# Algebraic properties of the ring $C(X)_{\mathcal{P}}$

Amrita Dey  Department of Pure Mathematics, University of Calcutta, 35,
Ballygunge Circular Road, Kolkata 700019, West Bengal, India
<EMAIL_ADDRESS>,  Sagarmoy Bag  Department of Mathematics, Bankim
Sardar College, Tangrakhali, West Bengal 743329, India
<EMAIL_ADDRESS>and  Dhananjoy Mandal  Department of Pure
Mathematics, University of Calcutta, 35, Ballygunge Circular Road, Kolkata
700019, West Bengal, India<EMAIL_ADDRESS>
# Algebraic properties of the ring $C(X)_{\mathcal{P}}$.

Amrita Dey  Department of Pure Mathematics, University of Calcutta, 35,
Ballygunge Circular Road, Kolkata 700019, West Bengal, India
<EMAIL_ADDRESS>,  Sagarmoy Bag  Department of Mathematics, Bankim
Sardar College, Tangrakhali, West Bengal 743329, India
<EMAIL_ADDRESS>and  Dhananjoy Mandal  Department of Pure
Mathematics, University of Calcutta, 35, Ballygunge Circular Road, Kolkata
700019, West Bengal, India<EMAIL_ADDRESS>
###### Abstract.

Our aim is to study certain algebraic properties of the ring
$C(X)_{\mathcal{P}}$ of real-valued functions on $X$ whose closure of
discontinuity set is in an ideal of closed sets. We characterize
$\mathcal{P}P$-spaces using $z$-ideals and essential ideals of
$C(X)_{\mathcal{P}}$ and also almost $\mathcal{P}P$-spaces using
$z^{0}$-ideals of $C(X)_{\mathcal{P}}$ and a topology finer than the original
topology on $X$. We deduce that each maximal ideal of $C(X)_{F}$ [13] (resp.
$T^{\prime}(X)$ [1]) is a $z^{0}$-ideal. We establish that the notions of
clean ring, weakly clean ring, semiclean ring, almost clean ring and exchange
ring coincide in the ring $C(X)_{\mathcal{P}}$. End of this paper, we also
characterize $\mathcal{P}P$-spaces and almost $\mathcal{P}P$-spaces using
certain ideals having depth zero. We exhibit a condition on $\mathcal{P}$
under which prime and essential ideals of $C(X)_{\mathcal{P}}$ have depth
zero.

###### Key words and phrases:

$\tau\mathcal{P}$-space, $\tau\mathcal{PU}$-space, $P$-completely separated,
$\tau\mathcal{P}$-compact, $\mathcal{P}P$-space

###### 2020 Mathematics Subject Classification:

Primary 54C30; Secondary 13C99

##  1\. Introduction

Let $\mathcal{P}$ be an ideal of closed subsets of a $T_{1}$ topological space
$(X,\tau)$ and we call the triplet $(X,\tau,\mathcal{P})$ a
$\tau\mathcal{P}$-space. The subring $C(X)_{\mathcal{P}}$ was introduced in
[10] as the subring of $\mathbb{R}^{X}$ consisting of those functions $f\colon
X\longrightarrow\mathbb{R}$ which satisfy the condition
$\overline{D_{f}}\in\mathcal{P}$, where $D_{f}=\\{x\in X\colon f\text{ is
discontinuous at }x\\}$. For a $\tau\mathcal{P}$-space $(X,\tau,\mathcal{P})$,
it is easy to check that the collection $coz_{\mathcal{P}}[X]=\\{coz(f)\colon
f\in C(X)_{\mathcal{P}}\\}$ of all cozero sets of functions in
$C(X)_{\mathcal{P}}$, forms a base for open sets for a topology
$\tau_{\mathcal{P}}$ on the set $X$, which is in general finer than $\tau$.
However, the two topologies, viz, $\tau_{\mathcal{P}}$ and $\tau$ coincide if
and only if $C(X)=C(X)_{\mathcal{P}}$. We use the following lemma to achieve
this equivalence.

######  Lemma 1.1.

[10] Let $\mathcal{P}$ be an ideal of all closed subsets of a $T_{1}$-space
$X$ and $\mathcal{P}^{\prime}$ be the ideal of all closed subsets of the set
of isolated points of $X$. Then $C(X)_{\mathcal{P}}=C(X)$ if and only if
$\mathcal{P}\subseteq\mathcal{P^{\prime}}$.

The following theorem is now obvious.

######  Theorem 1.2.

For a $\tau\mathcal{P}$-space $(X,\tau,\mathcal{P})$,
$\tau=\tau_{\mathcal{P}}$ if and only if $C(X)=C(X)_{\mathcal{P}}$.

######  Notation 1.3.

We simply write $X_{\mathcal{P}}$ for the topological space
$(X,\tau_{\mathcal{P}})$.

One may note that if $\chi_{\\{{a}\\}}\in C(X)_{\mathcal{P}}$ for all $a\in
X$, $\tau_{\mathcal{P}}$ is nothing but the discrete topology on $X$. So in
this case $C(X_{\mathcal{P}})=\mathbb{R}^{X}$ which is often not equal to
$C(X)_{\mathcal{P}}$ (for example, take $X=\mathbb{R}$, the real line and
$\mathcal{P}=\mathcal{P}_{f}$ [10]). Moreover, if $X=\mathbb{R}$ endowed with
cofinite topology and $\mathcal{P}=\\{\emptyset,\\{1\\}\\}$, then $(X,\tau)$
is a connected topological space but $(X,\tau_{\mathcal{P}})$ contains a
clopen set $\\{1\\}$ and is thus disconnected.

Throughout the article, for a subset $A\subseteq X$, we shall denote
$int_{X_{\mathcal{P}}}A$ to be the interior of $A$ in $X_{\mathcal{P}}$ and
$cl_{X_{\mathcal{P}}}A$ to be the closure of $A$ in $X_{\mathcal{P}}$.

The aim of this article is to study certain algebraic properties of
$C(X)_{\mathcal{P}}$ starting with the study of various types of ideals of the
ring.

In Section 2, we study $z$-ideal, $z^{0}$-ideal and essential ideal of
$C(X)_{\mathcal{P}}$. We have introduced the concept of
$z_{\mathcal{P}}$-ideals of $C(X)_{\mathcal{P}}$ in [10] and in this section
we establish that $z_{\mathcal{P}}$-ideals of $C(X)_{\mathcal{P}}$ and
$z$-ideals of $C(X)_{\mathcal{P}}$ are equivalent (Theorem 2.4). We
characterize $\mathcal{P}P$-spaces using $z$-ideals of $C(X)_{\mathcal{P}}$
(Theorem 2.8). We obtain that the nature of $z$-ideals of $C(X)_{\mathcal{P}}$
depends heavily on the choice of $\mathcal{P}$ (See Example 2.5). Furthermore,
we deduce that each ideal of $T^{\prime}(X)$ [1] is a $z$-ideal. In this
section, we also establish a topological characterization of $z^{0}$-ideals
([7]) of $C(X)_{\mathcal{P}}$ using $X_{\mathcal{P}}$. Using this
characterization, we show that all maximal ideals of the rings $C(X)_{F}$
[13], $T^{\prime}(X)$ [1] and $C(X)_{K}$ [10] are $z^{0}$-ideals. We introduce
the notion of almost $\mathcal{P}P$-spaces and characterize these spaces
algebraically using $z$-ideals and $z^{0}$-ideals of $C(X)_{\mathcal{P}}$
(Theorem 2.19). We end this section by discussing essential ideals of
$C(X)_{\mathcal{P}}$. We characterize these ideals using the topology
$X_{\mathcal{P}}$ and $\mathcal{P}P$-spaces using essential ideals of
$C(X)_{\mathcal{P}}$.

In Section 3, we discuss conditions under which the ring $C(X)_{\mathcal{P}}$
becomes a clean ring, semiclean ring, almost clean ring, weakly clean ring and
even exchange ring. The first interesting observation we make is that the
rings $C(X)_{F}$, $T^{\prime}(X)$ and $C(X)_{K}$ are clean rings provided $X$
contains at least three elements. We characterize clean elements and almost
clean elements of $C(X)_{\mathcal{P}}$ and even clean elements of
$C^{*}(X)_{\mathcal{P}}$, using the topology $X_{\mathcal{P}}$. We introduce
the notion of a $\tau_{\mathcal{P}}$-zero-dimensional space and use it to
characterize $C(X)_{\mathcal{P}}$ as a clean ring. We obtain that a Tychonoff,
extremally disconnected space is $\tau\mathcal{P}$-zero-dimensional, for any
choice of $\mathcal{P}$. We infer that $\beta\mathbb{N}$ is
$\tau\mathcal{P}$-zero-dimensional and $C(\beta\mathbb{N})_{\mathcal{P}}$ is a
clean ring, for any choice of $\mathcal{P}$. We end this section by
establishing that the notion of clean ring, weakly clean ring, semiclean ring,
almost clean ring and exchange ring, all coincide in case of the ring
$C(X)_{\mathcal{P}}$.

In the final section of this article, we study the depths of certain ideals of
$C(X)_{\mathcal{P}}$. We characterize $\mathcal{P}P$-spaces and almost
$\mathcal{P}P$-spaces using depth of ideals and essential ideals of
$C(X)_{\mathcal{P}}$ respectively (Theorem 4.5). We obtain that the depth of
each free ideal (and hence essential ideal), each prime ideal and each maximal
ideal of $C(X)_{\mathcal{P}}$ is zero whenever
$C(X)_{\mathcal{P}}\supseteq\\{\chi_{\\{{x}\\}}\colon x\in X\\}$ and as a
consequences we observe the same is true for the rings $C(X)_{F}$, $C(X)_{K}$
and $T^{\prime}(X)$. We can even say that the depth of each prime ideal of
$C(X)_{F}$ (also, $C(X)_{K}$ and $T^{\prime}(X)$) is zero. We note that $C(X)$
may have free ideals or maximal ideals having non-zero depths. At the end of
this section we point out that depth of all ideals of $C(X)_{\mathcal{P}}$ may
not be zero even if $C(X)_{\mathcal{P}}\supseteq\\{\chi_{\\{{x}\\}}\colon x\in
X\\}$.

##  2\. Ideals of $C(X)_{\mathcal{P}}$.

In this section, we deal with various types of ideals of $C(X)_{\mathcal{P}}$
which include $z$-ideal, $z^{0}$-ideal and essential ideal of
$C(X)_{\mathcal{P}}$.

Let $R$ be a commutative ring with unity. We denote $Max(R)$ and $P(R)$ to be
the set of all maximal ideals of $R$ and the set of all minimal prime ideals
of $R$ respectively. For each $a\in R$, we define $M(a)=\bigcap\\{M\in
Max(R)\colon a\in M\\}$ and $P(a)=\bigcap\\{I\in P(R)\colon a\in I\\}$. An
ideal $I$ of $R$ is said to be $z$-ideal if it satisfies the condition: $a\in
I\implies M(a)\subseteq I$ [16] and it is said to be a $z^{0}$-ideal if it
satisfies the condition: $a\in I\implies P(a)\subseteq I$ [7]. An ideal $E$ of
$R$ is said to be an essential ideal if it intersects all non zero ideals of
$R$ non-trivially. Intersection of all essential ideals of $R$ is said to be
the $Socle(R)$.

We first establish a characterization of essential ideals of an arbitrary
reduced ring $R$. (We note that $C(X)_{\mathcal{P}}$ is a reduced ring for any
$\tau\mathcal{P}$-space $(X,\tau,\mathcal{P})$.)

######  Lemma 2.1.

Let $E$ be a non-zero ideal of a reduced ring $R$. Then $E$ is an essential
ideal of $R$ if and only if $Ann(E)=\\{0\\}$, where $Ann(E)=\\{r\in R\colon
rx=0\text{ for all }x\in E\\}$.

###### Proof.

Let us assume $E$ to be an essential ideal of $R$ and let $a\in Ann(E)\cap E$.
Then $a^{2}=0\implies a=0$, as $R$ is a reduced ring. So $Ann(E)\cap
E=\\{0\\}$. Thus $Ann(E)=\\{0\\}$. Conversely, let $Ann(E)=\\{0\\}$ and $I$ be
a non-zero ideal of $R$. Also let $x\in I\setminus E$. Then there exists $r\in
E$ such that $rx\neq 0$. So $0\neq rx\in I\cap E$. Thus $E$ is an essential
ideal of $R$. ∎

###  2.1. $z$-ideals of $C(X)_{\mathcal{P}}$.

It is known that for a Tychonoff space $X$, an ideal $I$ of $C(X)$ is a
$z$-ideal if and only if for $f,g\in C(X)$, $Z(f)\subseteq Z(g)$ with $f\in I$
implies that $g\in I$ (See 4A(5) [14]). Also it is proved in [7] that an ideal
$I$ of $C(X)$ is a $z^{0}$-ideal if and only if for $f,g\in C(X)$, $int$
$Z(f)=int$ $Z(g)$ and $f\in I$ implies that $g\in I$.

######  Definition 2.2.

[10] An ideal $I$ of $C(X)_{\mathcal{P}}$ is called a $z_{\mathcal{P}}$-ideal
if $Z_{\mathcal{P}}^{-1}Z_{\mathcal{P}}[I]=I$. Equivalently, an ideal $I$ of
$C(X)_{\mathcal{P}}$ is a $z_{\mathcal{P}}$-ideal if and only if for $f,g\in
C(X)_{\mathcal{P}}$, $Z_{\mathcal{P}}(f)\subseteq Z_{\mathcal{P}}(g)$ with
$f\in I$ implies that $g\in I$.

We attempt to show that an ideal $I$ of $C(X)_{\mathcal{P}}$ is a
$z_{\mathcal{P}}$-ideal if and only if $I$ is a $z$-ideal of
$C(X)_{\mathcal{P}}$. For this, we need the following lemma.

######  Lemma 2.3.

If $f\in M(g)$, then $Z_{\mathcal{P}}(g)\subseteq Z_{\mathcal{P}}(f)$.

###### Proof.

If possible, let there exists $f\in M(g)$ such that
$Z_{\mathcal{P}}(g)\nsubseteq Z_{\mathcal{P}}(f)$. Then there exists $x\in
Z_{\mathcal{P}}(g)\setminus Z_{\mathcal{P}}(f)$. So $g\in M_{x}=\\{f\in
C(X)_{\mathcal{P}}\colon f(x)=0\\}$ ([10]) but $f\notin M_{x}$ which
contradicts that $f\in M(g)$. ∎

We now state and prove an equivalence between $z$-ideals and
$z_{\mathcal{P}}$-ideals of $C(X)_{\mathcal{P}}$.

######  Theorem 2.4.

An ideal $I$ of $C(X)_{\mathcal{P}}$ is a $z_{\mathcal{P}}$-ideal if and only
if $I$ is a $z$-ideal of $C(X)_{\mathcal{P}}$.

###### Proof.

Let $I$ be a $z$-ideal of $C(X)_{\mathcal{P}}$ and $Z_{\mathcal{P}}(f)\in
Z_{\mathcal{P}}[I]$ for some $f\in C(X)_{\mathcal{P}}$. Then there exists
$g\in I$ such that $Z_{\mathcal{P}}(f)=Z_{\mathcal{P}}(g)$. Let $M$ be an
arbitrary maximal ideal of $C(X)_{\mathcal{P}}$ containing $g$. Then $f\in
Z_{\mathcal{P}}^{-1}Z_{\mathcal{P}}[M]=M$. Since $M$ is an arbitrary maximal
ideal of $C(X)_{\mathcal{P}}$ containing $g$, $f\in M(g)$ and $M(g)\subseteq
I$ follows from our assumption.

Conversely, let $I$ be a $z_{\mathcal{P}}$-ideal of $C(X)_{\mathcal{P}}$ and
let $f\in I$. We need to show that $M(f)\subseteq I$. Let $g\in M(f)$. Then by
Lemma 2.3, $Z_{\mathcal{P}}(f)\subseteq Z_{\mathcal{P}}(g)$ and since $f\in
I$, we have $g\in I$. This completes the proof. ∎

Thus, from now onwards, we use the term $z$-ideal to describe a
$z_{\mathcal{P}}$-ideal of $C(X)_{\mathcal{P}}$.

We note that the nature of $z$-ideals of $C(X)_{\mathcal{P}}$ depends heavily
on the choice of $\mathcal{P}$ as can be seen in the following example.

######  Example 2.5.

Consider $X=\mathbb{R}$.

  1. (1)

For any ideal of closed sets $\mathcal{P}$ containing $\\{0\\}$, the ideal
generated by the identity function is a $z$-ideal of $C(X)_{\mathcal{P}}$ but
not of $C(X)$.

  2. (2)

Consider the function $f(x)=\sin\pi x$ for all $x\in X$ and the ideal
$I=<f^{2}>$. Then $I$ is not a $z$-ideal of $C(X)_{\mathcal{P}}$ for
$\mathcal{P}=\\{\emptyset\\}$, $\mathcal{P}=\mathcal{P}_{f}$ or even
$\mathcal{P}=\mathcal{K}$. But $I$ is a $z$-ideal of $C(X)_{\mathcal{P}}$ if
we take $\mathcal{P}=\mathcal{P}_{nd}$. Indeed if $I$ is a $z$-ideal of
$C(X)_{\mathcal{P}}$, then there exists $g\in C(X)_{\mathcal{P}}$ such that
$f=f^{2}g$ and so $D_{g}=\mathbb{Z}$. Thus $\mathbb{Z}\in\mathcal{P}$.

We need the following lemma to establish a characterization for
$\mathcal{P}P$-spaces. We recall that a $\tau\mathcal{P}$-space
$(X,\tau,\mathcal{P})$ is said to be a $\mathcal{P}P$ space [10] if the ring
$C(X)_{\mathcal{P}}$ is a Von-Neumann regular ring.

######  Lemma 2.6.

A function $f\in C(X)_{\mathcal{P}}$ is a Von-Neumann regular element (for a
commutative ring $R$ with unity, an element $a\in R$ is said to be a Von-
Neumann regular element if there exists $b\in R$ such that $a=a^{2}b$) if and
only if $Z_{\mathcal{P}}(f)=coz(g)$ for some $g\in C(X)_{\mathcal{P}}$.

###### Proof.

Let $f\in C(X)_{\mathcal{P}}$ be a Von-Neumann regular element. Then there
exists $g\in C(X)_{\mathcal{P}}$ such that $f=f^{2}g$ which implies that
$Z_{\mathcal{P}}(f)=coz(\boldsymbol{1}-fg)$. Conversely, let
$Z_{\mathcal{P}}(f)=coz(g)$ for some $g\in C(X)_{\mathcal{P}}$. Define
$h\colon X\longrightarrow\mathbb{R}$ by
$h(x)=\begin{cases}\dfrac{1}{f(x)}\text{ if }x\in coz(f)=Z_{\mathcal{P}}(g)\\\
0\text{ if }x\in Z_{\mathcal{P}}(f)\end{cases}.$ Then
$D_{h}\subseteq\overline{D_{f}}\cup\overline{D_{g}}\in\mathcal{P}$ which
ensures that $h\in C(X)_{\mathcal{P}}$ and $f=f^{2}h$. Thus $f$ is a Von-
Neumann regular element. ∎

The following corollary is an immediate consequence of above result.

######  Corollary 2.7.

A $\tau\mathcal{P}$-space $(X,\tau,\mathcal{P})$ is a $\mathcal{P}P$-space if
and only if $Z_{\mathcal{P}}[X]=coz_{\mathcal{P}}[X]$.

We are now able to characterize $\mathcal{P}P$-spaces as follows.

######  Theorem 2.8.

A $\tau\mathcal{P}$-space $(X,\tau,\mathcal{P})$ is a $\mathcal{P}P$-space if
and only if every ideal of $C(X)_{\mathcal{P}}$ is a $z$-ideal.

###### Proof.

Let $(X,\tau,\mathcal{P})$ be a $\mathcal{P}P$-space and $I$ an ideal of
$C(X)_{\mathcal{P}}$. Let $f\in C(X)_{\mathcal{P}}$ with
$Z_{\mathcal{P}}(f)\in Z_{\mathcal{P}}[I]$. Then there exists $g\in I$ such
that $Z_{\mathcal{P}}(f)=Z_{\mathcal{P}}(g)$. Since $f$ is Von-Neumann
regular, there exists $h\in C(X)_{\mathcal{P}}$ such that
$Z_{\mathcal{P}}(f)=coz(h)$ (by Lemma 2.6). Define
$\displaystyle{t(x)=\begin{cases}\dfrac{f(x)}{g(x)}\text{ if }x\notin
Z_{\mathcal{P}}(g)\\\ 0\text{ otherwise }\end{cases}}.$ Then $t$ is continuous
on $X\setminus(\overline{D_{f}}\cup\overline{D_{g}}\cup\overline{D_{h}})$ and
so $t\in C(X)_{\mathcal{P}}$ with $f=gt\in I$.
Conversely, let us assume that each ideal of $C(X)_{\mathcal{P}}$ is a
$z$-ideal and let $f\in C(X)_{\mathcal{P}}$ be a non-unit element. Then
$<f^{2}>$ is an ideal of $C(X)_{\mathcal{P}}$ and hence a $z$-ideal. Since
$Z_{\mathcal{P}}(f)=Z_{\mathcal{P}}(f^{2})$, it follows that $f\in<f^{2}>$.
Thus, $f=f^{2}g$ for some $g\in C(X)_{\mathcal{P}}$. This shows that
$(X,\tau,\mathcal{P})$ is a $\mathcal{P}P$-space. ∎

######  Remark 2.9.

The ring $T^{\prime}(X)$ is a Von-Neumann Regular ring (See [1]) and
$T^{\prime}(X)=C(X)_{\mathcal{P}_{nd}}$ (See [10]). This ensures that each
ideal of $T^{\prime}(X)$ is a $z$-ideal.

###  2.2. $z^{0}$-ideals of $C(X)_{\mathcal{P}}$.

In what follows, we establish a result analogous to Proposition 2.2 of [7]. We
need the following results to achieve this.

######  Lemma 2.10.

Let $f,g\in C(X)_{\mathcal{P}}$. Then
$int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f))\subseteq
int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(g))$ if and only if $Ann(f)\subseteq
Ann(g)$.

The following results are established in [8] for an arbitrary reduced ring
$R$.

######  Proposition 2.11.

Let $I$ be a proper ideal in a reduced ring $R$. Then the following statements
are equivalent.

  1. (1)

$I$ is a $z^{0}$-ideal.

  2. (2)

$P(a)=P(b)$ and $b\in I$, imply that $a\in I$.

  3. (3)

$Ann(a)=Ann(b)$ and $b\in I$, imply that $a\in I$.

  4. (4)

$a\in I$ implies that $Ann(Ann(a))\subseteq I$.

######  Lemma 2.12.

[8] Let $R$ be a reduced ring and $a\in R$. Then we have

$P(a)=\\{b\in R\colon Ann(a)\subseteq Ann(b)\\}.$

Using above results, we characterize $z^{0}$-ideals of the reduced ring
$C(X)_{\mathcal{P}}$.

######  Theorem 2.13.

If $I$ is a proper ideal of $C(X)_{\mathcal{P}}$, then the following
statements are equivalent.

  1. (1)

$I$ is a $z^{0}$-ideal.

  2. (2)

$P(f)=P(g)$ and $g\in I$, imply that $f\in I$.

  3. (3)

$Ann(f)=Ann(g)$ and $g\in I$, imply that $f\in I$.

  4. (4)

$f\in I$ implies that $Ann(Ann(f))\subseteq I$.

  5. (5)

$int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f))=int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(g))$
and $f\in I$, imply that $g\in I$.

  6. (6)

If $S\subseteq I$ is a finite set and $Ann(S)\subseteq Ann(g)$, then $g\in I$.

###### Proof.

It follows from Proposition 2.11 that statements $(1)$ to $(4)$ are
equivalent. Also by Lemma 2.10, Statements $(3)$ and $(5)$ are equivalent. We
next show that $(4)$ implies $(6)$. We assume $(4)$ is true and let
$Ann(S)\subseteq Ann(g)$, where $S\subseteq I$ is a finite set. Then
$S=\\{f_{1},f_{2},\cdots,f_{n}\\}$, where $f_{i}\in I$ for all
$i=1,2,\cdots,n$ for some $n\in\mathbb{N}$ which implies that $Ann(S)=Ann(k)$,
where $k=f_{1}^{2}+f_{2}^{2}+\cdots+f_{n}^{2}\in I$. By $(4)$,
$Ann(Ann(k))\subseteq I$. It is clear from $Ann(S)\subseteq Ann(g)$ that $g\in
Ann(Ann(k))$. So $g\in I$. This proves $(6)$. Finally we assume that $(6)$
holds and let $f\in I$ and $g\in Ann(Ann(f))$. Then $Ann(f)\subseteq Ann(g)$.
But $f\in I$ and hence by $(6)$, $g\in I$. ∎

The following definition can be found in [15].

######  Definition 2.14.

A ring $R$ is said to satisfy property A if each finitely generated ideal of
$R$ consisting of zero divisors has a non-zero annihilator.

If $I=<f_{1},f_{2},\cdots,f_{n}>$ is a finitely generated ideal of
$C(X)_{\mathcal{P}}$ consisting of zero divisors. Then it can be easily seen
that $Ann(I)=Ann(f)$, where $f=f_{1}^{2}+f_{2}^{2}+\cdots f_{n}^{2}\in I$.
This ensures that $C(X)_{\mathcal{P}}$ satisfies the property A. Thus any
ideal $I$ of $C(X)_{\mathcal{P}}$ consisting of zero divisors is contained in
a $z^{0}$-ideal (See Theorem 1.21 [8]). It follows that a maximal ideal $M$ of
$C(X)_{\mathcal{P}}$ consisting of zero divisors is a $z^{0}$-ideal. Also
since arbitrary intersection of $z^{0}$-ideals of a ring is itself a
$z^{0}$-ideal, if $I$ is an ideal of $C(X)_{\mathcal{P}}$ consisting of zero
divisors, then there is a smallest $z^{0}$-ideal containing $I$.

One must note that if $C(X)_{\mathcal{P}}$ contains all the characteristic
functions $\chi_{\\{{a}\\}}$, $a\in X$ (when $\mathcal{P}$ contains all
singleton subsets of $X$), each element of $C(X)_{\mathcal{P}}$ is either a
unit or a zero divisor. Thus, in this situation, every maximal ideal of
$C(X)_{\mathcal{P}}$ is a $z^{0}$-ideal of $C(X)_{\mathcal{P}}$. In particular
every maximal ideal of the rings $C(X)_{F}$ ([13]), $T^{\prime}(X)$ ([1] ) and
$C(X)_{K}$ ([10] ) are $z^{0}$-ideals of the respective rings.

Let $I$ be a $z^{0}$-ideal and $P$ be a prime ideal of $C(X)_{\mathcal{P}}$
which is minimal over $I$. Then we can show that $P$ is a $z^{0}$-ideal of
$C(X)_{\mathcal{P}}$. To see this, let
$int_{X_{\mathcal{P}}}Z_{\mathcal{P}}(f)=int_{X_{\mathcal{P}}}Z_{\mathcal{P}}(g)$,
where $f\in I$ and $g\in C(X)_{\mathcal{P}}$. If $f\in I$, then by Theorem
2.13 $g\in I$. If possible, let $f\in P\setminus I$ and $g\notin P$. We define
$S=(C(X)_{\mathcal{P}})\cup\\{hf^{n}\colon h\notin P,n\in\mathbb{N}\\}$. Then
$S$ is closed under multiplications and is disjoint from $I$, since if
$hf^{n}\in I$ for some $h\notin P$ and $n\in\mathbb{N}$, then $hg\in I$ which
is not possible. Therefore there exists a prime ideal $P_{1}$ of
$C(X)_{\mathcal{P}}$ containing $I$ and disjoint from $S$. It follows that
$P_{1}\subsetneq P$ which contradicts that $P$ is a prime ideal minimal over
$I$. Hence a minimal prime ideal of $C(X)_{\mathcal{P}}$ is a $z^{0}$-ideal of
$C(X)_{\mathcal{P}}$.

Moreover, it can be easily seen that if $S$ is a subset of
$C(X)_{\mathcal{P}}$ which is regular closed in $X_{\mathcal{P}}$, that is
$cl_{X_{\mathcal{P}}}(int_{X_{\mathcal{P}}}(S))=S$, then $M_{S}=\\{f\in
C(X)_{\mathcal{P}}\colon S\subseteq Z_{\mathcal{P}}(f)\\}$ is a $z^{0}$-ideal
of $C(X)_{\mathcal{P}}$. To show this, it is sufficient to show that
$S\subseteq Z_{\mathcal{P}}(g)$ when
$int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f))=int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(g))$
with $f\in I$. If not, then there exists $x\in S\cap coz_{\mathcal{P}}(g)$.
Since $coz_{\mathcal{P}}(g)$ is a neighbourhood of $x$ in $X_{\mathcal{P}}$
and $S$ is regular closed in $X_{\mathcal{P}}$, there exists $t\in
coz_{\mathcal{P}}(g)\cap int_{\mathcal{P}}S$ and $int_{\mathcal{P}}S\subseteq
int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f))=int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(g))\subseteq
Z_{\mathcal{P}}(g)$. Thus $t\in Z_{\mathcal{P}}(g)$ which is a contradiction.

Another easy yet interesting observation is that every $z^{0}$-ideal of
$C(X)_{\mathcal{P}}$ is a $z$-ideal. However the converse may not be true.
Consider $X=[0,1]\cup\\{2\\}$ endowed with the subspace topology of the real
line and $\mathcal{P}=\\{\\{2\\},\emptyset\\}$. Then $C(X)_{\mathcal{P}}=C(X)$
and $M_{0}=\\{f\in C(X)_{\mathcal{P}}\colon f(0)=0\\}$ is a $z$-ideal of
$C(X)_{\mathcal{P}}$ which is not a $z^{0}$-ideal. This makes us ask the
following question: When is every $z$-ideal of $C(X)_{\mathcal{P}}$ a
$z^{0}$-ideal? To answer this, we introduce the notion of almost
$\mathcal{P}P$-spaces.

######  Definition 2.15.

A $\tau\mathcal{P}$-space $(X,\tau,\mathcal{P})$ is said to be an almost
$\mathcal{P}P$-space if the zero set of every non-unit member of
$C(X)_{\mathcal{P}}$ has non-empty interior in $X_{\mathcal{P}}$. That is
whenever $Z_{\mathcal{P}}(f)\neq\emptyset$ for some $f\in C(X)_{\mathcal{P}}$,
$int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f))\neq\emptyset$.

It can be easily seen from the following lemma that a $\tau\mathcal{P}$-space
$(X,\tau,\mathcal{P})$ is an almost $\mathcal{P}P$-space if and only if for
every non-unit $f\in C(X)_{\mathcal{P}}$, $Z_{\mathcal{P}}(f)$ is a regular
closed set in $X_{\mathcal{P}}$ (i.e.
$cl_{X_{\mathcal{P}}}(int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f)))=Z_{\mathcal{P}}(f)$).

######  Lemma 2.16.

Every neighbourhood of a point $x\in X_{\mathcal{P}}$ contains a zero set
neighbourhood of $x$ in $X_{\mathcal{P}}$. In other words if $x\in U$, where
$U$ is open in $X_{\mathcal{P}}$, there exists $Z\in Z_{\mathcal{P}}[X]$ such
that $x\in int_{X_{\mathcal{P}}}Z\subseteq Z\subseteq U$.

###### Proof.

Let $x\in U$, where $U$ is open in $X_{\mathcal{P}}$. Then there exists a
$g_{1}\in C(X)_{\mathcal{P}}$ such that $x\in coz(g_{1})\subseteq U$ and
$g_{1}(x)=\\{\frac{3}{2}\\}$. Define
$g=(\boldsymbol{1}-g_{1})\vee\\{\boldsymbol{0}\\}\in C(X)_{\mathcal{P}}$. Then
$x\in\\{y\in X\colon g_{1}(y)>1\\}\subseteq\\{y\in X\colon g_{1}(y)\geq
1\\}=Z_{\mathcal{P}}(g)$. So $x\in
int_{X_{\mathcal{P}}}Z_{\mathcal{P}}(g)\subseteq Z_{\mathcal{P}}(g)\subseteq
U$. ∎

######  Lemma 2.17.

A function $f\in C(X)_{\mathcal{P}}\setminus\\{\boldsymbol{0}\\}$ is a regular
element (for a commutative ring $R$ with unity, an element $a\in
R\setminus\\{0\\}$ is said to be a regular element if it is not a zero
divisor) if and only if $int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f))=\emptyset$.

###### Proof.

This follows from the fact that $f\in
C(X)_{\mathcal{P}}\setminus\\{\boldsymbol{0}\\}$ is a zero divisor if and only
if $fg=\boldsymbol{0}$ for some $g\in
C(X)_{\mathcal{P}}\setminus\\{\boldsymbol{0}\\}$ if and only if $\emptyset\neq
coz(g)\subseteq Z_{\mathcal{P}}(f)$. ∎

It follows from the above lemma and the definition of almost
$\mathcal{P}P$-spaces that a $\tau_{\mathcal{P}}$-space $(X,\tau,\mathcal{P})$
is an almost $\mathcal{P}P$-space if and only if each element of
$C(X)_{\mathcal{P}}$ is either a unit or a zero-divisor. Such a ring is said
to be a classical ring or an almost regular ring.

Clearly, if $C(X)_{\mathcal{P}}$ contains all characteristic functions of the
form $\chi_{\\{{a}\\}}$ for all $a\in X$, then it is a classical ring and thus
in this case $(X,\tau,\mathcal{P})$ is an almost $\mathcal{P}P$-space. In
particular $(X,\tau,\mathcal{P}_{f})$ ($C(X)_{\mathcal{P}_{f}}=C(X)_{F}$
[13]), $(X,\tau,\mathcal{P}_{nd})$ ($C(X)_{\mathcal{P}_{nd}}=T^{\prime}(X)$
[1]) and $(X,\tau,\mathcal{K})$ ($C(X)_{\mathcal{K}}=C(X)_{K}$ [10]) are
examples of an almost $\mathcal{P}P$-spaces.

We establish another lemma which will help us to study almost
$\mathcal{P}P$-spaces.

######  Lemma 2.18.

Two sets $A$ and $B$ are $\mathcal{P}$-completely separated in $X$ if and only
if they are contained in disjoint zero sets. In fact $\mathcal{P}$-completely
separated sets in $X$ are contained in disjoint zero set neighbourhoods in
$X_{\mathcal{P}}$.

###### Proof.

This can be proved by closely following the proof of Theorem 1.15 in [14].

∎

The following theorem characterizes an almost $\mathcal{P}P$-space
algebraically.

######  Theorem 2.19.

The following statements are equivalent for a $\tau\mathcal{P}$-space
$(X,\tau,\mathcal{P})$.

  1. (1)

$X$ is an almost $\mathcal{P}P$-space.

  2. (2)

Every $z$-ideal in $C(X)_{\mathcal{P}}$ is a $z^{0}$-ideal.

  3. (3)

Every maximal ideal in $C(X)_{\mathcal{P}}$ is a $z^{0}$-ideal.

  4. (4)

Every maximal ideal in $C(X)_{\mathcal{P}}$ consists entirely of zero
divisors.

  5. (5)

The sum of any two ideals consisting of zero divisors is either
$C(X)_{\mathcal{P}}$ or consists of zero divisors.

  6. (6)

For each non-unit element $f\in C(X)_{\mathcal{P}}$, there exists a non-zero
$g\in C(X)_{\mathcal{P}}$ such that $P(f)\subseteq Ann(g)$.

###### Proof.

$(1)\implies(2)$ follows from Theorem 2.4, Theorem 2.13 and the discussion
following Definition 2.15. $(2)\implies(3)$ is evident from the fact that all
maximal ideals are $z$-ideals. It is clear from the definition of
$z^{0}$-ideals that a $z^{0}$-ideal consists entirely of zero divisors. This
proves $(3)\implies(4)$. $(4)\implies(5)$ is obvious. In order to show that
$(5)\implies(1)$, let $f\in C(X)_{\mathcal{P}}$ be a non-unit element. We need
to show that $int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f))\neq\emptyset$. If
$Z_{\mathcal{P}}(f)=X$ or $X\setminus\\{x\\}$ for some $x\in X$, then we are
done. Let $x,y\notin Z_{\mathcal{P}}(f)$ such that $x\neq y$. Then $x\notin
Z_{\mathcal{P}}(f)\cup\\{y\\}$. By Lemma 2.16 there exists $g\in
C(X)_{\mathcal{P}}$ such that $g\geq\boldsymbol{0}$ and $x\in
int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(g))\subseteq Z_{\mathcal{P}}(g)\subseteq
X\setminus(Z_{\mathcal{P}}(f)\cup\\{y\\})$. Again $y\notin
Z_{\mathcal{P}}(f)\cup Z_{\mathcal{P}}(g)$ implies that there exists $h\in
C(X)_{\mathcal{P}}$ such that $h\geq\boldsymbol{0}$ and $x\in
int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(h))\subseteq Z_{\mathcal{P}}(h)\subseteq
X\setminus(Z_{\mathcal{P}}(f)\cup Z_{\mathcal{P}}(g))$. Thus $fg$ and $fh$ are
non-zero elements in $C(X)_{\mathcal{P}}$ such that $fgh=\boldsymbol{0}$.
Therefore the ideals $<fg>$ and $<fh>$ are non-zero ideals consisting of zero
divisors. Since $f$ is not a unit, $<fg>+<fh>\neq C(X)_{\mathcal{P}}$. By
$(5)$, $fg+fh$ is a zero divisor. So
$int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f))=int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(fg+fh))\neq\emptyset$.
This proves $(1)$. To show that $(1)\implies(6)$, let $f\in
C(X)_{\mathcal{P}}$ be a non-unit element. Then by $(1)$, there exists $g\in
C(X)_{\mathcal{P}}\setminus\\{\boldsymbol{0}\\}$ such that
$fg=\boldsymbol{0}$. Thus $g\in Ann(f)$. It follows from Lemma 2.12 that
$P(f)\subseteq Ann(g)$. Finally we show that $(6)\implies(1)$. Let $f\in
C(X)_{\mathcal{P}}$ be a non-unit element. We need to show that
$int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f))\neq\emptyset$. It is enough to show
that $f$ is a zero divisor (by Lemma 2.17). By $(6)$, $P(f)\subseteq Ann(g)$
where $g$ is a non-zero element in $C(X)_{\mathcal{P}}$. This ensures that
$fg=\boldsymbol{0}$. ∎

###  2.3. Essential ideals of $C(X)_{\mathcal{P}}$.

We characterise essential ideals of $C(X)_{\mathcal{P}}$ by using Lemma 2.1
and the space $X_{\mathcal{P}}$. The following theorem works as an extension
to Theorem 3.1 of [5].

######  Theorem 2.20.

For a non-zero ideal $E$ of $C(X)_{\mathcal{P}}$, the following are
equivalent.

  1. (1)

$E$ intersects every non-zero $z$-ideal of $C(X)_{\mathcal{P}}$ non-trivially.

  2. (2)

$E$ is an essential ideal of $C(X)_{\mathcal{P}}$.

  3. (3)

$int_{X_{\mathcal{P}}}\bigcap Z_{\mathcal{P}}[E]=\emptyset$ (that is, $\bigcap
Z_{\mathcal{P}}[E]$ is nowhere dense in $X_{\mathcal{P}}$ ).

###### Proof.

(1)$\iff$(2): Let $(1)$ hold and let $I$ be a non-zero ideal of
$C(X)_{\mathcal{P}}$. Then $I_{z}=Z_{\mathcal{P}}^{-1}(Z_{\mathcal{P}}[I])$ is
a $z$-ideal of $C(X)_{\mathcal{P}}$. By (1), there exists $f\in E\cap I_{z}$.
So there exists $g\in I$ such that $Z_{\mathcal{P}}(f)=Z_{\mathcal{P}}(g)$.
Thus $0\neq fg\in I\cap E$. The converse is obvious

(2)$\iff$(3): Let $int_{X_{\mathcal{P}}}\bigcap
Z_{\mathcal{P}}[E]\neq\emptyset$. Then there exists $x_{0}\in
int_{X_{\mathcal{P}}}\bigcap Z_{\mathcal{P}}[E]$, which is open in
$X_{\mathcal{P}}$. So there exists $g\in C(X)_{\mathcal{P}}$ such that
$x_{0}\in coz(g)\subseteq int_{X_{\mathcal{P}}}\bigcap
Z_{\mathcal{P}}[E]\subseteq\bigcap Z_{\mathcal{P}}[E]$. So for every $f\in E$,
$fg=\boldsymbol{0}$ which implies that $\boldsymbol{0}\neq g\in Ann(E)$. By
Lemma 2.1, $E$ is not an essential ideal of $C(X)_{\mathcal{P}}$. Conversely,
let $int_{X_{\mathcal{P}}}\bigcap Z_{\mathcal{P}}[E]=\emptyset$ and let $I$ be
a non-zero ideal of $C(X)_{\mathcal{P}}$. Then there exists $g\in
I\setminus\\{\boldsymbol{0}\\}$. So $coz(g)\nsubseteq\bigcap
Z_{\mathcal{P}}[E]$ which implies that there exists $f\in E$ such that
$coz(g)\nsubseteq Z_{\mathcal{P}}(f)\implies fg\neq\boldsymbol{0}$. Thus
$\boldsymbol{0}\neq fg\in I\cap E$.

∎

It is obvious from the above theorem that free ideals of $C(X)_{\mathcal{P}}$
are essential ideals. Further it can be seen that for $f\in
C(X)_{\mathcal{P}}$, $<f>$ is essential if and only if $f$ is a non-zero, non-
unit, regular element. Thus, if $\mathcal{P}$ contains all singleton subsets
of $X$, $<f>$ is non-essential for any $f\in
C(X)_{\mathcal{P}}\setminus\\{\boldsymbol{0}\\}$. In particular, this is true
for the rings $C(X)_{F}$, $T^{\prime}(X)$ and $C(X)_{K}$. Further, it follows
from the fact that $C(X)_{\mathcal{P}}$ is a $pm$-ring [10] that if $P$ is a
prime ideal of $C(X)_{\mathcal{P}}$, $\bigcap Z_{\mathcal{P}}[P]$ is either a
singleton set or an empty set. (A commutative ring, $R$ with unity is a
$pm$-ring if every prime ideal of $R$ is contained in a unique maximal ideal
of $R$.) If $P$ is non-essential, then it follows from the above theorem that
$\bigcap Z_{\mathcal{P}}[P]=\\{x_{0}\\}$ for some $x_{0}\in X$ which is an
isolated point in $X_{\mathcal{P}}$. So there exists $g\in C(X)_{\mathcal{P}}$
such that $\emptyset\neq coz(g)\subseteq\\{x_{0}\\}$ which implies that
$g=g(x_{0})\chi_{\\{{x_{0}}\\}}$ and $\chi_{\\{{x_{0}}\\}}\in
C(X)_{\mathcal{P}}$. Thus $P\subseteq M_{x_{0}}=<e>$ where
$e=\boldsymbol{1}-\chi_{\\{{x_{0}}\\}}$. Since $P$ is a prime ideal,
$e(\boldsymbol{1}-e)=\boldsymbol{0}$ and $\boldsymbol{1}-e\notin M_{x_{0}}$,
$e\in P$. Thus $P=M_{x_{0}}=<e>$, which is also a minimal prime ideal. We
summarize these discussions in the following corollary.

######  Corollary 2.21.

A prime ideal of $C(X)_{\mathcal{P}}$ is either an essential ideal or a
maximal ideal which is also a minimal prime ideal.

Furthermore, if $X$ is a finite set, then $C(X)_{\mathcal{P}}$ has no free
(proper) ideals and for each $x\in X$, $\\{x\\}$ is open in $X$, and hence
open in $X_{\mathcal{P}}$. So, for any ideal $I$ of $C(X)_{\mathcal{P}}$,
$int_{X_{\mathcal{P}}}\bigcap Z_{\mathcal{P}}[I]\neq\emptyset\implies I$ is a
non-essential ideal of $C(X)_{\mathcal{P}}$.

Again, if each ideal of $C(X)_{\mathcal{P}}$ is non-essential, then $M_{x}$
(See [10]) is non-essential, for each $x\in X$ and so $\\{x\\}=\bigcap
Z_{\mathcal{P}}[M_{x}]$ is open in $X_{\mathcal{P}}$. If $X$ is an infinite
set, then $I=\\{f\in C(X)_{\mathcal{P}}\colon coz_{\mathcal{P}}(f)\text{ is
finite.}\\}$ is a free (proper) ideal of $C(X)_{\mathcal{P}}$. Indeed, if not,
then there exists $x_{0}\in\bigcap Z_{\mathcal{P}}[I]$ and since $\\{x_{0}\\}$
is open in $X_{\mathcal{P}}$, there exists $h\in C(X)_{\mathcal{P}}$ such that
$x_{0}\in coz_{\mathcal{P}}(h)\subseteq\\{x_{0}\\}$ which ensures that $h\in
I$ but $x_{0}\notin Z_{\mathcal{P}}(h)$, which is a contradiction. Hence $I$
is an essential ideal of $C(X)_{\mathcal{P}}$, which contradicts our
assumption. This establishes the following corollary.

######  Corollary 2.22.

Each ideal of $C(X)_{\mathcal{P}}$ is non-essential if and only if $X$ is a
finite set.

We provide a result stronger than Theorem 2.8 using essential ideals of
$C(X)_{\mathcal{P}}$ which can be proved by following closely the proof of
Proposition 3.7 of [5].

######  Corollary 2.23.

A $\tau\mathcal{P}$-space $(X,\tau,\mathcal{P})$ is a $\mathcal{P}P$-space if
and only if every essential ideal of $C(X)_{\mathcal{P}}$ is a $z$-ideal.

The following result, established by Azarpanah in [5], shows how rarely the
set of all essential ideals and the set of free ideals of $C(X)$ coincide.

######  Theorem 2.24.

The set of all essential ideals of $C(X)$ and the set of free ideals of $C(X)$
coincide if and only if $X$ is a discrete space.

######  Remark 2.25.

Note that $X_{\mathcal{P}}$ is the discrete space whenever
$C(X)_{\mathcal{P}}$ contains all characteristic functions of the form
$\chi_{\\{{x}\\}}$ for all $x\in X$. It follows from Theorem 2.20 that in this
case, the set of all essential ideals always coincides with the set of all
free ideals. In particular this property can be seen in rings $C(X)_{F}$
([13]) and $T^{\prime}(X)$ ([1]).

##  3\. When is $C(X)_{\mathcal{P}}$ clean?

W. K. Nicholson defined a commutative ring $R$ with unity to be clean if every
element in $R$ can be expressed as the sum of a unit and an idempotent in $R$
in [20] and an element $a\in R$ to be a clean element if it can be expressed
as the sum of a unit and an idempotent in $R$ in [21].

It is noted by Anderson and Camillo that for a commutative ring $R$ with
finitely many minimal prime ideals, $R$ is a clean ring if and only if it is a
$pm$-ring [3, Theorem 5]. In [10] we have shown that $C(X)_{\mathcal{P}}$ is a
$pm$-ring (Corollary 2.16). Further by Theorem 5.8 of [10], if $\mathcal{P}$
contains all singleton subsets of $X$, then the only minimal ideals of
$C(X)_{\mathcal{P}}$ are of the form $<\chi_{\\{{a}\\}}>$ for $a\in X$, where
$\chi_{\\{{a}\\}}(x)=\begin{cases}0\text{ when }x\neq a\\\ 1\text{ when
}x=a\end{cases}$ for all $x\in X$. But if $|X|\geq 3$ and $\mathcal{P}$
contains all singleton subsets of $X$, then $<\chi_{\\{{a}\\}}>$ is not a
prime ideal for any $a\in X$. Thus, if $|X|\geq 3$ and $\mathcal{P}$ contains
all singleton subsets of $X$, then $C(X)_{\mathcal{P}}$ has no minimal prime
ideals. Thus, under these conditions $C(X)_{\mathcal{P}}$ is a clean ring.
This is summarised in the following theorem.

######  Theorem 3.1.

Let $|X|\geq 3$ and $\mathcal{P}$ contains all singleton subsets of $X$, then
$C(X)_{\mathcal{P}}$ is a clean ring.

It follows from Theorem 3.1 that if $|X|\geq 3$, $C(X)_{F}$ is a clean ring.
Also, $T^{\prime}(X)=C(X)_{\mathcal{P}_{nd}}$ (See [10]) contains all
characteristic functions of the form $\chi_{\\{{a}\\}}$ for all $a\in X$ and
thus it is a clean ring, even though $\mathcal{P}_{nd}(=$the set of all closed
nowhere dense subsets of $X$) may not contain all singleton subsets of $X$.

######  Definitions 3.2.

A subset $U\subseteq X$ is said to be $\tau\mathcal{P}$-clopen if
$U=Z_{\mathcal{P}}(f)=X\setminus Z_{\mathcal{P}}(g)$ for some $f,g\in
C(X)_{\mathcal{P}}$. In other words, by a $\tau\mathcal{P}$-clopen set in $X$
we mean a subset of $X$ which is clopen in $X_{\mathcal{P}}$.

A $\tau\mathcal{P}$-space $X$ is said to be $\tau\mathcal{P}$-zero-dimensional
if for every pair of $\mathcal{P}$-completely separated sets $A,B\subseteq X$,
there exists a $\tau\mathcal{P}$-clopen set $U\subseteq X$ such that
$A\subseteq U\subseteq X\setminus B$.

If a subset $U$ of $X$ is such that $U=Z_{\mathcal{P}}(f)=X\setminus
Z_{\mathcal{P}}(g)$ for some $f,g\in C(X)_{\mathcal{P}}$, then
$U=Z_{\mathcal{P}}(e)=X\setminus Z_{\mathcal{P}}(\boldsymbol{1}-e)$, where
$e=\frac{|f|}{|f|+|g|}$ and $\boldsymbol{1}-e$ are idempotents in
$C(X)_{\mathcal{P}}$. This gives us the following observation.

######  observation 3.3.

A subset $U$ of $X$ is $\tau\mathcal{P}$-clopen if and only if there exists an
idempotent $e\in C(X)_{\mathcal{P}}$ such that
$U=Z_{\mathcal{P}}(e)=X\setminus Z_{\mathcal{P}}(\boldsymbol{1}-e)$.

We note that $\mathbb{R}$ with the usual topology is not
$\tau\mathcal{P}$-zero-dimensional if we take
$\mathcal{P}=\\{\emptyset,\\{0\\}\\}$.

######  Example 3.4.

Let us consider $X=\mathbb{R}$ and $\mathcal{P}=\\{\emptyset,\\{0\\}\\}$. Then
the function $\displaystyle{f(x)=\begin{cases}\sin\frac{\pi}{x}&\text{ if
}x\neq 0\\\ 0&\text{ otherwise}\end{cases}}$ is a member of
$C(X)_{\mathcal{P}}$. However since the only clopen sets in
$\mathbb{R}\setminus\\{0\\}$ are $\mathbb{R}\setminus\\{0\\}$, $\emptyset$,
$(-\infty,0)$ and $(0,\infty)$, there does not exist any idempotent $e\in
C(X\setminus\\{0\\})$ such that $A\setminus\\{0\\}\subseteq
Z(e)\subseteq(X\setminus B)\setminus\\{0\\}$. Thus there does not exist any
idempotent $e^{\prime}\in C(X)_{\mathcal{P}}$ such that $A\subseteq
Z(e)\subseteq(X\setminus B)$ which ensures that $X$ is not
$\tau\mathcal{P}$-zero-dimensional.
More over, one can also easily see that $\mathbb{R}$ is not
$\tau\mathcal{P}$-zero-dimensional if we take $\mathcal{P}=$the set of all
finite subsets of $\mathbb{R}$.

We look for some examples of $\tau\mathcal{P}$-zero dimensional spaces. We
know that a Tychonoff space is said to be strongly zero-dimensional [12] if
two disjoint zero sets can be separated by a clopen set. Clearly, strongly
zero-dimensional spaces are $\tau\mathcal{P}$-zero dimensional with
$\mathcal{P}=\\{\emptyset\\}$. We use the next observation to achieve a few
non-trivial examples (with various choices of $\mathcal{P}$).

######  observation 3.5.

A $C^{*}$-embedded subspace of a strongly zero-dimensional Tychonoff space is
also strongly zero-dimensional.

It is known that a Tychonoff, extremally disconnected space is strongly zero-
dimensional (Theorem 6.2.25 [12]). Not only this, but a Tychonoff, extremally
disconnected space is $\tau\mathcal{P}$-zero-dimensional for any choice of
$\mathcal{P}$ as is seen in the next theorem.

######  Theorem 3.6.

A Tychonoff, extremally disconnected space is $\tau\mathcal{P}$-zero-
dimensional.

###### Proof.

Let $X$ be a Tychonoff, extremally disconnected space and let $A,B\subseteq X$
be $\mathcal{P}$-completely separated. Then there exists $f\in
C^{*}(X)_{\mathcal{P}}$ such that $f(A)=\\{0\\}$ and $f(B)=\\{1\\}$. It
follows that $A\setminus\overline{D_{f}}$ and $B\setminus\overline{D_{f}}$ are
completely separated in $X\setminus\overline{D_{f}}$. Since
$X\setminus\overline{D_{f}}$ is an open subspace of $X$ and $X$ is extremally
disconnected, by Observation 3.5, there exists an idempotent $e\in
C(X\setminus\overline{D_{f}})$ such that $A\setminus\overline{D_{f}}\subseteq
Z(e)\subseteq(X\setminus\overline{D_{f}})\setminus(B\setminus\overline{D_{f}})$.
Define $e_{0}(x)=\begin{cases}e(x)\text{ when }x\in
X\setminus\overline{D_{f}},\\\ 0\text{ when }x\in A\cap\overline{D_{f}},\\\
1\text{ otherwise}\end{cases}.$ Since $X\setminus\overline{D_{f}}$ is open,
$D_{e_{0}}\subseteq\overline{D_{f}}$. Therefore $e_{0}\in
C^{*}(X)_{\mathcal{P}}$ is an idempotent and $A\subseteq
Z_{\mathcal{P}}(e_{0})\subseteq X\setminus B$. ∎

Another well-known result is that: a Tychonoff space $X$ is extremally
disconnected space if and only if $\beta X$ is extremally disconnected. Thus
we have the following corollary and example.

######  Corollary 3.7.

If $X$ is a Tychonoff, extremally disconnected space, then both $X$ and $\beta
X$ are $\tau\mathcal{P}$-zero-dimensional, for any choice of ideal,
$\mathcal{P}$ of closed sets.

######  Example 3.8.

$\beta\mathbb{N}$ is $\tau\mathcal{P}$-zero-dimensional, for any choice of
ideal, $\mathcal{P}$ of closed sets.

We now state a characterization of clean elements in $C(X)_{\mathcal{P}}$.

######  Theorem 3.9.

An element $f\in C(X)_{\mathcal{P}}$ is clean if and only if there exists a
$\tau\mathcal{P}$-clopen subset $U$ of $X$ such that $f^{-1}(\\{1\\})\subseteq
U\subseteq X\setminus Z_{\mathcal{P}}(f)$.

###### Proof.

This can be proved by following the proof of Lemma 2.1 [9]. ∎

The following corollary is immediate.

######  Corollary 3.10.

An element $f\in C(X)_{\mathcal{P}}$ is clean if and only if there exists a
$\tau\mathcal{P}$-clopen set $U\subseteq X$ such that either
$Z_{\mathcal{P}}(f)\subseteq U\subseteq X\setminus
Z_{\mathcal{P}}(\boldsymbol{1}-f)$ or
$Z_{\mathcal{P}}(\boldsymbol{1}-f)\subseteq U\subseteq X\setminus
Z_{\mathcal{P}}(f)$.

Throughout most of this paper, we shall use the above result to recognize
clean elements in $C(X)_{\mathcal{P}}$. It is obvious that every unit in
$C(X)_{\mathcal{P}}$ is clean. Also, for every idempotent $e\in
C(X)_{\mathcal{P}}$, $e=(\boldsymbol{1}-e)+(2e-\boldsymbol{1})$, where
$(\boldsymbol{1}-e)$ is an idempotent in $C(X)_{\mathcal{P}}$ and
$(2e-\boldsymbol{1})$ is a unit in $C(X)_{\mathcal{P}}$. So, every idempotent
in $C(X)_{\mathcal{P}}$ is clean. It is obvious that if for $f\in
C(X)_{\mathcal{P}}$ is such that $f^{-1}(\\{1\\})=\emptyset$, then $f$ is
clean. Thus if $|f|<\boldsymbol{1}$ or $|f|>\boldsymbol{1}$, then $f$ is
clean. Moreover, for a unit $f\in C(X)_{\mathcal{P}}$ with the property that
it is not an onto function, $rf$ is a clean element in $C(X)_{\mathcal{P}}$
for some $r\in\mathbb{R}$. Not only this, for a given clean element $f\in
C(X)_{\mathcal{P}}$, we have $|f|^{r}$ is clean for all $r>0$. Also,
$\frac{f^{2}}{f^{2}+\boldsymbol{1}}$ is clean for any $f\in
C(X)_{\mathcal{P}}$.

It is observed for $\mathcal{P}=\\{\emptyset\\}$ in Remark 2.3 of [9] that the
sum or product of two clean elements may not be clean. However we claim that
the product of an idempotent and a clean element in $C(X)_{\mathcal{P}}$ is
clean. Indeed, let $e,f\in C(X)_{\mathcal{P}}$ where $e$ is an idempotent and
$f$ is a clean element in $C(X)_{\mathcal{P}}$. So, by Theorem 3.9 and
Observation 3.3, there exists an idempotent $e^{\prime}\in C(X)_{\mathcal{P}}$
such that $f^{-1}(\\{1\\})\subseteq Z_{\mathcal{P}}(e^{\prime})\subseteq
X\setminus Z_{\mathcal{P}}(f)$. From this we have $(ef)^{-1}(\\{1\\})\subseteq
Z_{\mathcal{P}}(e_{0})\subseteq X\setminus Z_{\mathcal{P}}(ef)$, where
$e_{0}=e^{\prime}(\boldsymbol{1}-e)$ is an idempotent in $C(X)_{\mathcal{P}}$.
It follows from Theorem 3.9 and Observation 3.3 that $ef$ is a clean element
in $C(X)_{\mathcal{P}}$.

As we can see, Theorem 3.9 is a very useful tool to recognize clean elements
in $C(X)_{\mathcal{P}}$. However, the analog to this result fails in case of
$C^{*}(X)_{\mathcal{P}}$ if we take $X=(0,1)$ endowed with the subspace
topology inherited from $\mathbb{R}$ with usual topology and $f(x)=x$ for all
$x\in X$ and $\mathcal{P}=\\{\emptyset\\}$, then the only idempotents in
$C^{*}(X)_{\mathcal{P}}$ are $\boldsymbol{1}$ and $\boldsymbol{0}$. So if $f$
is the sum of a unit $u\in C^{*}(X)_{\mathcal{P}}$ and an idempotent in
$C^{*}(X)_{\mathcal{P}}$, then either $u(x)=x-1$ for all $x\in X$ or $u(x)=x$
for all $x\in X$, neither of which are units in $C^{*}(X)_{\mathcal{P}}$. Thus
$f$ is not a clean element in $C^{*}(X)_{\mathcal{P}}$ even though
$f^{-1}(\\{1\\})=\emptyset$ which is a $\tau\mathcal{P}$-clopen subset of
$X\setminus Z_{\mathcal{P}}(f)$. The necessary part of Theorem 3.9, however,
works even for clean elements in $C^{*}(X)_{\mathcal{P}}$. We have the next
sufficient condition that ensures that an element $f\in
C^{*}(X)_{\mathcal{P}}$ is clean.

######  Theorem 3.11.

Let $f\in C^{*}(X)_{\mathcal{P}}$ and define for each $a\in\mathbb{R}$,
$A_{a}=\\{x\in X\colon f(x)\geq a\\}$. If there exist $a,b\in\mathbb{R}$ with
the following properties that

  1. (1)

$0<a<b<1$, and

  2. (2)

there exists a $\tau\mathcal{P}$-clopen subset $U$ of $X$ such that
$A_{b}\subseteq U\subseteq A_{a}$,

then $f$ is a clean element of $C^{*}(X)_{\mathcal{P}}$.

###### Proof.

By Observation 3.3, there exists $e\in C^{*}(X)_{\mathcal{P}}$ such that
$U=Z_{\mathcal{P}}(e)$. We define $u(x)=f(x)-e(x)$. We show that $u$ is a unit
in $C^{*}(X)_{\mathcal{P}}$. Then for $x\in U$, $x\in A_{a}$ and so $u(x)\geq
a$. Again if $x\notin U$, then $x\notin A_{b}$ and so $u(x)<b-1$. Using the
fact that $f$ is bounded on $X$ and hence $u$ is bounded on $X$. Further
$|u(x)|\geq\max\\{a,1-b\\}$ for all $x\in X$. Thus $u$ is a unit in
$C^{*}(X)_{\mathcal{P}}$ and so $f$ is clean. ∎

F. Azarpanah established in [9] that $C(X)$ is clean if and only if $X$ is a
strongly zero-dimensional space. We attempt to achieve an analogous version of
this result in our setting.

######  Theorem 3.12.

The following are equivalent for a $\tau\mathcal{P}$-space
$(X,\tau,\mathcal{P})$.

  1. (1)

$C(X)_{\mathcal{P}}$ is a clean ring.

  2. (2)

$C^{*}(X)_{\mathcal{P}}$ is a clean ring.

  3. (3)

The set of clean elements in $C(X)_{\mathcal{P}}$ is a subring of
$C(X)_{\mathcal{P}}$.

  4. (4)

$X$ is a $\tau\mathcal{P}$-zero-dimensional space.

  5. (5)

Every zero divisor of $C(X)_{\mathcal{P}}$ is clean.

  6. (6)

$C(X)_{\mathcal{P}}$ has a clean prime ideal.

###### Proof.

(1)$\iff$ (2): Let us first assume that $C(X)_{\mathcal{P}}$ is a clean ring
and $f\in C^{*}(X)_{\mathcal{P}}$. Define $A=\\{x\in X\colon
f(x)\geq\frac{2}{3}\\}$ and $B=\\{x\in X\colon f(x)\leq\frac{1}{3}\\}$. Then
$A$ and $B$ are disjoint zero sets in $X$ and so are $\mathcal{P}$-completely
separated sets. So there exists $g\in C(X)_{\mathcal{P}}$ such that
$g(A)=\\{1\\}$ and $g(B)=\\{0\\}$. By Theorem 3.9, there exists a
$\tau\mathcal{P}$-clopen subset $U$ of $X$ such that $g^{-1}(\\{1\\})\subseteq
U\subseteq X\setminus Z_{\mathcal{P}}(g)$. Also, $A_{\frac{2}{3}}=A\subseteq
g^{-1}(\\{1\\})$ and $X\setminus Z_{\mathcal{P}}(g)\subseteq X\setminus
B\subseteq A_{\frac{1}{3}}$. By Theorem 3.11, $f$ is clean. Conversely, we
assume that $C^{*}(X)_{\mathcal{P}}$ is a clean ring. Let $f\in
C(X)_{\mathcal{P}}$ and define $g=(\boldsymbol{-1}\vee
f)\wedge\boldsymbol{1}\in C^{*}(X)_{\mathcal{P}}$. By (2) $g$ is clean. Also,
by Theorem 3.9 and using $f^{-1}(\\{1\\})\subseteq g^{-1}(\\{1\\})$ and
$Z_{\mathcal{P}}(f)=Z_{\mathcal{P}}(g)$, it is clear that $f$ is clean. This
ensures that the first two assertions are equivalent.

(1) $\implies$ (3) is obvious. Now, we assume (3) to be true and let
$A,B\subseteq X$ be two $\mathcal{P}$-completely separated subsets of $X$.
Then there exists $f\in C(X)_{\mathcal{P}}$ such that $|f|\leq\frac{1}{2}$,
$f(A)=\\{0\\}$ and $f(B)=\\{\frac{1}{2}\\}$. Since
$f^{-1}(\\{1\\})=\emptyset$, $f$ is a clean element in $C(X)_{\mathcal{P}}$.

(1)$\iff$(4): Let $(1)$ hold. Let $A$ and $B$ be two $\mathcal{P}$-completely
separated subsets of $X$. Then there exists $g\in C(X)_{\mathcal{P}}$ such
that $g(A)=\\{0\\}$ and $g(B)=\\{1\\}$. By (1), $g$ is a clean element. It
follows from Theorem 3.9 that there exists a $\tau\mathcal{P}$-clopen set $U$
such that $g^{-1}(\\{1\\})\subseteq U\subseteq X\setminus Z_{\mathcal{P}}(g)$.
But $A\subseteq Z_{\mathcal{P}}(g)$ and $B\subseteq g^{-1}(\\{1\\})$. This
ensures that $X$ is a $\tau\mathcal{P}$-zero-dimensional space. Conversely,
let $X$ be a $\tau\mathcal{P}$-zero-dimensional space and $f\in
C(X)_{\mathcal{P}}$. Then we need to show that there exists a
$\tau\mathcal{P}$-clopen set $U$ such that $f^{-1}(\\{1\\})\subseteq
U\subseteq X\setminus Z_{\mathcal{P}}$. We define $A=\\{x\in X\colon
f(x)\geq\frac{1}{2}\\}$ and $B=Z_{\mathcal{P}}(f)$. Then $A$ and $B$ are
disjoint zero sets and are thus $\mathcal{P}$-completely separated sets. The
rest follows directly since $f^{-1}(\\{1\\})\subseteq A$. This proves
(4)$\implies$(1).

Again (1)$\implies$(5) is obvious.

We next show that (5)$\implies$(6). Let every zero divisor of
$C(X)_{\mathcal{P}}$ be clean. If $P$ is a minimal prime ideal of
$C(X)_{\mathcal{P}}$. Then it follows from Lemma 2.12 that each $f\in P$ is a
zero divisor, as $P(f)\subseteq P$. Thus $P$ is a clean prime ideal. If
$C(X)_{\mathcal{P}}$ does not contain any minimal prime ideals, then it
follows from Theorem 5 in [3] that $C(X)_{\mathcal{P}}$ is a clean ring.

Finally, we need to show that (6)$\implies$(1). By (6), there exists a clean
prime ideal $P$ of $C(X)_{\mathcal{P}}$. Let $f\in C(X)_{\mathcal{P}}$. Define
$A=Z_{\mathcal{P}}(f)$ and $B=f^{-1}(\\{1\\})$. It follows from the
discussions following Corollary 3.10 that if either $A=\emptyset$ or
$B=\emptyset$, then we are done. So we assume that $A$ and $B$ are both non-
empty sets. Set two zero sets $C=\\{x\in X\colon f(x)\leq\frac{1}{3}\\}$ and
$D=\\{x\in X\colon f(x)\geq\frac{1}{3}\\}$. Since $A$ and $D$ are disjoint
zero sets, they are $\mathcal{P}$-completely separated. Analogously, $B$ and
$C$ are pairs of $\mathcal{P}$-completely separated sets. Therefore there
exists $g,h\in C(X)_{\mathcal{P}}$ such that $g(A)=\\{1\\}$, $g(D)=\\{0\\}$,
$h(B)=\\{1\\}$ and $h(C)=\\{0\\}$. Clearly $gh=\boldsymbol{0}\in P$. So either
$g\in P$ or $h\in P$ which implies that either $g$ or $h$ is clean. If $g$ is
clean, then there exists a $\tau\mathcal{P}$-clopen set $U$ such that
$g^{-1}(\\{1\\})\subseteq U\subseteq X\setminus Z_{\mathcal{P}}(g)$. But
$A\subseteq g^{-1}(\\{1\\})$ and $B\subseteq D\subseteq Z_{\mathcal{P}}(g)$.
Again if $h$ is clean, then there exists a $\tau\mathcal{P}$-clopen set $V$
such that $h^{-1}(\\{1\\})\subseteq V\subseteq X\setminus Z_{\mathcal{P}}(h)$.
But $B\subseteq g^{-1}(\\{1\\})$ and $A\subseteq C\subseteq
Z_{\mathcal{P}}(h)$. It follows from Corollary 3.10 that $f$ is clean. ∎

The following example follows directly from the above theorem and Example 3.8.

######  Example 3.13.

$C(\beta\mathbb{N})_{\mathcal{P}}$ is a clean ring for any choice of ideal
$\mathcal{P}$ of closed sets.

A ring $R$ is said to be an exchange ring if and only if for every $a\in R$
there exist $b,c\in R$ such that $c(1-a)(1-ba)=1-ba$ and $bab=b$. [19]. It is
stated in [9] that $C(X)$ is a clean ring if and only if it is an exchange
ring. It is natural to ask whether the analog to the above statement holds for
$C(X)_{\mathcal{P}}$. We answer this question in the affirmative.

######  Proposition 3.14.

$C(X)_{\mathcal{P}}$ is a clean ring if and only if it is an exchange ring.

###### Proof.

Let us assume that $C(X)_{\mathcal{P}}$ is a clean ring and let $f\in
C(X)_{\mathcal{P}}$. Then there exist $e,u\in C(X)_{\mathcal{P}}$ such that
$e$ is an idempotent and $u$ is a unit. Define $g(x)=\begin{cases}0,\text{
when }x\notin Z_{\mathcal{P}}(e)\\\ \frac{1}{f(x)},\text{ when }x\in
Z_{\mathcal{P}}(e)\end{cases}$ and $h(x)=\begin{cases}\frac{1}{1-f(x)}\text{
when }x\notin Z_{\mathcal{P}}(e)\\\ 0\text{ when }x\in
Z_{\mathcal{P}}(e)\end{cases}$. Then $g,h\in C(X)_{\mathcal{P}}$ as
$D_{g}\subseteq D_{f}$ and $D_{h}\subseteq D_{f}$. It is clear that $gfg=g$
and $h(\boldsymbol{1}-f)(\boldsymbol{1}-fg)=\boldsymbol{1}-fg$. This shows
that $C(X)_{\mathcal{P}}$ is an exchange ring.

Conversely, let $C(X)_{\mathcal{P}}$ be an exchange ring. Then there exists
$gh\in C(X)_{\mathcal{P}}$ such that $gfg=g$ and
$h(\boldsymbol{1}-f)(\boldsymbol{1}-fg)=\boldsymbol{1}-fg$. It follows that
$Z_{\mathcal{P}}(\boldsymbol{1}-fg)$ is a $\tau\mathcal{P}$-clopen set in $X$
and $f^{-1}({1})\subseteq U\subseteq X\setminus Z_{\mathcal{P}}(f)$. ∎

The following corollary follows from Theorem 3.12 and Proposition 3.14.

######  Corollary 3.15.

For a $\tau\mathcal{P}$-space $X$, $C(X)_{\mathcal{P}}$ is an exchange ring if
and only if $X$ is a $\tau\mathcal{P}$-zero-dimensional space.

Ye [22] defined an element $r$ of ring $R$ to be semiclean if $r=a+u$, where
$a$ is a periodic element, that is $a^{k}=a^{l}$ for some positive integers
$k$ and $l$ such that $k\neq l$ and $u$ is a unit in R. A ring $R$ is called a
semiclean ring if every element of $R$ is semiclean. Further, an element $a\in
R$ is said to be weakly clean if and only if $a$ can be expressed as $a=u+e$
or $a=u-e$ where $u$ is a unit and $e$ is an idempotent in $R$. A ring $R$ is
said to be a weakly clean ring if every element of $R$ is weakly clean [2].
Clearly a ring $R$ is semiclean if it is weakly clean and $R$ is weakly clean
if it is a clean ring. N. Arora and S. Kundu showed in [4] that in $C(X)$, the
concept of clean, semiclean and weakly clean are equivalent. We establish a
similar equivalence using Observation 3.3.

######  Theorem 3.16.

For a $\tau\mathcal{P}$-space $(X,\tau,\mathcal{P})$, the following statements
are equivalent.

  1. (1)

$C(X)_{\mathcal{P}}$ is clean.

  2. (2)

$C(X)_{\mathcal{P}}$ is weakly clean.

  3. (3)

$C(X)_{\mathcal{P}}$ is semiclean.

###### Proof.

$(1)\implies(2)\implies(3)$ is obvious. We just need to show that
$C(X)_{\mathcal{P}}$ is clean if it is semiclean. Let us assume that
$C(X)_{\mathcal{P}}$ is semiclean. Also let $f\in C(X)_{\mathcal{P}}$ and
$g=2f-\boldsymbol{1}$. Then there exists a unit $u\in C(X)_{\mathcal{P}}$ and
a periodic element $p\in C(X)_{\mathcal{P}}$ such that $g=u+p$. Define
$K=\\{x\in X\colon u(x)<0\\}=\\{x\in X\colon u(x)\leq 0\\}$. Then $K$ is a
$\tau\mathcal{P}$-clopen subset of $X$. By Observation 3.3, there exists an
idempotent element $e\in C(X)_{\mathcal{P}}$ such that
$K=coz(e)=Z_{\mathcal{P}}(\boldsymbol{1}-e)$. Since $p$ is a periodic elememt
in $C(X)_{\mathcal{P}}$, $p(X)\subseteq\\{0,1,-1\\}$. This ensures that
$u(x)\in\\{1,2\\}$ on $Z_{\mathcal{P}}(\boldsymbol{1}-f)$ and
$u(x)\in\\{-1,-2\\}$ on $Z_{\mathcal{P}}(f)$. Thus we have $f-e$ is a unit in
$C(X)_{\mathcal{P}}$. This proves $(1)$. ∎

McGovern introduced the notion of an almost clean element of a ring $R$ in
[17]. An element in a ring $R$ is said to be an almost clean element if it can
be expressed as the sum of a regular element (an element which is not a zero
divisor) and an idempotent. If every element of $R$ is almost clean, then $R$
is said to be an almost clean ring. He established that for a Tychonoff space
$X$, $C(X)$ is an almost clean ring if and only if it is clean. R. Mohamadian
later characterized almost clean elements in $C(X)$ (See [18]) and established
the same result ($C(X)$ is an almost clean ring if and only if it is clean)
using the characterization of almost clean elements. It is clear that if
$\mathcal{P}$ contains all singleton subsets of $X$, then each element in
$C(X)_{\mathcal{P}}$ is either a unit or a zero divisor. Hence in this case,
$C(X)_{\mathcal{P}}$ is almost clean if and only if it is clean.

######  Theorem 3.17.

An element $f\in C(X)_{\mathcal{P}}$ is an almost clean element if and only if
there exists a $\tau\mathcal{P}$-clopen subset $U$ of $X$ such that

$int_{X_{\mathcal{P}}}(f^{-1}(\\{1\\}))\subseteq U\subseteq
cl_{X_{\mathcal{P}}}(coz(f)).$

###### Proof.

Let $f\in C(X)_{\mathcal{P}}$ be an almost clean element. So there exist a
regular element $r\in C(X)_{\mathcal{P}}$ and an idempotent $e\in
C(X)_{\mathcal{P}}$ such that $f=r+e$. Thus $Z_{\mathcal{P}}(f)\cap
Z_{\mathcal{P}}(e)\subseteq Z_{\mathcal{P}}(r)$, where
$int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(r))=\emptyset$ (by Lemma 2.17). This
shows that $int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(f))\cap
Z_{\mathcal{P}}(e)=\emptyset$. Using this we get $Z_{\mathcal{P}}(e)\subseteq
cl_{X_{\mathcal{P}}}(coz(f))$. Again since $f=r+e$ and
$int_{X_{\mathcal{P}}}(Z_{\mathcal{P}}(r))=\emptyset$ (by Lemma 2.17), we have
$int_{X_{\mathcal{P}}}(f^{-1}(\\{1\\}))\cap coz(e)=\emptyset$. This shows that
$int_{X_{\mathcal{P}}}(f^{-1}(\\{1\\}))\subseteq Z_{\mathcal{P}}(e)$.

Conversely, let there exists a $\tau\mathcal{P}$-clopen subset $U$ of $X$ such
that
$int_{X_{\mathcal{P}}}(f^{-1}(\\{1\\}))\subseteq U\subseteq
cl_{X_{\mathcal{P}}}(coz(f))$. By Observation 3.3, there exists an idempotent
$e\in C(X)_{\mathcal{P}}$ such that
$U=Z_{\mathcal{P}}(e)=coz(\boldsymbol{1}-e)$. Let us suppose
$A=Z_{\mathcal{P}}(e)\cap Z_{\mathcal{P}}(f)$ and $B=f^{-1}(\\{1\\})\cap
coz(e)$. Then $A\cup B=Z_{\mathcal{P}}(h)$, where
$h=(e+|f|)(|\boldsymbol{1}-f|+(1-e))\in C(X)_{\mathcal{P}}$. Define $r\colon
X\longrightarrow\mathbb{R}$ by

$r(x)=\begin{cases}0&\text{, if }x\in Z(h|_{X\setminus\overline{D_{h}}})\\\
f(x)-e(x)&\text{, if }x\in cl_{X\setminus\overline{D_{h}}}(coz(h))\\\
0&\text{, if }x\in Z_{\mathcal{P}}(h|_{\overline{D_{h}}})\\\ f(x)-e(x)&\text{,
otherwise }\end{cases}.$

Then $r|_{X\setminus\overline{D_{h}}}\in C(X\setminus\overline{D_{h}})$. So
$D_{r}\subseteq\overline{D_{h}}$ and thus $r\in C(X)_{\mathcal{P}}$. It is to
note that $A$ and $B$ are disjoint closed sets in $X_{\mathcal{P}}$ such that
$int_{X_{\mathcal{P}}}A=\emptyset$ and $int_{X_{\mathcal{P}}}B=\emptyset$.
Thus we have $int_{X_{\mathcal{P}}}(A\cup B)=\emptyset$. But $A\cup
B=Z_{\mathcal{P}}(r)$. It follows from Lemma 2.17 that $r$ is a regular
element and $f=r+e$ is evident. ∎

We have now successfully characterized almost clean elements of the ring
$C(X)_{\mathcal{P}}$. Using this we achieve the following result.

######  Theorem 3.18.

$C(X)_{\mathcal{P}}$ is clean if and only if it is almost clean.

###### Proof.

Let us assume that $C(X)_{\mathcal{P}}$ is an almost clean ring and $f\in
C(X)_{\mathcal{P}}$. Let $A=\\{x\in X\colon f(x)\geq\frac{2}{3}\\}$ and
$B=\\{x\in X\colon f(x)\leq\frac{1}{3}\\}$. Then there exists $g\in
C(X)_{\mathcal{P}}$ such that $g(A)=\\{0\\}$ and $g(B)=\\{1\\}$. By Theorem
3.17 and Observation 3.3, there exists $e\in C(X)_{\mathcal{P}}$ such that
$int_{X_{\mathcal{P}}}(g^{-1}(\\{1\\}))\subseteq Z_{\mathcal{P}}(e)\subseteq
cl_{X_{\mathcal{P}}}(coz(g))$. This ensures that
$int_{X_{\mathcal{P}}}(f^{-1}(\\{1\\}))\subseteq f^{-1}(\\{1\\})\subseteq
f^{-1}(\frac{2}{3},\infty)\subseteq A\subseteq Z_{\mathcal{P}}(g)$. Thus we
have $int_{X_{\mathcal{P}}}(f^{-1}(\\{1\\}))\cap Z_{\mathcal{P}}(e)=\emptyset$
and $Z_{\mathcal{P}}(\boldsymbol{1}-e)\subseteq coz(f)$, that is,
$int_{X_{\mathcal{P}}}(f^{-1}(\\{1\\}))\subseteq
Z_{\mathcal{P}}(\boldsymbol{1}-e)\subseteq coz(f)$, where $\boldsymbol{1}-e$
is an idempotent. Thus $f$ is clean.

The rest is obvious. ∎

Thus in the ring $C(X)_{\mathcal{P}}$, the notions of a clean ring, a
semiclean ring, a weakly clean ring, an almost clean ring and an exchange ring
are all coincide. Furthermore, $C(X)_{\mathcal{P}}$ is any of the
aforementioned rings if and only if $(X,\tau,\mathcal{P})$ is
$\tau\mathcal{P}$-zero-dimensional.

In particular, $\boldsymbol{C(\beta\mathbb{N})_{\mathcal{P}}}$ is a clean
ring, a semiclean ring, a weakly clean ring, an almost clean ring and an
exchange ring for any choice of ideal $\mathcal{P}$ of closed sets (See
Example 3.8).

##  4\. Depth of ideals of $C(X)_{\mathcal{P}}$

Let $R$ be a commutative ring with unity and $M$ be an $R$-module. By a
$M$-regular element $a$, we mean an $a\in R$ such that $am\neq 0$ for all
$m\in M\setminus\\{0\\}$. A sequence $a_{1},a_{2},\cdots,a_{n}\in R$ is called
a $M$-regular sequence if $a_{1}$ is $M$-regular, $a_{2}$ is
$\dfrac{M}{a_{1}M}$-regular, $a_{3}$ is $\dfrac{M}{a_{1}M+a_{2}M}$-regular and
so on; and $\displaystyle{\sum_{i=1}^{n}a_{i}M\neq M}$. The length of a
maximal $M$-regular sequence is called the depth of $M$. [11]

Thus, for an ideal $I$ of $C(X)_{\mathcal{P}}$, an element $f\in
C(X)_{\mathcal{P}}$ is $I$-regular if and only if $Ann(f)\cap I=\emptyset$.
Clearly each unit in $C(X)_{\mathcal{P}}$ is $I$-regular for any ideal $I$ of
$C(X)_{\mathcal{P}}$. Also an element $f\in C(X)_{\mathcal{P}}$ is
$\dfrac{I}{gI}$-regular if and only if $fh\notin gI$ for every $h\notin I$.

In order to compute the depths of ideals of $C(X)_{\mathcal{P}}$, we first
need the following lemma.

######  Lemma 4.1.

[6] The following hold for a reduced ring $R$.

  1. (1)

If $a,b\in R$, then $a$ is $M(b)$-regular if and only if $b\in P(a)$.

  2. (2)

If $I$ is a $z$-ideal of $R$, then $a\in R$ is $I$-regular if and only if
$I\subseteq P(a)$.

The next results follow directly from Lemmas 2.12, 2.10 and 4.1.

######  Theorem 4.2.

Let $I$ be an ideal of $C(X)_{\mathcal{P}}$ and $f\in C(X)_{\mathcal{P}}$.
Then the following hold.

  1. (1)

For $g\in C(X)_{\mathcal{P}}$, $f$ is $M(g)$-regular if and only if
$int_{X_{\mathcal{P}}}Z_{\mathcal{P}}(f)\subseteq
int_{X_{\mathcal{P}}}Z_{\mathcal{P}}$ if and only if $g\in P(f)$.

  2. (2)

$f$ is $I$-regular if and only if it is $I_{z}$-regular, where
$I_{z}=Z_{\mathcal{P}}^{-1}(Z_{\mathcal{P}}[I])$ is a $z$-ideal of
$C(X)_{\mathcal{P}}$.

  3. (3)

$f$ is $I$-regular if and only if
$int_{X_{\mathcal{P}}}Z_{\mathcal{P}}(f)\subseteq\bigcap Z_{\mathcal{P}}[I]$
if and only if $I\subseteq P(f)$.

It is clear from above that for each $f\in C(X)_{\mathcal{P}}$, $f$ is
$<f>$-regular, $M(f)$-regular and $P(f)$-regular. Also, for $f\in
C(X)_{\mathcal{P}}$, the collection $\mathcal{R}_{f}=\\{I\colon I\text{ is an
ideal of }C(X)_{\mathcal{P}}\text{ such that }f\text{ is }I\text{-regular}\\}$
has a largest element, viz, $P(f)$. Further, we have the following corollary
using Theorem 4.2 and Lemma 2.12.

######  Corollary 4.3.

For any $g\in C(X)_{\mathcal{P}}$, the collection of $M(g)$-regular functions
coincide with that of $P(g)$-regular functions. That is, $f\in
C(X)_{\mathcal{P}}$ is $M(g)$-regular if and only if it is $P(g)$-regular.

A characterization of an essential ideal $I$ of $C(X)_{\mathcal{P}}$ can be
given using regular and $I$-regular elements of the ring as follows.

######  Theorem 4.4.

An ideal $I$ of $C(X)_{\mathcal{P}}$ is essential if and only if
$r_{I}(X)=r(X)$, where $r_{I}(X)$ (resp. $r(X)$) is the set of $I$-regular
(resp. regular) elements in $C(X)_{\mathcal{P}}$.

###### Proof.

It is clear that $f\in r_{I}(X)$ if and only if $I\cap
Ann(f)=\\{\boldsymbol{0}\\}$. So $I$ is an essential ideal if $Ann(f)$ is a
non-trivial ideal. Conversely, let $I$ be a non-essential ideal of
$C(X)_{\mathcal{P}}$. Then by Theorem 2.20, there exists $x_{0}\in
int_{X_{\mathcal{P}}}\bigcap Z_{\mathcal{P}}[E]$. Following the steps in the
proof of Lemma 2.16, there exists $g\in C(X)_{\mathcal{P}}$ such that
$x_{0}\in int_{X_{\mathcal{P}}}Z_{\mathcal{P}}(g)\subseteq
Z_{\mathcal{P}}(g)\subseteq int_{X_{\mathcal{P}}}\bigcap Z_{\mathcal{P}}[E]$
and $g(X\setminus int_{X_{\mathcal{P}}}\bigcap
Z_{\mathcal{P}}[E])=\\{\boldsymbol{1}\\}$. Thus it follows from Theorems
4.2(3) and Lemma 2.17 that $g\in r_{I}(X)\setminus r(X)$. ∎

The following theorem gives the depth of certain ideals of
$C(X)_{\mathcal{P}}$.

######  Theorem 4.5.

The following statements hold for a $\tau\mathcal{P}$-space
$(X,\tau,\mathcal{P})$.

  1. (1)

The depth of each ideal of $C(X)_{\mathcal{P}}$ is zero if and only if $X$ is
a $\mathcal{P}P$-space.

  2. (2)

The depth of each essential ideal of $C(X)_{\mathcal{P}}$ is zero if and only
if $X$ is an almost $\mathcal{P}P$-space.

###### Proof.

  1. (1)

Let us assume first that $X$ is a $\mathcal{P}P$-space and let $I$ be an ideal
of $C(X)_{\mathcal{P}}$. Then $I$ is a $z$-ideal (by Theorem 2.8). Also let
$f\in C(X)_{\mathcal{P}}$ be a non-unit element in $r_{I}(X)$. Then by Lemma
2.6, $Z_{\mathcal{P}}(f)=coz_{\mathcal{P}}(g)$ for some $g\in
C(X)_{\mathcal{P}}$. Define for each $i\in I$,
$h_{i}(x)=\begin{cases}\dfrac{i(x)}{f(x)}&\text{, if }x\in
Z_{\mathcal{P}}(g)\\\ 0&\text{, if }x\in Z_{\mathcal{P}}(f)\end{cases}.$ Then
$h_{i}|_{Y}\in C(Y)$, where
$Y=X\setminus(\overline{D_{f}}\cup\overline{D_{g}})$ is an open subset of $X$
and $\overline{D_{f}}\cup\overline{D_{g}}\in\mathcal{P}$. So $h_{i}\in
C(X)_{\mathcal{P}}$ and $i=h_{i}f$. Thus $I=fI$. Therefore $depth(I)=0$.
Conversely let the depth of every ideal of $C(X)_{\mathcal{P}}$ be zero and
let $f\in C(X)_{\mathcal{P}}$ be a non-unit. Clearly $f$ is $M(f)$-regular and
$depth(M(f))=0$. Thus $fM(f)=M(f)$. So there exists $g\in M(f)$ ($\implies$
$Z_{\mathcal{P}}(f)\subseteq Z_{\mathcal{P}}(g)$) such that $fg=f$ which
implies that $Z_{\mathcal{P}}(f)=coz(\boldsymbol{1}-g)$ and so by Lemma 2.6,
$f$ is Von-Neumann regular. Thus $X$ is a $\mathcal{P}P$-space.

  2. (2)

Let $I$ be an essential ideal of $C(X)_{\mathcal{P}}$ for an almost
$\mathcal{P}P$-space $X$. Then $r_{I}(X)=r(X)=U(X)$, where $U(X)$ is the
collection of all units in $C(X)_{\mathcal{P}}$. Thus there is no non-unit
$I$-regular element in $C(X)_{\mathcal{P}}$. Thus $depth(I)=0$. Conversely,
let us assume that $X$ is not an almost $\mathcal{P}P$-space. Then there
exists a non-unit zero divisor $f\in C(X)_{\mathcal{P}}$. Since
$Ann(f)=\\{\boldsymbol{0}\\}$, $I=<f>$ is essential and so $r_{I}(X)=r(X)$.
Thus $f$ is $I$-regular. If $depth(I)=0$, $fI=I$ which implies that there
exists $g\in I$ such that $f=f^{2}g$ and hence
$f(\boldsymbol{1}-fg)=\boldsymbol{0}$ which contradicts that $f$ is a non-unit
zero divisor. Thus $depth(I)\neq 0$.

∎

Note that since $(X,\tau,\mathcal{P}_{f})$ and $(X,\tau,\mathcal{P}_{nd})$ are
almost $\mathcal{P}P$-spaces, depth of each essential ideal of $C(X)_{F}$ (or
$T^{\prime}(X)$) is zero (See Remark 2.25). Furthermore, free ideals are
always essential and so depth of each free ideal of $C(X)_{F}$ (or
$T^{\prime}(X)$) is zero. In fact, if $\mathcal{P}$ is such that
$C(X)_{\mathcal{P}}$ contains the characteristic functions $\chi_{\\{{x}\\}}$
for all $x\in X$, then the depth of each free ideal (and hence essential
ideal) of $C(X)_{\mathcal{P}}$ is zero. Moreover, in this case, the depth of
each maximal ideal of $C(X)_{\mathcal{P}}$ is also zero. In fact, we get
something stronger, that is, depth of each prime ideal of $C(X)_{\mathcal{P}}$
is zero. These results are summarized as follows.

######  Theorem 4.6.

If $\mathcal{P}$ is such that $C(X)_{\mathcal{P}}$ contains the characteristic
functions $\chi_{\\{{x}\\}}$ for all $x\in X$, then

  1. (1)

the depth of each free ideal (and hence each essential ideal) of
$C(X)_{\mathcal{P}}$ is zero.

  2. (2)

the depth of each maximal ideal of $C(X)_{\mathcal{P}}$ is zero.

  3. (3)

the depth of each prime ideal of $C(X)_{\mathcal{P}}$ is zero.

###### Proof.

  1. (1)

Follows from the above discussions.

  2. (2)

Let $M$ be a maximal ideal of $C(X)_{\mathcal{P}}$. If $M$ is a free ideal,
$depth(M)=0$. For a fixed maximal ideal $M$, there exists $x_{0}\in X$ such
that $M=M_{x_{0}}$ (see [10]). Let $f\in C(X)_{\mathcal{P}}$ be $M$-regular.
Then $fg\neq\boldsymbol{0}$ for any $g\in M\setminus\\{\boldsymbol{0}\\}$.
Thus for each $x\in X\setminus\\{x_{0}\\}$,
$\chi_{\\{{x}\\}}f\neq\boldsymbol{0}$ and thus $f(x)\neq 0$. Therefore
$X\setminus\\{x_{0}\\}\subseteq coz(f)$.

    1. Case 1.

If $f(x_{0})=0$, then $f\in M$ and hence $fM=M$.

    2. Case 2.

If $f(x_{0})\neq 0$, then $f$ is a unit and so $fM=M$.

Thus there does not exist any $M$-regular sequence and hence $depth(M)=0$.

  3. (3)

Follows from Observations 4.6 and Corollary 2.21.

∎

In particular, depth of each prime ideal of $C(X)_{F}$ (and $T^{\prime}(X)$)
is zero.

######  Remark 4.7.

  1. (1)

It is important to note that the above observation may fail if
$C(X)_{\mathcal{P}}$ fails to contain the characteristic functions
$\chi_{\\{{x}\\}}$ for all $x\in X$. To see this let $X=[0,1]\cup\\{2\\}$
endowed with the subspace topology induced from the usual topology of
$\mathbb{R}$ and $\mathcal{P}=\\{\emptyset,\\{2\\}\\}$. Then $X$ contains non-
almost $P$-points (In a Tychonoff space $X$, a point $x\in X$ is said to be an
almost $P$-point if every zero set $Z\in Z[X]$ containing $x$ has non-empty
interior, i.e. $int_{X}Z\neq\emptyset$.) and $C(X)_{\mathcal{P}}=C(X)$. It
follows from Corollary 3.9 [6] that the depth of each maximal ideal of
$C(X)=C(X)_{\mathcal{P}}$ is $1$. Further, $C(X)_{\mathcal{P}}=C(X)$ contains
free maximal ideals and so there exists a free ideal (which is also essential)
of $C(X)_{\mathcal{P}}=C(X)$ having depth $1$. Indeed, there also exists a
prime ideal of $C(X)_{\mathcal{P}}=C(X)$ having depth $1$.

  2. (2)

In light of Theorem 4.6, we note that the depth of all ideals of
$C(X)_{\mathcal{P}}$ may not be zero even if
$C(X)_{\mathcal{P}}\supseteq\\{\chi_{\\{x\\}}\colon x\in X\\}$. For example,
take $X=\mathbb{R}$, the real line. Then $C(X)_{F}$ is not a Von-Neumann
Regular ring. Indeed, if $f(x)=sin$ $x$ for all $x\in X$, then there does not
exist a $g\in C(X)_{F}$ such that $f=f^{2}g$. Thus by Theorem 4.5, it follows
that there exists an ideal of $C(X)_{F}$ having non-zero depth.

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