text:
\section{Introduction}
\noindent
Throughout this paper, $R$ will denote a commutative ring with
identity and $\Bbb Z$ will denote the ring of integers.

An ideal $I$ of $R$ is said to be \emph{irreducible} if $I = J_1\cap J_2$ for ideals $J_1$ and
$J_2$ of $R$ implies that either $I=J_1$ or $I=J_2$. A proper ideal $I$ of $R$ is said to be
\emph{strongly irreducible} if for ideals $J_1, J_2$ of $R$, $J_1\cap J_2 \subseteq I$ implies that $J_1 \subseteq I$ or
$J_2 \subseteq I$ \cite{HRR02}. An ideal $I$ of $R$ is said to be \emph{2-irreducible} if whenever $I = J_1 \cap J_2 \cap J_3$ for ideals $J_1, J_1$ and $J_3$ of $R$, then
either $I = J_1 \cap J_2$ or $I= J_1 \cap J_3$ or $I = J_2 \cap J_3$. Clearly, any irreducible ideal is a
2-irreducible ideal \cite{DM16}.

A proper submodule $N$ of an $R$-module $M$ is said to be \emph{irreducible} (resp., \emph{strongly irreducible}) if for submodules $H_1$ and $H_2$ of $M$, $N=H_1\cap H_2$ (resp., $H_1\cap H_2 \subseteq N$) implies that  $N=H_1$ or $N=H_2$.( resp., $H_1 \subseteq N$ or $H_2 \subseteq N$).

The main purpose of this paper is to introduce the concept of 2-irreducible and strongly 2-irreducible submodules of an $R$-module $M$ as a generalization of irreducible and strongly irreducible submodules of $M$ and obtain some related results.

A submodule $N$ of an $R$-module $M$ is said to be a \textit{2-irreducible submodule} if whenever $N=H_1\cap H_2\cap H_3$ for submodules $H_1$, $H_2$ and $H_3$ of $M$, then either $N=H_1 \cap H_2$ or $N=H_2\cap H_3$ or $N=H_1\cap H_3$ (Definition \ref{11ld.1}).

A proper submodule $N$ of an $R$-module $M$ is said to be a \textit{strongly 2-irreducible submodule} if whenever $H_1\cap H_2\cap H_3 \subseteq N$ for submodules $H_1$, $H_2$ and $H_3$ of $M$, then either $H_1 \cap H_2 \subseteq N$ or $H_2\cap H_3 \subseteq N$ or $H_1\cap H_3 \subseteq N$  (Definition \ref{111ld.1}).

In Section 2 of this paper, for an $R$-module $M$, among other results, we prove that if $M$ is a Noetherian $R$-module and $N$ is a 2-irreducible submodule of $M$, then either $N$ is irreducible or $N$ is an intersection of exactly two irreducible submodules of $M$ (Theorem \ref{t1.5}). In Theorem \ref{t2.4}, we provide a characterization for strongly 2-irreducible submodules of $M$. Also, it is shown that if $M$ is a strong comultiplication $R$-module, then every non-zero proper submodule of $R$ is a  strongly sum 2-irreducible $R$-module if and only if every non-zero proper submodule of $M$ is a  strongly 2-irreducible submodule of $M$ (Theorem \ref{t42.1}).
Further, it is proved that if $N$ is a submodule of a finitely generated multiplication $R$-module $M$, then $N$ is a strongly 2-irreducible submodule of $M$ if and only if $(N:_RM)$ is a strongly 2-irreducible ideal of $R$ (Theorem \ref{p1.2}).
In Theorem \ref{t1.9} and \ref{tt1.9}, we provide some useful characterizations for strongly 2-irreducible submodules of some special classes of modules. Example \ref{e1.2} shows that the concepts of strongly irreducible submodules and strongly 2-irreducible submodules are different in general. Finally, let $R = R_1\times R_2\times \cdots \times R_n$ ($2\leq n < \infty$) be a decomposable ring and $M =M_1 \times M_2 \cdots \times M_n$ be an $R$-module, where for every $1\leq i \leq n$, $M_i$ is an $R_i$-module, respectively,  it is proved that a proper submodule $N$ of $M$ is a strongly 2-irreducible submodule of $M$ if and only if either $N =\times^n_{i=1}N_i$ such that for some $k \in \{1, 2, ..., n\}$, $N_k$ is a strongly 2-irreducible submodule of $M_k$, and $N_i =M_i$ for every $i\in \{1, 2, ..., n\}\setminus \{k\}$ or $N =\times^n_{i=1}N_i$ such that for some $k,m \in \{1, 2, ..., n\}$, $N_k$ is a strongly irreducible submodule of $M_k$, $N_m$ is a strongly irreducible submodule of $M_m$, and $N_i =M_i$ for every $i \in \{1, 2, ..., n\}\setminus \{k,m\}$ (Theorem \ref{t11.16}).

\section{Main results}
\begin{defn}\label{11ld.1}
We say that a submodule $N$ of an $R$-module $M$ is a \textit{2-irreducible submodule} if whenever $N=H_1\cap H_2\cap H_3$ for submodules $H_1$, $H_2$ and $H_3$ of $M$, then either $N=H_1 \cap H_2$ or $N=H_2\cap H_3$ or $N=H_1\cap H_3$.
\end{defn}

\begin{thm}\label{t1.5}
Let $M$ be a Noetherian $R$-module. If $N$ is a 2-irreducible submodule of $M$, then either $N$ is irreducible or $N$ is an intersection of exactly two irreducible submodules of $M$.
\end{thm}
\begin{proof}
Let $N$ be a 2-irreducible submodule of $M$. By \cite[Exercise 9.31]{SH90}, $N$ can be written
as a finite irredundant irreducible decomposition $N=N_1\cap N_2 \cap ... \cap N_k$. We show that
either $k=1$ or $k=2$. If $k > 3$, then since $N$ is 2-irreducible, $N=N_i \cap N_j$ for some
$1 \leq i,j \leq k$, say $i=1$ and $j=2$. Therefore $N_1 \cap N_2\subseteq N_3 $, which is a contradiction.
\end{proof}

\begin{cor}\label{c1.6}
Let $M$ be a Noetherian multiplication $R$-module. If $N$ is a 2-irreducible submodule of $M$, then $N$ a 2-absorbing primary  submodule of $M$.
\end{cor}
\begin{proof}
Let $N$ be a 2-irreducible submodule of $M$. By the fact that every irreducible submodule of a Noetherian $R$-module is primary and regarding Theorem \ref{t1.5}, we have either $N$ is a primary submodule or is a sum of two primary submodules. It is clear that every primary submodule is 2-absorbing  primary, also the sum of two primary submodules is a 2-absorbing
primary submodule, by \cite[Theorem 2.20]{mtoa16}.
 \end{proof}

\begin{defn}\label{111ld.1}
We say that a proper submodule $N$ of an $R$-module $M$ is a \textit{strongly 2-irreducible submodule} if whenever $H_1\cap H_2\cap H_3 \subseteq N$ for submodules $H_1$, $H_2$ and $H_3$ of $M$, then either $H_1 \cap H_2 \subseteq N$ or $H_2\cap H_3 \subseteq N$ or $H_1\cap H_3 \subseteq N$.
\end{defn}

\begin{ex}\label{e11ld.1}\cite[Corollary 2]{DM16}
Consider the $\Bbb Z$-module $\Bbb Z$. Then $n \Bbb Z$ is a strongly 2-irreducible submodule of  $\Bbb Z$ if $n=0$, $p^t$ or $p^rq^s$, where $p, q$ are prime integers and $t, r, s$ are natural numbers.
\end{ex}

\begin{prop}\label{p22.4}
The strongly 2-irreducible submodules of a distributive $R$-module are precisely the 2-irreducible submodules.
\end{prop}
\begin{proof}
This is straightforward.
\end{proof}

\begin{thm}\label{t2.4}
Let $N$ be a proper submodule of an $R$-module $M$. Then the following conditions are equivalent:
\begin{itemize}
\item [(a)] $N$ is a strongly 2-irreducible submodule;
\item [(b)] For all elements $x, y, z$ of $M$, we have
$(Rx + Ry) \cap (Rx + Rz) \cap (Ry+Rz)\subseteq N$ implies that either $(Rx + Ry) \cap (Rx + Rz)\subseteq N$ or  $(Rx + Ry) \cap (Ry+Rz)\subseteq N$ or $(Rx + Rz) \cap (Ry+Rz)\subseteq N$.
\end{itemize}
\end{thm}
\begin{proof}
$(a) \Rightarrow (b)$
This ia clear.

$(b) \Rightarrow (a)$
Let  $H_1\cap H_2\cap H_3 \subseteq N$ for submodules $H_1$, $H_2$ and $H_3$ of $M$. If  $H_1\cap H_2\not \subseteq N$,  $H_1\cap H_3 \not \subseteq N$, and  $H_2\cap H_3 \not \subseteq N$, then there exist  elements $x, y, z$ of $M$ such that $x \in H_2\cap H_3 $,  $y \in H_1\cap H_3$, and  $z \in H_1 \cap H_2$ but  $x \not \in N$, $y \not \in N$, and $z\not \in N$. Therefore,
$$
(Ry + Rz) \cap (Rx + Rz)\cap (Rx + Ry)\subseteq H_1\cap H_2 \cap H_3\subseteq  N.
$$
Hence by the part (a),  either $(Ry + Rz) \cap (Rx + Rz) \subseteq N$ or   $(Ry + Rz) \cap (Rx + Ry)\subseteq N$ or  $(Rx + Rz)\cap (Rx + Ry)\subseteq N$. Thus either $z \in N$ or $y \in N$ or $x \in N$. This contradiction completes the proof.
\end{proof}

Recall that a \textit{waist submodule} of an $R$-module $M$ is a submodule that is comparable to any other submodules of $M$.
\begin{prop}\label{p2.4}
Let $N$ be a proper submodule of an $R$-module $M$. Then we have the following.
\begin{itemize}
\item [(a)] If $N$ is a strongly 2-irreducible submodule of $M$, then it is also a 2-irreducible submodule of $M$.
\item [(b)] If $N$ is a strongly 2-irreducible submodule of $M$, then $N$ is a strongly 2-irreducible submodule of $T$ and $N/K$ is a strongly 2-irreducible submodule of $M/K$ for any $K \subseteq N \subseteq T$.
\item [(c)] If for all elements  $x, y, z$ of $M$ we have
$Rx \cap Ry \cap Rz \subseteq N$ implies that either $Rx \cap Ry \subseteq N$  or $Rx \cap Rz \subseteq N$  or $Ry \cap Rz \subseteq N$, then $N$ is a strongly 2-irreducible submodule of $M$.
\item [(d)] If $N$ is a waist submodule of $M$, then $N$ is strongly 2-irreducible submodule of $M$ if and only if $N$ is 2-irreducible module.
\item [(e)] If $N$ satisfies $(N + T) \cap (N +K) = N+ (T \cap K)$, whenever $T \cap K \subseteq N$, then $N$ is
strongly 2-irreducible submodule of $M$ if and only if $N$ is a 2-irreducible module.
\end{itemize}
\end{prop}

\begin{proof}
(a) Let $N$ be a strongly 2-irreducible submodule of $M$ and let $N=H_1\cap H_2 \cap H_3$ for submodules $H_1$, $H_2$ and $H_3$ of $M$. Then by assumption, either $H_1\cap H_2\subseteq N$ or  $H_1\cap H_3\subseteq N$ or  $H_2 \cap H_3\subseteq N$. Now the result follows from the fact that the reverse of inclusions are clear.

The parts (b), (d), and (e) are straightforward.

(c) Let  $ H_1 \cap H_2 \cap H_3 \subseteq N$ for submodules $H_1$, $H_2$ and $H_3$ of $M$. If  $ H_1 \cap H_2 \not \subseteq N$,  $ H_1 \cap H_3 \not \subseteq N$, and  $H_2 \cap H_3 \not \subseteq N$, then there exist elements $x, y, z$ of $M$ such that $x \in H_2\cap H_3$,  $y \in H_1\cap H_3$, and  $z \in H_1\cap H_2$ but  $x \not \in N$, $y \not \in N$, and $z \not \in N$. Now the result follows by assumption.
\end{proof}

An $R$-module $M$ is said to be a \emph{comultiplication module} if for every submodule $N$ of $M$ there exists an ideal $I$ of $R$ such that $N=(0:_MI)$, equivalently, for each submodule $N$ of $M$, we have $N=(0:_MAnn_R(N))$ \cite{AF07}.

An $R$-module $M$ satisfies the \emph{double annihilator
conditions} (DAC for short)  if for each ideal $I$ of $R$
we have $I=Ann_R(0:_MI)$ \cite{Fa95}.

An $R$-module $M$ is said to be a \emph{strong comultiplication module} if $M$ is
a comultiplication $R$-module and satisfies the DAC conditions \cite{AF09}.

A submodule $N$ of an $R$-module $M$ is said to be a \textit{strongly sum 2-irreducible submodule} if whenever $N\subseteq H_1+H_2+H_3$ for submodules $H_1$, $H_2$ and $H_3$ of $M$, then either $N\subseteq H_1+H_2$ or $N\subseteq H_2+H_3$ or $N\subseteq H_1+H_3$. Also, $M$ is said to be a \textit{strongly sum 2-irreducible module} if and only if $M$ is a strongly sum 2-irreducible submodule of itself \cite{FA05}.
\begin{thm}\label{t42.1}
Let $M$ be a strong comultiplication $R$-module. Then every non-zero proper submodule of $R$ is a strongly sum 2-irreducible $R$-module if and only if every non-zero proper submodule of $M$ is a  strongly 2-irreducible submodule of $M$.
\end{thm}
\begin{proof}
$"\Rightarrow"$
Let $N$ be a non-zero proper submodule of $M$ and let $H_1\cap H_2\cap H_3\subseteq N$ for submodules $H_1$, $H_2$ and $H_3$ of $M$. Then by using \cite[2.5]{HH19},
$$
Ann_R(N) \subseteq Ann_R(H_1)+Ann_R(H_2)+Ann_R(H_3).
$$
 This implies that either $Ann_R(N) \subseteq Ann_R(H_1)+Ann_R(H_2)$ or $Ann_R(N) \subseteq Ann_R(H_1)+Ann_R(H_3)$ or $Ann_R(N) \subseteq Ann_R(H_2)+Ann_R(H_3)$ since by assumption, $Ann_R(N)$ is a strongly sum 2-irreducible $R$-module. Therefore, either $H_1\cap H_2\subseteq N$ or $H_1\cap H_3\subseteq N$ or $H_2\cap H_3\subseteq N$ since $M$ is a comultiplication $R$-module.

$"\Leftarrow"$  Let $I$ be a non-zero proper submodule of $R$ and let $I \subseteq I_1+I_2+I_3$. Then
$$
(0:_MI_1)\cap (0:_MI_2)\cap (0:_MI_3) \subseteq (0:_MI).
$$
Thus by assumption,  either $(0:_MI_1)\cap (0:_MI_2) \subseteq (0:_MI)$ or $(0:_MI_1)\cap (0:_MI_3) \subseteq (0:_MI)$ or $(0:_MI_2)\cap (0:_MI_3) \subseteq (0:_MI)$. This implies that either $(0:_MI_1+I_2) \subseteq (0:_MI)$ or $(0:_MI_1+I_3) \subseteq (0:_MI)$ or $(0:_MI_2+I_3) \subseteq (0:_MI)$. Thus either $I \subseteq I_1+I_2$ or $I \subseteq I_1+I_3$ or $I \subseteq I_2+I_3$ since $M$ is a strong comultiplication $R$-module.
\end{proof}

An $R$-module $M$ is said to be a \emph{multiplication module} if for every submodule $N$ of $M$ there exists an ideal $I$ of $R$ such that $N=IM$ \cite{Ba81}.

\begin{thm}\label{p1.2}
Let $N$ be a submodule of a finitely generated multiplication $R$-module $M$. Then $N$ is a strongly 2-irreducible submodule of $M$ if and only if $(N:_RM)$ is a strongly 2-irreducible ideal of $R$.
\end{thm}
\begin{proof}
$"\Rightarrow"$ Let $N$ be a strongly  2-irreducible submodule of $M$ and let $ J_1\cap J_2 \cap J_3 \subseteq (N:_RM)$ for some ideals $J_1, J_2$, and $J_3$ of $R$. Then
$$
J_1M\cap J_2M \cap J_3M \subseteq (N:_RM)M=N
$$
 by \cite[Corollary 1.7]{BS89}. Thus by assumption, either  $J_1M\cap J_2M \subseteq N$ or  $J_1M\cap J_3M \subseteq N$ or $J_2M \cap J_3M \subseteq N$. Hence, either
$ (J_1\cap J_2)M \subseteq (N:_RM)M$ or $ (J_1\cap J_3)M \subseteq (N:_RM)M$ or $(J_2 \cap J_3)M \subseteq (N:_RM)M$.
Therefore, either $ J_1\cap J_2\subseteq (N:_RM)$ or  $ J_1\cap J_3 \subseteq (N:_RM)$ or  $J_2 \cap J_3 \subseteq (N:_RM)$ by \cite[Corollary of Theorem 9]{SM88}.

$"\Leftarrow"$
Let $(N:_RM)$ is a strongly 2-irreducible ideal of $R$ and let
$H_1\cap H_2 \cap H_3\subseteq N$ for some submodules $H_1$, $H_2$ and $H_3$ of $M$. Then we have
$$
(H_1\cap H_2 \cap H_3:_RM)M=((H_1:_RM)\cap (H_2 :_RM)\cap (H_3:_RM))M\subseteq (N:_RM)M.
$$
Thus $(H_1:_RM)\cap (H_2 :_RM)\cap (H_3:_RM)\subseteq (N:_RM)$
by \cite[Corollary of Theorem 9]{SM88}. Hence,  either  $(H_1:_RM)\cap (H_2 :_RM)\subseteq (N:_RM)$ or  $(H_1:_RM)\cap (H_3:_RM))\subseteq (N:_RM)$ or  $(H_2 :_RM)\cap (H_3:_RM)\subseteq (N:_RM)$
since $(N:_RM)$ is a strongly 2-irreducible ideal of $R$.
Therefore, either $H_1\cap H_2\subseteq N$ or $H_1\cap H_3\subseteq N$ or $H_2 \cap H_3\subseteq N$
 by \cite[Corollary 1.7]{BS89}.
\end{proof}

\begin{ex}
Consider the $\Bbb Z$-module $\Bbb Z_{p^tq^nr^m}$, where $p, q, r$ are prime integers and $t,n, m$ are natural numbers.
\begin{itemize}
\item [(a)] By using Theorem \ref{p1.2} and Example \ref{e11ld.1}, one can see that $\bar{p^t}\Bbb Z_{p^tq^nr^m}$ and $\bar{q^nr^m}\Bbb Z_{p^tq^nr^m}$ are strongly 2-irreducible submodules of  $\Bbb Z_{p^tq^nr^m}$.
\item [(b)] $\bar{pqr}\Bbb Z_{p^3qr}=\bar{pq}Z_{p^3qr} \cap \bar{pr}\Bbb Z_{p^3qr} \cap \bar{qr}\Bbb Z_{p^3qr}$ implies that $\bar{pqr}\Bbb Z_{p^3qr}$ is not  a  2-irreducible submodule of  $\Bbb Z_{p^3qr}$.
\end{itemize}
\end{ex}

The following example shows that the concepts of strongly irreducible submodules and strongly 2-irreducible submodules are different in general.
\begin{ex}\label{e1.2}
Consider the $\Bbb Z$-module $\Bbb Z_6$.  Then $0=\bar{3}\Bbb Z_6 \cap \bar{2}\Bbb Z_6$ implies that the $0$  submodule of $\Bbb Z_6$ is not strongly irreducible. But $(0:_{\Bbb Z}\Bbb Z_6)=6\Bbb Z$ is a strongly 2-irreducible ideal of $\Bbb Z$ by Example \ref{e11ld.1}. Since the $\Bbb Z$-module $\Bbb Z_6$  is a finitely generated multiplication $\Bbb Z$-module, $0$ is a strongly 2-irreducible submodule of $\Bbb Z_6$ by Theorem \ref{p1.2}.
\end{ex}

\begin{lem}\label{p1.4}
Let $M$ be an $R$-module. If $N_1$ and $N_2$ are strongly irreducible submodules of $M$, then $N_1\cap N_2$ is a  strongly
2-irreducible submodule of $M$.
\end{lem}
\begin{proof}
This is straightforward.
\end{proof}

A proper submodule $P$ of an $R$-module $M$ is said to be \emph{prime} if for any $r \in R$ and $m \in M$ with $rm \in P$, we have $m \in P$ or $r \in (P:_RM)$ \cite{Da78}.
\begin{prop}\label{p1.7}
Let $M$ be a multiplication $R$-module and let $N_1$, $N_2$, and $N_3$ be prime submodules of $M$ such that $N_1 + N_2=N_1 +N_3=N_2 + N_3=M$. Then $N_1\cap N_2 \cap N_3$ is not a  strongly 2-irreducible submodule of $M$.
\end{prop}
\begin{proof}
Assume on the contrary that $N_1\cap N_2 \cap N_3$ is a  strongly 2-irreducible submodule of $M$. Then $N_1\cap N_2 \cap N_3\subseteq N_1\cap N_2 \cap N_3$ implies that either $N_1\cap N_2 \subseteq N_1\cap N_2 \cap N_3$ or $N_1\cap N_3\subseteq N_1\cap N_2 \cap N_3$ or $N_2 \cap N_3\subseteq N_1\cap N_2 \cap N_3$. We can assume without loss of generality that $N_1\cap N_2 \subseteq N_1\cap N_2 \cap N_3$. Then $N_1\cap N_2 \subseteq N_3$. It follows that $(N_1:_RM)N_2 \subseteq N_3$. As $N_3$ is a prime submodule of $M$, we have  $N_2 \subseteq N_3$ or $(N_2:_RM) \subseteq (N_3:_RM)$. Thus $N_2 \subseteq N_3$ or $N_1 \subseteq N_3$ since $M$ is a multiplication $R$-module. Therefore, $N_3=M$, which is a contradiction.
\end{proof}

\begin{cor}\label{c1.8}
Let $M$ be a multiplication $R$-module such that every proper submodule of $M$ is  strongly 2-irreducible. Then $M$ has at most two maximal submodules.
\end{cor}
\begin{proof}
This follows from Proposition \ref{p1.7}
\end{proof}

Let $N$ be a submodule of an $R$-module $M$. The intersection of all prime submodules of $M$ containing $N$ is said to be the (\emph{prime}) \emph{radical} of $N$ and denote by $rad_MN$ (or simply by $rad(N)$). In case $N$ does not contained in any prime submodule, the radical of $N$ is defined to be $M$.
Also, $N \not =M$ is said to be a \emph{radical submodule} of $M$ if $rad(N)=N$ \cite{MM86}

\begin{lem}\label{l11.9}
Let $I$ be an ideal of $R$ and $N$ be a submodule of an $R$-module $M$. Then $rad(IN)=rad(N) \cap rad(IM)$.
\end{lem}
\begin{proof}
By \cite [Corollary of Theorem 6]{Lu89}, we have $rad(N\cap IM))= rad(N)\cap rad(IM)$. Since $IN \subseteq IM \cap N$, $rad(IN) \subseteq rad(IM \cap N)$. Thus $rad(IN) \subseteq rad(N)\cap rad(IM)$. Now let $P$ be a prime submodule of $M$ such that $IN\subseteq P$. As $P$ is prime, $N \subseteq P$ or $I \subseteq (P:_RM)$. Hence  $N \cap IM \subseteq P$. This in tourn implies that $rad(N)\cap rad(IM)\subseteq  rad(IN) $, as desired.
\end{proof}

A proper ideal $I$ of $R$ is said to be a \emph{2-absorbing ideal}
of $R$ if whenever $a, b, c \in R$ and $abc \in I$, then $ab \in I$ or
 $ac \in I$ or $bc \in I$ \cite{Ba07}.

A proper submodule $N$ of an $R$-module $M$ is said to be a \emph{2-absorbing primary submodule} of $M$ if whenever $a, b \in R$, $m \in M$, and $abm \in N$, then $am \in rad(N)$ or $bm \in rad(N)$  or $ab \in (N :_R M)$ \cite{mtoa16}.

A proper submodule $N$ of an $R$-module $M$ is called a \emph{2-absorbing submodule }of
 $M$ if whenever $abm \in N$
for some $a, b \in R$ and $m \in M$, then $am \in N$ or $bm \in N$ or
$ab \in (N :_R M)$ \cite{YS11} and  \cite{pb12}.

\begin{thm}\label{t1.9}
Let $M$ be a finitely generated multiplication $R$-module and $N$ be a radical submodule of $M$. Then the following
conditions are equivalent:
\begin{itemize}
\item [(a)] $N$ is a strongly 2-irreducible submodule of $M$;
\item [(b)] $N$ is a 2-absorbing submodule of $M$;
\item [(c)] $N$ is a 2-absorbing primary submodule of $M$;
\item [(d)] $N$ is either a prime submodule of $M$ or is an intersection of exactly two prime submodules of $M$.
\end{itemize}
\end{thm}
\begin{proof}
$(a) \Rightarrow (b)$
Let $I$, $J$ be ideals of $R$ and $K$ be a submodule of $M$ such that $IJK \subseteq N$. Then
 by using  Lemma \ref{l11.9},
$$
 K\cap IM \cap JM \subseteq rad(K)\cap rad(IM)\cap  rad(JM)=rad(IJK) \subseteq rad(N)=N
$$
Hence by part (a), either $ K\cap IM \subseteq N$ or $ K\cap JM \subseteq N$ or $IM \cap JM \subseteq N$. Thus either $IK\subseteq N$ or $JK\subseteq N$ or $IJM \subseteq N$ as needed.

$(b) \Rightarrow(c)$
This is clear.

$(c) \Rightarrow (b)$
This is clear by using \cite[Theorem 2.6]{mtoa16}.

$(b) \Rightarrow (d)$
Since $N$ is a 2-absorbing submodule of $M$,  $(N:_RM)$ is a 2-absorbing ideal of $R$ by \cite[Proposition 1]{DS12}.
Hence $\sqrt{(N:_RM)} = P$ is a prime ideal of $R$  or $\sqrt{(N:_RM)} =  P \cap Q$, where $P$ and $Q$ are distinct
prime ideals of $R$ that are minimal over $(N:_RM)$ by \cite[Theorem 2.4]{Ba07}.
We have $\sqrt{(N:_RM)}M=rad(N)$ by \cite[Theorem 4]{MM86}.
If $\sqrt{(N:_RM)} = P$, then $PM=rad(N)$. Since $M$ is a multiplication $R$-module, $PM$ is a prime submodule of $M$ by \cite[Corollary 2.11]{BS89}. Now let $\sqrt{Ann_R(N)} =  P \cap Q$, where $P$ and $Q$ are distinct
prime ideals of $R$. Then $(P\cap Q)M=rad(N)$. By \cite[Corollary 1.7]{BS89}, $(P\cap Q)M=PM\cap QM$.  Thus  by \cite[Corollary 2.11]{BS89}, $rad(N)$ is an intersection of two prime submodules of $M$. Now, we prove the claim by
assumption that $N$ is a radical submodule of $M$.

$(d) \Rightarrow (a)$
This follows from Lemma \ref{p1.4}.
\end{proof}

The following example shows that parts $(a)$ and $(b)$ of Theorem \ref{t1.9} are not equivalent in general.
\begin{ex}\label{e11.2}
Consider the submodule $G_t=\langle 1/p^t+\Bbb Z\rangle$ of the $\Bbb Z$-module $\Bbb Z_{p^\infty}$.  Then $G_t$ is a strongly 2-irreducible submodule of $\Bbb Z_{p^\infty}$.  But  $G_t$ is not a 2-absorbing submodule of $\Bbb Z_{p^\infty}$. It should be note that the $\Bbb Z$-module $\Bbb Z_{p^\infty}$ is not a finitely genrated multiplication $\Bbb Z$-module.
\end{ex}

A submodule $N$ of an $R$-module $M$ is said to be \emph{pure} if $IN=IM \cap N$ for every ideal $I$ of $R$ \cite{AF74}.
Also, an $R$-module $M$ is said to be \emph{fully pure} if every submodule of $M$ is pure \cite{AF122}.
\begin{thm}\label{tt1.9}
Let $M$ be a fully pure multiplication $R$-module and $N$ be a submodule of $M$. Then the following
conditions are equivalent:
\begin{itemize}
\item [(a)] $N$ is a strongly 2-irreducible submodule of $M$;
\item [(b)] $N$ is a 2-absorbing submodule of $M$;
\item [(c)] $N$ is a 2-irreducible submodule of $M$.
\end{itemize}
\end{thm}
\begin{proof}
$(a) \Rightarrow (b)$
Let $I$, $J$ be ideals of $R$ and $K$ be a submodule of $M$ such that $IJK \subseteq N$. Then since $M$ is fully pure,
$$
K \cap IM \cap JM=IJK \subseteq N.
$$
Hence by part (a), either $K \cap IM\subseteq N$ or $K \cap JM\subseteq N$ or $IM \cap JM\subseteq N$. Thus either $IK\subseteq N$ or $JK\subseteq N$ or $IJM \subseteq N$.

$(b) \Rightarrow (a)$
Let $H_1\cap H_2 \cap H_3\subseteq N$ for submodules $H_1$, $H_2$ and $H_3$ of $M$. Then
$$
(H_1:_RM)\cap (H_2:_RM) \cap (H_3:_RM)=(H_1\cap H_2 \cap H_3:_RM)\subseteq (N:_RM).
$$
Thus either $(H_1:_RM)(H_2:_RM)\subseteq (N:_RM)$ or  $(H_1:_RM)(H_3:_RM)\subseteq (N:_RM)$ or  $(H_2:_RM)(H_3:_RM)\subseteq (N:_RM)$ since $(N:_RM)$ is a 2-absorbing ideal of $R$ by \cite[Proposition 1]{DS12}.
We can assume without loss of generality that  $(H_1:_RM)(H_2:_RM)\subseteq (N:_RM)$. Thus as $M$ is fully pure, we have
$$
 (H_1:_RM)M \cap (H_2:_RM)M\subseteq (N:_RM)M \subseteq N.
$$
Therefore, $H_1 \cap H_2 \subseteq N$ since $M$ is a multiplication $R$-module.

$(a) \Leftrightarrow (c)$
By \cite[proof of Theorem 2.19]{AF122}, $M$ is a distributive $R$-module. Now the result follows from Proposition \ref{p22.4}.
\end{proof}

\begin{lem}\label{t1.5}
Let $M$ be an $R$-module, $S$ a multiplicatively closed subset of $R$, and $N$ be a finitely generated submodule of $M$. If $S^{-1}N \subseteq S^{-1}K$ for a submodule $K$ of $M$, then there exists $s \in S$ such that $sN \subseteq K$.
\end{lem}
\begin{proof}
 This is straightforward.
\end{proof}

\begin{prop}\label{p1.6}
Let $M$ be an $R$-module, $S$ be a multiplicatively closed subset of $R$ and $N$ be a finitely generated prime strongly 2-irreducible submodule of $M$ such that $(N:_RM) \cap S=\emptyset$. Then $S^{-1}N$ is a strongly 2-irreducible submodule of $S^{-1}M$ if $S^{-1}N\not =S^{-1}M$.
\end{prop}
\begin{proof}
Let $S^{-1}H_1\cap S^{-1}H_2 \cap S^{-1}H_3\subseteq S^{-1}N$ for submodules $S^{-1}H_1$, $S^{-1}H_2$ and $S^{-1}H_3$ of $S^{-1}M$. Then
$S^{-1}(H_1\cap H_2 \cap H_3)\subseteq S^{-1}N$.  By Lemma \ref{t1.5}, there exists $s \in S$ such that $s (H_1 \cap H_2 \cap H_3) \subseteq N$. This implies that $H_1 \cap H_2 \cap H_3 \subseteq N$ since $N$ is prime and $(N:_RM) \cap S=\emptyset$. Now as $N$ is a strongly 2-irreducible submodule of $M$, we have either $H_1 \cap H_2 \subseteq N$ or $H_1 \cap H_3 \subseteq N$ or $H_2 \cap H_3 \subseteq N$. Therefore, either  $S^{-1}H_1\cap S^{-1}H_2\subseteq S^{-1}N$ or $S^{-1}H_1\cap S^{-1}H_3\subseteq S^{-1}N$ or $S^{-1}H_2\cap S^{-1}H_3\subseteq S^{-1}N$, as needed.
\end{proof}

\begin{prop}\label{p8.21}
Let $M$ be an $R$-module and $\{K_i\}_{i \in I}$ be a chain of strongly
2-irreducible submodules of $M$. Then $\cap_{i \in I}K_i$ is a strongly 2-irreducible submodule of $M$.
\end{prop}
\begin{proof}
Let $H_1\cap H_2 \cap H_3\subseteq \cap_{i \in I}K_i$ for submodules $H_1$, $H_2$ and $H_3$ of $M$.
Assume that $H_1+H_2\not\subseteq \cap_{i \in I}K_i$, $H_1+H_3\not\subseteq \cap_{i \in I}K_i$, and $H_2+H_3\not\subseteq \cap_{i \in I}K_i$. Then there are $m,n,t \in I$, where
$ H_1\cap H_2 \not \subseteq K_m$, $H_1\cap H_3 \not\subseteq K_n$, and $H_2\cap H_3 \not\subseteq K_t$.  Since  $\{K_i\}_{i \in I}$ is a chain we can assume that $K_m \subseteq K_n\subseteq K_t$. But as  $H_1\cap H_2 \cap H_3\subseteq K_m$ and $K_m$ is a strongly sum 2-irreducible submodule of $M$, we have either $H_1\cap H_2 \subseteq K_m$ or  $H_1\cap H_3\subseteq K_m$ or $H_2 \cap H_3\subseteq K_m$.
In any case, we get a contradiction.
\end{proof}

\begin{thm}\label{t11.12}
Let $f : M \rightarrow \acute{M}$ be a epimorphism of R-modules. Then we have the following.
\begin{itemize}
\item [(a)] If $N$ is a strongly 2-irreducible submodule of $M$ such that $ker(f) \subseteq N$, then $f(N)$ is a  strongly 2-irreducible submodule  of $f(M)$.
\item [(b)] If $\acute{N}$ is a strongly 2-irreducible submodule of $\acute{M}$, then $f^{-1}(\acute{N})$ is a strongly 2-irreducible submodule of $M$.
 \end{itemize}
\end{thm}
\begin{proof}
(a) Let $N$ be a strongly 2-irreducible submodule of $M$. If $f(N) =\acute{M}$, then we have
$N+Ker(f)=f^{-1}(f(N))=f^{-1}(\acute{M})=f^{-1}(f(M))=M$. Now as $ker(f) \subseteq N$, we get that $N=M$, which is a contradiction. Therefore, $f(N) \not=\acute{M}$. Suppose that $\acute{H_1}\cap \acute{H_2}\cap \acute{H_3}\subseteq f(N)$ for submodules $\acute{H_1}$, $\acute{H_2}$ and $\acute{H_3}$ of $f(M)$. Then
$ f^{-1}(\acute{H_1}) \cap f^{-1}(\acute{H_2}) \cap f^{-1}(\acute{H_3})\subseteq f^{-1}(f(N))=N$ since $ker(f) \subseteq N$. Thus by assumption, either $ f^{-1}(\acute{H_1}) \cap f^{-1}(\acute{H_2})\subseteq N$ or $ f^{-1}(\acute{H_1}) \cap f^{-1}(\acute{H_3})\subseteq N$ or $f^{-1}(\acute{H_2}) \cap f^{-1}(\acute{H_3})\subseteq N$. Now as $f$ is epimorphism,  we have either $\acute{H_1}\cap \acute{H_2}\subseteq f(N)$ or $\acute{H_1}\cap \acute{H_3}\subseteq f(N)$ or $\acute{H_2}\cap \acute{H_3}\subseteq f(N)$, as needed.

(b) Let $\acute{N}$ be a strongly 2-irreducible submodule of $\acute{M}$.
Since $\acute{N} \not =\acute{M}$ and $f$ is a epimorphism, we have $f^{-1}(\acute{N}) \not =M$.
Now let $H_1\cap H_2 \cap H_3\subseteq f^{-1}(\acute{N})$ for submodules $H_1$, $H_2$ and $H_3$ of $M$. Then
$f(H_1)\cap f(H_2) \cap f(H_3)\subseteq f(f^{-1}(\acute{N}))= \acute{N}$. Thus by assumption, either $f(H_1)\cap f(H_2)\subseteq \acute{N}$ or $f(H_1)\cap f(H_3)\subseteq \acute{N}$ or $f(H_2) \cap f(H_3)\subseteq \acute{N}$. Now we have either $H_1\cap H_2\subseteq f^{-1}(\acute{N})$ or $H_1\cap H_3\subseteq f^{-1}(\acute{N})$ or $H_2 \cap H_3\subseteq f^{-1}(\acute{N})$, as required.
\end{proof}

\begin{thm}\label{t222.1}
Let $M$ be a finitely generated multiplication distributive $R$-module and let $N$ be a non-zero proper submodule of $M$. Then the following statements are equivalent:
\begin{itemize}
\item [(a)] $N$ is a strongly 2-irreducible submodule of $M$;
\item [(b)] $(N:_RM)$ is a strongly 2-irreducible ideal of $R$;
\item [(c)] $(N:_RM)$ is a 2-irreducible ideal of $R$.
\end{itemize}
\end{thm}
\begin{proof}
$(a)\Rightarrow (b)$
This follows from Theorem \ref{p1.2}.

$(b)\Rightarrow (c)$
This follows from \cite[Proposition 1]{DM16}.

$(c)\Rightarrow (a)$
Let $H_1\cap H_2 \cap H_3\subseteq N$ for submodules $H_1$, $H_2$ and $H_3$ of $M$. Then as $M$ is a distributive $R$-module, we have
$$
N=N +(H_1\cap H_2\cap H_3)=(N+ H_1)\cap (N\cap H_2)\cap (N\cap H_3).
$$
This implies that
$(N:_RM)=(N+ H_1:_RM) \cap (N+ H_2:_RM) \cap (N+ H_3:_RM)$. Thus by assumption, either$(N:_RM)=(N+ H_1:_RM) \cap (N+ H_2:_RM)$ or $(N:_RM)=(N+ H_1:_RM)\cap (N+ H_3:_RM)$ or $(N:_RM)=(N+ H_2:_RM) \cap (N+ H_3:_RM)$. Therefore, by \cite[Corollary 1.7]{BS89}, either $N=N+(H_1\cap H_2)$ or $N=N+(H_1\cap H_3)$ or $N=N+(H_2\cap H_3)$, since $M$ is a finitely generated multiplication $R$-module. Thus either, $H_1\cap H_2\subseteq N$ or  $H_1\cap H_3\subseteq N$  or  $H_2 \cap H_3\subseteq N$  as needed.
\end{proof}

Let $R_i$ be a commutative ring with identity and $M_i$ be an $R_i$-module, for $i = 1, 2$. Let $R = R_1 \times R_2$. Then $M = M_1 \times M_2$ is an $R$-module and each submodule of $M$ is in the form of $N = N_1 \times N_2$ for some submodules $N_1$ of $M_1$ and $N_2$ of $M_2$.

\begin{thm}\label{t11.15}
Let $R = R_1 \times R_2$ be a decomposable ring and $M = M_1 \times M_2$
be an $R$-module, where $M_1$ is an $R_1$-module and $M_2$ is an $R_2$-module. Suppose that $N = N_1 \times N_2$ is a proper submodule of $M$. Then the following conditions are equivalent:
\begin{itemize}
  \item [(a)] $N$ is a strongly 2-irreducible submodule of $M$;
  \item [(b)] Either $N_1 = M_1$ and $N_2$ strongly 2-irreducible submodule of $M_2$ or $N_2 = M_2$ and $N_1$ is a strongly 2-irreducible submodule of $M_1$ or $N_1$, $N_2$ are strongly irreducible submodules of $M_1$, $M_2$, respectively.
\end{itemize}
\end{thm}
\begin{proof}
$(a) \Rightarrow (b)$.
Let $N = N_1 \times N_2$ be a strongly 2-irreducible submodule of $M$ such that $N_2 = M_2$. From our hypothesis, $N$ is proper, so $N_1 \not =M_1$. Set $\acute{M}=M/(0\times M_2)$. One can see that $\acute{N}=N /(0\times M_2)$ is a strongly 2-irreducible submodule of $\acute{M}$. Also, observe that $\acute{M} \cong M_1$ and $\acute{N} \cong N_1$. Thus $N_1$ is a strongly 2-irreducible submodule of $M_1$. By a similar argument as in the previous case, $N_2$ is a strongly 2-irreducible submodule of $M_2$, where, $N_1 =M_1$. Now suppose that $N_1\not =M_1$ and $N_2 \not =M_2$. We show that $N_1$ is a irreducible submodule of $M_1$. Suppose that $H_1\cap K_1\subseteq N_1$ for some submodules $H_1$ and $K_1$ of $M_1$. Then
$$
(H_1\times M_2) \cap (M_1\times 0) \cap (K_1\times M_2)  \subseteq (H_1\cap K_1)\times 0 \subseteq N_1\times N_2.
$$
Thus by assumption, either $(H_1\times M_2) \cap (M_1\times 0) \subseteq N_1\times N_2$
or $(H_1\times M_2)\cap (K_1\times M_2)  \subseteq N_1\times N_2$ or $(M_1\times 0) \cap (K_1\times M_2)  \subseteq N_1\times N_2$. Therefore, $H_1 \subseteq N_1$ or $K_1 \subseteq N_1$ since $N_2 \not =M_2$. Thus $N_1$ is a strongly irreducible submodule of $M_1$. Similarly, we can show that $N_2$ is strongly irreducible submodule of $M_2$.

$(b) \Rightarrow (a)$.
Suppose that $N = N_1 \times M_2$, where $N_1$ is a strongly 2-irreducible submodule of $M_1$. Then it is clear that $N$ is a strongly 2-irreducible submodule of $M$. Now, assume that $N = N_1 \times N_2$, where $N_1$ and $N_2$ are strongly irreducible  submodules of $M_1$ and $M_2$, respectively. Hence $(N_1 \times M_2)\cap (M_1 \times N_2) = N_1 \times N_2 = N$ is a strongly 2-irreducible submodule of $M$, by Lemma \ref{p1.4}.
\end{proof}

\begin{thm}\label{t11.16}
Let $R = R_1\times R_2\times \cdots \times R_n$ ($2\leq n < \infty$) be a decomposable ring and $M =
M_1 \times M_2 \cdots \times M_n$ be an $R$-module, where for every $1\leq i \leq n$, $M_i$ is an $R_i$-module, respectively. Then for a proper submodule $N$ of $M$ the following conditions are equivalent:
\begin{itemize}
  \item [(a)] $N$ is a strongly 2-irreducible submodule of $M$;
  \item [(b)]  Either $N =\times^n_{i=1}N_i$ such that for some $k \in \{1, 2, ..., n\}$, $N_k$ is a strongly 2-irreducible submodule of $M_k$, and $N_i =M_i$ for every $i\in \{1, 2, ..., n\}\setminus \{k\}$ or $N =\times^n_{i=1}N_i$ such that for some $k,m \in \{1, 2, ..., n\}$, $N_k$ is a strongly irreducible submodule of $M_k$, $N_m$ is a strongly irreducible submodule of $M_m$, and $N_i =M_i$ for every $i \in \{1, 2, ..., n\}\setminus \{k,m\}$.
\end{itemize}
\end{thm}
\begin{proof}
We use induction on $n$. For $n = 2$ the result holds by Theorem \ref{t11.15}. Now
let $3\leq n < \infty$ and suppose that the result is valid when $K = M_1\times \cdots \times M_{n-1}$. We show that the result holds when $M = K \times M_n$. By Theorem \ref{t11.15}, $N$ is a strongly 2-irreducible submodule of $M$ if and only if either $N = L \times M_n$ for some strongly 2-irreducible submodule $L$ of $K$ or $N =K\times L_n$ for some strongly 2-irreducible submodule $L_n$ of $M_n$ or $N = L \times L_n$ for some strongly irreducible submodule $L$ of $K$ and some strongly irreducible submodule $L_n$ of $M_n$. Note that a proper submodule $L$ of $K$ is a strongly irreducible submodule of $K$ if and only
if $L = \times^{n-1}_{i=1}N_i$ such that for some $k \in \{1, 2, ..., n - 1\}$, $N_k$ is a strongly irreducible submodule of $M_k$, and $N_i =M_i$ for every $i\in \{1, 2, ..., n - 1\}\setminus \{k\}$. Consequently the claim is now verified.
\end{proof}

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


NEW HAVEN — Their first official basket, just 10 seconds into their existence, came on a John Winchester 3-pointer from the corner. The fun was just beginning for the Connecticut Topballerz.
Shortly thereafter came an alley-oop dunk from Bobby Moore to Roosevelt Lee, one of the many highlights of a thoroughly entertaining, high-scoring 48 minutes of basketball at Albertus Magnus College.
The Topballerz, the first-year New Haven-based entry into the American Basketball Association, cruised to a 122-104 victory over the Jersey Express in their inaugural game Friday night. Moore finished with 26 points, seven rebounds and five assists and was named the game's MVP, Lee and Anthony Davis each added 17 points and Joe Dorseet-Jeffreys netted 14.
More importantly, the Topballerz, comprised mainly by former high school and college stars from the area, seemed to truly delight the crowd of about 700 inside the Cosgrove Marcus Messer Athletic Center.
The night couldn't have gone much better, according to head coach A.J. Huston.
Keep in mind, this isn't your father's ABA. Julius Erving or George Gervin aren't walking through that door anytime soon. But the up-and-down style, combined with the local flavor of the team, makes the Topballerz a fun watch.
There are also some interesting quirks to the league. For one, any shot made from halfcourt or beyond is worth four points. Then there's the "3-D Rule," which adds an additional point to any basket made after the opponent turns the ball over on a pass from backcourt, or loses possession of the ball (violation of foul) before player control is established in the frontcourt.
Winchester's 3-pointer kicked off a 13-4 Topballerz start that was highlighted by Moore's alley-oop pass to Lee.
"Me and Rosey have been playing together for so many years, that's our specialty," said Moore.
The locals jumped out to as much as a 21-6 lead, but the Express battled back and were within eight (53-45) at halftime.
Winchester added 11 points for Connecticut and Lee and Davis each hauled in 11 rebounds. Japhet McNeil had four steals. Jersey's Darnell Braswell led all scorers with 37 points.
The evening started off with New Haven's Kaseem Johnson dedicating the season to the memory of T.J. Mathis, the former Hamden High star who was fatally shot in September. Flowers and a framed Topballerz jersey No. 10 were presented to his family.
Did the entire night go off without a hitch? No. There were some troubles with the microphone and P.A. system, and other assorted glitches associated with an opening game.
But overall, the Connecticut Topballerz' debut was a rousing success – both on the floor and off.


text:
Ecotourism, sustainable development, and conservation education: Development of a tour guide training program in Tortuguero, Costa Rica
@article{Jacobson1992EcotourismSD,
title={Ecotourism, sustainable development, and conservation education: Development of a tour guide training program in Tortuguero, Costa Rica},
author={Susan K. Jacobson and Rafael Azevedo Robles},
journal={Environmental Management},
S. Jacobson, R. Robles
Published 1 November 1992
A tour guide training program was developed for rural communities near Costa Rica's Tortuguero National Park to respond to the impacts of the 24-fold increase in park visitation in the past decade, to involve local communities in resource management, and to provide regional environmental education. The development of the training course involved a survey of scientists and park managers to ascertain resource management needs, priorities for information to be disseminated, and impacts of tourism…
widecast.org
View All 6 Figures & Tables
Mountain guiding in Kinabalu park: an evaluation of visitor satisfaction
Goh Hong Ching
Tourism development in protected area could potentially become one of the most significant sources of revenue and bring substantial benefits to the community living surrounding protected areas. Local…
Systems approach to design and development of a tourist guide training program: a case study of Sharjah tourism
Mohit Vij, A. Upadhya
ABSTRACT This study is a scientific synthesis of developing and conducting a tourist guide training program focused on destination – Sharjah, United Arab Emirates. The Sharjah Commerce and Tourism…
Linking commercial success of tour operators and agencies to conservation and community benefits in Costa Rica
Lisa Seales, T. Stein
SUMMARY Throughout the world tourism is a tool for economic growth, conservation and improved quality of life for local residents, yet negative environmental impacts and economic leakages are common.…
Sustainable tourism and the influence of privatization in protected area management. A case of Kinabalu Park, Malaysia
U. Bonn
Trends in environmental education for biodiversity conservation in Costa Rica
Amanda Jiménez, M. Monroe, N. Zamora, J. Benayas
Environment, Development and Sustainability
Costa Rica is internationally recognized for its abundant biodiversity and being a leader in the promotion of education strategies for biodiversity conservation. We interviewed staff from 16…
Case Study: The Evolution of Tourism and Sea Turtle Conservation at Tortuguero National Park, Costa Rica
Jimena Gutiérrez-Lince, Michelle Dorantes Palacios, R. Valverde
Abstract Sea turtle conservation and research efforts in Tortuguero, Costa Rica, began in the 1950s. Legislation protecting sea turtles and creating Tortuguero National Park (TNP) in the 1960 and…
Ecotourism as a Tool for Conservation of Coast Redwoods in Santa Cruz County, California
C. Rao
ECOTOURISM AS A TOOL FOR CONSERVATION OF COAST REDWOODS IN SANTA CRUZ, CALIFORNIA By Chandana Rao According to the World Tourism Organization, ecotourism is the fastest growing segment of the tourism…
Ecotourism and conservation in the Americas
A. Stronza, W. Durham
Ecotourism has been credited with conserving biological and cultural diversity, alleviating rural poverty, increasing public awareness of environmental concerns and strengthening ties between tourism…
Ecotourism Benefits and the Role of Local Guides at Masoala National Park, Madagascar
A. Ormsby, K. Mannle
An investigation of ecotourism at Masoala National Park, a forested coastal area in northeastern Madagascar and the country's largest national park, focused on ecotourism benefits and the role of…
Eco-Tourism Development Strategy Baluran National Park in the Regency of Situbondo, East Java, Indonesia
A. Siswanto, M. Moeljadi
Baluran National Park in the regency of Situbondo, East Java-Indonesia, highly prospective for development of sustainable tourism that can improve the welfare of local people. The suitable tourism…
Nature tourism and rural development in Tortuguero
S. Place
Abstract Costa Rica's national parks have profoundly affected local populations by causing rural economies to move from resource extraction to resource protection. This study focuses on the…
Conservation Education Programmes: Evaluate and Improve Them
S. Jacobson
Evaluation of the conservation education programme at Kinabalu Park in Malaysia demonstrated the importance of such evaluation in determining a programme's worth and providing due observation of…
Interpretation and Nonformal Environmental Education in Natural Resources Management
E. Olson, M. Bowman, R. Roth
Abstract A visitor education strategy was developed, implemented, and tested for its effectiveness in raising levels of knowledge and attitudes toward state nature preserve management. Visitor…
The concept of a tourist area cycle of evolution: implications for management of resources.
R. Butler
The concept of a recognizable cycle in the evolution of tourist areas is presented, using a basic s curve to illustrate their waving and waning popularity. Specific stages in the evolutionary…
Site Fixity in the Caribbean Green Turtle
A. Carr, M. H. Carr
The straying of nesting females as far as 7 km from previous nesting sites suggests how colony proliferation may occur and seems unlikely that olfaction could be an important cue in landfall and nest—site discrimination, and new delimitation of the homing goal reinforces that belief.
Tourism: Economic, physical, and social impacts
A. Mathieson, G. Wall
Economics, Geography
""1. introduction 2. conceptualization 3. economic impacts4. physical impacts5. social impacts6. conclusions"references and bibliography""index"""
Resident attitudes about a national park and conservation in Sabah , Malaysia
Resident populations and national parks in developing nations : Interdisciplinary perspectives and policy implications
The paradox of tourism in Costa Rica
Cultural Survival Quarterly
Nature travel in the tropics.
J. Laarman, P. Durst
Community conservation education program for the golden lion tamarin . Pages 8 - 16 in Proceedings of the building support for conservation in rural areas conference , 27 - 31 May 1986

redpajama_set_name: RedPajamaCommonCrawl


text:
UFC Interview
Chris Weidman says Conor McGregor will beat Jose Aldo due to belief in the Law of Attraction
If the featherweight championship fight between Conor McGregor and Jose Aldo still goes ahead, Chris Weidman believes 'The Notorious' will emerge victorious because of his unwavering beliefs.
By Lewis Mckeever@Mckeever89 Jun 24, 2015, 9:02am EDT
Share All sharing options for: Chris Weidman says Conor McGregor will beat Jose Aldo due to belief in the Law of Attraction
Joe Camporeale-USA TODAY Sports
With Jose Aldo's rib injury suffered during a sparring session with lightweight fighter Alcides Nunes, it's unknown whether the 145 lbs. champ will be able to defend his title against Conor McGregor at UFC 189. The medical results -- scheduled for release later today -- will determine the severity of Aldo's injury. If the suspicions over a rib fracture are true, it's expected that the Brazilian will have to sit out until October.
If, however, Jose Aldo will be able to soldier on for the UFC 189 main event, middleweight champ Chris Weidman believes Conor McGregor will take home the featherweight belt. In a recent off the cuff call with Chael Sonnen, the All-American believes McGregor will dethrone Aldo because he believes in the Law of Attraction:
"I'm going with Conor McGregor," Weidman said. "His mental mindset is the number one reason I'm with him because the way he talks - he believes it. It's something called the Law of Attraction and I believe in it. I think it's going to end up coming to fruition; all these dreams and goals he's had for himself for such a long time and he's been brave enough to talk about them out loud. I think they will come through on the night of the fight."
The Law of Attraction is based upon the principle of focusing on positive or negative ideas to manifest situations into reality. If someone is fixated on positive ideas, those ideas will eventually spawn into existence.
Chris Weidman could be seen utilizing the Law of Attraction prior to his championship clash with Vitor Belfort at UFC 187. The defending champ stuck a picture of Belfort on his fridge attached with the message 'I, Chris Weidman will defeat Vitor Belfort on May 23rd'.
Conor McGregor discussed his belief in the LOA in an in-depth interview with Ariel Helwani prior to his scrap with Diego Brandão at UFC Fight Night 46. If Jose Aldo is unable to defend his title at UFC 189, McGregor may have to utilize his self-affirming techniques against Frankie Edgar instead.
Jose Aldo injured, UFC 189 bout vs. Conor McGregor is off
McGregor vid: Aldo's emotions will be his downfall
Weidman: Conor champ due to law of attraction
In 2014, McGregor predicted Aldo would pull out
Askarov returns with Absolute Championship Akhmat


Most states understand robbery as theft or larceny of assets, that is, property or money, through the offender's use of physical force or fear against a victim. Where a deadly weapon such as a gun is used, or the victim suffers injury, the robbery may be charged as "armed" or "aggravated." Unlike burglary, the crime of robbery almost always requires the presence of a victim who suffers actual injury, or is threatened with harm.


Point, Counter-Point: Leader takes all, not.
Later this month, as the result of the Indian parliamentary elections will be announced, there will be— in all probability— a new leader in New Delhi. Notwithstanding, the multiple parties and factions involved, the average Indian voter is torn between the choice of economic development with Gujarat characteristics vs. a decadent, corrupt and unimpressive family-run political empire. As is also well known, the former is most likely to form the government in New Delhi this month.
Narendra Modi is different from his competitors. His forceful oratory skills, strong personality, humble origins, and escaping conviction in the 2002 Gujarat riots, make him a candidate to reckon with. The question however is: is that enough to warrant change in Indian foreign policy at the international and regional strategic levels?
The answer I propose is no.
In order to answer this question, I would not delve into the organizational character of a parliamentary democracy, and executive-legislative relations. This is because strong leadership personalities can transcend the executive in both parliamentary and presidential systems. Instead, I shall first cite some examples from the past 67 years of independent Indian foreign policymaking to my case. Moreover, I shall try and substantiate my viewpoint by briefly alluding to the Indian strategic culture debate, the question of the geopolitical threat of Hindu fundamentalism, and the aspect of US-Modi relations.
The historical examples I enlist demonstrate that in India, when a new leader from the opposition political party had attained power, s/he did not manifest a turnaround on larger foreign policy issues of the country. In other words, rhetoric had varied in its pitches, and explanatory rationales for state behavior had witnessed adjustments. Yet, at the larger foreign policy level, not much had altered.
a. The US-India civil nuclear agreement finalized in 2008 traced itself to the 2005 joint statement released by President George W. Bush and Prime Minister Manmohan Singh. This statement and the 2004 Next Steps in Strategic Partnership (NSSP) were preceded by the Bush-Vajpayee joint statement of September 2001. This statement underlined cooperation in defense, space and energy among others. In later years, Vajpayee had criticized the nuclear deal, but that the deal was a gradual materialization of improving US-India ties initiated during Vajpayee's prime ministerial stint is hard to deny.
b. As is known, India conducted five nuclear tests in May 1998. Prime Minister Vajpayee's letter famously underlined the Chinese threat as the cause for the tests. Yet, an attempt was made to test in December 1995, by then Prime Minister P.V. Narasimha Rao. That Rao was from the secular Congress Party and Vajpayee from the Hindu nationalist Bharatiya Janata Party (BJP) did not seem to matter in their propensity to test nuclear devices. Besides, the Indian nuclear program dating back to 1948 has been an endeavor of the Congress Party and its predecessors. The first nuclear test in 1974 also did not witness the involvement of the BJP. Then, why is May 1998 so often projected as an example of muscle-flexing Hindu nationalist foreign policy? A selective historical memory, I would say.
c. During the Cold War, US insistence that India accept full-scope IAEA safeguards on the Tarapur reactors, was not accepted by Prime Minister Indira Gandhi during the Emergency (June 1975-March 1977). Nor was it acceptable to the subsequent Morarji Desai government, much to the Carter administration's dismay. In fact, the Indian ambassador to the United States during the Desai government, Nani Palkhivala published an article in Newsweek entitled, "Disarming the Unarmed," which was reminiscent of V.C. Trivedi's 1966 speech at the Eighteen Nations Disarmament Committee in Geneva, in which the latter had underlined that a nuclear nonproliferation treaty must not lead to the "non-armament of the unarmed countries." In other words, New Delhi's position on the larger foreign policy issues related to the nuclear question remained largely unchanged. India's refusal to accept full-scope IAEA safeguards persisted when Indira Gandhi returned to power in 1980. In 1982, France eventually agreed to replace the United States as the fuel supplier to the Tarapur reactors.
Another interesting example is India's military-strategic ties with Israel. Although Indo-Israeli ties acquired greater visibility during the NDA government, the military strategic relations between the two countries had existed since at least the mid-1960s. One tends to remember National Security Advisor Brajesh Mishra's address at the annual dinner at the American Jewish Committee but forget that Israel supplied military hardware to India in both the 1965 and 1971 war against Pakistan.  This is more interesting because military and strategic cooperation preceded the formal establishment of diplomatic relations, which did not take place until early 1992. In 2009, during the UPA government, Israel became the largest defense supplier of India.
To my understanding, Indo-Israeli military ties have continued irrespective of the leader and the party in power. What has varied is the anti-Pakistan/anti-Muslim rhetoric concurrent to the ties. Such rhetoric is higher when a Hindu nationalist coalition is in power, and lower when a secular coalition hold the reins.
It is understandable that when a new leader is poised to ascend to power in New Delhi, he and his cohorts would underline his leadership style as different and unique. To be fair, there are often tangible changes in leadership styles. However, does that amount to massive alterations in Indian foreign policy? Not always, as the historical examples demonstrate.
Personally, I am expecting a renewed debate on Indian strategic culture in the next years since New Delhi will soon witness the ascent of a Hindu right nationalist government. It would be interesting to see however, if the debate is differentiating rhetoric from substance.
The geopolitical threat of Hindu fundamentalism?
Many people in India are worried about the rise of Hindu fundamentalism with the ascent of Narendra Modi to power. With respect to foreign policy however, the moot question is, will Modi's brand of Hindu fundamentalism be a geopolitical threat to India's neighbors?
a) Modi has to conduct his foreign policy through the Indian Foreign Service— the bureaucratic corps of Indian foreign policy. As a result, the institutional inertia will check his gung-ho calls.
b) In order to radically alter the foreign policymaking, especially with respect to military strategic issues, the nature of civil-military relations in New Delhi will have to undergo a major shift, which is unforeseeable in the near future.
c) The BJP election manifesto does not elaborate much on foreign policy issues. This might mean that there is less clear thinking on external issues of the country— a consistent post-independence era trend.
However, since "anti-Muslim" stance of the BJP in the domestic sphere is effortlessly equated with its anti-Pakistan position in the external sphere, tensions may be witnessed between South Asian adversaries during Modi's tenure. In fact, it is possible that the anti-Pakistan rhetoric might give the impression that the situation is far worse than it probably is. Moreover, aggression and armed conflict between India and Pakistan cannot be entirely ruled out since fundamentalist/right-wing nationalist forces often thrive on rallying around the flag. Yet, to my understanding, massive map-changing military campaigns will be avoided by Modi's New Delhi.
US-Modi relations a check on regional brinkmanship?
(a) Modi was boycotted by the United States for his involvement in the 2002 Gujarat riots and was denied visa to enter the country. As former US ambassador Nancy Powell's recent meeting with Modi demonstrates, the US is willing to reconsider and welcome Modi as the new leader of India. As the new Prime Minister, Modi himself will aspire for better relations with Washington.
(b) Modi is "good for business" and this economic diplomacy will boost Modi's aspirations for better ties with the United States, and US goals of having better relations with a highly market-friendly national leader.
Washington hence has the possibility of transforming its economic leverage into a political one, and thereby to ensure that Modi does not do anything rash in the regional strategic environment of South Asia, so much so that it deeply upsets US ally Pakistan.
In other words, I do not see any massive changes in foreign policymaking in India under Narendra Modi. I do however foresee a lot of fundamentalist rhetoric. In the next years, foreign policy analysts will thus be faced with the challenging task of differentiating all that noise from the actual action.
 For an in-depth analysis of Indo-Israeli relations see P. R. Kumaraswamy, India's Israel Policy (New York: Columbia University Press, 2010).
One would say the main driver of the Gujarat model is that for the first time since 1947 a leader has managed to terrorise the bureaucracy into doing what its paid to do.
Why do you think he won't be able to bulldoze/terrorise the IFS?
Thanks for the comment, Obhijeet.
I have more faith in institutional inertia. Besides, Gujarat is not the rest of India. A 'one-size-fits-all' policy, if it exists at all, in N. Modi's toolbox cannot be successful nationwide. Moreover, the IFS is not the only obstacle to Modi's radical foreign policy choices.
With this kind of majority (with the possibility of 2/3rds if Jayalalitha joins) I think all checks and balances calculations are out of the window.
That said looking at his modus operandi in Gujarat he will set up all those who can challenge him in powerful posts 'commensurate ' with their seniority… allow the Peter principle to take over…. purge them after three years, and that's when the real changes will happen…when people least expect it.
Yeah, with such a majority things are more unpredictable at the moment. Let's see where this heads.
PM-to-be Modi's first move — inviting all neighboring leaders — speaks volumes. Indian policy seems to be going from analog to digital.
You remember in my 26th March article here i had said NaMo would take the peace peace process forward? Today's attack on our Consulate in Herat notwithstanding, the BJP has expressed optimism that Sharif will attend. I am optimistic, and i do believe he will take the process forward like no one before has.
Sadly the two more alarmist trends i had spotted in that same article are also coming true, a lack of preparation for a response when a major terror strike attacks India (this is a matter of if not when IMO) and the fact that he seems to be following a nationalist defense policy based on procurement and spending, rather than good old fiscally responsible, right wing, effects based policy.
Note that in their presentations the three chiefs have been asked to categorize their shopping lists as "urgent" and "necessary", instead of justifying what they want to achieve and their grand strategy.
Euphoria came true, sadly i think Alarm and Exasperation will also come true.
Sorry for the belated response, Abhijit and MK.
P.S.: I absolutely love Jay-Z-ita!


text:
#ifndef __KERN_PROCESS_PROC_H__
#define __KERN_PROCESS_PROC_H__

#include <defs.h>
#include <list.h>
#include <trap.h>
#include <memlayout.h>
#include <skew_heap.h>

// process's state in his life cycle
enum proc_state {
    PROC_UNINIT = 0,  // uninitialized
    PROC_SLEEPING,    // sleeping
    PROC_RUNNABLE,    // runnable(maybe running)
    PROC_ZOMBIE,      // almost dead, and wait parent proc to reclaim his resource
};

// Saved registers for kernel context switches.
// Don't need to save all the %fs etc. segment registers,
// because they are constant across kernel contexts.
// Save all the regular registers so we don't need to care
// which are caller save, but not the return register %eax.
// (Not saving %eax just simplifies the switching code.)
// The layout of context must match code in switch.S.
struct context {
    uint32_t ra;
    uint32_t sp;
    uint32_t s0, s1, s2, s3, s4, s5, s6, s7;
    uint32_t gp;
    uint32_t fp;
};

#define PROC_NAME_LEN               50
#define MAX_PROCESS                 4096
#define MAX_PID                     (MAX_PROCESS * 2)

extern list_entry_t proc_list;

struct inode;
struct fs_struct;

struct proc_struct {
    enum proc_state state;                      // Process state
    int pid;                                    // Process ID
    int runs;                                   // the running times of Proces
    uintptr_t kstack;                           // Process kernel stack
    volatile bool need_resched;                 // bool value: need to be rescheduled to release CPU?
    struct proc_struct *parent;                 // the parent process
    struct mm_struct *mm;                       // Process's memory management field
    struct context context;                     // Switch here to run process
    struct trapframe *tf;                       // Trap frame for current interrupt
    pde_t* pgdir ;                              // Page Directroy Table(PDT)
    uint32_t flags;                             // Process flag
    char name[PROC_NAME_LEN + 1];               // Process name
    list_entry_t list_link;                     // Process link list
    list_entry_t hash_link;                     // Process hash list
    int exit_code;                              // exit code (be sent to parent proc)
    uint32_t wait_state;                        // waiting state
    struct proc_struct *cptr, *yptr, *optr;     // relations between processes
    struct run_queue *rq;                       // running queue contains Process
    list_entry_t run_link;                      // the entry linked in run queue
    int time_slice;                             // time slice for occupying the CPU
    skew_heap_entry_t lab6_run_pool;            // FOR LAB6 ONLY: the entry in the run pool
    uint32_t lab6_stride;                       // FOR LAB6 ONLY: the current stride of the process
    uint32_t lab6_priority;                     // FOR LAB6 ONLY: the priority of process, set by lab6_set_priority(uint32_t)
    struct files_struct *filesp;                // the file related info(pwd, files_count, files_array, fs_semaphore) of process
};

#define PF_EXITING                  0x00000001      // getting shutdown

#define WT_INTERRUPTED               0x80000000                    // the wait state could be interrupted
#define WT_CHILD                    (0x00000001 | WT_INTERRUPTED)  // wait child process
#define WT_KSEM                      0x00000100                    // wait kernel semaphore
#define WT_TIMER                    (0x00000002 | WT_INTERRUPTED)  // wait timer
#define WT_KBD                      (0x00000004 | WT_INTERRUPTED)  // wait the input of keyboard

#define le2proc(le, member)         \
    to_struct((le), struct proc_struct, member)

extern struct proc_struct *idleproc, *initproc, *current;

void proc_init(void);
void proc_run(struct proc_struct *proc);
int kernel_thread(int (*fn)(void *), void *arg, uint32_t clone_flags);

char *set_proc_name(struct proc_struct *proc, const char *name);
char *get_proc_name(struct proc_struct *proc);
void cpu_idle(void) __attribute__((noreturn));

struct proc_struct *find_proc(int pid);
int do_fork(uint32_t clone_flags, uintptr_t stack, struct trapframe *tf);
int do_exit(int error_code);
int do_yield(void);
int do_execve(const char *name, int argc, const char **argv);
int do_wait(int pid, int *code_store);
int do_kill(int pid);
//FOR LAB6, set the process's priority (bigger value will get more CPU time)
void lab6_set_priority(uint32_t priority);
int do_sleep(unsigned int time);
#endif /* !__KERN_PROCESS_PROC_H__ */

redpajama_set_name: RedPajamaGithub


text:
<?php

use_helper('I18N','rtTemplate');

slot('rt-title', ucwords(__(sfConfig::get('rt_shop_receipt_title', 'receipt'))));

$address = $rt_shop_order->getShippingAddress();
if(!$address)
{
  $address = $rt_shop_order->getBillingAddress();
}

$order_received = '
' . __('Thank you!') . '
';
$order_received .= '
'  . sprintf('%s%s%s',__('Your order has been received. The order reference is #'),$rt_shop_order->getReference(),__(' and a confirmation email has been sent to you.')) . '
';
?>

  <?php if(sfConfig::get('app_rt_templates_headers_embedded', true)): ?>

<?php echo ucwords(__(sfConfig::get('rt_shop_receipt_title', 'receipt'))) ?>

  <?php endif; ?>

    <?php echo $order_received ?>

    <?php if(sfConfig::get('app_rt_analytics_ecommerce_enabled', false)): ?>

    <?php endif; ?>