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\begin{document}

 \title{Monodromy and irreducibility criteria with algorithmic applications
  in zero--characteristic}
\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 $V$
which is the set of all common zeroes of homogeneous
polynomials of degrees less than $d$ in $n+1$ variables
in zero--characteristic. We suggest
an algorithm to decide whether two (or more) given points of $V$
belong to the same irreducible component of $V$.
Besides that we show how to construct for each $s<n$ an $(s+1)$--dimensional
plane in the projective space
such that the intersection of every irreducible component of dimension $n-s$
of $V$ with the constructed plane
is transversal and is an irreducible curve.
These algorithms are deterministic and polynomial in $d^n$ and
the size of input.
\end{abstract}




\newpage
 \section*{Introduction}

Let
$k$ be a field of zero--characteristic with algebraic closure
$\overline{k}$.
Denote by ${\Bbb P}^n(\overline{k})$, $n\ge 0$, the projective space over
the field
$\overline{k}$
with coordinates $X_0,\ldots , X_n$. We shall suppose that ${\Bbb
P}^n(\overline{k})$ is defined over $k$.
For arbitrary homogeneous polynomials
$g_1,\ldots ,g_m\in\overline{k}[X_0,\ldots , X_n]$ we shall denote by
${\cal Z}(g_1,\ldots ,g_m)$ the set of all common zeroes of polynomials
$g_1,\ldots ,g_m$ in
${\Bbb P}^n(\overline{k})$.
The similar notations will be used for the sets of zeroes
of ideals and polynomials with other fields of coefficients in affine and
projective spaces
(this will be seen from the context).



In the algorithmic part of the paper we shall
suppose that the field $k$ is finitely generated
over the field of rational numbers ${\Bbb Q}$.
Let $V$ be a
projective algebraic variety which is a
set of all common zeroes of homogeneous polynomials in $n+1$ variables of
degrees less than $d$
with coefficients from $k$.
Now let $z_1,\ldots , z_u\in V$ be arbitrary points.
We construct an algorithm to
decide whether
$z_1,\ldots , z_u$ belong to the same defined and
irreducible over $k$ (respectively
$\overline{k}$) component
of $V$ with the working time polynomial in $d^n$ and the size of input
(respectively the same working time for fixed $u$), see
Theorem~2. Besides that we construct linear forms $L_1,\ldots ,L_{n-1}$ in
$X_0,\ldots , X_n$ with integer coefficients of length
$O(n\,\log d)$ such that for every $0\le s\le n$ for every irreducible over
$\overline{k}$
component $W$ of $V$ with $\dim W=n-s$ the intersection $W\cap{\cal
Z}(L_{s+1},\ldots , L_{n-1})$ is transversal and is an
irreducible over $\overline{k}$ projective curve, see Theorem~1.

Note here that even the existence of the required
linear forms $L_1,\ldots ,L_{n-1}$ with length of integer coefficients
$O(n\,\log d)$ is not trivial.
In the algorithms of this paper we use essentially the results of
\cite{4}.
We apply also the first Bertini theorem,
see \cite{1}.
On the other hand, we need to attract the transcendental
methods of theory of analytical
functions similarly to \cite{5}, namely, the group of monodromy
of an algebraic function to substantiate the algorithm
(in \cite{5} we used these methods
not only for the proof of correctness
but in the algorithm itself), see Theorem~3, Section~1. As an immediate
consequence of
Corollary~2 from this theorem
we obtain the following nice result
(the main case here is $n\ge 3$).

\medskip\noindent
{\bf CRITERION FOR IRREDUCIBILITY}\quad
{\em Let $\Phi\in k[X_0,\ldots , X_n]$, $n\ge 1$,
be a separable homogeneous polynomial
over a field of zero--characteristic.
Let $\deg\Phi=\deg_{X_0}\Phi\ge 1$.
Let $R$ be the discriminant of $\Phi$ with respect to $X_0$ and $R_0$ be
square free part of
$R$ (i.e., $R_0$ is a
separable polynomial of the most degree dividing $R$).
Let $L$ be a nonzero linear form in $X_1,\ldots, X_n$ with
coefficients from $k$, and
$A=k[X_0,\ldots , X_n]/(L)$
(hence $A$ is isomorphic to the ring of polynomial in
$n$ variables).
Suppose that the polynomial $R_0\bmod L\in A$ is separable.
Then the irreducible over $k$ factors
of the polynomials $\Phi$ and $\Phi\bmod L\in A$ are in
the one--to--one correspondence
by the rule $\Psi\mapsto\Psi\bmod L$. In particular, if $\Phi$ is
irreducible over $k$ then
$\Phi\bmod L$ is irreducible over $k$.
Besides that for every linear subspace ${\cal L}$
of linear forms in $X_1,\ldots ,X_n$ with $\dim{\cal L}\ge 3$
there is $L\in{\cal L}$
satisfying the formulated conditions.}

\medskip
This criterion and Corollary~2
clarify the first Bertini theorem in
the important cases. In one of the next the next papers we shall
return to the first Bertini
theorem in the general case from algorithmic
point of view.
The results of the present paper, namely Theorem~1,
will be used there essentially.
We shall need to prove in the next paper
a more general version of Theorem~3 again.
Although this version can be given already in this paper we decided to make
the presentation more clear and omit the details
related to the case of not finite morphisms.

We leave to the reader to deduce
consequences of the given criterion for nonhomogeneous polynomials.
It is interesting to
prove analogs of this
criterion for nonzero characteristic for arbitrary $n$
and for nonhomogeneous
polynomials with integer coefficients with small $n$.

In \cite{6} we suggested a new method for representation of algebraic
varieties which is based on the representative systems of points.
The present paper is an important step to construct the algorithms for this
method in zero--characteristic.


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 $V={\cal Z}(f_1,\ldots ,f_m)$.




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 $f=f_i$ 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 $l(a)$ of an integer $a$ by the formula
 $l(a)=\min\{s \in {\Bbb Z}  : \: |a|<2^{s-1}\}$.
 The length of coefficients $l(f)$ 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 $l(F)$ 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, \;
 l(f_i)<M, \\
 \deg_Z (F)<d_1, \; \deg_{t_\gamma} (F)<d_1, \; l(F)<M_1 \: .
 \end{eqnarray*}
for all $1\le i\le m$.
 The size $L(f)$ of the polynomial $f$ is defined to be the product of
 $l(f)$ to the number of all the coefficients from ${\Bbb Z}$ of $f$ in
 the dense representation.
 We have
$$
 L(f_i)<({d+n \choose n}d_1+1)d_2^l M
$$
 Similarly $L(F)<d_1^{l+1} M_1$.  Unless
 we state otherwise, in what follows we suppose $l$ to be fixed.



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
$$
k(z)=k((X_1/X_{i_0})(z),\ldots , (X_n/X_{i_0})(z))\simeq
k[\eta]=k[Z]/(F_z)
\eqno (1)
$$
is given where $\sum_{0\le i\le n}c_i(X_i/X_{i_0})(z)\mapsto\eta$
under (1), the
coefficients $c_i\in {\Bbb Z}$ are given and $F_z\in k[Z]$ is
minimal polynomial of $\eta$ over $k$ with leading coefficient
$\mbox{\rm lc}_ZF_z=1$. So the point $z$ is defined up
to a conjugation over $k$ and, more precisely,
representation (1) gives
defined over $k$ and irreducible over $k$
algebraic variety of dimension $0$.

Let $a\in k[\eta]$ be an arbitrary element. Then $a=A(\eta)$ for the
uniquely defined polynomial
$A\in k[Z]$ such that $\deg_ZA<\deg_ZF_z$. The length of integer
coefficients $l(a)$,
the the size $L(a)$ and the degrees $\deg_{t_\alpha}a$,
$1\le \alpha\le l$, of $a$
are defined by the formulas
$$
l(a)=l(A),\quad L(a)=L(A), \quad  \deg_{t_\alpha}a=\deg_{t_\alpha}A.
$$
Let us define the size $L(z)$ of the point $z$ to be
$L(F_z)+\sum_{0\le i\le n}L(X_i/X_{i_0})$.
Using the algorithm from \cite{3}
one can construct a system of homogeneous polynomial
equations with coefficients from $k$ giving the least
defined over $k$ and irreducible over $k$
algebraic variety of dimension $0$ containing $z$
within the time polynomial in $L(z)$.

Let $W_1,W_2\subset{\Bbb P}^n(\overline{k})$ (respectively
$W_1,W_2\subset{\Bbb A}^n(\overline{k})$)
be two projective (respectively affine) algebraic varieties. Suppose that
all the irreducible components of $W_i$
have the same dimension
$n-s_i$, $i=1,2$. Let $W_3$ be an irreducible 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 $\dim W_3=n-s_1-s_2$
and 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., $\dim(T_{x,W_1}\cap T_{x,W_2})=n-s_1-s_2$. We shall
say that the intersection
$W_1\cap W_2$ is transversal if and only if it is transversal at every its
component.  In the similar way one defines
transversality for intersection of a finite number of
algebraic varieties.





\par\medskip\noindent{\bf THEOREM~1}\hspace{0.1em} {\it
Let the field $k$, the algebraic variety $V$
and polynomials $F$, $f_1,\ldots ,f_m$
be given as above. Then one can construct linear forms
$L_0,L_1,\ldots ,$ $L_{n+1}$ in $X_0,\ldots ,
X_n$ with integer coefficients of length $O(n\,\log d)$
and the finite defined over $k$ sets $A_s$, $V_{s,0}$, $0\le s\le n$,
of points of $V$ satisfying
the following properties. For every $0\le s\le n$
denote by $V_s$ the union of all
irreducible components of $V$ of dimension $n-s$.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item Then
for every $0\le s\le n$ the set
$A_s=V_s\cap{\cal Z}(L_{s+1},\ldots , L_n)$ is finite
(or empty if $V_s=\emptyset$), consists
of smooth points of $V$ and the intersections of tangent
spaces of $V_s$ and ${\cal Z}(L_{s+1},\ldots , L_n)$ at
all these points are transversal. The linear form $L_0$ does not vanish in
any point of
$A_s$, i.e., $0\not\in L_0(A_s)$, and
the number of elements $\#L_{n+1}(A_s)=\#A_s=\deg V_s$.
\item For every $0\le s\le n-1$ (the case $s=n-1$ is trivial)
irreducible over $\overline{k}$ (respectively $k$) components
of the algebraic varieties $V_s$ and $V_s\cap{\cal Z}(L_{s+1},\ldots ,
L_{n-1})$
are in the one--to--one correspondence by the rule
$W\mapsto W\cap{\cal Z}(L_{s+1},\ldots , L_{n-1})$. The last
intersection
is an irreducible over $\overline{k}$ (respectively irreducible over $k$)
curve. Besides that, if
$W$ is irreducible over $\overline{k}$ then the minimal fields of definition
containing $k$ of
$W$ and $W\cap{\cal Z}(L_{s+1},\ldots , L_{n-1})$ coincide.
\item For every $1\le s\le n-1$ denote by $\Phi_s\in
k[L_0,L_{s+1},L_{s+2},\ldots , L_{n+1}]=B_s$ (here all $L_j$ are considered
as variables)
the nonzero separable polynomial of minimal degree
such that $\Phi_s$ is vanishing on $V_s$.
Denote by $R_s$ the discriminant
of $\Phi_s$  with respect to of $L_{n+1}$
and by $R_{s,0}$ square free part of $R_s$ (i.e., $R_{s,0}$ is a
separable polynomial of the most degree dividing $R_s$). Then
 $\deg_{L_{n+1}}\Phi_s=\deg\Phi_s=\deg V_s$ and the polynomial
$$
R_{s,0}\bmod(L_{s+1},\ldots , L_{n-1})\in B_s/(L_{s+1},\ldots , L_{n-1})
$$
is nonzero separable. We choose and fix $\Phi_s$ such that the
leading coefficient of $\Phi_s$ with respect to $L_{n+1}$
is equal to is equal to $1$.
\item For every $0\le s\le n$ irreducible over
$k$ components
of the algebraic varieties $V_s$ and $V_{s,0}$
are in the one--to--one correspondence by the rule
$W\mapsto W\cap V_{s,0}$.
Besides that, $V_{s,0}\subset A_s$ and $V_{n,0}=A_n=V_n$.
\item For every $0\le s\le n-1$
for every $z\in V_{s,0}$ representation (1) is constructed. Denote
by $W_z$ the uniquely
defined irreducible over $\overline{k}$ component of $V$
such that $z\in W_z$,
denote by $k_z\supset k$ the minimal field of definition of the component
$W_z$.
Then $k(z)\supset k_z$ and the field $k_z$ is constructed by the algorithm.
It is given by its basis and
multiplication table. Two points from $V_{s,0}$ belong to $W_z$ if and only
if they are conjugated to $z$
over $k_z$. Hence one can easily define
a representative system (we do not construct it) $V'_{s,0}\subset V_{s,0}$
of points of irreducible over $\overline{k}$
components of $V_s$. If
$a\subset V_{s,0}$ is an arbitrary
defined and irreducible over $k$ component and $z\in a$
then $V'_{s,0}\cap a$ consists of $[k_z:k]$ points.
\end{enumerate}
The algorithm for constructing
linear forms $L_0,L_1,\ldots , L_{n+1}$, finite sets $A_s$, $V_{s,0}$,
$0\le s\le n$ and all fields $k_z\subset k(z)$, $z\in V_{s,0}$, is
polynomial in $d^n$, $d_1$, $d_2$, $M$, $M_1$ and $m$.
Note that for every
point $z\in A_s$ in representation (1) obtained by the
algorithm the degree $\deg_ZF_z\le\deg V_n\le d^n$ by the
B\'ezout theorem.
}\par\medskip


In the next theorem $u$ points $z_1,\ldots ,z_u\in V$ are given.
Namely, we replace in (1) $z,\eta,F_z$ by $z_i,\eta_i,F_{z_i}$
and get the representations for $z_i$, $1\le i\le u$.
Besides that in the assertion (b) of this theorem, we shall suppose that the
composite of fields
$k[\eta_1,\ldots ,\eta_u]$ is given explicitly (by its basis and
multiplication
table over $k$). One need to fix $u$ points here and hence
it is essential to
give not only the points $z_1,\ldots , z_u$ but also
the $u$--tuple of points $(z_1,\ldots , z_u)$ uniquely up to a conjugation
over $k$.
Note that one can construct any composite of the fields
$k[\eta_i]$, $1\le i\le u$, over $k$
within the time polynomial in $L(z_1),\ldots ,L(z_2)$ and
$\prod_{1\le i\le u}\deg F_{z_i}$, see \cite{3}. In addition
the size of the basis and the multiplication table  of each constructed
composite is
polynomial in the sum of sizes of basises and multiplication tables of all
fields $k[\eta_i]$, $1\le i\le u$, and the degree of this composite over $k$.
In what follows we shall suppose that this
estimate is always holds
for the given representation of the composite $k[\eta_1,\ldots ,\eta_u]$.



\par\medskip\noindent{\bf THEOREM~2}\hspace{0.1em} {\it  Let the field $k$,
the algebraic variety $V$
and polynomials $F$, $f_1,\ldots ,f_m$
be given as above. Let $u\ge 2$ points $z_1,\ldots ,z_u\in V$ be given.
\begin{enumerate} \renewcommand{\labelenumi}{(\alph{enumi})}
\item Then for every $0\le s\le n-1$ one can compute the number of
irreducible over $k$
components $W$ of $V$ with $\dim W=n-s$
such that $z_i\in W$ for all $1\le i\le u$.
The working time of this algorithm is polynomial in $d^n$, $d_1$, $d_2$,
$M$, $M_1$, $m$ and the sum of
the sizes of $\sum_{1\le i\le u}L(z_i)$, of the points $z_i$, $1\le i\le u$.
\item Suppose that the composite of fields $k[\eta_1,\ldots ,\eta_u]$
is given. Then for every $0\le s\le n-1$ one can compute the number of
irreducible over $\overline{k}$ components $W$ of $V$ with $\dim W=n-s$
such that $z_i\in W$ for all $1\le i\le u$.
The working time of this algorithm is polynomial in $d^n$, $d_1$, $d_2$,
$M$, $M_1$, $m$, the sum of
the sizes of $\sum_{1\le i\le u}L(z_i)$, of the points $z_i$, $1\le i\le u$
and the degree $[k[\eta_1,\ldots ,\eta_u]:k]$ of the extension of fields.
\end{enumerate}
}\par\medskip


\par\medskip\noindent{\bf REMARK~1}\hspace{0.1em} {\it  {\em Theorem~1 and
Theorem~2 are trivial for $s=0$. One can take
$L_i=X_i$, $0\le i\le n$ and $L_{n+1}=X_n$ for $s=0$.
So in what follows we shall assume without loss of
generality that the polynomials $f_1,\ldots, f_m$
are linearly independent over $k$. Hence $\dim
V\le n-1$ and we shall suppose without loss of
generality that $1\le s$ in the statements of Theorem~1 and Theorem~2.}
}\par\medskip


\section{Monodromy and irreducibility}\label{s1}

Let $n\ge 1$ and $1\le s\le n$ be integers.
Let $k$ be an arbitrary field of zero--characteristic, $V_s$ be a defined
over $k$
projective algebraic variety in ${\Bbb P}^n(\overline{k})$,
and all the irreducible components of $V_s$ have the same
dimension $n-s$.
Suppose that the degree $\deg V_s=D\ge 1$
(here and below  $D$ depends on $s$).
Let $T_0,
T_{s+1},T_{s+2},\ldots , T_{n+1}$ be linear forms in $X_0,\ldots ,X_n$ with
coefficients from $k$. Let
$$
A^{(s)}=V_s\cap{\cal Z}(T_{s+1},\ldots , T_n),\,
\&\, 0\not\in T_0(A^{(s)}) ,\,\&\, \#(T_{n+1}/T_0)(A^{(s)})=\deg
V_s,
\eqno (2)
$$
where the intersection is considered in
${\Bbb P}^n(\overline{k})$, and $\#$ denotes
the number of the elements of a set.
In particular, $\#A^{(s)}=D$ and the intersection
$V_s\cap{\cal Z}(T_{s+1},\ldots , T_n)$ is transversal at each point
by the B\'ezout theorem.


Consider the morphisms
\setcounter{equation}{2} \begin{eqnarray}
p\, :\, V_s\rightarrow{\Bbb P}^{n-s}(\overline{k}),\quad(X_0:\ldots : X_n)
\mapsto(T_0: T_{s+1},T_{s+2}:\ldots : T_n),\label{3}\\
V_s\rightarrow{\Bbb P}^{n-s+1}(\overline{k}),\quad(X_0:\ldots :
X_n)\mapsto
(T_0: T_{s+1},T_{s+2}:\ldots : T_{n+1}).
\label{4}
\end{eqnarray}
We shall suppose that $T_0, T_{s+1},T_{s+2},\ldots , T_n$, (respectively
$T_0, T_{s+1},T_{s+2},\ldots ,$ $T_{n+1}$ are coordinate functions on
${\Bbb P}^{n-s}(\overline{k})$,
(respectively ${\Bbb P}^{n-s+1}(\overline{k})$).
Then, cf., e.g., \cite{4}, the morphism $p$ is finite dominant, the
image of the morphism (4) is
${\cal Z}(\Phi)$
where $\Phi\in k[T_0, T_{s+1},T_{s+2},\ldots , T_{n+1}]$ is a homogeneous
separable
polynomial, the degree $\deg_{T_{n+1}}\Phi=D$ and leading coefficient
$\mbox{\rm lc}_{T_{n+1}}\Phi$
of $\Phi$ with respect to $T_{n+1}$ is equal to $1$.
Denote by $q\, :\, V_s\rightarrow
{\cal Z}(\Phi)$ the
morphism induced by (4). Hence $q(V_s)={\cal Z}(\Phi)$
and $\deg q(V_s)=D$.
Denote by $k(V_s)$ and $k(q(V_s))$ the algebras of defined over $k$ rational
functions on
$V_s$ and $q(V_s)$ respectively (by definition these
algebras are isomorphic to the direct products
of fields of rational functions of all defined over $k$ components of $V_s$
and $q(V_s)$ respectively).
The morphism $q$ is finite dominant and defined over $k$. Besides that,
$q$ is
birational in the sense that it induces the isomorphism $k(q(V_s))\simeq
k(V_s)$. Let $r\, :\, q(V_s)\rightarrow{\Bbb P}^{n-s}$ be the morphism
of the linear projection to the coordinates $T_0, T_{s+1}$,
$T_{s+2},\ldots , T_n$.
Hence $p=r\circ q$. The morphism $r$ is defined over $k$ and
finite dominant since $\mbox{\rm lc}_{T_{n+1}}\Phi=1$.

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

Let us identify ${\cal Z}(T_{s+1},\ldots , T_n)\subset{\Bbb
P}^n(\overline{k})$ (respectively in ${\Bbb
P}^{n-s+1}(\overline{k})$, ${\Bbb
P}^{n-s}(\overline{k})$) with ${\Bbb P}^{s+1}(\overline{k})$
(respectively ${\Bbb
P}^2(\overline{k})$, ${\Bbb
P}^1(\overline{k})$).
Consider the projective algebraic curve
$V_{s,*}=V_s\cap{\cal Z}(T_{s+1},\ldots , T_{n-1})\subset{\Bbb
P}^{s+1}(\overline{k})$.
Let us repeat the described construction to $V_{s,*}\subset{\Bbb
P}^{s+1}(\overline{k})$
in place of $V_s\subset{\Bbb P}^n(\overline{k})$. Now $n$ and $s$ are
replaced by $s+1$ and $s$ respectively.
The linear forms $T_0,T_{s+1},$ $T_{s+2},\ldots , T_{n+1}$ are
replaced by $T_0,T_n,T_{n+1}$.
The new set $A^{(s)}$ is the same as it was initially.
We get the objects
$p_*$, $q_*$, $r_*$, $\Phi_*$, $R_*$, $R_{0,*}$
which are similar to $p$, $q$, $r$, $\Phi$, $R$, $R_0$ and satisfy the
analogous
properties by the similar reasons.
Note that $\Phi_*=\Phi(T_0,0,\ldots ,0,T_n,T_{n+1})$,
$R_*=R(T_0,0,\ldots ,0,T_n)$.


The following lemma is known.
\par\medskip\noindent{\bf LEMMA~1}\hspace{0.1em} {\it
Let $Z$ be a new variable.
Let $g=Z^\nu+g_{\nu-1}Z^{\nu-1}+\ldots + g_0\in{\Bbb C}[Z]$ be a
polynomial of
degree $\nu\ge 1$ with
complex coefficients $g_i\in{\Bbb C}$, $0\le i\le \nu-1$.
Let $z_1,\ldots , z_\nu\in{\Bbb C}$ be the roots of the polynomial $g$
counting with the multiplicities.
Then for every $0<\varepsilon\in{\Bbb R}$ there is $0<\delta\in{\Bbb
R}$ satisfying the following property.
Let $\widetilde{g}_0,\ldots ,\widetilde{g}_{\nu-1}\in{\Bbb C}$. Suppose
that $|\widetilde{g}_i-g_i|<\delta$
for every $0\le i\le \nu-1$. Let $\widetilde{z}_1,\ldots ,
\widetilde{z}_\nu\in{\Bbb C}$ be the roots
of the polynomial $\widetilde{g}=Z^\nu+\widetilde{g}_{\nu-1}Z^{\nu-1}+\ldots
+\widetilde{g}_0\in{\Bbb C}[Z]$
counting with the multiplicities. Then there is a permutation $\sigma$ of
$1,\ldots , \nu$
such that $|\widetilde{z}_{\sigma(i)}-z_i|<\varepsilon$.
}\par\medskip


In what follows applying this lemma in the present paper
we shall suppose without loss of generality
that $\sigma=1$ is the identity permutation.
Now we can formulate our irreducibility criterion.
\par\medskip\noindent{\bf THEOREM~3}\hspace{0.1em} {\it
Let $n\ge 3$ and $1\le s\le n-2$ be integers.
Let $V_s$ be an equidimensional projective algebraic variety in ${\Bbb
P}^n(\overline{k})$ with $\dim V_s=n-s$.
Let $T_0,T_{s+1},$ $T_{s+2},\ldots , T_{n+1}$ be linear forms from
$k[X_0,\ldots , X_n]$
satisfying (2). Let the polynomial $\Phi$ and its
discriminant $R$ be as above. Suppose that the intersection
${\cal Z}(R)\cap{\cal Z}(T_{s+1},\ldots , T_{n-1})$ is transversal
(it is considered in ${\Bbb P}^{n-s}(\overline{k})$),
, i.e., that the polynomial
$R_0(T_0,0,\ldots , 0,T_n)$ is separable.
Then the irreducible over $\overline{k}$ (respectively $k$) components of
algebraic varieties $V_s$ and $V_{s,*}$ are
in the one--to--one correspondence by the rule $W\mapsto W\cap{\cal
Z}(T_{s+1},\ldots , T_{n-1})$.
}\par\medskip

\noindent {\bf PROOF} \quad
Since $V_s$ and $V_{s,*}$ are defined over $k$
it is sufficient to prove only
the assertion related to the irreducible over $\overline{k}$ components.
Denote by $N$ (respectively $N_*$) the number of all irreducible over
$\overline{k}$ components of $V_s$
(respectively $V_{s,*}$). Condition (2) implies that the intersection
$V_s\cap{\cal Z}(T_{s+1},\ldots , T_{n-1})$
is transversal at each its irreducible component. Hence by the B\'ezout
theorem the
inequality $N_*\ge N$ holds, and if $N_*=N$ then the required one--to--one
correspondence takes place. So it is sufficient to prove that $N=N_*$.
Note that $N$ (respectively $N_*$) is the number of irreducible over
$\overline{k}$ components of ${\cal Z}(\Phi)$ (respectively
${\cal Z}(\Phi_*)$) since $q$ (respectively $q_*$) is a finite birational
isomorphism.

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

Set $E=E_y$. Hence $\#E=D$.
Let $a\in E$. Denote by $\xi_a$ the analitical function in
$Z$ defined in a neighborhood of $y$
such that $\Phi_*(1,Z,\xi_a)=0$ and $\xi_a(y)=a$.
The result of the analitic continuation of $\xi_a$ along $\gamma_i$ is
$\xi_{\sigma_i(a)}$
where $\sigma_i$ is the uniquely defined permutation of $E$. Denote
by $S_D$ the group
of permutations of
$E$ and by $G$ the subgroup of $S_D$ generated by all $\sigma_i$,
$1\le i\le u$. Then it is known, see, e.g., \cite{2}, \cite{10},
that the number $N_*$ of irreducible over
$\overline{k}$ components of
$V_{s,*}$ coincides with the number
of classes of transitivity of $G$ on $E$.


Let $\mu_i,\nu_i\in k$, $s+1\le i\le n-1$ and $0<\delta\in{\Bbb R}$.
Denote by ${\cal A}_\delta$ the condition $\sum_{s+1\le i\le n-1}
(|\mu_i|^2+|\nu_i|^2)<\delta$.
Set
$\widetilde{T}_i=T_i$, $i=0,n,n+1$
and $\widetilde{T}_i=T_i-\mu_i T_0-\nu_iT_n$ for all
$s+1\le i\le n-1$. There is $0<\delta_0\in{\Bbb R}$ such that under
the condition ${\cal A}_{\delta_0}$
the linear forms $\widetilde{T}_i$
satisfy the same properties (considered above) as $T_i$,
$i\in\{0,s+1,s+2,\ldots , n+1\}$ do. In what follows
 we shall suppose that ${\cal A}_{\delta_0}$
holds. Then
for $\widetilde{T}_i$ the objects
$\widetilde{A}^{(s)}$, $\widetilde{\Phi}$, $\widetilde{R}$
and so on are defined.
They correspond to $A^{(s)}$, $\Phi$, $R$ and so on.
Note that $\Phi=\widetilde{\Phi}$, $R=\widetilde{R}$ (as polynomials in
$X_0,\ldots , X_n$), $N=\widetilde{N}$,
$u=\widetilde{u}=\deg R_0$ and
\setcounter{equation}{4} \begin{eqnarray}
&&\widetilde{\Phi}_*=\Phi(T_0,\mu_{s+1}T_0+\nu_{s+1}T_n,\ldots ,
\mu_{n-1}T_0+\nu_{n-1}T_n,T_n, T_{n+1}),\label{5}\\
&&\widetilde{R}_*=R(T_0,\mu_{s+1}T_0+\nu_{s+1}T_n,\ldots ,
\mu_{n-1}T_0+\nu_{n-1}T_n,T_n).\label{6}
\end{eqnarray}
Hence by Lemma~1 applied to the polynomial $R_*(1,Z)/\mbox{\rm
lc}_Z(R_*(1,Z))$
there is $0<\delta_1<\delta_0$ such that ${\cal A}_{\delta_1}$
entails $|\widetilde{y}_i-y_i|<m_1/2$ for every $1\le i\le u$.
We shall suppose in what follows that
${\cal A}_{\delta_1}$  holds.
Now we choose $\widetilde{y}=y$ and $\widetilde{m}_1=m_1/2$.

Let $m_2=\inf\{|a_1-a_2|\, :\, a_1,a_2\in E_z\,\&\,a_1\ne a_2\,\&\,z\in e\}$.
Then $m_2>0$ due to the compactness of $e$.
By
Lemma~1 applied to the polynomials $\Phi_*(1,z,Z)$,$z\in e$, by (5) and
due to the compactness of $e$ there is $0<\delta_2\le \delta_1$
such that ${\cal A}_{\delta_2}$
entails that for every $z\in e$ for every
$a\in E_z$ there is the unique $\widetilde{a}\in\widetilde{E}_z$ such that
$|a-\widetilde{a}|<m_2/3$.  We shall suppose
in what follows that ${\cal A}_{\delta_2}$
holds.

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

Denote by ${\cal L}_{n-s+1}$,
(respectively ${\cal L}_{n-s-1}$) the vector
space of linear forms with coefficients from
$\overline{k}$ with the basis $T_0,T_{s+1}
,\ldots , T_n$ (respectively $T_{s+1},
,\ldots ,$ $T_{n-1}$).
According to the first Bertini, see \cite{1}, theorem applied subsequently
$n-s-1$ times
in any neighborhood in the classic topology
of the vector subspace ${\cal L}_{n-s-1}\subset{\cal L}_{n-s+1}$
(the neighborhood of ${\cal L}_{n-s-1}$
is considered here in the corresponding
Grassmann variety)
there is an $(n-s-1)$--dimensional vector subspace
${\cal L}\subset{\cal L}_{n-s+1}$
such that $\dim q(V_s)\cap{\cal Z}({\cal L})=1$,
$\deg q(V_s)\cap{\cal Z}({\cal L})=D$, and
for every irreducible over $\overline{k}$ component $W$ of $q(V_s)$ the
algebraic variety $W\cap{\cal Z}({\cal L})$ is irreducible over
$\overline{k}$.
Hence one can choose ${\cal L}$ to be
a linear subspace generated by $\widetilde{T}_{s+1},\ldots ,
\widetilde{T}_{n-1}$
for some $\mu_i$, $\nu_i$, $s+1\le i\le n-1$,
satisfying ${\cal A}_{\delta_2}$. In this case $\widetilde{N}_*=
\widetilde{N}$. Thus,
$N=\widetilde{N}=\widetilde{N}_*=N_*$. The theorem is
proved.

\par\medskip\noindent{\bf COROLLARY~1}\hspace{0.1em} {\it  Let $1\le s\le
n-1$.
Let us identify the set of all $(n-s+1)$--tuples
$(L_0,L_{s+1},\ldots , L_n)$
of linear forms $L_j\in\overline{k}[X_0,\ldots , X_n]$ with the affine
space ${\Bbb A}^{(n-s)(n+1)}(\overline{k})$.
Then the set of all $(L_0,L_{s+1},\ldots , L_n)$ such that assertions (a)
and (b) of
Theorem~1 related to the ground field $\overline{k}$ hold for the considered
fixed $s$
is a nonempty defined over $k$ open in the Zariski topology
subset of ${\Bbb A}^{(n-s)(n+1)}(\overline{k})$.
}\par\medskip

\noindent {\bf PROOF} \quad The case $s=n-1$ is easy.
So we shall suppose that $s\le n-2$.
Consider the affine space of the
homogeneous polynomials in $n-s+2\ge 3$ variables with
coefficients from $\overline{k}$ of the given degree. It is known,
see, e.g., \cite{1}, that the set of all irreducible over
$\overline{k}$ homogeneous polynomials from this affine space is a
nonempty open in the Zariski topology subset of the considered
affine space. Now the corollary follows from the first paragraph
of the proof (we leave the details to the reader). The corollary
is proved.




\par\medskip\noindent{\bf COROLLARY~2}\hspace{0.1em} {\it  Let $n\ge 2$ and
$1\le s\le n-1$ be integers.
Let $V_s$ be an equidimensional projective algebraic variety in ${\Bbb
P}^n(\overline{k})$ with $\dim V_s=n-s$.
Let $T_0,T_{s+1},$ $T_{s+2},\ldots , T_{n+1}$ be linear forms from
$k[X_0,\ldots , X_n]$
satisfying (2). Let the polynomial $\Phi$ and its
discriminant $R$ be as above.
Let $L\in k[T_0,T_{s+1},$ $T_{s+2},\ldots , T_n]$ be a nonzero linear form.
Suppose that the intersection ${\cal Z}(R)\cap{\cal Z}(L)$
is transversal in ${\Bbb P}^{n-s}(\overline{k})$
(or which is equivalent $\deg{\cal Z}(R)=
\deg({\cal Z}(R)\cap{\cal Z}(L))$ in ${\Bbb
P}^{n-s}(\overline{k})$). Then
the intersection $V_s\cap{\cal Z}(L)$ is transversal, $\deg V_s=\deg
V_s\cap{\cal Z}(L)$ in ${\Bbb P}^n(\overline{k})$ and the
irreducible over $\overline{k}$ (respectively irreducible over $k$)
components of $V_s$ and $V_s\cap{\cal Z}(L)$ are in
the one--to--one correspondence by the rule $W\mapsto W\cap{\cal Z}(L)$
}\par\medskip

\noindent {\bf PROOF} \quad Let ${\Bbb P}^{n+1}(\overline{k})$ has
homogeneous coordinates $X_0,\ldots ,X_{n+1}$.
Let us identify ${\Bbb P}^n(\overline{k})={\cal
Z}(X_{n+1})\subset{\Bbb P}^{n+1}(\overline{k})$.
Hence $V_s\subset{\Bbb P}^{n+1}(\overline{k})$. Replacing if necessary
$n$ by $n+1$ and
$T_0,T_{s+1},\ldots , T_n$ by
$T_0,T_{s+1},\ldots , T_n$, $X_{n+1}$ we shall suppose without loss of
generality that $n\ge 3$ and
$1\le s\le n-2$. Set $\widetilde{T}_{s+1}=L$ (one should
not confuse the tilde here and below
with the one from the proof of the theorem).
We have $R\bmod L\ne 0$. Therefore, there are
linear forms
$\widetilde{T}_0,\widetilde{T}_{s+2},\ldots , \widetilde{T}_n$ in
$T_0,T_{s+1},$ $T_{s+2},\ldots , T_n$
with coefficients from $k$ such that
$\widetilde{T}_0,\widetilde{T}_{s+1},\ldots , \widetilde{T}_n$
are linearly independent, the number of elements $\#r^{-1}({\cal
Z}(\widetilde{T}_{s+1},$ $\ldots , \widetilde{T}_n))=\deg\Phi$,
$0\not\in\widetilde{T}_0(\#r^{-1}({\cal Z}(\widetilde{T}_{s+1},\ldots ,
\widetilde{T}_n)))$ and the intersection ${\cal Z}(R)\cap
{\cal Z}(\widetilde{T}_{s+1},\ldots ,
\widetilde{T}_{n-1})$ is transversal. Let us replace in (2)
$T_0,$ $T_{s+1},\ldots , T_n$ and $T_{n+1}$ by
$\widetilde{T}_0,$ $\widetilde{T}_{s+1},\ldots ,
\widetilde{T}_n$, and
$T_{n+1}$ respectively.
Then the obtained condition holds. We get the polynomials
$\widetilde{\Phi}$, $\widetilde{R}$ corresponding to
the new linear forms
according to the construction described before the Theorem~3.
Now applying
Theorem~3 in the considered situation
we get all the assertions of the corollary. The corollary is proved.



\section{Proof of Theorem~1}\label{s2}

At first our aim is to construct for every $1\le s\le n$ some linear forms
$Y_{s,0},\ldots ,$ $Y_{s,n+1}$ in $X_0,\ldots , X_n$. Let
$1\le s\le n$ be arbitrary but fixed for some time.
According to \cite{4}, Section 2 one can
construct
linear forms $L_{s,0},L_{s,s+1},L_{s,s+2},\ldots ,$ $L_{s,n}$ with integer
coefficients of length
$O(n\log d)$ and a finite set
$A'''_s$ such that $A'''_s=V_s\cap{\cal
Z}(L_{s,s+1},\ldots , L_{s,n})$, the last intersection is
transversal, consists of $D=\deg V_s$ smooth points of the algebraic
variety $V$ and $0\not\in L_{s,0}(A'''_s)$.
Let us construct a linear form $L_{s,n+1}$ with integer
coefficients of length
$O(n\log d)$ such that $\#(L_{s,n+1}/L_{s,0})(A'''_s)=D$. Let us
construct linear forms
$L_{s,i}$, $1\le i\le s$ with integer coefficients of length $O(n\,\log d)$
such that $L_{s,0},\ldots , L_{s,n}$ are linearly independent.
Denote for brevity $Z_j=L_{s,j}$ for all $j$.
Set $T_j=Z_j$ for all $0\le j\le n+1$.

Denote for brevity $T=(T_0,\ldots , T_{n+1})$.
Let ${\cal B}'_s(T)$ means that (2) holds for
$T$ and all the points of $A^{(s)}$ are smooth points of $V$, herewith
${\cal B}'_s$ is considered as a function of $T$ and all
$T_j$ may have coefficients from an extension of $\overline{k}$.

Suppose that $s\le n-2$ and $V_s\ne\emptyset$
for some time. In this case the linear forms
$T_j$ will be changed recursively in the algorithm.
Our aim is to construct finally linear forms $T_j$
with integer coefficients of length
$O(n\log d)$ such that
$T_0,T_{s+1},\ldots , T_{n+1}$ satisfy conditions of Theorem~3.
At all steps of the described recursion $T_0,T_{s+1},\ldots , T_n$ are
linear forms with integer coefficients in
$Z_0,Z_{s+1},\ldots , Z_n$, and $T_j=Z_j$ for $1\le j\le s$, $j=n,n+1$.

Let $\Phi$, $R$, $q$, $r$ be as in Section~1.
Using the algorithm from \cite{3} let us construct the curve $V_{s,*}$ and
the
polynomials
$\Phi_*$, $R_*$, see Section~1. Denote by $R_{*,0}$ the square free part of
 $R_*$.  Set
$$
v_s(T)=\deg R_0,\quad u_s(T)=\deg R_{*,0}.
$$
Hence $v_s(T)\ge u_s(T)$. Similarly to the proof of Theorem~3 we can
suppose without loss of generality that
$T_0$ does not divide $R_*$ replacing
if necessary $T_0$ by $T_0+\lambda T_n$ where
$\lambda$ is an integer of length
$O(n\,\log d)$. Now leading coefficient $\mbox{\rm
lc}_{T_n}R_*\in k$,
$\deg R_*=\deg R$ and $R$ is homogeneous.
Hence $\mbox{\rm lc}_{T_n}R\in k$.
Let us compute all the pairwise distinct roots
$y_1,\ldots , y_u$
of the polynomial $R_*(1,Z)$ (i.e., all the fields $k[y_i]$;
here one need to
factor $R_*(1,Z)$ over $k$ using \cite{3}). Hence $u_s(T)=u$.
Now ${\cal Z}(R_*,T_0)=\emptyset$
in ${\Bbb P}^1(\overline{k})$. Further, the leading coefficient
$\mbox{\rm lc}_{T_n}R_*\in k$,
$\deg R_*=\deg R$ and $R$ is homogeneous.
Hence $\mbox{\rm lc}_{T_n}R\in k$.
Therefore, the morphism $r_0\, :\,
{\cal Z}(R)\rightarrow{\Bbb P}^{n-s-1}(\overline{k})$ of the linear
projection to the coordinates
$T_0,T_{s+1},\ldots , T_{n-1}$ is finite dominant. Hence
$\deg r_0=\deg R_0$.

Let $\varepsilon_i>0$ be an infinitesimal with respect to the field
$k_{i-1}=k(\varepsilon_1,\ldots ,\varepsilon_{i-1})$,
$1\le i\le 5$. We shall suppose that the algebraic closure of the field
$k_4=k(\varepsilon_1,\ldots ,\varepsilon_4)$
is supplied with a real structure, see \cite{9}. Hence one can consider
systems of polynomial equations
and inequalities with squares of absolute values over $k_4$.
Let $X_{i,0},\ldots , X_{i,n}$, $1\le i\le 4$, be new variables.
Denote for brevity $X^{(i)}=(X_{i,0},\ldots , X_{i,n})$ for $1\le i\le 4$ and
$b(X^{(i)})=b(X_{i,0},\ldots , X_{i,n})$
for an arbitrary polynomial $b$ in $n+1$ variables.
Consider the system of equations and inequalities
$$
\left\{\begin{array}{ll}
\sum_{s+1\le i\le n-1}|T_i(X^{(1)})|^2
<\varepsilon_1, &\\
\varepsilon_3<|T_n(X^{(1)})-T_n(X^{(3)})|^2
<\varepsilon_2,&\\
|T_{n+1}(X^{(i)})-T_{n+1}(X^{(i+1)})|^2=\varepsilon_4,& i=1,3, \\
T_j(X^{(i_1)})-T_j(X^{(i_2)})=0,& 1\le i_1<i_2\le 4,\quad s+1\le j\le n-1,\\
T_n(X^{(i)})-T_n(X^{(i+1)})=0,& i=1,3,\\
T_0(X^{(i)})-1=0,& 1\le i\le 4, \\
f_j(X^{(i)})=0,& 1\le j\le m,\quad 1\le i\le 4.
\end{array}\right.
\eqno (7)
$$
For an arbitrary projective algebraic variety $E\subset{\Bbb
P}^n(\overline{k})$ we shall denote by $\mbox{\rm con}(E)$
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 $E$.

\par\medskip\noindent{\bf LEMMA~2}\hspace{0.1em} {\it \footnote{The
corresponding lemma from
the Russian journal version is slightly corrected here.
This required additional reference to \cite{7} in the algorithm.}
System (7) has a solution in $\mbox{\rm
con}(V_s(\overline{k}_4))^4\subset{\Bbb
A}^{4n+4}(\overline{k}_4)$ if and only if $u_s(T)<\deg r_0$.
}\par\medskip

\noindent {\bf PROOF} \quad
Let $u_s(T)<\deg r_0$. Then by Lemma~1 and
transfer principle, see \cite{8}, applied to the polynomial
$R_*(1,Z)/\mbox{\rm lc}_ZR_*(1,Z)$ there are
$z_{s+1},\ldots , $ $z_{n-1},z_{1,n},z_{3,n}\in \overline{k_3}$ such that
$\sum_{s+1\le i\le n-1}|z_i|^2
<\varepsilon_1$, $\varepsilon_3<|z_{1,n}-z_{3,n}|^2
<\varepsilon_2$ and $R(1,z_{s+1},\ldots ,z_{n-1},z_{n,\alpha})=0$,
$\alpha=1,3$. We have
$R(1,z_{s+1},\ldots ,z_{n-1},T_n)\ne 0$ since
leading coefficient $\mbox{\rm lc}_{T_n}R\in k$. Recall that
leading coefficient $\mbox{\rm lc}_{T_{n+1}}\Phi=1$.
Hence by Lemma~1 and transfer principle, see \cite{8}, applied to
two polynomials $\Phi(1,z_{s+1},\ldots , z_{n-1},z_{n,\alpha},Z)$,
$\alpha=1,3$, there are
$\widetilde{z}_{\alpha,n},\widetilde{z}_{\beta,n+1}
\in\overline{k_4}$, $\alpha=1,3$, $\beta=1,2,3,4$ such that
$\widetilde{z}_{\alpha,n}-z_{\alpha,n}$, $\alpha=1,3$,
are infinitesimal with respect to $k_3$,
$|\widetilde{z}_{\beta,n+1}-\widetilde{z}_{\beta+1,n+1}|^2=\varepsilon_4$,
$\beta=1,3$ and $\Phi(1,z_{s+1},\ldots ,
z_{n-1},\widetilde{z}_{\alpha,n},\widetilde{z}_{\beta,n+1})=0$
for $\alpha=1,3$, $\beta=\alpha,\alpha+1$. Set
$\widetilde{z}_{\beta,n}=\widetilde{z}_{\alpha,n}$ for $\alpha=1,3$,
$\beta=\alpha,\alpha+1$.
The morphisms $q$ is
finite dominant. Therefore, there is a solution of system (7) with
additional equations $T_j(X^{(\beta)})=z_j$, $s+1\le j\le n-1$,
$T_j(X^{(\beta)})=\widetilde{z}_{\beta,j}$, $j=n,n+1$, $\beta=1,2,3,4$.

Conversely, if system (7) has a solution then as it is seen from the
description of the algorithm, see below, $u_s(T)<\deg r_0$.
The lemma is proved.


\medskip  Let $\varphi_j=0$, $j\in J$ be the family of
all the equalities (not inequalities)
of system (7) apart of the  equations $f_j(X^{(i)})=0$.
Let us replace all the considered equations $\varphi_j=0$, $j\in J$ in
(7)
by one inequality $\sum_{j\in J}|\varphi_j|^2<\varepsilon_5$.
This new system has a solution from $\mbox{\rm
con}(V_s(\overline{k}_5))^4$ if and only if system (7) has a
solution from $\mbox{\rm con}(V_s(\overline{k}_4))^4$.
Now
using the algorithms from \cite{7}, \cite{4}
applied to $\mbox{\rm con}(V(\overline{k}_5))^4
\subset{\Bbb A}^{4n+4}(\overline{k}_5)$
we can construct a solution from
$\mbox{\rm con}(V_s(\overline{k}_5))^4$ of this new system
if it exists. For every $a\in\overline{k}_j$ that is not infinitely large
with respect to
$k_{j-1}$ denote, see \cite{9} Section~1, by $\mbox{\rm
st}_{\varepsilon_j}a\in\overline{k}_{j-1}$ the element
such that
$a-\mbox{\rm st}_{\varepsilon_j}a$ is an
infinitesimal with respect to $k_{j-1}$.
Computing the standard part $\mbox{\rm st}_{\varepsilon_5}$, see
\cite{9} Section~1, of this solution we construct a solution
$X_{i,j}=x_{i,j}\in\overline{k}_4$, $1\le i\le 4$, $0\le j\le n$ of system
(7) from $\mbox{\rm con}(V_s(\overline{k}_4))^4$
if it exists.

Suppose that (7) does not have a solution. Then $u_s(T)=\deg r_0=
\deg R_0$ by Lemma~2 and the linear forms
$T_i$ satisfying the conditions of Theorem~3 are constructed.
Suppose that (7) has a solution.
Let us compute
$x_j=\mbox{\rm st}_{\varepsilon_4}x_{1,j}$, $0\le
j\le n$.
Set $T'_j=T_j-T_j(x_0,\ldots , x_n)T_0$, $s+1\le j\le n-1$, and $T'_j=T_j$,
$j=0,\ldots ,s,\,n,n+1$.
Let the notation $T'$  for $T'_j$ be similar to $T$
defined for $T_j$. Then the property ${\cal B}'_s(T')$ is true.
Let the number $u(T')$ for $T'$ be similar to $u(T)$ defined for $T$.
From (7) we get $u'=u_s(T')>u_s(T)=u$.

Let $T''_0,T''_{s+1},\ldots , T''_{n+1}$
be linear forms in $Z_0,Z_{s+1},\ldots , Z_{n+1}$. Denote
$v_s=v_s(T)$.
Let $0\le a$ be
an integer.
The property ${\cal B}'_s(T'')\&(u_s(T'')\ge a)$
is a good property of the argument
$T''$ in a sense similar to \cite{9}, see also, \cite{4}.
Denote for brevity the considered
property by ${\cal B}_s(T'',a)$ and set ${\cal B}_s(T'')=
{\cal B}_s(T'',v_s)$.
Hence (the reader can see it also directly) one can replace subsequently the
nonzero coefficients in $Z_j$
of linear forms $T'_{s+1},\ldots ,
T'_{n-1}$ by integers of length
$O(n\log d)$ and obtain new linear forms $T''$
such that  ${\cal B}_s(T'',u')$ holds and
$T''_j=T_j$ for $j=0,\ldots ,s,\,n,n+1$.
Now we replace $T$ by $T''$ and
proceed to the next step of the recursion.
The number of such steps with the replacement $T$ by $T''$ at the end
is bounded from above by $d^{n-s}$ by the
B\'ezout theorem.

The new morphism $r_0$ is distinct from
the old one only by the automorphism of
${\Bbb P}^{n-s-1}(\overline{k})$. The morphism
$r_0$ is finite
dominant.
Hence $T_0$ may divide $R_*$ only in
the first step of the recursion.  So after the first step $T_j$,
$j=0,\ldots ,$ $s,$ $n,n+1$
are not replaced.

Thus, finally we obtain the linear forms
$T$ such that $T_0,T_{s+1},\ldots , T_{n+1}$ satisfy
conditions of Theorem~3. In what follows we shall not
suppose that $s\le n-2$ and $V_s\ne\emptyset$.
Define the property ${\cal B}_{n-1}={\cal B}'_{n-1}$.
Set $Y_{s,j}=T_j$
for all $0\le j\le n+1$.
In what follows we shall not suppose that $s$ is fixed.
 Denote
$Y_s=(Y_{s,0},\ldots ,Y_{s,n+1})$, $1\le s\le n$.
Then ${\cal B}_s(Y_s)$ is true for every $1\le s\le n-1$.
Let $0\le j\le n-1$.
Consider the property ${\cal C}_j={\cal
B}_1\&\ldots\&{\cal B}_j$ of linear forms in $X_0,\ldots , X_n$
(hence ${\cal C}_0$ is identically true). Let
$L'=(L'_0,\ldots , L'_{n+1})$
where all $L'_i$ are
linear forms in $X_0,\ldots , X_n$.
By definition ${\cal C}_n(L')$ is true if and only if ${\cal
C}_{n-1}(L')$ is true and the assertion (a)
of Theorem~1 holds for $L'_0,\ldots , L'_{n+1}$
in place of $L_0,\ldots , L_{n+1}$.
Note that ${\cal C}_j$ is a good property, see \cite{9}, \cite{4}, for
every $0\le j\le n$.


Set $Y^{(0)}_i=X_i$, $0\le i\le n$, and $Y^{(0)}_{n+1}=X_n$.
Let $0\le j\le n-1$ be an integer.
Suppose that we have constructed recursively
$Y^{(j)}=(Y^{(j)}_0,\ldots , Y^{(j)}_{n+1})$ where all $Y^{(j)}_i$ are
linear forms
in $X_0,\ldots , X_n$ with integer coefficients of length
$O(n\,\log d)$ such that
${\cal C}_j(Y^{(j)})$ is true. Let $\varepsilon$ be a transcendental
element over the field $k$.
Set $Y^{(j)}+\varepsilon Y_{j+1}=(Y^{(j)}_0+\varepsilon Y_{j+1,0},\ldots ,
Y^{(j)}_{n+1}+\varepsilon Y_{j+1,n+1})$
(we extend here $k$ till $k(\varepsilon)$).
Then ${\cal C}_{j+1}(Y^{(j)}+\varepsilon Y_{j+1})$ is true.
Since  ${\cal C}_{j+1}$ is a good property
one can replace subsequently
the nonzero coefficients in $X_i$, $0\le i\le n$,
of linear forms from $Y^{(j)}+\varepsilon Y_{j+1}$ by integers of length
$O(n\log d)$ and obtain new linear forms $Y^{(j+1)}$
such that ${\cal C}_{j+1}(Y^{(j+1)})$ is true.

At the end we check whether $v_s(Y^{(n)})=v_s$
for every $1\le s\le n-2$ such
that $V_s\ne\emptyset$. If it is not true then $v_s(Y^{(n)})>v_s$ for some
$s$.  In this case we replace $L_{s,i}=Y^{(n)}_i$ for all $1\le s\le n$,
$0\le i\le n+1$ and return to the beginning of the described construction.

Since $v_s\le d^s(d^s-1)$ for all $s$ finally we shall come to the case
when $v_s(Y^{(n)})=v_s$ for all $1\le s\le n-2$ such
that $V_s\ne\emptyset$. Since
$u_s(Y^{(n)})\ge v_s$ by the described construction we have
$u_s(Y^{(n)})=v_s(Y^{(n)})$
for all considered $s$ and one can apply Theorem~3 with $Y^{(n)}$
in place of $T$.
Finally we set $L=Y^{(n)}$. Now assertions (a), (b) and (c)
of Theorem~1 hold.

Let us prove (d) and (e).  Let $s=n$. Then we define $V_{n,0}=A_n$.
For every $1\le s<n-1$
using the algorithm of \cite{3} we construct all the
defined over $k$ and irreducible
over $k$ components of dimension $1$ of the
intersection $V\cap{\cal Z}(L_{s+1},\ldots , L_{n-1})$. Denote by
$W_j$, $j\in J_s$,
the obtained curves. Again using the algorithm from \cite{3} we choose
for every $j\in J_s$ a defined over $k$ and
irreducible over $k$ component $a_j\subset
A_s\cap W_j$.  Set $V_{s,0}=\cup_{j\in J_s}a_j$. Now the assertion (d) hold.

Let $z\in A_s$ and $\sigma$ be a $k$--automorphism of $\overline{k}$.
Then $\sigma(z)=z$
implies $\sigma(W_z)=W_z$ by our construction. Hence $k_z\subset k(z)$.
Besides that, $\sigma(W_z)=W_z$ if and only if
$\sigma(W_z\cap{\cal Z}(L_1,\ldots ,L_{n-1}))=W_z\cap{\cal
Z}(L_1,\ldots ,L_{n-1})$ due to (b). Hence the minimal
fields of definitions containing $k$ of $W_z$ and $W_z\cap{\cal
Z}(L_1,\ldots ,L_{n-1})$
coincide.

Using the algorithm from \cite{3}
we construct all the defined over $\overline{k}$ and irreducible over
$\overline{k}$ components of the algebraic varieties $W_j$, $j\in J_s$ and
their minimal fields of definition. More precisely, these minimal fields of
definitions are conjugated over
$k$ to the field $k_j$ which is constructed by the algorithm. The algorithm
also
constructs a finite family of polynomials $\psi_{j,i}\in k_j[X_0,\ldots ,
X_n]$, $i\in I_j$, such that
${\cal Z}(\psi_{j,i},\,i\in I_j)$ is conjugated over $k$ to each
irreducible over $\overline{k}$ component of $W_j$.
Let $z\in a_j\subset W_j$. For every $j\in J_s$
again using the algorithm from \cite{3} we construct all the embeddings
$\tau\, :\, k_j\rightarrow k(z)$ over $k$ (there are at most $[k_j:k]$ such
embeddings).
Let us extend $\tau$ to the ring of polynomials in $X_0,\ldots , X_n$
trivially, i.e., by the rule
$\tau(X_i)=X_i$ for every $i$. Then there is only one $\tau$ such that
$\tau(\psi_{j,i})(z)=0$ for every $i\in I_j$. We compute this $\tau$ and set
$k_z=\tau(k_j)$. Now the assertion of (e) hold.
The required estimation for the
working time of the described algorithm follows immediately from
ones of the applied algorithms.
Theorem~1 is proved.

\par\medskip\noindent{\bf REMARK~2}\hspace{0.1em} {\it  {\em Let $L$ be as
in the proof of Theorem~1 and $\varepsilon$ be a
transcendental element over $k$.
Let $L'_0,\ldots , L'_n\in k[X_0,\ldots , X_n]$ be arbitrary
linear forms.
Set $\widetilde{L}_j=L_j+\varepsilon L'_j$ for $0\le j\le n$ and
$\widetilde{L}_{n+1}=L_{n+1}$,
$\widetilde{L}=(\widetilde{L}_0,\ldots ,\widetilde{L}_{n+1})$. Let us replace
the field $k$ by $k(\varepsilon)$ and
$L$ by $\widetilde{L}$.
Then, see the statement of Theorem~1, the polynomials $\widetilde{\Phi}_s$,
$\widetilde{R}_s$,
$\widetilde{R}_{s,0}$ corresponding to $\widetilde{L}$ are defined.
They are similar to $\Phi_s$, $R_s$, $R_{s,0}$.
Let us consider $\widetilde{\Phi}_s$, $\widetilde{R}_s$,
$\widetilde{R}_{s,0}$ as
polynomials in $X_0,\ldots , X_n$ with coefficients from $k(\varepsilon)$.
Then the least common denominator from $k[\varepsilon]$ of the coefficients
of $\widetilde{\Phi}_s$ is not zero at $\varepsilon=0$.
Substituting $\varepsilon=0$ we get
$\widetilde{\Phi}_s|_{\varepsilon=0}=\Phi_s$ and
$\widetilde{R}_s|_{\varepsilon=0}=R_s$. Further,
$$
v_s(\widetilde{L})\ge v_s(L)
\eqno (8)
$$
for every $1\le s\le n-2$ such that $V_s\ne\emptyset$. If at least one of
inequalities (8)
is strict then extending the field $k$ till $k(\varepsilon)$ and
applying the construction from the proof of the theorem to
$L_{s,i}=\widetilde{L}_i$, $1\le s\le n$,
$0\le i\le n+1$, we get new linear forms $\overline{L}$ similar to $L$
such that $v_s(\overline{L})\ge v_s(L)$ for all
considered $s$ and at least one
of the last inequalities is strict.
Hence replacing $L$ by $\overline{L}$ one can iterate in the same
way until all the inequalities
from (8) become the equalities.

Suppose that (8) are equalities.
Then by our definitions $\widetilde{R}_{s,0}|_{\varepsilon=0}=R_{s,0}$.
Hence the linear forms $\widetilde{L}_j$
in place of $T_j$ (respectively $L_j$), $0\le j\le n+1$,
satisfy conditions of Theorem~3 (respectively Theorem~1) with
$k(\varepsilon)$
in place of $k$.}
}\par\medskip

\par\medskip\noindent{\bf REMARK~3}\hspace{0.1em} {\it  We do not construct
linear forms $L$ in general
position with maximal
possible $v_s(L)$. So we need to
formulate and use the previous remark as some
weak reduction to general position. Actually one can use also below
Corollary~1 in place of Remark~2.
}\par\medskip







\section{Proof of Theorem~2}\label{s3}

Let us prove assertion (b) of Theorem~2. Extending the field $k$ till
$k[\eta_1,\ldots ,\eta_u]$ we shall suppose in what follows that
$k=k[\eta_1,\ldots ,\eta_u]$.
Note that if $z_1,\ldots , z_u$
belong to an irreducible
over $\overline{k}$ component $W$ of $V$
then they belong to each conjugated over $k$ to $W$ component.
Let $\varepsilon$ be an infinitesimal with respect to the field $k$.
Hence, see \cite{9}, Section~1,
the mapping $\mbox{\rm st}_\varepsilon\, :\,{\Bbb
P}^n(\overline{k(\varepsilon)})\rightarrow{\Bbb P}^n(\overline{k})$
is defined such
that $\mbox{\rm st}_\varepsilon(z)$ is an
infinitely close to $z$ element.

Let us construct linear forms $L_0,\ldots ,L_{n+1}$ from Theorem~1.
Let $1\le s\le n-1$.
Let us construct for every $2\le i\le u$
linearly independent linear forms
$L_{i,0},\ldots , L_{i,n}$ in
$L_0,\ldots ,L_n$ with coefficients from $k$ of length
polynomial in $n$, $L(z_1)$, $L(z_i)$
such that $L_{i,j}(z_\alpha)=0$ for $1\le j\le n-1$,
$\alpha=1,i$ and $L_{i,0}(z_\alpha)\ne 0$ for $\alpha=1,i$.
For $\alpha=1,i$ put $\lambda_{i,\alpha}=
(L_{i,n}/L_{i,0})(z_\alpha)\in k$.
Set $L'_{i,j}=L_{i,j}+\varepsilon L_j$,
$0\le j\le n$. By Remark~2 applied to $\varepsilon^{-1}$ in place of
$\varepsilon$ we shall suppose without loss of generality that
$v_s(L)=
v_s(L'_{i,0},\ldots , L'_{i,n},L_{n+1})$
for all $2\le i\le u$ and $1\le s\le n-2$ such that $V_s\ne\emptyset$.
Then by Theorem~3 and Theorem~1
for every $1\le s\le n-1$
the irreducible over $k$ (respectively $\overline{k}$) components of $V_s$
and irreducible over $k(\varepsilon)$
(respectively $\overline{k}(\varepsilon)$) components of
dimension $1$ of
$V\cap{\cal
Z}(L'_{i,s+1},\ldots , L'_{i,n-1})$
are in the one--to--one correspondence by the rule
$$
W\mapsto W\cap{\cal
Z}(L'_{i,s+1},\ldots ,
L'_{i,n-1})=W_{s,i},
\eqno (9)
$$
and besides that, $\deg W=\deg W_{s,i}$ in (9)
for every irreducible over $\overline{k}$ component $W$ of $V_s$.
Besides that, by Theorem~1
for every $1\le s_1\le n$ the intersection $V_{s_1}\cap{\cal
Z}(L_{s+1},\ldots ,L_{n-1})$ is
nonempty if and only if $s_1\le s+1$ and $V_{s_1}\ne\emptyset$. In the last
case
$\dim  (V_{s_1}\cap{\cal Z}(L_{s+1},\ldots ,L_{n-1}))=s-s_1+1$.
Therefore, cf., e.g., \cite{3},
\begin{itemize}
\item[(*)] for every $1\le s_1\le n$ the intersection
$V_{s_1}\cap{\cal Z}(L'_{i,s+1},\ldots ,L'_{i,n-1})\ne\emptyset$
if and only if $s_1\le s+1$ and $V_{s_1}\ne\emptyset$. In the last case
$\dim(V_{s_1}\cap{\cal Z}(L'_{i,s+1},\ldots ,L'_{i,n-1}))=s-s_1+1$.
\end{itemize}


Suppose that $z_\alpha\in W$.
Then, see, e.g., \cite{9}, Proposition~6, the set $W_{s,i}\cap{\cal
Z}(L_{i,n}-\lambda_{i,\alpha} L_{i,0})$
is finite and $z_\alpha\in
\mbox{\rm st}_\varepsilon(W_{s,i}\cap{\cal
Z}(L_{i,n}-\lambda_{i,\alpha} L_{i,0}))$.
Conversely, if $z_\alpha\in\mbox{\rm st}_\varepsilon(W_{s,i})$
then $z_\alpha\in W$.

Now we are going to compute all $W_{s,i}$ and choose among them
such that $z_\alpha\in\mbox{\rm st}_\varepsilon(W_{s,i})$
for $\alpha=1,i$.
Consider $\varepsilon$ as an variable. Let ${\Bbb A}^{n+2}(\overline{k})$
be defined over $k$
and have the coordinates $X_0,\ldots ,X_{n+1}$, $\varepsilon$.
Using the algorithm of \cite{3} we construct all the
defined over $k$ and irreducible
over $k$ components $W''$ of dimension $3$ of the
intersection
${\cal Z}(f_1,\ldots ,f_m)\cap{\cal Z}(L'_{i,s+1},\ldots ,
L'_{i,n-1})$
in ${\Bbb A}^{n+2}(\overline{k})$.
For every $W''$ we construct a point from it and then
$\lambda_0\in\overline{k}$ (it depends on $W''$) such that
$W''\cap{\cal Z}(\varepsilon-\lambda_0)\ne\emptyset$. Further, using
\cite{3} we construct
$W''\cap{\cal Z}(\varepsilon-\lambda_0)$
and decide whether $\dim W''\cap{\cal Z}(\varepsilon-\lambda_0)=\dim
W''$.
Thus, we (can) choose among components $W''$ all that are
not contained in any finite
union of hyperplanes ${\cal Z}(\varepsilon-\lambda)$, $\lambda\in
\overline{k}$.
Denote by $\widetilde{W}_j$, $j\in J_{s,i}$, the family of obtained
algebraic varieties.
More precisely, for every $j\in J_{s,i}$ the algorithm construct
a finite family of polynomials
$\varphi_{j,v}\in k[X_0,\ldots ,X_n,\varepsilon]$, $v\in I_j$, such that
$\widetilde{W}_j={\cal Z}(\varphi_{j,v},\,v\in I_j)$ in ${\Bbb
A}^{n+2}(\overline{k})$. The algorithm construct also the number $\mu_j$
of all irreducible over $\overline{k}$
components of $\widetilde{W}_j$ for every $j\in J_{s,i}$.

Set $W_j={\cal Z}(\varphi_{j,v},\,v\in I_j)$ in ${\Bbb
P}^n(\overline{k(\varepsilon)})$.
Then, cf., e.g., \cite{3}, $W_j$, $j\in J_{s,i}$, is the family of
defined over $k(\varepsilon)$ and irreducible over
$k(\varepsilon)$ components of dimension $1$ of $V_s\cap
{\cal Z}(L'_{i,s+1},\ldots , L'_{i,n-1})$. Further, $\mu_j$ is the
number of defined over
$\overline{k}(\varepsilon)$ and
irreducible over $\overline{k}(\varepsilon)$ components of $W_j$, $j\in
J_{s,i}$.
Notice also that if $W_{s,i}=W_j$ in (9) then $\mu_j$ is the number of
irreducible over $\overline{k}$ component of $W$.
One sees that
$\mbox{\rm st}_\varepsilon(W_j)={\cal
Z}(\varphi_{j,v}|_{\varepsilon=0},\,v\in I_j)$, $j\in J_{s,i}$.
Let us construct for every $2\le i\le u$  the subset $J''_{s,i}
\subset J_{s,i}$ of all $j$
such that $(\varphi_{j,v}|_{\varepsilon=0})(z_\alpha)=0$ for all $v\in
I_j$, $\alpha=1,i$.

Denote by $W_j$, $j\in J_s$ the family of defined over $k$ components of
dimension $n-s$ of $V$.
Let $W=W_j$ in (9). Then $W_{s,i}=W_{j_1}$ for the uniquely
defined $j_1\in
J_{s,i}$. Denote $j_1=\gamma_{s,i}(j)$. Hence
$\gamma_{s,i}\, :\, J_s\rightarrow J_{s,i}$ are
bijections for all $2\le i\le u$.
Let us identify $J_s=J_{s,2}$, i.e., we suppose that $\gamma_{s,2}$
is the identity mapping. Now we are going to construct all
$\gamma_{s,i}$,
$2\le i\le u$. If $s=n-1$ then $J_s=J_{s,i}$ and
$\gamma_{s,i}$ is an identity
mapping for every $2\le i\le u$.

Let $s\le n-2$.
Set $i_2=2$. For all $3\le i_1\le u$ and $\nu=1,2$
we construct linearly independent
linear forms $L''_{i_1,i}$, $s+1\le i\le n-2$,  and a linear form
$L'''_{i_1,\nu}$
with coefficients from $k(\varepsilon)$ of lengths polynomial in $n$ and
$L(z_1)$, $L(z_2)$, $L(z_{i_1})$
such that
$$
\begin{array}{l}
{\cal Z}(L''_{i_1,i},\,s+1\le i\le n-2)\supset
{\cal Z}(L'_{i_\nu,i},\,s+1\le i\le n-1),\; \nu=1,2\\
{\cal Z}(L''_{i_1,i},\,s+1\le i\le n-2)\cap
{\cal Z}(L'''_{i_1,\nu})
={\cal Z}(L'_{i_\nu,i},\,s+1\le i\le n-1),\; \nu=1,2.
\end{array}
\eqno (10)
$$
Denote by
$V'_{s,i_1}$ the union of all irreducible over
$k(\varepsilon)$ components $W'$ of dimension $2$ of
$V\cap
{\cal Z}(L''_{i_1,i},\,s+1\le i\le n-2)$.
By (*) and (10) the algebraic variety $V'_{s,i_1}$ is
the union of all irreducible over
$k(\varepsilon)$ components $W'$ of dimension $2$ of
$V_s\cap
{\cal Z}(L''_{i_1,i},\,s+1\le i\le n-2)$.
The intersection
$V\cap{\cal
Z}(L'_{2,s+1},\ldots , L'_{2,n-1})$ is transversal at each curve and
(9) is a bijection for $i=2$. Hence the inclusion of (10)
with $\nu=2$
implies that the irreducible over $k$
components of $V_s$ and irreducible over
$k(\varepsilon)$ components of $V'_{s,i_1}$
are in the one--to--one correspondence by the rule
$$
W\mapsto W\cap
{\cal Z}(L''_{i_1,i},\,s+1\le i\le n-2)=W'_{s,i_1},
\eqno (11)
$$
and $W'_{s,i_1}\cap
{\cal Z}(L'''_{i_1,\nu})=W_{s,i_\nu}$,
$\nu=1,2$.
Using the algorithm of \cite{3} we construct all the
defined over $k(\varepsilon)$ and irreducible
over $k(\varepsilon)$ components of $V'_{s,i_1}$.
Denote by $W'_j$, $j\in J'_{s,i_1}$ the family of obtained
components.
More precisely, for every $j\in J'_{s,i_1}$ the algorithm construct a
family of polynomials
$\varphi_{j,v}\in k(\varepsilon)[X_0,\ldots , X_n]$, $v\in I_j$ such that
$W'_j={\cal Z}(\varphi_{j,v},\,v\in I_j)$.
For every $j\in J'_{s,i_1}$ the intersection
${\cal Z}(\varphi_{j,v},\,v\in I_j)\cap{\cal
Z}(L'''_{i_1,\nu})
={\cal Z}(\varphi_{j_\nu,i},\,i\in I_{j_\nu})$ for the uniquely
defined $j_\nu\in J_{s,i_\nu}$
for $\nu=1,2$. Using the algorithm from \cite{3} we construct these
$j_1,j_2$.
Now one sees that $(\gamma_{s,i_1}\circ\gamma^{-1}_{s,2})(j_2)=j_1$.
Thus, we can construct all $\gamma_{s,i}$, $2\le i\le u$.

Finally, for every $1\le s\le n-1$
we compute the intersection $\bigcap_{2\le i\le
u}\,\gamma^{-1}_{s,i}(J''_{s,i})$ $=J''_s$.
According to our construction the number of irreducible over $\overline{k}$
components of $V$ of dimension $n-s$ containing $z_1,\ldots , z_u$ is
$\sum_{j\in J''_s}\mu_j$ (recall $J''_s\subset J_s=J_{2,s}$)
for every $1\le s\le n-1$.
The algorithm for the assertion (b) is described. The required estimation
for the
working time of this algorithm follows immediately from
ones of the applied algorithms.
The assertion (b) of Theorem~2 is proved.

To prove (a) we construct the finite sets
$V_{s,0}$ from Theorem~1 and all its
defined over $k$ and irreducible over $k$
components $a_j$, $j\in J_s$.
Hence defined over $k$ and irreducible over $k$ components of $V_s$ and
$V_{s,0}$ are in the one--to--one by the rule
$W\mapsto W\cap V_{s,0}$.
Let $z^{(j)}\in a_j$ and $W\cap V_{s,0}=a_j$. Denote $W_j=W$.
We introduce the representation for
$z^{(j)}$, $j\in J_s$
replacing in (1) $z,\eta,F_z$ by $z^{(j)},\eta^{(j)},F_{z^{(j)}}$.
The field $k(z^{(j)})\simeq k[\eta^{(j)}]$. Using the algorithm from \cite{3}
we construct for every $1\le i\le u$, $j\in J_s$
all pairwise non--isomorphic composites $k_{j,w}$, $w\in I_{j,i}$, of the
fields
$k[\eta_i]$ and $k[\eta^{(j)}]$ over $k$. Now $z_i\in W_j$
if and only if there is $w\in I_{j,i}$ such that the points $z_i$ and
$z^{(j)}$ with
the field  $k[\eta_i,\eta^{(j)}]\simeq k_{j,w}$ belong to the same
irreducible over $\overline{k}$ component of $V_s$ (which is contained in
$W_j$).
Using the described algorithm for the assertion (b) one can
decide whether there is $w\in I_{j,i}$ satisfying the formulated
condition and,
therefore, construct the algorithm for (a). The required estimation for the
working time of this algorithm follows immediately from
ones of the applied algorithms. The theorem is proved.
















\newpage

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{\it ``Polynomial--time computation of the degrees
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Zap. Nauchn. Semin. St-Petersburg.
Otdel. Mat. Inst. Steklov (POMI) v. 258 (1999), p. 7--59
(in Russian)\footnote{Preliminary versions of \cite{4}, \cite{7} and \cite{6} in English
can
be found as Preprints (1999) of St.Petersburg
Mathematical Society, http://www.MathSoc.spb.ru}.

\bibitem{5} {\bf  Chistov A. L.:}
{\it ``Polynomial--time computation of the Galois
group over a zero--characteristic functional field
with algebraically closed constant field''},
In ``Mathematical methods of constructing
and analysis of algorithms'', Leningrad, ``Nauka'',
1990, p. 200--232 (in Russian).

\bibitem{6} {\bf  Chistov A. L.:}
{\it ``Efficient Smooth Stratification of an Algebraic Variety in
Zero--Characteristic  and its Applications''},
Zap. Nauchn. Semin. St-Petersburg.
Otdel. Mat. Inst. Steklov (POMI) v. 266 (2000) p.254--311 (in Russian).

\bibitem{7} {\bf  Chistov A. L.:}
{\it ``Strong version of the basic deciding algorithm for
  the existential theory   of real fields''},
Zap. Nauchn. Semin. St-Petersburg.  Otdel. Mat.
Inst. Steklov (POMI) 256 (1999) p. 168--211 (in Russian).



 \bibitem{8} {\bf Bochnak J., Coste M., Roy
 M.--F.:} {\it ``G\'eom\'etrie alg\'ebrique r\'eelle''},
 Springer--Verlag, Berlin, Heidelberg, New York, 1987.

 \bibitem{9} {\bf  Chistov A. L.:}  {\it ``Polynomial--time computation
of the dimensions of
 components of algebraic varieties in zero--characteristic''},
 Journal of Pure and Applied Algebra, 117 \& 118 (1997)
 145--175.



\bibitem{10}  {\bf  Jordan C.:} {\it ``Trait\'e des substitutions
et des \'equations alg\'ebriques''}, Paris 1870.





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