground field.
Denote the number of elements $\#S_i(z^{(2)})=N_i(z^{(2)})$, $0\le i\le 5$.

Let ${\mathfrak l}\subset{\Bbb P}^{n-s}(\overline{k_\alpha})$
be a line through a point $z^{(2)}$.
By the B\'ezout theorem there are at most a polynomial in $d^n(d')^{n-s}$
number of points
$z\in{\mathfrak l}$ such that $\rho^{-1}(z)=\emptyset$.
Denote by ${\mathfrak c}$ the union of all the irreducible components $E$
of $\rho^{-1}({\mathfrak l})$ such that
$E\cap S_1(z^{(2)})\ne\emptyset$. Suppose that $S_1(z^{(2)})\ne\emptyset$.
Then ${\mathfrak c}$ is a curve and
$N_1(z)\le d'\deg{\mathfrak c}\le (d')^{n-s}d^s$ for every $z\in{\mathfrak
l}$ by the definition of
${\mathfrak c}$ and
the B\'ezout theorem. Denote by $\rho|_{\mathfrak c}\, :\,
{\mathfrak c}\rightarrow{\mathfrak l}$ the restriction of
$\rho$ to ${\mathfrak c}$ and ${\mathfrak l}$.
By Lemma~8 and the B\'ezout theorem there is at most a polynomial in
$d^n(d')^{n-s}$ number of points
$z\in{\mathfrak l}$ such that
$\#\rho^{-1}(z)\cap{\mathfrak c}\ne\deg\rho|_{\mathfrak c}$
or $N_2(z)<N_2(z^{(2)})$. Let $x'\in S_3(z^{(2)})$.
Considering the local parameters of $W$ in the point $x'$
with degrees bounded from above by $d^s$, cf. the proof of Lemma~1,
one sees that there is a polynomial $\Delta_{x'}\in k[X_0,\ldots , X_n]$
(it is a minor of the corresponding
Jacobi matrix) such that $\deg\Delta_{x'}<sd^s+(n-s)(d')^{n-s}$,
$\Delta_{x'}(x')\ne 0$ and
$d_x\rho$ is an isomorphism for every $x\in W\setminus{\cal
Z}(\Delta_{x'})$.
Hence there is at most a polynomial in $d^s(d')^{n-s}$ number of points
$z\in{\mathfrak l}$ such that
$N_3(z)<N_3(z^{(2)})$ or $N_4(z)<N_4(z^{(2)})$.
Finally, let $\nu=\max\{\#p((\rho^{-1}(z)\cap{\mathfrak c}))\, :\,
z\in{\mathfrak l}\}$.
Hence $\nu\ge N_5(z^{(2)})$.
Then there is a linear form $L'_0\in k[X_0,\ldots ,X_r]$ such that $\nu=
\max\{\#(L'_0/X_0)(p(\rho^{-1}(z)\cap{\mathfrak c}))\, :\,
z\in{\mathfrak l}\setminus{\cal Z}(X_0)\}$.
Considering the corresponding resultant (we leave the details to the reader)
one sees that there are at most a polynomial in $d^s(d')^{n-s}$
number of point $z\in{\mathfrak l}\setminus{\cal Z}(X_0)$ such that
$\#(L'_0/X_0)(p(\rho^{-1}(z)\cap{\mathfrak c}))\ne\nu$.

Thus, ${\mathfrak l}$
contains at most a polynomial in $d^n(d')^{n-s}$
number of points $z^{(3)}$ such that
$N_i(z^{(3)})<N_i(z^{(2)})$ for some $1\le i\le 5$ or
$\rho^{-1}(z^{(3)})=\emptyset$.
Hence using the algorithm of reduction to integer coefficients
one can replace subsequently the coordinates $z^{(2)}_1,\ldots ,
z^{(2)}_{n-s}$  by
integers $z^{(4)}_1,\ldots , z^{(4)}_{n-s}$ with lengths $O(n\,\log
d+(n-s)\log(d'))$ and obtain a point
$z^{(4)}=(1:z^{(4)}_1:\ldots : z^{(4)}_{n-s})\in{\Bbb
P}^{n-s}(\overline{k})$ such that
$N_i(z^{(4)})\ge N_i(z^{(2)})$ for all $1\le i\le 5$ and
$\rho^{-1}(z^{(4)})\ne\emptyset$. We shall
refer to this construction with the
input $z^{(2)}$ and the output $z^{(4)}$ as to
{\it the second auxiliary algorithm.}





Let us return to the description of the algorithm.
Suppose that $N_0(z')=N_1(z')$
and system (18) has a solution (19)
for some $x\in S_1(z')$.  Then $N_1(z^*)>N_1(z')$. Besides that,
$N_i(z^*)\ge N_i(z')$ for $i=3,4,5$ since $z^*$ is infinitely close to $z'$.
Using the second
auxiliary algorithm with the input $z^*$ we get a point $z''$
such that $N_1(z'')>N_1(z)$ and $N_i(z'')\ge N_i(z')$ for $i=3,4,5$.
Then we replace $z'$ by $z''$ and proceed to the next step of the recursion.

Now suppose that $N_0(z')=N_1(z')$ but for every
$x\in S_1(z')$ system (18) does not have a solution (19).
Then $N_0(z')=N_1(z')=N_3(z')$.
Suppose that in the considered case $N_4(z')<N_3(z')$.
Then applying Theorem~2 \cite{8}
we construct a point $x'\in S_3(z')$ that belongs to more than
one irreducible over $\overline{k}$ component of $V$.
Similarly to the construction of the point $\widetilde{x}_s$ in the first
auxiliary algorithm
we compute a point $\widetilde{x}\in W$ that is a smooth point of $V$  and is
infinitely close to $x'$. Set $\widetilde{z}=\rho(\widetilde{x})$.
Then $N_i(\widetilde{z})\ge N_i(z')$ for $i=1,3,5$ and
$N_4(\widetilde{z})>N_4(z')$.
In this case as above applying the second
auxiliary algorithm we construct a point $z''$ such that
$N_i(z'')\ge N_i(z')$ for $i=1,3,5$ and  $N_4(z'')\ge
N_4(\widetilde{z})>N_4(z')$, then
replace $z'$ by $z''$ and proceed to the next step of the recursion.



Now suppose that $N_0(z')=N_1(z')=N_3(z')=N_4(z')$.
Then using Theorem~2 \cite{8} and applying Theorem~2 to $\mbox{\rm
con}(V)$ we decide whether
system (20) has solution (21) and construct this solution if it
exists. By Lemma~6 such a
solution exists if and only if $N_4(z')<\deg\rho$. Set
$x^*=(x^*_0:\ldots : x^*_n)\in W(\overline{k_2})$ and
$z^*=\rho(x^*)\in{\Bbb P}^{n-s}(\overline{k_2})$ for solution (21).

Suppose that solution (21) exists. Then we decide whether
$x^*\in S_1(z^*)$. Since
$\sum_{0\le i\le n}|x_i^*|^2$ is infinitely large and $e_0(x^*_0,\ldots
,x^*_n)=1$
the point $x^*$ is not infinitely
close to any point from $S_1(z')$. On the other hand, by the implicit
function theorem
there are $N_1(z')$ point of $S_1(z^*)$ which are infinitely close to the
corresponding points from $S_1(z')$.
Hence $N_1(z^*)\ge N_1(z')$ and $N_i(z^*)\ge N_i(z')$ for $i=3,4,5$.
Besides that, if $x^*\in S_1(z^*)$ then $N_1(z^*)>N_1(z')$.
In the last case as above applying the second
auxiliary algorithm we construct a point
$z''$ such that
$N_1(z'')> N_1(z')$ and $N_i(z'')\ge N_i(z')$ for $i=3,4,5$. Then
replace $z'$ by $z''$ and proceed to the next step of the recursion.




Now we are going to consider simultaneously two cases:
\begin{enumerate} \renewcommand{\labelenumi}{\arabic{enumi})}
\item $N_0(z')=N_1(z')=N_3(z')=N_4(z')<\deg\rho$ and
$x^*\not\in S_1(z^*)$ for the constructed $x^*$,
\item $N_0(z')\ne N_1(z')$.
\end{enumerate}
In the case 2) we set $z^*=z'$ and
choose an arbitrary point $x^*\in S_0(z')\setminus S_1(z')$
So now $x^*\in\rho^{-1}(z^*)
\setminus S_1(z^*)$ and $z^*\in{\Bbb P}^{n-s}(\overline{k_2})$
for the both cases 1) and 2).
The point $x^*$ is not infinitely
close to any point from $S_1(z^*)$.
Let us replace, see (16),
in the construction of the point $\widetilde{x}_s$ in the first
auxiliary algorithm $\widetilde{x}_s$
by $\widetilde{x}$ and the ground field $k$
by $k_2$ (so $k_2$ is replaced by $k_4$).
Then similarly to the considered construction
we compute a point $\widetilde{x}\in W(\overline{k_4})$
that is a smooth point of $V$  and is
infinitely close to $x^*$ with respect to the field $\overline{k_2}$.
Then $\widetilde{x}\not\in{\cal Z}(e_0,\ldots , e_r)$.
Set $\widetilde{z}=p(\widetilde{x})$. Then $\widetilde{z}$
is infinitely close to $z^*$ with respect to the field $k_2$.
Hence $N_i(\widetilde{z})\ge N_i(z^*)$ for $i=1,3,4,5$.
If $N_1(\widetilde{z})> N_1(z^*)$ then we apply the second
auxiliary algorithm to the point $\widetilde{z}$, obtain a point $z''$,
replace $z'$ by $z''$ and
proceed to the next step of the recursion.


Suppose that $N_1(\widetilde{z})=N_1(z^*)$. Hence the points of
$S_1(z^*)$ and $S_1(\widetilde{z})$ are in
the one--to--one correspondence by the implicit function theorem.
Therefore, $\widetilde{x}\not\in S_1(\widetilde{z})$.
Then replacing the ground field $k$ by $k_4$
and $x^*$ by $\widetilde{x}$ we shall suppose without loss of generality
that $x^*\in W(\overline{k_4})$ is a smooth point of $V$.
Let us choose a linear form $L_0$ such that $L_0(x^*)\ne 0$, see the
beginning of Section~2.
So we have at present linear forms $L_0,L_{s+1},\ldots , L_n$.
Let us apply the
first auxiliary algorithm to the point $x^*$
(now one needs to replace in
this algorithm $\varepsilon_1,\ldots , \varepsilon_5$ by
$\varepsilon_5,\ldots , \varepsilon_9$ respectively)
and get $x^{**}\in{\Bbb P}^{n-s}(\overline{k_6})$.
Put $z^{**}=\rho(x^{**})$.
Then $N_1(z^{**})>N_1(z^*)\ge N_1(z')$.
The point $x^{**}$ is infinitely close to $x^*$ with respect to the field
$k_4$. Hence $N_i(z^{**})\ge N_i(z^*)$, $i=3,4,5$.
As above applying the second
auxiliary algorithm to the input $z^*$
we construct a point $z''$ such that
$N_1(z'')>N_1(z')$, $N_i(z'')\ge N_i(z)$, $i=3,4,5$,
replace $z'$ by $z''$ and proceed to the next step of the recursion.

Thus, finally we shall come to the case when
$N_0(z')=N_1(z')=N_3(z')=N_4(z')$ and
system (20) does not have solution (21).
Then by Lemma~6 the number of elements
$\#\rho^{-1}(z')=\deg\rho=N_4(z')$.

Let $X_0,\ldots ,X_r$ be the
homogeneous coordinate functions of
${\Bbb P}^r(\overline{k})$. Denote by $\rho_2\, :\, W'\setminus{\cal
Z}(X_0,\ldots ,$ $X_{n-s})
\rightarrow{\Bbb P}^{n-s}(\overline{k})$, $(X_0,\ldots
,X_r)\mapsto(X_0:\ldots : X_{n-s})$ the linear projection.
Then according to our definitions $\rho=\rho_2\circ \rho_1$ where
$\rho_1$
is the restriction of $p'$ to
$W\setminus{\cal Z}(e_0,\ldots , e_{n-s})$. Now
$N_4(z')=\deg\rho=\deg\rho_2\deg\rho_1$.

Let $x=(x_0,\ldots , x_n)\in S_1(z')$ and all $x_j\in\overline{k}$.
Consider the system of polynomial equations and inequalities in
$X_0,\ldots , X_n$, $Y_0,\ldots , Y_n$ with coefficients from
$\overline{k_2}$:
$$
\left\{
\begin{array}{ll}
e_j=e_j(Y_0,\ldots , Y_n), & 1\le j\le n-s,\\
\sum_{1\le j\le n-s}|e_j-z'_j|^2\le\varepsilon_1,\\
\sum_{0\le j\le r}|e_j-e_j(Y_0,\ldots , Y_r)|^2=\varepsilon_2,\\
e_0=1,\\
e_0(Y_0,\ldots , Y_r)=1.
\end{array}
\right.
\eqno (23)
$$

\par\medskip\noindent{\bf LEMMA~9}\hspace{0.1em} {\it  Let
$N_4(z')=\deg\rho$. Then
the number of elements $\#S_5(z')=N_5(z')=\deg\rho_2$ if and only if
system (23) does not have a solution
$$
(x^*_0,\ldots , x^*_n, y^*_0,\ldots , y^*_n)\in \mbox{\rm
con}(W)\times\mbox{\rm con}(W)\subset{\Bbb
A}^{2n+2}(\overline{k_2}).
\eqno (24)
$$
Besides that, if
$N_5(z')=\deg\rho_2$
then $p(\rho^{-1}(z'))=\rho_2^{-1}(z')$.
Hence then
the number of elements $\#\rho_2^{-1}(z')=\deg\rho_2$, by Lemma~2
every $x'\in\rho_2^{-1}(z')$ is a smooth point of $W'$, and
the differential $d_{x'}\rho_2$ is an isomorphism. Further,
consider an arbitrary solution (24) of system  (23).
Set $z^*=\rho((x^*_0:\ldots : x^*_n))$. Then $N_5(z^*)>N_5(z')$.
}\par\medskip

\noindent {\bf PROOF} \quad
By Lemma~2 there is an open in the Zariski topology neighborhood $U$ of the
point $z'$ such that the morphism
$\rho^{-1}(U)\rightarrow U$ induced by $\rho$ is finite dominant, the number
of elements
$\#\rho^{-1}(z)=\deg\rho$ for every $z\in U$, and all the points from
$\rho^{-1}(U)$ are
smooth points of $V$. Hence the morphism $\rho_2^{-1}(U)\rightarrow U$
induced by $\rho_2$ is also finite
dominant. Therefore, $p(\rho^{-1}(z'))=\rho_2^{-1}(z')$
for every $z\in U$ and
$N_5(z)=\#\rho_2^{-1}(z)\le\deg\rho_2$ by Lemma~2.

If $N_5(z')=\deg\rho_2$ then by Lemma~2
the numbers of elements
$\#\rho_1^{-1}(z''')=\deg\rho_1=\deg p'$
for every $z'''\in\rho_2^{-1}(z')$. Hence by Lemma~2 and the implicit
function theorem
system (23) does not have solution (24).
The inverse implication is also true by Lemma~2.

On the other hand, consider an arbitrary solution (24). Denote
$y^*=(y^*_0:\ldots :
y^*_n)$.
The morphism $\rho_2$ is finite dominant and
$z^*$ is infinitely to close to $z'$.
Hence for every $y\in\rho^{-1}(z')$ there is $y'\in \rho^{-1}(z^*)$ such
that
$\rho_1(y')$ is infinitely close to $\rho_1(y)$.
Further, the points
$\rho_1(x^*),\rho_1(y^*)\in \rho_2^{-1}(z^*)$ are infinitely close to
$\rho_1(y)\in\rho_2^{-1}(z')$ for some $y\in\rho^{-1}(z')$.
Any two points from
$\rho_2^{-1}(z')$ are not infinitely close.
Hence $N_5(z')<N_5(z^*)\le\deg\rho_2$ since $z^*\in U$.
The lemma is proved.

\medskip
Let us return to the description of the general step of the recursion.
We consider the case when $N_i(z')=\deg\rho$ for $i=1,3,4$.
Let us decide using Theorem~2 whether system (23)
has solution (24) and construct such a solution if it exists.
Suppose that system (23) has a solution (24).
We construct the point $z^*=\rho((x^*_0:\ldots : x^*_n))$.
Then $N_i(z^*)=N_i(z')$ for $i=1,3,4$ by Lemma~2 since $z^*$ is
infinitely close to $z'$. Further, $N_5(z^*)>N_5(z')$ by Lemma~9.
We apply the second auxiliary algorithm to the point $z^*$ and obtain
$z''$ such that $N_i(z'')=N_i(z')$ for $i=1,3,4$ and $N_5(z'')\ge
N_5(z^*)>N_5(z')$. We replace $z'$ by $z''$ and proceed to the
next step of the recursion.

If system (23) does not have a solution then
$N_5(z')=\deg\rho_2$ by Lemma~9 and the recursion is finished.
The point $z'$ satisfy assertion (b) of Theorem~1 by Lemma~2.
Note that at every step of the described recursion
$N_i(z')\le N_0(z')$, $1\le i\le 5$,
and $N_0(z')$ is bounded from above by a polynomial in $(dd')^n$, see
above. Hence the total number of steps in the described recursion
is bounded from above by a polynomial in $(dd')^n$.
The algorithm for assertions (a) and (b) of
Theorem~1 is described.

\medskip Let us describe the algorithm for (c).
We use the recursion on $r$.
The case $r=n-s$ is trivial.
Let $z'=(1:z'_1:\ldots : z'_{n-s})$ be
the point constructed in the algorithm from (b). By Lemma~7 the degree
$\deg W'\le d^s(d')^{n-s}$.  Hence one can construct a
point $y=(1:z'_1:\ldots : z'_r)$ such that $y\not\in p(\rho^{-1}(z'))$
and all $z'_j$, $n-s+1\le j\le r$ are integers with lengths
$O(n\log d+(n-s)\log d')$.
Then by Lemma~2 the point $y\not\in W'$.

Denote by $p_y\, :\,{\Bbb P}^r(\overline{k})\rightarrow{\Bbb
P}^{r-1}(\overline{k})$ the linear projection with the center $y$. Then
$p_y(W')$ is closed in ${\Bbb P}^{r-1}(\overline{k})$ and $p_y$
induces the finite birational morphism $p'_y\, :\, W'\rightarrow p_y(W')$.

\par\medskip\noindent{\bf REMARK~5}\hspace{0.1em} {\it  Notice that $\deg
W'=\deg p_y(W')$.
If $r>n-s+1$ then one can choose $z'_r=0$.
}\par\medskip


Now we apply recursively the algorithm from (c)
for the morphism $p_y\circ p\, :\, W\rightarrow{\Bbb
P}^{r-1}(\overline{k})$.
So the analog of (7) holds for $p_y(W')$; $r-1$; and some constructed
linear forms $\overline{L}_0,
\overline{L}_{s+1},\ldots ,\overline{L}_n\in k[X_0,\ldots ,X_{r-1}]$
in place of
$W'$; $r$;
$L'_0,L'_{s+1},\ldots ,L'_n\in k[X_0,\ldots ,X_r]$
respectively. Set
$L''_j=\overline{L}_j(X_1-z'_1X_0,\ldots , X_r-z'_rX_0)$,
$j\in\{0,s+1,\ldots , n\}$.
Then
$L''_0,L''_{s+1},\ldots , L''_n$ are linear forms from
$k[X_0,\ldots ,X_r]$ with integer
coefficients such that
(7) is satisfied with $L''_0,L''_{s+1},\ldots ,$ $L''_n$
in place of $L'_0,L'_{s+1},\ldots , L'_n$.

Set $e''_i=L''_i(e_0,\ldots , e_r)$, $i\in\{0,s+1,\ldots ,
n\}$. Denote by
$$
\rho''\, :\,
W\setminus{\cal Z}(e''_0,e''_{s+1},\ldots ,
e''_n)\rightarrow{\Bbb P}^{n-s}(\overline{k}),\quad
(X_0,\ldots , X_n)\mapsto(e''_0:e''_{s+1}:\ldots : e''_n)
\eqno (25)
$$
the dominant morphism.
Now applying the algorithm from (b) for $\rho''$ in place of $\rho$
one computes $\deg\rho''$. By Lemma~1 \cite{6} this degree $\deg\rho''=
\deg p'\deg W'$. So $\deg W'$ is computed.


Let us identify ${\Bbb A}^{(r+1)(n-s+1)}(\overline{k})$ with the space
of all
$(n-s+1)$--tuples of linear forms from $\overline{k}[X_0,\ldots ,X_r]$,
cf. assertion (d) of Theorem~1. Then for
any line ${\mathfrak l}\subset{\Bbb A}^{(r+1)(n-s+1)}(\overline{k})$ such
that
$(L''_0,L''_{s+1},\ldots , L''_n)\in{\mathfrak l}$ there are at most a
polynomial in $\deg W'$ number of elements
$(L'''_0,L'''_{s+1},\ldots , L'''_n)\in{\mathfrak l}$ such that (7)
does not hold
for $L'''_0,L'''_{s+1},\ldots , L'''_n$. One can
decide whether (7) holds
for any $(L'''_0,L'''_{s+1},\ldots , L'''_n)\in{\mathfrak l}$ as follows.
Set $e'''_i=L'''(e_0,\ldots , e_r)$, $i\in\{0,s+1,\ldots ,
n\}$ and define $\rho'''$ replacing in the definition of $\rho''$ the linear
forms
$L''_0,L''_{s+1},\ldots , L''_n$ by $L'''_0,L'''_{s+1},\ldots , L'''_n$.
Let us apply the algorithm from (b) for $\rho'''$ in place of $\rho$.
Now by Lemma~1 \cite{6} and the B\'ezout theorem $\deg\rho'''\le\deg(p')\deg
W'$
and this inequality  is strict if and only if (7)
does not hold for $L'''_0,L'''_{s+1},\ldots , L'''_n$.

Hence using the algorithm of reduction to integer
coefficients one can replace subsequently all nonzero coefficients of
$L''_0,L''_{s+1},\ldots , L''_n$
by integers of length $O(n\,\log d+(n-s)\log(d'))$ and obtain
$L^*_0,L^*_{s+1},\ldots , L^*_n$ such that (7) holds for
$L^*_0,L^*_{s+1},\ldots , L^*_n$ in place of $L'_0,L'_{s+1},\ldots , L'_n$.
Set $e^*_i=L^*(e_0,\ldots , e_r)$, $i\in\{0,s+1,\ldots ,
n\}$ and define $\rho^*$ replacing in the definition of $\rho''$ the linear
forms
$L''_0,L''_{s+1},\ldots , L''_n$ by $L^*_0,L^*_{s+1},\ldots , L^*_n$.
Further, applying the algorithm from (b) for the
morphism $\rho^*$ in place of $\rho$
one obtains a point $z^*=(1:z^*_{s+1}:\ldots : z^*_n)$ similar to $z'$.
Hence all $z^*_j$ are integers of length $O(n\,\log d+(n-s)\log(d'))$.
Finally we set $L'_0=L^*_0$ and $L'_j=L^*_j-z^*_jL^*_0$.
Now (4) by Lemma~2, and (6), (3), (5)
hold. The algorithm for assertion (c) is described.
The required estimation for the working time of the
algorithm from (a)--(c) follows immediately from the ones
of the applied algorithms.

\medskip Let us prove (d).  Set $W'_1$ to
be the closure in the Zariski topology of
$p(W\cap(\,\overline{V\setminus W}\,)\setminus{\cal
Z}(e_0,\ldots , e_r))$. By Lemma~2 conditions (3)--(6)
holds if and only if (3) and (6) are satisfied,
$W'\cap{\cal Z}(L'_0,L'_{s+1},\ldots , L'_n)=\emptyset$, and
$W'_1\cap{\cal Z}(L'_{s+1},\ldots , L'_n)=\emptyset$. But, see, e.g.,
\cite{10}, from elimination theory it
follows that the set of all $(L'_0,L'_{s+1},\ldots , L'_n)$ such
that $W'\cap{\cal Z}(L'_0,L'_{s+1},\ldots , L'_n)=\emptyset$
(respectively $W'_1\cap{\cal Z}(L'_{s+1},\ldots ,
L'_n)=\emptyset$) is a nonempty open in the Zariski topology
subset of ${\Bbb A}^{(r+1)(n-s+1)}(\overline{k})$. Hence also
${\cal U}'_0$ is nonempty and open in the Zariski topology by
Lemma~2.

Let $(L'_0,L'_{s+1},\ldots , L'_n)\in{\mathfrak l}'\cap{\cal U}'_0$
be an arbitrary element (not necessary constructed above).
Set $e'_i=L'(e_0,\ldots , e_r)$, $i\in\{0,s+1,\ldots ,
n\}$ and define $\rho'$ replacing in the definition of $\rho''$ the linear
forms
$L''_0,L''_{s+1},\ldots , L''_n$ by $L'_0,L'_{s+1},\ldots , L'_n$.
Set
$$
{\mathfrak l}''=\bigcup_{(L''_0,L''_{s+1},\ldots , L''_n)\in{\mathfrak l}'}
{\cal Z}(L''_{s+1},\ldots , L''_n)
\subset{\Bbb P}^{n-s}(\overline{k}).
$$
Then ${\mathfrak l}''$ is a line or consists of one point. If ${\mathfrak
l}''$ is a point then
assertion (d) is trivial. Suppose that ${\mathfrak l}''$ is a line. Recall
that in the second
auxiliary algorithm for (b) we get the estimations for the number of bad
points of the line
${\mathfrak l}$, i.e., points $z$ such that $N_i(z)<N_i(z^{(2)})$
for some $0\le i\le 5$.
Let us apply these estimations from
the second auxiliary algorithm to $\rho'$, ${\mathfrak l}''$ and the point
from
${\cal Z}(L'_{s+1},\ldots , L'_n)$ in place of $\rho$, ${\mathfrak
l}$ and $z^{(2)}$ respectively.
Then we obtain immediately assertion (d). Assertion (d) is proved.


\section{Proof of assertions (e) and (f)}\label{s5}

Now our aim is to prove (f). Let $L'_j\in\overline{k}[X_0,\ldots ,X_r]$,
$j\in\{0,s+1,s+2,\ldots , n\}$, and (3) holds.
Recall that $e'_i=L'_j(e_0,\ldots , e_r)$, $j\in\{0,s+1,s+2,\ldots , n\}$.
Set
$$
\widetilde{W}_i=
\overline{(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap{\cal
Z}(e'_{s+1},\ldots , e'_i)}, \quad s\le i\le n.
\eqno (26)
$$
Put $W_i$, $s\le i\le n$, to be the union of all the irreducible over
$\overline{k}$ components $E$ of $\widetilde{W}_i$
such that $E$ contains a point from $A=W\cap{\cal Z}(e'_{s+1},\ldots ,
e'_n)\setminus{\cal Z}(e_0,\ldots , e_r)$. Then by (3),
Lemma~2 and the implicit function theorem
$\dim W_i=n-i$ and for every irreducible component $E$ of $W_i$ the
dimension $\dim \overline{p(E\setminus{\cal Z}(e_0,\ldots , e_r))}=n-i$.

\par\medskip\noindent{\bf LEMMA~10}\hspace{0.1em} {\it  Suppose that (3)
holds for linear forms
$L'_j\in\overline{k}[X_0,\ldots ,X_r]$,
$j\in\{0,s+1,s+2,\ldots , n\}$. Then the set
$W'\cap{\cal Z}(L'_{s+1},\ldots ,L'_n)=p(A)$ is finite.
If additionally $A\cap{\cal Z}(e'_0)=\emptyset$ then
$W'\cap{\cal
Z}(L'_0,L'_{s+1},\ldots , L'_n)=\emptyset$.
}\par\medskip

\noindent{\bf PROOF}\quad We can assume without loss of generality that
$A\cap{\cal Z}(e'_0)=\emptyset$.
Let us define the morphism $\rho'$ replacing in
(25) all $L''_j$, $e''_j$ by $L'_j$, $e'_j$.
Then by the theorem about dimensions of fibers, see \cite{11}, the morphism
$\rho'$ is dominant.
Hence the morphism $\rho'_2\, :\,
W'\setminus{\cal Z}(L'_0,L'_{s+1},\ldots , L'_n)
\rightarrow{\Bbb P}^{n-s}(\overline{k})$,
$(X_0:\ldots : X_r)\mapsto(L'_0,L'_{s+1}:\ldots : L'_n)$
is also dominant. By (3)
and
Lemma~2 there is a neighborhood $U$ with respect to the Zariski topology of
the point $z=(1:0:\ldots : 0)$
such that the morphism $(\rho')^{-1}(U)\rightarrow U$ induced by $\rho'$ is
finite dominant.
Hence the morphisms $(\rho')^{-1}(U)\rightarrow (\rho'_2)^{-1}(U)$ and
$(\rho'_2)^{-1}(U)\rightarrow U$ induced by $p$ and $\rho'_2$ are
also finite dominant.
This implies the required assertions. The lemma is proved.

\medskip
For $s\le i\le n$ denote by $W'_i$ the intersection
of $W'\cap{\cal Z}(L'_{s+1},\ldots ,
L'_i)$ in ${\Bbb P}^r(\overline{k})$.
So $p^{-1}(W'_i)$ is a subvariety of
(26).
Denote by $W''_i\subset W'_i$ the subset
of all points
$z\in W'_i$ such that the number of elements $\# p^{-1}(z)\ne \deg p'$
or $z$ is a singular point of $W'_i$.
By Lemma~10 the dimension $\dim W'_i=n-i$, and
for every irreducible component $E'$ of $W'_i$ the intersection $E'\cap\
p(A)\ne\emptyset$.

Let $s+1\le i\le n$. By (3) and the implicit function theorem
$\dim W_{i-1}\cap{\cal Z}(e'_i)=n-i$.
Denote by $W^{(3)}_i$ the union of all the irreducible components $E$ of
$W_{i-1}\cap{\cal Z}(e'_i)$ such that $E\not\subset{\cal
Z}(e_0,\ldots , e_r)$ and
$\dim\overline{p(E\setminus{\cal Z}(e_0,\ldots ,
e_r))}<n-i-1$.
Hence $W^{(3)}_i\subset W^{(4)}$ (in particular $W^{(3)}_n=\emptyset$).

{\it If an $(n-s+1)$--tuple of linear
forms $L'=(L'_0,L'_{s+1},\ldots , L'_n)$
is given and (3) holds then we denote also $W_i=W_i(L')$,
$\widetilde{W}_i=\widetilde{W}_i(L')$,
$W'_i=W'_i(L')$,
$W''_i=W''_i(L')$,
$W^{(3)}_i=W^{(3)}_i(L')$.}


Let $X_0,\ldots ,X_r$ and $X_0,X_{s+1},\ldots , X_n$ be
homogeneous coordinate functions on ${\Bbb
P}^r(\overline{k})$ and ${\Bbb
P}^{n-s}(\overline{k})$ respectively. Let us identify ${\Bbb
P}^{r-i}(\overline{k})
={\cal Z}(L'_{s+1},\ldots , L'_i)\subset{\Bbb P}^r(\overline{k})$ and
 ${\Bbb P}^{n-i}(\overline{k})={\cal Z}(X_{s+1},\ldots ,
X_i)\subset{\Bbb P}^{n-s}(\overline{k})$.
Let $E$ be an irreducible over $\overline{k}$ component
of the variety $\widetilde{W}_i$ for some $s\le i\le n$.
Set $E'=\overline{p(E\setminus{\cal
Z}(e_0,\ldots , e_r))}$ $\subset{\Bbb P}^r(\overline{k})$.
Recall that $\rho'$ is defined in the proof of Lemma~10.
Denote by
$p|_E\, :\, E\setminus{\cal Z}(e_0,\ldots , e_r)\rightarrow
{\cal Z}(L'_{s+1},\ldots , L'_i)$ and $\rho'|_E\, :\,
E\setminus{\cal Z}(e'_0,e'_{s+1},\ldots , e'_n)\rightarrow{\cal
Z}(X_{s+1},\ldots ,$ $X_i)$
the restrictions of $p$ and $\rho'$.
Denote by $p'|_E\, :\, E\setminus{\cal Z}(e_0,\ldots , e_r)$ $\rightarrow
E'$ the restriction of $p'$ to $E$ and $E'$.


\par\medskip\noindent{\bf LEMMA~11}\hspace{0.1em} {\it   Let linear forms
$L'_{s+1},\ldots ,
L'_n\in\overline{k}[X_0,\ldots ,X_r]$ satisfy conditions (3), and
let $s\le i\le n$ be an integer.
Then the following assertions hold.
\begin{enumerate} \renewcommand{\labelenumi}{\arabic{enumi})}
\item For an irreducible component $E$ of
$\widetilde{W}_i$
the dimension $\dim E=\dim E'=n-i$ if and only if $E$ is an irreducible
component of $W_i$. Hence
$$
W_{i-1}(L')\cap{\cal Z}(e'_i)\setminus{\cal Z}(e_0,\ldots ,
e_r)=W_i(L')\cup W^{(3)}_i(L')\setminus{\cal Z}(e_0,\ldots , e_r).
\eqno (27)
$$
\item The dimension $\dim W'_i=n-i$, $W'_i\cap{\cal Z}(L'_{i+1},\ldots ,
L'_n)=p(A)$,
the morphism $W_i\setminus{\cal Z}(e_0,\ldots , e_r)\rightarrow W'_i$ is
dominant.
In particular, for every irreducible component $E'$ of $W'_i$ there is an
irreducible component $E$ of $W_i$ such that the morphism
$E\setminus{\cal Z}(e_0,\ldots , e_r)\rightarrow E'$ is dominant.
Finally for every irreducible component $E$ of $W_i$
the number of elements
$$
\#\bigl(\,(E\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{i+1},\ldots , e'_n)\bigr)=\deg(p')\deg E',
\eqno (28)
$$
and hence $E'\cap p(A)\ne\emptyset$.
\item If (3) and (6) hold then
for every irreducible component $W'$ of $W'_i$ the number of elements
$$
\#\bigl(\,E'\cap{\cal Z}(L'_{i+1},\ldots , L'_n)\,\bigr)=\deg E'.
\eqno (29)
$$
\item Let $s+1\le j\le n-1$ be an integer.
Then (8) holds for all $s+1\le i\le j$ if and only if
$\widetilde{W}_i=W_i$ for all $s+1\le i\le j$, i.e., by (27)
if and only if $W^{(3)}_i=\emptyset$ for all $s+1\le i\le j$.
\end{enumerate}
}\par\medskip

\noindent {\bf PROOF} \quad
Assertion 4) follows from 1)--3) and (27).
Assertion 3) follows from
(3) and 1), 2) immediately.
By Lemma~10 the dimension $\dim W'_i=n-i$, and
every irreducible component of $W'_i$ contains a point from $p(A)$.
The morphism $(\rho')^{-1}(U)\rightarrow(\rho'_2)^{-1}(U)$ from the proof of
Lemma~10 is
finite and $A\subset(\rho'_2)^{-1}(U)$. This implies 2).

Suppose that 1) is not true. Then there is
an irreducible over $\overline{k}$ component
$E$ of $\widetilde{W}_i$ such that $E\cap A=\emptyset$.
Replacing $L'_0$ by a new linear form we shall assume
without loss of generality that $A\cap{\cal Z}(e'_0)=\emptyset$. Then
the morphism
$\rho'_2$ from the proof of Lemma~10
is finite dominant by the last assertion of Lemma~10.
Therefore the morphism $E\rightarrow{\Bbb P}^{n-i}(\overline{k})$ induced
by $\rho'_2$ is also finite dominant.
We claim that $E\subset W^{(4)}$. Indeed, otherwise, the restriction
$\rho'|_E$ is a dominant morphism and hence there is a point $z^*\in
{\Bbb P}^{n-i}(\overline{k})$ infinitely close to
$z'=(1:0:\ldots : 0)$ and such that
$(\rho')^{-1}(z^*)\cap E\ne\emptyset$.
Every point from
$(\rho')^{-1}(z')$ is not infinitely
close to any point from $E$. By the implicit function theorem and 2)
there are $\deg(p')\deg E'$
pairwise distinct isolated points from $(\rho')^{-1}(z^*)$ that are
infinitely close
to the corresponding points from
$(\rho')^{-1}(z')$. Hence the number
of the connected components of the inverse image
$\#(\rho')^{-1}(z^*)$ is more than $\deg(p')\deg E'$.
This contradicts to Lemma~2 and proves the required assertion.
The lemma is proved.



\medskip
Now under conditions of (f)
for every $s\le i\le n$ by assertions 3) and 4)
of Lemma~11 (9) holds, and the dimension
$\dim W''_i\le n-i-1$. For every $s\le i\le n-1$
each irreducible component of $W'_i\cap{\cal Z}(L'_{i+1})$ contains a
point from
$W'_i\cap{\cal Z}(L'_{i+1},\ldots , L'_n)$. All the points
from the last intersection are smooth points of $W'_i$ by (6) and
Lemma~2.
Hence each irreducible component of $W'_i\cap{\cal Z}(L'_{i+1})$ is not
an irreducible component of $W''_i$.
In particular, each irreducible component of $W'\cap{\cal Z}(L'_{s+1})$
is not an irreducible component of $W''$.

Further, Lemma~11 implies that each
irreducible component of the algebraic variety
$W^{(4)}\setminus{\cal Z}(e_0,\ldots , e_r)$
is not an irreducible component of $(W\setminus{\cal Z}(e_0,\ldots ,
e_r))\cap{\cal Z}(e'_{i+1})$.
Obviously condition (8) for $i=n-1$ implies (12).
Finally, if (5) holds then by Lemma~11
each irreducible component of
$(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap
{\cal Z}(e'_{s+1})$ is not an irreducible
component $(\overline{V\setminus W})\setminus{\cal Z}(e_0,\ldots , e_r)$.
Now the remaining part of assertion (f) follows from the following lemma.

\par\medskip\noindent{\bf LEMMA~12}\hspace{0.1em} {\it  Let linear forms
$L'_{s+1},\ldots ,
L'_n\in\overline{k}[X_0,\ldots ,X_r]$ satisfy condition (3)
and (8).
Then for every $s\le i\le n-1$ the algebraic variety
$W''_i\cap{\cal Z}(L'_{i+1})\subset W''_{i+1}$ and
hence (10), (11) hold.
}\par\medskip

\noindent {\bf PROOF} \quad
By the definitions of  $W''_{i+1}$ and  $W''_i$ it is sufficient to prove
that any singular point of
$W'_i$ is a singular point of $W'_i\cap{\cal Z}(L'_{i+1})$.
Suppose contrary, that $x'\in W'_i$ is singular but $x'\in
W'_i\cap{\cal Z}(L'_{i+1})$ is smooth.
Hence there is only one irreducible over $\overline{k}$ component $E$ of
$W'_i\cap{\cal Z}(L'_{i+1})$ such that $x'\in E$.
Then there are linearly independent linear forms $L''_j\in
\overline{k}[X_0,\ldots ,X_r]$, $i+2\le j\le n$,
such that the intersection $E\cap{\cal Z}(L''_{i+2},\ldots , L''_n)$ is
transversal at the point $x'$.
Set $L''_{i+1}=L'_{i+1}$.
Denote by $i(W'_i,{\cal Z}(L'_{i+1});E)$
(respectively $i(W'_i,{\cal Z}(L''_{i+1},\ldots , L''_n);x')$,
$i(E,{\cal Z}(L''_{i+2},\ldots , L''_n);x')$)
the index of intersection of  $W'_i$ and
${\cal Z}(L'_{i+1})$ at $E$ (respectively $W'_i$ and ${\cal
Z}(L''_{i+1},\ldots , L''_n)$ at $x'$,
$E$ and ${\cal Z}(L''_{i+2},\ldots , L''_n)$ at $x'$), see \cite{9}.
Then $i(E,{\cal Z}(L''_{i+2},\ldots , L''_n);x')=1$ since the
corresponding intersection is transversal.
The intersection $W'_i\cap{\cal Z}(L'_{i+1})$ is transversal at $E$
since the intersection
$W'_i\cap{\cal Z}(L'_{i+1})\cap{\cal Z}(L'_{i+2},\ldots , L'_n)$
is transversal by (6) and Lemma~2.
Therefore, $i(W'_i,{\cal Z}(L'_{i+1});E)=1$.
The  index $i(W'_i,{\cal Z}(L''_{i+1},\ldots , L''_n);$ $x')>1$ since
$x'$
is a singular point
of $W'_i$. On the other hand, by the property of indices of intersection
corresponding to the associativity of intersection,
see \cite{9},
$$
i(W'_i,{\cal Z}(L''_{i+1},\ldots , L''_n);x')=i(W'_i,{\cal
Z}(L'_{i+1});E)
i(E,{\cal Z}(L''_{i+2},\ldots , L''_n);x').
$$
This contradiction proves the lemma.

\medskip Let us prove (e).
\par\medskip\noindent{\bf LEMMA~13}\hspace{0.1em} {\it  Let us extend the
ground field $k$ till $k'$, where $k'$ is a
finitely generated field over $k$.
Let $L'=(L'_0,L'_{s+1},\ldots , L'_n)$, where all
$L'_j\in\overline{k'}[X_0,\ldots
,X_r]$ are linear forms. Suppose that $L'$
satisfies conditions (3), (4), (5) and
(6) (with $k'$ in place of $k$).
Let ${\mathfrak l}\subset{\Bbb A}^{(r+1)(n-s+1)}(\overline{k'})$ be a
line such that $L'\in{\mathfrak l}$.
Then for all $L''\in{\mathfrak l}$, except at most a
polynomial in $d^n(d')^{n-s}$ number,
conditions (3), (4), (5) and
(6) hold for $L''$
(with $L''$ and $k'$ in place of
$L'$ and $k$), and the inequality
$\deg W_i(L'')\ge\deg W_i(L')$
is satisfied for every $s\le i\le n-1$.
}\par\medskip

\noindent {\bf PROOF}\quad Recall that $e'_j=L'_j(e_0,\ldots , e_r)$ for
all $j\in
\{0,s+1,\ldots , n\}$. There are homogeneous polynomials $h_1,\ldots , h_s$
of
degrees bounded from above by a polynomial in $d^s$ such that $W$ is an
irreducible component of the algebraic variety
${\cal Z}(h_1,\ldots , h_s)$; all the points from
$(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap{\cal
Z}(e'_{s+1},\ldots , e'_n)$ are smooth points of ${\cal Z}(h_1,\ldots ,
h_s)$;
the intersection of $n$ hypersurfaces
${\cal Z}(h_1),\ldots ,$ ${\cal Z}(h_s)$, ${\cal
Z}(e'_{s+1}),\ldots ,{\cal Z}(e'_n)$ is transversal at every
point from $(W\setminus{\cal Z}(e_0,\ldots , e_r))\cap{\cal
Z}(e'_{s+1},\ldots , e'_n)$.

Further, there exists a linear subspace ${\cal L}_i\subset
{\Bbb P}^n(\overline{k'})$ with $\dim{\cal L}_i=i$ such that
the intersection $\widetilde{W}_i(L')\cap{\cal
L}_i$ is transversal, hence
the intersection $W_i(L')\cap{\cal
L}_i={\cal A}_i$ is transversal and finite, and all the elements of
${\cal A}_i$ are isolated points of $\widetilde{W}_i(L')\cap{\cal
L}_i$; the element
${\cal A}_i\cap{\cal Z}(e'_0)=\emptyset$,
and hence all the elements of
${\cal A}_i$ are isolated points of $W\cap{\cal
Z}(e'_{s+1},\ldots ,e'_i)\cap{\cal
L}_i$; for every point $z\in{\cal A}_i$
the intersection of $n$ hypersurfaces
${\cal Z}(h_1),\ldots ,{\cal Z}(h_s)$, ${\cal
Z}(e'_{s+1}),\ldots ,{\cal Z}(e'_i),$
${\cal Z}(e'_{i+1}-(e'_{i+1}/e'_0)(z)e'_0),\ldots ,{\cal
Z}(e'_n-(e'_{n}/e'_0)(z)e'_0)$
is transversal at $z$. All these are conditions of
general position which can be easily satisfied.

Let a
generic point $\widetilde{L}=(\widetilde{L}_0,\widetilde{L}_{s+1},\ldots ,
\widetilde{L}_n)$
of ${\mathfrak l}$ has the form
$\widetilde{L}_j=L'_j+\varepsilon L'''_j$ where $\varepsilon$ is a
transcendental element over $k'$ and $L'''_j\in
\overline{k'}[X_0,\ldots ,X_r]$ are linear forms
for all $j\in\{0,s+1,\ldots , n\}$.
Set $\widetilde{e}_j=\widetilde{L}_j(e_0,\ldots , e_r)$,
$e'''_j=L'''_j(e_0,\ldots , e_r)$,
$e''_j=L''_j(e_0,\ldots , e_r)$ for all $j$. Let $\varepsilon$ be the
coordinate function of ${\Bbb A}^1$, and
$\varepsilon$ be the coordinate of $\widetilde{L}$. We shall assume
that the coordinate of
$L''$ is $\varepsilon''\in\overline{k'}$.
Consider the algebraic variety $\widetilde{W}_{\varepsilon,i}=
(W\times{\Bbb A}^1(\overline{k'}))\cap
{\cal Z}(\widetilde{e}_{s+1},\ldots ,\widetilde{e}_i)\subset
{\Bbb P}^n(\overline{k'})\times{\Bbb A}^1(\overline{k'})$.
Notice that every point from ${\cal A}_i\times\{0\}$ is an isolated
point of
$\widetilde{W}_{\varepsilon,i}\cap({\cal L}_i\times\{0\})=
(W\cap{\cal Z}(e'_{s+1},\ldots , e'_i)\cap{\cal L}_i)\times\{0\}$.
Denote by $W_{\varepsilon,i}$ the union of all the irreducible components
$E$ of $\widetilde{W}_{\varepsilon,i}$
such that $E\cap({\cal L}_i\times\{0\})\cap
({\cal A}_i\times\{0\})\ne\emptyset$, and hence $\dim
E=n-i+1$. Obviously every irreducible component of $W_i(L')\times\{0\}$ is
an irreducible component of
$W_{\varepsilon,i}\cap
({\Bbb  P}^n(\overline{k'})\times\{0\})$.

Set $C_{\varepsilon,i}$ to be the union of all the irreducible components
$E_1$ of
$\widetilde{W}_{\varepsilon,i}\cap({\cal L}_i\times
{\Bbb A}^1(\overline{k'}))$ such that $E_1\cap({\cal
A}_i\times\{0\})\ne\emptyset$.
Hence every irreducible component $E_2$ of $C_{\varepsilon,i}$ is a curve,
and $E_2$ is an irreducible component of
$W_{\varepsilon,i}\cap({\cal L}_i\times
{\Bbb A}^1(\overline{k'}))$.
Finally, note that for every irreducible component $E$ of
$W_{\varepsilon,i}$ there is an irreducible
component $E_2$ of $C_{\varepsilon,i}$
such that $E_2\subset E_1$.

Let $\varepsilon''\in\overline{k'}$.
Denote $W_{\varepsilon,i}\cap{\cal
Z}(\varepsilon-\varepsilon'')=W_{\varepsilon'',i}\times\{\varepsilon''\}$
where
$W_{\varepsilon'',i}\subset{\Bbb P}^n(\overline{k'})$,
and $\dim W_{\varepsilon'',i}=n-i$.
Hence $(W_{\varepsilon'',i}\cap{\cal
L}_i)\times\{\varepsilon''\}\supset{\cal
Z}(\varepsilon-\varepsilon'')\cap\ C_{\varepsilon,i}$.
By the B\'ezout theorem and
Theorem~1 (d) (with the field $k'$
in place of $k$)
for all $\varepsilon''\in\overline{k'}$, except
at most a polynomial in $d^n(d')^{n-s}$ number,
the following assertions hold for all $s\le i\le n-1$:
conditions (3), (4) (5) and (6) hold for $L''$
(with $L''$ and $k'$ in place of
$L'$ and $k$);
the intersection $W_{\varepsilon'',i}$ is a union
of some irreducible components of dimension $n-i$ of $(W\cap
{\cal Z}(e''_{s+1},\ldots ,e''_i))\times\{\varepsilon''\}$; for
every $z\in W_{\varepsilon'',i}\cap{\cal L}_i$ if
$(z,\varepsilon'')\in C_{\varepsilon,i}$ then $z$ is
an isolated point of $W_{\varepsilon'',i}\cap{\cal L}_i$;
the number of elements $+\infty>
\#({\cal Z}(\varepsilon-\varepsilon'')\cap
C_{\varepsilon,i})\ge
\#({\cal Z}(\varepsilon)\cap C_{\varepsilon,i})=\#{\cal A}_i$;
the intersection of $n$ hypersurfaces
${\cal Z}(h_1),\ldots ,{\cal Z}(h_s)$, ${\cal
Z}(e''_{s+1}),\ldots ,{\cal Z}(e''_i),$
${\cal Z}(e''_{i+1}-(e''_{i+1}/e''_0)(z)e''_0),\ldots ,{\cal
Z}(e''_n-(e''_{n}/e''_0)(z)e''_0)$ is transversal at every
point $z\in W_{\varepsilon'',i}\cap{\cal L}_i$
such that $(z,\varepsilon'')\in C_{\varepsilon,i}$.
The last condition related  to the intersection of hypersurfaces
means that the restriction of the morphism $\rho'$ to
$W_{\varepsilon'',i}$ is dominant.
Therefore, for the considered $\varepsilon''$ the algebraic variety
$W_{\varepsilon'',i}$ is a union
of some irreducible components of $W_i(L'')$ and
$\deg W_{\varepsilon'',i}\ge\deg W_i(L')$.
The lemma is proved.



\medskip Now suppose that $L'\in{\cal U}'_0$, see assertion (d),
is constructed by (c).
So all the linear forms $L'_j$ have integer coefficients of length
$O(n\log d+(n-s)\log(d'))$.
We shall describe an algorithm recursive by $L'$. At the end of the
recursion $L'$ will satisfy
assertion (e).

Let us describe the general step of the recursion. At the beginning of the
step the
current $L'$ is given.
The recursive assumption is that an integer $s\le i\le n-2$ is known such
that
$\widetilde{W}_j(L')=W_j(L')$ for all $s\le j\le i$ (at the first step
$i=s$).
Let us apply Theorem~1 \cite{8} to the algebraic variety
$\widetilde{V}_i=(V\cap{\cal Z}(e'_{s+1},\ldots , e'_i))
\cup{\cal Z}(e'_{i+1})$ and construct linear forms $L_0,\ldots ,
L_{n+1}$ from this theorem. For every $s\le j\le n-1$
applying Theorem~5 \cite{6} (which allow to decide whether a morphism is
dominant) and
Theorem~2 \cite{8} to $V\cap{\cal Z}(e'_{s+1},\ldots , e'_j)$
we compute all the degrees $\deg W_j(L')$.
In what follows we shall compute in the similar way
the degrees $\deg W_j(L'')$, $s\le j\le n-1$, for $L''$ analogous to $L'$.

The morphism  $W_i(L')\setminus{\cal Z}(e'_0,e'_{i+1},\ldots
,e'_n)\rightarrow
{\cal Z}(L'_{s+1},\ldots , L'_i)$, $(X_0:\ldots
:X_n)\mapsto(e'_0:e'_{i+1}:\ldots : e'_n)$,
is dominant. Therefore, the
dimension $\dim W_i(L')\cap{\cal Z}(e'_{i+1})=n-i-1$.
Hence $e'_{i+1}$ does not vanish on any
irreducible component of $W_i(L')$.
Hence $W\cap{\cal Z}(L_{i+1},\ldots , L_n)\cap{\cal
Z}(e'_{i+1})=\emptyset$
according to the construction of $L_0,\ldots , L_{n+1}$.

Denote $L''_j=L_j$ for $i+2\le j\le n$.
We shall describe a recursive algorithm by $(L''_{i+2},\ldots , L''_n)$
(the element for the base of this second embedded recursion is defined).
The recursive assumption is that $L''_j=L_j-\mu_jL_{i+1}$ for some integers
$\mu_j$ for all $i+2\le j\le n$.
At the end of this recursion we shall obtain an
element $(L''_{i+2},\ldots , L''_n)$ such that $(W_{i+1}(L')\cup
W^{(3)}_{i+1}(L'))\cap{\cal Z}(L''_{i+2},\ldots ,L''_n)\cap{\cal
Z}(e_0,\ldots , e_r)=\emptyset$.

Let us describe the general step of this embedded recursion.
Let us construct using the algorithm from \cite{2} all the irreducible
components of dimension $1$
of the intersection $\widetilde{V}_i\cap
{\cal Z}(L''_{i+2},\ldots , L''_n)$
and
using Theorem~2 \cite{8} choose among them all the irreducible curves
$C_\gamma$, $\gamma\in\Gamma$ such that $C_\gamma\subset W_i(L')$.
By Theorem~1 \cite{8} each curve $C_\gamma$ contains a point
$y\in W_i(L')\cap{\cal Z}(L_{i+1},\ldots , L_n)$,
and every $y$ is a smooth point of $\widetilde{V}_i$.
Denote $C_i=\bigcup_{\gamma\in\Gamma}C_\gamma$.
Now $C_i=W_i(L')\cap{\cal Z}(L''_{i+2},\ldots , L''_n)$.

Let us construct by \cite{2} the finite set
$S_{i+1}=S(L''_{i+2},\ldots , L''_n)=C_i\cap{\cal Z}(e'_{i+1})$.
Every irreducible component of $W_i(L')\cap{\cal Z}(e'_{i+1})$
contains a point from
$S_{i+1}$, and the number of elements $\#S_{i+1}$ is bounded from above by a
polynomial in $d^n(d')^{n-s}$ by the B\'ezout theorem.
Set $S'_{i+1}=S'(L''_{i+2},\ldots , L''_n)=
S_{i+1}\setminus{\cal Z}(e_0,\ldots ,
e_r)$ (we shall use also the similar notation for linear forms with
coefficients from other fields).
Notice that $S(L''_{i+2},\ldots , L''_n)\cap{\cal
Z}(L_{i+1})=\emptyset$ for arbitrary $\mu_j$.

Let us enumerate the points $z\in S_{i+1}\setminus S'_{i+1}$.
Consider the following system of polynomial equations and inequalities with
coefficients from $\overline{k_2}$.
$$
\left\{\begin{array}{ll}
\sum_{0\le i\le r}|e_i|^2=\varepsilon_2,\\
\sum_{0\le i\le n}|X_i-(X_i/L_{i+1})(z)|^2\le\varepsilon_1,\\
e'_j=0, & s+1\le j\le i+1,\\
L_{i+1}=1.
\end{array}\right.
\eqno (30)
$$
Let us apply Theorem~2 to this system and the variety $\mbox{\rm
con}(V)$ 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 $x^*\in
S'''\cap W(\overline{k_2})$.
Obviously such a point exists if and only if $z\in W_{i+1}(L')\cap
W^{(3)}_{i+1}(L')$.
Suppose that such a point exists.
Then set $\mu^*_j=(L''_j/L_{i+1})(x^*)$ and
$L^*_j=L''_j-\mu^*_jL_{i+1}$
for all $i+2\le j\le n$.

Hence $x^*\in S'(L^*_{i+2},\ldots , L^*_n)$ and
for every $z_1\in S'(L''_{i+2},\ldots ,L''_n)$ there is $z^*_1\in
S'(L^*_{i+2},\ldots , L^*_n)$ which is infinitely close to $z_1$
by, e.g., Proposition~6 \cite{5}. Hence the number of elements
$\#S'(L^*_{i+2},\ldots , L^*_n)>\#S'(L''_{i+2},\ldots ,L''_n)$.
Using the algorithm of reduction to integer coefficients, see Proposition~2,
one can replace subsequently elements $\mu^*_j$
by integers $\mu^{(1)}_j$ (we find them by the enumeration)
of length $O(n\log d+(n-s)\log(d'))$ and
obtain linear forms $L^{(1)}_j=L''_j-\mu^{(1)}_jL_{i+1}$, $i+2\le j\le n$,
such that
$\#S'(L^{(1)}_{i+2},\ldots , L^{(1)}_n)\ge\#S'(L^*_{i+2},\ldots , L^*_n)$.
This follows from the B\'ezout theorem, cf. the
proof of Lemma~13. We leave the details to the reader here.
Now we replace $(L''_{i+2},\ldots ,L''_n)$ by $(L^{(1)}_{i+2},\ldots ,
L^{(1)}_n)$
and proceed to the next step of the recursion.

If $x^*\in S'''\cap W(\overline{k_2})$ does not exists
and not all the points from $S_{i+1}\setminus S'_{i+1}$ are enumerated then
we proceed to the next point
$z\in S_{i+1}\setminus S'_{i+1}$. If all the points from $S_{i+1}\setminus
S'_{i+1}$
are enumerated then we have obtained the required element $(L''_{i+2},\ldots
,L''_n)$. The second recursion is described.


\medskip Let us construct a linear form $L'''_{i+1}\in k[X_0,\ldots ,X_r]$
with integer coefficients of
length $O(n\log d+(n-s)\log(d'))$
such that $S'_{i+1}(L''_{i+2},\ldots
,L''_n)\cap{\cal Z}(L'''_{i+1}(e_0,\ldots , e_r))=\emptyset$.
Let $\varepsilon_1$ be a
be an infinitesimal with respect to the field $k$. Set
$k_1=k(\varepsilon_1)$ and $\widetilde{L}_{i+1}=L'_{i+1}+\varepsilon_1
L'''_{i+1}$.
Put $\widetilde{L}_\alpha=L'_\alpha$ for all $\alpha\in\{0,s+1,\ldots ,
n\}\setminus\{i+1\}$,
and $\widetilde{L}=(\widetilde{L}_0,\widetilde{L}_{s+1},\ldots
,\widetilde{L}_n)$.
Since ${\cal U}'_0$ is an open
in the Zariski topology subset, the element
$\widetilde{L}$ satisfies (3), (4),
(5) and (6) with $k_1$ in place of $k$.
We have $W^{(3)}_{i+1}(\widetilde{L})\subset W_i(L')\cap W^{(4)}$.
Hence $W^{(3)}_{i+1}(\widetilde{L})$ is a union of
some irreducible components
of dimension $n-i-1$ of the variety $W_i(L')\cap W^{(4)}$.
In particular, $W^{(3)}_{i+1}(\widetilde{L})$ is defined over $\overline{k}$.
Therefore, $W^{(3)}_{i+1}(L')\supset W^{(3)}_{i+1}(\widetilde{L})$.
Since $\widetilde{e}_{i+1}=
\widetilde{L}_{i+1}(e_0,\ldots , e_r)$ does not
vanish on any irreducible component $E$ of
$W^{(3)}_{i+1}(L')$ the algebraic variety
$W^{(3)}_{i+1}(\widetilde{L})=\emptyset$.
Hence $W_{i+1}(\widetilde{L})\setminus{\cal Z}(e_0,\ldots ,
e_r)=W_i(L')\cap{\cal Z}(\widetilde{e}_{i+1})
\setminus{\cal Z}(e_0,\ldots , e_r)$ over the field $\overline{k_1}$.
Now in Lemma~13
put $k'=k$ and ${\mathfrak l}$ to be the line containing $L'$ and
$L'''$. There are infinitely many isomorphic generic points
of this line. Hence extending $k$ till $k_1$ and applying
Lemma~13 we obtain that {\it the degrees
$\deg W_j(\widetilde{L})\ge\deg W_j(L')$ for all $s\le j\le n-1$.}



\par\medskip\noindent{\bf LEMMA~14}\hspace{0.1em} {\it  Suppose
$\widetilde{W}_j(L')=W_j(L')$ for all $s\le j\le i$
and there exists an irreducible component $E$
of  $W^{(3)}_{i+1}(L')$.
Then the degree $\deg W_{i+1}(\widetilde{L})>\deg W_{i+1}(L')$.
}\par\medskip

\noindent {\bf PROOF} \quad In the proof of this lemma we shall assume
without loss of generality that
the linear forms $L_{i+2},\ldots , L_n$ satisfy additionally the conditions:
$\#(W_i(L')\cap{\cal Z}(e'_{i+1})\cap{\cal Z}(L_{i+2},\ldots , L_n))
=\deg(W_i\cap{\cal Z}(e'_{i+1}))$; for every
irreducible component $E_1$ of $W_i(L')\cap{\cal Z}(e'_{i+1})$ such that
$E_1\not\subset{\cal Z}(e_0,\ldots , e_r)$ the intersection
$E_1\cap{\cal Z}(L_{i+2},\ldots , L_n)
\cap{\cal Z}(e_0,\ldots , e_r)=\emptyset$  (such linear forms
$L_{i+2},\ldots ,L_n$
exist in general position, and we do not need to construct them at present).

Let $z^{(0)}\in E\cap{\cal Z}(L_{i+2},\ldots , L_n)$.
Then $z^{(0)}\not\in W_{i+1}(L')\cap{\cal Z}(L_{i+2},\ldots , L_n)$ and
$z^{(0)}\not\in{\cal Z}(e_0,\ldots , e_r)$.

Since $\varepsilon$ is an
infinitesimal,
for every point $z\in W_i(L')\cap{\cal Z}(e'_{i+1})\cap{\cal
Z}(L_{i+2},\ldots ,$ $L_n)$
there is a point $z^*\in W_i(L')\cap{\cal
Z}(\widetilde{e}_{i+1})\cap{\cal Z}(L_{i+2},\ldots , L_n)$
such that $z^*$ is infinitely close to $z$ (here we leave the details to the
reader).
In particular, there is a point of $W_i(L')\cap{\cal
Z}(\widetilde{e}_{i+1})
\cap{\cal Z}(L_{i+2},\ldots , L_n)$ which is infinitely close
to $z^{(0)}$. This implies the assertion of the lemma. The lemma is proved.

\medskip Let us return to the description of the algorithm. Let us compute
all $\deg W_j(\widetilde{L})$, $s\le j\le n-1$
(similarly to the computation of $\deg W_j(L')$ above; here one should extend
the field $k$ till $k_1$). Suppose that $\deg W_{j_1}(\widetilde{L})>\deg
W_{j_1}(L')$ for some $s+1\le j_1\le n-1$.
Then by Lemma~13 one can apply the algorithm of reduction to integer
coefficients, see Proposition~2.  We replace
subsequently all the coefficients of the linear forms $\widetilde{L}_j$
by integers of
length $O(n\log d+(n-s)\log(d'))$ and obtain new
linear forms $L''_j$, $j\in\{0,s+1,\ldots , n\}$ such that
$L''=(L''_0,L''_{s+1},\ldots , L''_n)\in{\cal U}'_0$, the degrees
$\deg W_j(L'')\ge\deg W_j(\widetilde{L})$ for all $s\le j\le n-1$, and
hence $\deg W_{j_1}(L'')>\deg W_{j_1}(L')$.
In this case we replace $L'$ by $L''$ and proceed to the next step of the
recursion with the same $i$. By the B\'ezout theorem there might be at most
$d^s(d')^{n-j_1}$ steps when $\deg W_{j_1}(\widetilde{L})>\deg W_{j_1}(L')$.

Now suppose $\deg W_j(\widetilde{L})=\deg W_j(L')$ for all $s+1\le j\le n-1$.
Then by Lemma~14 and Lemma~11
we have $W_{i+1}(L')\setminus{\cal Z}(e_0,\ldots ,
e_r)=W_i(L')\cap{\cal Z}(e'_{i+1})
\setminus{\cal Z}(e_0,\ldots , e_r)$. Hence
$W_{i+1}(L')=\widetilde{W}_{i+1}(L')$.
In this case if $i+1\le n-2$ we replace $i$ by $i+1$
and proceed to the next step of the recursion with the same $L'$.

Finally we shall come to the case when
$\deg W_{i+1}(\widetilde{L})=\deg W_{i+1}(L')$ and $i=n-2$.
Then condition (8) holds for all $s+1\le i\le n-1$ by Lemma~14 and
Lemma~11.
This step is final.
The required estimation for the working time of the described
algorithm for assertion (e) follows immediately from the ones
of the applied algorithms.
Theorem~1 is proved completely.

\par\medskip\noindent{\bf REMARK~6}\hspace{0.1em} {\it  Let $k'$ be an
arbitrary finitely generated extension of $k$
(and $k'$ is given similarly to $k$, see the Introduction).
Let us extend the ground field $k$ till $k'$.
Suppose that $L'\in{\cal U}'_0$ is given,
and all linear forms $L'_j\in k'[X_0,\ldots ,X_r]$,
$j\in\{0,s+1,\ldots , n\}$.
Actually the algorithm for assertion (e) allows
to decide whether (8) holds for all
$s+1\le i\le n-1$ (with $k'$ in place of $k$).
Further, if (8) is not fulfilled then one can
construct an element $L''\in{\cal U}'_0$ such that all linear forms
$L''_j$ have
integer coefficients of length $O(n\log d+(n-s)\log d')$, condition (8)
holds for $L''$
(in place of $L'$), the inequalities $\deg W_j(L'')\ge\deg W_j(L')$ hold for
all $s\le j\le n-1$,
and at least one of the last inequalities is strict.
The working time of this modification of
the algorithm from (e) is polynomial in $d^n$,
$(d')^n$, $d_1$, $d_2$, $d'_2$,
$M$, $M'$, $M_1$, $m$, $r$, $N$,
${\rm L}(x^{(0)})$, the size ${\rm L}(L')$,
and the size of minimal polynomial of
the primitive element of $k'$ over a purely transcendental extension of $k$.
}\par\medskip




\newpage

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\end{document}
