text:
\section{Introduction}

The present paper is a sequel to \cite{BS} and so the notations
introduced in Introduction of \cite{BS} will remain in force. In
particular, all rings  are associative commutative with identity
$1\neq 0$ and modules are unital. Let $M$ be an $R$-module.  If
$N$ is a submodule (resp. proper submodule) of a module  $M$ we
write $N\leq M$ (resp. $N\lneqq M$). Also, we denote the
annihilator of the factor module $M/N$  by $(N:M)$ and we call $M$
faithful if $(0:M)=0.$

 Recall that the {\it spectrum }
Spec($R$) of a ring $R$ consists of all prime ideals of $R$ and is
non-empty. For each
 ideal $I$ of $R$, we set ${V(I)=\{{\cal{P}}\in {\rm Spec}(R) :
I\subseteq {\cal{P}}\}}$. Then the sets $V(I)$, where $I$ is an
ideal of $R$,  satisfy the axioms for the closed sets of a
topology on Spec($R$), called the {\it Zariski topology} of $R$.
The distinguished open sets of Spec($R$) are the open sets of the
form $D(f)=\{{\cal{P}}\in {\rm Spec}(R) : f\not\in
{\cal{P}}\}={\rm Spec}(R)\setminus V(f)$, where $V(f)=V(Rf)$.
These sets form a basis for the Zariski topology on Spec$(R)$ (see
for examples, Atiyah and Macdonald \cite{Atiyah} and Hartshorne
\cite{Hart}).

A proper submodule $P$ of $M$ is called a {\it prime submodule} if
$am\in P$ for  $a\in R$, and $m\in M$, implies that  $m\in P$ or
$aM\subseteq P$.  The prime spectrum Spec$(M)$ is defined to be
the set of all prime submodules of $M$.

  In the literature, there are many different
generalizations of the Zariski topology of rings to modules via
prime submodules. In  a series of papers (see for example,
\cite{Lu2,Lu3,MMS1,MMS2,MMS3,BH1,BH2,BN}) a group of algebraists
including mainly  Lu,  McCasland,  Moore, Smith and  Behboodi
carried out an intensive and systematic study of the spectrum of
prime submodules. For instance, as [13], for any submodule $N$ of
$M$ we define the {\it variety}
  of $N$, denoted by $V(N)$  to be the set of all prime submodules $P$ of
$M$ such that  $(N:M)\subseteq (P:M)$. Assume that $\zeta(M)$
denotes the collection of all subsets $V(N)$ of Spec$(M)$, then
$\zeta(M)$ contains the empty set and Spec$(M)$, and also,
$\zeta(M)$ is closed under arbitrary intersections and finite
union  and  under finite unions. Thus it is evident that for every
module $M$ there always exists a topology, $\tau$ say, on
Spec$(M)$ having  we conclude that $\zeta(M)$ as the family of all
closed sets. $\tau$ is called the {\it Zariski topology} on
Spec$(M)$. This topology on Spec$(M)$ is studied in
 [13].

It is well-known that for any commutative ring $R$, there is a
sheaf of rings on Spec$(R)$, called the {\it structure sheaf},
denoted by ${\cal{O}}_{{\rm Spec}(R)}$, defined as follows: for
each prime ideal ${\cal{P}}$ of $R$, let $R_{\cal{P}}$ be the
localization of $R$ at ${\cal{P}}$. For an open set $U\subseteq
{\rm Spec}(R)$, we define ${\cal{O}}_{{\rm Spec}(R)}(U)$ to be the
set of functions $s: U\rightarrow \coprod_{{\cal{P}}\in
U}R_{\cal{P}}$, such that $s({\cal{P}})\in R_{\cal{P}}$, for each
${\cal{P}}\in U$, and such that $s$ is locally a quotient of
elements of $R$: to be precise, we require that for each
${\cal{P}}\in U$, there is a neighborhood $V$ of ${\cal{P}}$,
contained in $U$, and elements $a,f\in R$, such that for each
${\cal{Q}}\in  V$, $f\not\in {\cal{Q}}$, and
$s({\cal{Q}})=\frac{a}{f}$ in $R_{\cal{Q}}$ (see for example
Hartshorne \cite{Hart},
 for definition and  basic properties of the sheaf
${\mathcal O}_{{\rm Spec}(R)}$).

For an $R$-module $M$ let $\psi: {\rm Spec}(M)\rightarrow {\rm
Spec}(R/{\rm Ann}(M))$ defined by $\psi(P)=(P:M)/{\rm Ann}(M)$ for
every $P\in {\rm Spec}(M)$. $\psi$ is called the natural map of
${\rm Spec}(M)$. An $R$-module $M$ is called {\it primeful}  if
either $M=(0)$ or $M\neq (0)$ and the natural map $\psi: {\rm
Spec}(M)\rightarrow {\rm Spec}(R/{\rm Ann}(M))$ is surjective.
This notion of primeful module has been extensively studied by Lu
\cite{Lu1}). Let $M$ be a primeful faithful $R$-module.   In
\cite{Tekir}, the author obtained  an $R$-module
$\mathcal{O}_X(U)$ for each open set $U$ in ${\rm Spec}(M)$ such
that $\mathcal{O}_{{\rm Spec}(M)}$ is a sheaf of modules over $X$.
In fact, $\mathcal{O}_{{\rm Spec}(M)}$ is a generalization of the
structure sheaf of rings to primeful faithful modules.

 The purpose of this paper  is to develop the
structure sheaf of  rings to a wider class of modules  called {\it
$\mathbb{P}$-radical modules} that was introduced in Part I (see
\cite{BS}).  Let $N$ be a proper submodule of an $R$-module $M$.
Then the {\it prime radical} $\sqrt[p]{N}$  is the intersection of
all prime submodules of $M$ containing $N$ or, in case there are
no such prime submodules, $\sqrt[p]{N}$  is $M$. Clearly
$V(N)=V(\sqrt[p]{N})$. We note that, for each ideal $I$ of $R$,
$\sqrt[p]{I}=\sqrt{I}$ (the intersection of all prime ideals of
$R$ containing $I$).  Let $M$ be  an $R$-module $M$. Then $M$ is
called  a {\it $\mathbb{P}$-radical module}  if $M$ satisfies
 the prime radical condition
``$(\sqrt[p]{{\cal{P}}M}:M)={\cal{P}}$'' for every prime  ideal
${\cal{P}}\supseteq {\rm Ann}(M)$.   In  \cite[Proposition 2.3 and
Example 2.4]{BS}
 it is shown that the  class of
$\mathbb{P}$-radical  modules contains the class of primeful
$R$-modules properly. Hence, all free modules, all finitely
generated modules and all homogeneous semisimple modules lie in
the class of $\mathbb{P}$-radical modules. Moreover, if $R$ is a
Noetherian ring (or an integral domain), then every projective
$R$-modules is $\mathbb{P}$-radical (see \cite[Theorem 2.5 and
Corollary 2.4(iii)]{BS}). In particular, if $R$ is an Artinian
ring, then
 all $R$-modules
are $\mathbb{P}$-radical (see \cite[Theorem 2.13]{BS}).

In this paper, we continue the study of this construction. In
Section 2, the behavior of $\mathbb{P}$-radical modules under
localization and direct sums  are studied. Section 3 is devoted to
the discussion of the main topic of this paper, the resemblance of
the theory of spectrum of $\mathbb{P}$-radical $R$-modules (with
the Zariski topology) and that of commutative  rings. In fact, for
each $R$-module $M$, we obtain an $R$-module ${\mathcal O}(U)$ for
each open set $U$ in Spec$(M)$. We show that ${\mathcal O}_{{\rm
Spec}(M)}$ is a sheaf of $R$-modules over Spec$(M)$. Moreover, the
following results are two fundamental
theorems in Section 3:\vspace{2mm}\\
\noindent{\bf Theorem 3.1} {\it Let $R$ be a ring and $M$ be an
$R$-module. For any prime submodule $P$ of $M$, the stalk
${{\mathcal O}_{{\rm Spec}(M)}}_P$ is isomorphic to
$M_{\mathcal{P}}$, where
$(P:M)={\mathcal{P}}$.}\vspace{2mm}\\
\noindent{\bf Theorem 3.2} {\it Let $R$ be a ring and  $M$ be a
$\mathbb{P}$-radical   $R$-module. Then for each $f\in R$, the
module ${\mathcal O}_{{\rm Spec}(M)}(D(fM))$ is isomorphic to the
 localized module  $M_f$. In particular, ${\mathcal O}_{{\rm Spec}(M)}({\rm Spec}(M))\cong M$.}

Finally, as an application in Section 4, we prove the following
interesting result:\vspace{2mm}\\
\noindent{\bf Theorem  4.1} {\it Let M be a $\mathbb{P}$-radical
$R$-module and $f,g\in R$. Then, $M_f\cong M_g$ as $R$-modules if
and only if $\sqrt{(fM:M)}=\sqrt{(gM:M)}$.} \

\section{The Behavior of $\mathbb{P}$-Radical Modules Under
Localization and Direct Sums }

 Now we are going to   investigate
  two important properties
of the $\mathbb{P}$-radical   Modules.
 In fact we show the affect of the localization on the $\mathbb{P}$-radical   property of modules.
 Moreover, we investigate the finite direct sum of the $\mathbb{P}$-radical   Modules. First, we  need the following two
lemmas  for our purpose:
\begin{Lem} Let $R$ be a ring and  $I$, $J$ and $K$ be ideals of $R$. Then,
$$\sqrt{(I+J)\cap (I+K)}=\sqrt{I+(J\cap K)}.$$
\end{Lem}
\noindent {Proof.} Let $r\in \sqrt{(I+J)\cap (I+K)}$. Then there
exist two integers $s$ and $t$ such that $r^s=i_1+j$ and
$r^t=i_2+k$ for some $i_1,i_2\in I$, $j\in J$ and $k\in K$. Then
we have $r^{s+t}=i_1i_2+i_1k+i_2j+kj$. So $r^{s+t}\in I+J\cap K$
and consequently $r\in \sqrt{I+(J\cap K)}$ i.e. $\sqrt{(I+J)\cap
(I+K)}\subseteq\sqrt{I+(J\cap K)}$. It is easy to check the
converse of this relation and so the equality satisfied.$~\square$

If $P$ is a prime  submodule of an $R$-module $M$, then
${\cal{P}}= (P:M)$  is a prime ideal and we shall say that $P$  is
a ${\cal{ P}}$-prime submodule.

 \begin{Lem} {\rm \cite [Lemma 4.6]{Lu1}} Let $R$ be a ring, $M_1$ and $M_2$ be $R$-modules  and $\mathcal P\in V(Ann(M))$. If $P$ is a
$\mathcal P$-prime submodule of $M_1$, then $P\bigoplus M_2$
 is a $\mathcal P$-prime submodule of $M$.
\end{Lem}
 \begin{The} Every  finite direct sum of $\mathbb{P}$-radical   modules, is a
$\mathbb{P}$-radical module.
\end{The}
\noindent{\bf Proof.} Let $M_1$ and $M_2$ be two
  $\mathbb{P}$-radical   $R$-modules. It is suffices to show that
  the $R$-module $M=M_1\bigoplus M_2$   is a $\mathbb{P}$-radical   module. Let ${\cal{P}}$
be a prime  ideal of $R$ containing ${\rm Ann}(M)={\rm
Ann}(M_1)\bigcap {\rm Ann}(M_2)$. Clearly
$V({\cal{P}}M_i)=V(({\cal{P}}+{\rm Ann}(M_i))M_i)$
for  $i=1,2$. Now using Lemmas 2.1 and 2.2, we have:\\
 $$(\sqrt[p]{{\cal{P}}M}:M)=\bigcap_{P\in V({\cal{P}}M)}(P:M)\subseteq(\bigcap_{P\in V({\cal{P}}M_1)}(P\oplus M_2:M))\bigcap(\bigcap_{P\in
V({\cal{P}}M_2)}(M_1\oplus P:M))$$
 $$~~~~~~~~~~~~~~~~~~=(\bigcap_{P\in V(({\cal{P}}+{\rm Ann}(M_1))M_1)}(P\oplus M_2:M))\bigcap(\bigcap_{P\in V(({\cal{P}}+{\rm Ann}(M_2))M_2)}(M_1\oplus
P:M))$$
 $$~~~=(\bigcap_{P\in V(({\cal{P}}+{\rm Ann}(M_1))M_1)}(P:M_1))\bigcap(\bigcap_{P\in V((I+{\rm Ann}(M_2))M_2)}(P:M_2))$$
 $$~=(\sqrt[p]{({\cal{P}}+{\rm Ann}(M_1))M_1}:M_1)\bigcap (\sqrt[p]{({\cal{P}}+{\rm Ann}(M_2))M_2}:M_2)$$
 $$=\sqrt{{\cal{P}}+{\rm Ann}(M_1)}\bigcap\sqrt{{\cal{P}}+{\rm Ann}(M_2)}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
 $$=\sqrt{({\cal{P}}+{\rm Ann}(M_1))\bigcap({\cal{P}}+{\rm Ann}(M_2))}~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
 $$=\sqrt{{\cal{P}}+({\rm Ann}(M_1\bigcap {\rm Ann}(M_2)))}=\sqrt{{\cal{P}}+{\rm Ann}(M)}={\cal{P}}.~~~~$$
Thus $M$ is a $\mathbb{P}$-radical   module.$~\square$\\

In \cite{Man}, the behavior of prime submodules under
localization, are studied. In this section, $R$ is a commutative
ring and $M$ is an   $R$-module. Suppose $S$ is a multiplicatively
closed subset   of $R$ which contains $1$ but not $0$. Let $M_S$
be the  localization of $M$ at $S$. Let $f: M\rightarrow M_S$ be
the   natural map. For any $R_S$-submodule  $Q$ of $M$, we define
  $Q^c=\{m\in M : f(m)\in Q\}$. The above notations are fixed  for this section.

  \begin{Lem} {\rm \cite[Corollary 2.2]{Man}}  Let $M$ be
an $R$-module and \vspace{2mm}\\
 \indent\indent $A=\big\{P~:~P~is~a~prime~R-submodule~of~M~with~M_S\neq P_S \big\}$\vspace{1mm}\\
\indent \indent $B=\big\{Q~:~Q~is~a~prime~R_S-submodule~of~M_S\big\}.$\vspace{1mm}\\
  Then the map $P\mapsto P_S$ is a bijective   order preserving map
  from $A$ to $B$. Its inverse map is given by  $Q\mapsto Q^c$.
  \end{Lem}

Thus by above lemma we conclude that if  $M$ is  an $R$-module and
$S$ is  a multiplicatively closed subset of $R$, then any prime
submodule of $M_S$ is of the form $P_S$ where $P$ is a prime
submodule of $M$ with $(P:M)\bigcap S=\emptyset$.

   We also need the following two lemmas.

 \begin{Lem} {\rm \cite[Corollary 1]{Lu2}} Let S be a multiplicatively closed subset of a
ring R and M an R-module. If P is a prime submodule of M, then
$(P:M)_S=(P_S:_{R_S}M_S)$
\end{Lem}
 \begin{Lem} Let $M$ be an $R$-module and $S$ be a multiplicatively closed subset
 of $R$. Then, for any ideal $I$ and prime ideal $\mathcal P$ of $R$ we
have,\vspace{2mm}\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\bigcap_{P\in V(IM)}(P:M)_S=\bigcap_{P_S\in
 V(I_SM_S)}(P_S:_{R_S}M_S)$.
 \end{Lem}
\noindent{\bf Proof.}  Let $P\in V(IM)$. Then, $P$ is a prime
submodule of $M$
 and $(P:M)\supseteq I$. Hence, $P_S$ is either equal to $M_S$ or a prime submodule
 of $M_S$. In the first case, $(P:M)_S=R_S$ and it doesn't affect on the intersections. In the second case,
 $(P:M)_S$ could appear in the right intersection by Lemma 2.5. Conversely, let $P_S\in
 V(I_SM_S)$. By Lemma 2.4, we can assume that
 $P$ is a prime submodule of $M$ with $(P:M)\bigcap S=\emptyset$. Thus
 $((P:M)_S)^c=(P:M)$. On the other hand,
 $((P:M)_S)^c=((P_S:_{R_S}M_S))^c\supseteq (I_S)^c\supseteq I$ and so $(P:M)\supseteq
 I$. i.e $P\in V(IM)$. Hence $(P:M)_S$ will appear
 in the left intersection.$~\square$\\
 \begin{Lem}
Let M be an R-module and S be a multiplicatively closed subset of
R such that for any prime ideal ${\mathcal P}_S\in V(Ann(M_S))$,
$\sqrt[p]{{\mathcal P}_SM_S}\neq M_S$. Then for any    ${\mathcal
P}\in V(Ann(M))$\vspace{2mm}\\
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\bigcap_{P\in V({\mathcal
P}M)}(P:M)_S=(\bigcap_{P\in V({\mathcal P}M)}(P:M))_S$.
 \end{Lem}
\noindent{\bf Proof.} First we note  that\vspace{2mm}\\
$~~~~~~~~~~~~~~~~~~\sqrt[p]{{\mathcal P}_SM_S}=\bigcap_{P_S\in
V({\mathcal P_S}M_S)}(P_S:M_S)=\bigcap_{P\in V({\mathcal
P}M)}(P:M)_S\neq
M_S.$\vspace{2mm}\\
 Hence there exists a  prime submodule $P\in V({\mathcal
P}M)$ such that $(P:M)\bigcap S=\emptyset$. Consequently we can
assume that the intersection $\bigcap_{P\in V({\mathcal
P}M)}(P:M)_S$ is constructed from such prime submodules $P\in
V({\mathcal P}M)$ with $(P:M)\bigcap S=\emptyset$. Now let
$\frac{r}{s}\in\bigcap_{P\in V({\mathcal P}M)}(P:M)_S$. Then for
any prime submodule $P$ in the intersection there exists $u\in
(P:M)$ and $t\in S$ such that $\frac{r}{s}=\frac{u}{t}$,    and so
there exists $h\in S$ with $h(rt-su)=0$. Hence since $0,u\in
(P:M)$ then $hrt\in (P:M)$. But $(P:M)\bigcap S=\emptyset$ and so
$r\in (P:M)$. Now since $P$ have been chosen arbitrary from the
intersection, we conclude that $\frac{r}{s}\in (\bigcap_{P\in
V({\mathcal P}M)}(P:M))_S$. Therefore $\bigcap_{P\in V({\mathcal
P}M)}(P:M)_S\subseteq(\bigcap_{P\in V({\mathcal P}M)}(P:M))_S$.
The converse of this    inequality is easy to check and hence the
equality satisfied.$~\square$\\
\begin{Pro}
Let $S$ be a multiplicatively closed subset of $R$ and  $M$ be a
$\mathbb{P}$-radical   $R$-module. If $\sqrt[p]{{\mathcal
P}_SM_S}\neq M_S$ for each  prime ideal ${\mathcal P}_S\in
V(Ann(M_S))$, then $M_S$    is a $\mathbb{P}$-radical
$R_S$-module.
 \end{Pro}
\noindent{\bf Proof.}  Let ${\mathcal P}_S\supseteq {\rm
Ann}(M_S)$ be a prime ideal of $R_S$ where ${\mathcal P}$ is a
prime ideal of $R$. Since ${\mathcal P}_S\supseteq {\rm
Ann}(M_S)\supseteq ({\rm Ann}(M))_S$ and $1\in S$, then ${\mathcal
P}={({\mathcal P}_S)}^c\supseteq (({\rm Ann}(M))_S)^c\supseteq
{\rm Ann}(M)$ where $({{\mathcal P}_S})^c$ is the contraction of
${\mathcal P}_S$. Using Lemmas 2.4 and 2.5, one
 can check that, for any $P\in V({\mathcal P}M)$, if $S\bigcap(P:M)=\emptyset$, then    $P_S\in
V({\mathcal P}_SM_S)$ and if $S\bigcap(P:M)\neq\emptyset$ then
$P_S=M_S$. Hence,\vspace{2mm}\\
 $~~~~~~~~~~~(\sqrt[p]{{\mathcal P}_SM_S)}:_{R_S}M_S)=\bigcap_{P_S\in V({\mathcal P}_SM_S))}(P_S:_{R_S}M_S)$\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\subseteq\bigcap_{P\in V({\mathcal P}M)}(P_S:_{R_S}M_S)$\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\bigcap_{P\in  V({\mathcal P}M)}(P:M)_S$\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(\bigcap_{P\in V({\mathcal P}M)}(P:M))_S$\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(\sqrt[p]{{\mathcal P}M}:M)_S.$\vspace{2mm}\\
  Since $M$    is a $\mathbb{P}$-radical module,  $(\sqrt[p]{{\mathcal P}M}:M)=\sqrt{{\mathcal P}}$ and so
  $(\sqrt[p]{{\mathcal P}_SM_S}:_{R_S}M_S)\subseteq
 \sqrt{{\mathcal P}}_S={\mathcal P}_S.$
 This implies that $M_S$    is a $\mathbb{P}$-radical   $R_S$-module.$~\square$\\

 As an important corollary of the above proposition  we have the following result:
 \begin{Cor}
 Let $M$ be a  $\mathbb{P}$-radical $R$-module and $\mathcal P$ be a prime ideal of $R$.
 If  for any prime ideal
 ${\mathcal Q}_{\mathcal P}\in V(Ann(M_{\mathcal P}))$,  $\sqrt[p]{{\mathcal Q}_{\mathcal P}M_{\mathcal P}}\neq M_{\mathcal P}$, then $M_{\mathcal P}$ is a $\mathbb{P}$-radical $R_{\mathcal P}$-module.
  \end{Cor}

 Now we investigate  the converse of the above process. In fact we
 investigate the existence of the    $\mathbb{P}$-radical   property on $M$ with
 the assumption of the existence of this property on its
 localizations.

 \begin{Pro}
Let $M$ be an $R$-module such that for any maximal ideal
${\mathcal M}\in V(Ann(M))$, $M_{\mathcal M}$ is a
$\mathbb{P}$-radical $R_{\cal{M}}$-module and $Ann(M_{\mathcal
M})=(Ann(M))_{\mathcal M}$. Then $M$    is a $\mathbb{P}$-radical
$R$-module.
 \end{Pro}
\noindent{\bf Proof.} Let ${\mathcal P}\supseteq {\rm Ann}(M)$ be
a prime ideal. Then there exists a maximal ideal ${\mathcal M}\in
V({\rm Ann}(M))$ with ${\mathcal P}\subseteq\mathcal M$. Now since
${\rm Ann}(M_{\mathcal M})=({\rm Ann}(M))_{\mathcal M}$ then we
have ${\mathcal P}_{\mathcal M}\supseteq {\rm Ann}(M_{\mathcal
M})$. Hence by the assumption and using Lemma 2.6, the followings
satisfied:\vspace{2mm}\\
 $~~~~~~~~~~~~(\sqrt[p]{{\mathcal P_{\mathcal M}}M_{\mathcal M}}:M_{\mathcal M})={\mathcal P}_{\mathcal M}~~~\Rightarrow~~~\bigcap_{(P_{\mathcal
M}:{M_{\mathcal M})}\supseteq {\mathcal P}_{\mathcal M}}({\mathcal
P}_{\mathcal M}:M_{\mathcal M})={\mathcal P}_{\mathcal M}$\vspace{2mm}\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~~~\bigcap_{(P:M)\supseteq\mathcal P}(P:M)_{\mathcal M}={\mathcal P}_{\mathcal M}$\vspace{2mm}\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~~~\bigcap_{(P:M)\supseteq\mathcal{P}}(P:M)_{\mathcal M}\subseteq{\mathcal P}_{\mathcal P}.$\vspace{2mm}\\
 It follows that\vspace{2mm}\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(\sqrt[p]{{\mathcal P}M}:M)=\bigcap_{(P:M)\supseteq\mathcal
 P}(P:M)$\vspace{1mm}\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\subseteq ((\bigcap_{(P:M)\supseteq\mathcal P}(P:M))_{\mathcal
 M})^c$\vspace{1mm}\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\subseteq (\bigcap_{(P:M)\supseteq\mathcal P}(P:M)_{\mathcal
 M})^c$\vspace{1mm}\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~={{\mathcal P}_{\mathcal P}}^c$\vspace{1mm}\\
 $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~={\mathcal P}$\vspace{1mm}\\
Therefore, $(\sqrt[p]{{\mathcal P}M}:M)={\mathcal P}$ and so  $M$
is a $\mathbb{P}$-radical $R$-module.$~\square$

 \begin{The} Let $M$ be an $R$-module such that $M_{\cal P}$ is a non-zero $\mathbb{P}$-radical   $R_{\mathcal P}$-module for
 any ${\cal P}\in V(Ann(M))$. Then $M$    is a $\mathbb{P}$-radical $R$-module.
\end{The}
\noindent{\bf Proof.} Let ${\cal{P}}\supseteq {\rm Ann}(M)$ be a
prime deal of $R$. At first we note that  since $M_{\mathcal
P}\neq 0$, ${\rm Ann}(M_{\mathcal P})\subseteq{\mathcal
P}_{\mathcal P}\neq R_{\mathcal P}$.  Thus   by the assumption we
have $(\sqrt[p]{{\cal{P}}_{\mathcal P}M_{\mathcal P}}:M_{\mathcal
P})={\cal{P}}_{\mathcal P}$.  Now by using Lemma 2.6, we
have:\vspace{2mm}\\
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(\sqrt[p]{{\cal{P}}M}:M)=\bigcap_{P\in
V({\cal{P}}M)}(P:M)$\vspace{2mm}\\
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\subseteq((\bigcap_{P\in
V({\cal{P}}M)}(P:M))_{\mathcal P})^c$\vspace{2mm}\\
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\subseteq(\bigcap_{P\in V({\cal{P}}M)}(P:M)_{\mathcal P})^c$\vspace{2mm}\\
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(\bigcap_{P_{\mathcal P}\in V({\cal{P}}_{\mathcal P}M_{\mathcal P})}(P_{\mathcal P}:_{R_{\mathcal P}}M_{\mathcal P}))^c$\vspace{2mm}\\
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=((\sqrt[p]{{\cal{P}}_{\mathcal
P}M_{\mathcal P}}:M_{\mathcal
P}))^c={{\mathcal P}_{\mathcal P}}^c$\vspace{2mm}\\
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\mathcal P.$\vspace{2mm}\\
Therefore, $(\sqrt[p]{{\mathcal P}M}:M)={\mathcal P}$ and so  $M$
is a $\mathbb{P}$-radical $R$-module.$~\square$

\section{ A Structure Sheaf for Modules}

Next we will define an structure sheaf of modules    on the
topological space Spec$(M)$. For each
 prime submodule $P$ of $M$, let ${\mathcal{P}}=(P:M)$
and $M_{\mathcal{P}}$ be the localization of $M$ at
${\mathcal{P}}$. For an open set $U\subseteq {\rm Spec}(M)$, we
define
$${\mathcal O}_{{\rm Spec}(M)}(U)=\{s:U\rightarrow\coprod_{P\in
U}M_{\mathcal{P}}~s.t.~s(P)\in M_{\mathcal{P}}\},$$ where the maps
$s$ are locally constant i.e. for any $P\in U$, there exists an
open neighborhood $V$ of $P$ where $V\subseteq U$ and $s$ is
constant on $V$.  Now it is clear that sums  of such functions
 as well as  the natural functions  $rs$ where  $r\in R$ and
$(rs)(P):=r(s(P))$ are again such. Thus ${\mathcal O}_{{\rm
Spec}(M)}(U)$ is a unitary  $R$-module. If $V\subseteq  U$ are two
open sets, the natural restriction map ${\mathcal O}_{{\rm
Spec}(M)}(U)\longrightarrow {\mathcal O}_{{\rm Spec}(M)}(U)$ is a
homomorphism of $R$-modules. Now, by definition, it is easy to
check that ${\mathcal O}_{{\rm Spec}(M)}$ is a sheaf of
$R$-modules.

Recall that for any sheaf $\mathcal O$ on a topological space $X$
and for any $x\in X$, the stalk of $\mathcal O$ at $x$, denoted by
${\mathcal O}_x$, is represented by all pairs $<U,s>$ where $U$ is
an open subset of $X$ containing $x$ and $s\in {\mathcal O}(U)$.
Two such pairs $<U,s>$ and $<V,t>$ define the same element of
${\mathcal O}_x$ if and only if there is an open neighborhood $W$
of $x$ with $W\subseteq U\cap V$ and $res_{UW}s=res_{VW}t$.
\begin{The} Let $R$ be a ring and $M$ be an
$R$-module. For any prime submodule $P$ of $M$, the stalk
${{\mathcal O}_{Spec(M)}}_P$ is isomorphic to $M_{\mathcal{P}}$,
where $(P:M)={\mathcal{P}}$.
\end{The}
\noindent {Proof.} First we define a homomorphism from ${{\mathcal
O}_{{\rm Spec}(M)}}_P$ to $M_{\mathcal{P}}$, by sending any local
section $s$ in a neighborhood of $P$ to its value $s(P)\in
M_{\mathcal{P}}$.  This gives a well- defined homomorphism
$\varphi$ from ${{\mathcal O}_{{\rm Spec}(M)}}_P$ to
$M_{\mathcal{P}}$. The map $\varphi$ is surjective, because any
element of $M_{\mathcal{P}}$, can be represented as a quotient
$m/f$ with $m\in M$ and $f\in R\setminus {\mathcal{P}}$.  Then
$D(fM)$ will be an open neighborhood of $P$ (since if $fM\subseteq
P$, then $f\in {\mathcal{P}}$, a contradiction),  and $m/f$
defines a section of ${{\mathcal O}_{{\rm Spec}(M)}}$ over $D(fM)$
whose value at $P$ is the given element. To show that $\varphi$ is
injective, let $U$ be a neighborhood of $P$, and let $s,t\in
{\mathcal{O}}(U)$ be
 elements having the same value $s(P)=t(P)$ at $P$. By the
 definition of our structure sheaf and
 shrinking $U$ if necessary, we may assume that $s=m/f$ and $t=n/g$
on $U$, where $m$, $n\in M$, and $f$, $g\in R\setminus
{\mathcal{P}}$. Since  $m/f$ and $n/g$  have the same image in
$M_{\mathcal{P}}$, it follows from the definition of localization
that there is an $h\in R\setminus {\mathcal{P}}$ such that
$h(gm-fn)=0$ in $M$. Therefore $m/f=n/g$ in every local module
$M_{\mathcal{P}'}$ such that $h,~f,~g\in R\setminus
{\mathcal{P}}'$. But the set of such ${\mathcal{P}}'$ is the open
set $D(fM)\cap  D(gM)\cap D(hM)$, which contains $P$. Hence $s=t$
in a whole neighborhood of $P$, so they have the same stalk at
$P$. So  $\varphi$  is an isomorphism, which completes the
proof.\\

Now we are going to prove an important property of the structure
sheaf of    $\mathbb{P}$-radical   modules. This property could
helps us to have a better investigation of the concept of spectrum
of modules in the algebraic geometry point of view. This property
is as follow.
\begin{The}
Let $R$ be a ring and  $M$ be a $\mathbb{P}$-radical   $R$-module.
Then for each $f\in R$, the module ${\mathcal O}_{Spec(M)}(D(fM))$
is isomorphic to the
 localized module  $M_f$. In particular, ${\mathcal O}_{Spec(M)}(Spec(M))\cong M$.
\end{The}
We prove this theorem in several steps. But at first we need the
following lemmas.
\begin{Lem}Suppose that $M$ be a    $\mathbb{P}$-radical   $R$-module. Then for two ideals $I_1$ and $I_2$ of $R$ containing
$Ann(M)$, $V(I_1M)\subseteq V(I_2M)$ implies that
$\sqrt{I_2}\subseteq\sqrt{I_1}$.
\end{Lem}
\noindent {Proof.}  The equality $V(I_1M)\subseteq V(I_2M)$
implies that $\cap_{P\in V(I_1M)}(P:M)\supseteq \cap_{P\in
V(I_2M)}(P:M)$. Thus $Rad_M(I_1)\supseteq Rad_M(I_2)$ and since
$M$    is $\mathbb{P}$-radical   then
$\sqrt{I_1}\supseteq\sqrt{I_2}$.
\begin{Lem}
Let $M$    be a $\mathbb{P}$-radical   $R$-module. Then for two
ideals $I$ and $J$ of $R$, $V(IM)\subseteq V(JM)$ implies that
$\sqrt{(IM:M)}\supseteq\sqrt{(JM:M)}$.
\end{Lem}
\noindent {Proof.} Let $V(IM)\subseteq V(JM)$, for two ideals $I$
and $J$ of $R$. Then $V((IM:M)M)\subseteq V((JM:M)M)$ and so
$\sqrt{(IM:M)}\supseteq\sqrt{(JM:M)}$  by Lemma 3.3.
\begin{Lem}
Let M be a    $\mathbb{P}$-radical   R-module and I be a radical
ideal of R. Then (IM:M)=I if and only if $Ann(M)\subseteq I$.
\end{Lem}
\noindent {Proof.} Let ${\rm Ann}(M)\subseteq I$. Then by
definition of the    $\mathbb{P}$-radical   property we have
$(IM:M)\subseteq(\sqrt[p]{IM}:M)=I$. On the other hand the
converse is obviously satisfied. Thus we have $(IM:M)=I$. The
converse of the statement is natural.
\begin{Lem} Let M be an R-module, $f\in R$ and $s\in{\mathcal O}_{Spec(M)}(D(fM))$. Then there exist some $h_i\in R$
and $m_i\in M$ such that $D(fM)=\bigcup_iD(h_iM)$ and
$s\equiv\frac{m_i}{h_i}$ on each $D(h_iM)$.
\end{Lem}
\noindent {Proof.} Let $s\in {\mathcal O}_{{\rm Spec}(M)}(D(fM))$.
By definition, there exist open subsets $V_i$ of $D(fM)$, on which
$s$ is represented by a quotient $\frac{m_i}{g_i}$, with
$g_i\notin (P:M)$ for all $P\in V_i$. Hence for each $i$,
$V_i\subseteq D(g_iM)$. Now, the open sets of the form $D(hM)$
form a base for the topological space ${\rm Spec}(M)$. Hence we
can assume that $V_i=D(h_iM)$ for some $h_i\in R$. Since
$D(h_iM)\subseteq D(g_iM)$, then we have $D(h_ig_iM)=D(h_iM)\cap
D(g_iM)=D(h_iM)$. On the other hand,
$\frac{m_i}{g_i}=\frac{h_im_i}{h_ig_iM}$. Thus, replacing $h_i$ by
$h_ig_i$ and $m_i$ by $h_im_i$, we can assume that $D(fM)$ is
covered by the open sets $D(h_iM)$ and that $s$ is represented by
$\frac{m_i}{h_i}$ on $D(h_iM)$.
\begin{Lem}Let M be a $\mathbb{P}$-radical   R-module, $f\in R$ and $D(fM)=\bigcup_iD(h_iM)$
for some $h_i\in R$. Then, there exists an integer $n\in\Bbb{N}$
such that $f^n=\sum_{i=1}^kr_ib_i$, for some $r_i\in (h_iM:M)$,
$b_i\in R$ and $k\in\Bbb{N}$.
\end{Lem}
\noindent {Proof.} Let $N:=\sqrt{(\sum(h_iM:M))}M$. Since
$D(fM)\subseteq\cup D(h_iM)$ we have:
\begin{eqnarray*}
&&V((fM:M)M)=V((f)M)\supseteq \cap V((h_i)M)=\\
&&\cap V((h_iM:M)M)=V((\sum(h_iM:M))M)=\\
&&V((\sqrt{\sum(h_iM:M)})M)=V(N)= V((N:M)M).
\end{eqnarray*}
 Hence
$\sqrt{(fM:M)}\subseteq\sqrt{(N:M)}$, by Lemma 3.3. Furthermore we
have $\sqrt{\sum(h_iM:M)}\supseteq$ {\rm Ann}$(M)$ and so
$(N:M)=\sqrt{\sum(h_iM:M)}$, by Lemma 3.5. Hence there exists
$n\in\Bbb{N}$ such that $f^n=b_1r_1+\ldots+b_kr_k$, for some
$b_i\in R$ and $r_i\in (h_iM:M)$.
\begin{Lem}Let M be a $\mathbb{P}$-radical R-module. By the notations of Lemmas
\emph{3.6} and \emph{3.7}, $D(fM)$ can be covered by the open
subsets $D(r_iM)$, $i=1,\ldots,k$. Furthermore, for any
$i=1,\ldots,k$, there exists $n_i\in M$ such that
$s\equiv\frac{n_i}{r_i}$ on each $D(r_iM)$.
\end{Lem}
\noindent {Proof.} Since $r_i\in (h_iM:M)$, for any
$i=1,\ldots,k$, we get $r_iM\subseteq h_iM$. Thus
$V(h_iM)\subseteq V(r_iM)$ and equivalently $D(r_iM)\subseteq
D(h_iM)$. Hence $D(fM)\supseteq \cup_{i=1}^{k} D(r_iM)$. For the
converse, we have to show that $V((f)M)\supseteq \cap_{i=1}^k
V((r_i)M)=V((\sum_{i=1}^k (r_i))M)$. If $P\in V((\sum_{i=1}^k
(r_i))M)$, then $(P:M)\supseteq \sum_{i=1}^k (r_i)$ and in
particular $r_1b_1+\ldots+r_kb_k\in (P:M)$. Hence, by Lemma 3.7,
$f^n\in (P:M)$ and so $P\in V((f^n)M)=V((f)M)$ as wanted.
Furthermore, $r_iM\subseteq h_iM$ implies that there exists
$n_i\in M$ such that $r_im_i=h_in_i$ and so
$\frac{m_i}{h_i}=\frac{n_i}{r_i}$ for $i=1,\ldots,k$. Since
$D(r_iM)\subseteq D(h_iM)$ and $s$ is represented by
$\frac{m_i}{h_i}$ on $D(h_iM)$, we can assume that $s$ is
represented by the quotient $\frac{n_i}{r_i}$ on $D(r_iM)$,
$i=1,\ldots,k$.\\

Now, we are ready to prove the Theorem 3.2. To do this, we define
the canonical map $\psi: M_f\rightarrow {\mathcal O}_{{\rm
Spec}(M)}(D(fM))$ by sending $\frac{m}{f^n}$ to the section $s\in
{\mathcal O}_{{\rm Spec}(M)}(D(fM))$ with $s(P)=\frac{m}{f^n}$,
for any $P\in D(fM)$. Clearly $\psi$ is an $R$-module
homomorphism. By this definition of $\psi$, we complete the proof
of Theorem 3.2 as follows:\vspace{.25cm}\\
\noindent{\bf Proof of Theorem 3.2} It is enough to show that the
homomorphism $\psi$ introduced as above is bijective. At firs we
prove the injectivity of this map. Let
$\psi(\frac{m_1}{f^{n_1}})=\psi(\frac{m_2}{f^{n_2}})$, then for
any $P\in D(fM)$, $\frac{m_1}{f^{n_1}}=\frac{m_2}{f^{n_2}}$ on
$M_{\cal P}$, where ${\mathcal P}=(P:M)$. Thus, for any $P\in
D(fM)$, there exists $h\notin (P:M)$ such that
$h(f^{n_1}m_2-f^{n_2}m_1)=0$. Let $\mathcal{A}$ be the ideal {\rm
Ann}$(f^{n_1}m_2-f^{n_2}m_1)$. Then for any $P\in D(fM)$, there
exists $h\in\mathcal{A}$ such that $h\notin (P:M)$. Thus
$\mathcal{A}\nsubseteq (P:M)$ and so $V((\mathcal{A}M:M)M)\cap
D(fM)=\emptyset$. Furthermore, one can check that
$V((\mathcal{A}M:M)M)=V((\sqrt{\mathcal{A}}M:M)M)$. Consequently,
$V((\sqrt{\mathcal{A}}M:M)M)\subseteq V((f)M)=V((fM:M)M)$. Hence
by Lemma 3.3, $\sqrt{(fM:M)}\subseteq
\sqrt{({\sqrt\mathcal{A}}M:M)}$ and so $f\in
\sqrt{(\sqrt{\mathcal{A}}M:M)}$. On the other hand, by Lemma 3.5,
$(\sqrt{\mathcal{A}}M:M)=\sqrt{\mathcal{A}}$. Thus $f\in
\sqrt{\mathcal{A}}$ and there exists $k\in\Bbb{N}$ with
$f^k(f^{n_1}m_2-f^{n_2}m_1)=0$. Therefore,
$\frac{m_1}{f^{n_1}}=\frac{m_2}{f^{n_2}}$ on $M_f$ and
consequently $\psi$
 is injective. On the other hand, $\psi$ is surjective. Indeed,  According to Proposition 3.7, it is
enough to show that the map $\psi$ considered as above is
surjective. Just as we saw in Lemmas 3.4 to 3.6, there exist
$r_i,b_i\in R$ and $n_i\in M$, $i=1,\ldots,k$, where
$D(fM)=\cup_{i=1}^kD(r_iM)$, $s|_{D(r_iM)}\equiv\frac{n_i}{r_i}$
and $f^n=\sum_{i=1}^kr_ib_i$ for some $n\in\Bbb{N}$. First, we
claim that for any $i,j=1,\ldots,k$, the equality $r_in_j=r_jn_i$
holds. Since $D(r_iM)\cap D(r_jM)=D(r_ir_jM)$, we have two
elements of $M_{r_ir_j}$, namely $\frac{n_i}{r_i}$ and
$\frac{n_j}{r_j}$ both of which represent $s$. Thus, according to
the injectivity of $\psi$, applied to $D(r_ir_jM)$, we have
$\frac{n_i}{r_i}=\frac{n_j}{r_j}$ in $M_{r_ir_j}$. Hence for some
$t\in\Bbb{N}$, the equality $(r_ir_j)^t(r_jn_i-r_in_j)=0$ holds.
Since there are only $k$ indices involved, we may pick $t$ so
large that it works for all $i,j$ at once. Rewrite this equation
as ${r_j}^{t+1}({r_i}^tn_i)-{r_i}^{t+1}({r_j}^tn_j)=0$. Replacing
each $r_i$ by ${r_i}^{t+1}$ and $n_i$ by ${r_i}^tn_i$, we still
have $s$ represented on $D(r_iM)$ by $\frac{n_i}{r_i}$ and
$r_in_j=r_jn_i$ for all $i,j=1,\ldots,k$. Now put
$m=\sum_{i=1}^kb_in_i$, then
$r_jm=\sum_{i=1}^kr_jb_in_i=\sum_{i=1}^kb_ir_in_j=f^nn_j$. This
implies that $\frac{n_j}{r_j}=\frac{m}{f^n}$ on $D(r_jM)$. So
$\psi(\frac{m}{f^n})=s$ everywhere, which shows that $\psi$ is
surjective. $\square$

\section{An application}

 As we saw in $\S1$, the results of this
paper could give the opportunity to consider the concept of
spectrum of modules in algebraic geometry point of view. But some
other interesting and directed results can yield from them. For
instance, the following theorem presents a criterion for the
existence of an isomorphism between two localizations of a
$\mathbb{P}$-radical module:
\begin{The}
Let M be a    $\mathbb{P}$-radical   R-module and $f,g\in R$.
Then, $M_f\cong M_g$ as R-modules if and only if
$\sqrt{(fM:M)}=\sqrt{(gM:M)}$.
\end{The}
\noindent {Proof.} Let $\varphi:M_f\rightarrow M_g$ be an
$R$-module isomorphism and $P\in D(fM)$. We claim that $P$ is an
element of $D(gM)$. Otherwise, if $P\notin D(gM)$ then
$gM_f\subseteq P_f$ and so $\varphi(P_f)\supseteq
g\varphi(M_f)=gM_g=M_g$ which implies that $\varphi(P_f)=M_g$. But
$P\in D(fM)$ and hence $P_f$ is a prime submodule of $M_f$. Thus,
$\varphi(P_f)$ must be a prime submodule of $M_g$ which is a
contradiction. Hence, $P\in D(gM)$ and so $D(fM)\subseteq D(gM)$.
Using the isomorphism $\varphi^{-1}$, we can prove the reverse
inclusion and hence $D(fM)=D(gM)$. Thus,
$\sqrt{(fM:M)}=\sqrt{(gM:M)}$ by
Lemma 3.4.\\
For the converse, let $\sqrt{(fM:M)}=\sqrt{(gM:M)}$. Then
$D(fM)=D(gM)$ and thus $M_f\cong {\mathcal O}_{{\rm
Spec}(M)}(D(fM))={\mathcal O}_{{\rm Spec}(M)}(D(gM))\cong M_g$ by
Theorem 3.2.$~\square$\\

We remark that there exist exemplars of modules such that the
above result is not true in those cases. For example:
\begin{Examp} {\rm
Suppose that $M$ is the Pr\"{u}fer group $\Bbb{Z}(p^{\infty})$ as
a $\Bbb{Z}$-module. $M$ is a torsion divisible module and so it is
primeless by \cite{MMS1}, Lemma 1.3. Hence it    is not a
$\mathbb{P}$-radical   module. Let $q\neq p$ be a prime integer.
Then we have, $\sqrt{(pM:M)}=\sqrt{(qM:M)}=\Bbb{Z}$. But, one can
check that $\frac{1}{q}+\Bbb{Z}$ is an element of rank $q$ in
$M_q$ Although $M_p$ doesn't have any element of this rank. Hence
$M_p\ncong M_q$ as two $\Bbb{Z}$-modules.}
\end{Examp}

\bigskip

redpajama_set_name: RedPajamaArXiv


> It announces the planned ABI changes for vhost-user multiple queues feature on v2.2.
> +* The ABI changes are planned for struct virtio_net in order to support vhost-user multiple queues feature. The upcoming release 2.1 will not contain these ABI changes, but release 2.2 will, and no backwards compatibility is planed due to the vhost-user multiple queues feature enabling. Binaries using this library build prior to version 2.2 will require updating and recompilation.


text:
Education and Story: Building Blocks of Community
August 25, 2015 nonprofitbarkadacreative
In keeping with Barkada's 2015 theme of education, I'd like to highlight an article that was recently published in the Chicago Sun-Times. Ameya Pawar, alderman of the 47th Ward on Chicago's North Side, wrote an opinion piece about Chicago's test-based high school system.
He begins by telling the story of his father living in Mumbai, India in the mid-1950s. His father and other students spent their childhood preparing for one test to secure one of a few coveted seats in the right high school. It meant the difference between spending the rest of their lives in poverty or winning the lottery for a chance of getting ahead. Twenty years later, his father immigrated to the United States so that when he started a family, they wouldn't be condemned to the same broken system. Fast forward to 2015 and we find the same broken system here in Chicago. Pawar describes what this looks like and feels like for the families in our City of Neighborhoods:
Children begin to prepare for Chicago's selective public high school entrance exam as early as 10 years old – spending hours on homework and stressing over earning straight As. Test-shaming and peer pressure are the norm. Parents scrounge together resources for tutors and admissions consultants. Parents peer pressure one another. To what end? Each year 25,000 eighth graders compete for 3,600 seats across eleven selective enrollment schools. A near-lifetime of preparation to get into the right high school. In a system which serves a few and neglects the whole.
And so, like many families, Pawar's father once again decided to move his elsewhere to seek out stability and equity, fleeing to the suburbs to find a stable neighborhood K-12 system.
According to Pawar, "Chicago is doing itself a tremendous disservice by replicating an educational pattern that helped force my family out of India. The narrow focus on selective enrollment high schools is stressing families, harming neighborhood high schools, and undermining long-term growth in Chicago."
Urban education reform treats schools like businesses, but we know that most businesses fail. Pawar's vision for Chicago's educational system is where schools are the starting point for neighborhoods to come together and make change happen, because everyone has some vested interest in making sure that their schools are doing well.
He describes this initiative as having three pillars:
Organize around the schools and change the dialogue in order to change the perception of neighborhood high schools
Generate conversations one household at a time and connect these conversations to a plan
Build community around the plan
For a community-driven initiative like this to be successful, the leaders have to be ready to do a lot of listening. By getting people to share stories with one another, they begin to realize that they are dealing with the same challenges and they all want the same thing for their children. They begin to realize that they will be more successful if they work together to make change happen around their schools to achieve stability and equity.
Leaders begin and end with stories.
In 2003, Harvard Business School interviewed screenwriting coach Robert McKee. He talks about why it is essential for a leader to be a skilled storyteller in order to successfully persuade members of her team to work toward common goals. Particularly in the nonprofit sector, a leader should be able to convince potential donors and funders that his organization is worthy of receiving their financial support. Cognitive research has shown that the human mind assembles the bits and pieces of an experience into a story in order to understand and remember. We tend to forget lists and bullet points. Instead, human desire and struggle are what stay with us.
Why does it make a world of difference to go beyond rhetoric and present your case in a story? According to McKee:
Essentially, a story expresses how and why life changes. You want to display the struggle between expectation and reality in all its nastiness. It demands vivid insight and storytelling skill to present an idea that packs enough emotional power to be memorable.
Our team at Barkada Circle has worked closely with leaders of nonprofits to help them realize that storytelling is a powerful catalyst that brings together key stakeholders of an organization to the same page. Stories reveal shared values and foster deeper understanding of common goals. The outcome is a shift in the collective consciousness of the organization to create a shared narrative about the mission that has more clarity and focus. From here, leaders can take their story and develop a more compelling message that resonates with their audience.
Jacqueline Novogratz and Simon Sinek
https://barkadacircle.files.wordpress.com/2015/08/nicecast-archived-audio-20150810-1018.mp3
LISTEN to Simon Sinek and Jacqueline Novogratz share stories that encourage us to take the initiative for inspiring others to change the way they think and see. Ultimately, the outcome will be something bigger than we had imagined.
Stories we tell define who we are
August 3, 2015 nonprofitChimamanda Adichie, collaborative efforts, sense of community, Thandie Newtonbarkadacreative
At Barkada Circle we believe that positive change begins with cultivating a sense of community around real collaborative efforts and allowing various ideas to converge. This is why we use storytelling as the main catalyst for realizing new ideas and shared goals.
As human beings, we are naturally wired for story. When you tell a story to another person, you will spark a story from that person. As an emergent form of communication, a compelling story conveys meaning that the other person can relate to because it reflects what they believe in. As a way to bring people together, stories reinforce the culture of the community. Stories build our identity and help to forge meaning for what it means to be a community, and ultimately, revealing who we are as individuals.
Chimamanda Adichie, Thandie Newton, Tavi Gevinson
LISTEN to three artists talk about how stories reflect our humanity and all its complexities––the multiplicity of the human experience.

redpajama_set_name: RedPajamaCommonCrawl


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Let the ocean breeze carry you away at 'BREEZY HILL' MAIN HOUSE and GUEST HOUSE, located on Great Exuma in the Bahamas, which are open 365 days a year, and are self-catering homes. AFFORDABLE Daily, Weekly, Monthly and LONG TERM rates are available.
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When former airline pilot Tony Royle came to see me last year to seek reassurance that it was OK to participate in an Ironman event, having stopped all his medications 18 months after suffering a heart attack, I was initially a little alarmed.
But after talking to him, I realized he had made an informed decision to stop the medication after suffering side effects, and instead had opted for a diet and lifestyle approach to manage his heart disease.
But our health care system has failed to keep to this gold standard of clinical practice for the most important goal of improving patient health outcomes.
The consequences have been devastating. Modern medicine, through over prescription, represents a major threat to public health. Peter Gøtzsche, co–founder of the reputed Cochrane Collaboration, estimates that prescribed medication is the third most common cause of death globally after heart disease and cancer.
In the UK, use of prescription drugs is at an all-time high, with almost half of adults on at least 1 drug and a quarter on at least 3 – an increase of 47% in the past decade. It's instructive to note that life expectancy in the UK has stalled since 2010, the slowdown being one of the most significant in the world's leading economies.
Contrary to popular belief, the cost of an aging population in itself is not a threat to the welfare system – an unhealthy ageing population is. A Lancet analysis revealed that if rising life expectancy means years in good health, then health expenditure is expected to increase by only 0.7% of GDP by 2060.
The greatest stress on Britain's National Health Service (NHS) comes from managing almost entirely preventable chronic conditions such as heart disease, high blood pressure and type 2 diabetes. Type 2 diabetes alone (demonstrated to be reversible in up to 60% of patients) takes up approximately 10% of the NHS budget. A disturbing report from the British Heart Foundation suggests that heart attacks and strokes are set to "surge" in England over the next 20 years as the prevalence of diabetes continues to increase.
Yet rather than address the root cause of these conditions through lifestyle changes, we prioritize drugs that give – at best – only a marginal chance of long-term benefit for individuals, most of whom will derive no health outcome improvement.
The reality is that lifestyle changes not only reduce the risk of future disease, their positive effects on quality of life happen within days to weeks. However, those patients unlucky enough to suffer side effects from prescribed medicines may find their quality of life will deteriorate in order to enjoy small longer term benefits from the medication.
Prof. Luis Correia, director of the Centre of Evidence Based Medicine in Brazil, says if a clinical decision is not in keeping with the patient's individual preferences and values, "it will not work".
A report commissioned by think tank the King's Fund in 2012 recommended putting patient preferences at the heart of decision making in medicine, suggesting it would not just be a victory for ethics and policy but for finance, too, as the data shows patients given all the information choose fewer treatments. But more than saving money, it will be about redistributing resources within the system to where they are needed most, in acute and social care.
This solution to the NHS financial crisis and giving patients the very best chance of improving their health will require a national public health campaign to reduce the amount of medications the population takes, improving lifestyle and adhering to the true principles of evidence-based medicine that make shared decision making the priority in clinical practice.
A few weeks ago, 4 years after his heart attack and 2 years after coming off all medications and dramatically changing his diet, Tony completed his first Ironman at the age of 58, revealing it's never too late to improve fitness. But the most important message remains clear: you can't drug people into being healthier.
Dr. Aseem Malhotra is a consultant cardiologist for the UK's National Health Health Service and visiting professor of evidence-based medicine, at the Bahiana School of Medicine and Public Health in Brazil.
This article was originally published by The Guardian. Reprinted by permission of the author.


Last week, we kicked off our MacGyver Challenge series by asking you to create a binder clip hack for your office. We received tons of great entries, but the winning hack belongs to reader Sam Terpstra.
Check out the description of Sam's winning entry below and read about some of our other favorite entries.
Take your two binder clips (medium and large) and remove one of the metal handles (it makes no difference which one). Hold the two so the "biting end" is pointing away from each other and slip a smaller clip arm around the larger one (you'll probably have to twist it). Now, take the other side (with the missing handle) and slide the removed handle into the same configuration and reconnect it to the clip. Voila, you're done.
When you're finished, you have a versatile clipping solution. Too see what Sam has done with this setup, check out the expanded instructions. Congratulations, Sam! And congratulation to Sam's mom, as well, who introduced him to Lifehacker.
Andres Maldonado uses the binder clips on the sides of his desk to hold his ink pen, wacom pen, and pair of chopsticks on the side. For holding the wacom pen, he increased the "hole size" of the binder clip using pliers. He did the same thing to reduce the hole so it would support the pen at the bottom.
We liked Jonathan Concepcion's simple smartphone dock because it clips to a vertical surface and uses both metal clips to hold the phone in place. It's very handy for clipping to the side of a bookshelf or cabinet.
Reader RCamyre sent us this very simple, but very effective, way of remembering to bring important papers along with you. A binder clip attached to your key ring means you won't forget to mail that check unless you also forget your keys.
Cable Organizers, Phone Hangers, and More!
We had so many great entries this time around that we wanted to share as many of them with you as we could. Check out all the other great binder clip hackery from our readers.


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