text:
\section*{Introduction}

In this paper we study monomial ideals
which are intersections of  powers of monomial prime ideals, and call them  {\em monomial ideals of intersection type}. Any squarefree monomial ideal is of intersection type since it is actually an intersection of monomial prime ideals. Obviously, among the non-radical monomial ideals, the monomial ideals of intersection type are closest to squarefree monomial ideals.  Hence one may expect that they are easier to understand than general monomial ideals. Indeed, various special types of such ideals have been studied in the literature, among them the so-called ideals of tetrahedral curves, see \cite{MN} and \cite{FMN}. In \cite{FV} the authors discuss the problem when a monomial ideal of intersection type is componentwise linear. In that paper monomial ideals of intersection type are called intersections of Veronese type ideals.

In Section 1 of the present paper we deal with the problem to characterize those monomial ideals which are of intersection type. The answer is given in Theorem~\ref{prime_intersection} where it is shown that a monomial ideal $I\subset S=K[x_1,\ldots,x_n]$ is of intersection type if and only if for each associated prime ideal $\pp$ of $I$ the minimal degree of a generator of the monomial localization $I(\pp)$ of $I$ is bigger than the maximal degree of a non-zero socle element of $S(\pp)/I(\pp)$. Moreover, it is shown that if $I$ is of intersection type,  its presentation as an intersection of powers of monomial prime ideals is uniquely determined. We call this presentation the {\em canonical primary decomposition} of $I$.  One should notice that an ideal of intersection type, if it has embedded prime ideals, may have primary decompositions, different from the canonical one. The exponents of the powers of the monomial prime ideals appearing in the intersection are bounded above. Indeed, if $I=\Sect_{\pp\in \Ass(S/I)}\pp^{d_\pp}$, then $d_\pp\leq \reg(I(\pp))$. We say that $I$ is of {\em strong intersection type}, if $d_\pp=\reg(I(\pp))$ for all $\pp\in \Ass(S/I)$, and show that this is the case if and only if $I(\pp)$ has a linear resolution for all $\pp\in \Ass(S/I)$. It is clear that any squarefree monomial ideal is of strong intersection type, while monomial ideals of intersection type with embedded prime ideals may or may not be of strong intersection type.

In Section 2 we consider classes of ideals which are of (strong) intersection type. It is shown in Proposition~\ref{polym_c} that any polymatroidal ideal is of strong intersection type, and in Theorem~\ref{primdec_polym} we show that the canonical primary decomposition of polymatroidal ideals  is given in terms of the rank function of the underlying polymatroid.  Unfortunately we do not have a uniform description of the associated prime ideals  in terms of the rank function.  However for polymatroidal ideals of Veronese type and for transversal polymatroidal ideals the canonical primary decomposition is made explicit, see Examples~\ref{known_polym}. On the other hand, a coherent primary decomposition of  any polymatroidal ideal, which however may not always be irredundant, is given in Theorem~\ref{redundant}. Among the ideals of Borel type, the principal Borel ideals are precisely those which are of strong intersection type, as shown in Proposition~\ref{borel_c}. In the remaining part of Section~2 we consider powers of edge ideals. In Theorem~\ref{depth_2power} we characterize,  terms of their underlying graph, those edge ideals   whose second power is of intersection type, and prove in Corollary~\ref{higher} that if the square of an edge ideal is not of intersection type, then so are all its higher powers.  It would be of interest to classify all graphs with the property that all its powers are of intersection type.

In the last section we list a few properties of ideals of intersection type. One nice property is that any ideal of intersection type is integrally closed, see Proposition~\ref{integrally_closed}.  In Corollary~\ref{supporting} the supporting hyperplanes of the Newton polyhedron of an ideal of intersection type are given in terms of its canonical primary decomposition. Finally in Theorem~\ref{symbolic_power} it is shown that if $I^k$ is of intersection type, then its  $\ell^{th}$ symbolic power $I^{(\ell)}$  of $I$ is contained in $I^k$, where $\ell = \reg(I^k)$. By using this fact and a result of Kodiyalam \cite{K}, and assuming that all powers of $I$ are of intersection type, we show in the second part of Theorem~\ref{symbolic_power} that $I^{(rk)}\subset I^k$ for all $k\gg 0$, where $r=\rho(I)+1$ and where  $\rho(I)$ is the minimum of $\theta(J)$  over all graded
reductions $J$ of $I$. Here $\theta(J)$ denotes the maximal degree of a generator in a minimal set of generators of $J$.

\section{When is a monomial ideal the intersection of  powers of monomial prime ideals? }

Let $S=K[x_1,\ldots,x_n]$ be the polynomial ring in $n$ variables over the field $K$ with graded maximal ideal $\mm=(x_1,\ldots,x_n)$, and let $I\subset S$ be a monomial ideal. In general, unless $I$ is squarefree,  the ideal $I$ can not be written as intersection of powers of monomial prime ideals. In the following we give a necessary and sufficient condition for $I$  to have such a presentation, and show that this presentation is unique once it is irredundant.
To describe the main result of this section we introduce some notation. Let $\pp$ be a monomial prime ideal, that is, $\pp$ is a prime ideal generated by a subset of the variables. As in \cite{HRV} we denote by $I(\pp)$ the monomial localization of $I$,  and by $S(\pp)$ the polynomial ring over $K$ in the variables which belong to $\pp$. Recall that $I(\pp)\subset S(\pp)$ is the monomial ideal which is obtained from $I$ as the image of the $K$-algebra homomorphism $\varphi\: S\to S(\pp)$ with $\varphi(x_i)=x_i$ if $x_i\in \pp$, and $\varphi(x_i)=1$,  otherwise. Monomial localizations are compatible with products and intersections. In other words, if $I$ and $J$ are monomial ideals, then $(IJ)(\pp)=I(\pp)J(\pp)$ and $(I\sect J)(\pp)=I(\pp)\sect J(\pp)$.

We denote by $V^*(I)$ the set of monomial prime ideals containing $I$. It is known that $\Ass(S/I) \subset V^*(I)$ (see for example \cite[Corollary 1.3.9]{HH}),  and that $\pp\in V^*(I)$ belongs to $\Ass(S/I)$ if and only if $\depth(S(\pp)/I(\pp))=0$. The latter is the case if and only if $I(\pp):\mm_\pp\neq I(\pp)$ where $\mm_\pp$ denotes the graded maximal ideal of $S(\pp)$. The $K$-vector space $(I(\pp):\mm_\pp)/ I(\pp)\subset S(\pp)/I(\pp)$ is called the {\em socle} of $S(\pp)/I(\pp)$, and denoted by $\Soc(S(\pp)/I(\pp))$. Thus $\pp\in \Ass(S/I)$ if and only if $\Soc(S(\pp)/I(\pp))\neq 0$. Since the colon ideal of monomial ideals is again a monomial ideal it follows that $\Soc(S(\pp)/I(\pp))$ is generated by the elements $u+I(\pp)$ where $u\in S(\pp)$ is a monomial with $u\not\in I(\pp)$ and  $\mm_\pp u\subset I(\pp)$. Since $\Soc(S(\pp)/I(\pp))$ is a finite dimensional graded $K$-vector space we may define the number
\[
\max(\Soc(S(\pp)/I(\pp)))=\max\{i\:\, (\Soc(S(\pp)/I(\pp)))_i\neq 0\}
\]
for any $\pp\in \Ass(S/I)$. We also set
\[
\min(M)=\min\{i\:\; M_i\neq 0\}
\]
for any finitely generated  graded $S(\pp)$-module $M$.  Finally we denote by $\wt{I(\pp)}$ the {\em saturation} of $I(\pp)$ which is defined to be the monomial ideal $\Union_k I(\pp): \mm_\pp^k$.

In the sequel we will also use the following notation: for a graded ideal $I$  and an integer $k\geq 0$ we denote by $I_{\geq k}$ the ideal generated by all graded components $I_i$ of $I$ with $i\geq k$.

\begin{Theorem}
\label{prime_intersection}
Let $I$ be a monomial ideal.
\begin{enumerate}
\item[(a)] The following conditions are equivalent:
\begin{enumerate}
\item[(i)] $I$ is an intersection of powers of monomial prime ideals;
\item[(ii)] for all $\pp\in\Ass(S/I)$ there exist positive integers $a_{\pp}$ such that $$I(\pp)=\wt{I(\pp)}\cap\mm_{\pp}^{a_{\pp}};$$
\item[(iii)] $\min(I(\pp))>\max(\Soc(S(\pp)/I(\pp)))$ for all $\pp\in\Ass(S/I)$.
\end{enumerate}
\item[(b)] Let $I=\Sect_{i=1}^r \pp_i^{d_i}$ be an irredundant presentation of $I$ as an intersection of powers of monomial prime ideals. Then  $\Ass(S/I)=\{\pp_1,\ldots,\pp_r\}$ and $$d_i=\max(\Soc(S(\pp_i)/I(\pp_i)))+1 \text{ for all } i=1,\ldots,r.$$
\end{enumerate}
\end{Theorem}
\begin{proof}
(a)  For the proof of (i) \implies (ii) let $I=\Sect_{i=1}^r \pp_i^{a_i}$. We may assume that this presentation is irredundant. Then $\Ass(S/I)=\{\pp_1,\ldots,\pp_r\}$. Fix some integer $j$. Then
\[
I(\pp_j)=\Sect_{i \atop \pp_i\subsetneq\pp_j}\pp_i(\pp_j)^{a_i}\cap\mm_{\pp_j}^{a_j}.
\]
It follows from this presentation that $\wt{I(\pp)}=\Sect_{i,\; \pp_i\subsetneq\pp_j}\pp_i(\pp_j)^{a_i}$.

The proof of (ii)\implies (iii) follows immediately once that it is noticed that (ii) is equivalent to saying that $I(\pp)=\wt{I(\pp)}_{\geq a_{\pp}}$ for all $\pp\in\Ass(I)$.  This identity implies that $\min(I(\pp))\geq a_{\pp}$,  and that the maximal degree of a non-zero element  $u+I(\pp)\in \Soc(S(\pp)/I(\pp))$, $u$ a monomial,  is at most $a_{\pp}-1$ since $u\in\wt{I(\pp)}\setminus I(\pp)$.

(iii)\implies (i): For all $\pp\in \Ass(S/I)$ let $d_\pp=\max(\Soc(S(\pp)/I(\pp)))+1$. We claim that $I=\Sect_{\pp\in \Ass(S/I)}\pp^{d_\pp}$. In order to prove this, we first show that $I\subset\Sect_{\pp\in \Ass(S/I)}\pp^{d_\pp}$. Indeed, let $\pp\in \Ass(S/I)$ and suppose that $I\not\subset \pp^{d_\pp}$. Then $I(\pp)\neq I(\pp)_{\geq d_p}$. It follows that $\min(I(\pp))< d_\pp$, a contradiction.

Next assume that $I$ is properly contained in $J=\Sect_{\pp\in \Ass(S/I)}\pp^{d_\pp}$. Then there exists $\pp\in\Min(J/I)\subset \Ass(S/I)$ such that $J(\pp)/I(\pp)=(J/I)(\pp)\neq 0$. Let $u\in J(\pp)\setminus I(\pp)$. Then $\deg u\geq d_\pp$ because $u\in\mm_\pp$. Since $\pp\in\Min(J/I)$, it follows that $J(\pp)/I(\pp)$ has finite length. Thus $\max(\Soc(S(\pp)/I(\pp)))\geq \deg u\geq d_\pp$, a contradiction.

\medskip
(b) Since this presentation is irredundant, it represents an irredundant primary decomposition of $I$. This shows that $\Ass(S/I)=\{\pp_1,\ldots,\pp_r\}$. Fix an integer $i$. Then $I(\pp_i)=\wt{I(\pp_i)}\cap\mm_{\pp_i}^{d_i}=\wt{I(\pp_i)}_{\geq d_i}$ and $I(\pp_i)$ is properly contained in $\wt{I(\pp_i)}$. Thus
\[
\max(\Soc(S(\pp_i)/I(\pp_i)))=\max(\Soc(\wt{I(\pp_i)}/I(\pp_i)))= \max(\Soc(\wt{I(\pp_i)}/\wt{I(\pp_i)}_{\geq d_i}))
\]
It follows that  $\max(\Soc(S(\pp_i)/I(\pp_i)))=d_i-1$, as desired.
\end{proof}

We call a monomial ideal satisfying the equivalent conditions of Theorem~\ref{prime_intersection} to be a monomial ideal of {\em intersection type}.
The presentation of an ideal of intersection type as given in Theorem~\ref{prime_intersection}(b) is called the {\em canonical primary decomposition} of $I$. Of course, a monomial ideal of intersection type, if it  has embedded prime ideals, may have also other primary decompositions than just the canonical one.

\begin{Example}
\label{stanleyreisner}
{\em Let $J$ be the Stanley-Reisner ideal associated to the natural triangulation of the real projective plane, that is,
\[
J = (xyz,xyt,xzu,xtv,xuv,yzv,ytu,yuv,ztu,ztv)\subset S=K[x,y,z,t,u,v].
\]
We apply our Theorem~\ref{prime_intersection} to show that the ideal $J^2$ is not of intersection type. For this, note that $\mm\in \Ass(S/J^2)$ with $xyztuv+J^2$ a non-zero socle element of $S/J^2$ of degree $6$. Since $J^2$ is generated in degree $6$,  Theorem~\ref{prime_intersection}(a) implies that $J^2$ is not of intersection type.}

\end{Example}

If the monomial ideal $I$ is of intersection type, then the powers of the prime ideals in the intersection are naturally bounded. More precisely we have

\begin{Theorem}\label{bounded}
Let $I$ be a monomial ideal of intersection type with presentation
$
I=\Sect_{\pp\in \Ass(S/I)}\pp^{d_\pp}.
$
Then the following statements hold:
\begin{enumerate}
\item[(a)] $d_\pp\leq \reg(I(\pp))$ for all $\pp\in \Ass(S/I)$;
\item[(b)] $d_\pp= \reg(I(\pp))$ if and only if $I(\pp)$ has a linear resolution.
\end{enumerate}
In particular, $I=\Sect_{\pp\in \Ass(S/I)}\pp^{\reg(I(\pp))}$  if and only if $I(\pp)$ has linear resolution for all $\pp\in \Ass(S/I)$.
\end{Theorem}

\begin{proof} (a) We consider the following short exact sequence
\begin{eqnarray}
\label{sequence}
0\To \wt{I(\pp)}/I(\pp) \To S(\pp)/I(\pp) \To S(\pp)/\wt{I(\pp)}\To 0.
\end{eqnarray}
Since $\wt{I(\pp)}/I(\pp)$ is a finite length module,  \cite[Corollary 20.19]{Ei} combined with Theorem~\ref{prime_intersection}  implies
\begin{eqnarray}
\label{last}
d_\pp&=&\max(\Soc(\wt{I(\pp)}/I(\pp)))+1=\reg(\wt{I(\pp)}/I(\pp))+1\\
&\leq & \reg(S(\pp)/I(\pp))+1=\reg(I(\pp)). \nonumber
\end{eqnarray}

(b) By Theorem~\ref{prime_intersection}  we have $I(\pp)=\wt{I(\pp)}_{\geq d_\pp}$. Thus if we assume that $I(\pp)$ has a linear resolution it follows that
$\reg(I(\pp))=\min(I(\pp))=d_\pp$.

Conversely, assume that $\reg(I(\pp))=d_\pp$. Then  $I(\pp)=\wt{I(\pp)}_{\geq \reg(I(\pp))}$. Thus by \cite[Theorem 1.2]{EG} it is enough to show that
$\reg(\wt{I(\pp)})\leq \reg(I(\pp))$. But this follows again from \cite[Corollary 20.19]{Ei} according to which
\[
\reg(S(\pp)/I(\pp))=\max\{\reg(\wt{I(\pp)}/I(\pp)),\reg(S(\pp)/\reg(\wt{I(\pp)})\}\geq \reg(S(\pp)/\reg(\wt{I(\pp)}).
\]
\end{proof}

A monomial ideal satisfying the equivalent conditions of Theorem~\ref{bounded} is said to be of {\em strong intersection type}. Obviously, any squarefree monomial ideal is of strong intersection type.

\begin{Example}
{\em  The ideal $$I=(x,y)\cap (x,z)\cap (x,t)\cap(x,y,z,t)^2=(x^2,xy,xz,xt,yzt)$$
of intersection type, and  $\mm=(x,y,z,t)\in \Ass(S/I)$.  But $I$ is not generated in a single degree, let alone  has a linear resolution. Therefore, $I$ is not of strong intersection type. }
\end{Example}

\section{Classes of monomial ideals of (strong) intersection type}

In this section we consider classes of ideals which are of intersection type or strong intersection type.

\medskip
\noindent
{\em  Polymatroidal ideals.} We fix a field $K$, and let $\mathcal P$ be a discrete polymatroid on the ground set $[n]$ of rank $d$, see \cite{HH2} where  these concepts are explained. The polymatroidal ideal associated with $\mathcal P$ is the monomial ideal $I\subset K[x_1,\ldots,x_n]$ generated by all monomials $\xb^\ub$ where $\ub\in B({\mathcal P})\subset \NN^n$. Here $B({\mathcal P})$ denotes the set of bases of $\MP$, namely  the set of all $\ub\in \MP$  whose modulus  $|\ub|=\sum_i u_i$ is maximal. All bases $\ub\in B(\MP)$ have the same modulus, namely $d$, which is defined to be the rank of $\MP$. In particular, $I$ is generated in the single degree $d$.

\begin{Proposition}\label{polym_c}
Any polymatroidal ideal is of strong intersection  type.
\end{Proposition}

\begin{proof}
We recall that a polymatroidal ideal has linear quotients \cite[Theorem 5.2]{CH} and thus has a linear resolution. Moreover, $I(\pp)$ is polymatroidal for any monomial prime ideal $\pp$ which contains $I$ (see \cite[Corollary 3.2]{HRV}). In particular, $I(\pp)$ has a linear resolution for all   $\pp \in \Ass(S/I)$. Thus Theorem~\ref{bounded} yields the desired conclusion.
\end{proof}

In \cite{BaHe} it is conjectured that a monomial ideal is polymatroidal, if all its monomial localizations have a linear resolution. The following example shows that  it does not suffice to require  that $I$ is of strong intersection type to conclude that $I$ is polymatroidal, that is, to require that $I(\pp)$ has a linear resolution for all $\pp\in\Ass(S/I)$.

\begin{Example}
\label{thirdpower} {\em
 Let $J$ be the monomial ideal of  Example~\ref{stanleyreisner}, and let $\pp\in V^*(J^3)$. We claim that $J^3(\pp)$ has a linear resolution if and only if $\pp\in \Ass(S/J^3)$. Calculations with Singular \cite{Si} show that $\Ass(S/J^3)$ consists of all the  prime ideals of height $3,5$ and $6$ which belong to $V^*(J^3)$, altogether these are $17$ prime ideals. Moreover we have the following irredundant primary decomposition of $J^3$
\[
J^3=(v,u,y)^3\cap (v,u,x)^3\cap (v,t,z)^3\cap (v,t,x)^3\cap (v,z,y)^3\cap (u,t,z)^3\cap (t,u,y)^3\cap (u,z,x)^3
\]
\[
\cap (t,x,y)^3\cap (z,x,y)^3\cap (x,y,z,t,v)^6\cap (x,y,z,t,u)^6\cap (x,y,z,v,u)^6\cap
\]
\[
(x,y,t,v,u)^6\cap (x,z,t,v,u)^6\cap (y,z,t,v,u)^6\cap (x,y,z,t,v,u)^9.
\]
We notice that $\mm\in\Ass(S/J^3)$, and by \cite[Corollary 3.3]{Bo} we know that $J^3(=J^3(\mm))$ has a linear resolution. Next consider any prime ideal $\pp\in V^*(J^3)$ of height $5$. Then $J^3(\pp)$ is just the third power of the edge ideal of the cycle of length $5$. For example, if $\pp=(x,y,z,t,u)$ then $$J^3(\pp)=J(\pp)^3=(xt,tz,zy,yu,ux)^3.$$ By Singular one can check that also in this case $J^3(\pp)$ has linear resolution. Finally let $\pp\in V^*(J^3)$ be  of height $3$. Then $\pp$ is a minimal prime ideal of $J^3$. Hence  $J^3(\pp)=\mm_{\pp}^3$, and thus  has a linear resolution. We conclude that  $J^3(\pp)$ has a linear resolution for any associated prime ideal $\pp\in\Ass(S/J^3)$, and hence  $J^3$ is an ideal of strong intersection type.

On the other hand, if $J^3$ would be a polymatroidal ideal, then $J^3(\pp)$ would be polymatroidal for any monomial prime ideal $\pp\in V^*(J^3)$, see \cite[Corollary 3.2]{HRV}. Any height $4$ monomial prime ideal $\pp$ belongs to $V^*(J^3)$ and contains exactly two minimal prime ideals associated to $J^3$. It follows that $J^3(\pp)$ is no longer generated in a single degree and thus can not be polymatroidal. To exemplify this, let $\pp=(x,y,z,t)$. Then $J^3(\pp)=(t,x,y)^3\cap (z,x,y)^3$ is minimally generated in degrees $3,4,5$ and $6$.

Our discussion showed that $\Ass(S/J^3)$ is a proper subset of $V^*(J^3)$, and  that $J^3(\pp)$ has a linear resolution if and only  $\pp\in\Ass(S/J^3)$. }
\end{Example}

Next we want to describe the canonical primary decomposition of a polymatroidal ideal.  Let $\mathcal P$ be a discrete polymatroid on the ground set $[n]$, and denote by $2^{[n]}$ the set of all subsets of $[n]$. For a vector $\ub\in \ZZ^n$ and  $F\in
2^{[n]}$ we set $\ub(F)=\sum_{i\in F}u_i$.

The ground set {\em rank function} of ${\mathcal  P}$ is the function $\rho\: 2^{[n]}\to \ZZ_+$  defined by
setting
\[
\rho(F) = \max\{\ub(F)\:\; \ub\in B({\mathcal  P})\}\quad \text{for all $F\in 2^{[n]}$, $F\neq \emptyset$, }
\]
together with $\rho(\emptyset) = 0$.

\begin{Lemma}
\label{rank}
Let $\mathcal P$ be a discrete polymatroid of rank $d$ on the ground set $[n]$ and $I$ the polymatroidal ideal associated with $\mathcal P$. Furthermore, let $F\subset [n]$ and $\pp_F$ be the monomial prime ideal generated by the variables $x_i$ with $i\in F$. Then $I(\pp_F)$ is generated in degree $d-\rho([n]\setminus F).$
\end{Lemma}

\begin{proof}
As observed before in the proof of Proposition~\ref{polym_c}, the ideal $I(\pp_F)$ is polymatroidal for all $\pp_F$ which contain $I$. If $I\not\subset \pp_F$, then $I(\pp_F)=S(\pp_F)$ which by definition we may also consider as a polymatroidal ideal. The ideal $I(\pp_F)$ is generated in a single degree, say $t_F$. By the definition of monomial localization we have that  $t_F=\min\{\ub(F)\:\; \ub\in B(\MP)\}$. Since
\[
\min\{\ub(F)\:\; \ub\in B(\MP)\}=d-\max\{\ub([n]\setminus F)\:\; \ub\in B(\MP)\},
\]
the  desired formula follows.
\end{proof}

\medskip
The canonical primary decomposition of a polymatroidal ideal is now given as follows:

\begin{Theorem}\label{primdec_polym}
Let $\mathcal P$ be a  discrete polymatroid of rank $d$ with rank function  $\rho$, and let $I$ be the  polymatroidal ideal associated with  $\mathcal P$.  Then
\[
I=\Sect_{\pp_F\in\Ass(S/I)}
\pp_F^{d-\rho([n]\setminus F)}.
\]
is the canonical  primary decomposition of $I$.
\end{Theorem}

\begin{proof} Let $\pp_F\in \Ass(S/I)$. Then  $\reg I(\pp_F)$ is equal to the common degree of the generators of $I(\pp_F)$, since $I(\pp_F)$ has a linear resolution. By Lemma~\ref{rank}, this common degree is $d-\rho([n]\setminus F)$. Since $I$ is of strong intersection type, the desired conclusion follows from Theorem~\ref{bounded}.
\end{proof}

In general it is not so easy to identify the associated prime ideals of a polymatroidal ideal. The complete answer is known for ideals of Veronese type as well as for polymatroidal ideals associated with transversal polymatroids. We describe these cases in the following examples.

\begin{Examples}\label{known_polym}
{\em (a) Let $d, a_1,\ldots,a_n$ be positive integers. The ideal $I_{d;a_1,\ldots,a_n}$ generated by all monomials $x_1^{c_1}\cdots x_n^{c_n}$  of degree $d$ with $c_i\leq a_i$ for $i=1,\ldots,n$ is called of {\em Veronese type}. Ideals of Veronese type are polymatroidal.

Let $I= I_{d;a_1,\ldots,a_n}$ be the ideal of Veronese type with the property that $d\geq a_i$ for all $i$ and $a_1,\ldots,a_n\geq 1$. The set of associated prime ideals is described in \cite[Proposition 3.1]{Vl} as follows
\[
\Ass(S/I)=\{\pp_F\:\ \sum_{i=1}^n a_i\geq d-1+|F| \ \text{and} \ \sum_{i\notin F} a_i\leq d-1\}.
\]
Thus  by applying Theorem~\ref{primdec_polym} the canonical  primary decomposition of $I$ is given as
\[
I=\Sect_{\pp_F\in\Ass(S/I)}\pp_F^{d-\sum_{i\notin F}a_i},
\]
since the rank function of the corresponding polymatroid $\MP$ is given by $\rho(F)=\min\{d,\sum_{i\in F}a_i\}$ for all $F\subset [n]$.

\medskip
Consider for example  $I=I_{4;3,2,1}\subset S=K[x_1,x_2,x_3]$. Then  $$I=(x_1^3x_2,x_1^3x_3,x_1^2x_2^2,x_1^2x_2x_3,x_1x_2^2x_3),$$ and the above formulas  yield
\[
\Ass(S/I)=\{(x_1),(x_2,x_3),(x_1,x_3),(x_1,x_2),(x_1,x_2,x_3)\},
\]
and  the canonical  primary decomposition
\begin{eqnarray*}
I=(x_1)\cap(x_2,x_3)\cap(x_1,x_3)^2\cap(x_1,x_2)^3\cap(x_1,x_2,x_3)^4.
\end{eqnarray*}

(b) Let $I=\pp_{F_1}\cdots \pp_{F_d}\subset K[x_1,\ldots,x_n]$ be a transversal polymatroidal ideal. Without loss of generality we may assume that $\Union_{i=1}^d F_i=[n]$. The set of associated prime ideals of $I$ is described in \cite[Exercise 3.9]{Ei} as well as in \cite[Theorem 4.7]{HRV}, and an irredundant primary decomposition for $I$ is given in \cite[Corollary 4.10]{HRV}. Both descriptions are given in terms of a graph $G_I$ attached to the ideal $I$, see \cite[Section 4]{HRV}. The graph $G_I$ is defined on the vertex set $[d]$ with  $\{i,j\}$  an edge of  $G_I$  if $F_i\cap F_j\neq\emptyset$. In \cite[Theorem 4.7]{HRV} it is shown  that the associated prime ideals  of $I$ corresponded to the trees of the graph $G_I$ in the following way:
$\pp_F\in\Ass(S/I)$ if and only if there exists a tree $\MT$ such that  $\pp_F=\pp_{\mathcal T}$.

In the following we show that the  irredundant primary decomposition given in terms of the rank function  as described in  Theorem~\ref{primdec_polym} coincides with  the one given in \cite[Corollary 4.10]{HRV}. There the irredundant  primary decomposition of $I$ is given as
\[
I=\Sect_{\pp_F\in\Ass(S/I)}\pp_F^{a_F},
\]
where $a_F$ is the number of vertices of a tree $\MT$ in $G_I$ which is maximal with respect to the property  that $\pp_F=\pp_{\mathcal T}$, where $\pp_{\mathcal T}=\sum_{i\in V(\mathcal T)}\pp_{F_i}$.

On the other hand,  let $\rho$ be the rank function of the transversal polymatroid attached  to $I$.  In \cite[Section 9]{HH2} it is shown that
\[
\rho(F)=|\{i:\ F\cap F_i\neq\emptyset\}| \quad \text{ for all } F\subset [n].
\]
It follows that
\begin{eqnarray}
\label{d}
d=|\{i:\  F_i\subset  F\}|+\rho([n]\setminus F).
\end{eqnarray}
Now let $\MT$ be a tree in $G_I$ which is maximal with respect to the property  that $\pp_F=\pp_{\mathcal T}$. Suppose  we have shown that
\[
|\{i:\  F_i\subset F\}|=|\{i\:\; i\in V(\MT)\}|,
\]
then it will follow from \eqref{d} that $a_F=d-\rho([n]\setminus F)$ for all $F$ which indeed  we want to show.

The inequality $|\{i:\  F_i\subset F\}|\geq |\{i\:\; i\in V(\MT)\}|$ is obvious, since $F_i\subset F$ for all $i\in V(\MT)$. Suppose this inequality is strict. Then there exists $j\not \in V(\MT)$ with $F_j\subset F$. Since $F=\Union_{i\in V(\MT)}F_i$, we then notice that $F_j\sect F_i\neq \emptyset$ for some $i\in V(\MT)$, contradicting  the maximality of  $\MT$.
}
\end{Examples}

A redundant primary decomposition of a polymatroidal ideal, which in some cases is even irredundant and which is given only in terms of the rank function of the corresponding polymatroid can be described  as follows:

\medskip
Let $\MP$ be a discrete polymatroid of rank $d$ on the ground set $[n]$ and $\rho\: 2^{[n]}\to \ZZ_+$ its rank function. Then $\rho$ satisfies the following conditions:
\begin{enumerate}
\item[(i)] $\rho(\emptyset)=0$;
\item[(ii)] $\rho(F)\leq \rho(G)$ for all $F\subset G\subset [n]$;
\item[(iii)] $\rho(F)+\rho(G)\geq \rho(F\sect G)+ \rho(F\union G)$ for all $F,G\in 2^{[n]}$.
\end{enumerate}

We define the {\em complementary rank function}  $\tau\: 2^{[n]}\to \ZZ_+$ by setting $\tau(F)=d-\rho([n]\setminus F)$ for all $F\in 2^{[n]}$. The function $\tau$ also satisfies the properties (i) and (ii), but instead of (iii) one has:
\begin{enumerate}
\item[(iii')] $\tau(F)+\tau(G)\leq \tau(F\sect G)+ \tau(F\union G)$ for all $F,G\in 2^{[n]}$.
\end{enumerate}

A subset $F\subset [n]$ will be called {\em  $\tau$-closed}, if $\tau (G)<\tau(F)$ for any proper subset $G$ of $F$, and $F$ will be called {\em $\tau$-separable} if there exist non-empty subsets $G$ and $H$ of $F$ with $G\sect H=\emptyset$ and $G\union H=F$ such that $\tau(G)+\tau(H)=\tau(F)$. If $F$ is not $\tau$-separable, then it is called $\tau$-inseparable. Corresponding concepts exist for $\rho$.

\begin{Theorem}
\label{redundant}
Let $I$ be a polymatroidal ideal associated with the discrete polymatroid $\mathcal P$ with complementary rank function $\tau$. Then
\[
I=\Sect_{F}\pp_F^{\tau(F)},
\]
where the intersection is taken over all $F\subset [n]$ which are $\tau$-closed and $\tau$-inseparable.
\end{Theorem}

\begin{proof}
We first show that
$I=\Sect_{F\subset [n]}\pp_F^{\tau(F)},$
where we take the intersection over all subsets $F$ of $[n]$.  Indeed, due to Theorem~\ref{primdec_polym} it suffices to show that $I\subset \pp_F^{\tau(F)}$ for all $F\subset [n]$. In order to show this let $\xb^{\ub}\in G(I)$. Then  $\ub([n])=d$ and $\ub(F)\leq\rho(F)$ for all subsets $F\subset [n]$. Therefore,  $d-\ub(F)=\ub([n]\setminus F)\leq\rho([n]\setminus F)$ and this  implies that $\ub(F)\geq \tau(F)$. In other words, $\xb^{\ub}\in \pp_F^{\tau(F)}$.

In the next step we show that whenever $F$ is not  $\tau$-closed or is  $\tau$-separable, then  $\pp_F^{\tau(F)}$ may be omitted in the intersection $\Sect_{F\subset [n]}\pp_F^{\tau(F)}$ without changing it. Indeed, if $F$ is not  $\tau$-closed, then there exists a proper subset $G$ of $F$ with $\tau(G)=\tau(F)$. It follows that $\pp_G^{\tau(G)}\subset \pp_F^{\tau(F)}$, and hence $\pp_F^{\tau(F)}$ may be omitted. On the other hand, if $F$ is separable, then there exist non-empty subsets  $G$ and $H$ of $F$ with $G\sect H=\emptyset$ and $G\union H=F$ such that $\tau(G)+\tau(H)=\tau(F)$. Thus we see that
\[
\pp_G^{\tau(G)}\sect \pp_H^{\tau(H)}= \pp_G^{\tau(G)} \pp_H^{\tau(H)}\subset (\pp_G+\pp_H)^{\tau(G)+\tau(H)}=\pp_F^{\tau(F)},
\]
and hence again $\pp_F^{\tau(F)}$ may be omitted in the intersection.
\end{proof}

\begin{Example}
{\em The intersection given in Theorem~\ref{redundant} may be an irredundant primary decomposition but also  very far of being irredundant. Consider for example the Veronese type ideal $I_{4;3,2,1}$ analyzed in Example~\ref{known_polym}(a). The rank function $\rho$ of the associated discrete polymatroid is given as follows: $\rho(\emptyset)=0$, $\rho(\{1\})=3$, $\rho(\{2\})=2$, $\rho(\{3\})=1$, $\rho(\{1,2\})=4$, $\rho(\{1,3\})=4$, $\rho(\{2,3\})=3$ and $\rho(\{1,2,3\})=4$. Thus the complementary rank function is the following: $\tau(\emptyset)=0$, $\tau(\{1\})=1$, $\tau(\{2\})=0$, $\tau(\{3\})=0$, $\tau(\{1,2\})=3$, $\tau(\{1,3\})=2$, $\tau(\{2,3\})=1$ and $\tau(\{1,2,3\})=4$. One can easily check that there are no $\tau$-inseparable subsets of $[3]$ and the $\tau$-closed subsets of $[3]$ are: $\{1\}, \{1,2\}, \{1,3\}, \{2,3\},\{1,2,3\}$. Applying now Theorem~\ref{redundant} we obtain that
\[
I_{4;3,2,1}=(x_1)\cap(x_2,x_3)\cap(x_1,x_3)^2\cap(x_1,x_2)^3\cap(x_1,x_2,x_3)^4.
\]
This is the canonical primary decomposition of $I_{4;3,2,1}$ presented already in Example~\ref{known_polym}(a). Thus in this case Theorem~\ref{redundant} yields an irredundant primary decomposition.

On the other hand if we consider the squarefree Veronese ideal $I=I_{n-1;1,\ldots,1}\subset K[x_1,\ldots,x_n]$ we will see that the intersection given in Theorem~\ref{redundant} is very far of being an irredundant primary decomposition. Indeed, the canonical primary decomposition of $I$ is
\[
I=\Sect_{1\leq i<j\leq n} (x_i,x_j).
\]
The rank function $\rho$ of the associated discrete polymatroid of $I$ can be easily computed:
\[
\rho(F)=|F| \quad \text{ for any } F\subsetneq [n],
\]
and $\rho([n])=n-1$. Thus we obtain that $\tau(\emptyset)=0$ and $\tau(F)=|F|-1$ for any nonempty subset $F\subset [n]$. This implies that every set $F\subset [n]$ with $|F|\geq 2$ is $\tau$-closed and thus the intersection given by Theorem~\ref{redundant} is the following
\[
I=\Sect_{i=2}^n (\Sect_{F\subset [n], |F|=i}\pp_F^{i-1}).
\]
}
\end{Example}

\medskip
\noindent
{\em Ideals of Borel type.}  We recall  (see \cite[Definition 2.1]{HPV}) that an {\em ideal of Borel type} is a monomial ideal $I\subset S$ such that
\[
I:x_j^{\infty}=I:(x_1,\ldots,x_j)^\infty \quad \text{ for all } j=1,\ldots,n.
\]
For such an ideal is known that the set $\Ass(S/I)$  is totally ordered. More precisely, if $\pp\in\Ass(S/I)$ then $\pp=(x_1,\ldots,x_j)$ for some $j=1,\ldots,n$,  see \cite[Proposition~4.2.9]{HH}.

The {\em principal Borel ideal} generated by the monomial $u$ is the smallest Borel ideal containing $u$, and is denoted $\langle u\rangle$.

\begin{Proposition}\label{borel_c}
Let $I$ be an ideal of Borel type. The following conditions are equivalent:
\begin{enumerate}
\item[(a)] $I$ is of strong intersection type;
\item[(b)] $I$ is of intersection type;
\item[(c)] $I$ is principal Borel.
\end{enumerate}
\end{Proposition}
\begin{proof}
(a) \implies (b) is obvious. For (b) \implies (c) assume that $I=\Sect_{i=1}^r \pp_i^{d_i}$ is an irredundant primary decomposition of $I$. Since $I$ is of Borel type it follows from the previous comments that we may assume that $\pp_1\subset\pp_2\subset\cdots\subset\pp_r$. Therefore we must have $d_1<d_2<\cdots<d_r$, otherwise the given primary decomposition would not be irredundant. We claim that $I$ is the principal Borel ideal generated by the monomial $$u:=x_{n_1}^{d_1}x_{n_2}^{d_2-d_1}\cdots x_{n_s}^{d_s-d_{s-1}},$$
where for all $i=1,\ldots,r$, $x_{n_i}$ is the variable of highest index appearing in the minimal system of generators of $\pp_i$. Indeed, note first that  $\langle u\rangle =\pp_1^{d_1}\pp_2^{d_2-d_1}\cdots \pp_s^{d_s-d_{s-1}}$. Thus the  ideal $\langle u\rangle$ is a transversal polymatroidal ideal and applying either \cite[Corollary 4.10]{HRV} or \cite[Theorem 4.3]{FMS} we obtain that $\langle u\rangle =\Sect_{i=1}^r \pp_i^{d_i}$ is the irredundant primary decomposition of $\langle u \rangle$.  Therefore,  $I=\langle u \rangle$. In order to prove the implication (c) \implies (a) we use the fact that a principal Borel ideal is polymatroidal and by Proposition~\ref{polym_c} we obtain the desired conclusion.
\end{proof}

\begin{Remark}
{\em (a) Every monomial ideal of the form $I=\Sect_{i=1}^r \pp_i^{a_i}$ with the property that $\pp_i\not\subset \pp_j$ for all $i\neq j$ is of strong intersection type. In particular all squarefree monomial ideals are of strong intersection type, as we already noticed before.

(b) Let $\pp_1\subset \pp_2\subset\cdots\subset \pp_s$ be a chain of monomial prime ideals in $S$ and $d_1<d_2<\cdots<d_s$ be a sequence of positive integers. Then the ideal $I=\Sect_{i=1}^r \pp_i^{d_i}$ is a principal Borel ideal after a relabeling of the variables and by Proposition~\ref{borel_c}, the ideal  $I$ is of strong intersection type.

}
\end{Remark}

\medskip
\noindent
{\em Powers of edge ideals.} Let $G$ be a simple graph on the vertex set $[n]$, that is,  a graph without loops and multiple edges. The edge ideal associated with $G$ is the monomial ideal $I(G)\subset K[x_1,\ldots,x_n]$ generated by the monomials $x_ix_j$ corresponding to the edges $\{i,j\}$ of $G$. Here we study the question under which conditions on $G$, powers of $I(G)$ are of (strong) intersection type. We recall that edge ideals have the persistence property, that is,  $\Ass(S/I(G)^k)\subset\Ass(S/I(G)^{k+1})$ for all $k\geq 1$. This is shown  in \cite[Theorem 2.15]{MMV}.

\medskip
We begin with some examples.

\begin{Example}{\em
(1) Let $G$ be a graph such that $\Ass(S/I(G))=\Ass(S/I(G)^k)$ for all $k$. Then $I(G)^k$ is of strong intersection type. This situation is given for example in the case that  $G$ is bipartite,  as shown in \cite[Theorem 5.9]{SVV}.

(2) Let $G$ be a graph with the property that $I(G)^k$ has a linear resolution for all $k\geq 2$ and such that $\Ass(S/I^k) \subset \Ass(S/I)\union \{\mm\}$ for all $k$. Then Theorem~\ref{bounded} implies that $I(G)^k$ is of strong intersection type for all $k$.
The 5-cycle  $C_5$ satisfies this condition. Indeed,  it follows from \cite[Theorem 6.12]{B} that $I(C_5)^k$ has a linear resolution for all $k\geq 2$, and from
\cite[Lemma 3.1]{CMS} that $\Ass(S/I(C_5)^k) \subset \Ass(S/I(C_5))\union \{\mm\}$ for all $k$.

(3) Let $G=C_{2k+1}$ be the odd cycle of length $2k+1$ with $k\geq 3$. Since $G^c$ contains an induced $4$-cycle, it follows that $I(G)^s$ has no linear resolution for any $s\geq 1$, see \cite[Proposition 1.8]{NP}. On the other hand it is  known that $\Ass(S/I(G)^s)=\Ass(S/I(G))$ for $s\leq k$ and $\Ass(S/I(G)^{s})=\Ass(S/I(G))\cup\{\mm\}$ for $s>k$, see \cite[Lemma 3.1]{CMS}. It follows that $I(G)^s$ is of strong intersection type for $s\leq k$, and not of strong intersection type for $s>k$. We do not know whether or not $I(G)^s$ is of intersection type for all $s\geq 1$.

}
\end{Example}

\medskip
Let $G$ be a graph. We call a $3$-cycle $C$ of $G$ {\em central} if all vertices of $G$ are neighbors of $C$.

A slight generalization of the \cite[Theorem 2.1]{HH3} is the following lemma, which we shall need in the proof of the next theorem.

\begin{Lemma}\label{depth_2}
Let $G$ be a graph on the vertex set $[n]$  and let $$I\subset R=K[x_1,\ldots,x_n,y_1,\ldots,y_k]$$ be the monomial ideal $I=(y_1,\ldots,y_k,I(G))$, where $I(G)\subset K[x_1,\ldots,x_n]$. The following conditions are equivalent:
\begin{enumerate}
\item[(a)] $\depth(R/I^2)=0$,
\item[(b)] $G$ is a connected graph containing a central $3$-cycle.
\end{enumerate}
If the equivalent conditions hold and $C$ is a central $3$-cycle of $G$ with $V(C)=\{i,j,k\}$, then $x_ix_jx_k+I^2$ is a non-zero socle element of $R/I^2$.
\end{Lemma}

\begin{proof}
(a)$\Rightarrow$(b): Since $\depth(R/I^2)=0$, there exists a monomial $u\notin I^2$ such that $u\cdot(x_1,\ldots,x_n,y_1,\ldots,y_k)\subset I^2$.  By \cite[Corollary 1.2]{HH3} we know that $u$ is a squarefree monomial. We claim that $u$ is not divisible by any $y_j$. Assuming the claim proved, it follows
that $u\notin I(G)^2$ and $u\cdot(x_1,\ldots,x_n)\subset I(G)^2$. Therefore $\depth(K[x_1,\ldots,x_n]/I(G)^2)=0$, and by \cite[Theorem 2.1.]{HH3} we obtain the desired conclusion.

In order to prove the claim we argue by contradiction. Without loss of generality we may assume that $y_1$ divides $u$. Since $u\notin I^2$ it follows that $u$ can not be further divisible by any $y_i$. Thus $u=x_{i_1}\cdots x_{i_s}y_1$ for some integers $1\leq i_1<\ldots<i_s\leq n$ with $s\geq 1$ because $y_1$ does not belong to the socle of $R/I^2$. Furthermore for every integers $p\neq q$ we have $\{i_p,i_q\}$ is not an edge of $G$, otherwise $u\in y_1I(G)$, a contradiction. This implies that $ux_{i_1}$ is not divisible by any of the minimal generators of $y_1I(G)$ or $I(G)^2$ and therefore $ux_{i_1}\notin I^2$, a contradiction. Hence our claim is proved and we are done.

(b)$\Rightarrow$(a): After a relabeling of the vertices of $G$ we may assume that the $3$-cycle has the edges $\{1,2\},\{1,3\}$ and $\{2,3\}$. It follows from the proof of \cite[Theorem 2.1]{HH3} that the monomial $u=x_1x_2x_3$ satisfies $u\notin I(G)^2$ and $u\in I(G)^2:(x_1,\ldots,x_n)$. It follows that $u\notin I^2$. Moreover, since $uy_i\in y_iI(G)\subset I^2$ for all $i=1,\ldots,k$ it follows that $u\cdot(x_1,\ldots,x_n,y_1,\ldots,y_k)\subset I^2$, and thus $\depth(R/I^2)=0$.
\end{proof}

We want to point out that the proof of Lemma~\ref{depth_2} shows  that if $\depth(R/I^2)=0$ then there exists a socle element of degree $3$.

\begin{Theorem}
\label{depth_2power}
Let $G$ be a graph on the vertex set $[n]$ with no isolated vertices, and let  $I=I(G)$ be the  edge ideal of $G$ in $S= K[x_1,x_2,\ldots,x_n]$. The following conditions are equivalent:
\begin{enumerate}
\item[(a)] $I^2$ is of intersection type,
\item[(b)] $\Ass(S/I^2)\subset \Ass(S/I)\cup\{\mm\}$,
\item[(c)] All $3$-cycles of the graph $G$ are central.
\end{enumerate}
\end{Theorem}

\begin{proof}
We prove the theorem by showing that (a)$\Rightarrow$(b)$\Rightarrow$(c)$\Rightarrow$(a).

(a)$\Rightarrow$(b): Since $\Ass(S/I)=\Min(S/I)\subset\Ass(S/I^k)$ for all $k\geq 1$, it follows that $\Ass(S/I)\subset\Ass(S/I^2)$. Assume now that there exists a monomial prime ideal $\pp\in\Ass(S/I^2)$ which does not belong to $\Ass(S/I)\cup\{\mm\}$. If $I(\pp)$ is only generated by variables, then $I(\pp)=\pp S(\pp)$, because otherwise  $\depth(S(\pp)/I(\pp)^2)>0$. It follows that   $\pp\in\Ass(S/I)$, a contradiction. Therefore, $I(\pp)$ contains also generators of degree $2$, and we may apply Lemma~\ref{depth_2}, and  conclude  that $\max(\Soc(S(\pp)/I(\pp)^2))\geq 3$. On the other hand,  by the assumption on $G$, each variable $x_i$ divides some generator of $I$.  Since at least one variable is mapped to 1 in $S(\pp)$, it follows that  $I(\pp)$ contains also variables, and hence  $\min(I(\pp)^2)=2$.  By  Theorem~\ref{prime_intersection} this contradicts our assumption that $I^2$ is of intersection type.

(b)$\Rightarrow$(c): Suppose that there exists a $3$-cycle $C$ of $G$  which is not central. Let $D$ be the set of vertices of $G$ which are not neighbors of $C$, and let $\pp$ be the monomial prime ideal generated by the variables $x_i$ with $i\not\in  D$.  Then $I(\pp)=I(H)+Q$ where $Q$ is generated by a set $X$ of variables,  and where $I(H)$ is an edge ideal in a set of variables disjoint from $X$ with  $H$ a connected graph containing the $3$-cycle $C$ which is central in $H$. Lemma~\ref{depth_2} implies that $\depth(S(\pp)/I(\pp)^2)=0$. It follows that $\pp\in \Ass(S/I^2)$. The prime ideal $\pp$ is not a  minimal prime ideal of $I$  since it contains monomial generators of $I$ of degree 2, and it is not maximal since $D\neq \emptyset$. This is a contradiction.

(c)$\Rightarrow$(a): Let $\pp\neq \mm$ be a monomial prime ideal containing $I^2$, and write  $I(\pp)=I(H)+Q$ as in the proof of (b)\implies (c).  This time however,  since all $3$-cycles of $G$  are central, it follows that $H$ contains no $3$-cycle at all. Thus from Lemma~\ref{depth_2} one deduces that $\depth(S(\pp)/I(\pp)^2)=0$ if and only if $I(\pp)=\pp S(\pp)$, in which case $\pp$ is a minimal prime ideal. On the other hand, if $\mm\in \Ass(S/I^2)$, then $\depth(S/I^2)=0$. By \cite[Corollary 1.2]{HH3},  $\Soc(S/I^2)$ is generated by elements of the form $u+I^2$ where $u$ is a squarefree monomial. Let $u+I^2$ be such a non-zero socle element, and let $H$ be the induced subgraph of $G$ whose vertex set is the support of $u$. In the proof of \cite[Theorem  2.1, (a)\implies (b)]{HH3} it is shown that $H$ is either a 3-cycle  or a line of length at most 2. It follows that $\deg u \leq 3$. Hence we have shown that $\max(\Soc(S/I^2))\leq 3$. Since $I^2$ is generated in degree $4$, Theorem~\ref{prime_intersection} yields  the desired conclusion.
\end{proof}

\begin{Corollary}
\label{higher}
Let $I$ be an edge ideal such that $I^2$ is not of intersection type. Then  for any $k\geq 2$,  the ideal $I^k$ is not of intersection type.
\end{Corollary}

\begin{proof}
Let $I=I(G)$. Since $I^2$ is not of intersection type, Theorem~\ref{depth_2power} implies that there  exists a $3$-cycle $C$ contained in $G$ which is not central. As in the proof of Theorem~\ref{depth_2power}(b)\implies (c) it follows that there exists a monomial prime ideal $\pp\neq \mm$ such that $I(\pp)=I(H)+Q$ where  $H$ is a connected graph containing the $3$-cycle $C$ which now is central in $H$, and where  $Q$ is generated in a disjoint set of variables, say $I(H)\subset K[x_1,\ldots,x_k]$ and $Q=(x_{k+1},\ldots,x_\ell)$. Furthermore we may assume that $V(C)=\{1,2,3\}$. It follows from Lemma~\ref{depth_2} that $u+I(\pp)^2$ with $u=x_1x_2x_3$ is a non-zero socle element of $S(\pp)/I(\pp)^2$. This implies that $(x_1x_2)^{k-2}u+I(\pp)^{k}\in \Soc(S(\pp)/I(\pp)^k).$

Assume that $(x_1x_2)^{k-2}u\in I(\pp)^k$. Then $(x_1x_2)^{k-2}u\in I(H)^{k-s}Q^s$ for some $s$ with $0\leq s\leq k$. Since $(x_1x_2)^{k-2}u$ is not divisible by any $x_i\in Q$, it follows that $s=0$, so that $(x_1x_2)^{k-2}u\in I(H)^k$. This is a contradiction, since $\deg (x_1x_2)^{k-2}u=2k-1$, while the generators of $I(H)^k$ are of degree $2k$.

We conclude that  $(x_1x_2)^{k-2}u+I(\pp)^{k}$ is a non-zero socle element of $S(\pp)/I(\pp)^k$, and hence $\max(\Soc(S(\pp)/I(\pp)^k)\geq 2k-1$. On the other hand $\min(I(\pp)^k)=k$. Thus Theorem~\ref{prime_intersection} implies that $I^k$ is not of intersection type.
\end{proof}

\begin{Examples}
{\em (1) A simple example to which Corollary~\ref{higher} applies is the following: let $I=(xy,xz,yz,zt,tu)$ be the edge ideal of the graph $G$ consisting of the  $3$-cycle $\{x,y,z\}$ and the edges $\{z,t\}$ and  $\{t,u\}$. The $3$-cycle $\{x,y,z\}$ is not central. Therefore by Theorem~\ref{depth_2power}, $I^2$ is not of intersection type,  and thus  by   Corollary~\ref{higher}, $I^k$ is not of intersection type for all $k\geq 2$.

(2) Let $I=I(G)=(xy,yz,xz,yt,zt,tu)$ be the edge ideal of the graph $G$. Note that $\Ass(S/I^2)=\Ass(S/I)\cup\{(x,y,z,t),\mm\}$ and  that the cycle $\{x,y,z\}$ is not central. Since $G^c$ is chordal, \cite[Theorem 10.2.6]{HH} implies that  $I^k$ has a linear resolution for all $k\geq 1$. On the other hand, it follows from Theorem~\ref{depth_2power} and Corollary~\ref{higher} that $I^k$ is not of intersection type for all $k\geq 2$.}
\end{Examples}

\section{Some properties of monomial ideals of (strong) intersection type}

We cannot expect that ideals of (strong) intersection type  have, compared with arbitrary monomial ideals, much better properties, since for example any squarefree monomial ideal is of strong intersection type. However we have

\begin{Proposition}\label{integrally_closed}
Monomial ideals of intersection type are integrally closed.
\end{Proposition}

\begin{proof}
Let $I=\Sect_{\pp\in \Ass(S/I)}\pp^{d_{\pp}}$, and let $u=x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$ be a monomial with $u^t\in I^t$ for some $t>0$. Then $u^t\in \pp^{td_\pp}$ for all $\pp\in \Ass(S/I)$. Thus for $\pp\in \Ass(S/I)$, we see  that $\sum_{i, x_i\in\pp}ta_i\geq td_\pp$. Therefore, $\sum_{i, x_i\in\pp}a_i\geq d_\pp$ which implies that $u\in \pp^{d_\pp}$. It follows that $u\in I$, and this proves that $I$ is integrally closed, see \cite[Theorem 1.4.2]{HH}.
\end{proof}

\begin{Corollary}\label{normality}
If $I^k$ is of intersection type for all $k\geq 1$,  then $I$ is normal.
\end{Corollary}

Not all integrally closed ideals are of intersection type. The simplest such example is $I=(x,y^2)$.

\medskip
For a monomial ideal $I\subset K[x_1,\ldots,x_n]$ we denote by $\con(I)$ the convex hull in $\RR^n$ of the  set $\{\ab\:\, \xb^\ab\in I\}$. The set $\con(I)$ is called the {\em Newton polyhedron} of $I$. It is well-known that $I$ is integrally closed if and only if $I$ is generated by the monomials $\xb^\ab$ with $\ab\in \con(I)$.

For a monomial ideal $I$ of intersection type, the supporting hyperplanes of $\con(I)$ can be easily described.  For a subset $F\subset [n]$ we let, as before,  $\pp_F$ be the monomial prime ideal generated by the variables  $x_i$ with $i\in F$. Associated with $\pp_F^a$ we consider the  hyperplane
\[
H_{F,a}=\{\bold{\xi}\in \RR^n\:\ \sum_{i\in F}\xi_i=a\}.
\]
It is clear that $\xb^{\cb}\in \pp_F^a$ if and only if $\cb$ belongs to the half space
\[
H_{F,a}^+=\{\bold{\xi}\in \RR^n\:\ \sum_{i\in F}\xi_i\geq a\}.
\]

Thus the following corollary follows directly from  the definition of ideals  of intersection type.

\begin{Corollary}
\label{supporting}
Let $I$ be a monomial ideal of  intersection type, and let
$$I=\Sect_{\pp_F\in \Ass(S/I)}\pp_F^{d_F}$$ be the canonical  primary decomposition of $I$. Then the hyperplanes $H_{F,d_F}$ with $\pp_F\in \Ass(S/I)$ together with the hyperplanes $H_i=\{\bold{\xi}\in \RR^n\:\; \xi_i\geq 0\}$ for which $(x_i)\not\in \Ass(S/I)$ are the supporting hyperplanes of $\con(I)$.
\end{Corollary}

\medskip
N.V.~Trung, in a discussion with the first author of this paper,  sketched an argument which shows that for any  squarefree monomial ideal $I\subset S$ one has that $I^{(rk)}\subset I^k$ for all $k\geq 1$. Here $r$ denotes the highest degree of a generator in the (unique) minimal set of monomial generators of $I$, and $I^{(t)}$ denotes the $t^{th}$ symbolic power of $I$ which for a monomial ideal we define as follows: Let $Q_1,\ldots, Q_s$ be the uniquely determined primary components corresponding to the minimal prime  ideals of $I$. Then $I^{(t)}=\Sect_{i=1}^sQ_i^t$. (It is also common to define the $t^{th}$ symbolic power of $I$ to be $S\sect \Sect_{\pp\in Ass(S/I)}I^tS_\pp$. This symbolic power is contained in the one defined here.)  In general we expect that $I^{(rk)}\subset I^k$ for all $k$ and all monomial ideals. On the other hand, it follows from a result of Ein, Lazarsfeld and Smith \cite[Theorem A]{ELS}, that for all $k$,  $J^{(ck)} \subset  J^k$ for any graded ideal $J \subset S$ of codimension $c$. Of course there are cases where $r>c$ and vice versa. In the squarefree case however both numbers are bounded by the dimension  of $S$.

Here we prove a related result for monomial ideals of intersection type.  Recall that an ideal $J\subset I$  is said to be a {\em reduction} of $I$,  if $I^k = JI^{k-1}$ for some $k\geq 2$. Following Kodiyalam we denote by $\rho(I)$ the minimum of $\theta(J)$  over all graded
reductions $J$ of $I$, where $\theta(J)$ denotes the maximal degree of a generator in a minimal set of generators of $J$.

\begin{Theorem}\label{symbolic_power}
Let $I\subset S$ be a monomial ideal. Then the following hold:
\begin{enumerate}
\item[(a)] if $I^k$ is of intersection type, then $I^{(\reg(I^k))}\subset I^k$;
\item[(b)] if $I^k$ is of intersection type for all $k\gg 0$, then $I^{(rk)}\subset I^k$ for all $k\gg 0$,
where $r=\rho(I)+1$. In particular, $r\leq \max(I)+1$. Moreover, if $I^k$ has a linear resolution for all $k\gg  0$, then $r$ can be chosen to be $\max(I)$.
\end{enumerate}
\end{Theorem}
\begin{proof}
(a) Set $s=\reg(I^k)$, and let $\qq\in \Ass(S/I^k)$.  There exists $\pp\in \Min(S/I)$ with $\pp\subset \qq$. It follows that $\pp^s\subset \qq^s\subset \qq^{d_{\qq}}$, because $d_\qq\leq \reg(I^k(\qq))\leq s$, see Theorem~\ref{bounded}. This implies that
\[
I^{(s)}=\Sect_{\pp\in \Min(S/I)}\pp^s\subset \Sect_{\pp\in \Ass(S/I^k)}\pp^{d_\pp}=I^k.
\]

(b) By  Kodiyalam \cite{K} we know that $\reg(I^k)=\theta(I)k+c$ for $k\gg 0$. If $I^k$ has a linear resolution for all $k\gg 0$, then $c=0$, and the assertion follows from (a). Otherwise $c>0 $, and for all $k\geq c$ one has that $(\theta(I)+1)k\geq \reg(I^k)$. Again (a) yields the desired conclusion.
\end{proof}

\begin{Corollary}
\label{symbolic_poly}
Let $I$ be a polymatroidal ideal, generated in degree $d$. Then $I^{(dk)}\subset I^k$ for all $k$.
\end{Corollary}

redpajama_set_name: RedPajamaArXiv


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AMSTERDAM, 3 September 2013 – One-year outcomes data from the first cohort of the Global Anticoagulant Registry in the FIELD (GARFIELD), an innovative, independent academic research initiative, provide insights into the elevated stroke risk among subpopulations of patients with atrial fibrillation (AF). The findings, from eight abstracts presented this week at the ESC Congress 2013, collectively show that anticoagulant therapy – which is known to significantly lower stroke risk in AF patients – is consistently under-utilised among those at-risk AF patients.
GARFIELD is led by an international steering committee under the auspices of the Thrombosis Research Institute (TRI), London. It is an international, observational, multicentre, prospective study designed to understand the global burden of AF, a common condition in which the two upper chambers of the heart (the atria) quiver rather than beat rhythmically and can lead to life-threatening complications, including stroke. Up to 2% of the population has AF.1 Despite the availability of highly effective preventive treatments, AF-related stroke remains a major and increasing clinical and societal burden.
The data presented at ESC Congress 2013 are from the first of five GARFIELD cohorts. The first cohort includes a total of 10,614 patients with non-valvular AF and at least one investigator-determined additional risk factor for stroke, recruited from 540 randomly-selected sites in 19 countries. Of these patients, 5,089 were recruited retrospectively as a validation cohort and 5,525 were recruited prospectively and comprise the study populations in these abstracts. ESC Guidelines for the management of atrial fibrillation recommend that all patients at high risk of stroke be prescribed anticoagulation therapy with vitamin K antagonists (VKAs), unless contraindicated. High stroke risk is defined as a score ≥2 on the CHA2DS2-VASc risk score. Previously reported baseline data showed that in Cohort 1, 82.6% of patients had CHA2DS2-VASc ≥2 but only 62% of these patients received anticoagulant therapy.
Data for the stroke-risk stratification research presented at ESC Congress 2013 were available in 5,523 patients enrolled prospectively between December 2009 and October 2011.
The 1-year data – which are preliminary and should be interpreted with caution – were included in one oral presentation and seven poster abstracts. The oral presentation was featured at the State of the Art: Acute coronary syndromes – current guidelines and future prospects, a session that spotlighted the four highest-rated abstracts in this topic.
The GARFIELD Registry is an observational, multicentre, international prospective study of men and women with newly diagnosed AF and one or more additional risk factors for stroke. It will prospectively follow 50,000 newly-diagnosed AF patients from at least 1,000 centres in 50 countries in the Americas, Eastern and Western Europe, Asia, Africa and Australia.
Included patients have been diagnosed with non-valvular AF within the past six weeks and have at least one additional risk factor for stroke, and as such, are candidates for anticoagulant therapy to prevent blood clots leading to stroke. It will be left to the investigator's clinical judgment to identify patient's stroke risk factor(s). Patients will be included whether or not they receive anticoagulant therapy so current and future treatment strategies and failures can be properly understood in relation to patients' risk profiles and co-morbidities.
Data will be collected over an extended follow-up period of up to 8 years, and will include the following outcomes: thromboembolic stroke; TIA ("mini-strokes"); MI/ACS; blood clots affecting other areas of the body; bleeding events; therapy persistence; rate of discontinuation; medical consultations and hospitalizations; need for urgent and elective interventions; cardiovascular morbidity and all-cause mortality.
Among patients treated with vitamin K antagonists, additional outcomes data will include the frequency and timing of monitoring required to maintain a safe and therapeutically effective level of anticoagulation and interventions needed to manage complications due to anticoagulation therapy.
Up to 2% of the global population has AF. Over 6 million Europeans suffer from this arrhythmia, and it is estimated that its prevalence will at least double by 2050 as the population ages. Around 4.5 million people in the European Union and 2.6 million people in the United States have AF, and estimates suggest that by 2014 more than 12 million people in the Asia-Pacific region will have AF.1,2,3,4 AF confers a 5-fold increase in the risk of stroke, and one in five of all strokes is attributed to this arrhythmia. Ischaemic strokes in association with AF are often fatal, and those patients who survive are left more frequently and more severely disabled and more likely to suffer a recurrence than patients with other causes of stroke. In consequence, the risk of death from AF-related stroke is doubled and the cost of care is increased by 50%.5 The condition occurs when parts of the atria emit uncoordinated electrical signals that cause the chambers to pump too quickly and irregularly, not allowing blood to be pumped out completely.6 As a result, blood may pool, clot and lead to thrombosis, which is the number one killer in both the developed and developing world.
Rockson SG, Albers GW. Comparing the guidelines: anticoagulation therapy to optimize stroke prevention in patients with atrial fibrillation. J Am Coll Cardiol 2004; 43(6):929-35.
American Heart Association. Why is AF a problem?. 8/16/13. Available at: http://www.heart.org/HEARTORG/Conditions/Arrhythmia/AboutArrhythmia/Why-is-AF-a-Problem_UCM_423776_Article.jsp.


The adenovirus outbreak at the Wanaque Center for Nursing and Rehabilitation has sickened 27 medically fragile children.
A 10th child died Wednesday night from adenovirus after an outbreak at a Wanaque long-term care center that has infected 27 medically fragile children, the state Health Department said.
The death came as the outbreak entered its sixth week at the Wanaque Center for Nursing and Rehabilitation. A second outbreak, believed to be a different type of adenovirus, was confirmed Wednesday at the Voorhees Pediatric Center in Camden County.
Four children have been affected in Voorhees with a milder form of illness, Dr. Shereef Elnahal, the state health commissioner, said Wednesday. Preliminary tests indicated it was not the same type of the virus as has swept through the Wanaque center.
All of the children infected in the two outbreaks have severely compromised immune systems, the Health Department said. The department has not identified the children by name, age or gender, citing privacy concerns. They have ranged from toddlers to young adults, the vast majority under 18. Most depend on ventilators or tracheostomies to breathe.
I am engaging in discussions with @CMSGov on standards for these facilities.
The Health Department has stationed a staff member at the Wanaque long-term care center full-time to monitor its implementation of infection-control procedures.
The center has stopped admitting new patients until the end of the outbreak. The diagnosis of the most recent case on Wednesday means the earliest endpoint for the outbreak would come in four weeks, or Nov. 24.
Additional diagnoses of adenovirus may be likely, because of the two-week incubation period for the disease, Elnahal said. While adenovirus typically causes cold- and flu-like symptoms in healthy people, it can be devastating to those with compromised immune systems.
Camden County health officials have been working with members of the state Health Department staff to control the Voorhees outbreak and were at the facility Tuesday and Wednesday to make sure infection protocols were followed, said Ron Tomasello, the Camden County public information officer. "The facility has been fully cooperative since [the state was] first in contact on Friday," he said.
A surprise inspection of the facility by the Health Department on Tuesday revealed no infection-control deficiencies and no citations were issued, Elnahal said.
"Even when you have a facility that has a perfect score on our infection-control survey, this can still happen," he said. "Not all outbreaks are preventable."
The center was able to "safely and swiftly separate sick, exposed and well individuals into separate areas," a step that is recommended to stop the spread of the disease, the state Health Department said. It has voluntarily curtailed new admissions.
The facility notified families with children at the Voorhees center about the Wanaque outbreak last week. Another letter was issued Wednesday to tell the Voorhees families about the outbreak there and the preventive steps that have been taken. The facility said it would notify parents of any changes in the status of their children.
The Voorhees facility is owned by a for-profit corporation, Forkid Care. The facility received an overall rating of two out of five stars on the federal government's "nursing home compare" website, scoring above average on staffing and much below average for health inspections. The telephone number provided for the owners on the state Health Department's website is the same as that for the facility.
Its patients may have degenerative diseases, complex seizure disorders, traumatic brain injuries and a variety of other conditions that require assistance in nutrition and respiration.
Teams from the state Health Department will be visiting the four pediatric long-term care facilities in New Jersey this month to discuss infection-control practices and the spread of disease. The two other facilities are branches of Children's Specialized Hospital in Toms River and Mountainside.


Conveniently located next door to the historic 19th-century Capitol Theatre in the Sydney CBD (Central Business District), Capitol Square Hotel offers budget accommodation. Darling Harbour, Chinatown and Central Station are just a 5-minute walk away.
Capitol Square Hotel guest rooms contain an en suite with bath/shower and individually controlled air-conditioning. Other features include tea/coffee making facilities, a mini refrigerator, flat screen TV, hairdryer, iron and complimentary toiletries.
Overlooking the Capitol Theatre shopping precinct, The Terrace Restaurant is open daily, serving buffet breakfast only. There are also a number of restaurants, cafés and bars within a 2-minute walk.
Capitol Square Hotel features laundry service, in-room modem access, minibar at reception, non-smoking rooms and 24-hour reception.
Clean and tidy small rooms and only a std d/bed.Would definatly stay there again.


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Buffed up Briony can also be used to stipple loose powder make-up to finish your look with highlighter, blush or contour.


Video Watch Online Rishton Ka Chakravyuh Star Plus Serial Full Episodes HD. Indian Channel Star Plus & HotStar Drama Serial Rishton Ka Chakravyuh HD Complete Episodes. Rishton Ka Chakravyuh First Episode Starts On 7th August 2017.


Thursday's epic spotlight for is the video for Manchester Orchestra's single, "I Can Barely Breathe". The five guys - Andy Hull, Jonathan Corley, Jeremiah Edmond, Chris Freeman, Robert McDowell - hail from Atlanta, Georgia. The band reminds me of indie rockers like The Format -- especially on their sweet cover of Annuals' "Brother". In "I Can Barely Breathe," Hull's vocals have a slightly more gritty feel which is balanced out by the ethereal feel of this amazing HD music video.
The video was designed and directed by Clay Lipsky and Jason Bognacki, also known as DestroyRockMusic. The guys utilized a RED ONE Camera for the video shoot.
There are additional behind the scenes photos from the video shoot available on Clay's Flickr. "I Can Barely Breathe" is available on iTunes, along with the rest of their debut album, "I'm Like a Virgin Losing a Child".


text:
Q: nginx reverse proxy with tomcat not working I am trying to proxy tomcat using nginx. I have this configuration below. Now the problem I am facing is if I access the url by IP(192.168.2.6) then it gets redirected to /auth_app is tomcat which is perfectly fine. The tomcat url redirect to a third party SAML2 provider and once it authenticated it redirected to http://localhost:8080/auth_app/ instead of http://192.168.2.6/auth_app/ and because of that my application doesn't open as tomcat is running on 192.168.2.6 and not localhost.
server {
    listen 80 default_server;
    server_name _;
    error_log /var/log/nginx/abc_error.log;
    access_log /var/log/nginx/abc_access.log;
    rewrite ^/?$ /auth_app;
    location /auth_app {
        proxy_redirect off;
        proxy_set_header X-Forwarded-Host $host;
        proxy_set_header X-Forwarded-Server $host;
        proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for;
        proxy_pass http://localhost:8080/auth_app/;
    }

}

A: There should be a config setting in your tomcat app to set the redirect URL to http://192.168.2.6/auth_app/.

redpajama_set_name: RedPajamaStackExchange


text:
La Dispute – Panorama (Album Review)
March 19, 2019 La Dispute – Panorama (Album Review)
Posted at 09:19h Katherine Szabo cryptic rock, CrypticRock, Music reviews, News, Reviews 0 Comments
Some things are worth waiting for. It has been nearly five years since Michigan-based La Dispute released their last record, but with the release of the band's fourth LP Panorama on the horizon, there is light at the end of the tunnel for those who eagerly waited to see what kind of magic La Dispute could weave this time. Panorama, set to release Friday, March 22nd, is the band's first record with Epitaph, and second with legendary producer Will Yip. This new record follows the phenomenal 2014 effort Rooms of the House and sounds like the logical next level from the foundations laid there.
La Dispute has an uncanny talent for defying definition. In doing so, the band has gradually carved out their own sacred space within multiple genres. There are truly no other bands who sound like La Dispute, or who can paint images with words and music in quite the same way. You will not find any catchy hooks or choruses in Panorama, or in any of the band's catalog.
All that in mind, La Dispute's power rests within an undefinable combination of vocalist Jordan Dreyer's essay-like lyrics and the band's medley of Hardcore, Jazz, Noise, and countless other genres. Something like Spoken Word, something like Hardcore, and something else entirely make up the expanse of Panorama, an aptly titled record chronicling Dreyer's reflections on a drive across the Midwest and all the tragedies that took place on those roads, coupled with his relationship with his partner.
Whilee Panorama is just that—a macro shot of a Midwestern landscape—it also finds power in its pinpoint focus on particular moments or objects. Where Rooms of the House was Fiction, Panorama is rooted in reality and experience. Still, songs like the explosive "FULTON STREET II" and "VIEW FROM OUR BEDROOM WINDOW" blur the lines between music and literature, with focus shifting between crashing drums, heavy melodic guitars, and Dreyer's vocals. Occasionally, Dreyer's vocals become lost and drown among the music, but even that has a particular emotional effect.
While some tracks find their emotion in the raw rhythms and melodies, others shine in their subtleties. "RHODONITE AND GRIEF" and "IN NORTHERN MICHIGAN" draw on Dreyer's vocal delivery and reflective lyrics, opting for calmer melodies over pure power. Overall, Panorama still draws from Hardcore and Post-Hardcore, but has softer edges than any of the band's previous releases.
Even so, tracks like "ANXIETY PANORAMA" and "FOOTSTEPS AT THE POND" draw from a well of dazzling intensity, proving that La Dispute can bridge gaps between genres and feelings with ease while demanding your undivided attention throughout.
In that same vein, "YOU ASCENDANT" is deeply reflective and reminiscent of Rooms of the House closer "Objects in Space." An eerie melody plays behind Dreyer's narration, slowly building to a desperate crescendo before fading Panorama to a quiet conclusion. Though this track is among the most subdued, it is also its most profound, and possibly Panorama's standout.
The way Dreyer can conjure up images and emotions with such an atypical vocal delivery is a true testament to his writing. How those lyrics pair so well with the band's blend of music genre can feel like magic at times, and it is no wonder that La Dispute has amassed such a devout following. Panorama's reflective tracks can sit neatly alongside the band's already stellar catalog, and it will no doubt be charming listeners for years to come. For these reasons, Cryptic Rock gives Panorama 5 out of 5 stars.
Purchase Panorama:
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Like the in-depth, diverse coverage of Cryptic Rock? Help us in support to keep the magazine going strong for years to come with a small donation.
2019 Album Review, Album review, Emo, Epitaph, experimental rock, hardcore, Jordan Dreyer, La Dispute, Michigan, music, Post-hardcore, Spoken Word, Will Yip
Katherine Szabo
Katherine has been living for music since she was a young teen. Using her B.A. in English Literature and (almost complete) M.A. in English and Creative Writing, she hopes to combine her penchant for Punk music and live shows with her passion for writing in order to make exciting content for fellow fans. On the side, she writes about her two other passions: books and video games.

redpajama_set_name: RedPajamaCommonCrawl


Browse through Melissa d'Arabian's most-recent family trip photos from Paris.
I met and married my husband, Philippe, who is from a small village in the South of France, while we were both living and working in Paris. We enjoy traveling to France not just to visit Philippe's family, but also to spend time in Paris, where Philippe and I started our lives together. Here my family is in downtown Aix-en-Provence, walking around and enjoying the quaint city.
On our "Mom and Dad before kids" tour, we took the girls to the hotel in Aix-en-Provence where I stayed the night before our wedding. I walked down these stairs as a bride back in 2003, so the girls practiced walking down the stairs as brides, too.
Valentine gives me a huge hug!
Charlotte walks the very stairs that I walked as a bride.
Our standard cheese platter — always served at the end of a meal (I love cheese, so you can see why I loved living in France!).
Mont-d'Or Vacherin is my favorite cheese of all time. It's a seasonal cheese, though, and available only in winter.
At my in-laws' house, the table is decorated with the exact same tablecloth that was used 30 years ago (we saw the photos that proved it).
Margaux, Oceane and Charlotte put on a show. I have no idea where they get the performance bug (ha ha).
My father-in-law is famous for making huge paellas for family gatherings, and we served paella at our rehearsal dinner years ago.
Seeing Valentine spend sweet moments with her great-grandmother made the 22-hour trip completely worth it.
The Provence version of King Cake is based on marzipan in a flaky pastry — it's my favorite!
We take the four girls on the Metro in Paris — always a sardine-like experience!
This is a must-do pose for anyone going to Paris (get off the Metro at Trocadero to re-create this pose).
Unbelievable pastry at Angelina's on Rue de Rivoli — not only a Paris legend, but also where I had my bridal luncheon a week before my wedding.
Margaux's middle name is Capucine, so a trip to her namesake street in the 9th arrondisement of Paris was a trip highlight for her.
Here's a shot of the girls with me at the bottom of the stairs inside the Eiffel Tower. We climbed more than 700 stairs that day!
We took the girls to the exact spot where Philippe and I had our first kiss 12 years ago. They insisted we re-create it. And then each girl wanted to give her dad a smooch there, too.
I snapped a photo of the girls running around and playing in the Marais (4th arrondissement of Paris), right outside the apartment where Philippe and I lived when we were newlyweds.
Melissa is known for transforming everyday meals into budget-friendly dishes the whole family will enjoy on her show, Ten Dollar Dinners. Browse through her recipes now.


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View Through My Camera: Busy Weekend!!
Wow, what a busy weekend on Sand Mountain. It all started Saturday night when I got called out to an attempted robbery and stabbing in Collinsville at the Dollar General Store. The store clerk was stabbed in the stomach several times, but luckily she is doing well. Police caught the suspect in about 30 minutes. So after going to the bureau and feeding the video back to Huntsville, I go home. I'm there about half an hour when the pager goes off again. There has been a shooting in Geraldine. That's the top picture of the trailer where a 17 year old boy was shot at least twice with a shotgun. They airlifted him to Huntsville Hospital, and he is still alive the last time I checked. Neighbors heard the shots and saw someone run from the trailer and get into a green car and leave, but DeKalb County deputies have no idea who did the shooting at this time.
Then on Sunday, I get called because an Albertville police car is on fire. Get to the scene and find out it was a K9 unit. Officer was chasing a runner when he heard his car explode. Went running back to find his car totally engulfed in flames. He then ran up to the car and was able to get his dog, Kilo, out before he was seriously hurt. The car was a total wreck.
So this is what we call the "waves rolling in". It has been pretty quite around here for awhile, but the last month has seen at least 3 murders, 2 or 3 accidental shootings, lots of robberies, and the stabbing. So grab your surfboard and hold on tight. We are going for a ride!
WHEN IT RAINS IT POURS!!!!!
You got that right, Zac!


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text:
Article - 'Elements of the Arcade' by Kosuke
An item about Game Design posted on Feb 21, 2004
Volume 1: Balance
ELEMENTS OF THE ARCADE
Written by: KosukeGW (MaceDX)
-Introduction
'For those of you that don't know me, my name is KosukeGW. I recently changed my name to MaceDX but you can call me either or.
In the title of this article, the word "Arcade" doesn't mean a place in the mall where you and your friends go after lunch to see how badly you guys could whoop each other's ass in Tekken after eating some of the greasiest chili dogs at Bob's Hotdog Stand you've ever seen. The "Arcade" as a metahpor is what I like to call the pulse of the game whatever it may be. The rhythm that game gives persuading you and your friends to come back to the mall the very next weekend for lunch. Same arcade, same greasy assed chilidog.
"Elements of the Arcade" is a series of articles that simply defines some of the key tones of gaming in your game's "Arcade" and how you can use them the right way in your game in order to keep the player wanting to come back to your "Arcade". Seems simple enough, eh? Also, I'll explain to you the key essential points that most GW game makers tend to overlook when in production of their game. It's a balance. You want to make sure your game balances out perfectly whether it's the right percentage of gameplay with storyline or vise versa, accurately. Too much of one quality and a too much blanked out quality can equal an unprofessional outlook and a bad impression on your game. You must also consider the community and what's common in it. Too much of what's common in the community becomes what many like to call "cliché".
This first chapter explains what exactly I mean by "balance", how it affects your game and how it alters your "Arcade".
-BALANCE
Balance has much to do with how the game flows and how steady the rhythm of the arcade beats. Before you can understand how to balance your game you must know 'what to balance in your game.
First Impression-
*One thing you don't want is for the player to lose interest and send your game to a trip to the recycling bin. If that happens then all your efforts to make the rest of the game would be futile*
Without argue, first impression is THE MOST important part of the game. If you want to impress the GamingW community one thing is for sure, DO NOT start off with a 'supposing deep dramatic dialogue explaining your story in a prologue. Especially the "The World is not what it seems" method. That has been overused so many times over its gotten uninteresting/too common or "cliché". Another first impression "don't" you want to avoid is the "dungeons, dragons and legendary hero" method. I'm quite skeptical on this one because such a method can be so customizable, turned and twisted to the point that if used the right way it can have the potential to be an original idea. However, sticking with the basics such as "unlikely hero saves the world from evil demon from another dimension" stuff is definitely a no. If you be original, creative and give out the same "want to keep playing/watching" effects as a cliffhanger (not an ACTUAL cliffhanger but the same "want to know what happens next" effect) then you have the player right where you want 'em. How do you do that? I can't give specifics. If I did then that wouldn't be your own original idea.
Introductory-
Not like the intro or introduction. The part in the game where you get to know most of the key characters, obviously the main character, or in an Rpg's case, some of the beginning party members. Now the player isn't really "in" to the game yet so technically this is still part of the first impression. You have to keep this part as interesting as possible. This is the beginning of the game and you cannot afford to have the player lose interest but at this point it is VERY easy to do so. Keep it interesting and the most effective way is to have the game "rise up" to the plot. One method to keeping interest is to start of with gameplay, like a puzzle. You want the player to feel like their actually doing something, like *cough* playing a game instead of reading miles upon miles of text and moving from one NPC to another just to read a few more miles of text. How do you stick a puzzle in the beginning of the game? Be creative. It's your game, you do whatever you want. Don't feel limited or try to follow the trend or the "fad". Remember, though, If you use this puzzle method then purposely it has be easy as hell. Don't want to get the player stuck with such a small interest level do ya. You're game goes STRAIGHT to the recycling bin if we're talking most people because they don't see any purpose or reason why, exactly, they should continue playing this game.
Rising Action/Climax-
This would be somewhat in the middle of the game. If you've gotten the player this far then you're good to go (unfortunately there's no way of telling unless you get a beta tester who has absolutely no guarantee's they'll finish the game if they don't like it). Don't think you have the player wrapped around your fingers and no matter what crap you give them they'll still finish the game though. The only (and I mean only REAL) reason they've gotten this far is because you've got them interested enough to get here. ENOUGH, meaning there's a dead point and depending on your game you can be lucky enough to have just the required minimal amount of reasons to keep them interested. This is an interest keeping dangerzone. They can lose their interest easily at this "minimal" point. 'OR, However depending on your game if you've gotten the player at a very high point of interest then you don't have much interest keeping dangers here to worry or your reputation in previous good games you've made can serve as an interest keeper (however this also works in the negative…if you have a bad reputation. AND YOU KNOW WHO YOU ARE!). This covers most of the game so it's very important. Your plot must keep leading/heating up to your biggest deadline point "the climax". This is what I mean:
>There's a stick man named Steve
>Steve the Stick had a wife and two stick children
>Steve the stick had a very jealous and angry brother, Stan
>Stan kidnapped Steve's stick wife and his two stick children
>Steve ventures various stick worlds in order to save his stick family
>Steve confronts Stan on a stick hilltop where his stick wife and stick kids were held hostage
>Steve fights Stan and wins but not before Stan reveals a shocking secret
>Steve the stick's children were actually Stan the stick's children
>Steve's stick wife had kept it secrete because she was too embarrassed to tell Steve the Stick that she had a night out with Stan the stick once.
>Stan Kidnapped Steve's wife and stick kids because he wanted his stick children back and to know the truth
>It was a shock to Steve the stick because all his life he had believed that Stan the stick was gay
>Steve the stick kills Stan the stick with a stick blow torch but doesn't know if he can forgive his stick wife.
-ends with cliffhanger*
Well, not the best example of a totally original, interesting and complex plot but it does have something that a lot of unprofessional games leave out. The plot is constantly rising. It never goes down for a second before the climax. That's how you keep a players interest. That's how you keep the rhythm of your arcade flowing steadily. Now lets say something like this happens-
-prologue introducing Steve the stick using a "The world is not what you think method" explaining the supernatural (Something that has absolutely NOTHING to do with Steve's Stick world)
>Steve the stick ventures and finds Stan the Stick
>Stan the stick fights Steve the stick
>Stan the stick gets run over by a stick pickup truck
>the stick woman in the stick pickup truck is Steve's wife
>Steve and his stick wife come home with their stick children at last and lives happily ever after
>Stan the stick rises from the dead and nocks Steve the stick's stick door down
>Steve the stick fights Stan the stick
>Steve had to burn down his stick home in order to kill Stan the stick
THE END-
Now as absurd as the plot was in the first place there's still something besides that which just doesn't sound right in this version of Steve the Stick. The plot raised once then came back down again and then rose once more making it look as if it was a sequel in the middle of the game. Then if that wasn't enough, you don't know where exactly the climax resides. The peak point of the rising action is what CREATES the climax. Also you must have a FALLING action to give finishing touches to where the climax was. This is VERY important. Subconsciously, people who play games/watch movies like to (Subconsciously mind you) know exactly how far they are from the end of the movie/game. What they really above all want to see is how everything turns out when they beat the game. That's why people play rpg/adventure games in the first place. It's not always comfortable not knowing when the climax will come because you shouldn't have to look too hard for it (that's why some people can't watch "The Lord of the Rings". Not because it's long and at some points tedious. It's because the just can't put their tongue on the climax since it was based off a book and progresses like one instead of a cinema).
Climax/Final Confrontation/Boss stage-
When it comes to RPGs or adventure games they are one and the same. I know you're tired of hearing about climax but this is a different angle. Remember plot and game play difficulty rise in unity when you're talking about most Rpgs/adventure games (that's why I can't stand some rpgs where they make you fight and you have to lose in order to progress. That makes entirely no sense and it keeps only storyline in mind). Since these are games they have to have a boss. The Boss has to be the final battle/confrontation. Since the boss is the final fight then fighting the boss HAS to be the climax or something close by it. Definition wise there's no other way. I know I talk of originality but you can't really try anything original to escape this fact. If you have no boss you'd probably be thinking after you beat the game: "Oh……the games over…..okay", because that's how videogames itself had evolved from. Even one of the very first games "ping pong" had a boss. No matter how much you played that game you knew that pixel that flies from one 6 pixel paddle to the other would only get faster, faster, and fastest (Boss). However, times have changed and plot is incorporated with rpg/adventure games. You know when climax comes because the rising action rises fastest at this point. You don't just get to the final room in the final dungeon (a dungeon that's no different from all the other dungeons) that the final boss happens to be in. There are (or should be) things at stake.
Boom- something happens and you life is at stake
Boom-something happens and the world is at stake
Boom-something happens and now the dimension is at stake
Boom-something happens and now the universe is at stake
All in the climax. Not exactly that basic and cliché as the example but my point is in the climax it should only get worse really fast. For example lets the Disney flick Aladdin (you can really learn from other things too. It doesn't always have to be revolving around stuff like Final Fantasy). The final confrontation to Aladdin was the evil guy revealed Aladdin wasn't a prince. To make it worse, he sent him off in the middle of nowhere. Somehow (forgot how) he made his way back to the palace just to fight an all powerful sorcerer (same evil dude). To make it even worse he put the princess in an hour glass so she only had certain period of time to live. To make it EVEN worse, Aladdin called the evil dude a snake or something and the evil dude turned into a giant 50 foot cobra. To make it EVEN worse Aladdin convinced the giant cobra to turn into a genie so now the evil guy has phenomenal cosmic power. See how during the climax it gets worse the fastest during this point? This is "suspense" and should be a great and possibly the best influence over the final confrontation/boss/climax in your game.
Falling Action-
Here's the "the day is saved, we all live happily ever after" point. As long as the action IS NOT rising then this part is pretty much customizable all around. It could be a sad, it could be a happy, and it could be whatever you want. Not much I could say about this part. It just all depends on the way your climax is resolved.
Final impression-
The third most important thing. You've gotten them through your game, now what? This part is the part the player is will be walking away with. It's the influence of this final impression that most greatly helps the player choose if he/she recommends this game to others or not.
Story-
Sweatn' the unnecessary stuff is how you get distracted from trying to work out what really matters. The bigger more important flukes like can throw your game off balance and really throw the rhythm of you arcade off beat.


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ARIA Music Teacher of the Year – Nominees Announced!
At a special event on Tuesday 10 October, at Sydney's Art Gallery of New South Wales, ARIA announced the nominees for the 2017 ARIA Awards with Apple Music across all categories, including the inaugural ARIA Music Teacher of the Year Award, presented in conjunction with The Song Room and sponsored by Telstra.
The Song Room are extremely proud to be co-presenting this Award with ARIA, to celebrate the exceptional influence music teachers have on our kids, setting them up for a bright future.
Creative kids are brighter kids, and this award highlights the great work done by music teachers around the country, as well as the fact that despite this, three out of four Australian children still miss out on music education in schools.
Meet the nominees below and vote for your favourite today!
Alex Manton – Asquith Girls High School, NSW
Alex Manton credits her programs at Asquith Girls with encouraging students to find their passion for music. "I go to great lengths to ensure all students are capable of playing or singing in ensemble situations. Differentiating to suit my students' abilities is essential… I encourage and support each student's personal development and journey through music."
Vote for Alex here!
Julie Layt – Crescent Lagoon State School, QLD
Because of the inclusive approach to her music programs, which incorporate Auslan for students with hearing impairments, Julie Layt says students are happy to come to music class and love to perform. "For me, the obvious thing is the confidence and joy that is instilled by performing especially in a group situation. On several levels, music at 'Cresso' helps our students become better people."
Vote for Julie here!
Renee McCarthy – Woodcroft College – SA
"Music keeps students involved in positive activities, establishes connections with others and develops confidence." says Renee McCarthy. "Performing within the community (local nursing homes and the Adelaide Town Hall) has raised the profile of the College and our program. The buzz about music at Woodcroft is a talking point around the community."
Vote for Renee here!
Stephen McEwan – Bellarine Secondary College – VIC
Stephen McEwan's music program has created a safe learning environment where students have expressed that music is the only thing that keeps them at school. "We make music fun and work with passion. We learn together, create together, play together, sing together, and explore the many possibilities that music has to offer together." Stephen says.
Vote for Stephen here!
The benefits of music education have been consistently proven by decades of research, and they are experienced every day by Australian teachers and students who actively engage and participate in music making. Research into The Song Room's programs in particular, proves that children involved in music at school achieve better grades, attend school more often, improve their self-esteem and confidence, and also experience less stress, depression, loneliness and anger.
Help us shine a spotlight on all the benefits of music education by voting for your favourite nominee today! The music teacher with the most public votes will be presented with an ARIA at the 2017 ARIA Award with Apple Music ceremony held in Sydney on 28 November, 2017 and broadcast nationally on Channel Nine.
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