% !TEX TS-program = xelatex % !TEX encoding = UTF-8 \documentstyle[12pt]{article} %\usepackage{amsthm,amssymb,amsmath,amsfonts} \begin{document} \newtheorem{coun}{}[section] %\def\Ann{{\rm Ann }} %\def\ker{{\rm ker}} \def\be{\begin{enumerate}} \def\ee{\end{enumerate}} %\def\defi{\noindent{\bf Definition. }} \def\proof{\noindent{\bf Proof. }} \def\exa{\noindent{\bf Example. }} \newtheorem{ttheo}{Theorem}[section] \newtheorem{ccoro}[ttheo]{Corollary} \newtheorem{llem}[ttheo]{Lemma} \newtheorem{rrem}[ttheo]{Remark} \newtheorem{ppro}[ttheo]{Proposition} \newtheorem{ddefi}[ttheo]{Definition} \def\defi#1{\begin{ddefi}#1\end{ddefi}} \def\theo#1{\begin{ttheo}#1\end{ttheo}} \def\pro#1{\begin{ppro}#1\end{ppro}} \def\coro#1{\begin{ccoro}#1\end{ccoro}} \def\lem#1{\begin{llem}#1\end{llem}} \def\rem#1{\begin{rrem}#1\end{rrem}} \newenvironment{pproof}{\noindent{\bf Proof. }}{} \def\proof#1{\begin{pproof}#1\end{pproof}} %\def\proof#1{\noindent{\bf Proof. }{#1}\par} \newcommand{\Om}{\Omega} \newcommand{\om}{\omega} \newcommand{\al}{\alpha} %\newcommand{\be}{\beta} \newcommand{\la}{\lambda} \newcommand{\La}{\Lambda} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\vp}{\varphi} \newcommand{\spq}{\supseteq} \newcommand{\dis}{\displaystyle} \newcommand{\les}{\leqslant} \newcommand{\ex}{\exists} \newcommand{\ov}{\overline} \newcommand{\fo}{\forall} \newcommand{\blk}{\blacksquare} \newcommand{\un}{\underline} \newcommand{\Roo}{\Longrightarrow} \newcommand{\Loo}{\Longleftarrow} \newcommand{\RTA}{\Rightarrow} \newcommand{\LTA}{\leftarrow} \newcommand{\BLRTA}{\Longleftrightarrow} \newcommand{\LRTA}{\Leftrightarrow} \newcommand{\TOO}{\longrightarrow} \newcommand{\tohi}{\varnothing} \newcommand{\all}{\forall} \newcommand{\sumin}{\sum_{i=1}^n} \newcommand{\sumjn}{\sum_{j=1}^n} \newcommand{\sumkn}{\sum_{k=1}^n} \newcommand{\sumion}{\sum_{i=0}^n} \newcommand{\sumjon}{\sum_{j=0}^n} \newcommand{\sumkon}{\sum_{k=0}^n} \newcommand{\sumim}{\sum_{i=1}^m} \newcommand{\sumjm}{\sum_{j=1}^m} \newcommand{\sumkm}{\sum_{k=1}^m} \newcommand{\sumno}{\sum_{n=0}^\infty} \newcommand{\sumio}{\sum_{i=0}^\infty} \newcommand{\sumjo}{\sum_{j=0}^\infty} \newcommand{\sumko}{\sum_{k=0}^\infty} \newcommand{\sumn}{\sum_{n=1}^\infty} \newcommand{\sumi}{\sum_{i=1}^\infty} \newcommand{\sumj}{\sum_{j=1}^\infty} \newcommand{\sumk}{\sum_{k=1}^\infty} \def\iff {if and only if\ \ } \def\sub {\subseteq} \def\nsub {\nsubseteq} \def\subn {\subsetneq} \def\set {\setminus } \def\RTA {\Rightarrow } \def\LTA {\Leftarrow } \def\TOO {\longrightarrow } \def\TO {\rightarrow } \def\ov {\overline} \def\un {\underline } \def\tohi {\varnothing } \def\RE {{\Bbb R}} \def\C {{\Bbb C}} \def\Q {{\Bbb Q}} \def\N {{\Bbb N}} \def\Z {{\Bbb Z}} \title{{\rm P}-Ideals and {\rm PMP}-Ideals in Commutative Rings} \author{A.R. Aliabad, J. Hashemi, R. Mohamadian\\ {\small Department of Mathematics, Shahid Chamran University, Ahvaz, Iran.}\\ {\small {aliabady\_r@scu.ac.ir, jhashemi@scu.ac.ir, mohamadian\_r@scu.ac.ir}}} \maketitle \input{amssym} \begin{center} {\bf ABSTRACT} \end{center} Recently, P-ideals studied in ${\rm C}(X)$ by some authors. In this article we investigate {\rm P}-ideals and a new concept as PMP-ideals in commutative rings. We show that $I$ is a {\rm P}-ideal (resp., {\rm PMP}-ideal) in $R$ if and only if every prime ideal of $R$ which does not contain $I$ is a maximal (resp., minimal prime) ideal of $R$. Also, we characterize largest {\rm P}-ideals (resp., PMP-ideals) in commutative rings, specially in ${\rm C}(X)$. Furthermore, we study relation between these ideals and pure ideals. %\end{abstract} Finally we prove that ${\rm C}(X)$ is a von Neumann regular ring if and only if every pure ideal of it is P-ideal. \vspace{0.3cm} {\bf Key words}: P-ideal, PMP-ideal, pure ideal, von Neumann regular ideal, P-space.\\\vspace{2mm} \vspace{0.3cm} {\bf MSC}: Primary 13A18; Secondary 54C40.\vspace{2mm} \section{\noindent{\bf Introduction}} Throughout this paper the notation $R$ stands for a commutative ring with unity and $X$ stands for a topological Tychonoff space. We denote by Spec($R$), Max($R$) and Min($R$) the set of all prime ideals, maximal ideals and minimal prime ideals of $R$, respectively. Also, by Jac($R$) and Rad($R$) we mean the Jacobson radical and the prime radical of $R$, respectively. If $S\sub R$, then by ${\cal A}(S)$ we mean the set of all annihilators of $S$; briefly, we use ${\cal A}(a)$ instead of ${\cal A}(\{a\})$. For each $a\in R$, let $aR$, ${\bf M}_a$ and ${\bf P}_a$ be the ideal generated by $a$, the intersection of all maximal ideals containing $a$ and the intersection of all minimal prime ideals containing $a$, respectively. If $A\subseteq R$, then we briefly use the notations $$V(A)=\{P\in {\rm Spec}(R): A\subseteq P\}~~,~~D(A)={\rm Spec}(R)\set V(A).$$ For more information about commutative rings and topological spaces, one can refer to \cite{KAP} and \cite{WIL}, respectively. Assuming that $I$ is an ideal of $R$, the set $\{a\in R: a\in aI\}$ is denoted by $mI$ which is called the pure part of $I$. It is well-known that $m(I)$ is an ideal of $R$ and $m(I)=\{a\in R:I+{\cal A}(a)=R\}$. An ideal $I$ is said to be pure if $I=m(I)$. One can easily see that a maximal ideal $M$ of a reduced ring $R$ is pure if and only if $M\in {\rm Min}(R)$. For more information about the pure ideals, refer to \cite{AB1}, \cite{AB2} and \cite{DM}. The ring of all continuous functions on a topological space $X$ is denoted by ${\rm C}(X)$. By $A^{\circ}$ and $\overline{A}$ we mean the interior and the closure of $A$, respectively. Also if $f\in {\rm C}(X)$ and $A\subseteq X$, then we define $$Z(f)=\{x\in X:f(x)=0\}~~~~~~~~,~~~~~~~{\rm Coz}(f)=X\setminus Z(f)$$ $$O_{A}(X)=\{f\in {\rm C}(X): A\subseteq Z^{\circ}(f)\}~~~~,~~~~M_{A}(X)=\{f\in {\rm C}(X): A\subseteq Z(f)\}.$$ In particular, if $A=\{x\}$, then we use $O_{x}(X)$ and $M_{x}(X)$ instead of $O_{\{x\}}(X)$ and $M_{\{x\}}(X)$, respectively. In \cite{GIL}, one can see more information about these notions. \vspace{3mm} In Section 1, first we deal with the connection between the set of ideals of a ring $R$ and the set of ideals contained in a fixed ideal of $R$. Next, we give some statements about von Neumann regular elements and ideals (briefly, we use regular instead of von Neumann regular), see \cite{AB3} and \cite{GOO}, for more information about regular ideals. In the sequential, we see that, under some conditions (for example, in reduced rings) regular ideals coincide with $\rm P$-ideals. Section 2 is devoted to ${\rm P}$-ideals and ${\rm PMP}$-ideals in a ring $R$. ${\rm P}$-ideals in ${\rm C}(X)$ is introduced and considered in \cite{RUD}, but ${\rm PMP}$-ideal is a new concept. In this section, we find some equivalent conditions for these notions and then we obtain some new results. For instance, we show that an ideal $I$ of $R$ is a ${\rm P}$-ideal if and only if $D(I)\subseteq {\rm Max}(R)$; also, it is shown that an ideal $I$ of $R$ is a ${\rm PMP}$-ideal if and only if $D(I)\subseteq {\rm Min}(R)$. We find that in any commutative ring $R$, the largest $P$-ideal (resp., $PMP$-ideal) exists. Indeed in Proposition 2.6, we characterize these ideals via intersection of prime ideals. In section 3, we prove that if $R$ is a reduced ring, then $I$ is a $P$-ideal \iff $I$ is regular and also we prove that a proper ideal $I$ of a reduced ring $R$ is ${\rm PMP}$-ideal if and only if $R$ is a regular ring or $R$ is a local ring with $\rm dim(R)=1$. In addition, in this section, we find an equivalent condition for a ${\rm PMP}$-ideal to be a ${\rm P}$-ideal. In Section 4, we deal with two above notions in ${\rm C}(X)$, and show that these are equivalent in ${\rm C}(X)$. Also, we obtain some new results, for instance, we show that every pure ideal in ${\rm C}(X)$ is a ${\rm P}$-ideal. \vspace{3mm} We need the following lemma to find in Proposition 1.3, a one-one correspondence between the set of prime ideals not containing a given ideal $I$ of $R$ and the set of prime ideals of $I$ as a ring. \vspace{3mm} The following lemma is useful in the sequel. %\label{1.1} \lem {Let $I$ be an ideal of $R$ and $H$ be a semiprime ideal in the ring $I$, then $H$ is an ideal in $R$.} \begin{proof} {Suppose that $a\in H $ and $r\in R$, hence $r^2a\in I$ which implies that $(ra)^2=(r^2a)a\in H$. This shows that $ra\in H$.\hfill$\square$} \end{proof} \vspace{3mm} %\label{1.2} \defi{Let $I$ be an ideal of $R$. A maximal prime ideal of $I$ is a prime ideal of $I$ which is maximal with this property.} \vspace{3mm} In the following proposition ${\rm Maxp}(I)$ and $D_{M}(I)$ denotes the set of all maximal prime ideal of $I$ and $D(I)\cap {\rm Max}(R)$, respectively. Note that part (a) of the proposition is given in Lemma 3.8 of \cite{AL2}. %\label{1.3} \pro {Let I be an ideal of R and $\varphi$ be the mapping from $D(I)$ to ${\rm Spec}(I)$ with $\varphi(P)= P\cap I$. Then\\ {\rm(a)} $\varphi$ is an order-preserving isomorphism.\\ {\rm(b)} $H$ is a prime and maximal ideal of $I$ if and only if $\varphi^{-1}(H)\in D_M(I)$. In other words we have $\varphi (D_M(I))={\rm Maxp}(I)\cap {\rm Max}(I)$.} {\proof (a). It is clear that $\varphi$ is well-define. We claim that $\varphi$ is onto. To see this, Let $H\in {\rm Spec}(I)$. Clearly, $S=I\setminus H$ is a multiplicatively closed set in $R$ and $S\cap H=\varnothing$. Hence, there exists a prime ideal $P$ of $R$ containing $H$ such that $P\cap S=\varnothing$. Furthermore, it is clear that $P\cap I=H$. Now, suppose that $P, Q \in D(I)$ and $P\cap I\subseteq Q\cap I$. Therefore, $P\cap I\subseteq Q$ and $I\nsubseteq Q$ which imply that $P\subseteq Q$. (b). Suppose that $H\in {\rm Maxp}(I)\cap {\rm Max}(I)$. By part (a), there exists $P\in D(I)$ such that $P\cap I=H$. It is sufficient to show that $P\in {\rm Max}(R)$. Let $a\notin P$ and $i\in I\setminus P$, hence $ai\in I\setminus P$ and then $(P+aiR)\cap I=I$. Therefore, there exist $p\in P$ and $r\in R$ such that $i=p+rai$. Thus, $i(1-ar)=p\in P$, hence $1-ar\in P$ and consequently $P+aR=R$. Conversely, suppose that $M=\varphi^{-1}(H)\in D_M(I)$, we must show that $H\in {\rm Max}(I)$. Let $a\in I\setminus H$, then $a\notin M$ and hence $M+aR=R$. Therefore, $I=IR=I(M+aR)=IM+aI\sub(M\cap I)+aI=H+aI$. Thus, $I=H+aI$ and we are done.\hfill$\square$} %\label{1.4} \coro {Let $I$ be an ideal of $R$, $S$ be a subring of $R$ and $I\sub S$. Then there exists an order preserving isomorphism between the set of prime ideals of $R$ not containing $I$ and the set of all prime ideals of $S$ not containing $I$.} \proof {By the previous proposition, it is clear.\hfill$\square$} \vspace{3mm}Recall that an element $a\in R$ is called a regular element whenever there exists $b\in R$ such that $a=a^2b$. An ideal $I$ of $R$ is called a regular ideal if each of its elements is regular. If $I=R$, then we say that $R$ is a regular ring, see \cite{GOO} and \cite{AB3}. %\vspace{3mm} In the following discussion we observe that there %exists a close relation between regular ideals and $P$-ideals in %a reduced ring. \vspace{3mm} The following proposition is well-known. %\label{1.5} \pro {Let $a\in R$, then the following statements are equivalent:\\ {\rm(a)} $a$ is a regular element.\\ {\rm(b)} There exists an idempotent $e\in R$ such that $aR=eR$.\\ {\rm(c)} ${\cal A}(a)$ is generated by an idempotent.\\ {\rm(d)} ${\cal A}^2(a)$ is generated by an idempotent.\\ {\rm(e)} ${\cal A}(a)\oplus {\cal A}^2(a)=R$.\\ {\rm(f)} ${\cal A}(a)\oplus aR=R$.} %\label{1.6} \lem { Let $R$ be a reduced ring and $e\in R$ be an idempotent element. Then ${\bf P}_e=eR$. Furthermore, if {\rm Jac}$(R)=(0)$, then ${\bf M}_e={\bf P}_e=eR$.} \proof {By [6, Theorem 1.4], we have ${\bf P}_a={\cal A}^2(a)$. Hence, ${\bf P}_e={\cal A}^2(e)=eR$. To show the second part, by [3, Theorem 2.9] , we have ${\bf M}_a\subseteq {\bf P}_a$ for any $a\in R$ and so ${\bf M}_e={\bf P}_e=eR$.\hfill$\square$} %\label{1.7} %\defi {Let $a, b\in R$, if there exists the natural number $n\geq %2$, such that $a=b^n$, then we say that $b$ is a {\rm nth} root of %$a$.} %\label{1.8} \lem {Let $R$ be a reduced ring and $a, b\in R$ such that $a=b^n$ for a natural number $n\geq 2$ %(i.e., $b$ is an {\rm nth} root of $a$) . Then the following statements are equivalent:\\ {\rm(a)} $a$ is a regular element.\\ {\rm(b)} $aR$ is a semiprime ideal.\\ {\rm(c)} ${\bf P}_a=aR$.\\ Furthermore, if ${\rm Jac}(R)=(0)$, then\\ {\rm(d)} ${\bf M}_a=aR$.\\ {\rm(e)} $aR$ is an intersection of maximal ideals.} {\bf Proof.} First we prove the implications (a) $\Rightarrow$ (b), (c), (d), (e). By part (b) of Proposition 1.5, there exists an idempotent element $e\in R$ such that $aR=eR$, so ${\bf P}_a={\bf P}_e=eR=aR$. Also, if ${\rm Jac}(R)=(0)$, then ${\bf M}_a={\bf M}_e=eR=aR$. (b) $\Rightarrow$ (a). By our hypothesis, we have $b\in aR$ and so there exists $c\in R$ such that $b=ac$. Clearly, $a=b^n=(ac)^n=a^2d$ in which $d=c^na^{n-2}$. Hence, $a$ is a regular element. (c) $\Rightarrow$ (b). It is clear.\\ Furthermore, if ${\rm Jac}(R)=(0)$, then (d) $\Rightarrow$ (b) and (e) $\Rightarrow$ (b) are clear.\hfill$\square$ %\label{1.9} \pro {Let $R$ be a reduced ring and $I$ be an ideal of $R$. Then the following statements are equivalent:\\ {\rm(a)} $I$ is a regular ideal.\\ {\rm(b)} $aR$ is a semiprime ideal for any $a\in I$.\\ {\rm(c)} ${\bf P}_a=aR$ for any $a\in I$.\\ Furthermore, if ${\rm Jac}(R)=(0)$, then\\ {\rm(d)} ${\bf M}_a=aR $ for any $a\in I$.\\ {\rm(e)} $aR$ is an intersection of maximal ideals for any $a\in I$.} {\proof Except the implication ($b\Rightarrow a$), which is clear, the reminder of the proof is an immediate conclusion of Lemma 1.7.\hfill$\square$} \section{${\rm P}$-ideals and ${\rm PMP}$-ideals in commutative rings} The concept of ${\rm P}$-ideal has already been defined in the context of ${\rm C}(X)$ and the concept of ${\rm PMP}$-ideal is new. We study the properties of these ideals. For more details about ${\rm P}$-ideals in the ring ${\rm C}(X)$, see \cite {RUD}. %\label{2.1} \defi {Let $R$ be a ring and $I$ be an ideal of $R$. Then $I$ is called ${\rm P}$-ideal, whenever every prime ideal in the ring $I$ is a maximal ideal in $I$. Also, $I$ is called ${\rm PMP}$-ideal, whenever every prime ideal in the ring $I$ is a maximal prime ideal in $I$.} Obviously, the zero ideal is a ${\rm P}$-ideal and ${\rm PMP}$-ideal and also every ${\rm P}$-ideal is a ${\rm PMP}$-ideal; but a ${\rm PMP}$-ideal is not a ${\rm P}$-ideal in general. To see this, consider the reduced local ring $R=\Z_{2\Z}$, then clearly, the unique maximal ideal of $R$ is a ${\rm PMP}$-ideal which is not a ${\rm P}$-ideal. In some rings such as ${\rm C}(X)$, these concepts coincide. At the end of this section, we show that ${\rm PMP}$-ideals and ${\rm P}$-ideals are the same, for pure ideals. %\vspace{3mm} In the following example, we give a ${\rm PMP}$-ideal which %is not a ${\rm P}$-ideal. %\vspace{3mm} {\exa \vspace{3mm} In the next proposition, we find necessary and sufficient condition for an ideal $I$ to be a ${\rm P}$-ideal (resp., ${\rm PMP}$-ideal ). The first part of the following theorem is well-known in the context of ${\rm C}(X)$, see \cite {RUD}. %\label{2.2} \theo {Let $R$ be a ring and $I$ be an ideal of $R$. Then\\ {\rm(a)} $I$ is a ${\rm P}$-ideal if and only if $D(I)\subseteq {\rm Max}(R)$.\\ {\rm(b)} $I$ is a ${\rm PMP}$-ideal if and only if $D(I)\subseteq {\rm Min}(R)$.} {\proof (a $\RTA$). Assume that $P\in D(I)$, then $P\cap I\in {\rm Spec}(I)$. Hence, $H=P\cap I$ is a maximal ideal in $I$. Now, by part (b) of Proposition 1.3, we have $P=\varphi^{-1}(H)\in {\rm Max}(R)$. (a $\LTA$). Suppose that $H\in {\rm Spec}(I)$. By part (a) of Proposition 1.3, we have $\varphi^{-1}(H)=P\in D(I)$. By our hypothesis, $P\in {\rm Max}(R)$ and so by part (b) of Proposition 1.3, we have $H\in {\rm Maxp}(I)\cap {\rm Max}(I)$. (b). Suppose that $P\in D(I)$ and $Q$ is a prime ideal contained in $P$, hence $Q\in D(I)$. Consequently, $P\cap I\in {\rm Maxp}(I)$ and $Q\cap I\in {\rm Maxp}(I)$. Hence, $P\cap I=Q\cap I$ implies that $P=Q$ and consequently, $P\in {\rm Min}(R)$. The converse is clear.\hfill$\square$} %\label{2.3} \pro {Let $I$ and $J$ be two ideals of a ring $R$ and $I\sub J$. If $J$ is a ${\rm P}$-ideal (resp., ${\rm PMP}$-ideal), then $J/I$ is a ${\rm P}$-ideal (resp., ${\rm PMP}$-ideal) of the ring $R/I$. The converse is true if $I$ is a ${\rm P}$-ideal (resp., ${\rm PMP}$-ideal).} {\bf Proof.} The proof is straightforward.\hfill$\square$ \vspace{3mm} In the following proposition, we observe that for investigating ${\rm P}$-ideals and ${\rm PMP}$-ideals it is enough to consider semiprime ideals. %\label{2.4} \pro {Let $R$ be a ring and $I$ be an ideal of $R$. Then\\ {\rm(a)} $D(I)=D(\sqrt{I})$ hence $I$ is a ${\rm P}$-ideal (resp., ${\rm PMP}$-ideal) if and only if $\sqrt {I}$ is a ${\rm P}$-ideal (resp., ${\rm PMP}$-ideal).\\ {\rm(b)} $D(J)\sub D(I)$ for every ideal $J\sub I$ and consequently $I$ is a ${\rm P}$-ideal (resp., ${\rm PMP}$-ideal) \iff every ideal contained in $I$ is too, and this is equivalent to the fact that $aR$ is a ${\rm P}$-ideal for any $a\in I$.} {\bf Proof.} It is clear.\hfill$\square$ %\label{2.5} \pro{ The sum of any family of {\rm P}-ideals (resp., {\rm PMP}-ideals) of a ring $R$ is a {\rm P}-ideal (resp., {\rm PMP}-ideal).} {\bf Proof.} By the inclusion $D(\sum_{\lambda\in\Lambda}I_\lambda)\sub\cup_{\lambda\in\Lambda}D(I_\lambda)$, the proof is clear.\hfill$\square$\\ By the previous proposition, it follows that the largest P-ideal (resp., PMP-ideal) of $R$ exists. We denote this largest ideal by P$(R)$ (resp., PMP$(R)$-ideal). It is obvious to see that if $I$ is an ideal of $R$, then $I\cap {\rm P}(R)$, (resp., $I\cap {\rm PMP}(R)$) is the largest P-ideal (resp., PMP-ideal) contained in $I$. Also, $J$ is the largest P-ideal of a ring $R$ \iff $J$ is a P-ideal (resp., PMP-ideal) and $R/J$ has no any nonzero P-ideal (resp., PMP-ideal). \vspace{3mm} Here, a natural question arises: Is the largest P-ideal (resp., PMP-ideal) in a ring $R$ (or in an ideal of $R$) a prime ideal? The answer is no. To see this, suppose that the topological space $X$ has not any P-point. It is enough to prove ${\rm PMP(C}(X))=(0)$. Assume that $I$ is a nonzero ideal of ${\rm C}(X)$. By our hypothesis, there exists $0\neq f\in I$. Thus, there exists $x\in {\rm Coz}(f)$. Clearly, $I\nsub M_x(X)$; i.e., $M_x(X)\in D(I)$. Since $x$ is not a P-point, we infer that $M_x(X)$ is not a minimal prime ideal and hence $I$ is not a PMP-ideal. This example, also, shows that if $I$ is a P-ideal (resp., PMP-ideal) and $P\in {\rm Min}(I)$, then $P$ is not necessarily a P-ideal (resp., PMP-ideal). %\label{2.7} \pro {Let $R$ be a ring. Then\\ {\rm(a)} ${\rm P}(R)=\cap_{P\in {\rm Min}(R)\set {\rm Max}(R)}P$.\\ {\rm(b)} ${\rm PMP}(R)=\cap_{P\in {\rm Spec}(R)\set {\rm Min}(R)}P$.} {\bf Proof.} (a). We show that $J_0=\cap_{P\in {\rm Min}(R)\set {\rm Max}(R)}P$ is ${\rm P}$-ideal. Clearly, $D(J_0)\sub {\rm Max}(R)$ and so by part (a) of Theorem 2.2, $J_0$ is a ${\rm P}$-ideal. Now, suppose that $I$ is ${\rm P}$-ideal. Thus, $D(I)\sub {\rm Max}(R)$. It follows that ${\rm Min}(R)\set {\rm Max}(R)\sub V(I)$ and so $I\sub \cap_{P\in {\rm Min}(R)\set {\rm Max}(R)}P=J_0$. (b). It is similar to the proof of part (a).\hfill$\square$ %\label{2.8} \rem {Let $I$ and $J$ be two ideals of $R$. Then\\ {\rm(a)} $IJ=(0)$ \iff $D(J)\sub V(I)$.\\ {\rm(b)} If $M_1,\cdots, M_n\in {\rm Max}(R)$, then ${\cal A}(\cap_{i=1}^n M_i)$ is a P-ideal.} %\label{2.9} \coro{ If $R/{\cal A}(I)$ is a regular ring, then $I$ is a ${\rm P}$-ideal.} {\bf Proof.} Since $R/{\cal A}(I)$ is a regular ring, it follows that $V({\cal A}(I))\sub {\rm Max}(R)$ and so by part (a) of the above remark we are done.\hfill$\square$\\ The following example shows that the converse of Corollary 2.9 is not true. Note that if $R=\prod_{i=1}^nR_i$ and $I_i$ is an ideal of $R_i$ for every $i=1,\cdots, n$, then $I$ is a ${\rm P}$-ideal of $R$ if and only if $I_i$ is so in $R_i$ for every $i=1,\cdots, n$. \vspace{3mm} \exa Assuming that $F$ is any field, set $R=F\times F\times\Bbb{Z}$. Now, if we let $I=(\circ)\times F\times\Bbb{Z}$ and $J=F\times (\circ)\times\Bbb{Z}$, then $K={\cal A}(I\cap J)$ is a ${\rm P}$-ideal, but $\frac{R}{{\bf\cal A}(K)}$ is not a regular ring. \vspace{3mm} The following result shows that the converse of the above corollary is true, even if $I$ is a summand. %\label{2.10} \coro{ Suppose that an ideal $I$ of $R$ is summand. Then the following statements are equivalent:\\ {\rm(a)} $I$ is a ${\rm P}$-ideal.\\ {\rm(b)} $R/{\cal A}(I)$ is a regular ring.\\ {\rm (c)} $I$ is a regular ideal.} {\bf Proof.} Since $I$ is summand, it follows that $I\simeq R/{\cal A}(I)$. Thus, it suffices to show (a) $\LRTA$ (b). To see this, by our hypothesis, there exists an ideal $J$ of $R$ such that $R=I\oplus J$. Clearly, $D(I)=V(J)=V({\cal A}(I))$ and by this fact the proof is easy.\hfill$\square$ %\label{2.11} \coro{Let $R$ be a reduced ring. If $R$ has a maximal ideal which is ${\rm PMP}$-ideal, then every prime ideal is minimal or maximal ideal. (i.e., ${\rm dim}(R)\leq 1$). } Now, we investigate some connections between annihilator ideals and ${\rm P}$-ideals. First, we recall the following well-known fact. For the proof see [12, Lemma 11.40]. %\label{2.12} \lem {Let $R$ be a reduced ring and $I$ be an ideal of $R$, then ${\cal A}(I)=\bigcap_{_{P\in {\rm Min}(R)\cap D(I)}} P=\bigcap_{_{P\in D(I)}} P$.} \pro{Let $R$ be a ring and $I$ be an ideal of $R$. \\ {\rm(a)} If ${\cal A}(I)$ is the intersection of finitely many maximal ideals, then $I$ is ${\rm P}$-ideal.\\ {\rm(b)} If $R$ is reduced and $I$ is a ${\rm P}$-ideal, then ${\cal A}(I)$ is the intersection of a family of maximal ideals.} {\bf Proof.} {(a). Suppose that ${\cal A}(I)=\cap_{i=1}^n M_i$ where $M_i\in {\rm Max}(R)$ for any $i=1,\cdots, n$. Let $P\in D(I)$, since ${\cal A}(I)=\cap_{i=1}^n M_i\subseteq P$, there exists $1\leq i\leq n$, such that $M_i\subseteq P$. Hence by the maximality of $M_i$, it follows that $P=M_i$. Therefore, by part (a) of Theorem 2.2, $I$ is a ${\rm P}$-ideal.\\ The proof of part (b) is clear, by the above lemma. \hfill$\square$ } %\label{2.14} \coro{ Suppose that $R$ is a semilocal reduced ring, then $I$ is a ${\rm P}$-ideal if and only if ${\cal A}(I)$ is the intersection of finitely many maximal ideals.} %\label{2.15} {The converse of part \rm{(a)} of Proposition 2.12 is not true in general (even if ${\cal A}(I)$ is also an intersection of finitely many minimal prime ideal). For instance assume that $I$ is a nonzero ideal of the ring $\Z$. Then ${\cal A}(I)=(0)$ is a minimal prime ideal and also is the intersection of infinitely many maximal ideal; while $I$ is not a ${\rm P}$-ideal.} \section{von Neumann regularity, pure ideals and ${\rm P}$-ideals (${\rm PMP}$-ideals)} In this section we observe that in every reduced ring ${\rm P}$-ideals and regular ideals coincide. We also show that every ${\rm P}$-ideal in a reduced ring is a $z^{\circ}$-ideal. Finally, we prove that an ideal $I$ in a reduced ring is a ${\rm P}$-ideal if and only if it is a pure ${\rm PMP}$-ideal. %\label{3.1} \pro {For a reduced ring $R$ the following statements are equivalent:\\ {\rm(a)} $R$ is a regular ring.\\ {\rm(b)} Every ideal $I$ of $R$ is a ${\rm P}$-ideal and $\frac{R}{I}$ is a regular ring .\\ {\rm(c)} There exists an ideal $I$ such that $I$ is ${\rm P}$-ideal and $\frac{R}{I}$ is a regular ring.} \proof{It is evident.\hfill$\square$} %\label{3.2} \pro {Let $R$ be a ring, $a\in R$ and $S=\{a^n: n\in \Bbb{N}_0\}$. Then the ideal $aR$ is a ${\rm P}$-ideal if and only if ${\rm Spec}(S^{-1}R)={\rm Max}(S^{-1}R)$.} {\bf Proof.} Since there exists an order isomorphism between $D(aR)$ and ${\rm Spec}(S^{-1}R)$, the proof is obvious.\hfill$\square$ \vspace{3mm} Let $P\in {\rm Spec}(R)$, we define $O(P)=\{a\in R: {\cal A}(a)\nsubseteq P\}$. The following theorem shows that this concept is closely related to the concept of pure ideal. %\label{3.3} \theo {Suppose that $R$ is a ring, $Q\in {\rm Spec}(R)$ and ${\cal B}=\{P\in {\rm Min}(R):P\subseteq Q\}$. Then\\ {\rm(a)} $m(Q)\subseteq O(Q)\subseteq \cap_{P\in \cal{B}}P$.\\ {\rm(b)} If $Q$ is a pure ideal, then $Q\in {\rm Min}(R)$.\\ {\rm(c)} If $Q$ is a maximal ideal, then $m(Q)=O(Q)$.\\ Furthermore, if $R$ is reduced, then\\ {\rm(d)} $O(Q)= \cap_{P\in \cal{B}}P$.\\ {\rm(e)} If $Q\in {\rm Max}(R)$, then $m(Q)= O(Q)= \cap_{P\in \cal{B}}P$.\\ {\rm(f)} If $Q\in{\rm Max}(R)$, then $Q$ is a pure ideal if and only if $Q\in {\rm Min}(R)$.} {\bf Proof.} (a). Let $a\in m(Q)$, then there exists $q\in Q$ such that $a=aq$, hence $a(1-q)=0$. Therefore, ${\cal A}(a)\nsubseteq Q$ and so $a\in O(Q)$. Now, suppose that $a\in O(Q)$, hence ${\cal A}(a)\nsubseteq P$ for any $P\in \cal{B}$. This implies that $a\in P$ for any $P\in \cal{B}$, and consequently $a\in \cap_{P\in \cal{B}}P$. (b). By part (a), it is clear. (c). Suppose that $Q\in {\rm Max}(R)$ and $a\in O(Q)$. Clearly $$a\in O(Q)\Leftrightarrow {\cal A}(a)\nsubseteq Q\Leftrightarrow Q+{\cal A}(a)=R\Leftrightarrow a\in m(Q).$$ (d). Let $a\in \cap_{P\in \cal{B}}P$ and $S=R\setminus Q$. It is clear that $\frac {a}{1}\in {\rm Rad}(S^{-1}R)$. This implies that there exists a natural number $n$ such that $(\frac{a}{1})^n=0$. Hence there exists $s\in S$ such that $sa^n=0$. Therefore, ${\cal A}(a)={\cal A}(a^n)\nsubseteq Q$ and so $a\in O(Q)$. (e) and (f) are obvious.\hfill$\square$ \vspace{3mm} The next proposition is the same as Theorem 2.4 in \cite {AB3}. %\label{3.4} \pro {An element $a\in R$ is regular if and only if for every $M\in {\rm Max}(R)$ with $a\in M$, we have $a\in m(M)$.} \vspace{3mm} In the following theorem, we show that in reduced ring, regular ideals and ${\rm P}$-ideals coincide. \vspace{3mm} Recall that an ideal in a ring $R$ is called $z$-ideal (resp., $z^{\circ}$-ideal) whenever ${\bf M}_a\subseteq I$ (resp., ${\bf P}_a\subseteq I$) for any $a\in I$. For more details and examples of $z$-ideals and $z^{\circ}$-ideals in reduced commutative rings and in ${\rm C}(X)$ the reader is referred to \cite{AL1}, \cite{AL3} and \cite{AZ}. %\label{3.5} \theo{ Let $R$ be a reduced ring and $I$ is an ideal of $R$. Then the following statements are equivalent:\\ {\rm(a)} $I$ is a ${\rm P}$-ideal.\\ {\rm(b)} $I$ is a regular ideal.\\ {\rm(c)} $aR$ is a semiprime ideal for any $a\in I$.\\ {\rm(d)} ${\bf P}_a=aR$ for any $a\in I$.\\ {\rm(e)} $aR$ is a $z^{\circ}$-ideal for any $a\in I$.\\ Furthermore if $Jac(R)=(0)$, then\\ {\rm(f)} ${\bf M}_a=aR$ for any $a\in I$.\\ {\rm(g)} $aR$ is an intersection of maximal ideals for any $a\in I$.\\ {\rm(h)} $aR$ is a $z$-ideal for any $a\in I$.} %\end{theo} {\bf Proof.} By Proposition 1.8 and definitions of $z$-ideal and $z^\circ$-ideal, it suffices to prove (a) $\LRTA$ (b).\\ (a) $\RTA$ (b). Suppose that $M\in {\rm Max}(R)$ and $a\in I\cap M$. By Proposition 3.4, it suffices to show that $a\in m(M)$. On the other hand, by Theorem 3.3, we have $m(M)=O(M)=\cap\{P\in {\rm Min}(R): P\subseteq M\}$. thus, it is enough to show that $a\in \cap\{P\in {\rm Min}(R): P\subseteq M\}$. Let $P\in {\rm Min}(R)$ and $P\subseteq M$, we must show that $a\in P$. This is clear, for on the contrary, we have $P\in D(I)$ and consequently $P\in {\rm Max}(R)$ which is a contradiction. (b) $\RTA$ (a). Suppose that $P\in D(I)$, by Theorem 2.2, we must show that $P\in {\rm Max}(R)$. Let $a\notin P$; since $P\in D(I)$, there exists an $i\in I\setminus P$. Clearly, $ai\in I\setminus P$ and by assumption, there exists $r\in R$ such that $ai=(ai)^2r$. Hence, $ai(1-air)=0\in P$ and so $1-air\in P$ which imlies that $P+aR=R$.\hfill$\square$ %\label{3.6} \coro {Every ${\rm P}$-ideal in a reduced ring is a $z^{\circ}$-ideal. } \vspace{3mm} The following proposition and theorem show the connection between ${\rm P}$-ideals, ${\rm PMP}$-ideals and pure ideals. %\label{3.7} \pro {Let $R$ be a reduced ring. Then\\ {\rm(a)} $a\in R$ is regular if and only if $aR$ is a pure ideal.\\ {\rm(b)} $I$ is a ${\rm P}$-ideal if and only if every ideal contained in $I$ is a pure ideal.} {\bf Proof.} (a $\RTA$). Suppose that $x=ar\in I=aR$. By our hypothesis, there exists $s\in R$ such that $a=a^2s$ and so $x=ar=a^2sr\in aI$. (a $\LTA$). Since $I=aR$ is pure and $a\in I$, it follows that $a=a(ar)=a^2r$ for some $r\in R$. (b). By part (a), it is easy.\hfill$\square$ %\label{3.8} \theo {Let $R$ be a reduced ring and $I$ be an ideal of $R$. Then $I$ is a ${\rm P}$-ideal if and only if it is a pure ${\rm PMP}$-ideal.} {\proof ($\Rightarrow$). It is clear. ($\Leftarrow$). By Theorem 3.5, it is enough to show that $I$ is a regular ideal. To see this, let $a\in I$, by Proposition 3.4, if we prove $a\in M\in {\rm Max}(R)$ implies that $a\in m(M)$, then the proof is complete. For this, let $a\in M$, if $M\in D(I)$, then by Theorem 2.2, we have $M\in {\rm Min}(R)$ and so by part (f) of Proposition 3.3, $M$ is a pure ideal. Hence, $a\in M=m(M)$. If $M\notin D(I)$, then $I\subseteq M$ and so by the purity of $I$, we have $a\in I=m(I)\subseteq m(M)$.\hfill$\square$} \vspace{3mm} It is clear that a reduced ring $R$ is regular if and only if every ideal of $R$ is a ${\rm P}$-ideal. In the next theorem we give a similar assertion for ${\rm PMP}$-ideals. %\label{3.9} \theo {Every proper ideal in reduced ring $R$ is a ${\rm PMP}$-ideal if and only if $R$ is regular or a local ring with ${\rm dim}(R)=1$.} {\proof ($\Rightarrow$). Assume that $R$ is not regular. Hence, there exist $M_0\in {\rm Max}(R)$ and $P\in {\rm Spec}(R)$ such that $P\varsubsetneq M_0$. It is enough to show that ${\rm Max}(R)=\{M_0\}$ and $P\in {\rm Min}(R)$. Let $M\in {\rm Max}(R)$, since $M$ is a ${\rm PMP}$-ideal, by part (b) of Theorem 2.2, we have $M\subseteq M_0$ and hence $M=M_0$. This implies that ${\rm Max}(R)=\{M_0\}$. Now, suppose that $Q\in {\rm Spec}(R)$ and $Q\subseteq P$. Since $M_0$ is a ${\rm PMP}$-ideal, by part (b) of Theorem 2.2, we conclude that $P=Q$. This implies that $P\in {\rm Min}(R)$. ($\Leftarrow$). It is clear.\hfill$\square$} %\label{3.10} \theo {Let $R$ be a reduced ring. Then the following statements are equivalent:\\ {\rm(a)} $R$ is a regular ring.\\ {\rm(b)} There exists an ideal $M\in {\rm Max}(R)$ which is a ${\rm P}$-ideal.\\ {\rm(c)} There exists a pure ideal $M\in {\rm Max}(R)$ which is a ${\rm PMP}$-ideal.} {\bf Proof.} The implications (a) $\Rightarrow$ (b) $\Rightarrow$ (c) are clear. ($c\Rightarrow a$). Suppose that $M\in {\rm Max}(R)$ is a pure ${\rm PMP}$-ideal and $M\neq N\in {\rm Max}(R)$. Clearly, $N \in D(M)$ and so $N\in {\rm Min}(R)$.\hfill$\square$ \section{${\rm P}$-ideals and ${\rm PMP}$-ideals in ${\rm C}(X)$} In \cite{RUD}, ${\rm P}$-ideals have studied in ${\rm C}(X)$. Theorem 3.5 shows that P-ideals and regular ideals are the same in ${\rm C}(X)$. Also, we show that the concepts ${\rm P}$-ideal and ${\rm PMP}$-ideal in ${\rm C}(X)$ are the same and then we obtain a new equivalent condition for ${\rm P}$-spaces. First, we need several propositions and lemmas. %\label{4.1} \lem {Let $X$ be a topological space and $f\in {\rm C}(X)$. Then\\ $${\rm(a)}~ P_f={\cal A}^2(f)=\cap_ {x\in Z^{\circ}(f) } O_{x}(X)=\{g\in {\rm C}(X):Z^{\circ}(f)\subseteq Z^{\circ}(g)\}=O_{Z^{\circ}(f)}(X).$$ ${\rm(b)}~ M_f=\cap_{x\in Z(f)}M_{x}(X)=\{g\in {\rm C}(X):Z(f)\subseteq Z(g)\}=M_{Z(f)}(X).$} {\proof The proof is straightforward.\hfill$\square$} %\label{4.2} \lem {Let $X$ be a topological space and $A$ be a closed subset of $X$. Then $A$ is a regular closed subset (i.e., $\overline {A^{\circ}}=A$) if and only if $O_{A^{\circ}}(X)=M_A(X)$.} {\proof ($\Rightarrow$). By our hypothesis $A=\overline {A^{\circ}}$ and consequently we can write $$O_{A^{\circ}}(X)=\{g\in {\rm C}(X):A^{\circ}\subseteq Z^{\circ}(g)\}=\{g\in {\rm C}(X): A\subseteq Z(g)\}=M_A(X).$$ ($\Leftarrow$). On the contrary, let $x\in A\setminus \overline {A^{\circ}}$. Hence, there exists $h\in {\rm C}(X)$ such that $x\in {\rm Coz}(h)$ and $\overline {A^{\circ}}\cap {\rm Coz}(h)=\varnothing$. Thus, $A\nsubseteq Z(h)$ and $A^{\circ}\subseteq \overline {A^{\circ}}\subseteq Z(h)$. Therefore, $h\in O_{A^{\circ}}(X)\setminus M_A(X)$ which gives a contradiction.\hfill$\square$ } \vspace{3mm} The following result is an immediate consequence of Lemmas 4.1 and 4.2. %\label{4.3} \coro {Let $X$ be a topological space and $f\in {\rm C}(X)$. Then $Z(f)$ is a regular closed subset if and only if $M_f=P_f$.} \vspace{3mm} The proof of the following result is evident. %\label{4.4} \pro {Let $X$ be a topological space and $f\in {\rm C}(X)$. The following statements are equivalent:\\ {\rm(a)} $Z(f)=Z^{\circ}(f)$.\\ {\rm(b)} $f$ is a regular element.\\ {\rm(c)} There exists an idempotent $e\in {\rm C}(X)$ such that $f{\rm C}(X)=e{\rm C}(X)$.\\ {\rm(d)} $f{\rm C}(X)$ is a semiprime ideal.\\ {\rm(e)} $P_f=f{\rm C}(X)$.\\ {\rm(f)} $M_f=f{\rm C}(X)$.\\ {\rm(g)} $f{\rm C}(X)$ is a ${\rm P}$-ideal.\\ {\rm(h)} $f{\rm C}(X)$ is a pure ideal.\\ {\rm(i)} $f{\rm C}(X)$ is an intersection of maximal ideals.} %{\bf Proof.} By ....... it is evident. %\label{4.5} \theo {Let $I$ be an ideal in ${\rm C}(X)$. Then $I$ is a ${\rm P}$-ideal if and only if it is a ${\rm PMP}$-ideal.} {\bf Proof.} ($\Rightarrow$). It is clear. ($\Leftarrow$). By Theorem 3.5, it is enough to show that $I$ is a regular ideal. Assume that $I$ is not regular, hence by Proposition 4.4, there exists $f\in I$ such that $Z(f)\neq Z^{\circ}(f)$ and so there exists $x\in Z(f)\setminus Z^{\circ}(f)$; i.e., $f\in M_x(X)\setminus O_x(X)$. Now, by exercise I6 of chapter 4 of \cite{GIL}, there exists a prime ideal $P$ such that $O_x(X)\subseteq P$, $f\notin P$ and $P$ is not a $z$-ideal. Clearly, $P\in D(I)$ and by Theorem 2.2, and our hypothesis we have $P\in {\rm Min}({\rm C}(X))$. This implies that $P$ is a $z$-ideal which is a contradiction.\hfill$\square$ \vspace{3mm} In the following, we give an example of pure ideal which is not a ${\rm P}$-ideal. In fact, as we see later, ${\rm P}$-ideals and pure ideals are the same if and only if $X$ is a ${\rm P}$-space. \vspace{3mm} \exa {Suppose that $X=\{0\}\cup [1, \infty)$. One can easily see that $M_0(X)$ is a pure ideal. But $M_0(X)$ is not a ${\rm P}$-ideal, To see this, assume that $f\in M_0(X)$ such that $Z(f)=\{0, 1\}$. Hence, $Z(f)\neq Z^{\circ}(f)$ and so by Propositon 4.4, $f$ is not a regular element. This implies that $M_0(X)$ is not a ${\rm P}$-ideal.} \vspace{3mm} Recall that $X$ is a ${\rm P}$-space whenever every prime ideal of ${\rm C}(X)$ is a maximal ideal; i.e., ${\rm C}(X)$ is a regular ring or equivalently every zeroset in $X$ is open, for other equivalent, see \cite{GIL}. In the next theorem we give a new equivalent for ${\rm P}$-spaces. %\label{4.6} \theo {A space $X$ is a ${\rm P}$-space if and only if every pure ideal in ${\rm C}(X)$ is a ${\rm P}$-ideal.} {\proof ($\Rightarrow$). By Proposition 4.4, it is evident. ($\Leftarrow$). {Suppose that $X$ is not a ${\rm P}$-space. Hence, there exist a zeroset $Z_1\varsubsetneq X$ and $x\in X$ such that $x\in Z_1\setminus Z_1^\circ$. Now, we consider $Z_2=Z(g)$ with $g\in O_y(X)$ and $Z_1\cap Z_2=\varnothing$. Put $Z(f)=Z_1\cup Z_2$, clearly, $f\in O_y(X)$ and $x\in Z(f)\setminus Z^\circ(f)$. Therefore, by Proposition 4.4, $f$ is not a regular element. This means that $O_y(X)$ is not a ${\rm P}$-ideal; while, $O_y(X)$ is a pure ideal.\hfill$\square$} \vspace{3mm} The above theorem is not true in general. In other words, if every pure ideal of $R$ is a ${\rm P}$-ideal, then we can not conclude that $R$ is a regular ring. For example, in the ring $\Z$ every pure ideal is a ${\rm P}$-ideal but $\Z$ is not a regular ring. We should remind the reder that the zero ideal of $\Bbb{Z}$ is only its pure ideal. \vspace{3mm} We conclude this paper with the characterization of largest P-ideal in ${\rm C}(X)$. %\label{4.7} %\lem {For every ideal $J$ of $C(X)$ there exists a unique closed subset $A$ of $\beta X$ such that $O^A(X)\sub J\sub M^A(X)$.} %{\bf Proof.} Taking $A=\theta(J)=\cap_{g\in J}cl_{\beta X}Z(g)$, we prove that $O^A(X)\sub J\sub M^A(X)$. Clearly, $J\sub M^A(X)$. To complete the proof, it is enough to show that for every $f\in O^A(X)$, there exists $g\in J$ such that $cl_{\beta X}Z(g)\sub int_{\beta X}cl_{\beta X}Z(f)$. On the contrary, let $f\in O^A(X)$ be such that $cl_{\beta X}Z(g)\cap(\beta X\set int_{\beta X}cl_{\beta X}Z(f))\neq\tohi$ for every $g\in J$. Since $\beta X$ is compact, there exists an element $p\in \cap_{g\in J}cl_{\beta X}Z(g)\cap(\beta X\set int_{\beta X}cl_{\beta X}Z(f))$. It follows that $p\in A$ and $f\notin O^p(X)$ and this is a contradiction. Now, suppose that $B$ is closed subspace of $\beta X$ and $O^B(X)\sub J\sub M^B(X)$. It is enough to prove that $B=\theta(J)$. Obviously, $B\sub\theta(J)$. Suppose $p\notin B$, there exsists $f\in C(X)$ such that $f\notin M^p$ and $B\sub int_{\beta X}cl_{\beta X}Z(f)$. Thus $f\in O^B(X)\set M^p(X)\sub J\set M^p(X)$. Therefore, $p\notin\theta(J)$. %\label{4.8} %The following lemma is not true %\lem {If $J$ is a $P$-ideal in $C(X)$, then there exists a unique closed subset of $\beta X$ such that $J=O^A(X)=M^A(X) $.} %{\bf Proof.} Suppose that $J$ is a $P$-ideal of $C(X)$. By the above lemma, there exsits a closed set $A$ in $\beta X$ such that $O^A(X)\sub J\sub M^A(X)$. It is enough to show that $O^A(X)=M^A(X)$. By ........ every element of $J$ is regular and so by ........ $Z(f)=Z^\circ(f)$ for every $f\in O^A(X)$. Therefore, $cl_{\beta X}Z(f)=int_{\beta X}cl_{\beta X}Z(f)$ for every $f\in O^A(X)$ and clearly it follows that $O^A(X)=M^A(X)$.} \vspace{3mm} %The following lemma is not necessary. %\label{4.9} %\lem {Let $A$ and $B$ be two closed subset of $\beta X$. Then the following statements hold.\\ %\rm{(a)} $M^A(X)\sub M^B(X)$ \iff $B\sub A$.\\ %\rm{(b)} if $O^A(X)=M^A(X)$, then $B\sub A$, then $B\sub A$ \iff $O^A(X)\sub O^B(X)$.} %{\bf Proof.} Part (a) is well-known and part (b) is an immediate consequence of part (a). %\label{4.10} \theo {Let $B$ be the set of all non $P$-points of $\beta X$ with respect to $X$. Then ${\rm P}({\rm C}(X))=O^B(X)$.} {\bf Proof.} By Proposition 2.6, we can write $${\rm P}({\rm C}(X))=\bigcap\{P\in {\rm Min}({\rm {\rm C}}(X)):~O^p(X)\sub P~{\rm{and}}~p\in B\}$$ \hspace{3.3cm}$=\cap_{p\in B}O^p(X)=O^B(X).$\hfill$\square$ \begin{thebibliography}{19} \bibitem{AB1} Abu Osba, E.A., Al-Ezeh, H., {\em The pure part of the ideals in ${\rm C}(X)$}, Math. J. Okayama Univ. 45 (2003), 73-82. \bibitem{AB2} Abu Osba, E.A., Al-Ezeh, H. {\em Some properties of the ideal of continuous functions with pseudocompact support}, IJMMS, 27:3 (2001), 169-176. \bibitem{AB3} Abu Osba, E.A., Henriksen, M., Alkan, O., {\em Combining local and von Neumann regular rings}, Comm. in Algebra 32 (2004), 2639-2653. \bibitem{AL1} Aliabad, A.R, {\em $z^{\circ}$-ideals in ${\rm C}(X)$}, Ph.D. thesis, 1996. \bibitem{AL2} Aliabad, A.R., Azarpanah, F., Namdari, M. {\em Rings of cotinuous functions vanishing at infinity}, Comment. Math. Univ. 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