text:
\section*{Introduction}
\markboth{Introduction}{Introduction}

Let $X=V(f)\subset\mathbb{P}^N_{\mathbb{K}}$ be a hypersurface, where the underlying
field $\mathbb{K}$ has characteristic $0$; the \emph{Hessian determinant of
  $f$} (which we call \emph{the Hessian of $f$} or
\emph{the Hessian of $X$}) is the \emph{determinant of the Hessian matrix
  of $f$}.

Hypersurface with vanishing Hessian were studied for the first time in
1851 by O. Hesse; he wrote two papers (\cite{OH:1,OH:2}) according to
which these hypersurfaces should be necessarily cones. In  1876
Gordan and Noether (\cite{G:N}) proved that Hesse's claim is true for
$N\leq3$, and it is false for $N\geq4$. They and Franchetta classified
all the counterexamples to Hesse's claim in $\mathbb{P}^4$
(see~\cite{G:N,F:A,GR,G:R:S}). In 1900, Perazzo classified cubic
hypersurfaces with vanishing Hessian for $N\leq6$ (\cite{P:U}). This
work was studied and generalized in~\cite{GR:RF}, and the problem is
still open in  spaces of  higher dimension.

Hessians of higher degree were introduced in~\cite{M:W} and used to
control the so called \emph{Strong Lefschetz Properties} (for short,
\emph{SLP}). The Lefschetz properties have attracted a great attention
in the last years. The basic papers of the algebraic theory of
Lefschetz properties were the original ones of
Stanley~\cite{SR:1,SR:2,SR:3} and the book of Watanabe and
others~\cite{HMMNWW}.

An algebraic tool that occurs frequently in these papers is the
\emph{Nagata Idealization}: it is a tool to convert any module $M$
over a (commutative) ring (with unit) $R$  to  an ideal of another ring
$R\ltimes M$. The starting point is the isomorphism between the
idealization of an ideal $I=(g_0,\dotsc,g_n)$ of $\mathbb{K}[u_1,\dotsc,u_m]$
and its level algebra see~\cite[Definition 2.72]{HMMNWW}. In this way,
the new ring is a \emph{Standard Graded Artinian Gorenstein Algebra}
(\emph{SGAG algebra}, for short). An explicit formula for the Macaulay
generator $f$ is:
\begin{equation*}
  f=x_0g_0+\dotsm+x_n g_n\in\mathbb{K}[x_0,\dotsc,x_n,u_1,\dotsc,u_m]_{(1,d)}.
\end{equation*}
A generalization of this construction is to consider polynomials
of the form:
\begin{equation*}
  f=x_0^dg_0+\dotsm+x_n^dg_n\in\mathbb{K}[x_0,\dotsc,x_n,u_1,\dotsc,u_m]_{(d,d+1)};
\end{equation*}
these are called \emph{Nagata polynomials of degree $d$}. The
Lefschetz properties for the relevant associated algebras $A$, the
geometry of \emph{Nagata hypersurfaces of degree $d$}, the interaction
between the combinatorics of $f$ and the structure of $A$ were studied
in~\cite{CGIM}, where the $g_i$'s are square free monomials, using a
simplicial complex associated to $f$.

In this paper we use the \emph{CW-complexes}, to study Nagata polynomials of bidegree
$(d_1,d_2)$. We study the Hilbert vector and we give a complete
description of the ideal $I$ for every case, also if the $g_i$'s are
not square free monomials.

The geometry of the Nagata hypersurface is very similar to the
geometry of the hypersurfaces with vanishing Hessian.

More precisely, we introduce a new Construction \ref{con:1} which
allows us to identify each (monic) monomial of degree $d$ in $m$
variables with an element of the $(d-1)$-skeleton of a
CW-complex that we  call $P(m)$. This CW-complex is constructed
by generalizing the construction introduced in \cite{F:S} which
associates to a (monic) square-free monomial in $m$ variables of
degree $d$ a unique $(d-1)$-cell of the simplex of dimension $m-1$,
and vice versa.   We consider an $h$-power $u_i^h$ as a
product of $h$ linear forms: $\tilde{u}_1\dotsm\tilde{u}_h$; this
corresponds to a $(h-1)$-simplex, and we identify all the
$\delta$-faces, with $\delta<h-1$, of this simplex to just one
$\delta$-face, recursively, starting from $\delta=0$ to $\delta=h-2$:
for $\delta=0$ we identify all the points to one, then if $\delta=1$
we obtain a bouquet of $h$-circles, and we identify all these circles,
and so on. Generalizing this construction to a general monic monomial
and attaching the corresponding CW-complexes along the common
skeletons, we obtain $P(m)$.

The paper is organized as follows: in Section \ref{sec:1} we recall
some generalities about graded Artinian Gorenstein Algebras and
Lefschetz Properties, with their connections with the vanishings of
higher order Hessians. In Section \ref{sec:2} we recall  what the
Nagata idealization is, what we intend for a higher Nagata idealization
and we show its connection with the Lefschetz Properties for
bihomogeneous polynomials. Section \ref{sec:3} is the core of this
article. After recalling
generalities about bigraded algebras and the topological definitions
that we need, we give the construction of the CW-complex $P(m)$; then,
we apply it to the  Nagata polynomials ( Definition
\ref{def:nag}) in Theorems \ref{thm:1} and \ref{thm:2}, which give
Theorem \ref{thm:1} a precise description of the Artinian Gorenstein
Algebra associated to a Nagata polynomial and Theorem \ref{thm:2} the
generators of the annihilator of the polynomial. We show that from
these theorems  a generalization of the principal results of
\cite{CGIM} follows:  Corollaries \ref{cor:1} and \ref{cor:2}.

We think that the study of the Nagata hypersurfaces can be---among
other things---a  useful tool for the classification of the
hypersurfaces with vanishing Hessian in $\mathbb P^n$.
\begin{acks}
  We thank the anonymous referee for the careful reading and the valuable suggestions.
\end{acks}
\begin{notations}
  In this the paper we fix the following notations and assumptions:
  \begin{itemize}
  \item  $\mathbb{K}$ is  a field  of characteristic $0$.
  \item $R:=\mathbb{K}[x_0,\dotsc,x_n]$ will always be the ring of
    polynomials in $n+1$ variables $x_0,\dotsc,x_n$.
  \item $Q:=\mathbb{K}[X_0,\dotsc,X_n]$ will be the the
    ring of differential operators of $R$,
    where $\displaystyle X_i=\pdv{x_i}$.
  \item The subscript of a graded $\mathbb{K}$-algebra will indicate  the part of that degree;  $R_d$ is the
    $\mathbb{K}$-vector space of the homogeneous polynomials of degree $d$,
    and $Q_{\delta}$ the $\mathbb{K}$-vector space of the homogeneous
    differential operators of order $\delta$.
  \end{itemize}
\end{notations}

\section{Graded Artinian Gorenstein Algebras and Lefschetz Properties}\label{sec:1}
\markboth{Graded Artinian Gorenstein Algebras and Lefschetz
  Properties}{Graded Artinian Gorenstein Algebras and Lefschetz
  Properties}

\subsection{Graded Artinian Gorenstein Algebras are Poincar\'e
  Algebras}

\begin{definition}
  Let $I$ be a homogeneous ideal of $R$ such that
  $\displaystyle A=R/I=\bigoplus_{i=0}^d A_i$ is a graded Artinian
  $\mathbb{K}$-algebra, where $A_d\neq 0$.
   The integer $d$ is the \emph{socle degree of $A$}. The algebra $A$ is said \emph{standard}
  if it is generated in degree $1$.  Setting
  $h_i=\dim_{\mathbb{K}}A_i$, the vector $\Hilb(A)=(1,h_1,\dotsc,h_d)$ is
  called \emph{Hilbert vector of $A$}.  Since $I_1=0$, then $h_1=n+1$
  is called \emph{codimension of $A$}.
\end{definition}
We also recall the following definitions.
\begin{definition}
  A graded Artinian $\mathbb{K}$-algebra $A=\bigoplus_{i=0}^d A_i$ is a
  \emph{Poincar\'e algebra} if $\cdot\colon A_i\times A_{d-i}\to A_d$
  is a \emph{perfect pairing} for $i\in\{0,\dotsc,d\}$.
\end{definition}

\begin{definition}\label{th2.1}
  A graded Artinian $\mathbb{K}$-algebra $A$ is \emph{Gorenstein} if (and
  only if) $\dim_{\mathbb{K}}A_d=1$ and it is a Poincar\'e algebra.
\end{definition}

\begin{remark}
  The Hilbert vector of a Poincar\'e algebra $A$ is \emph{symmetric
    with respect to
    $h_{\textstyle\left\lfloor\frac{d}{2}\right\rfloor}$}, that is
  $\Hilb(A)=(1,h_1,h_2,\dotsc,h_2,h_1,1)$.
  \begin{flushright}
    $\Diamond$
  \end{flushright}
\end{remark}

\subsection{Graded Artinian Gorenstein Quotient Algebras of $Q$}

For any $d\geq\delta\geq0$ there exists a natural $\mathbb{K}$-bilinear map
$B\colon R_d\times Q_{\delta}\to R_{d-\delta}$ defined by
differentiation
\begin{equation*}
  B(f,\alpha)=\alpha(f)
\end{equation*}
\begin{definition}
  Let $I=\gen{f_1,\dotsc, f_\ell}$---where $f_1,\dotsc, f_\ell$ are
  forms in $R$---be a finite dimensional $\mathbb{K}$-vector subspace of $R$.
  The \emph{annihilator of $I$ in $Q$} is the following homogeneous
  ideal
  \begin{equation*}
    \Ann(I):=\{\alpha\in Q\mid\forall f\in I,\alpha(f)=0\}.
  \end{equation*}
  In particular, if $I$ is generated by a homogeneous element $f$, we
   write  $\Ann(I)=\Ann(f)$.
\end{definition}
Let $A=Q/\Ann(f)$, where $f$ is homogeneous. By construction $A$ is a
standard graded Artinian $\mathbb{K}$-algebra; moreover $A$ is Gorenstein.
\begin{theorem}[\cite{Mac2}, \S 60ff, \cite{M:W}, Theorem 2.1]\label{th2.2}
  Let $I$ be a homogeneous ideal of $Q$ such that $A=Q/I$ is a
  standard Artinian graded $\mathbb{K}$-algebra. Then $A$ is Gorenstein if
  and only if there exist $d\geq1$ and $f\in R_d$ such that
  $A\cong Q/\Ann(f)$.
\end{theorem}
\begin{remark}
  Using the notation as above, $A$ is called the \emph{SGAG
    $\mathbb{K}$-algebra associated to $f$}. The socle degree $d$ of $A$ is
  the degree of $f$ and the codimension is $n+1$, since $I_1=0$.
  \begin{flushright}
    $\Diamond$
  \end{flushright}
\end{remark}

\subsection{Lefschetz Properties and the Hessian Criterion}

Let $\displaystyle A=\bigoplus_{i=0}^d A_i$ be a graded Artinian
$\mathbb{K}$-algebra.
\begin{definition}
  If there exists an $L\in A_1$ such that:
  \begin{enumerate}
  \item The multiplication map $\cdot L\colon A_i\to A_{i+1}$ is of
    maximal rank for all  $i$,
    then
    $A$ has the \emph{Weak Lefschetz Property} (\emph{WLP}, for
    short);
  \item The multiplication map $\cdot L^k\colon A_i\to
    A_{i+k}$ is of maximal rank for all $i$
    and $k$,
    then
    $A$ has the \emph{Strong Lefschetz Property} (\emph{SLP}, for
    short);
  \end{enumerate}
\end{definition}
\begin{definition}
  Let $A$ be the SGAG $\mathbb{K}$-algebra associated to an element $f\in
  R_d$, and let $\mathcal{B}_k=\{\alpha_j\in A_k\mid
  j\in\{1,\dotsc,\sigma_k\}\}$ be an ordered $\mathbb{K}$-basis of
  $A_k$. The \emph{$k$-th Hessian matrix} of
  $f$ with respect to $\mathcal{B}_k$ is
  \begin{equation*}
    \Hess_f^k=\left(\alpha_i\alpha_j(f)\right)_{i,j=1}^{\sigma_k}.
  \end{equation*}
  The \emph{$k$-th Hessian} of $f$ with respect to $\mathcal{B}_k$ is
  \begin{equation*}
    \hess_f^k=\det\left(\Hess_f^k\right).
  \end{equation*}
\end{definition}

\begin{theorem}[\cite{W:J} Theorem 4]
  An element $L=a_0X_0+\dotsm+a_n X_n\in
  A_1$ is a \emph{strong Lefschetz element} of
  $A$ if and only if
  $\hess_f^k(a_0,\dotsc,a_n)\neq0$ for all $\displaystyle
  k\in\left\{0,\dotsc,\left\lfloor\frac{d}{2}\right\rfloor\right\}$. In
  particular, if for some $k$ one has $\hess_f^k=0$, then
  $A$ does not have  SLP.
\end{theorem}

\section{Higher Order Nagata Idealization}\label{sec:2}
\markboth{Higher Order Nagata Idealization}{Higher Order Nagata
  Idealization}

\subsection{Nagata Idealizations}

\begin{definition}
  Let $A$ be a ring and let $M$ be an $A$-module. The \emph{Nagata
    idealization $A\ltimes M$} of $M$ is the ring with underlying set
  $A\times M$ and operations defined as follow:
  \begin{equation*}
    (r,m)+(s,n)=(r+s,m+n),\,(r,m)\cdot(s,n)=(rs,sm+r n).
  \end{equation*}
\end{definition}

\subsubsection{Bigraded Artinian Gorenstein Algebras}
Let $\displaystyle A=\bigoplus_{i=0}^d A_i$ be a SGAG $\mathbb{K}$-algebra,
it is \emph{bigraded} if:
\begin{equation*}
  A_d=A_{(d_1,d_2)}\cong\mathbb{K},\,A_i=\bigoplus_{h=0}^i A_{(i,h-i)}\,\text{\,for\,}\,i\in\{0,\dotsc,d-1\},
\end{equation*}
since $A$ is a Gorenstein ring, and the pair $(d_1,d_2)$ is said the
\emph{socle bidegree of $A$}. In this case we call  $A$  an \emph{SBAG
  algebra}.
\begin{remark}\label{rmk:41}
  By Definition~\ref{th2.1}, $A_i\cong A_{d-i}^{\vee}=\Hom_{\mathbb{K}}( A_{d-i},\mathbb{K})$ and since the
  duality commutes with direct sums, one has
  $A_{(i,j)}\cong A_{(d_1-i,d_2-j)}^{\vee}$ for any pair $(i,j)$.
  \begin{flushright}
    $\Diamond$
  \end{flushright}
\end{remark}
We fix notation as in Theorem~\ref{th3.1}:
  \begin{itemize}
  \item
    $S:=R\otimes_\mathbb{K}\KK[u_1,\dotsc,u_m]=
    \mathbb{K}[x_0,\dotsc,x_n,u_1,\dotsc,u_m]$ is the bigraded ring of
    polynomials in $m+n+1$ variables $x_0,\dotsc,x_n,u_1,\dotsc,u_m$;
 \item We have chosen the natural bigrading of $S$: $x_i$ has bidegree $(1,0)$ and $u_j$ has bidegree $(0,1)$;
  \item Define $S_{(d_1,d_2)}$ to be the $\mathbb{K}$-vector space of bihomogeneous
    polynomials $f$ of bidegree $(d_1,d_2)$; that is, $f$ can be
    written as $\displaystyle\sum_{i=0}^pa_i b_i$, where
    $a_i\in R_{d_1}=\mathbb{K}[x_0,\dotsc,x_n]_{d_1}$ and
    $b_i\in\mathbb{K}[u_1,\dotsc,u_m]_{d_2}$.
  \item
    $T:=Q\otimes_\mathbb{K}\KK[U_1,\dotsc,U_m]=\mathbb{K}[X_0,\dotsc,X_n,U_1,\dotsc,U_m]$
    is the (bigraded) ring of differential operators of $S$, where
    $\displaystyle X_i=\pdv{x_i}$ and $\displaystyle U_j=\pdv{u_j}$; $X_i$ has bidegree $(1,0)$ and $U_j$ has bidegree $(0,1)$.
  \end{itemize}

A homogeneous ideal $I$ of $S$ is a \emph{bihomogeneous ideal} if:
\begin{equation*}
  I=\bigoplus_{i,j=0}^{\infty}I_{(i,j)},\text{\,where\,}\,\forall i,j\in\mathbb{N}_{\geq0},\,I_{(i,j)}=I\cap S_{(i,j)}.
\end{equation*}
Let $f\in S_{(d_1,d_2)}$, then $I=\Ann(f)$ is a bihomogeneous ideal and
using Theorem~\ref{th2.2}, $A=T/(\Ann(f))$ is a SBAG $\mathbb{K}$-algebra of
socle bidegree $(d_1,d_2)$ (and codimension $m+n+1$).
\begin{remark}
  Using the above notations, one has:
  \begin{equation*}
    \forall i>d_1,j>d_2,\,I_{(i,j)}=T_{(i,j)}.
  \end{equation*}
  Indeed, for all $\alpha\in T_{(i,j)}$ with $i>d_1,j>d_2$,
  $\alpha(f)=0$; as a consequence:
  \begin{equation*}
    \forall k\in\{0,\dotsc,d_1+d_2\},\,A_k=\bigoplus_{\substack{0\leq i\leq d_1\\
        0\leq j\leq d_2\\
        i+j=k}}A_{(i,j)}.
  \end{equation*}
  Moreover, the evaluation map
  $\alpha\in T_{(i,j)}\mapsto \alpha(f)\in A_{(d_1-i,d_2-j)}$ provides
  the following short exact sequence:
  \begin{equation}\label{SES}
    \xymatrix{
      0\ar[r] & I_{(i,j)}\ar[r] & T_{(i,j)}\ar[r] & A_{(d_1-i,d_2-j)}\ar[r] & 0.
    }
  \end{equation}
  \begin{flushright}
    $\Diamond$
  \end{flushright}
\end{remark}

The following theorem, which links Nagata idealizations with bihomogeneous polynomials, holds.
\begin{theorem}[\cite{HMMNWW}, Theorem 2.77]\label{th3.1}
  Let $S^{\prime}:=\mathbb{K}[u_1,\dotsc,u_m]$ and
  $S:=R\otimes_\mathbb{K} S^{\prime}$ be rings of polynomials, let
  $T^{\prime}=\mathbb{K}[U_1,\dotsc,U_m]$ and $T:=Q\otimes_\mathbb{K} T^{\prime}$ be
  the associated ring of differential operators, where
  $\displaystyle X_i=\pdv{x_i}$ and $\displaystyle U_j=\pdv{u_j}$. Let
  $g_0,\dotsc,g_n$ be homogeneous elements of $S^{\prime}$ of degree
  $d$, let $I$ be the $T^{\prime}$-submodule of $S^{\prime}$ generated
  by $\{\partial(g_i)\in R\mid\partial\in T,i\in\{0,\dotsc,n\}\}$ and
  let $A^{\prime}:=T^{\prime}/\Ann(I)$.  Define
  $f=x_0g_0+\dotsm+x_n g_n\in R$, it is a bihomogeneous polynomial of
  bidegree $(1,d)$, and let $A:=T/\Ann(f)$. Considering $I$ as an
  $A^{\prime}$-module,  $A^{\prime}\ltimes I\cong A$.
\end{theorem}

\subsection{Lefschetz Properties for Higher Nagata Idealizations}

\begin{definition}\label{def:nag}
  A bihomogeneous polynomial
  \begin{equation*}
    f=\sum_{i=0}^n x_i^{d_1}g_i\in S_{(d_1,d_2)}
  \end{equation*}
  is called a \emph{CW-Nagata polynomial of degree $d_1\geq1$} if
  $g_i\in \mathbb{K}[u_1,\dotsc,u_m]$, $i=0,\dotsc,n$, are linearly
  independent monomials of degree $d_2\geq 2$.
\end{definition}
\begin{remark}
  One needs $\displaystyle n\leq\binom{m+d_2-1}{d_2}$ otherwise the
  $g_i$  cannot be linearly independent.

  From now on, we assume that $n$ satisfies this condition.
  \begin{flushright}
    $\Diamond$
  \end{flushright}
\end{remark}

We will need  the following propositions.
\begin{proposition}[\cite{GR} Proposition 2.5]
  Let  $n+1\geq m\geq 2,d_2>d_1\geq1$ and
  $\displaystyle s>\binom{m+d_1-1}{d_1}$; for any
  $j\in\{1,\dotsc,s\}$, let $f_j\in S_{(d_1,0)},\,g_j\in
  S_{(0,d_2)}$. Then the form $f=f_1g_1+\dotsm+f_s g_s$ of degree
  $d_1+d_2$ satisfies
  \begin{equation*}
    \hess_f^{d_1}=0;
  \end{equation*}
  that is, $A=T/\Ann(f)$ does not have the SLP condition.
\end{proposition}
\begin{proposition}[\cite{CGIM} Proposition 2.7]
  Let  $n+1\geq m\geq2,d_1\geq d_2$.  Then
  $\displaystyle L=\sum_{i=0}^n X_i$ is a \emph{Weak Lefschetz
    Element}; that is, $A=T/\Ann(f)$ has the WLP condition.
\end{proposition}

\section{CW-complex Nagata Idealization of Bidegree $(d_1,d_2)$}\label{sec:3}
\markboth{CW-complex Nagata Idealization of Bidegree
  $(d_1,d_2)$}{CW-complex Nagata Idealization of bidegree $(d_1,d_2)$}

Let  $S$ and $T$  be as in the previous subsection.

\begin{definition}
  A bihomogeneous CW-Nagata polynomial
  \begin{equation*}
    f=\sum_{i=0}^n x_i^{d_1}g_i\in S_{(d_1,d_2)}
  \end{equation*}
  is called a \emph{simplicial Nagata polynomial of degree $d_1$} if
  the monomials $g_i$ are square free.
\end{definition}
\begin{remark}
  One needs $\displaystyle n\leq\binom{m}{d_2}$ otherwise the $g_i$
  cannot be square free.
  \begin{flushright}
    $\Diamond$
  \end{flushright}
\end{remark}

\subsection{CW-complexes and bihomogeneous polynomials}

\subsubsection{Abstract finite simplicial complexes}
\begin{definition}
  Let $V=\{u_1,\dotsc,u_m\}$ be a finite set. An  \emph{abstract
    simplicial complex $\Delta$ with vertex set $V$} is
  a subset of $2^V$ such that
  \begin{enumerate}
  \item $\forall u\in V\Rightarrow\{u\}\in\Delta$,
  \item
    $\forall
    \sigma\in\Delta,\tau\subsetneqq\sigma,\tau\neq\emptyset\Rightarrow\tau\in\Delta$.
  \end{enumerate}
\end{definition}
The elements $\sigma$ of $\Delta$ are called \emph{faces} or
\emph{simplices}; a face with $q+1$ vertices is called \emph{$q$-face}
or \emph{face of dimension $q$} and one writes $\dim\sigma=q$; the
maximal faces (with respect to the inclusion) are called
\emph{facets}; if all facets have the same dimension $d\geq1$ then one
says that $\Delta$ is \emph{of pure dimension $d$}.  The set
$\Delta^k$ of faces of dimension at most $k$ is called
\emph{$k$-skeleton of $\Delta$}.  $2^V$ is called \emph{simplex} (of
dimension $m-1$).
\begin{remark}\label{rem1}
  \
  \begin{enumerate}
  \item (cfr.~\cite[Remark 3.4]{CGIM}) There is a natural bijection,
     introduced  in \cite{F:S}, between the
    square free monomials, of degree $d$, in the variables
    $u_1,\dotsc,u_m$ and the $(d-1)$-faces of the simplex $2^V$, with
    vertex set $V=\{u_1,\dotsc,u_m\}$.  In fact, a square free
    monomial $g=u_{i_1}\dotsm u_{i_d}$ corresponds to the subset
    $\{u_{i_1},\dotsc,u_{i_d}\}$ of $2^V$.  Vice versa, to any subset
    $F$ of $V$ with $d$ elements one associates the free square
    monomial $\displaystyle m_F=\prod_{u_i\in F}u_i$ of degree $d$.
  \item Let
    $\displaystyle f=\sum_{i=0}^n x_i^{d_1}g_i\in S_{(d_1,d_2)}$ be a
    simplicial Nagata polynomial; by hypothesis there is bijection
    between the monomials $g_i$  and the indeterminates $x_i$.
    From  this, we can associate to $f$ a simplicial complex
    $\Delta_f$ with vertices $u_1,\dotsc,u_m$ where the facet which
    corresponds to $g_i$ is identified with $x_i^{d_1}$.
  \end{enumerate}
  \begin{flushright}
    $\Diamond$
  \end{flushright}
\end{remark}

\subsubsection{CW-complexes}

For the topological background, we refer to~\cite{H:AE}. We start by fixing some notations.
\begin{definition}\label{cell}
  Let $k\in\mathbb{N}_{\geq1}$. A topological space $e^k$ homeomorphic
  to the open (unitary) ball

$\{(x_1,\dotsc,x_k)\in\mathbb{R}^k\mid x_1^2+\dotsm+x_k^2<1\} $
of dimension $k$ (with the  natural topology induced by
$\mathbb{R}^{k+1}$) is called a \emph{$k$-cell}. Its boundary, i.e. the
\emph{$(k-1)$-dimensional sphere} will be denoted by
$\mathbb{S}^{k-1}=\{(x_1,\dotsc,x_k)\in\mathbb{R}^k\mid x_1^2+\dotsm+x_k^2=1\}$ and
its closure, \ i.e. the closed (unitary) $k$-dimensional disk will be denoted by
$\mathbb{D}^k:=\{(x_1,\dotsc,x_k)\in\mathbb{R}^k\mid x_1^2+\dotsm+x_k^2\leq1\}$.
\end{definition}

We recall the following
\begin{definition}
  A \emph{CW-complex} is a topological space $X$ constructed in the
  following way:
  \begin{enumerate}
  \item There exists a fixed and discrete set of points
    $X^0\subset X$, whose elements are called \emph{$0$-cells};
  \item Inductively, the \emph{$k$-skeleton $X^k$ of $X$} is
    constructed from $X^{k-1}$ by attaching $k$-cells $e_{\alpha}^k$
    (with index set $A_k$) via continuous maps
    $\varphi^k_{\alpha}\colon\mathbb{S}^{k-1}_{\alpha}\to X^{k-1}$
    (the \emph{attaching maps}).  This means that $X^k$ is a quotient
    of
    $\displaystyle Y^k=X^{k-1}\bigcup_{\alpha\in
      A_k}\mathbb{D}^k_{\alpha}$ under the identification
    $x\sim\varphi_{\alpha}(x)$ for
    $x\in\partial\mathbb{D}^k_{\alpha}$; the \emph{elements of the
      $k$-skeleton} are the (closure of the) attached $k$ cells;
  \item $\displaystyle X=\bigcup_{k\in\mathbb{N}_{\geq0}}X^k$ and a
    subset $C$ of $X$ is closed if and only if $C\cap X^k$ is closed
    for any $k$ (\emph{closed weak topology}).
  \end{enumerate}
\end{definition}

\begin{definition}
  A subset $Z$ of a CW-complex $X$ is a \emph{CW-subcomplex} if it is
  the union of cells of $X$, such that the closure of each cell is in
  $Z$.
\end{definition}

\begin{definition}
  A CW-complex is \emph{finite} if it consists of a finite number
  of cells.
\end{definition}

We will be interested mainly in finite CW-complexes.

\begin{example}[Geometric realization of an abstract simplicial
  complex]\label{ex:1}
  It is an obvious fact that to any simplicial complex $\Delta$ one can associate a
  finite CW-complex $\widetilde{\Delta}$  via the \emph{geometric
    realization} of $\Delta$ as a
  simplicial complex (as a topological space) $\widetilde{\Delta}$.

  \begin{flushright}
    $\triangle$
  \end{flushright}
\end{example}

  In what follows we will always identify abstract simplicial
  complexes with their corresponding simplicial complexes.

\begin{construction}\label{con:1}
  In  Remark~\ref{rem1}, we saw that to any
  degree $d$ square-free monomial
  $u_{i_1}\dotsm u_{i_d}\in\mathbb{K}[u_1,\dotsc,u_m]_d$ one can associate
  the $(d-1)$-face $\{u_{i_1},\dotsc,u_{i_d}\}$ of the abstract
  $(m-1)$-dimensional simplex $\Delta(m):=2^{\{u_1,\dotsc,u_m\}}$, and
  vice versa: if we call
  \begin{align*}
    \rho_d&:=\{f\in \mathbb{K}[u_1,\dotsc,u_m]_d\mid \textup{$f\neq 0$ is a square-free monic monomial}\}\\
    D(m)_d&:=\Delta(m)^d\setminus\Delta(m)^{d-1},
  \end{align*}
  we have a bijection
  \begin{align*}
    \sigma_d\colon\rho_d & \to D(m)_d\\
    u_{i_1}\dotsm u_{i_d}&\mapsto\left\{u_{i_1},\dotsc,u_{i_d}\right\}.
  \end{align*}
  Alternatively, we can associate to $u_{i_1}\dotsm u_{i_d}$ the
  element of the $(d-1)$-skeleton
  $\overline{\left\{u_{i_1},\dotsc,u_{i_d}\right\}}\in\widetilde{\Delta(m)}^{d-1}$,
  so  we have a bijection
  \begin{align*}
    \sigma_d\colon\rho_d &\to\widetilde{\Delta(m)}^{d-1}\\
    u_{i_1}\dotsm u_{i_d}&\mapsto\overline{\left\{u_{i_1},\dotsc,u_{i_d}\right\}}
  \end{align*}
  between the square-free monomials and the $(d-1)$-faces of the
  (topological) simplex $\widetilde{\Delta(m)}$.

  Using CW-complexes, we will  extend this construction to the
  \emph{non-square-free monic monomials}.  We  proceed as
  follows. Let $g:=u_1^{j_1}\dotsm u_m^{j_m}$ be a generic degree
  $d:=j_1+\dotsb+j_m$ monomial.  Consider the following finite set:
  $W:=\left\{u^1_1,\dotsc,u^{j_1}_1,\dotsc,u^1_m,\dotsc,u^{j_m}_m\right\}$,
  and if $\Delta(d):=2^W$ is the abstract associated (finite) simplex,
  we  consider the corresponding
  (topological) simplex (which is a CW-complex)
  $\widetilde{\Delta(d)}$.

  If  $j_k\leq1$ we do nothing, while if $j_k\ge 2$, we
  recursively identify, for $\ell$ varying from $0$ to $j_k-2$, the
  $\ell$-faces of the subsimplex
  $\widetilde{2^{\left\{u^1_k,\dotsc,u^{j_k}_k\right\}}}\subset
  \widetilde{\Delta(d)}$: start with $\ell=0$, and we identify all the
  $j_k$ points to one point---call it $u_k$. Then, for $\ell=1$, we
  obtain a bouquet of $\binom{j_k+1}{2}$ circles, and we identify them
  in just one circle $\mathbb S^1$ passing through $u_k$, and so on,
  up to the facets of
  $\widetilde{2^{\left\{u^1_k,\dotsc,u^{j_k}_k\right\}}}$, i.e. its
  $j_k+1$ $(j_k-1)$-faces, which, by the construction, have all their
  boundary in common, and we identify all of them.

  Make all these identifications for all $j_1,\dotsc ,j_m$; in this
  way, we obtain a finite CW-complex $X=X_g$ of dimension $d-1$, with
  $0$-skeleton $X^0=\{u_i\mid j_i\neq0\}\subset \{u_1,\dotsc,u_m\}$,
  obtained from the $(d-1)$-dimensional simplex
  $\widetilde{\Delta(d)}$, with the above identification.

  In this way, we obtain a finite CW-complex
  $X=X_g$
  of dimension $d-1$, with $0$-skeleton
  $X^0=\{u_i\mid j_i\neq0\}\subset \{u_1,\dotsc,u_m\}$, obtained from
  the $(d-1)$-dimensional simplex $\widetilde{\Delta(d)}$, with the
  above identification. Under this identification  each closure of a $(j_k-1)$-cell
  $\overline{\left\{u_k^1,\dotsc,u_k^{j_k}\right\}}$ becomes  a
    point if $j_k=1$, a circle $\mathbb{S}^{1}$ if $j_k=2$, a
  topological space with fundamental group $\mathbb Z_3$ if $j_k=2$
  (i.e. it is not a topological surface), etc. We will denote these
  spaces in what follows by $\epsilon^{j_k-1}_k$,
  i.e. $\epsilon^{j_k-1}_k$ corresponds to $u_k^{j_k}$, and vice
  versa:
    \begin{proposition}
    Every power in $u_1^{j_1}\dotsm u_m^{j_m}$ (up to a permutation of
    the variables) corresponds to a $\epsilon^{j_k-1}_k$, and vice
    versa.
  \end{proposition}
    We can see $X_g$ as a $(d-1)$-dimensional
  \emph{join} between these spaces $\epsilon^{j_k-1}_k$ and the span
  of the $0$-skeleton $X^0$ i.e. the simplex
  $S_X\subset\widetilde{\Delta(m)}$ associated to it;
  $S_X\cong \widetilde{\Delta(\ell)}$, where $\ell=\# X^0\le m$.
  \begin{remark}
    This last observation suggests we  consider an alternative
    construction:  recall that the cellular decomposition of the
    real projective space is obtained attaching a single cell at each
    passage; indeed, $\mathbb{P}_{\mathbb R}^n$ is obtained from
    $\mathbb{P}_{\mathbb R}^{n-1}$ by attaching one $n$-cell with the
    quotient projection
    $\varphi^{n-1}\colon \mathbb S^{n-1}\to \mathbb{P}_{\mathbb R}^{n-1}$ as
    the attaching map.

    Then, to each power
    $u_k^{j_k}$ we associate a real projective space of dimension
    $j_k-1$
    $\mathbb{P}_k^{j_k-1}$ and immersions
    $i_{k-1}\colon\mathbb{P}_k^{j_k-1}\hookrightarrow
    \mathbb{P}_k^{j_k}$; so $\mathbb{P}_k^0=u_k\in \mathbb{P}_k^{j_k-1}$.

    Finally, to $g=u_1^{j_1}\dotsm u_m^{j_m}$ we associate the join
    between the $\mathbb{P}_k^{j_k-1}$ and the $S_X$ defined above; if we
    call this join by $X_g$, we can proceed in an equivalent way, by
    changing $\epsilon^{j_k-1}_k$ with $\mathbb{P}_k^{j_k-1}$.
    \begin{flushright}
      $\Diamond$
    \end{flushright}
  \end{remark}
  It is  clear how to glue two of these finite CW-complexes---say
  $X=X_{u_1^{j_1} \dotsm u_m^{j_m}}$ and
  $Y=Y_{u_1^{k_1} \dotsm u_m^{k_m}}$, of degree $d=j_1+\dotsb+j_m$ and
  $d^{\prime}=k_1+\dotsb+k_m$---along $\widetilde{\Delta(m)}$: we
  simply attach $X$ and $Y$ via the inclusion maps
  $S_X\subset \widetilde{\Delta(m)}$ and
  $S_Y\subset \widetilde{\Delta(m)}$, where $S_X$ and $S_Y$ are the
  simplexes associated to, respectively, $X$ and $Y$.

  Finally, taking all these finite CW-complexes together, we obtain a
  CW-complex $P$ in the following way:
  \begin{align*}
    C&:=\bigsqcup_{ u_1^{j_1} \dotsm u_m^{j_m}\in \mathbb{K}[u_1,\dotsc,u_m]}X_{u_1^{j_1} \dotsm u_m^{j_m}} & P(m):={C} /{\sim}
  \end{align*}
  where $\sim$ is the equivalence relation induced by the above
  gluing.

  \begin{proposition}
    There is bijection between the monomials of degree $d$ in
    $\mathbb{K}[u_1,\dotsc,u_m]$ and the elements of the $(d-1)$-skeleton of
    $P(m)$.
  \end{proposition}
  In other words, if we define
  \begin{equation*}
    \rho^{\prime}_d:=\{f\in \mathbb{K}[u_1,\dotsc,u_m]_d\mid \textup{$f\neq 0$ is a monic monomial}\}
  \end{equation*}
  we have a bijection, using the above notation
  \begin{align*}
    \sigma^{\prime}_d\colon\rho^{\prime}_d & \to P(m)_{d-1}\\
    u_1^{j_1} \dotsm u_m^{j_m}&\mapsto X_{u_1^{j_1} \dotsm u_m^{j_m}}.
  \end{align*}
  \begin{proposition} \label{prop:xx}
    $X_{u_1^{j_1} \dotsm u_m^{j_m}}\subset X_{u_1^{k_1} \dotsm
      u_m^{k_m}}$ if and only if $u_1^{j_1} \dotsm u_m^{j_m}$ divides
    $u_1^{k_1} \dotsm u_m^{k_m}$.
  \end{proposition}
  Let $\displaystyle f=\sum_{i=0}^n x_i^{d_1}g_i\in S_{(d_1,d_2)}$ be
  a CW-Nagata polynomial; by hypothesis there is bijection between the
  monomials $g_i$ and the indeterminates $x_i$. From  this, we
  can associate to $f$ a finite $(d_2-1)$-dimensional, CW-subcomplex
  of $P(m)$, $\Delta_f$ where the $(d_2-1)$-skeleton is given by the
  $X_{g_i}$'s glued together with the above procedure. Each $X_{g_i}$
  can be identified with $x_i^{d_1}$ as before.
\end{construction}
The previous construction generalizes the analogous one given
in~\cite{CGIM}.

\subsection{The Hilbert Function of SBAG Algebras}

The first main result of this paper is the following general theorem.
\begin{remark}\label{rmk:1}
  In order to state it, we observe that the canonical bases of
  \begin{displaymath}
    S_{(d_1,d_2)}=\mathbb{K}[x_0,\dotsc,x_n]_{d_1}\otimes\mathbb{K}[u_1,\dotsc,u_m]_{d_2}
  \end{displaymath}
  and
  \begin{displaymath}
    T_{(d_1,d_2)}=\mathbb{K}[X_0,\dotsc,X_n]_{d_1}\otimes\mathbb{K}[U_1,\dotsc,U_m]_{d_2}
  \end{displaymath}
  given by monomials are dual bases each other, i.e.
  \begin{equation*}
    X_0^{k_0}\dotsm X_n^{k_n}U_1^{\ell_1}\dotsm U_m^{\ell_m} (x_0^{i_0}\dotsm x_n^{i_n}u_1^{j_1}\dotsm u_m^{j_m})=\delta^{i_0,\dotsc,i_n,j_1,\dotsc,j_m}_{k_0,\dotsc,k_n,\ell_1,\dotsc,\ell_m}
  \end{equation*}
  where $i_0+\dotsm+i_n=k_0+\dotsm+k_n=d_1$,
  $j_1+\dotsm+j_m=\ell_1+\dotsm+\ell_m=d_2$ and
  $\delta^{i_0,\dotsc,i_n,j_1,\dotsc,j_m}_{k_0,\dotsc,k_n,\ell_1,\dotsc,\ell_m}$
  is the Kronecker delta.

  This simple observation allows us to identify---given a CW-Nagata
  polynomial
  $\displaystyle f=\sum_{r=0}^n x_r^{d_1}g_r\in S_{(d_1,d_2)}$---
  the \emph{dual differential operator $G_r$} of the monomial
  $g_r$---i.e. the monomial $G_r\in \mathbb{K}[U_1,\dotsc,U_m]_{d_2}$ such
  that $G_r(g_r)=1$ and $G_r(g)=0$ for any other monomial
  $g\in \mathbb{K}[u_1,\dotsc,u_m]_{d_2}$---with the same element of the
  $(d_2-1)$-skeleton of
  {\Delta}_f$ associated to $g_r$. In other words, we associate to
  $g_r=u_{1}^{j_1}\dotsm u_{m}^{j_m}$ and to
  $G_r=U_{1}^{j_1}\dotsm U_{m}^{j_m}$ the CW-subcomplex of
  $\Delta_f\subset P(m)$, $X_{u_{1}^{j_1}\dotsm u_{m}^{j_m}}$.
\end{remark}

\begin{theorem}\label{thm:1}
  Let $\displaystyle f=\sum_{r=0}^nx^{d_1}_rg_r\in R_{(d_1,d_2)}$,
  with $g_r=u_{1}^{j_1}\dotsm u_{m}^{j_m}$, be a
  CW-Nagata polynomial of (positive) degree $d_1$, where
  $\displaystyle n\leq\binom{m}{d_2}$, let $\Delta_f$ be the
  CW-complex associated to $f$ and let $A=Q/\Ann(f)$. Then
  \begin{equation*}
    A=\bigoplus_{h=0}^{d=d_1+d_2}A_h
  \end{equation*}
  where
  \begin{equation*}
    A_h=A_{(h,0)}\oplus\dotsb\oplus A_{(p,q)}\oplus\dotsb\oplus A_{(0,h)},\,p\leq d_1,\,q\leq d_2,\,A_d=A_{(d_1,d_2)}
  \end{equation*}
  and moreover, $\forall j\in\{0,1,\dotsc,d_2\}$,
  \begin{equation*}
    \dim A_{(i,j)}=a_{i,j}=\begin{cases}
      f_j & i=0\\
      \displaystyle\sum_{r=0}^n f_{j,r} & i\in\{1,\dotsc,d_1-1\},\\
      f_{d_2-j} & i=d_1
    \end{cases}
  \end{equation*}
  where:
  \begin{itemize}
  \item $f_j$ is the number of the elements of the $(j-1)$-skeleton of
    the CW-complex $\Delta_f$ (with the convention that $f_0=1$);
  \item $f_{j,r}$ is the number of the elements of the
    $(j-1)$-skeleton of the CW complex $X_{G_r}$ (with the convention
    that $f_{0,r}=1$, so that $\dim A_{(i,0)}=n+1$).
  \end{itemize}
  More precisely, a basis for $A_{(i,j)}$,
  $\forall j\in\{0,1,\dotsc,d_2\}$, is given by
  \begin{enumerate}
  \item\label{1:1} If $i=0$, $\{\Omega_1,\dotsc,\Omega_{f_j}\}$, where
    any $\Omega_s:=U_{1}^{s_1}\dotsm U_m^{s_{m}}$, with
    $s_1+\dotsm+s_m=j$, is associated to the element
    $X_{u_{1}^{s_1}\dotsm u_m^{s_{m}}}$ of the $(j-1)$-skeleton of
    $\Delta_f$;
  \item\label{1:2} If $i=1,\dotsc, d_1-1$, $\left\{\Omega_s^{i,s_1,\dotsc,s_m}\right\}_{\substack{s\in\{0,\dotsc,n\}\\
        s_k\le,r_k,k=1,\dotsc,m\\ \sum_k s_k=j}}$ where
    $\Omega_s^{i,s_1,\dotsc,s_m}:=X_s^i\cdot U_{1}^{s_1}\dotsm
    U_m^{s_{m}}$ is associated to the element
    $X_{u_{1}^{s_1}\dotsm u_m^{s_{m}}}$ of the $(j-1)$-skeleton of
    $X_{g_s}$;
  \item\label{1:3} If
    $i=d_1,\,\left\{X_0^{d_1}\Omega_1(f),\dotsc
      X_n^{d_1}\Omega_{f_{d_2}-j}(f)\right\}$, where
    $\left\{\Omega_1,\dotsc\Omega_{f_{d_2-j}}\right\}$ is the basis
    for $A_{(0,d_2-j)}$ of case \eqref{1:1}.
  \end{enumerate}
  In the cases \eqref{1:1} and \eqref{1:2} the basis are given by
  monomials, in the case \eqref{1:3}, in general, not.
\end{theorem}

\begin{proof}
  We divide the proof into computing the dimension of $A_{(i,j)}$ and
  find a basis for it, as $i$ varies:

  \begin{itemize}

  \item[$i=0$:]  $A_{(0,0)}\cong\mathbb{K}$.

    Then, by definition, if $j\in\{1,\dotsc,d_2\}$, $A_{(0,j)}$ is
    generated by the  (canonical images of the) monomials
    $\Omega_s\in Q_j=\mathbb{K}[U_1,\dotsc,U_m]_j\cong Q_{(0,j)}$ that do not
    annihilate $f$. This means that, if we write
    \begin{align*}
      \Omega_s&=U_{1}^{s_1}\dotsm U_m^{s_{m}} & s_1+\dotsm+s_m&=j,
    \end{align*}
    there exists an $r_s\in\{0,\dotsc,n\}$ such that
    $g_{r_s}=u_{1}^{s_1}\dotsm u_m^{s_{m}}g^{\prime}_{r_s}$, where
    $g^{\prime}_{r_s}\in R_{d_2-j}$ is a (nonzero) monomial; this
    means that $X_{u_{1}^{s_1}\dotsm u_m^{s_{m}}}$ is an element of
    the $(j-1)$-skeleton of the CW-complex $\Delta_f$ by Proposition
    \ref{prop:xx}.

    We need  to prove that these monomials are linearly independent
    over $\mathbb{K}$: let $\{\Omega_1,\dotsc,\Omega_{f_j}\}$ be a system of
    monomials of $Q_{(0,j)}$, where any
    $\Omega_s=U_{1}^{s_1}\dotsm U_m^{s_{m}}$ with $s_1+\dotsm+s_m=j$,
    is associated to an element of the $(j-1)$-skeleton of the
    CW-complex $\Delta_f$; take a linear combination of them and apply
    it to $f$:
    \begin{equation*}
      0=\sum_{s=1}^{f_j}c_s\Omega_s(f)=\sum_{s=1}^{f_j}c_s\sum_{r=0}^nx_r^{d_1}\Omega_s(g_r)=\sum_{r=0}^nx_r^{d_1}\sum_{s=1}^{f_j}c_s\Omega_s(g_r).
    \end{equation*}
    By the linear independence of the $x_r^{d_1}$'s
    \begin{align}\label{eq:omega}
      \sum_{s=1}^{f_j}c_s\Omega_s(g_r)&=0,&\forall r&\in\{0,\dotsc,n\}.
    \end{align}
    By hypothesis, for any index $s$ there exists an
    $r_s\in\{0,\dotsc,n\}$ such that
    $\Omega_{s}(g_{r_s})=g^{\prime}_{r_s}\in R_{d_2-j}\setminus\{0\}$,
    then for any index $s$ one has $c_s=0$, since the linear
    combinations in \eqref{eq:omega} are formed by linearly
    independent monomials ($g_r$ is fixed in each linear
    combination!). In other words, $\dim A_{(0,j)}=f_j$.

  \item[$0<i<d_1$:]  Observe that $X_a X_b(f)=0$ if
    $a\neq b$.  Therefore $A_{(i,j)}$ is generated by the only
    (canonical images of) the monomials
    $\Omega_s^{i,s_1,\dotsc,s_m}:=X_s^iU_{1}^{s_1}\dotsm
    U_{m}^{s_m}\in Q_{(i,j)}$, with $s_1+\dotsm+ s_m=j$, that do not
    annihilate $f$. In particular, a basis for $A_{(i,0)}$ is given by
    $X_0^i,\dotsc,X_n^i$ and we can suppose from now on that
    $j>0$. Since
    \begin{equation*}
      \Omega_s^{i,s_1,\dotsc,s_m}(f)=x_s^{d_1-i}\left(U_{1}^{s_1}\dotsm U_{m}^{s_m}\right)(g_s),
    \end{equation*}
    in order to obtain that this is not zero, we must have that
    $g_s=u_{1}^{s_1}\dotsm u_{m}^{s_m}g^{\prime}_s$, where
    $g^{\prime}_{s}\in R_{d_2-j}$ is a nonzero monomial.  This means
    $X_{u_{1}^{s_1}\dotsm u_{m}^{s_m}}\subset X_{g_s}$ by Proposition
    \ref{prop:xx}.

    As above, we can prove that these monomials are linearly
    independent over $\mathbb{K}$:
    let $$\left\{\Omega_s^{i,s_1,\dotsc,s_m}\right\}_{\substack{s\in\{0,\dotsc,n\}\\ s_k\le,r_k, k=1,\dotsc,m\\
        \sum_k s_k=j}}$$ be a system of monomials of $Q_{(i,j)}$, where
    any
    $\Omega_s^{i,s_1,\dotsc,s_m}=X_s^i\cdot U_{1}^{s_1}\dotsm
    U_{m}^{s_m}$ is associated to the element
    $X_{u_{1}^{s_1}\dotsm u_{m}^{s_m}}$ of the $(j-1)$ skeleton of
    $X_{g_s}\subset\Delta_f$, i.e.
    $X_{u_{1}^{s_1}\dotsm u_{m}^{s_m}}\subset X_{g_s}\subset\Delta_f$
    by Proposition \ref{prop:xx}.

    Take a linear combination of them and apply it to $f$:
    \begin{equation}\label{eq:tanti}
      0=\sum_{\substack{s\in\{0,\dotsc,n\}\\ s_k\le,r_k, k=1,\dotsc,m\\
          \sum_k s_k=j}}c_s^{i,s_1,\dotsc,s_m}\Omega_s^{i,s_1,\dotsc,s_m}(f)
      =\sum_{s=0}^{n}x^{d_1-i}\sum_{\substack{s_k\le,r_k, k=1,\dotsc,m\\
          \sum_k s_k=j}}c_s^{i,s_1,\dotsc,s_m}g_s^{i,s_1,\dotsc,s_m}
    \end{equation}
    where $g_s^{i,s_1,\dotsc,s_m}\in R_{d_2-j}$ is the nonzero
    monomial such that
    $g_s=u_{1}^{s_1}\dotsm u_{m}^{s_m}g_s^{i,s_1,\dotsc,s_m}$.  From
    \eqref{eq:tanti} we deduce, as in the preceding case, that
    \begin{align}\label{eq:bis}
      \sum_{\substack{s_k\le,r_k, k=1,\dotsc,m\\
      \sum_k s_k=j}}c_s^{i,s_1,\dotsc,s_m}g_s^{i,s_1,\dotsc,s_m}&=0 & s&=0,\dotsc, n;
    \end{align}
    as before, given one choice of $s_1,\dotsc,s_m$ there exists an
    $s\in\{0,\dotsc,n\}$ such $g_s^{i,s_1,\dotsc,s_m}(f)$ is a nonzero
    monomial, and the (nonzero) $g_s^{i,s_1,\dotsc,s_m}$'s in
    \eqref{eq:bis} are linearly independent since are obtained by a
    fixed $g_s$.

  \item[$i=d_1$:] By duality, see Remark \ref{rmk:41},
    $A_{(d_1,j)}\cong A^{\vee}_{(0,d_2-j)}$ so
    $\dim A_{(d_1,j)}=f_{d_2-j}$.  To find a basis for $A_{(d_1,j)}$,
    we consider the exact sequence \eqref{SES} given by evaluation at
    $f$, which in this case reads
    \begin{equation}\label{eq:seq}
      0\to I_{(0,d_2-j)}\to Q_{(0,d_2-j)}\to A_{(d_1,j)}\to 0,
    \end{equation}
    then a basis for $A_{(d_1,j)}$ is obtained in the following way:
    if $\{\Omega_1,\dotsc\Omega_{f_{d_2-j}}\}$ is the basis for
    $A_{(0,d_2-j)}\cong Q_{(0,d_2-j)}/I_{(0,d_2-j)}$ of the case
    $i=0$, then a basis for $A_{(d_1,j)}$ is
    $\left\{X_0^{d_1}\Omega_1,\dotsc,X_n^{d_1}\Omega_{f_{d_2}-j}(f)\right\}$.
  \end{itemize}
\end{proof}

As a corollary of Theorem \ref{thm:1} we see that we can deduce the
general case of the simplicial Nagata polynomial, which is a
slight improvement of the first part of \cite[Theorem 3.5]{CGIM}.

\begin{corollary}\label{cor:1}
  Let $\displaystyle f=\sum_{r=0}^nx^{d_1}_rg_r\in R_{(d_1,d_2)}$,
  with $g_r=x_{r_1}\dotsm x_{r_{d_2}}$, be a simplicial Nagata
  polynomial of (positive) degree $d_1$, where
  $\displaystyle n\leq\binom{m}{d_2}$, let ${\Delta}_f$ be the
  simplicial complex
  associated to $f$ and let $A=Q/\Ann(f)$. Then
  \begin{equation*}
    A=\bigoplus_{h=0}^{d=d_1+d_2}A_h
  \end{equation*}
  where
  \begin{equation*}
    A_h=A_{(h,0)}\oplus\dotsb\oplus A_{(p,q)}\oplus\dotsb\oplus A_{(0,h)},\,p\leq d_1,\,q\leq d_2,\,A_d=A_{(d_1,d_2)}
  \end{equation*}
  and moreover,
    $\forall j\in\{0,1,\dotsc,d_2\}$,
  \begin{equation*}
    \dim A_{(i,j)}=a_{i,j}=\begin{cases}
      f_j & i=0\\
      \displaystyle\sum_{r=0}^n f_{j,r} & i\in\{1,\dotsc,d_1-1\},\\
      f_{d_2-j} & i=d_1
    \end{cases}
  \end{equation*}

  where:
  \begin{itemize}
      \item $f_j$ is the number of $(j-1)$-cells of the
    ${\Delta}_f$ (with the convention that $f_0=1$);
  \item $f_{j,r}$ is the number of $(j-1)$-subcells of $\Delta_{g_r}$,
    i.e. the $(d_2-1)$-cell
    of the
    ${\Delta}_f$ associated to $g_r$ (with the convention that
    $f_{0,r}=1$, so that $\dim A_{(i,0)}=n+1$).
  \end{itemize}
  More precisely, a basis for $A_{(i,j)}$,
  $\forall j\in\{0,1,\dotsc,d_2\}$, is given by
  \begin{enumerate}
  \item\label{c1:1} If $i=0$,
    $\{\Omega_1,\dotsc,\Omega_{f_j}\}$,
    where any $\Omega_s:=U_{s_1}\dotsm U_{s_{j}}$ is associated to the
    $(j-1)$-subcell $\{u_{s_1},\dotsc, u_{s_{j}}\}$ of ${\Delta}_f$;
  \item\label{c1:2} If $i=1,\dotsc, d_1-1,\,\left\{\Omega_s^{i,s_1,\dotsc,s_j}\right\}_{\substack{s\in\{0,\dotsc,n\}\\
        s_1,\dotsc,s_j\in\left\{r_1,\dotsc,r_{d_2}\right\}}}$ where
    $\Omega_s^{i,s_1,\dotsc,s_j}:=X_s^iU_{s_1}\dotsm U_{s_j}$ is
    associated to the $(j-1)$-subcell $\{u_{s_1},\dotsc, u_{s_{j}}\}$
    of
    $\Delta_{g_s}(\subset\Delta_f)$;
  \item\label{c1:3} If $i=d_1$,
    $\left\{X_0^{d_1}\Omega_1(f),\dotsc
      X_n^{d_1}\Omega_{f_{d_2}-j}(f)\right\}$, where
    $\{\Omega_1,\dotsc\Omega_{f_{d_2-j}}\}$ is the basis for
    $A_{(0,d_2-j)}$ of case \eqref{1:1}.
  \end{enumerate}
  In the cases \eqref{c1:1} and \eqref{c1:2} the bases are given by
  monomials, in the case \eqref{c1:3},
    in general, not.
\end{corollary}

\begin{theorem}\label{thm:2}
  Let $\displaystyle f=\sum_{r=0}^n x^{d_1}_r g_r\in S_{(d_1,d_2)}$,
  with $g_r=x_{1}^{r_1}\dotsm x_{m}^{r_m}$ such that
  $r_1+\dotsm+r_m=d_2$, be a
  CW-Nagata polynomial whose associated CW-complex is $\Delta_f$, as in
  the preceding theorem.

  Then $I:=\Ann(f)$ is generated by:
  \begin{enumerate}
  \item\label{2:1} $X_i X_j$ and $X_k^{d_1+1}$, for
    $i,j,k\in\{0,\dotsc,n\}$, $i<j$;
  \item\label{2:2} $\gen{U_1,\dotsc,U_m}^{d_2+1}$, i.e. all the
    (monic) monomials of degree $d_2+1$;

  \item\label{2:3} The monomials $U_{1}^{s_1}\dotsm U_{m}^{s_m}$ such
    that $s_1+\dotsm+ s_m=j$, where
    $X_{u_{1}^{s_1}\dotsm u_{m}^{s_m}}$ is a (minimal) element of the
    $(j-1)$-skeleton of $P(m)$ not contained in $\Delta_f$ (for
    $j\in\{1,\dotsc,d_2\}$);
  \item\label{2:4} The monomials $X_rU_i$, where
    $u_i$
    does not divide $g_r$ (i.e. $\{u_{i}\}$ is not an element of the
    $0$-skeleton
    of $X_{g_r}$);
      \item\label{2:5} The monomials $X_s U_{1}^{r_1}\dotsm U_{m}^{r_m}$
        such that $r_1+\dotsm+ r_m=j$, where
    $u_{1}^{r_1}\dotsm u_{m}^{r_m}$ is minimal among those that do not
    divide $g_s$ (i.e. the (minimal) element of the $(j-1)$-skeleton
    of $P(m)$, $X_{u_{1}^{r_1}\dotsm u_{m}^{r_m}}$, is not contained
    in $X_{g_s}$),
        for
    $j\in\{1,\dotsc,d_2\}$;
  \item\label{2:6} The binomials
    $X_r^{d_1}U_1^{\rho_1}\dotsm
    U_m^{\rho_m}-X_s^{d_1}U_1^{\sigma_1}\dotsm U_m^{\sigma_m}$ with
    $\rho_1+\dotsm+\rho_m=\sigma_1+\dotsm+\sigma_m=j$ such that
    $g_{r,s}=\GCD(g_r,g_s)$ and
    $g_r=u_1^{\rho_1}\dotsm u_m^{\rho_m}g_{r,s}$,
    $g_s=u_1^{\sigma_1}\dotsm u_m^{\sigma_m}g_{r,s}$
    (i.e. $X_{g_{r,s}}$ is the element of the $(d_2-j-1)$-skeleton of
    $\Delta_f$ which represents the intersection of $X_{g_r}$ and
    $X_{g_s}$: $X_{g_{r,s}}=X_{g_r}\cap X_{g_s}$).
  \end{enumerate}
\end{theorem}
\begin{proof}
    Let  $A:=T/I$, where
  $T=\mathbb{K}[X_0,\dotsc,X_n,U_1,\dotsc,U_m]$.

  By Theorem \ref{thm:1}, \eqref{1:1} a basis for $A_{(0,j)}$,
  $\forall j\in\{1,\dotsc,d_2\}$, is
  $\{\Omega_1,\dotsc,\Omega_{f_j}\}$, where
  $\Omega_s:=U_{1}^{s_1}\dotsm U_m^{s_{m}}$, with $s_1+\dotsm+s_m=j$,
  is associated to the element $X_{u_{1}^{s_1}\dotsm u_m^{s_{m}}}$ of
  the $(j-1)$-skeleton of $\Delta_f$.
    Therefore, using the identification introduced in Remark
  \ref{rmk:1},a basis
  for
  $I_{(0,j)}$ is given by
    the monomials $U_{1}^{s_1}\dotsm U_{m}^{s_m}$ such that
  $s_1+\dotsm+ s_m=j$, where $X_{u_{1}^{s_1}\dotsm u_{m}^{s_m}}$ is an
  element of the $(j-1)$-skeleton of $P(m)$ not contained in
  $\Delta_f$ (for $j\in\{1,\dotsc,d_2\}$);

   Observe that $X_i X_j(f)=0$ if $i\neq j$ and
  $X_k^{d_1+1}(f)=0=U_1^{i_1}\dotsm U_m^{i_m}(f)$ with
  $\sum_{j=1}^mi_j=d_2+1$, for degree reasons.  Set $$\beta:
  (X_0X_1,\dotsc,X_{n-1}X_n,X_0^{d_1+1},\dotsc,X_n^{d_1+1},\gen{U_1,\dotsc,U_m}^{d_2+1});$$
  this is a homogeneous ideal such that
  \beta\subset I$ and
  $\displaystyle A\cong\frac{T}{\beta}/\frac{I}{\beta}$.

  By Theorem \ref{thm:1}, \eqref{1:2}, if $i=1,\dotsc, d_1-1$, a basis
  for $A_{(i,j)}$ $\forall j\in\{1,\dotsc,d_2\}$, is given by
  $$\left\{\Omega_s^{i,s_1,\dotsc,s_m}\right\}_{\substack{s\in\{0,\dotsc,n\}\\
      s_k\le,r_k,k=1,\dotsc,m\\ \sum_k s_k=j}}$$ where
  $\Omega_s^{i,s_1,\dotsc,s_m}:=X_s^i\cdot U_{1}^{s_1}\dotsm
  U_m^{s_{m}}$ is associated to the element
  $X_{u_{1}^{s_1}\dotsm u_m^{s_{m}}}$ of the $(j-1)$-skeleton of
  $X_{g_s}$.

   Again using the identification introduced in Remark
  \ref{rmk:1}, a basis for
  $\displaystyle\left(\frac{I}{\beta}\right)_{(i,j)}$ is given by
  \begin{itemize}
  \item\label{2:4} The monomials $X_r^i U_{1}^{s_1}\dotsm U_{m}^{s_m}$
    such that $s_1+\dotsm+ s_m=j$, with $r\neq s$, where
    $u_{1}^{s_1}\dotsm u_{m}^{s_m}$ divides $g_s$
    (i.e. $X_{u_{1}^{s_1}\dotsm u_{m}^{s_m}}$ is an element of the
    $(j-1)$-skeleton of $X_{g_s}$),
        for $i=1,\dotsc, d_1-1$, and
  \item\label{2:5} The monomials $X_s^i U_{1}^{r_1}\dotsm U_{m}^{r_m}$
    such that $r_1+\dotsm+ r_m=j$, where
    $u_{1}^{r_1}\dotsm u_{m}^{r_m}$ does not divide $g_s$ (i.e. the
    element of the $(j-1)$-skeleton of $P(m)$,
    $X_{u_{1}^{r_1}\dotsm u_{m}^{r_m}}$, is not contained in
    $X_{g_s}$),
      \end{itemize}
  for $j\in\{1,\dotsc,d_2\}$.

  It remains to find the generators of $I$ of bidegree $(d_1,j)$, with
  $j\in\{1,\dotsc,d_2\}$.  This is more complicated since the
  generators of $A_{(d_1,j)}$ are not monomials.  Let $\gamma$ be the
  homogeneous ideal generated by the monomials of the cases
  \eqref{2:1}, \eqref{2:2},
  \eqref{2:3}, \eqref{2:4} and \eqref{2:5}, i.e. the generators that
  we have found so far.  We have
  \beta\subset\gamma\subset I$ and the exact sequence \eqref{SES}
  given by evaluation at $f$ becomes
    \begin{equation*}
    0\to \left( \frac{I}{\gamma}\right)_{(d_1,j)}\to \left(\frac{T}{\gamma}\right)_{(d_1,j)}\to A_{(0,d_2-j)}\to 0,
  \end{equation*}
  since we identify
  $\displaystyle A\cong\frac{T}{\gamma}/\frac{I}{\gamma}$.  Then, if
  $\rho_1+\dotsm+\rho_m=\sigma_1+\dotsm+\sigma_m=j$,
  $X_r^{d_1}U_1^{\rho_1}\dotsm
  U_m^{\rho_m}-X_s^{d_1}U_1^{\sigma_1}\dotsm
  U_m^{\sigma_m}\in\left(\frac{T}{\gamma}\right)_{(d_1,j)}$ is in
  $\displaystyle\left(\frac{I}{\gamma}\right)_{(d_1,j)}$ if and only
  if
  $X_r^{d_1}U_1^{\rho_1}\dotsm
  U_m^{\rho_m}=X_s^{d_1}U_1^{\sigma_1}\dotsm U_m^{\sigma_m}\in
  A_{(0,d_2-j)}$, which means
  $U_1^{\rho_1}\dotsm U_m^{\rho_m}(g_r)=U_1^{\sigma_1}\dotsm
  U_m^{\sigma_m}(g_s)$.  Since $A_{(0,d_2-j)}$ is generated by the
  monomials $\Omega_s:=U_{1}^{s_1}\dotsm U_m^{s_{m}}$, with
  $s_1+\dotsm+s_m=d_2-j$, associated to the elements of the
  $(d_2-j-1)$-skeleton of $\Delta_f$, we obtain case \eqref{2:6}.

  \end{proof}

As we have done for Theorem \ref{thm:1}, we give, as a corollary of
Theorem \ref{thm:2} the case of the simplicial Nagata polynomial,
giving an improvement of the second part of \cite[Theorem 3.5]{CGIM}; we
also correct that statement, since the authors forgot the
generators $X_iX_j$, $i\neq j$.

\begin{corollary}\label{cor:2}
  Let $\displaystyle f=\sum_{r=0}^nx^{d_1}_rg_r\in R_{(d_1,d_2)}$,
  with $g_r=x_{r_1}\dotsm x_{r_{d_2}}$, be a simplicial Nagata
  polynomial whose associated
  simplicial complex is ${\Delta}_f$, as in the preceding theorem.

  Then $I:=\Ann(f)$ is generated by:
  \begin{enumerate}
  \item\label{c2:1} $X_iX_j$ and $X_k^{d_1+1}$, for
    $i,j,k\in\{0,\dotsc,n\}$, $i<j$;
  \item\label{c2:2} $U_1^2,\dotsc,U_m^2$;
  \item\label{c2:3} The monomials $U_{s_1}\dotsm U_{s_j}$, where
    $\{u_{s_1},\dotsc, u_{s_j}\}$ is a (minimal) $(j-1)$-cell of
    $2^{\{u_1,\dotsc,u_m\}}$ not contained in ${\Delta}_f$ (for
    $j\in\{1,\dotsc,d_2\}$);
  \item\label{c2:4} The monomials $X_r U_i$, where
    $u_i$
    does not divide $g_s$ (i.e. $\{u_i\}\notin \Delta{g_r}$
    );
      \item\label{c2:6} The binomials $X_r^{d_1}U_{\rho_1}\dotsm
    U_{\rho_j}-X_s^{d_1}U_{\sigma_1}\dotsm
    U_{\sigma_j}$ such that
    $g_{r,s}\GCD(g_r,g_s)$, $g_r=u_{\rho_1}\dotsm
    u_{\rho_j}g_{r,s}$, $g_s=u_{\sigma_1}\dotsm
    u_{\sigma_j}g_{r,s}$ (i.e.
    $g_{r,s}$ represents the
    $(d_2-j-1)$-face given by the intersection
    $\Delta_{g_r}\cap\Delta_{g_s}$ of the facets of $g_r$ and
    $g_s$: $\Delta_{g_{r,s}}= \Delta_{g_r}\cap\Delta_{g_s}$).

  \end{enumerate}
\end{corollary}

\begin{proof}
  We note only that we have to add the squares of case \eqref{c2:2}
  although they do not correspond to cells, since the polynomials
  $g_i$  are square-free. The rest follows  from Theorem
  \ref{thm:2}. We observe that these squares are in case \eqref{2:2} of Theorem
  \ref{thm:2}.
\end{proof}

\begin{example}
  Let
  \begin{displaymath}
    f=x_0^du_1u_2u_3+x_1^du_1u_2u_4+x_2^du_1u_4u_5+x_3^du_1u_3u_5+x_4^du_2u_3u_6+x_5^du_2u_4u_6+x_6^du_4u_5u_6+x_7^du_3u_5u_6
  \end{displaymath}
  be a bihomogeneous bidegree $(d,3)$ polynomial with $d\geq1$; it is
  a simplicial Nagata polynomial, whose associated simplicial complex
  is in the following figure:
  \begin{center}
    \begin{tikzpicture}[scale=0.50]
      \path (0,0) edge[dashed, left] (2,2); \path (2,2) edge[dashed,
      right] (6,2); \path (0,0) edge[ left] (4,0); \path (4,0) edge[
      right] (6,2); \path (4,0) edge[ left] (3,-4); \path (3,-4) edge[
      right] (6,2); \path (4,0) edge[ right] (3,5); \path (3,5) edge[
      left] (6,2); \path (0,0) edge[ left] (3,5); \path (0,0) edge[
      right] (3,-4); \path (2,2) edge[dashed] (3,-4); \path (2,2)
      edge[dashed, right] (3,5); \fill (0,0) circle(3pt); \fill (2,2)
      circle(3pt); \fill (4,0) circle(3pt); \fill (6,2) circle(3pt);
      \fill (3,5) circle(3pt); \fill (3,-4) circle(3pt); \node at
      (3.2,5.25) {$u_1$}; \node at (6.35,2.25){$u_2$}; \node at
      (3.5,-0.35){$u_3$}; \node at (1.5,2.25){$u_4$}; \node at
      (-0.4,-0.2){$u_5$}; \node at (3.2,-4.35){$u_6$}; \node at
      (4.50,2.5){$x^d_0$}; \node at (5,4.9){$x^d_1$}; \node at
      (0,3){$x^d_2$}; \node at (2.5,1.50) {$x^d_3$}; \node at
      (4.25,-0.75) {$x^d_4$}; \node at (7.3,0) {$x^d_5$}; \node at
      (0.45,-3){$x^d_6$}; \node at (2.5,-1.5){$x^d_7$}; \path
      (4.65,4.65) edge[->,bend right] (4.25,3.45); \path (0.2,3)
      edge[->, bend left] (1,2.5); \path (7,0) edge[->,bend left]
      (5.2,0); \path (0.7,-3) edge[->,bend right] (1.3,-2);
    \end{tikzpicture}
  \end{center}
  We have:
  \begin{displaymath}
    A=A_0\oplus A_1\oplus\hdots\oplus A_{d+3}.
  \end{displaymath}
We want firstly to compute the Hilbert vector by applying
  Corollary \ref{cor:1}; first of all,
  \begin{align*}
    a_{1,0}&=8 & a_{0,1}&=6,
  \end{align*}
  and therefore
  \begin{align*}
    h_0&=h_{d+3}=1 \\
    h_1&=h_{d+2}=a_{1,0}+a_{0,1}=8+6=14.
  \end{align*}
  Then, we analyze the possible cases depending on the degree $d$:
  \begin{itemize}
  \item If $d=1$, then
    \begin{align*}
      a_{1,1}&=8\cdot3=24\\
      a_{0,2}&=12\\
      h_2&=a_{1,1}+a_{0,2}=36
    \end{align*}
    and the Hilbert vector is $(1,14,36,14,1)$.
  \item If $d=2$, then, recalling bigraded Poincar\'e duality,
    \begin{align*}
      a_{2,0}&=a_{0,3}=8 & a_{2,1}&=a_{0,2}=12
    \end{align*}
    and therefore
    \begin{align*}
      h_2&=a_{2,0}+a_{1,1}+a_{0,2}=8+8\cdot3+12=44,\\
      h_3&=0+a_{2,1}+a_{1,2}+a_{0,3}=8+8\cdot3+8=44
    \end{align*}
    in accordance with Poincar\'e duality; so the Hilbert vector is
    $(1,14,44,44,14,1)$ (cfr.~\cite[Example 3.6]{CGIM}).
  \item If $d=3$, then, again by bigraded Poincar\'e duality,
    \begin{align*}
      a_{3,0}&= a_{0,3}=8, & a_{2,1}&=a_{1,2}=8\cdot3=24, & a_{3,1}&=a_{0,2}=12, &a_{2,2}&=a_{1,1}=24, &a_{1,3}&=a_{2,0}=8,
    \end{align*}
    therefore
    \begin{align*}
      h_2&=a_{2,0}+a_{1,1}+a_{0,2}=44,\\
      h_3&=a_{3,0}+a_{2,1}+a_{1,2}+a_{0,3}=64\\
      h_4&=0+a_{3,1}+a_{2,2}+a_{1,3}=44
    \end{align*}
    $h_2=h_4$ in accordance with Poincar\'e duality and the Hilbert
    vector is $(1,14,44,64,44,14,1)$.
  \item In general, let $d\geq4$; by hypothesis
    \begin{equation*}
      h_{d+1}=h_2=a_{2,0}+a_{1,1}+a_{0,2}=44,
    \end{equation*}
    and
    \begin{align*}
      h_k&=a_{k,0}+a_{k-1,1}+a_{k-2,2}+a_{k-3,3} & \forall k&\in\{3,\dotsc,d\},
    \end{align*}
    where
    \begin{align*}
      a_{k,0}&=8 & a_{k-1,1}&=8\cdot3=24 & a_{k-2,2}&=8\cdot3=24 & a_{k,3}&=8.
    \end{align*}
    Again using the Poincar\'e duality we have:
    \begin{align*}
      h_{d+3-k}&=h_k=64 & \forall k&\in\left\{3,\hdots,\left\lfloor\frac{d+3}{2}\right\rfloor\right\}
    \end{align*}
    and the Hilbert vector is $(1,14,44,64,\dotsc,64,44,14,1)$.
  \end{itemize}

  Now, we want to find the generators of $\Ann(f)$, by applying
  Corollary \ref{cor:2}. Behaviour depends on  $d$:

  \begin{itemize}
  \item If  $d=1$, by Corollary \ref{cor:2} $\Ann(f)$ is (minimally)
    generated by:
    \begin{enumerate}
    \item $\langle X_0,\hdots,X_7\rangle^2=X_0^2,X_0X_1,\dotsc;$
    \item $U_1^2,\hdots,U_6^2$;
    \item $U_1U_6,U_2U_5,U_3U_4$;
    \item $X_0U_4,X_0U_5,X_0U_6,X_1U_3,X_1U_5,X_1U_6,X_2U_2,X_2U_3,X_2U_6,X_3U_2,X_3U_4,X_3U_6,\\
      X_4U_1,X_4U_4,
      X_4U_5,X_5U_1,X_5U_3,X_5U_5,X_6U_1,X_6U_2,X_6U_3,X_7U_1,X_7U_2,X_7U_4$;
    \item $X_0U_3-X_1U_4,X_0U_2-X_3U_5,X_0U_1-X_4U_6,X_1U_2-X_2U_5,X_1U_1-X_5U_6,\\
      X_2U_4-X_3U_3,X_2U_1-X_6U_6,X_3U_1-X_7U_6,X_4U_3-X_5U_4,X_4U_2-X_7U_5,\\
      X_5U_2-X_6U_5,X_6U_4-X_7U_3$.
    \end{enumerate}

  \item If $d\geq2$, by Corollary \ref{cor:2} $\Ann(f)$ is
    (minimally) generated by
    \begin{enumerate}
    \item $\langle X_0,\hdots,X_7\rangle^{d+1}$ and $X_hX_k$ where
      $h,k\in\{0,\hdots,7\}$, $h<k$;
    \item $U_1^2,\hdots,U_6^2$;
    \item $U_1U_6,U_2U_5,U_3U_4$;
    \item $X^d_0U_4,X^d_0U_5,X^d_0U_6,X^d_1U_3,X^d_1U_5,X^d_1U_6,X^d_2U_2,X^d_2U_3,X^d_2U_6,X^d_3U_2,X^d_3U_4,X^d_3U_6,\\
      X^d_4U_1,X^d_4U_4,X^d_4U_5,X^d_5U_1,X^d_5U_3,X^d_5U_5,X^d_6U_1,X^d_6U_2,X^d_6U_3,X^d_7U_1,X^d_7U_2,X^d_7U_4$;
    \item $X^d_0U_3-X^d_1U_4,X^d_0U_2-X^d_3U_5,X^d_0U_1-X^d_4U_6,X^d_1U_2-X^d_2U_5,X^d_1U_1-X^d_5U_6,\\
      X^d_2U_4-X^d_3U_3,X^d_2U_1-X^d_6U_6,X^d_3U_1-X^d_7U_6,X^d_4U_3-X^d_5U_4,X^d_4U_2-X^d_7U_5,\\
      X^d_5U_2-X^d_6U_5,X^d_6U_4-X^d_7U_3$.
    \end{enumerate}
  \end{itemize}
\end{example}

\begin{example}
  Let
  \begin{displaymath}
    f=x_0^du_1u_2+x_1^du_1^2+x^d_2u_2u_3
  \end{displaymath}
  be a bihomogeneous bidegree $(d,2)$ polynomial, with $d\geq1$; it is
  a CW-Nagata polynomial whose CW-complex is the following:
  \begin{center}
    \begin{tikzpicture}[scale=0.5]
      \path (0,0) edge[left] (1,1); \path (1,1) edge[right] (2,2);
      \path (0,0) edge[right] (3,0); \node at (0.65,1.35) {$x_0^d$};
      \draw (2,3) circle (1cm); \node at (2,4.5) {$x_1^d$}; \fill
      (2,2) circle(3pt); \node at (2.5,1.75) {$u_1$}; \fill (0,0)
      circle(3pt); \node at (0,-0.5) {$u_2$}; \fill (3,0) circle(3pt);
      \node at (3.275,0.25) {$u_3$}; \node at (2,-0.5) {$x_2^d$};
    \end{tikzpicture}
  \end{center}
  We have:
  \begin{displaymath}
    A=A_0\oplus A_1\oplus\hdots\oplus A_{d+2}
  \end{displaymath}
  and we want to find its Hilbert vector; first of all,
  \begin{align*}
    a_{1,0}&=3 & a_{0,1}&=3
  \end{align*}
  and therefore
  \begin{align*}
    h_0&=h_{d+2}=1 & h_1&=h_{d+1}=a_{1,0}+a_{0,1}=6.
  \end{align*}
  Therefore, if $d=1$, then Hilbert vector is $(1,6,6,1)$.

  If $d=2$, we have
  \begin{equation*}
    a_{1,1}=2+1+2=5,
  \end{equation*}
  so
  \begin{equation*}
    h_2=\dim A_2=a_{2,0}+a_{1,1}+a_{0,2}=3+5+3=11
  \end{equation*}
  and the Hilbert vector is $(1,6,11,6,1)$.

  If $d=3$ then, by bigraded  Poincar\'e duality
  \begin{align*}
    a_{3,0}&= a_{0,2}=3 & a_{0,3}&=3
  \end{align*}
  so
  \begin{align*}
    h_2&=a_{2,0}+a_{1,1}+a_{0,2}=11\\
    h_3&=a_{3,0}+a_{2,1}+a_{1,2}+a_{0,3}=3+5+3=11
  \end{align*}
  and the Hilbert vector is $(1,6,11,11,6,1)$.

  In general, let $d\geq4$; by hypothesis
  \begin{equation*}
    h_d=h_2=a_{2,0}+a_{1,1}+a_{0,2}=11,
  \end{equation*}
  and
  \begin{align*}
    h_k&=\dim A_{(k,0)}+\dim A_{(k-1,1)}+\dim A_{(k-2,2)}& \forall k&\in\{3,\dotsc,d\},
  \end{align*}
  so, since
  \begin{align*}
    a_{k,0}&=3 & a_{k-1,1}&=5 & a_{k-2,2}&=3
  \end{align*}
  using Poincar\'e duality we have:
  \begin{align*}
    h_{d+2-k}&=h_k=a_{k,0}+a_{k-1,1}+a_{k-2,2}=11 & \forall k&\in\left\{3,\hdots,\left\lfloor\frac{d+2}{2}\right\rfloor\right\},
  \end{align*}
  and the Hilbert vector is $(1,6,11,\hdots,11,6,1)$.

 Let $d=1$, by Theorem \ref{thm:2} $\Ann(f)$ is (minimally)
  generated by:
  \begin{itemize}
  \item $\langle X_0,X_1,X_2\rangle^2,U_2^2,U_3^2,U_1U_3,U_1^3$;
  \item $X_0U_1^2,X_0U_3,X_1U_2,X_1U_3,X_2U_1$;
  \item $X_0U_2-X_1U_1,X_0U_1-X_3U_3$.
  \end{itemize}
  Let $d\geq2$, by Theorem \ref{thm:2} $\Ann(f)$ is
  (minimally) generated by:
  \begin{itemize}
  \item
    $\langle
    X_0,X_1,X_2\rangle^{d+1},X_0X_1,X_0X_2,X_1X_2,U_2^2,U_3^2,U_1U_3,U_1^3$;
  \item $X^d_0U_1^2,X^d_0U_3,X^d_1U_2,X^d_1U_3,X^d_2U_1$;
  \item $X^d_0U_2-X^d_1U_1,X^d_0U_1-X^d_3U_3$.
  \end{itemize}
\end{example}

redpajama_set_name: RedPajamaArXiv


text:
Barrick outlines agreement with Tanzania aimed at ending Acacia gold export ban – by Niall McGee (Globe and Mail – February 21, 2019)
February 21, 2019 in Africa Mining, Barrick Gold Corporation, Canada Mining, Canadian Media Resource Articles, Gold and Silver
https://www.theglobeandmail.com/
Barrick Gold Corp. has reached a new agreement with Tanzania that may finally end a punishing multiyear gold-export ban at its subsidiary Acacia Mining PLC that has weighed heavily on the share prices of both companies.
The development comes about six weeks after skilled African operator Mark Bristow took over as the new chief executive officer of Toronto-based Barrick.
The latest proposal would see Acacia split "economic benefits," including taxes and royalties from its Tanzanian mines, 50/50 with the East African country. Acacia would also pay Tanzania US$300-million to resolve a long-running tax dispute. While the agreement is similar to one announced in late 2017, the tax penalty can be paid over time, instead up front. Barrick says it will present a proposal to Acacia in the "near future."
Unusually, Barrick, which owns 63.9 per cent of London-based Acacia, has been negotiating with the Tanzanian government on its behalf. Acacia's management team was locked out of discussions partly because of its poor relationship with Tanzania.
"Significant amounts of real value have been destroyed by this dispute," Mr. Bristow said in a statement on Wednesday.
For the rest of this article: https://www.theglobeandmail.com/business/article-barrick-outlines-agreement-with-tanzania-aimed-at-ending-acacia-gold/
Kirkland Lake expects a million ounces in 2019 – by Staff (Mining.com – February 21, 2019)
Barrick eyes hostile bid as Newmont set to become No. 1 gold producer – by Niall McGee and Rachelle Younglai (Globe and Mail – February 21, 2019)

redpajama_set_name: RedPajamaCommonCrawl


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text:
What Is Equal Opportunity in Education?
Equality in educational opportunity - Check this out now
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Check out Equality in educational opportunity on Betabuzz.com. Find Equality in educational opportunity here
Equal opportunity in education means that each state must provide the same opportunities for everyone who attends school regardless of gender, race, color or nationality. It also means that school districts must take action to overcome barriers to students' equal participation.
Equal opportunity in education became law in 1974 with the passing of the Equal Educational Opportunities Act. The ground work for this was laid with the Civil Rights Act of 1964 that banned segregation of African Americans and women. The EEOA takes this a step further by adding faculty and staff to the list of protected people, allowing people who have suffered discrimination to file suit against whomever is responsible.
What Is the Legal Age to Quit School?
What Are 10 Rights Included in the Universal Declaration of Human Rights?
What Does "discrimination" Mean?
What Does the Sociology Term Ascribed Status Mean?


Contemporary psychoanalytic thinking about the interdependence of subjectivity and intersubjectivity has reenvisioned the analytic process, and with it the very nature of creative and engaged psychoanalytic listening. Yet few systematic writings on psychoanalytic listening or technique provide comprehensive instruction that would prepare the analyst for the kind of analytic listening needed to participate imaginatively in this sort of intersubjective experience.Offering a short course in analytic listening, Creative Listening and the Psychoanalytic Process provides a guide for the clinical uses of imaginative literature.
Outside the psychoanalytic literature, extraordinary pieces of imaginative literature exist that provide the kind of experience in analytic listening that can guide clinicians in their work with patients. Certain works of fiction create textured, sensory worlds in which complex characters possessing shifting states of consciousness live within fluid emotional atmospheres. In this book, Fred Griffin demonstrates that by entering the worlds that original writers create in their texts, the psychoanalytic therapist will learn to attend more closely to varying emotional states that generate nuanced, multidimensional views of the analysand's internal and relational worlds. He illustrates how these works capture more fully the sensory experience encountered by psychoanalysts when taking in what the patient communicates within the analytic space.


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text:
Industrial Alliance supports the Musée national des beaux-arts du Québec expansion project
Quebec City, October 6, 2011
Industrial Alliance Insurance and Financial Services Inc. is pleased to support the Musée national des beaux-arts du Québec (MNBAQ) in its project to develop a new pavilion, the details of which will be disclosed to the public today. The support will take the form of a 1 million dollar donation. This is the largest financial contribution that the company has ever made to the visual arts domain.
"Few projects transform a city so dramatically," said Yvon Charest, President and CEO of Industrial Alliance. "In Quebec City, the expansion of the Musée national des beaux-arts du Québec, directly on the Grande Allée, will enable us to rediscover a museum housing extraordinary collections."
Industrial Alliance has promoted cultural life in Quebec City for many years. An outstanding example was the company's commissioning of a large mural by Jean Dallaire, titled Quebec City under the French Regime. This work, which greatly influenced Dallaire's artistic direction, has graced a wall of the head office for 60 years.
A company with the wellbeing of the community at heart, lndustrial Alliance supports numerous cultural organizations across Canada, including the Calgary Opera and the Vancouver, Montreal and Quebec City symphony orchestras.
About Industrial Alliance
Founded in 1892, Industrial Alliance Insurance and Financial Services Inc. is a life and health insurance company with operations in all regions of Canada as well as in the United States. The company offers a wide range of life and health insurance products, savings and retirement plans, RRSPs, mutual and segregated funds, securities, auto and home insurance, mortgage loans and other financial products and services. The fourth largest life and health insurance company in Canada, Industrial Alliance contributes to the financial wellbeing of over three million Canadians, employs more than 3,900 people and manages and administers more than $70 billion in assets. Industrial Alliance stock is listed on the Toronto Stock Exchange under the ticker symbol IAG.
Office: 418 684-5000, ext 1660
Email: pierre.picard@inalco.com


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text:
Le Club Patí Vilanova ou CP Vilanova est un club espagnol de rink hockey de la ville Vilanova i la Geltrú en Catalogne. Il fut officiellement fondé en 1951 et évolue actuellement au sein du championnat de Primera División.

Histoire

La première tentative
En 1942, les garçons qui patinaient dans la rue ont eu l'idée de créer un club à Vilanova et ont loué la piste de . Les matchs se disputaient avec un palet au lieu d'une balle. Ils prirent part au championnat de Catalogne sous le nom de . L'équipe était composée de César Bernat, Robert, Granda, Cucullera, Bellmunt, Solà y Ferrando.

Malgré de bons résultats avec notamment une troisième place lors du championnat régional, les coûts importants et le manque de sponsors ont fait que le club n'a duré que deux saisons avant de disparaître.

Fondation
En 1950, Carlos Albet Olivella, Antoni Castells Gibert, Vicente Coll Claramunt, Antonio Ferrer Font, Josep Jiménez Nin, José O. Mas Planas, J. M. Tello Vilalta, Miguel Tello Vilalta, Francisco Bernad Claramunt, José María Coll Claramunt, Joaquim Durán Mulà, Pedro Ferrer Solé, José R. Julià Tomàs, Isidro Roset Ventosa y Francisco Vilaseca Roselló se réunirent afin de promouvoir le rink hockey à Vilanova. De cette réunion est ressorti un conseil qui a conduit à la création officielle du club la saison suivante. Nin Josep i Jiménez, Joaquim Duran ont été désignés comme secrétaires et les frères Coll en tant que membres.

Une fois que tout fut organisé, l'enregistrement fédéral effectué, les joueurs trouvés, alors le premier match fut joué le , contre J. F. de Sitges. Depuis, de nombreux titres ont fait de ce club l'un des plus importants d'Espagne.

Identité
L'hymne du club est le .

Palmarès
 Champion de la coupe du Roi (3): 1963-64; 1967-68 y 1975-76
 Finaliste de la coupe du Roi: 1965-66, 1966-67 y 2009-10
 Champion de la Coupe CERS (1): 2006-07
 Finaliste de la Coupe CERS: 2005-06 y 2010-11
 Finaliste de la Coupe d'Europe: 1976-77
 Champion du championnat de Catalogne: 1963-64 y 1965-66
 Deuxième du championnat de Catalogne: 1966-67
 Champion de la Coupe de la Generalidad Catalana: 1996-97
 Deuxième de la Ligue Nationale espagnole/Division d'Honneur: 1965-66; 1967-68 y 1975-76
 Champion de Première Division Nationale: 1972-73 y 1981-82
 Champion du tournoi de San Juan (Argentine): 1964
 Champion du tournoi de Sintra (Portugal): 1967
 Champion du tournoi de Monza (Italie): 1967

Notes et références

Voir aussi

Liens externes
 Site officiel du club

Club sportif fondé en 1951
Sport en Espagne en 1951
Rink hockey en 1951
1951 en Catalogne

redpajama_set_name: RedPajamaWikipedia


text:
Beyond The Knife: Martin Falck
© Martin Falck
Words by: Steven Dores
Martin Falck is surely most widely known for the playfully gauche artwork that graces The Knife's boundary pushing 2013 album Shaking The Habitual.
The Swedish designer has a bit of a penchant for the kind of confrontational work that challenges the archaeic, inflexible approach he was taught at the notoriously hardcore Rietveld Academie in Amsterdam. Crack caught up with him over Skype to talk boring design, national stereotypes and the wisdom of Oprah Winfrey.
First up, The Knife. How did you come to work with them?
I'm a huge fan of The Knife! I was working a lot with this idea of 'Norm Critical Form' and I've done a lot of collective projects that included feminist and queer ideas, so since this was what The Knife were addressing with their record, Olof and Karin asked me to do it. I had no idea they were going to ask me and I was super duper happy when they did.
Can you elaborate on the term 'Norm Critical Form'?
In Swedish we call it "Norm-Kritisk". It's the idea that we try to find different ways of doing things in different constellations. Open processes, non-professionals working within a non-hierarchy etc. You sort of include everyone in it; themes of intersectionality, feminism, the whole works! I think this is a big thing now in Sweden.
There's a preconception about Swedes being quite withdrawn as a nation. Do you think your work is a reaction against that stereotype?
I think graphic designers are in general. It's quite a non-social profession, but I feel like it's changing a bit now, which makes me really excited. Graphic design has developed a lot and gets more and more attention which allows more space to work. I really like the idea of sharing work with a lot of people, or making it loud and visible so people can really react to it.
How do you go about translating the ideas and themes of the music into something visual?
I start by listening to the music A LOT, and then I have this thing when I hear music. I see a lot of images, videoclips, colours, typefaces. I talk a lot to the artist and slowly build this reference library, a sort of 'moodboard' – but I really don't like that word. Maybe building them their own world is better, like 'here is your new world and this record will be your national anthem'.
I really like spontaneous and honest ideas. When things get too thought-out and complicated they become less interesting. Have you seen the Oprah talk show? She talks a lot about lightbulb moments. This one moment of clarity, or this one thing you want to say. In my mind I always see images moving: this square is moving in this direction, this font is dripping down into that corner. There's a lot of energy in what you see.
Your work could be considered 'anti' or 'against' traditional notions of design. Why did you choose this as an approach? Is it because so much of that stuff is just pretty boring?
When I was studying I was trying to find my own approach towards graphic design; to really learn, or feel, or understand what I could do with it. When I started to do this 'anti' design I was just breaking all the rules to learn why they are there. I felt like a lot of them were just Stone Age rules that didn't apply to our reality anymore. Now I have a much more relaxed approach to them, and I even sometimes use those old rules that I was constantly breaking.
But I really don't like the idea that 'we can only use this if it has a function', because that means that visuals are secondary and the function is the norm. With music it's just another language, one that's much more based on colours and moods. I feel like music and graphic design have so much in common, music is more free though. It's funny because everyone can make sounds and everyone could draw something, but it's still the visual part that has all the elitism. It says a lot about how important aesthetics and visual expressions are in our culture that having good taste is the ultimate symbol of status.
Would you feel OK about being misunderstood in the present if it means that in the future people will say you were ahead of your time?
Yes, but I think it's nice to have a balance. One foot in the now and one in the future… I felt very misunderstood in the beginning, like people really didn't get what I was doing. But I got a lot of energy from that. Then seeing things change and other people doing what I did; that was great. I love the idea of the internet in that sense. No copyright rules exist anymore. Graphic design is like fashion; trends change every week. It's almost as if there is this one big brain doing graphic design and posting it all on Tumblr.


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Bill Withers - Bill Withers' Greatest Hits
(Numbered Limited Edition *Awaiting Repress*)
AMOB 445
This item is Back Ordered and currently unavailable.
Please inform me of any status changes for this product regarding its upcoming availability.
• Vinyl Record
• Hybrid Stereo SACD
Numbered Limited Edition
Bill Withers' Greatest Hits on numbered, limited edition 180-gram LP
Withers' smooth soul voice, mellow grooves and gospel accents showcased
Booker T. Jones, Al Jackson and Stephen Stills guest instrumentalists
Bill Withers created mellow, downhome-style soul for barely more than a decade before retreating from the industry to pursue craftsman interests. Yet over the course of the handful of albums he made for Sussex and CBS, the Appalachian native struck lasting emotional chords in legends ranging from Booker T. Jones to Stephen Stills — not to mention the millions of listeners that fell under the spell of now-standard tracks such as "Lean on Me," "Use Me," and "Ain't No Sunshine." The antithesis of the sweaty R&B shouter that prowled the edge of stages, Withers dealt in calm and vulnerability, qualities that come to fore on the seminal Bill Withers' Greatest Hits.
Pressed at RTI, Mobile Fidelity's vinyl reissue of the 1981 compilation provides a transparent view of Withers' relaxing timbre and the subtle grooves underlining his arrangements. Characteristics ranging from the tension of the guitars, funky bends of the bass, whisper-soft coo of the formal strings, airiness of the backing harmonies, and sharpness of the snare drum emerge with utmost clarity and lifelike presence. Always prized for its naked honesty and pure conviction, Withers' music positively caresses the senses on this LP, the unadulterated production and beautiful soundscapes revealed anew with each listen.
Incredibly, Withers almost never got a shot at recording. After receiving a deal from Sussex while laboring as a toilet-seat manufacturer, he landed in the care of Stax bandleader Booker T. Jones, who gathered most of his fellow MG members — along with Stills and Jim Keltner — to assist in Withers' debut. Two takes from those sessions, the Grammy-winning "Ain't No Sunshine" and devotional "Grandma's Hands," are included here, and showcase the vocalist's incredible breath control, folksy drawl, and restrained phrasing. They also indicate his penchant for converting biographical experience into eminently catchy combinations of pop, gospel, blues, and soul.
Evidence of Withers' crossover appeal is found not only via his chart successes on both the pop and R&B charts, but in the DNA of his best work. There's not one forgettable moment on Greatest Hits. Covered by the likes of Fiona Apple, Mick Jagger, Isaac Hayes, Alicia Keys, and dozens more, "Use Me" rides a riveting clarinet riff and urban tempo to express the double-edged feelings associated with an edgy relationship.
On the loping "Who Is He (And What Is He to You?)," Withers again splits the bitter and sweet. For the Latin-tinged "Lovely Day," Withers holds a single note for 18 seconds and composed a song that's been sampled and rerecorded countless times by artists spanning Diana Ross to R. Kelly. And with "Lean on Me," the West Virginian devised a masterpiece that's been for more than four decades part of the classic American songbook, an updated hymn carried to immortality by members of the Charles Wright & 103rd Street Rhythm Band.
A veritable template for British soul contemporaries like Sam Smith and an extension of the timeless fare explored by Van Morrison, Curtis Mayfield, and Al Green, Bill Withers' Greatest Hits belongs in every music lover's library.
Lovely Day
I Want to Spend the Night
Soul Shadows
Lean on Me
Grandma's Hands
Hello Like Before
Who Is He (And What Is He to You)?
Customer Reviews (5.00 Stars) 2 person(s) rated this product.
Reviewer: Greg
Music 5/5 Sound 5/5. The sound is very smooth with nice strong bass. Despite the fact that the songs are taken from different albums (different recording years) the overall sound is consistent. Recommended.
truly great sound
Reviewer: EW
in this hobby for over 20 years but this is my first review. Why, simply because this is a truly amazing recording. Typically I am not a fan of Greatest Hits albums. This however is one of the really good ones.
Write a review and let people know what you think of this product OR just rate it.
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The latest figures highlight how student debt—which has tripled over the past decade to $1.19 trillion, according to the Federal Reserve Bank of New York—has quickly become a crushing burden for more Americans.


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text:
Gritsyuk grabs shootout winner
by Andy Potts|18 FEB 2022
ROC's Arseni Gritsyuk scores the 2-1 game-winning goal against Sweden in the shootout.
photo: Andre Ringuette / HHOF-IIHF Images
Arseni Gritsyuk, team ROC's 20-year-old forward, kept his nerve to grab a shoot-out winner for the defending champion after a hard-fought semi-final against Sweden. The game was deadlocked at 1-1 after 60 minutes and neither team could decide it in overtime.
Sweden came close to claiming that shoot-out verdict. Lucas Wallmark and Joakim Nordstrom both managed to get the better of their CSKA Moscow clubmate Ivan Fedotov in the ROC net, meaning Yegor Yakovlev had to score to keep the defending champion in the tournament. The 2018 gold medallist did just that, and in sudden death Gritsyuk stepped up after two failures from Nikita Gusev. The Avangard Omsk youngster fired in off the post to give his team the verdict and send it to back-to-back Olympic finals for the first time since the Soviet era.
"I was to blame for the first goal, I lost the puck in their zone and then lost their player behind me," said Gritsyuk. "In my heart, I felt like my mistake had hurt the whole team and I really wanted to put it right. I'm happy that Vanya Fedotov played so well and got us to the shoot-out.
"When it came to my turn, I went out and knew straight away what I was going to do – and you saw the result. If you keep turning over your shots in your thoughts, you can lose your head."
Many head coaches would rely on experience in such circumstances, but Alexei Zhamnov never doubted that Gritsyuk could do it. "It was a collective decision," he said. "Of course I understood the risk, and I took responsibility as the head coach.
"But we drew our conclusions from what we saw in practice, who looked good on penalty shots. We know who can score and who can't. We thought that against the Swedish goalie this might work, and it did."
And the youngster added that it wasn't hard to keep his hands from trembling, pointing out that Yakovlev's do-or-die attempt came under far greater pressure. "I love penalty shots, I take a lot of them," he added. "I treat it like a one-on-one game, just me and the goalie. One of us wins, one of us loses."
Yakovlev himself credited his psychologist with an assist on his shoot-out success. "To be honest, I don't understand how I kept my nerves in check," he said. "I just went out and took a shot. Thanks to my sports psychologist!"
Team ROC goes on to face Finland on Sunday for gold; Sweden must try to recover in time for Saturday's bronze-medal game against Slovakia.
Captain Anton Lander was succinct. "It sucks, it really sucks. Right now, it sucks."
His fellow forward Wallmark had more to say. "It was a tight game but I still think there were some scoring chances there," he said. "It was two good teams and it sucks to lose it like that.
"I think both teams were playing their own systems and didn't give up that many chances, but we had a couple and a chance to win the game and when there aren't many, you have to take advantage of them."
There were few changes from the quarter-final line-ups. Sweden put forward Gustav Rydahl onto the fourth line in place of defenceman Linus Hultstrom. Team ROC scratched Stanislav Galiev, with Kirill Semyonov taking his place alongside Vadim Shipachyov and Andrei Chibisov on the first line. Artur Kayumov returned to the team on the fourth.
Prior to the game, both teams promised a hard-fought battle with the emphasis on minimizing defensive errors and ensuring that nobody left the ice with any regrets. As a consequence, scoring chances were at a premium throughout this semi-final clash. There was particular attention for Wallmark, the CSKA Moscow forward who had scored in every game in Beijing so far. His KHL clubmate Fedotov snapped that streak and the Swedish sniper's shoot-out tally was not enough for the Tre Kronor.
Team ROC shaded the opening stages, with Anton Slepyshev getting on a break down the left and firing in an angled shot that Johansson turned around the corner. However, that early momentum was disrupted by Andrei Chibisov's offensive zone trip on Nordstrom which gave the Swedes the first power play of the game.
That two-minute spell offered little evidence of Swedish attacking prowess and the next big opportunity came when Chibisov jumped out of the box to get clear and test Johansson once again. For the most part, though, it was more about energy than creativity in the first 10 minutes.
Sweden came close to grabbing the lead in the 12th minute when a Lukas Bengtsson point shot came through traffic and took a deflection onto the post with Fedotov beaten. Max Friberg and Alexander Nikishin were battling on the slot for the vital touch, and the ROC goalie was grateful to hear the 'ding' of rubber on iron. The Tre Kronor had another good chance when an ROC attack broke down. Theodor Lennstrom led a two-on-one rush but Fredrick Olofsson's shot went straight to Fedotov's glove.
Team ROC ended the first period on a high, with Slava Voinov firing in a testing shot that caused confusion on the Swedish crease. And the men in red maintained that momentum through the intermission, returning to the game to open the scoring 15 seconds after the break.
As in the opening stages of the first period, it was Slepyshev collecting Pavel Karnaukhov's long pass and moving down the left-hand channel. This time, the CSKA Moscow forward beat Christian Folin and walked to the net, where he executed a neat little backhand-forehand shuffle to beat Johansson at last. That was the first time Sweden had fallen behind in regulation time action here in Beijing.
Subsequently, Team ROC concentrated on further tightening its defensive effort and Sweden was limited to a few half chances as it looked to get back into the game. There was an anxious moment for Fedotov when Carl Klingberg went around the back and slung the puck right across the paint, but Mathieu Brome could not steer it home at the back door. However, for the most part, Alexei Zhamnov's team did a good job of closing down the play and protecting its lead.
The Swedes are at their most dangerous when the defence is able to link up with the forwards. Going into the semi-final, Jonathan Pudas and Henrik Tommernes were among the team's leading playmakers. And it was those two who combined to deliver the tying goal in the 47th minute. Pudas fired in a point shot that Tommernes redirected into Fedotov's pads from close range. As the puck bounced free, Lander reacted first and the Swedish captain gleefully slotted home a vital goal.
"I think we played a good game," said defeated coach Johan Garpenlov. "We're happy with the way we played, we felt we were better during the game, we had more chances but we didn't find a way to score than one goal. Then again, [ROC] is good defensively and have lots of gifted players, and they found a way.
"I thought we had a few [chances] there, especially in the third period but we didn't score. That's the way it goes sometimes."
Now, discipline was crucial for both teams. Chances were even harder to come by and the puck spent much time up against the boards, attended by a scrum of players battling for every inch of the ice. As time ticked down, it became clear that either a moment of brilliance or an unexpected individual error was the most likely source of a winning goal in regulation. Shipachyov came close to delivering the former, bursting through on his own to force a big pad save from Johansson.
Team ROC has faced criticism at home for a lack of invention on offence and it's fair to say that the class of 2022 is some way from the swashbuckling teams of old. However, defenceman Nikita Nesterov is more than happy with performances at both ends of the ice.
"It was a good game," he said. "We played our game. I wouldn't say we are just a defensive team. We play aggressive but we need to score more. Every game we've played better and we hope our best game will be the final."
Overtime followed and Sweden was first to show with Lander carving out the first opening of the extras. Then Karnaukhov crashed the net - literally - as he bore down on Johansson at high speed, slamming his whole 96kg frame into the goal but leaving the puck behind.
The biggest chance of the extras went to Sweden. Only a stick thrust out at the vital moment by Vladimr Tkachyov prevented Brome from shooting at an open net as Fedotov got caught out of position by a swift raid on his goal.
Tkachyov later missed in the shoot-out, but Gritsyuk's winner ensured he got full value for that brave defensive play. And ROC captain Shipachyov paid tribute to the youngster. "Even before the Olympics, I said that he was ready," Shipachyov said. "Clearly, the kid has no nerves."
Olympic Winter Games Men
Slafkovsky voted MVP
Teen star takes Beijing plaudits
Finns win historic gold in Beijing
Bjorninen gets winner, ROC team can't defend PyeongChang title


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Blog Solving the Murder at the Cheatin' Heart Motel
Solving the Murder at the Cheatin' Heart Motel
Longtime readers of this blog will know I have a somewhat unresolved relationship with my home region of West Tennessee. It's not the most scenic area: the state of Tennessee slopes downhill from Appalachia in the east, so the western end is the lowest, muddiest and flattest part. Except for Memphis, there are no notable cities (I suppose you could count Jackson, but it's always felt like a city consumed by its own worst interests). And the people? Well, let's just say that when I lived there, they thought nothing of beating up a kid for reading a book. Because reading was weird.
So the last thing I expected to do was to find that this dull area had inspired hard-boiled genre literature. But damned if it didn't.
I ran across Art Bourgeau's Murder at the Cheatin' Heart Motel in the late Eighties, when I lived in Huntsville, AL. It was written in 1985, and concerned Claude "Snake" Kirlin, a freelance reporter for Ultra Suave magazine, and his buddy F.T. Zervich, trying to solve the murder of Snake's aunt, proprietor of the titular motel. The establishment is located on Chocktaw Lake, which Bourgeau describes thus:
"Anyway, in the winter of 1811 three earthquakes hit right where you're sitting. Each one was several times worse than the famous San Francisco earthquake. It was so bad that a land area of about a hundred miles simply fell into the earth. The banks of the Mississippi broke down, and the water rushed in to fill it. That's how Chocktaw Lake was formed."
(p. 13)
Wow, I thought. That sounds familiar. It sounds, in fact, like Reelfoot Lake. According to this entry on the ever-reliable Wikipedia:
"Popular history says that the lake was formed when the region subsided after the New Madrid earthquakes of 1811–1812, and that the Mississippi River flowed backward for 10–24 hours to fill it."
Someone wrote a mystery set on a fictionalized version of Reelfoot Lake, a place I'd gone fishing and on picnics and visited my whole life! It was, for me, a world-shifting realization. And it got better.
Bougreau introduced the book's villain, Sheriff Casper Denny, with this:
"I'd heard of him before. Everyone had heard of him. He was a genuine, bona fide legend. A twentieth-century Wyatt Earp who had single-handedly taken on the west Tennessee mob, a group whose roots went all the way back to Jean Boquin. They had tried to move into his county, and in the process, the sheriff had been shot, had his house bombed, and his wife and son had been killed."
(pp. 17-18)
The resemblance to Sheriff Buford Pusser seemed unmistakable.
Recently I spoke to Art Bourgeau about his book, and to my surprise he said, "The sheriff character wasn't modeled on Buford Pusser. Sheriff Pusser was a heroic figure, my character wasn't. The two things that could make you think it was Sheriff Pusser was his haircut and him fighting the mob. I used the Glen Campbell haircut which he and many other sheriffs of the time wore as a metaphor to show he was a very uptight, stressed-out, anal type of person. The job didn't make my character this way, it was his nature. You can tell a lot about a man by his haircut…As to the nastiness of the personality of Sheriff Casper Denny, that was not a reflection on Sheriff Pusser. Quite the contrary, I never met the man. As far as I know, he was a saint. The character of Sheriff Casper Denny was an extension of my own life. My father was a Tennessee Deputy Sheriff and he was a shit."
So for twenty years, I'd had that wrong. But by the time I found out, I'd learned to look at the world of my youth with a writer's detachment instead of a ex-pat's ambivalence. Hints of that world have shown up throughout my work, and will probably always do so.
And at least I was right about Reelfoot Lake. Bourgeau told me, "The idea of the lake comes from reading about Reelfoot Lake and thinking it must be one of the neatest places on earth, but I've never seen it. However, I have seen the bayou of Louisiana and the wetlands of New Jersey, so I am familiar with how it might look. Still, it is darn fascinating. The New Madrid Fault and all that. It is larger than history. It belongs in fantasy."
Fantasy, huh? Hey, I write fantasy. Hmm….