text:
\section*{Introduction}
The class of polymatroidal ideals is one of the rare classes of monomial ideals with the property that all powers of an ideal in this class have a linear resolution. This is due to the fact that the powers of a polymatroidal ideal are again polymatroidal \cite[Theorem 5.3]{CH} and that polymatroidal ideals have linear quotients \cite[Lemma 1.3]{HT} which implies that they have linear resolutions. Recall that a monomial ideal is called polymatroidal, if its monomial generators correspond to the bases of a discrete polymatroid, see \cite{HH}. Since the set of bases of a discrete polymatroid is characterized by the so-called exchange property, it follows that a polymatroidal ideal may as well be characterized as follows: let $I\subset S=K[x_1,\ldots,x_n]$ be a monomial ideal generated in a single degree. We denote, as usual by $G(I)$ the unique minimal set of monomial generators of $I$. Then $I$ is said to be polymatroidal, if for any two elements $u,v\in G(I)$ such that $\deg_{x_i}(u)> \deg_{x_i}(v)$ there exists an index $j$  with $\deg_{x_j}(u)< \deg_{x_j}(v)$ such that $x_j(u/x_i)\in I$.

Recently it has been observed that a monomial localization of a polymatroidal is again polymatroidal \cite[Corollary 3.2]{HRV}.
The monomial localization  of a monomial ideal $I$ with respect to a monomial prime ideal $P$ is the monomial ideal $I(P)$ which is obtained from $I$ by substituting the variables $x_i\not \in P$ by $1$. Observe that $I(P)$ is the unique monomial ideal with the property that $I(P)S_P=IS_P$. The monomial localization $I(P)$ can also be described as the saturation
$I\: (\prod_{x_i\not\in P}x_i)^\infty$.  Thus in the case that the polymatroidal ideal $I$  is squarefree, in which case it is called matroidal, we see  that $I(P)=I\:u$ where $u=\prod_{x_i\not\in P}x_i$.

By what we have explained so far it follows that all monomial localizations of polymatroidal ideals have a linear resolution. The natural question arises whether this property characterizes  polymatroidal ideals. The main purpose of this paper is to discuss this question. In Theorem~\ref{easier} we give an affirmative answer requiring however more that just the condition that all monomial localizations have a linear resolution. To be precise we show, that a monomial ideal $I$ is polymatroidal if and only $I\: u$ has a linear resolution for all monomials $u$ in $S$. In fact, among a few other equivalent conditions, we also  show that $I$ is polymatroidal if we only require that $I\: u$ is generated in a single degree for all monomials $u\in S$.   Since for a squarefree  monomial ideal $I$,  the colon ideal $I\: u$  is  a monomial localization for any monomial $u$, it follows (see Corollary~\ref{newyear}) that a squarefree monomial ideal $I$ is matroidal if and only if $I(P)$ is generated in a single degree for all monomial prime ideals $P$. It turns out that this characterization of matroidal ideals corresponds to a well-known characterization of matroids which  says that a simplicial complex is a matroid if and only if all its induced subcomplexes are pure, see  \cite[Proposition 3.1]{S}.

Even though  matroidal ideals are characterized by the property that all its monomial localizations have a linear resolution, we don't know whether the corresponding statement is true for polymatroidal ideals. There are simple examples of monomial ideals which show that all monomial localizations are generated in a single degree but the ideals themselves  are not polymatroidal. However due to computational evidence we are lead to conjecture that the  monomial ideals with the property that all monomial localizations have a linear resolution are precisely the polymatroidal ideals. In Section~\ref{polymatroidalsection} we discuss several special cases which support this conjecture. In fact we give an affirmative answer to the conjecture in the following cases: 1.\ $I$  is generated in degree 2 (Proposition~\ref{somayehbelievesit}), 2.\ $I$ contains at least $n-1$ pure powers (Proposition~\ref{purepowers}), 3.\ $I$ is monomial ideal in at most 3 variables (Corollary~\ref{dimtwo} and Proposition~\ref{dimthree}), 4.\  $I$ has no embedded prime ideal and either $|\Ass(S/I)|\leq 3$  or $\height(I)=n-1$ (Proposition~\ref{anothercase}).

We would like to point out that in each of the special cases mentioned above we use completely   different arguments for  the proof of our conjecture.  For the moment we do not have a general strategy to prove it.

In Section~\ref{extension} we introduce componentwise polymatroidal ideals, namely those monomial ideals with the property that each of its components is generated by  a polymatroidal ideal. In contrast to polymatroidal ideals, powers of componentwise polymatroidal ideals need not to be componentwise polymatroidal, unless the  ideal is generated in at most two degrees, see Proposition \ref{veryspecial}. On the other, it might be that powers of componentwise linear ideals are componentwise linear. For this we could not find a counter example.

One would expect that an exchange property of its generators characterizes componentwise polymatroidal ideals.  For that purpose we introduce the so-called non-pure exchange property and show in Proposition \ref{exchange} that componentwise polymatroidal ideals enjoy the non-pure exchange property. On the other hand, we show by an example that an ideal with the non-pure exchange property need not to be componentwise polymatroidal.

It is natural to ask whether componentwise polymatroidal ideals have linear quotients. We expect that this is the case and prove it for  ideals which are componentwise of Veronese type. It is also an open question whether ideals satisfying the non-pure exchange property have linear quotients, even they are not componentwise polymatroidal.

 \section{An algebraic characterization of polymatroidal ideals and monomial localizations of  matroidal ideals}
\label{algebraicsection}
 Let $K$ be a field, $S=K[x_1,\ldots,x_n]$ the polynomial ring in the indeterminates  $x_1,\ldots, x_n$ and $I\subset S$ a monomial ideal. We first show

\begin{Theorem}
\label{easier}
Let $I$ be a monomial ideal. The following conditions are equivalent:
\begin{enumerate}
\item[(a)] $I$ is polymatroidal.
\item[(b)] $I\:u$ is polymatroidal  for all monomials $u$.
\item[(c)] $I\:u$  is generated in a single degree for all monomials $u$ and has linear quotients with respect to the reverse lexicographic order of the generators.
\item[(d)] $I\:u$ has a linear resolution for all monomials $u$.
\item[(e)] $I\:u$ is   generated in a single degree for all monomials $u$.
\end{enumerate}
\end{Theorem}

\begin{proof}
(a)\implies (b): It is enough to show that for variable $x_i$, $I\:x_i$ is polymatroidal. Let $I=\sum_{j=0}^dI_jx_i^j$,
where for all $u\in G(I_j)$, $x_i\nmid u$. Then $I\:x_i=I_0+\sum_{j=1}^dI_jx_i^{j-1}$. Set $J=\sum_{j=1}^dI_jx_i^{j-1}$. Then
$I=I_0+x_iJ$, and $I\:x_i=I_0+J$. If $J=0$, then $I\:x_i=I_0=I$, and there is nothing to prove.

Now let $J\neq 0$.
We want to show that $I_0\subseteq J$. Let  $u$ be monomials with  $u\in I_0$. Since $J\neq0$ there exists a monomial $v\in I$ such that  $v\in x_iJ$.  Since $I$ is polymatroidal it satisfies the symmetric exchange property, see \cite[Theorem 12.4.1]{HH}.  Therefore, since  $x_i$ does not divide $u$ but does divide $v$, it follows that  there exists a variable $x_t$ with $t\neq i$  such that
$ux_i/x_t\in I$. Hence $ux_i/x_t\in x_iJ$, so $u/x_t\in J$. This  implies that $u\in J$. Thus we conclude that $I\:x_i=J$.

Let $u,v\in G(J)$ . So $x_iu,x_iv\in x_iJ\subseteq I$. If $\deg_{x_i}(u)=\deg_{x_i}(v)$,  since $I$ is polymatroidal,  it follows that $x_iu,x_iv$ satisfies  exchange property. Hence exchange property is satisfied for $u$ and $v$.

Let $\deg_{x_i}(u)>\deg_{x_i}(v)$, so $x_i|u$. Now for variable $x_l$ with $\deg_{x_l}(u)>\deg_{x_l}(v)$, we want to show that there exists variable $x_j$ such that $\deg_{x_j}(v)>\deg_{x_j}(u)$
and $(u/x_l)x_j\in G(J)$. Since $\deg_{x_l}(x_iu)>\deg_{x_l}(x_iv)$ and $I$ is polymatroidal, it follows that there exists variable $x_j$ such that $\deg_{x_j}(x_iv)>\deg_{x_j}(x_iu)$
and $(x_iu/x_l)x_j\in G(I)$. Since $x_i|u$, we have that $(x_iu/x_l)x_j\in I_tx_i^t$  for $t\geq 1$. Hence $(u/x_l)x_j\in I_tx_i^{t-1}\subseteq J$. Also we have that $\deg_{x_j}(v)>\deg_{x_j}(u)$.

(b)\implies (c): Any polymatroidal ideal is generated in a single degree   and has linear quotients with respect to the reverse lexicographic order of the generators, as shown in \cite[Lemma 1.3]{HT}. Therefore (b) implies (c) trivially.

(c)\implies (d) follows from the general fact that ideals generated in a single with  linear quotients have a linear resolution (see \cite[Lemma 4.1]{CH}), and (d)\implies (e) is trivial.

(e)\implies (a): Let $v,w\in G(I)$ with $\deg_{x_i}(v)>\deg_{x_i}(w)$. We want to show that there exists variable $x_j$
such that $\deg_{x_j}(w)>\deg_{x_j}(v)$ and $(v/x_i)x_j\in G(I)$. By assumption $I\:\frac{v}{x_i}$ is generated in a single degree. Hence, since $x_i\in G(I\:v/x_i)$ it follows that
$I\:v/x_i$ is generated in degree $1$. Hence, since $w/\gcd(w,v/x_i)\in  I\:v/x_i$, there exists $z\in G(I)$ such that $x_j=z/\gcd(z,v/x_i)$ for some $j$ and such that $x_j$ divides $w/\gcd(w,v/x_i)$. Then $\deg_{x_j}(w)>\deg_{x_j}(v/x_i)$. So since $\deg_{x_i}(v)>\deg_{x_i}(w)$ it follows  that $x_j\neq x_i$. Hence $\deg_{x_j}(w)>\deg_{x_j}(v/x_i)=\deg_{x_j}(v)$. Our  assumption (for $u=1$) implies that  $I$ is generated in a single degree.  Hence $\deg(z)=\deg(v)$. On the other hand,  it follows from   $x_j=z/\gcd(z,v/x_i)$ that $\deg_{x_l}(z) \leq \deg_{x_l}(v/x_i)=\deg_{x_l}(v)$ for all $l\neq i,j$ and $\deg_{x_j}(z)=\deg_{x_j}(v/x_i)+1=\deg_{x_j}(v)+1$ and also $\deg_{x_i}(z) \leq \deg_{x_i}(v/x_i)=\deg_{x_i}(v)-1$.  Therefore, $z=(v/x_i)x_j$.
\end{proof}

 We denote the set of monomial prime ideals of $S=K[x_1,\ldots,x_n]$ by $\P(S)$. Let $P\in \P(S)$ be a monomial prime ideal. Then $P=P_C$ for some subset $C\subset [n]$, where $P_C=(\{x_i\:\; i\not\in C\})$ and  $IS_P=JS_P$ where $J$ is the monomial ideal  obtained from $I$ by the substitution $x_i\mapsto 1$ for all $i\in C$. We call $J$ the monomial localization of $I$ with respect to $P$ and denote it by $I(P)$.

 For example, if $I=(x_1x_2x_3, x_2x_3x_4, x_3x_5x_6)\subset K[x_1,\ldots,x_6]$ and $C=\{4\}$, then $I(P_C)=(x_2x_3, x_3x_5x_6)$.

 Let $C\subset [n]$ and set $x_C=\prod_{i\in C}x_i$. Then $I(P_C)=I\: x_C^\infty =I\: x_C^k$ for $k$ large enough. In particular, if $I$ is a squarefree monomial ideal we have  that $I(P_C)=I\: x_C$. Therefore we obtain

 \begin{Corollary}
 \label{newyear}
 Let $I$ be a squarefree monomial ideal. The following conditions are equivalent:
 \begin{enumerate}
 \item[(a)] The ideal $I$ is a matroidal.
 \item[(b)] For all $P\in \P(S)$ the ideal  $I(P)$ is matroidal.
 \item[(c)] For all $P\in \P(S)$ the ideal  $I(P)$  is generated in a single degree and has linear quotients with respect to the reverse lexicographic order of the generators.
 \item[(d)] For all $P\in \P(S)$ the ideal $I(P)$ has  a linear resolution.
 \item[(e)] For all $P\in \P(S)$ the ideal $I(P)$  is   generated in a single degree.
 \end{enumerate}
 \end{Corollary}

 \begin{Corollary}
 \label{easyproof}
 Let $I$ be a squarefree monomial ideal. The following conditions are equivalent:
 \begin{enumerate}
 \item[(a)] The ideal $I$ is a matroidal.
 \item[(b)] For all $P\in \P(S)$ and all integers $k>0$ the ideal  $I^k(P)$ has a linear resolution.
 \item[(c)] For all $P\in \P(S)$ there exists an integer $k>0$ such that the ideal  $I^k(P)$ has a linear resolution.
 \item[(d)] For all $P\in \P(S)$ there exists an integer $k>0$ such that the ideal  $I^k(P)$ is  generated in a single degree.
 \item[(e)] For all $P\in \P(S)$ and all integers $k>0$ the ideal $I^k(P)$  is generated in a single degree.
 \end{enumerate}
 \end{Corollary}

 \begin{proof}
 (a)\implies (b): Since $I$ is a matroidal,  $I^k$ is polymatroidal for all $k$ (see \cite[Theorem 5.3]{CH}). Hence by \cite[Corollary 3.2]{HRV}, $I^k(P)$ is polymatroidal for all $P\in \P(S)$. So   $I^k(P)$ has a linear resolution for all $P\in \P(S)$ and all $k$.

 The implications (b)\implies (c) \implies (d), and (b)\implies (e)\implies (d) are trivial.

(d)\implies (a): By Corollary \ref{newyear} it is enough to show that $I(P)$ is generated in a single degree for all $P$.  By assumption we know that  $ (I(P))^k$ (which is equal to $I^k(P)$) is generated in a single degree. Thus, since $I(P)$ is a squarefree, the desired conclusion follows once we have shown  that  if $J$ is squarefree monomial ideal and $J^k$ is generated in a single degree, then $J$ is generated in a single degree as well. Let $s$ be  the  smallest degree of a generator of $J$  and assume  that there exists $v\in G(J)$ with $\deg(v)=t$, $t>s$.  Then our assumption implies that  $J^k$ is generated in degree $sk$. Since $v^k\in J^k$ and $\deg(v^k)=tk>sk$,  there exist $u_1,\ldots,u_k\in G(J)$ such that $\prod_{i=1}^ku_i$ divides $v^k$ and $\deg(u_i)=s$ for each $i=1,\ldots,k$. Then  $u_1$ divides $v^k$, so since $u_1$ and $v$ are squarefree monomials, it follows that $u_1$ divides $v$, a contradiction.
 \end{proof}

\section{Monomial localizations of polymatroidal ideals}
\label{polymatroidalsection}
One would expect that Corollary~\ref{newyear} remains true if we replace  in its  statements  ``matroidal" by ``polymatroidal". This is the case for the equivalence of (a) and (b). However the following example shows that (a) is not equivalent to (e) if we replace ``matroidal" by ``polymatroidal" in statement (a).

Indeed, let $I=(x_1^2, x_1x_2, x_3^2,x_2x_3)$. Then $I$ is not polymatroidal, but all monomial localizations are generated in a single degree. On the other hand, the ideal $I$ in this example does not have a linear resolution. So one may expect that polymatroidal ideals can be characterized by the properties (c) and  (d) of Corollary~\ref{newyear}.

In the following  special cases we can prove this.

\begin{Proposition}
\label{somayehbelievesit}
Let $I\subset K[x_1,\ldots,x_n]$ be a monomial ideal generated in degree $2$. Then the following conditions are equivalent:
\begin{enumerate}
\item[(a)] The ideal $I$ is a polymatroidal.
 \item[(b)] For all $P\in \P(S)$ the ideal  $I(P)$ is polymatroidal.
 \item[(c)] For all $P\in \P(S)$ the ideal  $I(P)$  is generated in a single degree and has linear quotients with respect to the reverse lexicographic order of the generators.
 \item[(d)] For all $P\in \P(S)$ the ideal $I(P)$ has  a linear resolution.
\item[(e)] After relabeling of the variables there exist integers $0\leq k\leq m\leq n$ such that
\[
I=((x_1,\ldots,x_k)(x_1,\ldots,x_m), J),
\]
where $J$ is a  squarefree monomial ideal in the variables $x_{k+1},\ldots,x_m$ satisfying the following property:
\[
\text{$(*)\hspace{0.5cm} $If  $x_ix_j\in J$ and  $k+1\leq l\leq m$ with $l\neq i,j$, then $x_ix_l\in J$ or $x_jx_l\in J$.}
\]
\end{enumerate}
\end{Proposition}

\begin{proof}
The implication (a)\implies (b)  is known (\cite[Corollary 3.2]{HRV}) and the implications (b)\implies (c) \implies (d) are known.

(d)\implies (e): After a relabeling of the variables we may assume the $x_i^2\in I$ if and only if $i\in[k]$. Suppose that $k\geq 2$ and let $1\leq i, j\leq k$ and $i\neq j$. Since $I$ is generated in degree 2 and has a linear resolution it  is known by \cite[Theorem 3.2]{HHZ} that $I$ has linear quotients with respect to a suitable order of the generators. We may assume that $x_i^2$ comes before $x_j^2$ in this order. Hence, since   $(x_i^2)\:x_j^2= (x_i^2)$,  there exists a monomial $u\in G(I)$ coming before $x_j^2$ such $(u)\:x_j^2=(x_i)$. It follows that $u=x_ix_j$. This shows that $(x_1,\ldots,x_k)^2\subset I$.

Let $\MI$ be the subset of elements $j\in [n]$ with the property that $j>k$ and $x_j|u$ for some $u\in G(I)$. After a relabeling of the variables $x_{k+1},\ldots,x_n$ we may assume that $\MI=\{k+1,\ldots,m\}$. Let $u=x_ix_j$ with $j\in \MI$ and $i\in[k]$. Then $x_i\in I(P_{\{j\}})$, and since $I(P_{\{j\}})$ has a linear resolution, all generators of  $I(P_{\{j\}})$ are of degree $1$. In particular, for any $t\in [k]$ we must have that $x_t \in G(I(P_{\{j\}}))$. This implies that $x_tx_j\in G(I)$. Thus we have shown that $(x_1,\ldots,x_k)(x_1,\ldots,x_m)\subset I$.

Let $J$ be the ideal generated by all $u\in G(I)$ which do not belong to the ideal  $(x_1,\ldots,x_k)(x_1,\ldots,x_m)$. Then  $J$ is a squarefree monomial ideal in the variables $x_{k+1},\ldots,x_m$. Let $x_ix_j\in J$ and $l$ an integer with  $k+1\leq l\leq m$ and $l\neq i,j$.

If $k=0$, then $x_lx_h\in J$ for some $h$ and  $J$ is matroidal by Corollary~\ref{newyear}.  Comparing $x_lx_h$ with $x_ix_j$ we see $x_ix_l\in J$ or $x_jx_l\in J$.

If $k>0$, then $x_1x_l\in I$. Therefore $x_1\in  I(P_{\{l\}})$, and hence  $I(P_{\{l\}})$ is generated in degree $1$, since it has a linear resolution. This implies that $x_ix_l\in J$ or $x_jx_l\in J$.

(e)\implies (a):  Let $u,v\in G(I)$. We have  to show that this pair satisfies the polymatroidal  exchange property. Since $(x_1,\ldots,x_k)(x_1,\ldots,x_m)$ is polymatroidal and $J$ is matroidal because of $(*)$, we may assume that $u\in  (x_1,\ldots,x_k)(x_1,\ldots,x_m)$ and $v\in J$.

Let $u=x_tx_l$ and $v=x_ix_j$, then the exchange property is satisfied because  $x_sx_i\in G(I)$ or $x_sx_j\in G(I)$ for all $s\neq i,j$, due to
$(*)$.
\end{proof}

For the proof of the next result we recall the following well-known fact.

\begin{Lemma}
\label{wellknown}
Let $J\subset S$ be a graded ideal with linear resolution and such that $\length (S/J)<\infty$. Then $J=(x_1,\ldots,x_n)^k$ for some $k$.
\end{Lemma}

\begin{proof}
Since $\length (S/J)<\infty$ it follows that  $\reg(S/J)=\max\{j\:\; (S/J)_j\neq 0\}$, see \cite[Lemma 1.1 ]{CH}. We may assume that  $J$ has a $k$-linear resolution. Therefore,   $\reg(S/J)=k-1$, and hence   $(S/J)_j=0$ for $j\geq k$. It follows that $J=(x_1,\ldots,x_n)^k$.
\end{proof}

\begin{Definition}
 {\em Given  positive integers $d, a_1,\ldots, a_n$. We let $I_{(d;a_1,\ldots,a_n)}\subset  S=K[x_1,\ldots,x_n]$ be the monomial ideal generated by the monomials $u\in S$ of degree $d$ satisfying  $\deg_{x_i}(u)\leq a_i$ for  all $i=1,\ldots,n$. Monomial ideals of this type are called ideals of {\em Veronese type}.}
\end{Definition}
Obviously, monomial ideals of Veronese type are polymatroidal.

\begin{Proposition}
\label{purepowers}
Let $I\subset K[x_1,\ldots,x_n]$ be a monomial ideal generated in degree $d$ and suppose that $I$ contains at least $n-1$ pure powers of the variables, say $x_1^d,\ldots,x_{n-1}^d$. Then the following conditions are equivalent:
\begin{enumerate}
\item[(a)] The ideal $I$ is a polymatroidal.
 \item[(b)] For all $P\in \P(S)$ the ideal $I(P)$ has  a linear resolution.
 \item[(c)] The ideals $I$ and $I(P_{\{n\}})$ have a linear resolution.
 \item[(d)] $I=I_{(d;d,\ldots,d,k)}$ for some $k$.
 \end{enumerate}

\end{Proposition}
\begin{proof}
The implication (a)\implies (b) is known and the implications (b)\implies (c) and (d)\implies (a) are trivial. Thus it remains to show that (c) implies (d).

To this end   we write
\[
I=I_0+I_1x_n+\cdots +I_kx_n^k,
\]
where $I_j$ is a monomial ideal in $S'=K[x_1,\ldots,x_{n-1}]$ for all $j$.

Several times in our proof we will apply the following fact, which is an immediate consequence of \cite[Theorem 2.1]{GHP}: let $J\subset S$ be a monomial ideal with linear resolution, and let $a_1, \ldots, a_n$ be positive integers. Then the monomial  ideal $J'$ generated by the monomials $u\in G(J)$ with $\deg_{x_i}u\leq a_i$ for $i=1,\ldots, n$ has linear resolution as well.  We refer to this result as to the `restriction lemma'.

Applying the restriction lemma to $I$ it follows that $I_0$ has a $d$-linear resolution. Our assumption implies that $x_1^d,\ldots x_{n-1}^d\in I_0$. In particular, it follows that $\length(S'/I_0)<\infty$. Thus Lemma~\ref{wellknown} implies that $I_0=\nn^d$ where $\nn=(x_1,\ldots,x_{n-1})$.

Next we show by induction on $j$ that $I_{k-j}=\nn^{d-k+j}$. For $j=0$, we have to show that $I_k=\nn^{d-k}$. Indeed, by assumption the ideal $I(P_{\{n\}})=I_0+I_1+\cdots +I_k$ has a linear resolution. Since $I_j$ is generated in degree $d-j$, it follows that $I(P_{\{n\}})=I_k$ and moreover, that $\nn^d=I_0\subset I_k$. Hence $I_k$ has a $(d-k)$-linear resolution and contains  $x_i^{d-k}$ for $i=1,\ldots,n-1$. Again applying Lemma~\ref{wellknown}, it follows that $I_k=\nn^{d-k}$. This completes the proof of the induction begin.

Now assume that $j>0$ (and $\leq k-1$), and assume that $I_{k-l}=\nn^{d-k+l}$ for $l=0,\ldots,j-1$. We set
\[
J=I_0+I_1x_n+\cdots +I_{k-j}x_n^{k-j} \quad \text{and}\quad L=\nn^{d-k+j-1}x_n^{k-j+1}+\cdots +\nn^{d-k}x_n^k.
\]
The ideal $L$ is polymatroidal, and hence has a $d$-linear resolution. Applying the restriction lemma to $I$ we see that $J$ has a $d$-linear resolution.
We have $$J\cap L=(I_0\cap L)+(I_1x_n\cap L)+\cdots +(I_{k-j}x_n^{k-j}\cap L)=$$
$$I_0x_n^{k-j+1}+I_1x_n^{k-j+1}+\cdots+I_{k-j}x_n^{k-j+1}=
(I_0+I_1+\cdots+I_{k-j})x_n^{k-j+1}.$$
So $\reg(J\cap L)\geq d+1$. On the other hand by  the exact sequence
\[
0\rightarrow J\cap L\rightarrow J\oplus L\rightarrow I\rightarrow 0
\]
we have that $\reg(J\cap L)\leq \max\{\reg(J\oplus L),\reg(I)+1\}=d+1$. Then $\reg(J\cap L)=d+1$. Hence $J\cap L=(I_0+I_1+\cdots+I_{k-j})x_n^{k-j+1}=I_{k-j}x_n^{k-j+1}$. So $I_{k-j}$ has a $(d-k+j)$-linear resolution and contains $x_i^{d-k+j}$ for $i=1,\ldots,n-1$, because $\nn^d=I_0\subset I_{k-j}$. By Lemma~\ref{wellknown},  $I_{k-j}=\nn^{d-k+j}$.
Altogether we have shown that $I=\nn^d+\nn^{d-1}x_n+\cdots +\nn^{d-k}x_n^k=I_{(d;d,\ldots,d,k)}$, as desired.
\end{proof}

\begin{Corollary}
\label{dimtwo}
Let $I\subset K[x_1,x_2]$ be a monomial ideal. The following conditions are equivalent:
\begin{enumerate}
\item[(a)] $I$ is polymatroidal.
\item[(b)]  For all $P\in \P(S)$ the ideal $I(P)$ has  a linear resolution.
\item[(c)] $I$ has a linear resolution.
\end{enumerate}
\end{Corollary}

\begin{proof}
The conditions (b) and (c) are equivalent, because $I(P)$ is a principal ideal for $P\neq (x_1,x_2)$, and the implication (a) \implies (b) is known. For the proof of the implication (b) \implies (a) we write $I=uJ$, where $u$ is the greatest common divisor of the generators of $I$. $I$ is polymatroidal if and only if $J$ is polymatroidal, and $I$ satisfies (b) if and only if $J$ does. So we assume from the very beginning that greatest common divisor of  the generators $I$ is $1$. This implies that $I$ contains a pure power of $x_1$ or a pure power of  $x_2$. Thus the desired conclusion follows from Proposition~\ref{purepowers}.
\end{proof}

\begin{Definition}{\em
Let $I$ be a monomial ideal. We say that $I$ satisfies the {\em strong exchange property} if $I$ is generated in a single degree and
for all $u, v\in G(I)$ and for all $i,j$ with $\deg_{x_i}(u)> \deg_{x_i}(v)$ and $\deg_{x_j}(u)< \deg_{x_j}(v)$, one has
$x_j(u/x_i)\in I$.}
\end{Definition}

\begin{Proposition}
\label{dimthree}
Let $I\subset S=K[x_1,x_2,x_3]$ be a monomial ideal. The following conditions are equivalent:
\begin{enumerate}
\item[(a)] $I$ is polymatroidal.
\item[(b)] $I$ is polymatroidal satisfying the strong exchange property.
\item[(c)]  For all $P\in \P(S)$ the ideal $I(P)$ has  a linear resolution.
\end{enumerate}
\end{Proposition}

\begin{proof}
The implications (b)\implies (a) and (a) \implies (c) are known. This it remains to be shown that (c) implies (b). Let $I=uJ$ where $u$ is the greatest common divisor of the generators of $I$. It is known \cite[Theorem 1.1]{HHV} that $I$ is polymatroidal satisfying the strong exchange property, if and only if $J$ is of Veronese type. Since $I(P)$  has  a linear resolution for all $P\in P(S)$ if and only if the same holds true for all $J(P)$, we may assume from the very beginning that $u=1$, and then have to show that $I$ is of Veronese type. Let $a_i=\max\{\deg_{x_i}(u)\:\; u\in G(I)\}$ for $i=1,\ldots,3$. We claim that $I=I_{(d;a_1,a_2,a_3)}$ where $d$ is the common degree of the generators of $I$. We first show that for each $i$  the set of monomials
\[
\mathcal{A} =\{u\in K[x_1,x_2,x_3]\:\; \deg(u)=d, \deg_{x_i}(u)=a_i \text{ and } \deg_{x_j}(u)\leq a_j \text{ for } j\neq i \}
\]
belongs to $I$.

Indeed, (c) implies that $I(P_{\{i\}})$ is generated by the monomials $v\in K[x_j,x_k]$ such that $vx_i^{a_i}\in I$ and has a linear resolution. Therefore, by Corollary~\ref{dimtwo}, $I(P_{\{i\}})$ is polymatroidal. Hence there exist numbers $0\leq e\leq f\leq d-a_i$ such that
\[
I(P_{\{i\}})=(x_j^rx_k^s\:\; r+s=d-a_i, r\leq a_j, s\leq a_k \text{ and }e\leq r\leq f).
\]
Assume now that $\mathcal{A}\not\subset I$. Then it follows $e>0$ or $f<d-a_i$. We may assume that $e>0$. Therefore, $x_k^{d-a_i}x_i^{a_i}\not \in I$. On the other hand, since the greatest common divisor of the elements of $G(I)$ is equal to $1$, it follows that there exists monomial $x_k^{d-b}x_i^{b}\in I$  with $b<a_i$. Hence $x_k^{d-b}\in I(P_{\{i\}})$, a contradiction because $I(P_{\{i\}})$ does not contain a pure power of $x_k$.

In order to complete the proof of the claim, we introduce the following ideals $J_{b_1,b_2,b_3}$ with $a_i\leq b_i\leq d$ for $i=1,2,3$. The ideal  $J_{b_1,b_2,b_3}$ is generated by all generators of $I$ and all monomials $x_1^{r_1}x_2^{r_2}x_3^{r_3}$ of degree $d$ such that $r_j\leq b_j$ for all $j$ and there exists $i\in[3]$ with $a_i\leq r_i\leq b_i$. We will show by induction on $b_1+b_2+b_3$  that  $J_{b_1,b_2,b_3}$  has a linear resolution for all $b_i$. In particular,  $J_{d,d,d}$ has a linear resolution.  Hence by  Lemma~\ref{wellknown},  $J_{d,d,d}=(x_1,x_2,x_3)^d$  since  $J_{d,d,d}$  contains the pure powers $x_i^d$. This then implies that $I=I_{(d;a_1,a_2,a_3)}$.

The induction begin with $b_1+b_2+b_3=a_1+a_2+a_3$ is trivial because in that case  $a_i=b_i$ and $J_{a_1,a_2,a_3}=I$, which by assumption has a linear resolution. Now assume that $b_1+b_2+b_3>a_1+a_2+a_3$. Then $b_i>a_i$ for some $i$, say for $i=1$. By induction hypothesis the ideal $H=J_{b_1-1,b_2,b_3}$ has a $d$-linear resolution. Let $J=J_{b_1,b_2,b_3}$, and consider the exact sequence
\[
0\To H \To J\To J/H\To 0.
\]
The module $J/H$ is annihilated by $x_2$ and $x_3$. Therefore, $J/H$ is an $S/(x_2,x_3)$-module generated by the residue classes of the elements $vx_1^{b_1}$ with $v\in K[x_2,x_3]$ of degree $d-b_1$. Since no power of $x_1$ annihilates the generators of $J/H$ it follows that $J/H$ is a free $S/(x_2,x_3)$. It follows that $J/H$ has a $d$-linear resolution. Therefore we conclude from the above exact sequence that $J$ has a $d$-linear resolution.
\end{proof}

\begin{Proposition}
\label{anothercase}
Let $I\subset S=K[x_1,\ldots,x_n]$ be a monomial ideal with no embedded prime ideals such that $I(P)$ has a linear resolution for all $P\in V^*(I)$, and let $\Ass(S/I)=\{P_1,\ldots,P_r\}$. Let $\mm=(x_1,\ldots,x_n)$ be the graded maximal ideal of $S$.  Then the following holds:
\begin{enumerate}
\item[(a)] If $P_i+P_j=\mm$ for all $i\neq j$, then $I$ is polymatroidal.
\item[(b)] If  $r\leq 2$, then  $I$ is a transversal polymatroidal ideal.  If $r=3$, then either $I$ is again  a transversal polymatroidal ideal or $I$ is a matroidal ideal generated in degree $2$ of the form $I=P_1\sect P_2\sect P_3$ such that  $\Sect_{i=1}^3G(P_i)=\emptyset$ and  $G(P_i)\union G(P_j)=\{x_1,\ldots,x_n\}$ for all $i\neq j$.
\item[(c)] If $\height(I)=n-1$, then $I$ is polymatroidal.
\end{enumerate}
\end{Proposition}

\begin{proof}
Let $P\in\Ass(S/I)$. Since $I$ is a monomial ideal with no embedded prime ideals, it follows that $P$ is a minimal prime ideal of $I$. Therefore,  $\length(S(P)/I(P))<\infty$. Since $I(P)$ has a linear resolution, it follows from Lemma \ref{wellknown}, that $I(P)=P^k$ for some $k$.
Therefore $I=P_1^{a_1}\cap\cdots\cap P_r^{a_r}$.

(a) Since $I$ is generated in a single degree and $P_i+P_j=\mm$ for all $i\neq j$, it follows from  a result of Francisco and Van Tuyl \cite[Theorem 3.1]{FV} that $I$ is polymatroidal.

(b) If $r=1$, then $I=P_1^{a_1}$ is a transversal  polymatroidal.

If $r=2$, then $I=P_1^{a_1}\sect P_2^{a_2}$. Since $I$ is generated in a single degree we conclude that $G(P_1)\sect G(P_2)=\emptyset$. Therefore, $I=P_1^{a_1}P_2^{a_2}$, and the assertion follows.

Now let $r=3$, then $I=P_1^{a_1}\cap P_2^{a_2}\cap P_3^{a_3}$. We may assume that $I$ is full supported, i.e., $\{x_1,\ldots,x_n\}= \Union_{u\in G(I)}\supp(u)$.

First  assume that $P_i\nsubseteq P_j+P_k$ for all $i,j,k$.  Then,  since $I(P_j+P_k)=P_j^{a_j}\cap P_k^{a_k}$ is generated in a single degree, it follows that $G(P_j)\cap G(P_k)=\emptyset$ for $j\neq k$. Hence $I=P_1^{a_1} P_2^{a_2} P_3^{a_3}$
is a transversal polymatroidal ideal.

Next we may assume that $P_1\subseteq P_2+P_3$. In particular,  $P_2+P_3=\mm$, since $I$ is full supported. We claim that $P_i+P_j=\mm$ for all $i\neq j$ and hence by part (a), $I$ is polymatroidal. It remains  to be shown that $P_1+P_2=\mm$ and $P_1+P_3=\mm$.
Assume that $P_1+P_2\neq\mm$ and set $P=P_1+P_2$. Then $I(P)=P_1^{a_1}\cap P_2^{a_2}$. Since $I(P)$ is generated in a single degree, we have that $G(P_1)\cap G(P_2)=\emptyset$. So since
$P_1\subseteq P_2+P_3$, it follows that $P_1\subseteq P_3$, a contradiction.
Therefore $P_1+P_2=\mm$. Similarly we can see that $P_1+P_3=\mm$.

Now we want to show that  $G(P_i)\cap G(P_j)\nsubseteq G(P_k)$ for distinct $i$, $j$ and $k$. Assume  $G(P_i)\cap G(P_j)\subseteq G(P_k)$ for some $i,j$ and $k$. Let $x_\ell$ be a variable.  If $x_\ell\in G(P_i)\cap G(P_j)$, then $x_\ell\in G(P_k)$, and if $x_\ell\not\in G(P_i)\cap G(P_j)$, then we may assume that  $x_\ell\not\in G(P_i)$. In that case it follows that  $x_\ell\in G(P_k)$,  since $P_i+P_k=\mm$.  Therefore $P_k=\mm$, a contradiction.

Now we claim that $a_1=a_2=a_3$.
We may assume that $a_1\geq a_2\geq a_3$ and  that $I$ is generated in degree $d$. Let $x_i\in G(P_1)\cap G(P_2)\setminus G(P_3)$ and $x_j\in G(P_3)\setminus G(P_1)$. Then since $a_1\geq a_2$, it follows that $x_i^{a_1}x_j^{a_3}\in I$. So there exist integers $s\leq a_1$ and $t\leq a_3$ such that $x_i^sx_j^t\in G(I)$. Since
$x_i^sx_j^t\in P_3^{a_3}$ and $x_i\not\in P_3$, we have  $x_j^t\in P_3^{a_3}$, and so $t=a_3$. On the other hand, since $x_i^sx_j^t\in P_1^{a_1}$ and $x_j\not\in P_1$,
 it follows that $x_i^s\in P_1^{a_1}$, and so $s=a_1$. Hence $x_i^{a_1}x_j^{a_3}\in G(I)$. Therefore $d=a_1+a_3$. Now let $x_i\in G(P_1)\cap G(P_3)\setminus G(P_2)$ and $x_j\in G(P_2)\setminus G(P_1)$.
Then similarly $x_i^{a_1}x_j^{a_2}\in G(I)$, so $d=a_1+a_2$. Therefore  $a_2=a_3$. Set $a=a_2=a_3$. Next we show that $a_1<2a$. Assume $a_1\geq2a$. Let $x_i\in G(P_1)\cap G(P_2)$
and $x_j\in G(P_1)\cap G(P_3)$, then $x_i^{a_1-a}x_j^a\in I$. Hence $a_1=\deg(x_i^{a_1-a}x_j^a)\geq d=a_1+a$, so $a\leq 0$, a contradiction.
Now let $x_i\in G(P_1)\cap G(P_2)$, $x_j\in G(P_1)\cap G(P_3)$ and $x_k\in  G(P_3)$. Then  $x_i^ax_j^{a_1-a}x_k^{2a-a_1}\in I$. Therefore,
$2a=\deg(x_i^ax_j^{a_1-a}x_k^{2a-a_1})\geq d=a_1+a$, hence $a\geq a_1$, and so $a_1=a$.

Now we have  $I=P_1^a\cap P_2^a\cap P_3^a$. We claim  that $a=1$. The claim  implies that $I=P_1\cap P_2\cap P_3$. Hence, since $I$ is generated in a single degree we conclude that $G(P_1)\cap G(P_2)\cap G(P_3)=\emptyset$.

In order to prove the claim, assume to contrary that  $a>1$. Let $x_i\in G(P_1)\cap G(P_2)$, $x_j\in G(P_1)\cap G(P_3)$ and $x_k\in G(P_2)\cap G(P_3)$. Then $x_i^{a-1}x_j^{a-1}x_k\in I$, because $x_i^{a-1}x_j^{a-1}\in P_1^a$, $x_i^{a-1}x_k\in P_2^a$ and $x_j^{a-1}x_k\in P_3^a$.
So $2a-1=\deg(x_i^{a-1}x_j^{a-1}x_k)\geq d=2a$, a contradiction.

(c) If $r=1$, then  $I=P_1^{a_1}$ is  polymatroidal, and if $r>1$, the assertion follows from (a).
\end{proof}

Based on Proposition~\ref{somayehbelievesit},  Proposition~\ref{purepowers}, Corollary~\ref{dimtwo}, Proposition~\ref{dimthree} and Proposition~\ref{anothercase} and based on experimental evidence we are inclined   to make the following

\begin{Conjecture}
\label{mustbetrue}{\em
A monomial ideal $I$ is polymatroidal if and only if $I(P)$ has a linear resolution for all monomial prime ideals $P$.}
\end{Conjecture}

\medskip
The following examples show  that the localization condition of Conjecture~\ref{mustbetrue}  can not be weakened.

\begin{Example}
\label{threexamples}
{\em (a) The ideal  $I=(x_1x_3^2,x_1^2x_3,x_1x_2x_3, x_2^2x_3)$  and all $I\: x_i$ have a linear resolution, but $I$ is not polymatroidal.

(b)  The ideal  $I=(x_1^3,x_1^2x_2, x_1^2x_3,x_2x_3x_4, x_1x_2x_3, x_1x_3x_4, x_1^2x_4)$ and all $I(P_{\{i\}})$   have a linear resolution, but $I$ is not polymatroidal.

(c) The ideal $I=(x_1^3,x_1^2x_2,x_1^2x_3, x_2^3,x_1x_2^2,x_2^2x_3, x_3^3,x_1x_3^2,x_2x_3^2)$ has  linear relations, and all $I(P_{\{i\}})$
are polymatroidal, but $I$ is not polymatroidal. }
\end{Example}

\medskip
For the proof of Proposition~\ref{anothercase}(c) one could skip the assumption that $I$ has no embedded components, if one could prove the following statement: $(*)$ Let $I\subset S$ be a monomial ideal with linear resolution and such that $I\mm$ is polymatroidal. Then $I$ is polymatroidal.

Indeed, assuming $(*)$ the following can be shown: Let $I=J\sect Q$ and assume that $I$ has  a linear resolution, $J$ is componentwise polymatroidal and $Q$ is $\mm$-primary, then $I$ is polymatroidal. To see this, observe that $I\mm^{j-d}= I_{\langle j\rangle}=J_{\langle j\rangle}$ for $j\gg 0$, where $d$ is the degree of the generators of $I$. Here, for any graded ideal $L$, we denote  by $L_{\langle j\rangle}$ the ideal generated by the $j$th graded component of $L$.  Since $J$ is componentwise polymatroidal it follows that $I\mm^{j-d}$ is polymatroidal. The assertion now follows by induction on $j-d$ and by using $(*)$.

\medskip
Observe  that $(*)$ holds if our Conjecture~\ref{mustbetrue} is satisfied, because $I(P)=(I\mm)(P)$ for all $P\neq \mm$.

\medskip
We believe that if $I$ is a polymatroidal ideal generated in degree $d$, then $(I:\mm)_{\langle d-1\rangle}$ is polymatroidal. It can be shown that this is the case at least when $I$ is a polymatroidal ideal satisfying the strong exchange property. Assuming this is true in general, the above condition $(*)$ follows, because $I=I\mm:\mm$, if $I$ has a linear resolution. Obviously we have $I\subseteq I\mm:\mm$. Assume the inclusion is strict. Then there exists a homogeneous element   $f\in  I\mm:\mm\setminus I$. Thus the residue class  of $f$ in $S/I$ is a non-zero socle element of $S/I$. Say,  $I$ has a $d$-linear resolution.  Then it follows that $\deg(f)=d-1$. On the other hand, $I\mm$ has $(d+1)$-linear resolution. Therefore $I\mm:\mm$ is generated in degree $\geq d$, a contradiction since $f\in I\mm:\mm$.

\medskip
Note that  our conjecture is equivalent to the following statement: let $I$ be monomial ideal with linear resolution. Then  $I$ is polymatroidal if and only if  $I(P_{\{i\}})$ is polymatroidal for all $i$. We prove this version of Conjecture~\ref{mustbetrue} under additional assumptions.

\begin{Proposition}
\label{biequalsai}
Let $I$ be a monomial ideal with $d$-linear resolution, and assume that $I(P_{\{i\}})=I_{(d-a_i; a_1,\ldots,a_{i-1},a_{i+1},\ldots, a_n)}$ for $i=1,\ldots,n$. Then $I=I_{(d; a_1,\ldots, a_n)}$.
\end{Proposition}

\begin{proof}
For $k=1,\ldots,n$, let $I_k=x_k^{a_k}I(P_{\{k\}})$ and set $J=\sum_{k=1}^nI_k$. Then $J\subseteq I$. We first show that $(J:\mm^\infty)_{\langle d\rangle}=I$. In fact, by the definition of $J$ it follows that $(I/J)_{x_k}=0$ for $k=1,\ldots,n$. Therefore,  $I/J$ is a module of finite length, and hence we get
\begin{eqnarray}
\label{stupidr}
I\subseteq (J:\mm^\infty)_{\geq d}\subseteq (I:\mm^\infty)_{\geq d}.
\end{eqnarray}
Here for any graded ideal $L$ we set $L_{\geq d}=\Dirsum_{i\geq d}L_i$.

Since  $I$  has $d$-linear resolution, it follows that  $(I:\mm^\infty)_{\geq d}=I$. Indeed, our assumption on $I$ implies that $I:\mm^\infty=I+H$ where $H$ is generated in degree $\leq d-1$. This follows from \cite[Corollary 20.19]{Eis}. Thus  $(I:\mm^\infty)_{\geq d}= I+\mm H_{\langle d-1\rangle}$. Since $\mm H_{\langle d-1\rangle}\subset I$, the desired conclusion follows. Thus in combination with (\ref{stupidr}) we see that $(J:\mm^\infty)_{\geq d}=I$. Since $I$ is generated in degree $d$, we even get $(J:\mm^\infty)_{\langle d\rangle}=I$.

Now we want to show that $(J:\mm^\infty)_{\langle d\rangle}=I_{(d; a_1,\ldots, a_n)}$. Let $u\in(J:\mm^\infty)_{\langle d\rangle}$ such that
$\deg(u)=d$ and $u\mm^r\subseteq J\subseteq I_{(d; a_1,\ldots, a_n)}$ for some integer $r\geq 0$. Then $ux_i^r\in I_{(d; a_1,\ldots, a_n)}$ for all $i\in[n]$. Hence
for all $i\in [n]$ there exists $v_i\in G(I_{(d; a_1,\ldots, a_n)})$ such that $v_i|ux_i^r$. Therefore, $\deg_{x_j}(v_i)\leq \deg_{x_j}(u)$ for all $j\neq i$. Since  $\deg(u)=\deg(v_i)=d$, it follows that $\deg_{x_i}(u)\leq \deg_{x_i}(v_i)\leq a_i$. This shows that $u\in I_{(d; a_1,\ldots, a_n)}$.  Hence we proved that $(J:\mm^\infty)_{\langle d\rangle}\subseteq I_{(d; a_1,\ldots, a_n)}$.

Now let $x_1^{b_1}\cdots x_n^{b_n}\in G(I_{(d; a_1,\ldots, a_n)})$, then $\sum_{i=1}^nb_i=d$ and $b_i\leq a_i$ for all $i\in[n]$. We claim that $x_1^{b_1}\cdots x_n^{b_n}\mm^s\subseteq J$ with $s=\sum_{i=1}^na_i-d$.
Let $x_1^{c_1}\cdots x_n^{c_n}\in G(\mm^s)$. Then $x_1^{b_1+c_1}\cdots x_n^{b_n+c_n}\in x_1^{b_1}\cdots x_n^{b_n}\mm^s$. If $b_i+c_i<a_i$ for all $i$, then $\sum_{i=1}^na_i=d+s=\sum_{i=1}^nb_i+\sum_{i=1}^nc_i<\sum_{i=1}^na_i$, a contradiction. Hence for convenience we may  assume that $b_1+c_1\geq a_1$, and show that $x_1^{b_1+c_1}\cdots x_n^{b_n+c_n}\in I_1=x_1^{a_1}I_{(d-a_1; a_2,\ldots, a_n)}$. Since  $b_1+c_1\geq a_1$, it is enough to show that $x_2^{b_2+c_2}\cdots x_n^{b_n+c_n}\in I_{(d-a_1; a_2,\ldots, a_n)}$.

We may assume that $b_i+c_i>a_i$ for $i=2,\ldots, t$  and $b_i+c_i\leq a_i$ for $i=t+1,\ldots, n$ with $1\leq t\leq n$.
Since $b_i\leq a_i$ for all $i$, it follows that
\[
\sum_{i=1}^ta_i+\sum_{i=t+1}^n(b_i+c_i)=\sum_{i=1}^ta_i+d-\sum_{i=1}^tb_i+\sum_{i=t+1}^nc_i=\sum_{i=1}^t(a_i-b_i)+\sum_{i=t+1}^nc_i+d\geq d.
\]
Hence $\sum_{i=2}^ta_i+\sum_{i=t+1}^n(b_i+c_i)\geq d-a_1$. This  implies that
\[
x_2^{a_2}\cdots x_t^{a_t}x_{t+1}^{b_{t+1}+c_{t+1}}\cdots x_n^{b_n+c_n}\in I_{(d-a_1; a_2,\ldots, a_n)}.
 \]
Therefore $x_2^{b_2+c_2}\cdots x_n^{b_n+c_n}=w(x_2^{a_2}\cdots x_t^{a_t}x_{t+1}^{b_{t+1}+c_{t+1}}\cdots x_n^{b_n+c_n})\in I_{(d-a_1; a_2,\ldots, a_n)}$, as desired.
\end{proof}

\section{Componentwise polymatroidal ideals}
\label{extension}

In this section we extend the notion of polymatroidal ideals to monomial ideals which are not necessarily generated  in a single degree.

Let $I$ be a monomial ideal. We denote by $I_{\langle j\rangle}$ the monomial ideal generated by all monomial of degree $j$ in $I$. The ideal $I$ is called {\em componentwise linear}, if  $I_{\langle j\rangle}$ has a linear resolution for all $j$. Basic properties about componentwise linear ideals can be found in \cite{HH}.

\begin{Definition}
{\em Let $I$ be a monomial ideal. We say that $I$ is {\em componentwise polymatroidal}, if $I_{\langle j\rangle}$ is polymatroidal for all $j$.}
\end{Definition}

Observe that if  $d$ is the highest degree of a generator of $I$, then $I$ is componentwise polymatroidal if and only if $I_{\langle j\rangle}$ is polymatroidal for all $j\leq d$. Indeed, $I_{\langle j\rangle}=I_{\langle d\rangle}\mm^{j-d}$ for $j\geq d$. Moreover, all powers of $\mm$ are polymatroidal and products of polymatroidal ideals are again polymatroidal, see \cite[Theorem 5.3]{CH}

\medskip
 It is easy to see that  $I$ is componentwise polymatroidal if and only if $I\: u$ is componentwise polymatroidal for all monomials $u$. However if we only assume that  $I\: u$  is componentwise linear for all monomials $u$, it does not necessarily  follow that $I$ is componentwise polymatroidal. Indeed, let  $I=(x_1x_2,x_1x_3^2,x_2x_3^2).$
Then $I\:u $ is componentwise linear for all monomials $u$, but $I$ is not componentwise polymatroidal.

\medskip

It is natural to ask whether powers of componentwise  polymatroidal ideals are again componentwise  polymatroidal. There is a positive answer to this question in the following case.

\begin{Proposition}
\label{veryspecial}
Let $I$ be a componentwise polymatroidal ideal generated in at most 2 degrees. Then $I^k$ is componentwise polymatroidal for all $k$.
\end{Proposition}

\begin{proof} The statement is trivial if $I$ is generated in a single degree. So now assume that $I$ is generated in 2 degrees, say, in degree $d$ and $d+t$ with $t>0$. Then $I=I_{\langle d\rangle} +I_{\langle d+t\rangle}$. Hence
\begin{eqnarray}
\label{idealsum}
I^k=  \sum_{j=0}^k(I_{\langle d\rangle})^{k-j}(I_{\langle d+t\rangle})^{j}.
\end{eqnarray}
Since $I^k$ is generated in degree $\geq dk$ it remains to be shown that $(I^k)_{\langle kd+r\rangle}$ is polymatroidal for all $r\geq 0$.  It follows from
(\ref{idealsum}) that
\[
(I^k)_{\langle kd+r\rangle}=  \sum_{j=0}^\ell (I_{\langle d\rangle})^{k-j}(I_{\langle d+t\rangle})^{j}\mm^{r-tj},
\]
where  $\ell =\min\{k,\lfloor r/t\rfloor\}$.

Observe that for $j< \ell$ we have
\begin{eqnarray*}
(I_{\langle d\rangle})^{k-j}(I_{\langle d+t\rangle})^{j}\mm^{r-tj}&=&(I_{\langle d\rangle})^{k-j-1}(I_{\langle d+t\rangle})^{j}I_{\langle d\rangle}\mm^t\mm^{r-t(j+1)}\\
&\subseteq &(I_{\langle d\rangle})^{k-(j+1)}(I_{\langle d+t\rangle})^{j+1}\mm^{r-t(j+1)}.
\end{eqnarray*}
It follows that $(I^k)_{\langle kd+r\rangle}=(I_{\langle d\rangle})^{k-\ell}(I_{\langle d+t\rangle})^{\ell}\mm^{r-t\ell}$. Since products of polymatroidal ideals are polymatroidal the desired conclusion  follows.
\end{proof}

In general powers of componentwise polymatroidal ideals are not componentwise polymatroidal.

\begin{Example}
\label{polymatroidal}
{\em Let $I=(x_1^2, x_2^2x_3,x_1x_2x_3, x_1x_2^2, x_1x_3^3, x_2x_3^3)$. By using Proposition~\ref{dimthree} it is easy to see that  $I$ is componentwise polymatroidal. However $(I^2)_{\langle 6\rangle}$ is not polymatroidal, because  $(I^2)_{\langle 6\rangle}(P_{\{3\}})=(x_1x_2^3,x_2^4,x_1^2x_2,x_1^3)$ is not generated in a single  degree.}
\end{Example}

One would expect that componentwise polymatroidal ideals can also be  characterized by an exchange property of its minimal set of monomial generators. Suppose for a monomial ideal $I$  we require that  for all monomials $u,v\in G(I)$ the following condition holds: $(*)$ if $\deg_{x_i}(u)>\deg_{x_i}(v)$ for some $i$, then there exists an integer $j$ such that $\deg_{x_j}(v)>\deg_{x_j}(u)$ and $x_j(u/x_i)\in I$. Then it is easily checked that $I$ is necessarily generated in a single degree and hence polymatroidal.

Therefore  we give the following

\begin{Definition}
{\em Let $I$ be a monomial ideal. We say that $I$ satisfies the {\em non-pure exchange property},  if for all $u,v\in G(I)$ with $\deg(u)\leq\deg(v)$ and for all $i$ such that $\deg_{x_i}(v)>\deg_{x_i}(u)$, there exists $j$ such that
 $\deg_{x_j}(v)<\deg_{x_j}(u)$ and $x_j(v/x_i)\in I$.}
\end{Definition}

\begin{Proposition}
\label{exchange}
If $I$ is componentwise polymatroidal, then $I$ has the non-pure exchange property.
\end{Proposition}

\begin{proof}
Let $u,v\in G(I)$ with $\deg(u)\leq \deg(v)=t$ and  $\deg_{x_i}(v)>\deg_{x_i}(u)$ for some $i$. We may assume that  $\deg(u)< \deg(v)$, since $I_{\langle t\rangle}$ is polymatroidal.  By using the fact that  $u$ does not divide $v$, it follows that
there exists $l\neq i$ such that
\begin{eqnarray}
\label{star}
\deg_{x_l}(v)<\deg_{x_l}(u).
\end{eqnarray}
Since  $\deg(u)<\deg(v)$, there exists integer $a$ such that
$\deg(ux_l^a)=\deg(v)$. Then there exists $j$ such that
\begin{eqnarray}
\label{starstar}
\deg_{x_j}(v)<\deg_{x_j}(ux_l^a),
\end{eqnarray}
since $I_{\langle t\rangle}$ is polymatroidal and since   $\deg_{x_i}(v)>\deg_{x_i}(u)=\deg_{x_i}(ux_l^a)$. Moreover,   $x_j(v/x_i)\in I$.
If $j=l$, then by (\ref{star}),  $\deg_{x_j}(v)<\deg_{x_j}(u)$ and $x_j(v/x_i)\in I$. If $j\neq l$,  then (\ref{starstar}) implies
that  $\deg_{x_j}(v)<\deg_{x_j}(ux_l^a)=\deg_{x_j}(u)$ and $x_j(v/x_i)\in I$.
\end{proof}

Unfortunately, the converse of Proposition \ref{exchange} is not true. Indeed, let $I=(x_1x_2,x_1x_3^2,x_2x_3^2).$
Then $I$ has the non-pure exchange property but $I_{\langle 3\rangle}$ is not polymatroidal. On the other hand,  $I$ has linear quotients. Thus the question arises whether any monomial ideal satisfying the  non-pure exchange property has linear quotients. In view of Proposition~\ref{exchange} a positive answer to this question would imply that any componentwise polymatroidal ideal has linear quotients. In the following we show that ideals which are componentwise of Veronese type have linear quotients.

The following concept is needed for the next results: Let $I\subset J$ be monomial ideals with $G(I)\subset G(J)$. We say that $I$ can be {\em extended by linear quotients} to $J$, if the set $G(J)\setminus G(I)$ can be ordered $v_1,\ldots,v_m$  such that $(G(I),v_1, \ldots,v_i)\: v_{i+1}$ is generated by variables for $i=1,\ldots,m-1$. In a particular a monomial ideal $L$ has linear quotients, $(0)$ can be extended to $L$ by linear quotients.

It is known (\cite[Corollary 2.8]{SZ}) that an ideal with linear quotients is componentwise linear. In particular, if $I$ has linear quotients and $I$ can be extended to $J$ by linear quotients, then $J$ has linear quotients and hence a linear resolution.

\begin{Theorem}
\label{extend}
Let $I$ be an ideal of Veronese type generated in degree $d$, and  $J$ an ideal of Veronese type generated in degree $d+1$ such that $I\mm\subseteq J$.  Then $I\mm$ can be extended by linear quotients to $J$.
\end{Theorem}

\begin{proof}
Let $I=I_{(d;a_1,\ldots,a_n)}$. In the first step of the proof we assume that $J=I_{(d+1;a_1+1,\ldots,a_n+1)}$.
Let $u=x_1^{h_1} \cdots x_n^{h_n} \in G(J)$.
We define the set $$S_u=\{i\in [n]\;|\;h_i=a_i+1\}$$
and  the monomial $\bar{u}=\prod_{i\in S_u}x_i^{h_i}$.

\medskip
Now we consider the following order for elements of $G(J)\setminus G(I\mm)$:  we say that $u>v$, if either $|S_u|<|S_v|$,  or  $|S_u|=|S_v|$ and $\bar{u}>_{\lex}\bar{v}$, or  $|S_u|=|S_v|$, $\bar{u}=\bar{v}$ and $u>_{\lex}v$. We also set $u>v$ for all $v\in G(J)\setminus G(I\mm)$
and all $u\in G(I\mm)$.

We claim that with this order,  $I\mm$ can be extended to $J$ by linear quotients.
We have  to show that for all $u=x_1^{h_1}\cdots x_n^{h_n}\in G(J)$ and all $v=x_1^{t_1}\cdots x_n^{t_n}\in G(J)\setminus G(I\mm)$  with $u>v$  there exists $w\in G(J)$ with $w>v$  such that $(w)\:v=(x_j)$ and $x_j$ divides $u/\gcd(u,v)$. We distinguish several cases.

Case (a):  $u\in G(I\mm)$ and $v\in G(J)\setminus G(I\mm)$.  Since $u\in G(I\mm)$,  there exists $r\in [n]$ such that  $h_j\leq a_j$  for $j\neq r$. On the other hand,  since $v\in G(J)\setminus G(I\mm)$,  there exists $l\in[n]$ such that $t_l=a_l+1$. If there exists $p\neq r$ such that $x_p$ divides $u/\gcd(u,v)$, then $t_p<h_p\leq a_p$. Let $w=(v/x_l)x_p$; then $(w)\:v=(x_p)$ and $w\in G(J)$ with $w>v$, because $|S_w|<|S_v|$.  Next we consider the case that $x_p$ does not divide $u/\gcd(u,v)$ for all  $p\neq r$.  Then $(u)\:v=(x_r^c)$ for some integer $c$. If the $c=1$, then there is nothing to prove.  Otherwise,    $t_r+1<h_r\leq a_r+1$. Let  $w=(v/x_l)x_r$; then  $(w)\:v=(x_r)$ and $w\in G(J)$ with $w>v$, because $|S_w|<|S_v|$.

Case (b):  $u,v\in G(J)\setminus G(I\mm)$ and $|S_u|<|S_v|$. Since  $v\in G(J)\setminus G(I\mm)$, it follows that there exists $l\in[n]$ such that $t_l=a_l+1$. If there exists $r\in S_u$ with
$t_r<a_r$, we set  $w=(v/x_l)x_r$. Then $(w)\: v=(x_r)$  and $w\in G(J)$ with $w>v$, because $|S_w|<|S_v|$.
Next we consider the case that $t_r\geq a_r$ for all $r\in S_u$. Since $\deg(u)=\deg(v)$ and $|S_u|<|S_v|$, it follows that there exists $s\in [n]\setminus S_u$ such that $h_s>t_s$. We set $w=(v/x_l)x_s$. Then again $(w)\: v=(x_s)$,  and $w\in G(J)$ with $w>v$, because $|S_w|<|S_v|$.

Case (c):  $u,v\in G(J)\setminus G(I\mm)$, $|S_u|=|S_v|$ and $\bar{u}>_{\lex}\bar{v}$. There exist $l,r$ such that $r<l$, $h_r=a_r+1>t_r$ and $h_l<t_l=a_l+1$.
Let $w=(v/x_l)x_r$; then $(w)\: v=(x_r)$ and $w\in G(J)$ with $w>v$. Indeed, if  $|S_w|<|S_v|$ then $w>v$, and if  $|S_w|=|S_v|$, then $\bar{w}>_{\lex}\bar{v}$ and again  $w>v$.

Case (d):  $u,v\in G(J)\setminus G(I\mm)$, $|S_u|=|S_v|$ and $\bar{u}=\bar{v}$ and $u>_{\lex}v$. There exist $l,r$ such that $r<l$, $t_r<h_r\leq a_r$ and $h_l<t_l\leq a_l$. We set  $w=(v/x_l)x_r$.  Then   $(w)\: v=(x_r)$ and $w\in G(J)$ with $w>v$, because   $|S_w|=|S_v|$, $\bar{w}=\bar{v}$ and $w>_{\lex}v$.

\medskip
In the next step we consider the general case where  $J=I_{(d;b_1,\ldots,b_n)}$ and $I\mm\subseteq J$. By the first step we can extend $I\mm$ to $L=I_{(d+1;a_1+1,\ldots,a_n+1)}$ by linear quotients. Since $I\mm\subseteq J$ it follows that  $a_i+1\leq b_i$ for  $i=1,\ldots,n$. Therefore, $L\subseteq J$, and hence it suffices to extend $L$ to $J$ by linear quotients.

Set $c_i=a_i+1$ for $i=1,\ldots,n$. It is enough to show that $L=I_{(d+1;c_1,\ldots,c_n)}$ can be extended to $K=I_{(d+1;c_1,\ldots, c_{s-1},c_s+1,c_{s+1},\ldots c_n)}$ for some $s\in[n]$. For monomials $u,v\in G(K)$, we say $u>v$, if $u\in G(L)$ and $v\in G(K)\setminus G(L)$ or $u,v\in G(K)\setminus G(L)$ and $u>_{\lex}v$.

We claim that with this order,  $L$ can be extended to $K$ by linear quotients.
We have  to show that for all $u=x_1^{h_1}\cdots x_n^{h_n}\in G(K)$ and all $v=x_1^{t_1}\cdots x_n^{t_n}\in G(K)\setminus G(L)$  with $u>v$  there exists $w\in G(K)$ with $w>v$  such that $(w)\:v=(x_j)$ and $x_j$ divides $u/\gcd(u,v)$.
 We distinguish two cases.

 (i)  $u\in G(L)$ and $v\in G(K)\setminus G(L)$. Since $v\in G(K)\setminus G(L)$, it follows that $t_s=c_s+1$, so $t_s>h_s$. On the other hand since $\deg(u)=\deg(v)$,  there exists $r\in[n]$ such that $h_r>t_r$. Let $w=(v/x_s)x_r$, then $(w)\:v=(x_r)$ and $w>v$ because $w\in G(L)$.

(ii)  $u,v\in G(K)\setminus G(L)$ and $u>_{\lex}v$. So there exist $l,r$ such that $r<l$, $t_r<h_r$ and $h_l<t_l$.
Let $w=(v/x_l)x_r$, then $(w)\:v=(x_r)$ and $w\in G(K)$ with $w>v$, because $w>_{\lex}v$.
\end{proof}

A monomial ideal $I$ is called {\em componentwise of  Veronese type}, if $I_{\langle j\rangle}$ is of Veronese type for all $j$.

\begin{Corollary}
\label{componentwiseveronese}
 Let $I$ be an ideal which is componentwise of Veronese type. Then $I$ has linear quotients.
\end{Corollary}

\begin{proof}
 It follows from Theorem \ref{extend}  that  $I_{\langle j\rangle}\mm$ can be extended to $I_{\langle j+1\rangle}$ by linear quotients for all $j$. Hence by \cite[Proposition 2.9]{SZ}  $I$ has linear quotients.
\end{proof}

redpajama_set_name: RedPajamaArXiv


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62-year old Mahmoud Badavam holds a degree in electronic engineering from Tehran Polytechnic University and a postgraduate degree from the Massachusetts Institute of Technology, U.S.A. The manager of civil engineering projects and a lecturer for the Baha'i Institute for Higher Education, Mr. Badavam is married with one child. In 1986, Mr. Badavam was incarcerated for a period of three years for being a Baha'i. He has been in prison since his arrest on 22 May 2011, and first appeared in court on Tuesday 27 September 2011. He was released in April 2015 after serving a four-year prison sentence.
Nasim Bagheri Tari, 32, a graduate of the Bahá'í Institute for Higher Education (BIHE), has both a bachelor's and a master's degree a psychology. During the years following her graduation she did her best to serve BIHE administratively as well as academically. On 22 May 2011 her house, along with 38 residences of other BIHE collaborators and students, was raided. Having refused to sign an undertaking to stop serving BIHE, Nasim was taken to court in the autumn of 2013. Her four-year prison sentence was confirmed in early 2014 and she was abruptly taken to prison on 27 April 2014.
Negin Ghedamian, 34, a graduate of the BIHE, has both a bachelor's and a master's degree of Persian Literature and Iranian Culture. During the years following her graduation, she served BIHE as a literature and writing skills instructor, and worked as a freelance copy editor. On 22 May 2011, her house was raided and a few days later she was interrogated. Then, on March 10, 2013, she was summoned to Evin prison's public prosecutor's office. Having refused to sign an undertaking to stop serving BIHE, Negin was arrested on the spot. The judge ordered her to pay bail for release pending their trials. On June 8, 2014, the judge of Branch 28 of the Revolutionary Court sentenced her to five years in prison though she was absent. She appealed, but in Feb 9, 2015, before the court of appeal had issued its verdict, Negin was arrested in her workplace and taken to Evin Prison. Her lawyer, however, managed to secure her release on bail pending a decision by the appeals court. In 2016, her sentence was approved. On 16 December 2017, she was arrested in Tabriz airport and taken to Tehran to start serving her five-year sentence in Evin prison from 18 December 2017.
Faran Hesami, 41, worked as a psychology instructor with BIHE and has also been involved in private practice. After completing their undergraduate education at BIHE, she and her husband – Kamran Rahimian – graduated in December 2003 with Master's degrees in Educational Counseling from the University of Ottawa, Canada. Mrs. Hesami was summoned to court and arrested along with her husband on 13 September 2011. She was told that her Master's degree is illegal and therefore her work as a counselor is also illegal. She was sentenced to four-years imprisonment. She was released in April 2016 at the conclusion of her sentence.
Nooshin Khadem, 52, gained a General Studies degree – the only major available via correspondence course from Indiana University, U.S.A. She also later received a postgraduate Masters of Business Administration from Carleton University, Canada. Ms. Khadem worked in an administrative capacity with BIHE. She was arrested on 22 May 2011 and first appeared in court on Tuesday 27 September 2011. She was released in April 2015 after serving a four year prison sentence.
Payman Koushk-Baghi, 38, was born in the city of Gorgan in Iran. He graduated with a bachelor's degree in computer engineering from the Bahá'í Institute for Higher Education (BIHE) in 2004 and has served BIHE since 2008. On 22 May 2011, government agents raided his home, but he was not arrested. The intent was only to search his home. A few days later, he was summoned to the neighborhood investigation office for questioning. The issue remained suspended until 13 March 2013, when his charges were explained to him at Evin's Shahíd Muqaddas public prosecutor's office. He was then released on bail. Finally, in May 2015, he was sentenced to five years of imprisonment under ta'zír law in Branch 28 of the Islamic Revolutionary Court. On 28 February 2016, Payman was arrested whilst on a visit to his wife, Azita Rafizadeh, who was imprisoned in is in Evin in October 2015. Payman and Azita have a five year old son.
Vahid Mahmoudi, 52, studied sociology at BIHE and had been working as a director of the Institute. He is married with two children. In 1982, at the age of 19, he was arrested and served six years in prison for being a Baha'i. Sentenced in October 2011 to five years in prison, he was released on 8 January 2012 after his sentence was reportedly suspended.
Foad Moghaddam, 66, has a Medical degree in general medicine from the Medical school of Tabriz. He practiced medicine for 33 years and has been involved with BIHE for 17 years. Dr. Moghaddam was arrested on 22 May 2011. On 25 June 2011 he was released on bail. On 16 June 2012, he was summoned to court and was sentenced to five years' imprisonment. He is now serving his sentence. Dr. Moghaddam suffers heart problems. He is married and has three children. He was released in November 2017.
Hassan Momtaz, 56, worked as a faculty member in BIHE department of Iranian Culture and Persian Literature. Hassan's home in Tehran was among several BIHE teachers' homes to be raided and searched by security agents on May 22 and May 23 in 2011. Then on March 10, 2013, he was summoned to Evin prison's public prosecutor's office for his first interrogation. He was told that to secure his freedom, he could sign a pledge to stop working with the BIHE. Having refused to sign, the judge ordered him to pay bail for release pending their trials. On February 25, 2014, of the judge of Branch 28 of the Revolutionary Court sentenced him to five years in prison. Three years later, on 15 November 2017, he was arrested at home in Shiraz and taken to Adelabad Prison for serving his 5-year imprisonment.
Kamran Mortezaie, 63, was sentenced to serve a five year prison sentence. He holds a degree in electronic engineering from Áryámihr University – now the Sharif University of Technology – in Iran, as well as a postgraduate degree from George Washington University in the U.S.A. Having been denied the right to practice his profession for being a Baha'i, he worked in the building industry. He was a director of BIHE and a lecturer in computing. He was among 36 members of BIHE's faculty and staff who were arrested during a series of raids carried out in 1998 by the Iranian authorities. He is the father of one child. He was released in April 2016 after serving a five-year sentence.
Amanollah Mostaghim, 67, completed his early education in Shiraz and went to the United States for higher education; he holds a BSc degree in Civil engineering from Texas Tech University. He returned to Iran in 1980, and worked in the area of civil engineering in several provinces, before eventually settling in Shiraz. He was arrested on 22 May 2011 and released on bail after 38 days. On 16 June 2012, he was summoned to court and was sentenced to five years' imprisonment. Arrested in May 2013, he carried out his sentence until November 2015, when he was released due to a medical condition. He is married with three children.
Fattaneh Nabilzadeh, 55, served BIHE as a volunteer proctor and host during the exam periods. She was arrested on 13 August 2013 at her home in the city of Mashhad while she was administering a final exam to her son and another Baha'i student on behalf of BIHE. After the arrest, security agents took Mrs. Nabilzadeh to Mashhad's Intelligence Bureau and then to the Vakilabad Prison. After two months, she was released on bail. Four years later, in 2017, Mrs. Nabilzadeh was tried at Branch Three of Mashhad's Revolutionary Court. On 28 January 2018, the judge sentenced Mrs. Nabilzadeh to one year in prison due to "propaganda against the regime." She appealed the sentence, but the appeals court upheld the lower court's verdict and, as a result, on 16 July 2018, she was summoned to start serving her one-year sentence in the city's Central Prison in Vakilabad.
Shahin Negari, 48, is a BIHE graduate in pharmaceutical science. He also received his M.SC in Microbiology and Immunology from the University of Ottawa, Canada. Until his arrest, Mr. Negari had worked in Tehran as a technical advisor, while being involved with the operation of BIHE. He was arrested on 22 May 2011 and was released on bail after a month. He was sentenced to four years of imprisonment. On 13 January 2013, Mr. Negari was taken into custody without prior notification. He is married and has two children. He was released in November 2016 at the conclusion of his four-year sentence.
Azita Rafizadeh, 35, completed a bachelor's degree in computer engineering at the Bahá'í Institute for Higher Education (BIHE) in 2003. In 2005, she received her master's degree in computer engineering at the University of Pune in India. Since 2003, she has served BIHE in both administrative and academic capacities. On 22 May 2011 her house, along with 38 residences of other BIHE collaborators and students, was raided. Having refused to sign an undertaking to stop serving BIHE, Azita was summoned by authorities for questioning the following day and interrogated. In the subsequent years, she has been harassed and summoned on a number of occasions and ultimately convicted in February 2015. After an appeal process, she was summoned to Evin prison on 25 October 2015 to begin a four-year sentence. Azita is married to Peyman Koushk-Baghi, who is appealing a five-year sentence. The couple have a five year old son.
Along with his wife Faran Hesami, Kamran Rahimian, 46, worked as a psychology instructor with BIHE. After completing his undergraduate education at BIHE, he graduated with Master's degrees in Educational Counseling from the University of Ottawa, Canada in December 2003. He was summoned to court with his wife and two other Baha'is on 13 September 2011. The other two were released on bail soon afterwards. Mr. Rahimian was sentenced to four-years imprisonment and served his full sentence at Gohardasht prison. He was released on 17 August 2015.
Kayvan Rahimian, 52, is a BIHE graduate in psychology and had been involved in private practice as a counselor. He also worked as a psychology instructor with BIHE. He was arrested on 14 September 2011 and was released on bail on 21 September 2011. Mr. Rahimian was told that his Master's degree is illegal and therefore his work as a counselor is also illegal. On 30 September 2012, Mr. Rahimian was summoned to begin his five years' imprisonment sentence. He is the father of a 13-year old daughter and has recently lost his wife, Fereshteh Sobhani, to cancer. He was released on 25 August 2017 after serving a five-year prison sentence.
Mr. Azízu'lláh Samandarí, 43, has a diploma in mathematics and physics, and received his bachelor's degree in management from the Bahá'í Institute for Higher Education (BIHE). He is an expert on computer networking and, for five years, was a member of a national institute for English language studies while assisting with BIHE's online studies and providing technical support to BIHE's website. He was originally arrested on 14 January 2009 and then released before being subsequently arrested and tried in November/December 2012 for "activities against national security through membership in Bahá'í administration". He was sentenced to five years' imprisonment. The order was executed on 7 July 2012, and he was taken to Evin Prison and later transferred to the Rajaeishahr Prison. During this period he was not granted any permission for leave. He was released on 14 April 2017 after serving a five-year prison sentence.
Farhad Sedghi, 68, has a degree in accountancy and financial management. He was expelled from his postgraduate mechanical engineering studies for being a Baha'i. A lecturer with BIHE, he also worked as an accountant and financial adviser. He is married with three children. He was arrested on 22 May 2011 and first appeared in court on Tuesday 20 September. He was released in April 2015 after serving a four-year prison sentence.
Arrested on 14 June 2011, Riaz Sobhani holds a post-graduate diploma certificate in building industry. He worked for BIHE as a building industry consultant and manager. Married with three children, he appeared in court on 1 October 2011. He was given a four-year jail term which came to an end on 25 May 2015.
Ramin Zibaie, 48, has a postgraduate degree in psychology from BIHE where he worked as a director and a lecturer in psychology. He is also an educational counselor. Married with two children, Mr. Zibaie reportedly first appeared in court on 1 October 2011. He was released in April 2015 after serving a four year prison sentence.


Scientists and researchers estimate that 75-90% of all visits to doctors are for stress related complaints.
Matthias Burisch, professor of psychology at the University of Hamburg has listed 130 non-specific symptoms for burnout.
The following is only a short overview of symptoms we can start showing when suffering from too much stress and/or burnout. However, knowing about the symptoms can help you to better understand what is going on with you.
Mental and behavioural: lack of concentration and focus, learning difficulties, procrastination, perfectionism, over-eating or under-eating, negative thoughts, drug abuse (including alcohol and smoking), stop listening to others, stop asking questions, assuming, taking risky decisions, making more mistakes, etc.
Physical: Sweating, breathing difficulties, tightness in the stomach or chest, cold and clammy hands, stiff neck and shoulders, headaches, migraine, acute hearing loss, back pain, gastro-intestinal problems, decreased immunity, insomnia and fatigue, type 2 diabetes, high blood pressure, cardiovascular diseases, heart attack, stroke, etc.
Psychological: forgetfulness, feeling helpless, hopeless and out of control, uneasiness, lack of sense of humour, being cynical, being impatient, listless, tearful, resentful, having low self-esteem and lack of self-confidence, sadness, anger, increased testiness and aggressiveness, depression, anxiety, etc.
If you show one or more of these signs and you are not sure it is stress or burnout related, or if you have already taken steps to control and relieve stress but the symptoms persist, please arrange an appointment and consult your GP.
Stress is too much of everything, a little bit like drowning; and you usually are consciously aware of being under stress.
Burnout is not enough anymore, like being dried out, a complete lack of energy. Very often we do not even realise that we are completely burned out already.


text:
Board of Water Engineers
Laurie E. Jasinski General Entry
The Board of Water Engineers, established by the Texas legislature in 1913, was composed of three members appointed by the governor for six-year terms. One of the members was then made chairman and served as the chief administrative officer. The major duties of the board were to approve plans concerning the organization of irrigation and water-supply districts, approve the issuance of bonds by such districts, issue permits for storage and diversion of water, and make plans for storage and use of floodwater. In 1949 the legislature gave the board the power to define and designate underground water reservoirs and authorize underground water-conservation districts. In 1955 the board was granted the authority to approve federal projects requiring the governor's approval. This included any related construction done by the United States Army Corps of Engineers. The board was also required to employ a chief engineer at that time. The Board of Water Engineers dealt with matters concerning state surface-water appropriation and the collecting of related data. It conducted groundwater examinations and studied the effects of silt in streams, water runoff, and evaporation. In 1956 the board employed eighty-five full-time employees who worked within the various divisions of the agency: administrative, statistical and publications, permits and appropriations, surface water, and ground water. In 1962 the title of the Board of Water Engineers was changed to the Texas Water Commission, and the agency was reorganized three years later to become the Texas Water Rights Commission.
Laurie E. Jasinski, "Board of Water Engineers," Handbook of Texas Online, accessed January 20, 2021, https://www.tshaonline.org/handbook/entries/board-of-water-engineers.

redpajama_set_name: RedPajamaCommonCrawl


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				$id= (int)$_GET['id'];
				echo $id;
				$sql_lock = "update topherballschedule set `lock` = 1 where gameID = $id";
				if(mysql_query($sql_lock))
				{
					$sql_lock = "update topherballpicks set `lock` = 1 where gameID = $id";
					if(mysql_query($sql_lock))
					{
					?>

					<?php
					}
				}
				else
				{
					echo "Error while setting the lock";
				}
			}

			if(isset($_GET['week']))
			{
				$week= $_GET['week'];
				if(($week > 0)&&($week<18))
				{	echo 'Week Number: '.$week."
";
				$sql_team="select * from topherballschedule where weekNum=$week and `lock` = 0";
				$query_team= mysql_query($sql_team);
				if(mysql_num_rows($query_team)>0)
				{
					echo '

Match ID	Home Team	Visitor Team	Time
';
					while($result_team = mysql_fetch_array($query_team))
					{	echo '
';
						$homeID=$result_team['homeID'];
						$visitorID=$result_team['visitorID'];
						$gameID=$result_team['gameID'];
						$time=$result_team['gameTimeEastern'];
						$visitor_team = getTeamName($visitorID);
						$home_team = getTeamName($homeID);
						echo ''.$gameID."	".$home_team."	 ".$visitor_team."	".$time; ?>	 <button onclick="lock('<?php echo $gameID; ?>')">Lock the Match</button>	<?php echo "
";
					}
						echo '
';
				}
				else
				{
					echo " All Matches for the week is locked!";
				}

				}
				else
				{
					echo "Please enter the week number in between 1 to 17 only.";

				}
			}

		}
		else
		{
			echo "Error in connection with the Database";
		}

	}
	else
	{
		echo $err_week;
	}
}
	else
	{
?>
<?php
	}
?>

<title>
Lock The Match
</title>

redpajama_set_name: RedPajamaGithub


Our 6th year class spent the day over at Leinster Driving school and had a super day. All students took th theory test and from there it was out into the cars for a drive. The Leinster deriving school provides a great facility for learner drivers and we thank them for looking after us so well.


Are you curious about zines*? Have you always wanted to self-publish your own radical alternative media on a topic which you are passionate about? Join the 24-hour-zine challenge and make your own zine or mini-zine in just 24 hours (sleep included :-))! A small friendly workshop in the safe space of LeSpace, everyone welcome, also just for an hour or two if you can't make the whole event!
*Zines = self-made, non-commercial booklets or magazines, often on radical and political topics, which you can write/draw/make however you want and copy to give or swap with friends.
The next 24 hour zine event will take place at Le Space in Brussels. Everyone welcome, from experienced zinesters to newbies! For people who've never made zines before, we explain the zine basics and teach you how you can make your own zine!
If possible, bring some basic materials such as: a pencil, a black pen, an eraser, scissors, a glue stick, and paper. Additional materials such as stickers, old magazines to cut up, wrapping paper, stamps, photos, and yarn can be fun to use too. If you prefer to work digitally, you can bring your laptop and use open office, Scribus, or InDesign to make your zine.
This entry was posted in Activities, International Zine Month, News, Workshops and tagged 24 hour zine thing, international zine month, le space, lespace, zine workshop. Bookmark the permalink.


Thank you for a wonderful sailing week on an equally wonderful boat in wonderful Galicia!
What Leon really accomplished was to create a familiar atmosphere, despite the fact that we didn't know each other before.
The tenor and spirit onboard was wonderful from the very beginning and we had so much fun together! How can Leon create such an atmosphere, I wonder?
Not to forget my three wonderful sailing companions!
Leon's joy of sailing is simply contagious. All I experienced and learned during the week I will now slowly practice during my forthcoming sailing trips onboard my own boat with an increased self-confidence.
Small picturesque villages with nice local people, tasty food and delightful Albarino wines!
During our sailing trip we mainly had good winds, but also calms and fog are part of the local weather, which gave good practice in radar navigation.
I would love to come sailing with Leon again. Anytime!
Antje Paetzold, Germany, sailed on the all-women Leg 6-2017 in Galicia, Spain.


King Solomon used them in the construction of the temple that would bear his name, the Phoenicians used them to build their merchant ships, and the ancient Egyptians used their resin in the mummification process. But now Lebanon's cedar trees (Cedrus libani), described in the Scriptures as "the glory of Lebanon" and by the 19th-century French Romantic poet Alphonse de Lamartine as "the most famous natural monuments in the world", face a new threat in the form of climate change.
Emblazoned on the national flag, currency, and the country's national airline, the cedar is the one great unifying emblem of this small Mediterranean nation. Centuries of deforestation have already seen the tree's 500,000 hectares decimated to its current 2,000 hectares, and it has been added to the IUCN's red list of threatened species, albeit at the lowest level of threat.
The cedars, some up to 3,000 years old and almost all of which are now protected, need a minimum amount of snow and rain for natural regeneration. But global warming has meant that Lebanon's cedars are being subjected to shorter winters and less snow, and the Lebanese government estimates that snow cover could be cut by 40 percent by 2040.
Elsa Sattout, whose doctorate entailed a multidisciplinary study of Lebanon's cedar forests, defended the country's approach and noted the cedar is protected by law, but said more could be done.
"There are many reforestation projects undertaken by nongovernmental organizations and funded by international agencies, which have and are taking place, but, for me, these could have been more successful," she said. Since 2009, for example, Starbucks and the Association for Forest Development and Conservation have partnered to plant several thousand cedars in areas of Lebanon afflicted by forest fires.
She also highlighted a mass spraying project against Cephalcia tannourinensis in a joint effort between the American University of Beirut and the ministries of agriculture and environment.


Issues based blog about Hoboken politics and its impact on residents.
Daniel Bryan, who formally served as Chief of Staff to former Hoboken Mayor Dawn Zimmer has become Press Secretary and Deputy Communications Director for Governor Phil Murphy. Bryan was a senior adviser on the Murphy campaign early on so this should come as no surprise. Congratulations Dan! Well done and you put in a lot of hard work in to deserve this position.
Still no announcement has been made regarding former Mayor Dawn Zimmer who was an early Phil Murphy ally. It has been speculated by some that there was the potential for her to be given a position on the Phil Murphy team as well but so far no dice.
No. Keep Winner Take All No Majority Requirement.
Yes. Change to Instant Runoff or Ranked Choice Voting.
Yes. Reinstate Runoff in December For Top 2 if No Majority.


text:
Q: Dynamic Entity Manager in EJB I have an application using Spring MVC and Spring Security.
The application project has ejbs and web application.
I have a common database for users and one database by users' group.
So, when a user authenticate, Spring Security use the common database to check login/password and when a controller calls an ejb, the entity manager connects to the user's group database.
My goal is : create users/groups and their associated database on the fly.
Example :
The common database localhost:3306 database_schema1 contains
  USERS
----------------------
| NAME    | GROUP_ID |
----------------------
| Pierre  |        1 |
| Paul    |        1 |
| Jacques |        2 |
----------------------

  GROUPS
------------------------------------------
| ID | URL            | SCHEMA_NAME      |
------------------------------------------
|  1 | localhost:3307 | database_schema1 |
|  2 | localhost:3308 | database_schema1 |
------------------------------------------

When Pierre authenticate, the home page call an ejb. This ejb has an entity manager which has to connects to group 1 database localhost:3307 with schema database_schema1
If Jacques authenticate, it must be group 2 database.
How can I do to tell to the entity manager to use a given database that I don't know when I build the project?
I mean, I don't want to use persistence.xml in my ejb but dynamically initialize entity manager depending on the user authenticated.
Does Persistence.createEntityManagerFactory() will work in this context ? And if it does, is it the best solution?

A: You can do this by removing the database URL from persistence.xml and specifying it at runtime:
Map<String, String> params = new HashMap<String, String>();
// put the correct URL here
// you can also specify any other parameter from 'persistence.xml'
String dbUrl = "jdbc:mysql://localhost:3308/database_schema1";
params.put("javax.persistence.jdbc.url", dbUrl);
EntityManagerFactory factory =
  Persistence.createEntityManagerFactory("...", params);
Then, use this factory object to create EntityManager instances as needed.

redpajama_set_name: RedPajamaStackExchange


"It's a DFS crapshoot": Strategy is key in the penultimate round of 2018.
You know it's a tricky game when you are liking the safety of a 17k ruckman!
Hedge and Roody run through all the DFS strategy ahead of the two massive $40,000 Prelims on Draftstars.


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text:
eBA Basketball Statistics Analysis System
El Sistema eBA de Análisis de las Estadísticas del Baloncesto
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Analyzing the Basketball Rating Pct. Index
by Jean Louis Trezeguet, Belgium - 11.27.04 ~
Updated by: Prof. Roberto Azar ~ 01.19.2010 & 01.05.2015
The RPI is usually defined as the sum of 25 percent of a basketball team's winning percentage (WP), 50 percent of the team's opponents' winning percentage (OWP), and 25 percent of the team's Opponents' Opponents' Winning Percentage (OOWP). Sometimes it is described as 25 percent winning percentage and 75 percent strength of schedule, with SOS defined as 2/3 × OWP + 1/3 × OOWP.
RPI Ratings are (approximately) those used by the NCAA to determine teams and seeding for the NCAA Tournament?. Anyone who has studied statistics knows that these rankings are extremely flawed, but for whatever reason, the NCAA uses them. eBA System doesn't condone them as a ranking system, just compute them.
Most discussion of the RPI involves the weaknesses of the system. The RPI is a tool in the tournament selection process, so it's not meant to be used until March.
We can agree that the best thing the RPI has going for it is its simple formula. For those who don't know, it's
25% x your winning percentage (WP) + 50% x your opponents' WP
+ 25% x your opponents' opponents' WP
More simply its 25% x WP + 75% x strength of schedule (SOS)
It's not something one can compute in their head. But any dork with a computer can calculate the RPI. While many fans know the formula, nobody really knows how it works. Which leads us to the first of the three main complaints about the RPI:
Complaint #1: ¾ of the RPI is out of a team's control.
While it seems like an obvious truth, that's not how it works at all. To illustrate this let's look at the range of values for both winning percentage and strength of schedule among all teams in recent end-of-season RPI's.
Year      Max WP     Min WP      Diff         Max SOS   Min SOS      Diff
2001      .9286      .0385      .8901         .6127      .4080      .2047
Avg.                            .8827                               .2140
So the portion in the team's control has a range of values of roughly .8827, while the portion out of a teams control only has a range of .2140. Even when one accounts for the fact the winning percentage is just 25% of the formula, it still turns out to have a bigger impact than SOS:
WP: .25 x .8827 = .2207
SOS: .75 x .2140 = .1605
SOS plays an important role, but unless your SOS is in the bottom third of college Basketball, a poor schedule can be overcome with a great record. For instance last year Weber St. had a schedule ranked 178 out of 327, but with a 25-3 record they were able to have an RPI rank of 41. This is the beauty of the RPI, you can't really schedule your way into a good rating as most people think. The more difficult your schedule, the harder it is to maintain a good record and therefore a good RPI.
The range of values in the SOS is also part of the reason the RPI is useless early in the year. SOS has a much greater range this time of year, so it does control the ratings. But as the year progresses, everybody's SOS gravitates towards .500 and winning percentage becomes more important.
Till here we related the myth that strength of schedule controls the RPI, while winning percentage is relatively insignificant. Now I'll examine the common complaint that merely playing "so and so" can lower or raise your RPI, regardless of the outcome
This is a statement that can't be refuted because it's true. And this article isn't really a defense, but a demonstration as to how the RPI works.
The effect of an RPI fluctuation merely due to whom one plays is obvious early in the year. By the end of the season this effect is reduced, but not eliminated. The RPI is only meaningful at the end of the year, so that's the perspective one should take when poking holes in it. And the RPI really only matters to a few teams at the end of the year, and those are the teams in the 35-70 range that are under consideration for the NCAA tournament. So for an example we should examine a bubble team's data at the end of the season.
Therefore let's look at the #43 team from last year, Gonzaga. They weren't really a bubble team, but their non-conference schedule provides more variety than anyone else, so they are a great team to examine. Gonzaga played #1 Kentucky, but also #308 Long Beach St. Here's how Gonzaga's RPI rank would have looked with certain games removed from their schedule:
Loss vs. #1 Kentucky............43 (no change)
Win  vs. #308 Long Beach........38 (+5)
Win  vs. #232 Washington St.....40 (+3)
Win  vs. #161 Washington........43 (no change)
As the game against Kentucky indicates, losing to a highly ranked team doesn't help a bubble team's RPI much. So Gonzaga was not helped by losing to Kentucky, yet they did suffer for beating LBSU and WSU. The RPI is a mish-mash of results, where sometimes winning hurts (but you knew this). The thing is, almost everybody has games against weak teams on their schedule and feels this effect to some degree.
And this effect might not be so bad. One point to consider is this: how would you replace the stinker teams on the schedule? One way is to schedule a non-D1 team, a game that won't even exist is the eyes of the NCAA. Supposedly the committee frowns on such tactics, but to what degree who knows. The other option is to schedule a better team.
The problem is, the better the team you play, the more likely you are to lose, and a loss will certainly hurt your RPI. Gonzaga's loss to #82 San Diego had as much negative impact as their win against LBSU. Had they scheduled anyone worse than #82 and lost, the effect would have been worse than the win against LBSU. And the risk of losing to a team ranked 101-200 is twice that of playing someone ranked 201 or higher.
Teams ranked 40-60 in the RPI - the bubble teams - were 130-13 (.909 win %) against teams ranked 201+ in the RPI, but they were 146-36 (.802) against teams 101-200.
So in exchange for playing a nearly risk-free game, why shouldn't a team be penalized? True, sometimes one can't anticipate just how bad a scheduled team will be in the upcoming year. But for the most part, I'd say a team knows when they're scheduling a win. And a few teams seem to do this better than others (I'm looking at you Georgetown, Virginia, and Pitt). A little incentive to avoid this practice isn't a bad thing.
There are some improvements that could be made to the formula to make it more fair to teams that can't help playing bad teams because of conference affiliation.
Add margin of victory, or throw out the games against the 1 or 2 (or more) worst rated teams on the schedule. But any change would make a simple formula more complicated. The best thing the RPI has going for it is its simplicity. And unlike the BCS, the RPI is only a tool that humans use to make their own decisions, so whether Gonzaga finished 43rd or 38th wasn't a big deal.
This one deals with how a lack of including home court advantage affects the formula. For the most part the effects are negligible. For one, most of a team's schedule is made up of conference games that involve an equal distribution of home and away games. So the smaller non-conference part of the schedule is where a team can be unfairly rewarded or penalized for an imbalance of home or road games.
Sure, it's easy to pick on a team like Georgetown that constantly plays a non-conference schedule of weak opponents at home. However, even if they played all of their games on the road, the competition has been so weak you might only expect one additional loss if that. Of their three losable non-conference games (Penn St., Duke, Temple), two are on the road. So they are not a good example of a team that takes advantage of this weakness in the RPI.
Pitt's schedule is similar to Georgetown, they've had a lot of no-risk wins on their schedule. Those are games that have been at home, but it didn't matter where they were played, they were Pitt wins regardless. It's hard to find any tournament worthy team where the imbalance in home (or road) games could make a significant difference in their RPI. Vanderbilt might be an example, but if they are really a top 10 team, then even the Indiana game being played at Bloomington wouldn't have made much difference.
There's always a cry from the mid-majors about the lack of "away-game disadvantage" in the system. But take a team like Toledo or Western Michigan that has tourney aspirations. Their non-conference schedules are loaded with road games. But as of 12/23 both have one loss. Even if all the games were at home, their record could only be improved by one game so far (Toledo still has to play at Louisville).
Overall the complaints about home court advantage not being factored into the RPI are a lot of bluster.
The team that does the best job, year-in year-out, of abusing the home court advantage is Hawaii. It's really no fault of theirs because they never have to schedule non-conference road games and they usually get some decent competition to come to the islands. This year, 2003, they will have to head to the mainland to play in the Bracket Buster in February. But for the other 325 teams, the distribution of game locations doesn't make enough difference to justify making a simple formula more complicated.
Considering its simplicity, the RPI is pretty good at what it's supposed to do. Winning games is good, playing a tough schedule is good. But more fundamentally, just being a good team is good. There's really no easy way to cheat the system. And even if you could, there is a bunch of humans waiting at the end of the season to try to smooth out the problems.
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NCS Counselors provide the best possible treatment for each individual situation. Consultation is provided for a set fee, and, at times, a reduced fee based on the client's income. An amount of 70% of the counseling fees is donated to the less privileged in the region. The LEAF Foundation has various projects to build balance in people's lives. LEAF helps people with micro-financing, coaching and education.
It is our desire to help all people in need of help to receive it. Inquiries and referrals are welcome, whether from individuals, concerned family members and friends, counselors, or physicians and other professionals. Workshops and seminars are held on upon request.
NCS is a member of the association for counselors (ACC) in the United Kingdom.


text:
With 2.35 Average Overnight Trips Per Year, Sikhs Lead the Travellers Among all Religions
Highest incidence of trips is seen in J&K, with 836 per 100 households, followed by Himachal. Delhi at the bottom.
NEW DELHI: Indians are going places—travelling fast and furious. Domestic tourism is not only driving economy, it is also an indicator of consumption pattern of the country's citizens.
A tourism ministry study has put the Sikhs at the top of the heap when it comes to making most number of trips by a person in a year, followed by the Jains.
"The overall average number of overnight trips made by a person was estimated at 2.09. The averages differed among the religions. The average overnight trips by the Sikhs was (2.35), followed by Jains (2.19), Hindus (2.13), Christians (1.94), Muslims (1.82)," says the study conducted by the National Council of Applied Economic Research (NCAER) for the ministry.
New Delhi: The average number of nights spent on overnight trips was about three for the rural population and about four for the urban. However, the average duration of overnight trips undertaken by the urban population varied from two nights in case of shopping trips to 5.6 in case of medical trips.
As far as spending is concerned, Jains have the highest pertrip expenditure (`2,444 per trip) among all religions, followed by Christians (`1,740), Hindus (`965), Sikhs (`906) and Muslims (`879).
The report of the study, 'How households of different socioeconomic background spend on tourism', submitted recently, relied on the 200809 National Sample Survey (NSS) data to make its observations.
"The average number of overnight trips made by a rural person was estimated at 2.10, while for urban person, it was 2.07. Incidence of overnight trips is observed as the highest among Sikh persons in rural areas (2.37) and among Jains (2.34) in urban areas. While analysing incidence of trips by age groups of the visitors, highest trips per person is made by those belonging to 3059 years of agegroups and this is followed by persons belonging to 1529 years agegroup," the report says.
The study took into account domestic travel for various purposes, ranging from social reasons, religious, medical and leisure.
The urban Indian on an average made overnight trips of two nights for shopping; the number was 5.6 for medical trips. In rural areas, the least average duration was for shoppingrelated trips (1.5 nights), while the highest average duration pertained to leisure trips (4.3 nights).
"In India, shopping is the most important item of expenditure for Scheduled Tribe, Scheduled Caste and Other Backward Class," the report states.
Legislators, senior officials and managers among all professionals record the highest per overnight trip expenditure in India, of about `2,294.9.
In another study, NCAER, took into account how states performed on the domestic tourism front. Average trips made by a household are four during a year.
The highest incidence of trips is seen in Jammu & Kashmir, with 836 per 100 households, followed by Himachal Pradesh with 769 trips per 100 households. Delhi is at the bottom with only 232 average number of trips per 100 households.
On an average, at allIndia level, contribution of social trips to total trips works out to be 74 per cent. This is followed by religious trips and medical trips, which account for 10 per cent and 6.2 per cent respectively of the total overnight trips. Business and leisure trips together account for only 5.3 per cent of the total trips.
Leisure or holiday is among the least popular purposes of domestic overnight trips across all the states. The highest share (6.7 per cent) of leisure trips in total is seen in Delhi.
Social trips are undertaken the most in Uttar Pradesh, while Maharashtra, Andhra Pradesh and Tamil Nadu together account for 44 per cent of total religious trips. Tamil Nadu is the most soughtafter by domestic leisure tourists (12.5 per cent of total interstate leisure trips), according to the tourism ministry study.
Jains spend highest per trip with `2,444, followed by Christians (Rs 1,740), Hindus (Rs 965), Sikhs (Rs 906) and Muslims (Rs 879).
Published in: The Sunday Standard, April 10, 2016