$\delta$-ideals of p-algebras

Abstract

In p-algebras, the concepts of $\delta$-ideals and principal $\delta$-ideals are presented, and some of their respective properties are discussed. It is observed that the set $I^{\delta }(L)$ of all $\delta$-ideals of a p-algebra L is a bounded lattice, and the class $I_{p}^{\delta }(L)$ of all principal $\delta$-ideals forms a bounded sublattice of $I^{\delta }(L)$ and a Boolean algebra on its own. A characterization of a $\delta$-ideal in terms of principal $\delta$-ideals, in p-algebras, is given. Also, the concept of comaximality of $\delta$-ideals is discussed in p-algebras. After that, a number of properties of the homomorphic image of $\delta$-ideals are considered.

1 Introduction

In distributive lattices and semi-lattices, Birkhoff (1948) and Frink (1962) presented the notion of pseudo-complements. Many authors later, such as Balbes and Horn (1970); Frink (1962) and Grätzer (1971), have characterized pseudo-complements in Stone algebras. In distributive p-algebras, Sambasiva Rao (2012) introduced and characterized the notion of $\delta$-ideals which was extended by Badawy (2016a) to the concept of MS-algebras.

In this article, we continue studying $\delta$-ideals (principal $\delta$-ideals) in p-algebras and present a number of their main important properties. The contents of our work are organized as follows: in Sect. 2, we list some notions and notations that are needed for this topic. In Sect. 3, we introduce the notion of $\delta$-ideals and establish some of their properties. It is proven that the set $I^{\delta }(L)$ of all $\delta$-ideals of a p-algebra L forms a bounded lattice. In Sect. 4, we introduce the notion of principal $\delta$-ideals. It is proven that the set $I_{p}^{\delta }(L)$ of all principal $\delta$-ideals of a p-algebra L forms a bounded sublattice of $I^{\delta }(L)$ and a Boolean algebra on its own. A characterization of a $\delta$-ideal via principal $\delta$-ideals, in p-algebras, is given. Section 5 is devoted to the comaximality of $\delta$-ideals in p-algebras and some related properties. Finally, in Sect. 6, some properties of the homomorphic images and inverse homomorphic images of $\delta$-ideals are studied.

2 Preliminaries

Here are some definitions and important results that we will use for the development of the paper.

Definition 2.1

(Davey and Priestley 2002) Let L be a lattice. Then,

• (i) A nonempty subset J of L is called an ideal of L if

  • (1) $z, y\in J$ implies $z\vee y\in J,$

  • (2) $for\ a\in L,\ z\in J,\ a\le z$ imply $a\in J$. Moreover, J is called a proper ideal of L if $J\ne L$. The set of all ideals of L is denoted by I(L). Also, $(z]=\{c\in L: c\le z\}$ is called the principal ideal generated by z.

  • (ii) Dually, a nonempty subset G of L is called a filter of L if

    • (1) $z, y\in G$ implies $z\wedge y\in G$,

    • (2) $for\ a\in L,\ z\in G,\ a\ge z$ imply $a\in G$.

Moreover, G is called a proper filter of L if $G\ne L$. F(L) denotes to the set of all filters of L. Also, $[z)=\{c\in L:c\ge z\}$ is called the principal filter generated by z.

For a lattice L, $(I(L);\wedge ,\vee )$ is the lattice of all ideals of L which is called the ideal lattice of L, where $J_{1}\wedge J_{2}=J_{1}\cap J_{2}$ and

$$\begin{aligned} J_{1}\vee J_{2}= & {} \{z\in L: ~z\le j_{1}\vee j_{2}~ for~some~ j_{1}\in J_{1},j_{2}\in J_{2}\}\\{} & {} \forall \, J_{1},J_{2} \in I(L). \end{aligned}$$

Dually, $(F(L);\wedge ,\vee )$ is the filter lattice of L(the lattice of all filters of L), where $G_{1}\wedge G_{2}=G_{1}\cap G_{2}$ and

$$\begin{aligned} G_{1}\vee G_{2}= & {} \{z\in L: ~z\ge g_{1}\wedge g_{2}~ for~some~g_{1}\in \\{} & {} G_{1},g_{2}\in G_{2}\},\forall \, G_{1},G_{2} \in F(L). \end{aligned}$$

Furthermore, $(I(L);\wedge ,\vee )$ and $(F(L);\wedge ,\vee )$ are distributive (modular) lattices if and only if L is a distributive (modular) lattice.

An ideal $J_{1}$ of a lattice L is called a prime ideal if $z\wedge y\in J_{1}$ implies $z\in J_{1}$ or $y\in J_{1}$.

Definition 2.2

(Blyth 2005; Grätzer 1971) The element $a^{*}$ is the pseudo-complement of the element a of the lattice L with 0 if

$$\begin{aligned} a \wedge x=0~~iff~~ x\le a^{*}. \end{aligned}$$

A lattice L with 0 is called a pseudo-complemented lattice (or a p-algebra) if each element of L has a pseudo-complement. A modular (distributive) p-algebra L is a p-algebra, whenever L is a modular (distributive) lattice. If L satisfies the Stone identity, $c^{*}\vee c^{**}=1$ for all $c\in L$, then L is called an S-algebra. A Stone algebra L is an S-algebra, whenever L is a distributive lattice.

Theorem 2.3

(Blyth 2005; Katriňák and Mederly 1974) Let $L_{1}$ be a p-algebra. For $z_{1},y_{1}\in L_{1}$. We have

• (1) $z_{1}\le y_{1}\Rightarrow y_{1}^{*}\le z_{1}^{*} \Rightarrow z_{1}^{**}\le y_{1}^{**}$,

• (2) $z_{1}^{*}=z_{1}^{***}$,

• (3) $z_{1}\wedge (z_{1}\wedge y_{1})^{*}=z_{1}\wedge y_{1}^{*}$,

• (4) $(z_{1}\wedge y_{1})^{**}=z_{1}^{**}\wedge y_{1}^{**}$,

• (5) $(z_{1}\vee y_{1})^{*}=z_{1}^{*}\wedge y_{1}^{*}$. Also, if $L_{1}$ is an S-algebra, then

• (6) $(z_{1}\wedge y_{1})^{*}=z_{1}^{*}\vee y_{1}^{*}$,

• (7) $(z_{1}\vee y_{1})^{**}=z_{1}^{**}\vee y_{1}^{**}$.

A subset $B(L_{1})=\{c\in L_{1} : c=c^{**}\}$ of a p-algebra $L_{1}$ consists of all closed elements of $L_{1}$, and a subset $D(L_{1})=\{z\in L_{1} : z^{*}=0\}$ consists of all dense elements of $L_{1}$.

Lemma 2.4

(Blyth 2005; Haviar 1995) Let $L_{1}$ be a p-algebra. Then,

• (1) $(B(L_{1});\wedge , \,\bigtriangledown ,^{*}, 0,1)$ is a Boolean algebra, where $z_{1}\bigtriangledown z_{2}=(z_{1}\vee z_{2})^{**}=(z_{1}^{*}\wedge z_{2}^{*})^{*}, \forall \,z_{1},z_{2}\in B(L_{1})$.

• (2) $D(L_{1})$ is a filter of $L_{1}$.

A minimal prime ideal $J_{1}$ of a distributive p-algebra L is a prime ideal $J_{1}$ of L such that for $z\in J_{1}$ there exists $y\notin J_{1}$, $z\wedge y=0$.

Definition 2.5

(Grätzer 1971) Let L and $L_{1}$ be two bounded lattices. A map $h:L\longrightarrow L_{1}$ is said to be a (0, 1)-lattice homomorphism if it preserves $0,1,\wedge$ and $\vee$.

Definition 2.6

Let $h:L\longrightarrow L_{1}$ be a (0, 1)-lattice homomorphism from a bounded lattice L into a bounded lattice $L_{1}$. The Kernel of h (briefly Ker h) and Cokernel of h (briefly Coker h) are defined by

Ker $h=\{z\in L : h(z)=0\},$ and Coker $h=\{z\in L : h(z)=1\}$, respectively.

Definition 2.7

A (0, 1)-lattice homomorphism $h:B_{1}\longrightarrow B_{2}$, between Boolean algebras $B_{1}=(B_{1};\vee ,\wedge ,^{\prime },0,1)$ and $B_{2}=(B_{2};\vee ,\wedge ,^{\prime },0,1)$ is said to be a Boolean homomorphism if

$$\begin{aligned} h(z^{\prime })=h(z)^{\prime },\ \forall \,z\in B_{1}. \end{aligned}$$

For more information about ideals, filters, intervals and p-algebras, we refer the readers to Badawy (2016a, 2016b, 2017, 2018), Badawy and Atallah (2015, 2019), Badawy and Sambasiva Rao (2014), Badawy and Shum (2014), Sambasiva Rao (2012), Sambasiva Rao and Badawy (2014, 2017); Badawy and Helmy (2023); Badawy and Shum (2017).

3 $\delta$-ideals of p-algebras

Definition 3.1

For any filter G of a p-algebra L, the set $\delta (G)$ is defined as follows:

$$\begin{aligned} \delta (G)=\{z\in L:z^{*}\in G\}. \end{aligned}$$

Now, we study the properties of $\delta (G)$.

Lemma 3.2

Let L be a p-algebra. Then for any filter G of L, $\delta (G)$ is an ideal of L.

Proof

Since $0^{*}=1\in G$, then $0\in \delta (G)$. Let $a,b\in \delta (G)$. Then, $a^{*},b^{*}\in G$. Hence, $(a\vee b)^{*}=a^{*}\wedge b^{*}\in G$. Thus, $a\vee b\in \delta (G)$. Now, let $x\in L,~x\le a$ for $a\in \delta (G)$. Then, $a^{*}\in G$. Hence $x^{*}\ge a^{*}\in G$. Thus $x\in \delta (G)$. Then, we conclude that $\delta (G)$ is an ideal of L. $\square$

Lemma 3.3

For any two filters F, G of a p-algebra L, we have

• (1) (G is a proper filter of $L)\Rightarrow$ $G\cap \delta (G)=\varnothing ,$

• (2) $x\in \delta (G)\Rightarrow x^{**}\in \delta (G)$,

• (3) $x\in G\Rightarrow x^{*}\in \delta (G)$,

• (4) $G\subseteq F\Rightarrow \delta (G)\subseteq \delta (F)$,

• (5) $\delta (D(L))=\{0\}$,

• (6) $G\subseteq D(L)\Rightarrow \delta (G)=\{0\}$,

• (7) $G=L\Leftrightarrow \delta (G)=L$.

Proof

• (1) Assume that $y\in G\cap \delta (G)$. Then, $y\in G$ and $y^{*}\in G$. Thus, $0=y\wedge y^{*}\in G$ and hence $G=L$, which is a contradiction. Therefore, $G\cap \delta (G)=\phi$.

• (2) Let $y\in \delta (G)$. Since $y^{***}=y^{*}\in G$, then $y^{**}\in \delta (G)$.

• (3) Let $y\in G$. Then, $y^{**}\in G$. Thus, $y^{*}\in \delta (G)$ by definition of $\delta (G)$.

• (4) Let $F\subseteq G$ and $z\in \delta (F)$. Then, $z^{*}\in F\subseteq G$ and hence $z^{*}\in G$. Thus, $z\in \delta (G)$. Therefore, $\delta (F)\subseteq \delta (G)$.

• (5) $\delta (D(L))=\{z\in L:z^{*}\in D(L)\}=\{z\in L:z^{**}=0\}=\{z\in L:z^{*}=1\}=\{0\}$.

• (6) Let $G\subseteq D(L)$. By (4) and (5), we get $\delta (G)\subseteq \delta (D(L))=\{0\}$. Thus, $\delta (G)=\{0\}$.

• (7) Let $L=G$. Then,

  $$\begin{aligned} L=G\Leftrightarrow 0^{**}=0\in G\Leftrightarrow 1=0^{*}\in \delta (G)\Leftrightarrow \delta (G)=L. \end{aligned}$$

$\square$

Definition 3.4

Assume that L is a p-algebra. An ideal J of L is said to be a $\delta$-ideal if $J=\delta (G)$ for some filter G of L.

Lemma 3.5

In a p-algebra, every prime ideal without a dense element is a $\delta$-ideal.

Proof

Let J be a prime ideal without a dense element. Then, $(L-J)$ is a prime filter. Let $z\in J$. Clearly, $z\wedge z^{*}=0\in J$, and $z\vee z^{*}$ is a dense element. Hence, $z\vee z^{*}\notin J$. Since J is an ideal of L and $z\in J$, we get $z^{*}\notin J$ and hence $z^{*}\in L-J$. Thus, $z\in \delta (L-J)$. Therefore, $J\subseteq \delta (L-J)$.

Conversely, let $z\in \delta (L-J)$. Then, $z^{*}\in L-J$. Thus, $z^{*}\notin J$. Since $0=z\wedge z^{*}\in J$, and J is a prime ideal, we get $z\in J$. Hence, $\delta (L-J)\subseteq J$. Thus, $J=\delta (L-J)$. Therefore, J is a $\delta$-ideal. $\square$

Lemma 3.6

A proper $\delta$-ideal J of a p-algebra L contains no dense element.

Proof

Assume that J is a proper $\delta$-ideal of L and $z\in J$. Suppose that $z\in D(L)$, since $z\in J=\delta (G_{1})$ for some filter $G_{1}$ of L. Hence, $0=z^{*}\in G_{1}$, which is a contradiction. Thus, $J\cap D(L)=\phi$. $\square$

Let $I^{\delta }(L)$ denote the set of all $\delta$-ideals of L. The following example shows that $I^{\delta }(L)$ is not a sublattice of I(L).

Example 3.7

Consider a p-algebra $L=B_{4}\oplus M_{3}$ in Fig. 1, where $B_{4}=\{0<x,y<d\}$ is the four Boolean lattice and $M_{3}=\{z,a,b,c,1\}$ is the diamond lattice, and $\oplus$ stands for the ordinal sum. Let $J_{1}=\{0,x\}$, $J_{2}=\{0,y\}$ be two ideals of L, and $G_{1}=\{y,d,z,a,b,c,1\}$, $G_{2}=\{x,d,z,a,b,c,1\}$ two filters of L. Clearly, $\delta (G_{1})=J_{1}$ and $\delta (G_{2})=J_{2}$. Therefore, $J_{1}$ and $J_{2}$ are $\delta$-ideals of L. Now, we observe that $\delta (G_{1})\vee \delta (G_{2})=\{0,x,y,d\}$, which is not a $\delta$-ideal of L but $\delta (G_{1})\cap \delta (G_{2})=J_{1}\cap J_{2}=\{0\}=\delta (D(L))$ is a $\delta$-ideal of L. Consequently, $(I^{\delta }(L);\cap ,\vee )$ is not a sublattice of I(L), but $(I^{\delta }(L);\cap )$ is a $\wedge$-subsemilattice of the semi-lattice $(I(L);\cap )$.

Fig. 1

$L=B_{4} \oplus M_{5}$ is a modular p-algebra

Theorem 3.8

For a p-algebra L, $(I^{\delta }(L);\cap ,\sqcup )$ forms a bounded lattice, where

$~~~~~~~~~~~~~~~~\delta (G_{1})\cap \delta (G_{2})=\delta (G_{1}\cap G_{2})~and~\delta (G_{1})\sqcup \delta (G_{2})=\delta (G_{1}\vee G_{2}).$

Proof

Let $G_{1}$ and $G_{2}$ be two filters of a p-algebra L. We prove that the infimum and supremum of both $\delta (G_{1})$ and $\delta (G_{2})$ in $I^{\delta }(L)$ are $\delta (G_{1}\cap G_{2})$ and $\delta (G_{1}\vee G_{2})$, respectively, that is,

$$\begin{aligned} \delta (G_{1})\cap \delta (G_{2})= & {} \delta (G_{1}\cap G_{2})~~and~~\delta (G_{1})\sqcup \delta (G_{2})\\ {}= & {} \delta (G_{1}\vee G_{2}). \end{aligned}$$

Since $G_{1}\cap G_{2}\subseteq G_{1},G_{2}$, $\delta (G_{1}\cap G_{2})\subseteq \delta (G_{1}),\delta (G_{2})$. Thus, $\delta (G_{1}\cap G_{2})$ is a lower bound of $\delta (G_{1})$ and $\delta (G_{2})$. Assume $\delta (H)$ is another lower bound of $\delta (G_{1})$ and $\delta (G_{2})$. Then, $\delta (H)\subseteq \delta (G_{1}),\delta (G_{2})$. Let $z\in \delta (H)$. Then, $z\in \delta (G_{1}),\delta (G_{2})$ and hence $z^{*}\in G_{1}\cap G_{2}$. Thus, $z\in \delta (G_{1}\cap G_{2})$. So $\delta (H)\subseteq \delta (G_{1}\cap G_{2})$ and hence $\delta (G_{1}\cap G_{2})$ is the greatest lower bound of both $\delta (G_{1})$ and $\delta (G_{2})$ in $I^{\delta }(L)$.

Clearly, $\delta (G_{1}\vee G_{2})$ is an ideal. Now, we prove that $\delta (G_{1}\vee G_{2})$ is the least upper bound of $\delta (G_{1})$ and $\delta (G_{2})$ in $I^{\delta }(L)$. Since $G_{1},G_{2}\subseteq G_{1}\vee G_{2}$, then $\delta (G_{1}),\delta (G_{2})\subseteq \delta (G_{1}\vee G_{2})$. Thus, $\delta (G_{1}\vee G_{2})$ is an upper bound of $\delta (G_{1})$ and $\delta (G_{2})$. Let $\delta (H)$ be another upper bound of $\delta (G_{1})$ and $\delta (G_{2})$. Then, $\delta (G_{1}),\delta (G_{2})\subseteq \delta (H)$. Let $z\in \delta (G_{1}\vee G_{2})$. Then, $z^{*}\in G_{1}\vee G_{2}$ and hence $z^{*}\ge g_{1}\wedge g_{2}$ for some $g_{1}\in G_{1}$ and $g_{2}\in G_{2}$. This implies $g_{1}^{*}\in \delta (G_{1})$ and $g_{2}^{*}\in \delta (G_{2})$. Since $\delta (G_{1}),\delta (G_{2})\subseteq \delta (H)$, then $g_{1}^{*}, g_{2}^{*}\in \delta (H)$. Now,

$$\begin{aligned}&g_{1}^{*}\in \delta (H), g_{2}^{*}\in \delta (H)\Rightarrow g_{1}^{*}\vee g_{2}^{*}\in \delta (H)\\&\quad \Rightarrow (g_{1}^{*}\vee g_{2}^{*})^{**}\in \delta (H)\\&\quad \Rightarrow (g_{1}^{**}\wedge g_{2}^{**})^{*}\in \delta (H)\\&\quad \Rightarrow z^{**}\le (g_{1}\wedge g_{2})^{*}=(g_{1}\wedge g_{2})^{***}=(g_{1}^{**}\wedge g_{2}^{**})^{*}\in \delta (H)\\&\quad \Rightarrow z^{**}\in \delta (H)\\&\quad \Rightarrow z\in \delta (H). \end{aligned}$$

Thus, $\delta (G_{1}\vee G_{2})\subseteq \delta (H)$. Therefore, $\delta (G_{1}\vee G_{2})$ is the least upper bound of $\delta (G_{1})$ and $\delta (G_{2})$ in $I^{\delta }(L)$. Then, sup $\{\delta (G_{1}),\delta (G_{2})\}=\delta (G_{1}\vee G_{2})$. It is clear that $\delta (L)=L$ and $\delta ([1))=\{0\}=(0]$ are the greatest and smallest members of $I^{\delta }(L)$, respectively. Thus, $(I^{\delta }(L);\cap ,\sqcup )$ is a bounded lattice. $\square$

4 Principal $\delta$-ideals of a p-algebra

In this section, we introduce and investigate the basic properties of principal $\delta$-ideals of a p-algebra L.

Lemma 4.1

Let L be a p-algebra. Then for each $z\in L,~(z^{*}]$ is a $\delta$-ideal of L.

Proof

We prove that $(z^{*}]=\delta ([z))$. To this end, let $a\in (z^{*}]$. Then, $a\le z^{*}$ and hence $a^{*}\ge z^{**}\ge z\in [z)$. So $a\in \delta ([z))$. Therefore, $(z^{*}]\subseteq \delta ([z))$. On the other hand, let $a\in \delta ([z))$. Then, $a^{*}\in [z)$. Thus, $a^{*}\ge z$. Since $a\le a^{**}\le z^{*}\in (z^{*}]$, we get $a\in (z^{*}]$. Thus, $\delta ([z))\subseteq (z^{*}]$. Therefore, $(z^{*}]=\delta ([z))$. $\square$

Definition 4.2

A $\delta$-ideal of the form $\delta ([z))=(z^{*}]$ for $z\in L$ is called a principal $\delta$-ideal of L.

Theorem 4.3

For a p-algebra L, we have the following statements:

• (1) $\delta ([z))=\delta ([z^{**}))$, $\forall$ $z\in L$.

• (2) $\delta ([d))=\{0\}$, $\forall$ $d\in D(L)$.

• (3) if $z\le y$, then $\delta ([y))\subseteq \delta ([z))$, but the converse is not hold.

• (4) for a filter G of L , $\delta ([z))\subseteq \delta (G)$ for all $z\in G$.

Proof

• (1) From Lemma 4.1, $\delta ([z))=(z^{*}]=(z^{***}]=\delta ([z^{**}))$.

• (2) for all $d\in D(L)$, $\delta ([d))=(d^{*}]=(0]=\{0\}$.

• (3) Let $z\le y$. Then, $[y)\subseteq [z)$. By (4) of Lemma 3.3, we get $\delta ([y))\subseteq \delta ([z))$. For the converse, consider the p-algebra in Fig.  2. Clearly, $\{0\}=\delta ([c))\subseteq \delta ([e))=\{0,a\}$, but $c\nleq e$ and $e\nleq c$.

• (4) For all $z\in G$, suppose that $a\in \delta ([z))$. Now,

  $$\begin{aligned} a\in \delta ([z))&\Rightarrow a^{*}\in [z)\\&\Rightarrow a^{*}\ge z\in G\\&\Rightarrow a^{*}\in G\\&\Rightarrow a\in \delta (G). \end{aligned}$$

  Thus, $\delta ([z))\subseteq \delta (G)$.

$\square$

Fig. 2

L is a stone algebra

Theorem 4.4

Let L be a p-algebra. Then, the following are equivalent:

• (1) L is an S-algebra.

• (2) For any $z_{1},z_{2}\in L,~(z_{1}\wedge z_{2})^{*}=z_{1}^{*}\vee z_{2}^{*}$.

• (3) For any two filters $G_{1},G_{2}$ of L, $\delta (G_{1})\vee \delta (G_{2})=\delta (G_{1}\vee G_{2})$.

• (4) $I^{\delta }(L)$ is a bounded sublattice of I(L).

Proof

$(1)\Rightarrow (2)$: It is clear from (6) of Theorem 2.3. $(2)\Rightarrow (3)$: Assume (2) holds and $G_{1},G_{2}\in L$. Since $G_{1},G_{2}\subseteq G_{1}\vee G_{2}$, then $\delta (G_{1}),\delta (G_{2})\subseteq \delta (G_{1}\vee G_{2})$. Thus, $\delta (G_{1})\vee \delta (G_{2})\subseteq \delta (G_{1}\vee G_{2})$.

Conversely, let $z\in \delta (G_{1}\vee G_{2})$. Then, $z^{*}\in G_{1}\vee G_{2}$ and hence $z^{*}\ge g_{1}\wedge g_{2}$ for some $g_{1}\in G_{1}$ and $g_{2}\in G_{2}$ implies $g_{1}^{*}\in \delta (G_{1})$ and $g_{2}^{*}\in \delta (G_{2})$. Now,

$$\begin{aligned}&z^{*}\ge g_{1}\wedge g_{2}\Rightarrow z^{**}\le (g_{1}\wedge g_{2})^{*}=g_{1}^{*}\vee g_{2}^{*}\in \delta (G_{1})\vee \delta (G_{2})\\&\quad \Rightarrow z^{**}\in \delta (G_{1})\vee \delta (G_{2})\\&\quad \Rightarrow z\in \delta (G_{1})\vee \delta (G_{2})\qquad \qquad \qquad (\text {as} \,\,z\le z^{**}) \end{aligned}$$

Thus, $\delta (G_{1}\vee G_{2})\subseteq \delta (G_{1})\vee \delta (G_{2})$. Therefore, $\delta (G_{1})\vee \delta (G_{2})=\delta (G_{1}\vee G_{2})$.

$(3)\Rightarrow (4)$: Assume (3) holds. It is clear that $\delta (L)$ and $\delta ([1))$ are the greatest and smallest members of $I^{\delta }(L)$, respectively. Let $\delta (G_{1}),~\delta (G_{2})\in I^{\delta }(L)$. Then, $\delta (G_{1})\vee \delta (G_{2})=\delta (G_{1}\vee G_{2})\in I^{\delta }(L)$ and $\delta (G_{1})\cap \delta (G_{2})=\delta (G_{1}\cap G_{2})\in I^{\delta }(L)$. Thus, $I^{\delta }(L)$ is a bounded sublattice of I(L).

$(4)\Rightarrow (1)$: Assume (4). Since $(z^{*}]$ and $(z^{**}]$ are $\delta$-ideals of L, we have

$$\begin{aligned}&(z^{*}\vee z^{**}]=(z^{*}]\vee (z^{**}]\\&\quad =\delta ([z))\vee \delta ([z^{*}))\\&\quad =\delta ([z)\vee [z^{*}))\qquad \qquad \qquad (\text {by}\,\, (3))\\&\quad =\delta ([z\wedge z^{*}))\\&\quad =\delta ([0))\\&\quad =(0^{*}]\\&\quad =(1]. \end{aligned}$$

Thus, $z^{*}\vee z^{**}=1$. Therefore, L is an S-algebra. $\square$

Now, we characterize the concept of $\delta$-ideals in terms of principal $\delta$-ideals.

Theorem 4.5

Let J be a $\delta$-ideal of a p-algebra L. Then,

• (1) $j\in J\Leftrightarrow j^{**}\in J$.

• (2) $\delta ([j^{*}))\subseteq J, \forall j\in J$.

• (3) $J=\bigcup _{j\in J}\delta ([j^{*}))$.

Proof

• (1) Let $j \in J$. Since J is a $\delta$-ideal of L, then $j \in J=\delta (G)$ for some filter G of L. Hence, $j^{***}=j^{*}\in G$. Thus, $j^{**}\in \delta (G)=J$. The converse implication follows from the fact that $j\le j^{**}$.

• (2) Let j be an element of J and let $x\in \delta ([j^{*}))$. Then, $x^{*}\in [j^{*})$ and hence $x\le x^{**}\le j^{**}\in J($ by (1)). Thus, $x\in J$. Therefore, $\delta ([j^{*}))\subseteq J$.

• (3) Since J is a $\delta$-ideal, we get $J=\delta (G)$ for some filter G of L. Let $z\in J$. Then,

  $$\begin{aligned}&z\in J=\delta (G)\Rightarrow z^{*}\in G\\&\quad \Rightarrow z^{*}\in [z^{*})\subseteq G\\&\quad \Rightarrow z^{**}\in \delta ([z^{*}))\subseteq \delta (G)\\&\quad \Rightarrow z\le z^{**}\in \delta ([z^{*}))\\&\quad \Rightarrow z\in \delta ([z^{*}))\subseteq \bigcup _{j\in J}\delta ([j^{*}))\qquad \qquad \qquad (\text {as}\,\, z\in J) \end{aligned}$$

  Then, $J\subseteq \bigcup _{j\in J}\delta ([j^{*}))$.

Conversely, let $z\in \bigcup _{j\in J}\delta ([j^{*}))$. Then,

$$\begin{aligned}&z\in \bigcup _{j\in J}\delta ([j^{*}))\Rightarrow z\in \delta ([k^{*})),~k\in J\\&\quad \Rightarrow z\in \delta ([k^{*}))=(k^{**}]\subseteq J\qquad \qquad \qquad (\text {as}\,\, k^{**} \in J)\\&\quad \Rightarrow z\in J. \end{aligned}$$

Thus, $\bigcup _{j\in J}\delta ([j^{*}))\subseteq J$. Therefore, $J=\bigcup _{j\in J}\delta ([j^{*})).$ $\square$

Now, a characterization of $\delta$-ideals is given.

Theorem 4.6

Let L be a p-algebra. Then, for an ideal J of L, the following statements are equivalent:

• (1) J is a $\delta$-ideal.

• (2) For any $a,b\in L$, $\delta ([a^{*}))=\delta ([b^{*}))$ and $a\in J$ imply $b\in J$.

• (3) $J=\bigcup _{c\in J}\delta ([c^{*}))$.

Proof

$(1)\Rightarrow (2)$: Let J be a $\delta$-ideal of L. Suppose that $\delta ([a^{*}))=\delta ([b^{*}))$ and $a\in J$. Then, $a\in J=\delta (G)$ for some filter G of L. Hence, $a^{*}\in G$. $(a^{**}]=\delta ([a^{*}))=\delta ([b^{*}))=(b^{**}]$ implies $a^{**}=b^{**}$. Then, $b^{*}=a^{*}\in G$. Therefore, $b\in \delta (G)=J$.

$(2)\Rightarrow (3)$: Assume (2) holds and $a\in J$. Since $a^{**}\in (a^{**}]$, we get $a\le a^{**}\in (a^{**}]=\delta ([a^{*}))$. Thus, $a\in \delta ([a^{*}))\subseteq \bigcup _{c\in J}\delta ([c^{*}))~($ as $a\in J~)$. Therefore, $J\subseteq \bigcup _{c\in J}\delta ([c^{*}))$.

Conversely, let $a\in \bigcup _{c\in J}\delta ([c^{*}))$. Then, $a\in \delta ([b^{*}))$ for some $b\in J$. Since $a\in \delta ([b^{*}))=\delta ([b^{***}))$ and $b\in J$, then $b^{**}\in J$ by (2). Since $a\in \delta ([b^{*}))$, we get $a^{*}\in [b^{*})$ and hence $a^{*}\ge b^{*}$ which implies $a\le a^{**}\le b^{**}\in J$. Thus, $a\in J$. Therefore, $\bigcup _{c\in J}\delta ([c^{*}))\subseteq J$. So $J=\bigcup _{c\in J}\delta ([c^{*})).$

$(3)\Rightarrow (1)$: At first, we need to prove that $\bigcup _{c\in J}[c^{*})$ is a filter of L. Since $1=0^{*}\in [0^{*})\subseteq \bigcup _{c\in J}[c^{*})$, we get $1\in \bigcup _{c\in J}[c^{*})$.Let $a,b\in \bigcup _{c\in J}[c^{*})$. Then, $a\in [x^{*}),b\in [y^{*})$ for some $x,y\in J$ and hence $a\wedge b\ge x^{*}\wedge y^{*}=(x\vee y)^{*}$. Thus, $a\wedge b\in [(x\vee y)^{*})\subseteq \bigcup _{c\in J}[c^{*})$ ( as $x\vee y\in J~)$. Now, let $z\ge b\in \bigcup _{c\in J}[c^{*})$. Then, $z\ge b\in [x^{*})$ for some $x\in J$. Thus, $z\in [x^{*})\subseteq \bigcup _{c\in J}[c^{*})$. Therefore, $\bigcup _{c\in J}[c^{*})$ is a filter of L.

Secondly, we prove that $\delta (\bigcup _{c\in J}[c^{*}))=\bigcup _{c\in J}\delta ([c^{*}))$. Since $[c^{*})\subseteq \bigcup _{c\in J}[c^{*})$, then $\delta ([c^{*}))\subseteq \delta (\bigcup _{c\in J}[c^{*}))$. Thus, $\bigcup _{c\in J}\delta ([c^{*}))\subseteq \delta (\bigcup _{c\in J}[c^{*}))$.

Conversely, let $a\in \delta (\bigcup _{c\in J}[c^{*}))$. Then,

$$\begin{aligned}&a\in \delta (\bigcup _{c\in J}[c^{*}))\Rightarrow a^{*}\in \bigcup _{c\in J}[c^{*})\\&\quad \Rightarrow a^{*}\in [x^{*}) \qquad \qquad \qquad (\text {for some} \,\, x\in J)\\&\quad \Rightarrow a\in \delta ([x^{*}))\subseteq \bigcup _{c\in J}\delta ([c^{*}))\qquad \qquad \qquad (\text {as} \,\, x\in J) \end{aligned}$$

Then, $\delta (\bigcup _{c\in J}[c^{*}))\subseteq \bigcup _{c\in J}\delta ([c^{*}))$. Thus, $\delta (\bigcup _{c\in J}[c^{*}))=\bigcup _{c\in J}\delta ([c^{*}))$. Now, we prove $(3)\Rightarrow (1)$. Assume that $J=\bigcup _{c\in J}\delta ([c^{*}))$. Then, $J=\bigcup _{c\in J}\delta ([c^{*}))=\delta (\bigcup _{c\in J}[c^{*}))$. Hence, J is a $\delta$-ideal of L. $\square$

Let $I_{p}^{\delta }(L)=\{(z^{*}]:z\in L\}=\{\delta ([z)):z\in L\}$ be the set of all principal $\delta$-ideals of L.

Theorem 4.7

Let L be a p-algebra. Then,

• (1) $(I_{p}^{\delta }(L);\wedge ,\sqcup ,(0],L)$ is a bounded sublattice of $I^{\delta }(L)$.

• (2) $I_{p}^{\delta }(L)$ is a Boolean algebra.

• (3) $I_{p}^{\delta }(L)$ is a homomorphic image of L.

• (4) B(L) is isomorphic of $I_{p}^{\delta }(L)$.

Proof

• (1) Let $(x^{*}],(y^{*}]\in I_{p}^{\delta }(L)$. Then,

  $$\begin{aligned} (x^{*}]\wedge (y^{*}]=(x^{*}\wedge y^{*}]=((x\vee y)^{*}]\in I_{p}^{\delta }(L) \end{aligned}$$

  and

  $$\begin{aligned} (x^{*}]\sqcup (y^{*}]= & {} \delta ([x))\sqcup \delta ([y))=\delta ([x)\vee [y))\\ {}= & {} \delta ([x\wedge y))=((x\wedge y)^{*}]\in I_{p}^{\delta }(L). \end{aligned}$$

  We observe that L, $(0]\in I_{p}^{\delta }(L)$ which are the greatest and least elements of $I_{p}^{\delta }(L)$, respectively. Thus, $I_{p}^{\delta }(L)$ is a bounded sublattice of $I^{\delta }(L)$.

• (2) Let $(x^{*}],(y^{*}],(z^{*}]\in I_{p}^{\delta }(L)$. We observe that

  $$\begin{aligned} (x^{*}]\sqcup (y^{*}]&=\delta ([x))\sqcup \delta ([y))\\&=\delta ([x)\vee [y))\\&=\delta ([x\wedge y))\\&=((x\wedge y)^{*}]\\&=((x\wedge y)^{***}] \\ ~~~~~~~~~~&=((x^{**}\wedge y^{**})^{*}]\\&=(x^{*}\triangledown y^{*}]. \end{aligned}$$

  Now, we prove that $I^{\delta }(L)$ is a distributive lattice.

  $$\begin{aligned}&(x^{*}]\cap ((y^{*}]\sqcup (z^{*}])=(x^{*}]\cap (y^{*}\triangledown z^{*}]\\&\quad =(x^{*}\wedge (y^{*}\triangledown z^{*})]\\&\quad =((x^{*}\wedge y^{*})\triangledown (x^{*}\wedge z^{*})] \qquad \qquad \qquad (\text {as} \,\, x^{*},y^{*},z^{*}\in B(L))\\&\quad =((x\vee y)^{*}\triangledown (x\vee z)^{*}]\\&\quad =((x\vee y)^{*}]\sqcup ((x\vee z)^{*}]\\&\quad =(x^{*}\wedge y^{*}]\sqcup (x^{*}\wedge z^{*}]\\&\quad =((x^{*}]\cap (y^{*}])\sqcup ((x^{*}]\cap (z^{*}]). \end{aligned}$$

  Thus, $I_{p}^{\delta }(L)$ is a bounded distributive sublattice of $I^{\delta }(L)$. We have $(z^{*}]\wedge (z^{**}]=(z^{*}\wedge z^{**}]=(0]$ and $(z^{*}]\sqcup (z^{**}]=\delta ([z))\sqcup \delta ([z^{*})) =\delta ([z)\vee [z^{*}))=\delta ([z\wedge z^{*}))=\delta ([0))=L$. Thus, $(z^{**}]$ is the complement of $(z^{*}]$, and we can write $((z^{*}])^{\prime }=(z^{**}]$. Hence, $(I_{p}^{\delta }(L);\wedge ,\sqcup ,^{\prime },(0],L)$ is a Boolean algebra.

• (3) Define $\alpha :L\longrightarrow I_{p}^{\delta }(L)$ by $\alpha (z)=(z^{**}]$. Clearly, $\alpha$ is a well-defined map, and $\alpha (0)=(0]$, $\alpha (1)=L$. Let $z,y\in L$. Then,

  $$\begin{aligned} \alpha (z\wedge y)&=((z\wedge y)^{**}]\\&=(z^{**}\wedge y^{**}]\\&=(z^{**}]\wedge (y^{**}]\\&=\alpha (z)\wedge \alpha (y) \end{aligned}$$

  and

  $$\begin{aligned} \alpha (z\vee y)&=((z\vee y)^{**}]\\&=((z^{*}\wedge y^{*})^{*}]\\&=\delta ([z^{*}\wedge y^{*}))\\&=\delta ([z^{*})\vee [y^{*}))\\&=\delta ([z^{*}))\sqcup \delta ( [y^{*}))\\&=(z^{**}]\sqcup (y^{**}]\\&=\alpha (z)\sqcup \alpha (y). \end{aligned}$$

  Now, $\alpha (z^{*})=(z^{***}]=((z^{**}])^{\prime }=(\alpha (z))^{\prime }$. Thus, $\alpha$ is a homomorphism of L into $I_{p}^{\delta }(L)$. Now, for every $(z^{**}]\in I_{p}^{\delta }(L)$, there exists $z\in L$ such that $\alpha (z)=(z^{**}]$. Thus, $\alpha$ is an onto map. Moreover, $\alpha$ is not a one-to-one map, because of $\delta ([a))=\delta ([b))$ implies $a^{*}=b^{*}$ and $a\ne b$.

• (4) Define $h:B(L)\longrightarrow I_{p}^{\delta }(L)$ by $h(z)=(z]$. Clearly, h is a well-defined map. Let $z,y\in B(L)$. Then,

  $$\begin{aligned} h(z\wedge y)&=(z\wedge y]\\&=(z]\wedge ( y]\\&=h(z)\wedge h(y) \end{aligned}$$

  and

  $$\begin{aligned}&h(z\triangledown y)=(z\triangledown y]\\&\quad =((z\vee y)^{**}]\\&\quad =((z^{*}\wedge y^{*})^{*}]\\&\quad =\delta ([z^{*}\wedge y^{*}))\\&\quad =\delta ([z^{*})\vee [y^{*}))\\&\quad =\delta ([z^{*}))\sqcup \delta ( [y^{*}))\\&\quad =(z^{**}]\sqcup (y^{**}]\\&\quad =(z]\sqcup (y]\qquad \qquad \qquad (\text {as} \,\,z,y\in B(L))\\&\quad =h(z)\sqcup h(y). \end{aligned}$$

We have $h(0)=(0],~h(1)=L$, and $h(z^{*})=(z^{*}]=(z^{***}]=((z^{**}])^{\prime }=((z])^{\prime }=(h(z))^{\prime }$. For every $(z^{**}]\in I_{p}^{\delta }(L)$, there exists $a\in B(L)$ such that $a=z^{**}$. Then, $h(a)=(a]=(z^{**}]=(z]$. Also, let $h(z)=h(y)$. Then, $(z]=(y]$ and hence $z=y$. Thus, h is one to one. Therefore, h is an isomorphism from B(L) into $I_{p}^{\delta }(L)$. $\square$

Corollary 4.8

Assume that L is an S-algebra. Then, $(I_{p}^{\delta }(L);\wedge ,\vee ,(0],L)$ is a bounded sublattice of I(L).

5 Comaximality of $\delta$-ideals

This section is devoted to introducing the notion of comaximality of $\delta$-ideals and studying some related properties.

Let us recall that two ideals $J_{1}$ and $J_{2}$ of a p-algebra L are called comaximal if $J_{1}\vee J_{2}=L.$

Now, we introduce the $\sqcup$-comaximality of $\delta$-ideals of a p-algebra L.

Definition 5.1

Two $\delta$-ideals $J_{1}$ and $J_{2}$ of a p-algebra L are called $\sqcup$-comaximal if $J_{1}\sqcup J_{2}=L.$

Lemma 5.2

Any two comaximal $\delta$-ideals of a p-algebra L are $\sqcup$-comaximal.

Proof

Let $J_{1}$ and $J_{2}$ be two comaximal $\delta$-ideals of a p-algebra L. Then, $J_{1}\vee J_{2}=L$. Since $J_{1}=\delta (G_{1})$ and $J_{2}=\delta (G_{2})$ for some filters $G_{1}$ and $G_{2}$ of L, we get $L=\delta (G_{1})\vee \delta (G_{2})$. Now,

$$\begin{aligned} L=\delta (G_{1})\vee \delta (G_{2})\subseteq \delta (G_{1}\vee G_{2})=\delta (G_{1})\sqcup \delta (G_{2})=J_{1}\sqcup J_{2}. \end{aligned}$$

Thus, $J_{1}$ and $J_{2}$ are $\sqcup$-comaximal. $\square$

The following example shows that the converse of the above lemma is not true.

Example 5.3

Consider a p-algebra $L=B_{4}\oplus N_{5}$ in Fig.  3, where $B_{4}=\{0<a,b<d\}$ is the four Boolean lattice and $N_{5}=\{c,x,y,z,1\}$ is the Pentagon lattice. We observe that (a] and (b] are $\sqcup$-comaximal $\delta$-ideals, but they are not comaximal as $(a]\sqcup (b]=(b^{*}]\sqcup (a^{*}]=\delta ([b))\sqcup \delta ([a))=\delta ([b)\vee [a))=\delta ([b\wedge a))=\delta ([0))=L$, and $(a]\vee (b]=(a\vee b]=(d]\ne L$.

Fig. 3

$L=B_{4} \oplus N_{5}$ is a non-modular p-algebra

The converse of the above lemma holds in the following special case:

Corollary 5.4

Any two $\sqcup$-comaximal $\delta$-ideals of an S-algebra L are comaximal.

Lemma 5.5

Let L be a p-algebra. If $z,y\in L$ such that $z\wedge y=0$, then $\delta ([z))$ and $\delta ([y))$ are $\sqcup$-comaximal in L.

Proof

Let $z,y\in L$ with $z\wedge y=0$. Then,

$$\begin{aligned} \delta ([z))\sqcup \delta ([y))=\delta ([z)\vee [y))=\delta ([z\wedge y))=\delta ([0))=L. \end{aligned}$$

Therefore, $\delta ([z))$ and $\delta ([y))$ are $\sqcup$-comaximal in L. $\square$

Theorem 5.6

Let L be a distributive p-algebra. Then,

• (1) Every prime $\delta$-ideal of L is a minimal prime ideal.

• (2) Any two distinct prime $\delta$-ideals of L are $\sqcup$-comaximal.

Proof

• (1) Let $J_{1}$ be a prime $\delta$-ideal of L. Then, $J_{1}=\delta (G_{1})$ for some filter $G_{1}$ of L, let $z\in J_{1}=\delta (G_{1})$. Then, $z^{*}\in G_{1}$. We have $0=z\wedge z^{*}\in J_{1}$. Suppose that $z^{*}\in J_{1}$. Then, $z^{*}\in G_{1}\cap \delta (G_{1})\ne \phi$ which contradicts with (1) of Lemma 3.3. Hence, $z^{*}\notin J_{1}$, that is, for $z\in J_{1}$, there exists $y=z^{*}\notin J_{1}$ such that $z\wedge y=0$. Thus, $J_{1}$ is a minimal prime ideal.

• (2) Let $J_{1}$ and $J_{2}$ be two distinct prime $\delta$-ideals of L. Then by (1), $J_{1}$ and $J_{2}$ are minimal prime ideals. Let $a\in J_{1}-J_{2}$ and $b\in J_{2}-J_{1}$. Since $J_{1}$ and $J_{2}$ are two minimal prime ideals, then there exist $x\notin J_{1}$ and $y\notin J_{2}$ such that $a\wedge x=0=b\wedge y$. Since $x\notin J_{1}$ and $b\notin J_{1}$, then $b\wedge x\notin J_{1}($ as $J_{1}$ is a prime ideal ), similarly $a\wedge y\notin J_{2}$. By definition of pseudo-complement and the fact that $J_{1}$ is a prime ideal, we get $(b\wedge x)^{*}\in J_{1}$. Thus, $\delta ([b\wedge x))=((b\wedge x)^{*}]\subseteq J_{1}$. Similarly, $\delta ([a\wedge y))\subseteq J_{2}$. Now, $(b\wedge x)\wedge (a\wedge y)=(a\wedge x)\wedge (b\wedge y)=0\wedge 0=0$. Then by Lemma 5.5, we get $\delta ([a\wedge y))$ and $\delta ([b\wedge x))$ are $\sqcup$-comaximal. Hence,

  $$\begin{aligned} L=\delta ([a\wedge y))\sqcup \delta ([b\wedge x))\subseteq J_{1}\sqcup J_{2}. \end{aligned}$$

  Thus, $J_{1}\sqcup J_{2}=L$. Therefore, $J_{1}$ and $J_{2}$ are $\sqcup$-comaximal.

$\square$

Let J be an ideal of a p-algebra L. For any $z\in L$, consider

$$\begin{aligned} \delta _{J}([z))=\{j\in J: j^{*}\in [z)\}. \end{aligned}$$

Lemma 5.7

Let J be an ideal of p-algebra L. Then,

• (1) $\delta _{J}([z))=J\cap \delta ([z))$ is an ideal of J.

• (2) $\delta _{J}([z))$ is a $\delta$-ideal of L, whenever J is a $\delta$-ideal of L.

Proof

• (1) It is clear that $\delta _{J}([z))=\{j\in J: j\in \delta ([z))\}=J\cap \delta ([z))$, and hence, $\delta _{J}([z))$ is an ideal of J.

• (2) Since $\delta _{J}([z))=J\cap \delta ([z))$ and J, $\delta ([z))$ are two $\delta$-ideals of L, then $\delta _{J}([z))$ is a $\delta$-ideal of L.

$\square$

Theorem 5.8

• (1) Let J be a principal $\delta$-ideal of a p-algebra L. Then for any $z,y\in L$ with $z\wedge y=0,~\delta _{J}([z))$ and $\delta _{J}([y))$ are $\sqcup$-comaximal in J.

• (2) Let J be a $\delta$-ideal of a distributive p-algebra L. Then for any $z,y\in L$ with $z\wedge y=0,~\delta _{J}([z))$ and $\delta _{J}([y))$ are $\sqcup$-comaximal in J.

Proof

• (1) Let $J=\delta ([a))$ be a principal $\delta$-ideal of a p-algebra L and $z,y\in L$ with $z\wedge y=0$. Then by Lemma 5.5, we have $\delta ([z))\sqcup \delta ([y))=L$. Now,

  $$\begin{aligned}&\delta _{J}([z))\sqcup \delta _{J}([y))=(J\cap \delta ([z)))\sqcup (J\cap \delta ([y)))\\&\quad =(\delta ([a))\cap \delta ([z)))\sqcup (\delta ([a))\cap \delta ([y)))\qquad \qquad \qquad (\text {as} \,\,J=\delta ([a)))\\&\quad =((a^{*}]\cap (z^{*}])\sqcup ((a^{*}]\cap (y^{*}])\\&\quad =(a^{*}]\cap ((z^{*}]\sqcup (y^{*}])\qquad \qquad \qquad (\text {as} \,\, I_{p}^{\delta }(L) \,\,\text {is distributive })\\&\quad =\delta ([a))\cap (\delta ([z))\sqcup \delta ([y)))\\&\quad =\delta ([a))\cap L\qquad \qquad \qquad (\text {as} \,\,\delta ([z))\sqcup \delta ([y))=L)\\&\quad =J\cap L\\&\quad =~~J \end{aligned}$$

  Therefore, $\delta _{J}([z))$ and $\delta _{J}([y))$ are $\sqcup$-comaximal in J.

• (2) Using a similar way of (1), one can prove that $\delta _{J}([z))$ and $\delta _{J}([y))$ are $\sqcup$-comaximal in J.

$\square$

6 Homomorphic images of $\delta$-ideals

This section discusses the properties of images and the inverse images of $\delta$-ideals ( principal $\delta$-ideals ) with respect to a homomorphism of two p-algebras. By a homomorphism on a p-algebra L, we mean a lattice homomorphism h satisfying $(h(x))^{*}= h(x^{*})$ for all $x\in L$.

Theorem 6.1

Assume that $h:L_{1}\longrightarrow L_{2}$ is an onto homomorphism of a p-algebra $L_{1}$ to a p-algebra $L_{2}$. Then,

• (1) The image of a principal $\delta$-ideal is a principal $\delta$-ideal, that is, for any $z\in L_{1},~h(\delta [z))=\delta ([h(z)))$.

• (2) for any filter $G_{1}$ of $L_{1},~h(\delta (G_{1}))=\delta (h(G_{1}))$.

• (3) for any $\delta$-ideal $J_{1}$ of $L_{1}$, $h(J_{1})$ is a $\delta$-ideal of $L_{2}$.

• (4) for any $\delta$-ideal J of $L_{1}$, $h(J)=\bigcup _{j\in J}\delta ([(h(j))^{*}))$.

Proof

• (1) For all $z\in L_{1}$, we get

  $$\begin{aligned} h(\delta ([z)))=h((z^{*}])&=h\{c\in L_{1}:c\le z^{*}\}\\&=\{h(c)\in L_{2}:h(c)\le h(z^{*})\}\\&=\{h(c)\in L_{2}:h(c)\le h(z)^{*}\}\\&=(h(z)^{*}]=\delta ([h(z))). \end{aligned}$$

• (2) For any filter $G_{1}$ of $L_{1}$, let $z\in \delta (h(G_{1}))$. Then,

  $$\begin{aligned}&z\in \delta (h(G_{1}))\Rightarrow z^{*}\in h(G_{1})\\&\quad \Rightarrow z^{*}=h(j_{1}) ~for~j_{1}\in G_{1}\\&\quad \Rightarrow z\le z^{**}=(h(j_{1}))^{*}=h(j_{1}^{*})\in h(\delta (G_{1}))\\&\quad \Rightarrow z\in h(\delta (G_{1}))\qquad \qquad \qquad (\text {as} \,\,h(\delta (G_{1})) \text {is an ideal})\\&\quad \Rightarrow \delta (h(G_{1}))\subseteq h(\delta (G_{1})). \end{aligned}$$

  Conversely, let $z\in h(\delta (G_{1}))$. Then,

  $$\begin{aligned} z\in h(\delta (G_{1}))&\Rightarrow z=h(y)~for~y \in \delta (G_{1})\\&\Rightarrow y^{*}\in G_{1}\\&\Rightarrow z^{*}=(h(y))^{*}=h(y^{*})\in h(G_{1})\\&\Rightarrow z\in \delta (h(G_{1}))\\&\Rightarrow h(\delta (G_{1}))\subseteq \delta (h(G_{1})) \end{aligned}$$

  Thus, $h(\delta (G_{1}))=\delta (h(G_{1}))$.

• (3) Let $J_{1}$ be a $\delta$-ideal of $L_{1}$. Then, $J_{1}=\delta (G_{1})$ for some filter $G_{1}$ of $L_{1}$. Now,

  $$\begin{aligned} h(J_{1})&=h(\delta (G_{1}))\\&=h\{z\in L_{1}:z^{*}\in G_{1}\}\\&=\{h(z)\in L_{2}:h(z^{*})\in h(G_{1})\}\\&=\{h(z)\in L_{2}:h(z)^{*}\in h(G_{1})\}\\&=\delta (h(G_{1})). \end{aligned}$$

  Thus, $h(J_{1})$ is a $\delta$-ideal of $L_{2}$.

• (4) For any $\delta$-ideal J of $L_{1}$, $J=\bigcup _{j\in J}\delta ([j^{*}))$ from (3) Theorem 4.6. Let $z\in h(J)$. Then, $z=h(j)$ for some $j\in J$. Then, $z\le z^{**}\in (z^{**}]=\delta ([z^{*}))=\delta ([(h(j))^{*}))\subseteq \bigcup _{j\in J}\delta ([(h(j))^{*}))$. Thus, $h(J)\subseteq \bigcup _{j\in J}\delta ([(h(j))^{*}))$. Conversely, let $z\in \bigcup _{j\in J}\delta ([(h(j))^{*}))$. Now,

  $$\begin{aligned}&z\in \bigcup _{j\in J}\delta ([(h(j))^{*}))\Rightarrow z\in \delta ([(h(c))^{*})),~c\in J\\&\quad \Rightarrow z\in ((h(c))^{**}]\\&\quad \Rightarrow z\le ((h(c))^{**}=h(c^{**})\in h(J) \qquad \qquad \qquad (\text {as} \,\,c^{**}\in J)\\&\quad \Rightarrow z\in h(J) \qquad \qquad \qquad (\text {as} \,\,h(J) \text {is an ideal}) \end{aligned}$$

  Thus, $\bigcup _{j\in J}\delta ([(h(j))^{*}))\subseteq h(J)$. Therefore, $h(J)=\bigcup _{j\in J}\delta ([(h(j))^{*}))$.

$\square$

Theorem 6.2

Let $h:L_{1}\rightarrow L_{2}$ be a homomorphism of a p-algebra $L_{1}$ into a p-algebra $L_{2}$. Then,

• (1) Ker h is a $\delta$-ideal of $L_{1}$.

• (2) For any $\delta$-ideal K of $L_{2},h^{-1}(K)$ is a $\delta$-ideal of $L_{1}$ containing Ker h.

Proof

• (1) Since h is a homomorphism of a p-algebra $L_{1}$ into a p-algebra $L_{2}$, then Ker $h=\{z\in L_{1}:h(z)=0\,\}$ and Coker $h=\{z\in L_{1}:h(z)=1\,\}$ are an ideal and a filter of $L_{1}$, respectively. We show that Ker $h=\delta ($Coker h). Let $z\in$ Ker h. Then,

  $$\begin{aligned} z\in ~Ker~h&\Leftrightarrow h(z)=0\\&\Leftrightarrow (h(z))^{*}=h(z^{*})=1\\&\Leftrightarrow z^{*}\in Coker~h\\&\Leftrightarrow z\in \delta (Coker~h) \end{aligned}$$

  Thus, Ker $h=\delta ($Coker h).

• (2) Let K be a $\delta$-ideal of $L_{2}$. Then, $K=\delta (G_{1})$ for some filter $G_{1}$ of $L_{2}$. Since $h^{-1}(K)$ is an ideal of $L_{1}$. We prove that $h^{-1}(K)=\delta (h^{-1}(G_{1}))$. So let $z\in h^{-1}(K)$. Then,

  $$\begin{aligned}&z\in h^{-1}(K)\Leftrightarrow h(z)=c,c\in K=\delta (G_{1})\\&\quad \Leftrightarrow h(z)^{*}=h(z^{*})=c^{*}\in G_{1} \quad \quad \quad (\text {as} \,\,c\in \delta (G_{1})\Rightarrow c^{*}\in G_{1})\\&\quad \Leftrightarrow z^{*}\in h^{-1}(\{c^{*}\})\subseteq h^{-1}(G_{1})\\&\quad \Leftrightarrow z\in \delta (h^{-1}(G_{1})) \end{aligned}$$

  Thus, $h^{-1}(K)=\delta (h^{-1}(G_{1}))$. Therefore, $h^{-1}(K)$ is a $\delta$-ideal of $L_{1}$. Let $z\in Ker~h$. Then, $h(z)=0\in K=\delta (G_{1})$, and hence, $h(z^{*})=h(z)^{*}=1\in G_{1}$ implies $z^{*}\in h^{-1}(G_{1})$. Thus, $z\in \delta (h^{-1}(G_{1}))$. Therefore, Ker $h\subseteq \delta (h^{-1}(G_{1}))$.

$\square$

Theorem 6.3

Let $h:L_{1}\longrightarrow L_{2}$ be an onto homomorphism of a p-algebra $L_{1}$ to a p-algebra $L_{2}$. Then,

• (1) $\alpha :I_{p}^{\delta }(L_{1})\longrightarrow I_{p}^{\delta }(L_{2}),\ \alpha (\delta ([z)))=\delta ([h(z))),\,\forall \,\delta ([z))\in I_{p}^{\delta }(L_{1})$ is a Boolean homomorphism.

• (2) $\alpha :I^{\delta }(L_{1})\longrightarrow I^{\delta }(L_{2}),\ \alpha (\delta (G))=\delta (h(G)),\,\forall \,\delta (G)\in I^{\delta }(L_{1})$ is a homomorphism.

Proof

• (1) Define $\alpha :I_{p}^{\delta }(L_{1})\longrightarrow I_{p}^{\delta }(L_{2})$ by $\alpha (\delta ([z)))=\delta ([h(z)))$. Clearly, $\alpha$ is well-defined. $\alpha (L_{1})=\alpha (\delta ([0_{L_{1}})))=\delta ([h(0_{L_{1}}))) =\delta ([0_{L_{2}}))=\delta (L_{2})=L_{2}$ and $\alpha \{0_{L_{1}}\}=\alpha (\delta ([1_{L_{1}})))=\delta ([h(1_{L_{1}}))) =\delta ([1_{L_{2}}))=\{0_{L_{2}}\}$. Let $\delta ([a)), \delta ([y))\in I_{p}^{\delta }(L_{1})$. Then,

  $$\begin{aligned} \alpha (\delta ([z))\wedge \delta ([y)))&=\alpha (\delta ([z)\wedge [y)))\\&=\alpha (\delta ([z\vee y)))\\&=\delta ([h(z\vee y))\\&=\delta ([h(z)\vee h(y)))\\&=\delta ([h(z))\wedge [h(y)))\\&=\delta ([h(z)))\wedge \delta ([h(y)))\\&=\alpha (\delta ([z)))\wedge \alpha (\delta ([y))) \end{aligned}$$

  and

  $$\begin{aligned} \alpha (\delta ([z))\sqcup \delta ([y)))&=\alpha (\delta ([z)\vee [y)))\\&=\alpha (\delta ([z\wedge y)))\\&=\delta ([h(z\wedge y))\\&=\delta ([h(z)\wedge h(y)))\\&=\delta ([h(z))\vee [h(y)))\\&=\delta ([h(z)))\sqcup \delta ([h(y)))\\&=\alpha (\delta ([z)))\sqcup \alpha (\delta ([y))). \end{aligned}$$

  Since $I_{p}^{\delta }(L_{1})$ and $I_{p}^{\delta }(L_{2})$ are Boolean algebras, we get $(\alpha (\delta ([z)))^{*}=(\alpha ((z^{*}]))^{*}=(\alpha ((z^{*}]))^{\prime }$. Thus,

  $$\begin{aligned} \alpha ((\delta ([z)))^{*})=&\alpha (((z^{*}])^{*})\\&=\alpha ((z^{**}])\\&=\alpha (\delta ([z^{*})))\\&=\delta ([h(z^{*}))\\&=\delta ([(h(z))^{*}))\\&=((h(z))^{**}]\\&=(((h(z))^{*}])^{*}\\&=(\delta ([h(z))))^{*}\\&=(\alpha (\delta ([z))))^{*}. \end{aligned}$$

  Therefore, $\alpha$ is a Boolean homomorphism of $I_{p}^{\delta }(L_{1})$ into $I_{p}^{\delta }(L_{2})$.

• (2) Define $\alpha :I^{\delta }(L_{1})\longrightarrow I^{\delta }(L_{2})$ by $\alpha (J_{1})=\delta (h(G))$ where $J_{1}=\delta (G)$. Clearly, $\alpha$ is well-defined and $\alpha \{0_{L_{1}}\}=\{0_{L_{2}}\}$, $\alpha (L_{1})=L_{2}$. Let $J_{1},J_{2}\in I^{\delta }(L_{1})$. Then, $J_{1}=\delta (G_{1})$ and $J_{2}=\delta (G_{2})$ for some filters $G_{1}$ and $G_{2}$ of $L_{1}$. Then, we get

  $$\begin{aligned} \alpha (J_{1}\wedge J_{2})&=\alpha (\delta (G_{1})\wedge \delta (G_{2}))\\&=\alpha (\delta (G_{1}\wedge G_{2}))\\&=\delta (h(G_{1}\wedge G_{2}))\\&=\delta (h(G_{1}))\wedge \delta (h(G_{2}))\\&=\alpha (J_{1})\wedge \alpha (J_{2}) \end{aligned}$$

  and

  $$\begin{aligned} \alpha (J_{1}\sqcup J_{2})&=\alpha (\delta (G_{1})\sqcup \delta (G_{2}))\\&=\alpha (\delta (G_{1}\vee G_{2}))\\&=\delta (h(G_{1}\vee G_{2}))\\&=\delta (h(G_{1}))\sqcup \delta (h(G_{2}))\\&=\alpha (J_{1})\sqcup \alpha (J_{2}). \end{aligned}$$

  Therefore, $\alpha$ is a homomorphism of $I^{\delta }(L_{1})$ into $I^{\delta }(L_{2})$.

$\square$