UDC 512
© 2002
T. R. Seifullin
Extension of bounded root functionals of a system
of polynomial equations

(Presented by Corresponding Member of the NAS of Ukraine A. A. Letichevsky)
The notion of a root functional of a system of polynomials or ideal of
polynomials is a generalization of the notion of a root, in particular, for a
multiple root. We consider bounded root functionals and their extension
operation for a system of polynomial equation at which the number of equations
is equal to the number of unknows.

Let ${\bf R}$ be a commutative ring with unity $1$ and zero $0$.

Let $x=(x_{1},\ldots,x_{n})$ be variables, ${\bf R}[x]$ is the ring of all
polynomials in variables $x$ with coefficients in the ring ${\bf R}$.

The degree of a monom $x^{\alpha}=x^{{\alpha}_{1}}_{1}\cdot\ldots\cdot
x^{{\alpha}_{n}}_{n}$ is called $|{\alpha}|={\alpha}_{1}+\ldots+{\alpha}_{n}$,
where ${\alpha}=({\alpha}_{1},\ldots,{\alpha}_{n})$. The degree of a
polynomial $F(x)$ is called the maximal degree of a monom with a nonzero
coefficient, and such a degree is denoted by $\deg(F)$; if $F(x)=0$, then put
$\deg(F)=-\infty$.

Definition 1. Let $x=(x_{1},\ldots,x_{n})$ be variables; we denote by ${\bf
R}[x^{\leq d}]$ the set of all polynomials of degree $\leq d$. Note that ${\bf
R}[x^{\leq\infty}]={\bf R}[x]$ and if $d<0$, then ${\bf R}[x^{\leq
d}]=\\{0\\}$.

Definition 2. Let $x=(x_{1},\ldots,x_{n})$ be variables; we denote by ${\bf
R}[x]_{*}$ the set of all maps from ${\bf R}[x]$ to ${\bf R}$ that are linear
over ${\bf R}$, write such maps as $l(x_{*})$, where
$x_{*}=(x^{1}_{*},\ldots,x^{n}_{*})$, and call such maps linear functionals or
simply functionals. We denote the action of $l(x_{*})$ on $F(x)\in{\bf R}[x]$
by $l(x_{*}).F(x).$

Definition 3. Let $x=(x_{1},\ldots,x_{n})$ be variables, and let
$f(x)=(f_{1}(x),\ldots,f_{s}(x))$ be polynomials.

For a covector of polynomials $g(x)=(g^{1}(x),\ldots,g^{s}(x))^{\top}$, we
denote $f(x)g(x)=\sum\limits^{s}_{i=1}f_{i}(x)g^{i}(x)$.

Denote $(f(x))^{\leq d}_{x}=\\{\sum\limits^{s}_{i=1}f_{i}(x)g^{i}(x)|\forall
i=1,s:g^{i}(x)\in{\bf R}[x]$ and $\deg(f_{i})+\deg(g^{i})\leq d\\}$.

Denote $(f(x))_{x}=\\{\sum\limits^{s}_{i=1}f_{i}(x)g^{i}(x)|\forall
i=1,s:g^{i}(x)\in{\bf R}[x]\\}$.

Note that $(f(x))^{\leq\infty}_{x}=(f(x))_{x}$, and if $d<0$, then
$(f(x))^{\leq d}_{x}=\\{0\\}$.

We call a functional in ${\bf R}[x]_{*}$ that annuls $(f(x))_{x}$ a root
functional, and a functional in ${\bf R}[x]_{*}$ that annuls $(f(x))^{\leq
d}_{x}$ a bounded root functional.

Definition 4. Let $x=(x_{1},\ldots,x_{n})$ be variables, and let
${\lambda}=({\lambda}_{1},\ldots,{\lambda}_{n})\in{\bf R}^{n}$; we denote by
${\bf 1}_{x}({\lambda})={\bf
1}_{(x_{1},\ldots,x_{n})}({\lambda}_{1},\ldots,{\lambda}_{n})$ the map such
that ${\bf 1}_{x}({\lambda}).F(x)=F({\lambda})$ for any $F(x)\in{\bf R}[x]$.

Definition 5. Let $x=(x_{1},\ldots,x_{n})$, $y=(y_{1},\ldots,y_{n})$, and
$\hat{u}=(\hat{u}_{1},\ldots,\hat{u}_{n})$ be variables. We call a difference
derivative of a polynomial $F(x)\in{\bf R}[x]$ a covector $\hat{u}\nabla
F(x,y)=\sum\limits^{n}_{k=1}\hat{u}_{k}\nabla^{k}F(x,y)$, such that

$\hat{u}\nabla
F(x,y)=\sum\limits^{n}_{k=1}\hat{u}_{k}\nabla^{k}F(x,y)\mapsto(x-y)\nabla
F(x,y)=\sum\limits^{n}_{k=1}(x_{k}-y_{k})\nabla^{k}F(x,y)=F(x)-F(y),$

where $\forall k=1,n:\nabla^{k}F(x,y)\in{\bf R}[x,y]$.

We call a difference derivative monotonous if the degree of $\nabla F(x,y)$ in
$(x,y)$ is $\leq\deg(F)-1$.

We call a mapping that linear over ${\bf R}$ and assign, to a polynomial
$F(x)\in{\bf R}[x]$, a covector of a difference derivative $\nabla F(x,y)$, an
operator of difference derivative and denote it by $\nabla_{x}(x,y)$; then, we
have $\nabla_{x}(x,y).F(x)=\nabla F(x,y)$. Moreover,

$\hat{u}\nabla_{x}(x,y)=\sum\limits^{n}_{k=1}\hat{u}_{k}\nabla^{k}_{x}(x,y)\mapsto(x-y)\nabla_{x}(x,y)=\sum\limits^{n}_{k=1}(x_{k}-y_{k})\nabla^{k}_{x}(x,y)={\bf
1}_{x}(x)-{\bf 1}_{x}(y).$

We call an operator of difference derivative monotonous if, for any polynomial
$F(x)\in{\bf R}[x]$, the degree of $\nabla_{x}(x,y).F(x)$ is $\leq\deg(F)-1$.

Lemma 1. Let $x=(x_{1},\ldots,x_{n})$ and $y=(y_{1},\ldots,y_{n})$ be
variables. A difference derivative of a polynomial $F(x)$ exists, for example,
$\forall k=1,n:$

$\nabla^{k}F(x,y)=\frac{F(y_{1},\ldots,y_{k-1},x_{k},x_{k+1},\ldots,x_{n})-F(y_{1},\ldots,y_{k-1},y_{k},x_{k+1},\ldots,x_{n})}{x_{k}-y_{k}},$

its degree is $\leq\deg(F)-1$. A mapping that assigns, to any polynomial
$F(x)$, a covector $\nabla F(x,y)$ is linear over ${\bf R}$. Thus, there
exists a monotonous difference derivative and a monotonous operator of
difference derivative.

Lemma 2. Let $x=(x_{1},\ldots,x_{n})$, $y=(y_{1},\ldots,y_{n})$, and
$\hat{u}=(\hat{u}_{1},\ldots,\hat{u}_{n})$ be variables.

1\. For any polynomial $F(x)$, a covector
$\hat{u}\nabla^{\prime}F(x,y)=\hat{u}\nabla F(y,x)$ is a difference derivative
of the polynomial $F(x)$, and
$\hat{u}\nabla^{\prime}_{x}(x,y)=\hat{u}\nabla_{x}(y,x)$ is an operator of
difference derivative.

2\. Let $V(x)=F(x)\cdot G(x)$, then $\hat{u}\nabla F(x,y)\cdot
G(y)+F(x)\cdot\hat{u}\nabla G(x,y)$ is a difference derivation of the
polynomial $V(x)$.

3\. Let $F(x)\in{\bf R}[x^{\leq d}]$, and let $\nabla^{\prime}F(x,y)$ and
$\nabla^{\prime\prime}F(x,y)$ be two difference derivatives of the polynomial
$F(x)$ of degrees $\leq d-1$; then

$\hat{u}\nabla^{\prime}F(x,y)=\hat{u}\nabla^{\prime\prime}F(x,y)+\sum\limits_{k,l}\left((x_{k}-y_{k})\cdot\hat{u}_{l}-(x_{l}-y_{l})\cdot\hat{u}_{k}\right)\cdot
T^{kl}(x,y),$

where $k<l$ and $\deg(T^{kl})\leq d-2$.

Proof 1.

$\displaystyle(x-y)\cdot\nabla^{\prime}F(x,y)$ $\displaystyle=(x-y)\cdot\nabla
F(y,x)=$ $\displaystyle=-(y-x)\cdot\nabla F(y,x)$ $\displaystyle=-(F(y)-F(x))\
\ =F(x)-F(y)$

It follows from the first part of Statement 1 that
$\hat{u}\nabla^{\prime}_{x}(x,y)=\hat{u}\nabla_{x}(y,x)$ assigns, to any
polynomial $F(x)\in{\bf R}[x]$, its a difference derivative. The linearity of
the map $\hat{u}\nabla_{x}(x,y)$ over ${\bf R}$ implies the linearity of the
map $\hat{u}\nabla^{\prime}_{x}(x,y)=\hat{u}\nabla_{x}(y,x)$ over ${\bf R}$.
We finally obtain that $\hat{u}\nabla^{\prime}_{x}(x,y)$ is an operator of
difference derivative.

Proof 2.

$\displaystyle\left((x-y)\cdot\nabla F(x,y)\right)\cdot
G(y)+F(x)\cdot\left((x-y)\cdot\nabla G(x,y)\right)$ $\displaystyle=$
$\displaystyle=\left(F(x)-F(y)\right)\cdot
G(y)+F(x)\cdot\left(G(x)-G(y)\right)$ $\displaystyle=$
$\displaystyle=F(x)\cdot G(x)-F(y)\cdot G(y)$ $\displaystyle=V(x)-V(y).$

Proof 3. Set $W^{k}(x,y)=\nabla^{\prime k}F(x,y)-\nabla^{\prime\prime
k}F(x,y)$, and set

$T^{kl}(x,y)=\nabla^{k}_{x}(x,y).W^{l}(x,y)=\frac{1}{x_{k}-y_{k}}\cdot(W^{l}(y_{<k},x_{k},x_{>k},y)-W^{l}(y_{<k},y_{k},x_{>k},y)).$

It is directly verified that the equality in the statement is true. Further,
since the degrees of difference derivatives $\nabla^{\prime}F(x,y)$ and
$\nabla^{\prime\prime}F(x,y)$ are $\leq d-1$, then we have $\deg(W^{l})\leq
d-1$, hence, $\deg(T^{kl})\leq\deg(W^{l})-1\leq d-2$.

Assumption 1. In the sequel, unless otherwise stated, we will consider only
monotonous difference derivatives of polynomials and only monotonous operators
of difference derivative.

If $x=(x_{1},\ldots,x_{n})$ are variables, then by $y\simeq x$ we mean
$y=(y_{1},\ldots,y_{n})$.

Theorem 1. Let $x=(x_{1},\ldots,x_{n})$ and $y\simeq x$ be variables, let
$f(x)=(f_{1}(x),\ldots,f_{n}(x))$ be polynomials, let
${\delta}_{f}=\sum\limits^{n}_{i=1}(\deg(f_{i})-1)$, and let $F(x)\in{\bf
R}[x^{\leq d}]$; then we have

$\det\left\|\begin{matrix}\nabla f(x,y)&\nabla F(x,y)\cr
f(x)&F(x)\end{matrix}\right\|=\det\left\|\begin{matrix}\nabla f(x,y)&\nabla
F(x,y)\cr f(y)&F(y)\end{matrix}\right\|.$

Denote this polynomial by $R(x,y)$; then we have the following:

$R(x,y)$ have a degree $\leq{\delta}_{f}+d$,

$R(x,y)$ is uniquely determined up to an addend of the form

$\sum\limits_{i,j}\left(f_{i}(x)\cdot f_{j}(y)-f_{i}(y)\cdot
f_{j}(x)\right)\cdot{\omega}^{ij}(x,y)$

indepedently of the choice of $\nabla F(x,y)$, where $i<j$ and
$\deg(f_{i})+\deg(f_{j})+\deg({\omega}^{ij})\leq{\delta}_{f}+d$,

$R(x,y)$ is uniquely determined up to an addend of the form

$\sum\limits_{i,j}\left(f_{i}(x)\cdot f_{j}(y)-f_{i}(y)\cdot
f_{j}(x)\right)\cdot{\omega}^{ij}(x,y)+\sum\limits_{i}\left(f_{i}(x)\cdot
F(y)-f_{i}(y)\cdot F(x)\right)\cdot{\Omega}^{i}(x,y)$

indepedently of the choice of $\nabla f(x,y)$, where $i<j$ and
$\deg(f_{i})+\deg(f_{j})+\deg({\omega}^{ij})\leq{\delta}_{f}+d$ for the first
summand, and $\deg(f_{i})+\deg({\Omega}^{i})\leq{\delta}_{f}$ for the second
summand.

Proof. Since
$f_{i}(x)-\sum\limits^{n}_{k=1}(x_{k}-y_{k})\cdot\nabla^{k}f_{i}(x,y)=f_{i}(x)-(f_{i}(x)-f_{i}(y))=f_{i}(y)$
and
$F(x)-\sum\limits^{n}_{k=1}(x_{k}-y_{k})\cdot\nabla^{k}F(x,y)=F(x)-(F(x)-F(y))=F(y)$,
by adding, to the last row, the linear combination of the rest rows of the
first determinant matrix, we obtain the second determinant matrix. It implies
the equality of determinants.

It follows from the monotony of a difference derivative that the degree of
$\nabla F(x,y)$ is $\leq d-1$; and the degree of $\nabla f_{i}(x,y)$ is
$\leq\deg(f_{i})-1$ for any $i$, then the degree of the polynomial $R(x,y)$ is
$\leq\sum\limits^{n}_{i=1}(\deg(f_{i})-1)+(\deg(F)-1)+1\leq{\delta}_{f}+d$.

Since the degree of $\nabla F(x,y)$ is $\leq d-1$, by Statement 3 of Lemma 2,
variation of $\nabla F(x,y)$ is of the form

$\hat{u}\cdot\nabla^{\prime}F(x,y)=\hat{u}\cdot\nabla
F(x,y)+\sum\limits_{k,l}\left((x_{k}-y_{k})\cdot\hat{u}_{l}-(x_{l}-y_{l})\cdot\hat{u}_{k}\right)\cdot
T^{kl}(x,y),$

where $k<l$, and $\deg(T^{kl})\leq d-2$. Then $R(x,y)$ is uniquely determined
up to the addend

$\displaystyle\sum\limits_{k,l}\pm\det\left\|\begin{matrix}\nabla^{\not=k,l}f(x,y)&0\cr\nabla^{k}f(x,y)&-(x_{l}-y_{l})\cr\nabla^{l}f(x,y)&(x_{k}-y_{k})\cr
f(x)&0\end{matrix}\right\|\cdot
T^{kl}(x,y)=\hphantom{hhhhhhhhhhhhhhhhhhhhhhhhhh}$
$\displaystyle\qquad=\sum\limits_{k,l}\pm\det\left\|\begin{matrix}\nabla^{\not=k,l}f(x,y)\cr(x_{k}-y_{k})\cdot\nabla^{k}f(x,y)+(x_{l}-y_{l})\cdot\nabla^{l}f(x,y)\cr
f(x)\end{matrix}\right\|\cdot T^{kl}(x,y)=$
$\displaystyle\qquad=\sum\limits_{k,l}\pm\det\left\|\begin{matrix}\nabla^{\not=k,l}f(x,y)\cr-f(y)\cr
f(x)\end{matrix}\right\|\cdot T^{kl}(x,y)=\sum\limits_{i,j}\left(f_{i}(x)\cdot
f_{j}(y)-f_{i}(y)\cdot f_{j}(x)\right)\cdot{\omega}^{ij}(x,y),$

where $i<j$. The second equality is true since $\forall i=1,n:$

$\displaystyle-f_{i}(y)$
$\displaystyle=(x_{k}-y_{k})\cdot\nabla^{k}f_{i}(x,y)+(x_{l}-y_{l})\cdot\nabla^{l}f_{i}(x,y)\
+$
$\displaystyle\qquad+\sum\limits_{m\not=k,l}(x_{m}-y_{m})\cdot\nabla^{m}f_{i}(x,y)-f_{i}(x),$

i. e., the last but one row of the third determinant matrix is the sum of the
last but one row and the lineare combination of the rest row of the second
determinant matrix. The last equality is obtained by decomposition of the
determinant into minors of the two last rows. Moreover, we have
$\deg(f_{i})+\deg(f_{j})+\deg({\omega}^{ij})\leq{\delta}_{f}+d$.

Permuting $f_{t}(x)$ and $F(x)$ in the statement proved above, we obtain that
$R(x,y)$ is uniquely determined up to an addend

$\sum\limits_{i,j\not=t}\left(f_{i}(x)\cdot f_{j}(y)-f_{i}(y)\cdot
f_{j}(x)\right)\cdot{\omega}^{ij}(x,y)+\sum\limits_{i\not=t}\left(f_{i}(x)\cdot
F(y)-f_{i}(y)\cdot F(x)\right)\cdot{\Omega}^{i}(x,y)$

under lack of uniqueness of $\nabla f_{t}(x,y)$, where $i<j$ and
$\deg(f_{i})+\deg(f_{j})+\deg({\omega}^{ij})\leq{\delta}_{f}+d$ for the first
summand, and $\deg(f_{i})+\deg(F)+\deg({\Omega}^{i})\leq{\delta}_{f}+\deg(F)$,
hence, $\deg(f_{i})+\deg({\Omega}^{i})\leq{\delta}_{f}$, for the second
summand. Summing the additional addends appearing on changing $\nabla
f_{t}(x$,y$)$ for all $t=1,n$, we obtain that $R(x,y)$ is uniquely determined
up to an addend of the form

$\sum\limits_{i,j}\left(f_{i}(x)\cdot f_{j}(y)-f_{i}(y)\cdot
f_{j}(x)\right)\cdot{\omega}^{ij}(x,y)+\sum\limits_{i}\left(f_{i}(x)\cdot
F(y)-f_{i}(y)\cdot F(x)\right)\cdot{\Omega}^{i}(x,y)$

under lack of uniqueness of $\nabla f(x,y)$, where $i<j$ and
$\deg(f_{i})+\deg(f_{j})+\deg({\omega}^{ij})\leq{\delta}_{f}+d$ for the first
summand, and $\deg(f_{i})+\deg({\Omega}^{i})\leq{\delta}_{f}$ for the second
summand.

Theorem 2. Let $x=(x_{1},\ldots,x_{n})$ and $y\simeq x$ be variables,
$f(x)=(f_{1}(x),\ldots,f_{n}(x))$ be polynomials, let
${\delta}_{f}=\sum\limits^{n}_{i=1}(\deg(f_{i})-1)$, let a functional
$L(x_{*})$ annuls $(f(x))^{\leq{\delta}_{f}+{\delta}}_{x}$, where
${\delta}\geq 0$, and let $F(x)\in{\bf R}[x^{\leq d}]$. We set

$H(x)=L(y_{*}).\det\left\|\begin{matrix}\nabla f(x,y)&\nabla F(x,y)\cr
f(x)&F(x)\end{matrix}\right\|=L(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla F(x,y)\cr f(y)&F(y)\end{matrix}\right\|.$

Then we have the following:

1\. $H(x)\in{\bf R}[x^{\leq\max({\delta}_{f},d-{\delta}-1)}]$.

2\. $H(x)$ is uniquely determined up to an addend in
$(f(x))^{\leq\max({\delta}_{f},d-{\delta}-1)}_{x}$, indepedently of the choice
of $\nabla f(x,y)$, and uniquely determined up to an addend in $(f(x))^{\leq
d-{\delta}-1}_{x}$, indepedently of the choice of $\nabla F(x,y)$.

3\. If $F(x)\in(f(x))^{\leq d}_{x}$, then $H(x)\in(f(x))^{\leq
d-{\delta}-1}_{x}$.

4\. $H(x)$ is uniquely determined up to an addend in $(f(x))^{\leq
d-{\delta}-1}_{x}$, indepedently of the determination of $L(x_{*})$ outside
${\bf R}[x^{\leq{\delta}_{f}+{\delta}}]$.

Proof 1. We have

$\displaystyle H(x)$ $\displaystyle=L(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla F(x,y)\cr f(y)&F(y)\end{matrix}\right\|=$
$\displaystyle=L(y_{*}).F(y)\det\left\|\nabla
f(x,y)\right\|+L(y_{*}).\det\left\|\begin{matrix}\nabla f(x,y)&\nabla
F(x,y)\cr f(y)&0\end{matrix}\right\|.$

The first summand $\in{\bf R}[x^{\leq{\delta}_{f}}]$, and the second summand
$\in L(y_{*}).\sum\limits_{{\alpha},{\beta}}(f(y))^{\leq{\alpha}}_{y}\cdot{\bf
R}[x^{\leq{\beta}}]$, where ${\alpha}+{\beta}\leq{\delta}_{f}+d$. Since
$L(y_{*})$ annuls $(f(y))^{\leq{\delta}_{f}+{\delta}}_{y}$, without changing
of the sum we can retain only these terms for which
${\alpha}\geq{\delta}_{f}+{\delta}+1$, this means that
$-{\alpha}\leq-({\delta}_{f}+{\delta}+1)$, and, hence, for the remaining
terms, we have
${\beta}=({\alpha}+{\beta})-{\alpha}\leq({\delta}_{f}+d)-({\delta}_{f}+{\delta}+1)=d-{\delta}-1$.
Hence, the second summand $\in\sum\limits_{\beta}{\bf
R}[x^{\leq{\beta}}]\subseteq{\bf R}[x^{\leq d-{\delta}-1}]$, where
${\beta}\leq d-{\delta}-1$. Then the sum of the both summands $\in{\bf
R}[x^{\leq{\delta}_{f}}]+{\bf R}[x^{\leq d-{\delta}-1}]\subseteq{\bf
R}[x^{\leq\max({\delta}_{f},d-{\delta}-1)}]$. Hence, we have $H(x)\in{\bf
R}[x^{\leq\max({\delta}_{f},d-{\delta}-1)}]$.

Proof 2. Under lack of uniqueness of $\nabla F(x,y)$, by Theorem 1, $H(x)$ is
uniquely determined up to an addend

$L(y_{*}).\sum\limits_{i,j}f_{i}(x)\cdot f_{j}(y)\cdot{\omega}^{ij}(x,y)\in
L(y_{*}).\sum\limits_{{\alpha},{\beta}}(f(y))^{\leq{\alpha}}_{y}\cdot(f(x))^{\leq{\beta}}_{x},$

where ${\alpha}+{\beta}\leq{\delta}_{f}+d$. The last inclusion is true since
$\forall i:\deg(f_{i})+\deg(f_{j})+\deg({\omega}^{ij})\leq{\delta}_{f}+d$.
Since $L(y_{*})$ annuls $(f(y))^{\leq{\delta}_{f}+{\delta}}_{y}$, without
changing the sum we can retain only these terms for which
${\alpha}\geq{\delta}_{f}+{\delta}+1$; this means that
$-{\alpha}\leq-({\delta}_{f}+{\delta}+1)$, and, hence, for the remaining
terms, we have
${\beta}=({\alpha}+{\beta})-{\alpha}\leq({\delta}_{f}+d)-({\delta}_{f}+{\delta}+1)=d-{\delta}-1$.
Hence, this addend
$\in\sum\limits_{\beta}(f(x))^{\leq{\beta}}_{x}\subseteq(f(x))^{\leq
d-{\delta}-1}_{x}$, where ${\beta}\leq d-{\delta}-1$. We finally obtain that,
under lack of uniqueness of $\nabla F(x,y)$, $H(x)$ is uniquely determined up
to an addend in $(f(x))^{\leq d-{\delta}-1}_{x}$.

Under lack of uniqueness of $\nabla f(x,y)$, by Theorem 1, $H(x)$ is uniquely
determined up to an addend of the form

$L(y_{*}).\sum\limits_{i,j}f_{i}(x)\cdot
f_{j}(y)\cdot{\omega}^{ij}(x,y)+L(y_{*}).\sum\limits_{i}\left(f_{i}(x)\cdot
F(y)-f_{i}(y)\cdot F(x)\right)\cdot{\Omega}^{i}(x,y),$

where $\forall
i,j:\deg(f_{i})+\deg(f_{j})+\deg({\omega}^{ij})\leq{\delta}_{f}+d$, $\forall
i:\deg(f_{i})+\deg({\Omega}^{i})\leq{\delta}_{f}$. As shown above, the first
summand $\in(f(x))^{\leq d-{\delta}-1}_{x}$. Since
$\sum\limits_{i}\left(-f_{i}(y)\cdot
F(x)\cdot{\Omega}^{i}(x,y)\right)\in(f(y))^{\leq{\delta}_{f}}_{y}\cdot{\bf
R}[x]$, it is annuled by $L(y_{*})$. The polynomial
$\sum\limits_{i}\left(f_{i}(x)\cdot F(y)\cdot{\Omega}^{i}(x,y)\right)\in{\bf
R}[y]\cdot(f(x))^{\leq{\delta}_{f}}_{x}$. Acting by $L(y_{*})$ on this
polynomial, we obtain a polynomial $\in(f(x))^{\leq{\delta}_{f}}_{x}$.

We finally obtain that this sum $\in$ $(f(x))^{\leq
d-{\delta}-1}_{x}+(f(x))^{\leq{\delta}_{f}}_{x}\subseteq(f(x))^{\leq\max({\delta}_{f},d-{\delta}-1)}_{x}$.
Hence, under lack of uniqueness of $\nabla f(x,y)$, $H(x)$ is uniquely
determined up to an addend in
$(f(x))^{\leq\max({\delta}_{f},d-{\delta}-1)}_{x}$.

Proof 3. In the proofs of Theorems 1 and 2, we use a weaker condition than the
condition under which a difference derivative of the polynomial $F(x)$ is
monotonous, namely, the condition under which its degree is $\leq d-1$. Hence,
these theorems are true if the last condition is satisfied instead the first
condition.

Let $F(x)=f(x)g(x)\in(f(x))^{\leq d}_{x}$. By Statement 2 of Lemma 2, $F(x)$
have two difference derivatives $\nabla F(x,y)$ and $\nabla
f(x,y)g(y)+f(x)\nabla g(x,y)$, and their degrees are $\leq d-1$, although the
second difference derivative may be not monotonous when $\deg(F)<d$ . We have

$\displaystyle H(x)$ $\displaystyle=L(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla_{x}(x,y).f(x)g(x)\cr
f(y)&f(y)g(y)\end{matrix}\right\|\buildrel(f(x))^{\leq
d-{\delta}-1}_{x}\over{\equiv}$ (by Statement 2 on the uniqueness of $H(x)$
under lack of uniqueness of $\nabla F(x,y)$ ) $\displaystyle\equiv
L(y_{*}).\det\left\|\begin{matrix}\nabla f(x,y)&\nabla f(x,y)g(y)+f(x)\nabla
g(x,y)\cr f(y)&f(y)g(y)\end{matrix}\right\|=$
$\displaystyle=L(y_{*}).\det\left\|\begin{matrix}\nabla f(x,y)&f(x)\nabla
g(x,y)\cr f(y)&0\end{matrix}\right\|\in
L(y_{*}).\sum\limits_{{\alpha},{\beta}}(f(y))^{\leq{\alpha}}_{y}\cdot(f(x))^{\leq{\beta}}_{x},$

where ${\alpha}+{\beta}\leq{\delta}_{f}+d$. Since the functional $L(y_{*})$
annuls $(f(y))^{\leq{\delta}_{f}+{\delta}}_{y}$, without changing the sum, we
can retain only these terms for which ${\alpha}\geq{\delta}_{f}+{\delta}+1$,
this means that $-{\alpha}\leq-({\delta}_{f}+{\delta}+1)$, and, hence, for the
remaining terms, we have
${\beta}=({\alpha}+{\beta})-{\alpha}\leq({\delta}_{f}+d)-({\delta}_{f}+{\delta}+1)=d-{\delta}-1$.
Hence, the obtained polynomial
$\in\sum\limits_{\beta}(f(x))^{\leq{\beta}}_{x}\subseteq(f(x))^{\leq
d-{\delta}-1}_{x}$, where ${\beta}\leq d-{\delta}-1$. Since difference of
$H(x)$ and the obtained polynomial $\in(f(x))^{\leq d-{\delta}-1}_{x}$, we
have $H(x)\in(f(x))^{\leq d-{\delta}-1}_{x}$.

Proof 4. Let $L^{\prime}(x_{*})=L(x_{*})$ in ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}}]$, then it, as well as the functional
$L(x_{*})$, annuls $(f(x))^{\leq{\delta}_{f}+{\delta}}_{x}\subseteq{\bf
R}[x^{\leq{\delta}_{f}+{\delta}}]$, and $l(x_{*})=L^{\prime}(x_{*})-L(x_{*})$
annuls ${\bf R}[x^{\leq{\delta}_{f}+{\delta}}]$. We have

$\displaystyle L^{\prime}(y_{*}).\det\left\|\begin{matrix}\nabla f(x,y)&\nabla
F(x,y)\cr
f(x)&F(x)\end{matrix}\right\|-L(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla F(x,y)\cr f(x)&F(x)\end{matrix}\right\|=$
$\displaystyle\qquad=l(y_{*}).\det\left\|\begin{matrix}\nabla f(x,y)&\nabla
F(x,y)\cr f(x)&F(x)\end{matrix}\right\|=$ $\displaystyle\qquad=F(x)\cdot
l(y_{*}).\det\left\|\nabla
f(x,y)\right\|+l(y_{*}).\det\left\|\begin{matrix}\nabla f(x,y)&\nabla
F(x,y)\cr f(x)&0\end{matrix}\right\|.$

Since $l(y_{*})$ annuls ${\bf R}[y^{\leq{\delta}_{f}+{\delta}}]\supseteq{\bf
R}[y^{\leq{\delta}_{f}}]$ and $\det\left\|\nabla f(x,y)\right\|\in{\bf
R}[y^{\leq{\delta}_{f}}]\cdot{\bf R}[x]$, the first summand is equal to $0$.
The second summand $\in l(y_{*}).\sum\limits_{{\alpha},{\beta}}{\bf
R}[y^{\leq{\alpha}}]\cdot(f(x))^{\leq{\beta}}_{x}$, where
${\alpha}+{\beta}\leq{\delta}_{f}+d$. Since $l(y_{*})$ annuls ${\bf
R}[y^{\leq{\delta}_{f}+{\delta}}]$, without changing the sum we can retain
only these terms for which ${\alpha}\geq{\delta}_{f}+{\delta}+1$, this means
that $-{\alpha}\leq-({\delta}_{f}+{\delta}+1)$, and, hence, for the remaining
terms
${\beta}=({\alpha}+{\beta})-{\alpha}\leq({\delta}_{f}+d)-({\delta}_{f}+{\delta}+1)=d-{\delta}-1$.
Hence, the obtained polynomial
$\in\sum\limits_{\beta}(f(x))^{\leq{\beta}}_{x}\subseteq(f(x))^{\leq
d-{\delta}-1}_{x}$, where ${\beta}\leq d-{\delta}-1$. We finally obtain that
$H(x)$ is uniquely determined up to an addend in $(f(x))^{\leq
d-{\delta}-1}_{x}$, indepedently of the determination of $L(x_{*})$ outside
${\bf R}[x^{\leq{\delta}_{f}+{\delta}}]$.

Theorem 3. Let $x=(x_{1},\ldots,x_{n})$ and $y\simeq x$ be variables, let
$f(x)=(f_{1}(x),\ldots,f_{n}(x))$ be polynomials, and let
${\delta}_{f}=\sum\limits^{n}_{i=1}(\deg(f_{i})-1)$. Let $\forall
i=1,2:L_{i}(x_{*})$ annuls $(f(x))^{\leq{\delta}_{f}+{\delta}_{i}}_{x}$, where
${\delta}_{i}\geq 0$. We set

$\displaystyle L(x_{*})$
$\displaystyle=L_{1}(x_{*}).L_{2}(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla_{x}(x,y)\cr f(x)&{\bf 1}_{x}(x)\end{matrix}\right\|=$
$\displaystyle=L_{1}(x_{*}).L_{2}(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla_{x}(x,y)\cr f(y)&{\bf 1}_{x}(y)\end{matrix}\right\|.$

Then we have the following:

1\. $L(x_{*})$ is uniquely determined in ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}]$, indepedently of the
choice of $\nabla f(x,y)$ and the choice of $\nabla_{x}(x,y)$.

2\. $L(x_{*})$ annuls
$(f(x))^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}_{x}$.

3\. $L(x_{*})$ is uniquely determined in ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}]$, indepedently of the
determination of $L_{1}(x_{*})$ outside ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}}]$, and the determination of
$L_{2}(x_{*})$ outside ${\bf R}[x^{\leq{\delta}_{f}+{\delta}_{2}}]$.

Proof. Since $\nabla_{x}(x,y)$ is an operator linear over ${\bf R}$,
$L(x_{*})$ is a map that linear over ${\bf R}$, i. e., it is a linear
functional. Let a polynomial $F(x)\in{\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}]$. Set
$d={\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1$ and ${\delta}={\delta}_{2}$. Then
$L_{2}(x_{*})$ annuls $(f(x))^{\leq{\delta}_{f}+{\delta}}_{x}$ and
$F(x)\in{\bf R}[x^{\leq d}]$. Also, we have
$d-{\delta}-1=({\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1)-{\delta}_{2}-1={\delta}_{f}+{\delta}_{1}$,
$\max({\delta}_{f},d-{\delta}-1)=\max({\delta}_{f},{\delta}_{f}+{\delta}_{1})={\delta}_{f}+{\delta}_{1}$
since ${\delta}_{1}\geq 0$. Set

$H(x)=L_{2}(y_{*}).\det\left\|\begin{matrix}\nabla f(x,y)&\nabla F(x,y)\cr
f(y)&F(y)\end{matrix}\right\|,$

then

$L(x_{*}).F(x)=L_{1}(x_{*}).L_{2}(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla F(x,y)\cr f(y)&F(y)\end{matrix}\right\|=L_{1}(x_{*}).H(x).$

By Statement 1 of Theorem 2, $H(x)\in{\bf
R}[x^{\leq\max({\delta}_{f},d-{\delta}-1)}]={\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}}]$.

Proof 1. By Statement 2 of Theorem 2, the polynomial $H(x)$ is uniquely
determined up to an addend in $(f(x))^{\leq
d-{\delta}-1}_{x}=(f(x))^{\leq{\delta}_{f}+{\delta}_{1}}_{x}$, indepedently of
the choice of $\nabla F(x,y)$, and is uniquely determined up to an addend in
$(f(x))^{\leq\max({\delta}_{f},d-{\delta}-1)}_{x}=(f(x))^{\leq{\delta}_{f}+{\delta}_{1}}_{x}$,
indepedently of the choice of $\nabla f(x,y)$. Since $L_{1}(x_{*})$ annuls
$(f(x))^{\leq{\delta}_{f}+{\delta}_{1}}_{x}$,
$L(x_{*}).F(x)=L_{1}(x_{*}).H(x)$ is uniquely determined, indepedently of the
choice of $\nabla_{x}(x,y).F(x)=\nabla F(x,y)$, and the choice of $\nabla
f(x,y)$. From the arbitrariness of $F(x)\in{\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}]$, we obtain that
$L(x_{*})$ is uniquely determined in ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}]$, indepedently of the
choice of $\nabla f(x,y)$ and the choice of $\nabla_{x}(x,y)$.

Proof 2. Let
$F(x)\in(f(x))^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}_{x}=(f(x))^{\leq
d}_{x}$; then, by Statement 3 of Theorem 2, the polynomial
$H(x)\in(f(x))^{\leq
d-{\delta}-1}_{x}=(f(x))^{\leq{\delta}_{f}+{\delta}_{1}}_{x}$, and since
$L_{1}(x_{*})$ annuls $(f(x))^{\leq{\delta}_{f}+{\delta}_{1}}_{x}$, we have
$L(x_{*}).F(x)=L_{1}(x_{*}).H(x)=0$. From the arbitrariness of
$F(x)\in(f(x))^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}_{x}$, we obtain
that $L(x_{*})$ annuls
$(f(x))^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}_{x}$.

Proof 3. By Statement 4 of Theorem 2, $H(x)$ is uniquely determined up to an
addend in $(f(x))^{\leq
d-{\delta}-1}_{x}=(f(x))^{\leq{\delta}_{f}+{\delta}_{1}}_{x}$, indepedently of
the determination of $L_{2}(x_{*})$ outside ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}}]={\bf R}[x^{\leq{\delta}_{f}+{\delta}_{2}}]$.
Since $L_{1}(x_{*})$ annuls $(f(x))^{\leq{\delta}_{f}+{\delta}_{1}}_{x}$,
$L(x_{*}).F(x)=L_{1}(x_{*}).H(x)$ is uniquely determined indepedently of the
determination of $L_{2}(x_{*})$ outside ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{2}}]$.

Since the polynomial $H(x)\in{\bf R}[x^{\leq{\delta}_{f}+{\delta}_{1}}]$,
$L(x_{*}).F(x)=L_{1}(x_{*}).H(x)$ is uniquely determined indepedently of the
determination of $L_{1}(x_{*})$ outside ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}}]$.

Hence, it follows from the arbitrariness of $F(x)\in{\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}]$ that the functional
$L(x_{*})$ is uniquely determined in ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}]$, indepedently of the
determination of $L_{1}(x_{*})$ outside ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}}]$, and the determination of
$L_{2}(x_{*})$ outside ${\bf R}[x^{\leq{\delta}_{f}+{\delta}_{2}}]$.

Theorem 4. Let $x=(x_{1},\ldots,x_{n})$ and $y\simeq x$ be variables, let
$f(x)=(f_{1}(x_{1},\ldots,f_{n}(x))$ be polynomials, and let
${\delta}_{f}=\sum\limits^{n}_{i=1}(\deg(f_{i})-1)$. Let $\forall
i=1,2:L_{i}(x_{*})$ annuls $(f(x))^{\leq{\delta}_{f}+{\delta}_{i}}_{x}$, where
${\delta}_{i}\geq 0$; then we have

$L_{1}(x_{*}).L_{2}(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla_{x}(x,y)\cr f(x)&{\bf
1}_{x}(x)\end{matrix}\right\|=L_{2}(x_{*}).L_{1}(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla_{x}(x,y)\cr f(x)&{\bf 1}_{x}(x)\end{matrix}\right\|$

in ${\bf R}[x^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}]$.

Proof.

$\displaystyle L_{1}(x_{*}).L_{2}(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla_{x}(x,y)\cr f(x)&{\bf
1}_{x}(x)\end{matrix}\right\|=L_{1}(x_{*}).L_{2}(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla_{x}(x,y)\cr f(y)&{\bf 1}_{x}(y)\end{matrix}\right\|=$ (permuting
$L_{1}(x_{*})$ and $L_{2}(y_{*})$ and substituting $x\mapsto y$, $y\mapsto x$)
$\displaystyle\qquad=L_{2}(x_{*}).L_{1}(y_{*}).\det\left\|\begin{matrix}\nabla
f(y,x)&\nabla_{x}(y,x)\cr f(x)&{\bf 1}_{x}(x)\end{matrix}\right\|=$
$\displaystyle\qquad=L_{2}(x_{*}).L_{1}(y_{*}).\det\left\|\begin{matrix}\nabla^{\prime}f(x,y)&\nabla^{\prime}_{x}(x,y)\cr
f(x)&{\bf 1}_{x}(x)\end{matrix}\right\|=$
$\displaystyle\qquad=L_{2}(x_{*}).L_{1}(y_{*}).\det\left\|\begin{matrix}\nabla
f(x,y)&\nabla_{x}(x,y)\cr f(x)&{\bf 1}_{x}(x)\end{matrix}\right\|.$

By Statement 1 of Lemma 2, $\nabla^{\prime}_{x}(x,y)=\nabla_{x}(y,x)$ is
operator of a difference derivative, $\forall
i=1,n:\nabla^{\prime}f_{i}(x,y)=\nabla f_{i}(y,x)$ is a difference derivative
of the polynomial $f_{i}(x)$. The last equality is true in ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{2}+{\delta}_{1}+1}]$ by Statement 1 of
Theorem 3, since this functional is uniquely determined in ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}]$, indepedently of the
choice of $\nabla_{x}(x,y)$ and the choice of $\nabla f(x,y)$. Hence, the both
functionals is coincide in ${\bf
R}[x^{\leq{\delta}_{f}+{\delta}_{1}+{\delta}_{2}+1}]$.

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V. M. Glushkov Institute of Cybernetics of the NAS of Ukraine, Kiev Received
06.07.2001