Note on $p$-ideals set of orthomodular lattices

1.   Introduction

The concept of orthomodular lattices (referred to as OML for convenience) originated from von Neumann algebraic theory. In 1936, Birkhoff and von Neumann proposed using lattices of closed subspaces in Hilbert spaces as the fundamental mathematical model for studying quantum mechanics. However, the projection lattice of a Hilbert space is modular only if the Hilbert space is finite dimensional. However, the projection lattice of type Ⅱ von Neumann algebras related to continuous geometries—especially those favoured by John von Neumann—is modular. Therefore, if the projection lattice is required to be a modular lattice, the case of infinite-dimensional Hilbert spaces is excluded. As a remedy, Husimi proposed to study the following orthomodular law instead:

this law, indeed, holds in arbitrary Hilbert spaces.

Finch ^[1] shows that a congruence relation in an OML $L$ is completely determined by its kernel and that an ideal $I$ of $L$ is called a $p$-ideal or orthomodular ideal. Let $I_P(L)$ denote the set of all $p$-ideals of an orthomoular lattice. Then, $I_p(L)$ is a complete distributive lattice. In the literature ^[2], it is raised: when are $p$-ideals of an irreducible orthomodular lattice well ordered under set inclusion? In the paper, we exhibit three big classes of orthomodular lattices where the answer is positive.

During the past decade, $L$-algebras occurred in various contexts, including algebraic logic, combinatorial group theory, operator algebras, and number theory (see, e.g., ^[3,4,5,6]). Wu and Yang provide a succinct description of orthomodular lattices ^[7], which is called $OM$-$L$-algebra. The ideals of an $OM$-$L$-algebra $L$ are equivalent to $p$-filters of orthomodular $L$. The paper provides a sufficient and necessary condition for the $p$-ideals set of orthomodular lattices to be a chain under set inclusion from the perspective of $L$-algebra.

The paper is organized as follows: In Section 2, we recall some basic definitions and facts related to OML $L$ and we exhibit three big classes of OMLs where its $p$-ideals are set to be a chain under the set inclusion relation. In Section 3, we focus on the $OM$-$L$-algebras. A sufficient and necessary condition for the $L$-ideals set of $OM$-$L$-algebra to be a chain is given.

2.   $P$-Ideals set of orthomodular lattices

At the beginning of this section, we introduce our basic terminology and present some results.

Definition 2.1. ^[2] An ortholattice is a structure (L, ≤, ', 0, 1), where $(L, \leq, ', 0, 1)$ is a lattice with maximum 1 and minimum 0, and $'$ is a unary operation on $L$ such that the following conditions are satisfied:}

An orthomodular lattice (OML) is an ortholattice $(L, \leq, ', 0, 1)$, that satisfies the following condition for $\forall \; a, b\in L$:

In an ortholattice $L$, we write $aCb$ (in words, $a$ commutes with $b$) if $a = (a\wedge b)\vee(a\wedge b'), a, b\in L$. Let $L$ be an OML. It follows that $aCb$, $aCb'$, $a'Cb$ and $a'Cb'$ are equivalent. In particular, $aCb$ is characterized by $a = (a\vee b)\wedge (a\vee b')$. Let $L$ be a lattice. For $x, y, z\in L$, we call $\{ x, y, z \}$ is a distributive triple, if the elements of a triple $\{ x, y, z \}$ can generate a distributive sublattice. In any orthomodular lattice $L$, let $aCb$ and $aCc$. Then $\{a, b, c\}$ generate a distributive sublattice of $L$ ^[8]. If $M$ is a subset of an orthomodular lattice $L$, the set $C(M) = \{x\in L\mid xCb, \forall b\in M\}$ is called the commutant of $M$ in $L$. The elements of $C(L)$ are called central and $C(L)$ the centre of $L$. For every subset $M$ of $L$, the set $C(M)$ is a subalgebra of $L$ containing arbitrary joins and meets, provided they exist in $L$. In an OML $L$, $x\in L$, then $x$ has a unique complement if and only if $x \in C(L)$ ^[2].

Definition 2.2. Let $a, b$ be elements of an orthomodular lattice $L$ ^[2].

(ⅰ). $a, b\in L$, $a$ is said to be perspective to $b$, $a\thicksim b$, if $a$ and $b$ have a common complement, i.e., if there exists $c$ in $L$ such that $a\vee c = 1 = b\vee c$, $a\wedge c = 0 = b\wedge c$ hold; elements $a, b$ are said to be strongly perspective, $a \sim_{s} b$, if $a$ and $b$ have a common complement $c$ in the sublattice $[0, a \vee b]$, the element $c$ then is called a relative complement of $a$ and $b$; it is easy to see that strong perspectivity implies perspectivity. Note, by ^[2] (the parallelogram law for $\sim$), we have $\forall a, b \in L$, $a\wedge(a'\vee b)\sim b \wedge (b' \vee a)$.

(ⅱ). $x$ is projective to $y$, denoted $x\approx y$, if there exists a sequence of elements $x_{1}, ..., x_{n}$ such that $x_{1} = x$, $x_{n} = y$, and $x_{i} \sim x_{i+1}$ hold for $1\leq i \leq n-1$.

(ⅲ). A $p$-ideal in $L$ is an ideal $I$ closed under perspectivity, i.e., $a\in I$ and $a\thicksim b$ imply $b\in I$. Dually, a $p$-filter in $L$ is a filter $F$ closed under perspectivity, i.e., $a\in F$ and $a\thicksim b$ imply $b\in F$.

Let $L$ be an OML; $a, b, c\in L$. ^[2] presents: If $c \leq a$ and $a\sim b$, then there exists $d\in L$ such that $d\leq b$ and $c \sim d$ holds. Dual, we have the following result:

Lemma 2.1. Let $L$ be an orthomodular lattice. Assume that $a, b, c \in L$ satisfies $c \sim a$ and $a \leq b$. Then, there exists $d$ such that $c \leq d$ and $d \sim b$ holds.

Proof. Let $y$ be the common complement of $c$ and $a$, satisfy $y \vee c = y \vee a = 1, y \wedge c = y \wedge a = 0$. Define $z = y\wedge (b' \vee y') \leq y$, $d = c \vee (c' \wedge z')\geq c$. Then, $z$ is the common complement of $d$ and $b$, i.e., $d \vee z = b \vee z = 1, d \wedge z = b \wedge z = 0$. In fact, $d \vee z = (c \vee (c' \wedge z')) \vee z = (c \vee z)\vee (c' \wedge z') = (c \vee z)\vee (c \vee z)' = 1$. Since $b', y' \leq b' \vee y'$, $\{ b, y, b'\vee y' \}$ is a distributive triple. We have $b \vee z = b \vee (y \wedge (b' \vee y'))$ = $(b \vee y) \wedge (b' \vee y' \vee b)$ = $b \vee y \geq a \vee y = 1$, so $b \vee z = 1$. Since $c' \wedge z'\leq c', z'$, $\{c, z, c'\wedge z' \}$ is a distributive triple. We have $d \wedge z = (c \vee (c' \wedge z')) \wedge$ $z = (c \wedge z) \vee (c' \wedge z' \wedge z)$ = $(c \wedge z) \vee 0$ = $c \wedge z \leq c \wedge y = 0$, $b \wedge z = b \wedge (y \wedge (b' \vee y'))$ = $(b \wedge y)\wedge (b' \vee y')$ = $(b \wedge y) \wedge (b \wedge y)' = 0$. □

Corollary 2.1. Let L be an orthomodular lattice. For $a, b, c \in L$ with $a \approx b \leq c$. Then, there exists $d$ such that $a \leq d \approx c$ holds.

Proof. Assume first that $a \approx b \leq c$. There exist $z_{1}, z_{2}, \cdots, z_{n}$ such that $a = z_{1} \sim \cdots \sim z_{n-2} \sim z_{n-1} \sim z_{n} = b \leq c$. For $z_{n-1} \sim z_{n} = b \leq c$. By Lemma 2.1, $\exists\; d_{n} \in L$, we have $a = z_{1} \sim \cdots \sim z_{n-2} \sim z_{n-1} \leq d_{n} \sim c$. Repeat to use the lemma 2.1, $\exists\; d_{n-1}, d_{n-2}, \cdots, d_{2}$, such that $a = z_{1} \leq d_{2} \cdots \sim d_{n-2} \sim d_{n-1} \sim d_{n} \sim c$, i.e., $\exists\; d = d_{2}$, $a \leq d \approx c$. □

Let $L$ be an orthomodular lattice. In ^[1], Finch shows that $Con(L)$ is isomorphic to $I_{p}(L)$, where $Con(L)$ represents the set of all congruence relations on $L$, $I_{p}(L)$ represents the set of all $p$-ideals on $L$, i.e., congruence relations in an OML $L$ is completely determined by its kernel, and an ideal $I$ of $L$ is the kernel of a congruence relation if and only if it is $p$-ideal (or orthomodular ideal). There are a lot of other conditions that can be used to define $p$-ideal. More precisely, we have:

Lemma 2.2. ^[2] If $I$ is an ideal of orthomodular lattice$L$, then, the following statements are equivalent:

(ⅰ) $I$ is a $p$-ideal;

(ⅱ) $a\in I$ implies $x\wedge (x'\vee a)\in I$ for all $x\in L$.

Dual to Lemma 2.2, we have the following result:

Lemma 2.3. ^[9] If $F$ is a filter of orthomodular lattice $L$, then, the following statement are equivalent:

(ⅰ) $F$ is a $p$-filter;

(ⅱ) $a\in F$ implies $x \vee (x' \wedge a)\in F$ for all $x\in L$.

Remark 2.1. Let $L$ be an orthomodular lattice. If $I$ be an ideal of $L$, $I' = \{ x' \in L \mid x \in I\}$, by Lemmas 2.2 and 2.3, we can obtain $I$ is a $p$-ideal of $L$ if and only if $I'$ is a $p$-filter of $L$, and vice versa.

Lemma 2.4. Let $L$ be an orthomodular lattice, $y\in L$. Then, $[0, y]$ is a $p$-ideal if and only if $y\in C(L)$.

Proof. Let $y\in C(L)$. Clearly, $[0, y]$ is the principle ideal of $L$. Set $a \in [0, y]$, i.e., $0\leq a \leq y$. Since $y\in C(L)$, then $\forall x \in L$, $xCy$. Then, $0 \leq x \wedge (x' \vee a)$ $\leq x \wedge (x' \vee y)$ $\leq (x \vee y) \wedge (x' \vee y) = y$. Consequently, $x \wedge (x' \vee a)\in [0, y]$. By Lemma 2.2, we have $[0, y]$ is a $p$-ideal.

Conversely, if $[0, y]$ is a $p$-ideal, $\forall x\in L$, since $x', y \leq x' \vee y$, then $x'C(x' \vee y), yC(x' \vee y)$. Then, we have $(x' \vee y)Cx, (x' \vee y)Cy$, so $\{x, y, x'\vee y\}$ is a distributive triple; hence, $(x' \vee y) \wedge (x \vee y) = ((x' \vee y) \wedge x) \vee ((x' \vee y) \wedge y) = ((x' \vee y) \wedge x) \vee y$. Since $[0, y]$ is a $p$-ideal and $y \in[0, y]$, by Lemma 2.2, we have $x \wedge (x' \vee y)\in [0, y]$, i.e., $x \wedge (x' \vee y)\leq y$. So, $(x' \vee y) \wedge (x \vee y) = ((x' \vee y) \wedge x) \vee y = y$, i.e., $y\in C(L)$. □

The properties of commute listed above, together with Lemma 2.4, imply.

Corollary 2.2. Let $L$ be an orthomodular lattice, $[y, 1]$ is a $p$-filter if and only if $y\in C(L)$.

Proposition 2.1. Let $L$ be an orthomodular lattice. Then, $c\in C(L)$ if and only if $\forall a \in L, c\sim a$ implies $a = c$.

Proof. Let $c\in C(L)$, $c\sim a$. Then, $a$ and $c$ have a common complement $c'\in L$. Since $aCc \Leftrightarrow a = (c \vee a)\wedge (c' \vee a) = (c \vee a)\wedge 1 = c \vee a$, then $c\leq a$. Similarly, $cCa \Leftrightarrow c = (c \vee a)\wedge (c \vee a') = (c \vee a)\wedge(c' \wedge a)' = (c \vee a)\wedge 0' = (c \vee a)\wedge 1 = c \vee a$, which implies $a \leq c$. Thus, $a = c$.

Conversely, assuming $c\notin C(L)$, we have $c'\notin C(L)$. Then, $c'$ has at least two different complements, $a, c$. Obviously, if $c\sim a$, which is contrary to $a = c$. Hence, $c \in C(L)$. □

Corollary 2.3. Let $L$ be an orthomodular lattice. Then, $c\in C(L)$ if and only if $\forall a \in L, c\approx a$ implies $a = c$.

Proof. Let $c\in C(L)$ and $c\approx a$. Then, $\exists\; d_{1}, d_{2}, ..., d_{n} \in L$, $c = d_{1}\sim d_{2}\sim...\sim d_{n} = a$. Since $c\sim d_{2}$, By proposition 2.1, we have $c = d_{2}$, and repeat to use proposition 2.1, we have $c = d_{1} = d_{2} = ... = d_{n} = a$.

Conversely, Let $\forall a \in L, c\approx a$ implies $a = c$ and $c\notin C(L)$. proposition 2.1 yields $\exists\; a \in L, c\sim a$ and $a \neq c$. Clearly, $c\sim a$ implies $c\approx a$, it can be inferred that $a = c$, which is a contradiction. Hence, $c \in C(L)$. □

An order ideal in a lattice $L$ is a subset $I$, such that $x\in L, y \in I$, and $x \leq y$ imply $x\in I$.

Lemma 2.5. ^[2] Let $L$ be an orthomodular lattice and $M$ be an order ideal of $L$ that is closed under perspectivity:

(ⅰ) The smallest $p$-ideal containing $M$ is the set $B = \{ c \in L \mid$, there exists a finite subset C of M with c = $\bigvee C\}$,

(ⅱ) The set $N_{a} = \{d\in L\mid d\approx e, e \leq a \}$ is an order ideal and is closed under perspectivity. $P_a$ denotes the smallest $p$-ideal containing $N_{a}$.

Proposition 2.2. Let $L$ be an orthomodular lattice. Then, $c\in C(L)$ if and only if $N_{c} = [0, c]$.

Proof. Let $c \in C(L)$. Lemma 2.5 gives $[0, c]\subseteq N_{c}$. Let $\forall d \in N_{c}$. Then, $\exists\; e \leq c$, $d \approx e$. By Corollary 2.1, we have $\exists\; d_{1} \in L$, $d \leq d_{1}$ and $d_{1} \approx c$. By Corollary 2.3, we have $d_{1} = c$. Hence, $d \leq d_{1} = c$, and then $d \in [0, c]$, hence $N_{c}\subseteq [0, c]$, so $N_{c} = [0, c]$.

If $c\notin C(L)$, we have $c'\notin C(L)$. Then, $c'$ has at least two different complements $a, c$, whence $a \in N_{c} = [0, c]$. Thus, $a \leq c$, whence $aCc$. Then, $a = (a \vee c)\wedge (a \vee c') = (a \vee c)\wedge 1 = c \vee a = c$, which is a contradiction. Hence, $c \in C(L)$. □

Proposition 2.3. Let $L$ be an orthomodular lattice. Then, $N_{c}\cap N_{c'} = \{0\}$ if and only if $c\in C(L)$.

Proof. Let $c\sim a$. Then, $c'\sim a'$. Since $a \sim c \in N_{c}, a'\sim c' \in N_{c'}$. By Lemma 2.5, we have $a\in N_{c}, a' \in N_{c'}$. Then, $c\wedge (c' \vee a') \sim a'\wedge (a \vee c)$, since $a'\wedge (a \vee c)\leq a' \in N_{c'}$, then $c\wedge (c' \vee a') \in N_{c'}$. $c\wedge (c' \vee a') \leq c$ yields $c\wedge (c' \vee a')\in N_{c}$. Thus, $c\wedge (c' \vee a')\in N_{c} \cap N_{c'}$. Similarly, $a \wedge (a' \vee c')\in N_{c} \cap N_{c'} = \{ 0\}$. Then, $c\wedge (c' \vee a') = c\wedge (c \wedge a)' = 0$, since $N_{c} \cap N_{c'} = \{ 0 \}$. Then, $c \leq a$. Similarly, $a \leq c$, i.e., $a = c$. Thus, proposition 2.1 implies that $c \in C(L)$.

Conversely, let $c \in C(L)$. By Proposition 2.2, we get $N_{c} = [0, c], N_{c'} = [0, c']$. Assume that $d\in N_{c} \cap N_{c'}$, then $d\leq c, c'\Leftrightarrow d \leq c\wedge c' = 0$, i.e., $d = 0$. Hence, $N_{c} \cap N_{c'} = \{ 0 \}$. □

Proposition 2.4. Let $L$ be an orthomodular lattice. $M$ and $N$ are order ideals and closed under perspectivity; $P_{M}$ and $P_{N}$ are the smallest $p$-ideals containing $M$ and $N$, respectively. Then, $M\cap N = \{0\} \Leftrightarrow P_{M}\cap P_{N} = \{0\}$.

Proof. Let $d\in P_{M}\cap P_{N}$; we have $d \in P_{M}, P_{N}$. Then, Lemma 2.5 yields that there exists a finite subset $C$ of $M$, with $d = \vee C$. Let $e_{i} \in C, \forall i \in I$, then $e_{i} \leq \vee C = d \in P_{M}\cap P_{N}$. Since $P_{M}\cap P_{N}$ is an ideal, then $e_{i} \in P_{M}\cap P_{N}\subseteq P_{N}$. By Lemma 2.5, there exists a finite subset $D_{i}$ of $N$, with $e_{i} = \vee D_{i}$. Let $f_{ij} \in D_{i}, \forall j \in J$, we have $f_{ij} \in D_{i} \subseteq N$ and $f_{ij} \leq \vee D_{i} = e_{i} \in C \subseteq M$. Then, $f_{ij} \in M \cap N = \{ 0\}$, i.e., $f_{ij} = 0$. Hence, $d = \vee C = \underset{i \in I}{\vee} e_{i} = \underset{i \in I}{\vee} \underset{j \in J}{\vee} f_{ij} = 0$. Therefore, $P_{M} \cap P_{N} = \{0\}$.

Conversely, since $M\subseteq P_{M}$, $N\subseteq P_{N}$, $M \cap N \subseteq P_{M}\cap P_{N} = \{ 0\}$. So, $M\cap N = \{0\}$. □

Recall that a poset $P$ satisfies the ascending condition $(ACC)$ and is called Noetherian when every nonvoid subset of $P$ has a maximal element. It satisfies the descending chain condition (DCC) when its dual satisfies the ACC. Nonvoid chains that satisfy the $DCC$ are called well-ordered sets. The length $l(P)$ of a poset $P$ is defined as the least upper bound of the lengths of the chains in $P$. When $l(P)$ is finite, $P$ is said to be finite length. A poset $P$ is chain-finite if all chains in $L$ are finite. It is easy to show that a finite length lattice must be chain-finite; a chain-finite lattice must be Noetherian. The $dual$ of a poset $P$ is that poset $\widetilde{P}$ defined by the converse partial ordering relation on the same elements.

Lemma 2.6. ^[8] The following conditions about a poset $P$ are equivalent:

(ⅰ) $P$ is Noetherian,

(ⅱ) every chain in the dual $\widetilde{P}$ of $P$ is well-ordered,

(ⅲ) every well-ordered subset of $P$ is finite.

Lemma 2.7. ^[10] In every infinite boolean algebra, there is an infinite pairwise disjoint family, a strictly increasing infinite sequence is and a strictly decreasing infinite sequence.

Next, we give a characterization of Noetherian orthomodular lattice and boolean lattice.

Proposition 2.5. Let $L$ be a boolean lattice. Then, $L$ is Noetherian if and only if it is finite.

Proof. Let $L$ be a Noetherian boolean lattice. Assume that $L$ is infinite. By Lemma 2.7, $L$ has a strictly increasing sequence $A$. Since $L$ is a Noetherian lattice, then $A' = \{x\in L | x'\in A\}$ is also a Noetherian lattice. Let $B$ be any subset of $A$, then $B'$ is a subset of $A'$, so $B'$ has a maximal element $b'$, which is equivalent to $B$ having a minimal element, i.e., $A$ satisfies the descending chain condition, thus $A$ is well-ordered. By Lemma 2.6, $A$ is finite, which is a contradiction. Thus, $L$ is finite. Conversely, any well ordered subset of a finite lattice is finite. By Lemma 2.6, a finite lattice is a Noetherian lattice. □

Proposition 2.6. Let $L$ be an orthomodular lattice. Then, $L$ is Noetherian if and only if it is chain-finite.

Proof. Assume that $L$ has an infinite chain, say $A$. Then $A$ generates a boolean subalgebra $B$ of $L$; $B$ is also Noetherian. By Proposition 2.5, $B$ is finite; it follows that $A$ is finite, which is a contradiction. Then $L$ is chain-finite. The converse is trivial. □

Corollary 2.4. Let $L$ be an orthomodular lattice. Then, $L$ is Noetherian $\Leftrightarrow$ $L$ satisfies DCC.

Proof. $L$ is Noetherian $\Leftrightarrow$ $L$ is chain-finite $\Leftrightarrow$ $\widetilde{L}$ is chain-finite $\Leftrightarrow$ $\widetilde{L}$ is Noetherian $\Leftrightarrow$ $L$ satisfies DCC. □

Let $(P; \leq)$ be a partially ordered set, $a, b\in P$ and $a \leq b$. If $a < b$ and $[a, b] = \{a, b\}$, then $b$ covers $a$ (or $a$ is covered by $b$). We use the symbol $b\succ a$ (or $a \prec b$)to indicate that $b$ covers $a$ (or $a$ is covered by $b$). Let $P$ have a minimum element of 0. We call $a \in L$ an atom of $P$ if $0\prec a$. Let $L$ be a lattice. if for all $x \in L \setminus \{0 \}$, there exists an atom $a$ of $L$ such that $a \leq x$. Then, $L$ is called an atomic lattice. Dually, let $L$ be a lattice with a maximum element 1. And we call $b \in L$ is a coatom of $L$ if $b\prec 1$. For all $y \in L \setminus \{1 \}$, there exists a coatom $b$ of $L$ such that $y \leq b$. Then, $L$ is called a coatomic lattice. Let $L$ be an atomic orthomodular lattice. For $\forall x\in L -\{1\}$, we have $x = \bigvee \{ a \in L \mid a \leq x, a\; is\; an\; atom\}$. Dually, let $L$ be a coatomic orthomodular lattice. For $\forall y\in L- \{0\}$, we have $x = \bigwedge \{b \in L \mid b \geq x, b\; is\; a\; coatom\}$. An element $b$ of an atomic orthomodular lattice is finite if it is the join of finitely many atoms.

Corollary 2.5. Let $L$ be a Noetherian orthomodular lattice. Then, $L$ is atomic.

Proof. $L$ is Noetherian. Corollary 2.4 implies that $L$ satisfies DCC (Descending Chain Condition), i.e., every nonvoid subset of $P$ has a minimal element. Thus, for $\forall x\in L$ and $0 < x$, $(0, x]$ has a minimal element, which is an atom. This proves that $L$ is atomic. □

We are now ready to have a look at the $p$-ideals set of irreducible OML. First, however, we shall need some additional machinery.

Definition 2.3. ^[2] Every orthomodular lattice $L$ is irreducible, i.e., $L$ is isomorphism to a product $M\times N$ of orthomodular lattices implies that $|M| = 1$ or $|N| = 1$, where $|M|$ denotes cardinality of $M$.

Remark 2.2. It is easy to see that $C(L) = \{0, 1\}$ is equivalent to the irreducibility of $L$.

Lemma 2.8. ^[2] Let $L$ be an orthomodular lattice. If $L$ is chain-finite. Then, $Con(L)$ is isomorphic to $C(L)$. In particular, $Con(L)$ is a boolean algebra.

Theorem 2.1. Let $L$ be an irreducible orthomodular lattice and $\textrm{Noetherian}$. Then, the $p$-ideals are well ordered under set inclusion.

Proof. Let $L$ be Noetherian. Proposition 2.6 implies that $L$ is chain-finite. Since $L$ is irreducible, Remark 2.2 implies that $C(L) = \{0, 1\}$. By Lemma 2.8 and the fact that $Con(L)$ and $I_{p}(L)$ are isomorphic, we have $I_{p}(L) = \{\{0\}, L \}$, Hence, the $p$-ideals set is well ordered under set inclusion. □

Next, we introduce our basic terminology and notations, which are borrowed from ^[2]:

Let $L$ be an OML. The binary relation $\nabla, S^{0}, S^{1}, S^{\infty}$ are defined in an orthomodular lattice $L$ by:

Immediately, we can obtain $aS^{\infty}b \Rightarrow aS^{1}b \Rightarrow aS^{0}b$. Furthermore, we have the following results:

Lemma 2.9. ^[2] Let $L$ be an orthomodular lattice. For $\forall a, b \in L$, $a S^{0} b \Leftrightarrow aSb$.

Lemma 2.10. ^[2] For elements $a, b$ of an orthomodular lattice $L$, the following statements are equivalent:

(ⅰ) $a\nabla b$,

(ⅱ) $aS^{1}b$,

(ⅲ) $a \vee x = 1$ implies $b \leq x$,

(ⅳ) $(a \vee x)\wedge(b \vee x) = x, \forall x \in L$.

Proposition 2.7. Let $L$ be an orthomodular lattice, $a, b\in L$. Then, $N_{a}\cap N_{b} = \{0\}$ if and only if $aS^{\infty}b$.

Proof. For $c, d\in L$, let $c \leq a, d \leq b, c\approx d$, then $c, d \in N_{a}\cap N_{b} = \{0 \}$, whence $c = 0 = d$. Hence, $aS^{\infty} b$.

Conversely, let $d\in N_{a}\cap N_{b}$. There exists $c, e$, such that $d \approx c \leq a, d\approx e \leq b$. Then, $c \approx e$. Since $aS^{\infty} b$, we have $c = 0 = e$. Thus, $d = 0$, which proves that $N_{a}\cap N_{b} = \{ 0 \}$. □

Corollary 2.6. Let $L$ be an orthomodular lattice, $a\in L$. Then, $a \nabla a' \Leftrightarrow a S^{1} a' \Leftrightarrow aS^{0}a' \Leftrightarrow aSa'\Leftrightarrow aS^{\infty}a' \Leftrightarrow a\in C(L)$.

Proof. By Lemma 2.9, we have $a S^{0} b \Leftrightarrow aSb$. By Lemma 2.10, this gives $a\nabla a' \Leftrightarrow aS^{1}a' \Leftrightarrow a\in C(L)$ and $aS^{\infty}b \Rightarrow aS^{1}b \Rightarrow aS^{0}b$. It suffices to prove $a\in C(L)\Rightarrow aS^{\infty}a'$ and $aSa'\Rightarrow a\in C(L)$. For $a\in C(L)\Rightarrow aS^{\infty}a'$, by Propositions 2.3 and 2.7. We have $N_{a} \cap N_{a'} = \{ 0 \}$ and $aS^{\infty}a'$, then $a\in C(L)\Rightarrow aS^{\infty}a'$. Let $aSa'$, we obtain $a\in C([0, a\vee a']) = C([0, 1]) = C(L)$. So, $aS^{\infty}a' \Rightarrow aS^{0}a' \Rightarrow aSa' \Rightarrow a\in C(L)$. □

Let $L$ be an orthomodular lattice, $a\in L$, $M = \{z\in C(L)\mid a \leq z \}$. If the infimum of $M$ exists, we denote $v(a) = \bigwedge M$ and say $v(a)$ is the center cover of $a$. If for $\forall a \in L$, $C([0, a]) = \{a\wedge c\mid c \in C(L)\}$, we say that the relative centre property holds.

Lemma 2.11. ^[2] Assume that the orthomodular lattice $L$ has the relative centre property. For elements $a, b$ of $L$, the following statements are equivalent:

(ⅰ) There exists $c\in C(L)$ with $a\leq c$ and $b\leq c'$,

(ⅱ) $a S^{1} b$,

(ⅲ) $aS^{0}b$.

Lemma 2.12. ^[2] Let $L$ be a complete orthomodular lattice. For $\forall a, b \in L$, there exists $c, d, e, f \in L$ such that the following conditions hold:

(ⅰ) $a = c\vee d, b = e\vee f$,

(ⅱ) $c \leq d', e\leq f'$,

(ⅲ) $c \sim_{s}e$,

(ⅳ) $dS^{0}f$.

Theorem 2.2. Let $L$ be a complete irreducible orthomodular lattice with the relative centre property. Then, the $p$-ideals set $I_{p} (L)$ is a chain under set inclusion.

Proof. If $L$ is irreducible, then by Remark 2.2, we get $C(L) = \{0, 1\}$. If $L$ has the relative centre property, by Lemma 2.11, we have $aS^{0}b \Leftrightarrow a = 0$ or $b = 0$. Next, we prove $\forall a, b\neq 0, 1$, $P_{a}\subseteq P_{b}$ or $P_{b}\subseteq P_{a}$. In fact, Lemma 2.12 implies that there exists $c = a, d = 0, b = e\vee f, e\leq f', c \sim_{s}e$, or $b = e, f = 0, a = c\vee d, c\leq d', c \sim_{s}e$. In the first case, $a = c\sim e\leq b$ implies $a\in N_{b}$ and thus $N_{a}\subseteq N_{b}$. Similarly, $N_{b}\subseteq N_{a}$, which yields $P_{a}\subseteq P_{b}$ or $P_{b}\subseteq P_{a}$. Let $I, J\in I_{p} (L), I\nsubseteq J, J\nsubseteq I$. Then, $\exists\; x\in I-J, y\in J-I$. If $P_x\subseteq P_y$, we have $x\in P_y\subseteq J$, thus $x\in J$, which is a contradiction; if $P_y\subseteq P_x$, we obtain $y\in P_x\subseteq I$, which is impossible. Then, $I_{p} (L)$ is a chain. □

Now, we consider the OML with exchange axiom, which states the following:

We say that the atomic orthomodular lattice $L$ satisfies the exchange axiom if one of the equivalent conditions of the following conditions holds for $L$:

(ⅰ) For atoms $p, q$ of $L$, the conditions $p\leq q\vee a$ and $p\wedge a = 0$ imply $q\leq p\vee a$;

(ⅱ) If for an atom $p$ of $L$ holds $p\neq a$, then $p\vee a$ covers $a$;

(ⅲ) If $a$ covers $a\wedge b$, then $a\vee b$ covers $b$.

Lemma 2.13. ^[2] Let $L$ be an atomic complete orthomodular lattice satisfying the exchange axiom. Then, $L$ is irreducible if and only if every two atoms of $L$ are perspective.

Theorem 2.3. Let $L$ be an atomic irreducible complete orthomodular lattice satisfying the exchange axiom. Then 1 is finite $\Leftrightarrow P_{r} = L \Leftrightarrow \forall x\in L, x$ is finite, where $\forall r \in A(L)$, $A(L)$ is the set of all atoms of $L$. In particular, in the case, $I_p(L)$ is well ordered under set inclusion relations.

Proof. First, we show that $p$ is an atom, $p\sim q$, then $q$ is an atom. In fact, let $a$ be a common complement of $p, q$. Then, $p\wedge a = 0\prec p \Rightarrow$ $a\prec p\vee a = 1 = q\vee a \Rightarrow$ $0 = q\wedge a\prec q$, i.e., $q$ is an atom. Then, $\forall$ $r\in A(L)$, $N_{r} = \{x | x\approx d\leq r\} = \{0\}\cup A(L)$. By Lemma 2.5, we have $P_{r} = \{x = q_1\vee q_2\vee \cdots\vee q_n | q_i \in A(L), \; i = 1, 2, \cdots, n\}$. Then, 1 is finite $\Rightarrow 1\in P_{r} \Rightarrow P_{r} = L \Rightarrow \forall x\in L, x\in P_{r} \Rightarrow x$ is finite $\Rightarrow 1$ is finite. For $\forall 0\neq a\in L, \forall$ $r\in A(L)$, we have $r\in N_{a}$. In fact, since $0 < a$, then $\exists\; s\in A(L)$, $s\leq a$. Thus, $r\sim s\leq a$ implies $r\in N_{a}$. It is a routine to show $P_{r}\subseteq P_{a}, \forall 0\neq a\in L, \forall r\in A(L)$. So, $I_p(L) = \{\{0\}, L\}$ is well ordered, as desired. □

3.   $L$-Ideals set of $OM$-$L$-algebras

Definition 3.1. ^[7] An $L$-algebra is an algebra $(L, \rightarrow, 1)$ of type $(2, 0)$ satisfying:

For $\forall x, y, z \in L$, condition (3.1) states that 1 is a logical unit. Note that a logical unit is always unique. There is a partial ordering by [7,P. 2332,Prop.2].

Let $L$ be an $L$-algebra. By ^[3], $x\leq y$ implies that $z\rightarrow x \leq z\rightarrow y$ for $\forall x, y, z\in L$. An $L$-algebra $L$ is a $KL$-algebra. If it satisfies:

Definition 3.2. ^[11] Let $L$ be an $L$-algebra. A subset $Y$ of $L$ is said to be an $L$-subalgebra if $\forall x, y \in Y$ implies $x\rightarrow y \in Y$. If $x\rightarrow y \in Y$ holds for $\forall x \in L, y\in Y$, we call $Y$ an invariant $L$-subalgebra.

Definition 3.3. ^[3] Let $(L, \rightarrow, 1)$ be an $L$-algebra. We call $I\subseteq L$ an ideal if the following holds for all $x, y \in L$:

We call the ideal of $L$-algebra is an $L$-ideal. For an element $x\in L$, $\langle x \rangle$ represents the $L$-ideal generated by $x$. We will use $I(L)$ to denote the set of all $L$-ideals of an $L$-algebra $L$.

Regarding the relevant background of $L$-algebra, we refer to ^[3].

We call an $L$-algebra $(L, \rightarrow, 1)$ self-similar if, for $\forall x\in L$, the left multiplication $\varepsilon_x: y\mapsto x\rightarrow y$ induces a bijection $\downarrow x \rightarrow L$. The inverse of $\varepsilon_x$ gives rise to a multiplication $xy: = \varepsilon_{y}^{-1}(x)$ that happens to be associative. This allows to describe a self-similar $L$-algebra equationally as a monoid with an operation $\rightarrow$ satisfying:

Axiom (3.6) implies that as a monoid; a self-similar $L$-algebra is right cancellative. Let $L$ be a self-similar $L$-algebra. For $a, b \in L$, we define

Then, (3.9) is an infimum, and for $a, b, c \in L$, we have

Let $L$ be a self-similar $L$-algebra. Then $L$ satisfies

Every $L$-algebra $L$ admits a self-similar closure, a self-similar $L$-algebra $S(L)$ with $L$ as an $L$-subalgebra such that as a monoid, $S(L)$ is generated by $L$. Up to isomorphism of left hoops, $S(L)$ is the unique self-similar closure of $L$.

In this article, we regard $(L, \rightarrow, 1, 0)$ as $OM$-$L$-algebra; $S(L)$ is the self-similar closure of $L$; for $\forall a \in S(L)$, define $a' = a\rightarrow 0$. We call $\mathcal{C}(L) = \{a\in L\mid ab = ba, \forall b\in L\}$ the centre of $L$.

Definition 3.4. ^[7] We call $(L, \rightarrow)$ an $OM$-$L$-algebra if it is an $L$-algebra with 0 (i.e., $L$ admits the smallest element 0) and satisfies the following condition, where $x': = x \rightarrow 0$.

Lemma 3.1. ^[7] Let $L$ be an $OM$-$L$-algebra. Then, $L$ is an orthomodular lattice and satisfies the following conditions, where $x': = x \rightarrow 0$.

According to the proof process of ^[7], (3.12) is equivalent to $x' \leq y \Leftrightarrow x \rightarrow y = y \Leftrightarrow y \rightarrow x = x$. In particular, $\forall x, y \in L, x' \leq x \rightarrow y \Leftrightarrow x' \rightarrow (x \rightarrow y) = 1$, whence $x \rightarrow (x \rightarrow y) = x \rightarrow y$, and $(x \rightarrow y)\rightarrow x = x$. This shows that $OM$-$L$-algebra is sharp ^[12].

Lemma 3.2. ^[7] Every orthomodular lattice $L$ gives rise to an $OM$-$L$-algebra $(L, \rightarrow, 1, 0)$, where

Lemma 3.3. ^[7] Let $L$ be an $OM$-$L$-algebra. Then, $0 \leq a \Leftrightarrow a \in L$.

Lemma 3.4. ^[7] Let $L$ be an $OM$-$L$-algebra, $a, b\in L$. The equivalence $aCb \Leftrightarrow ab = ba$ is valid.

Lemma 3.5. ^[7] Let $L$ be an $OM$-$L$-algebra. $a, b\in L$. The equivalence $ab = ba \Leftrightarrow a \leqslant b \rightarrow a$ holds. In particular, $a\leq b$ implies $ab = ba$.

Lemma 3.6. ^[7] Let $L$ be an $OM$-$L$-algebra. For all $a, b\in L$, the following are satisfied:

Remark 3.1. Let $L$ be an $OM$-$L$-algebra. Then, for all $a, x\in L, a\leq x\rightarrow a\; or\; a\parallel x\rightarrow a$, where $x \parallel y$ denotes $x, y$ are incomparable, i.e., $x\nleq y$ and $y \nleq x$. In fact, if not, then there exists $b\in L$ such that $b\rightarrow a < a$, i.e., $b'\vee (b\wedge a) < a$, which implies $b' < a$. Thus, $a > b'\vee (b\wedge a) = a$, which is impossible.

Proposition 3.1. Let $L$ be an $OM$-$L$-algebra. For $x, y, z \in L$, we have the following properties:

(ⅰ) Let $x \leq y, z$ and $y \rightarrow x = z \rightarrow x$. Then, $y = z$.

(ⅱ) Let $x \leq y$. Then, there exists a unique element $z \in L$ such that $x \leqslant z$ and $y = z\rightarrow x$.

Proof. (ⅰ) Set $x \leqslant y, z$ and $y \rightarrow x = z \rightarrow x$; we have $(y \rightarrow x) \rightarrow x = (z \rightarrow x) \rightarrow x$. Then, Lemma 3.5 and (3.18) imply that $y = z$.

(ⅱ) Set $x \leq y$. By Lemma 3.5, we get $xy = yx \Leftrightarrow x \leq y \rightarrow x$. Then, $\exists\; z = y \rightarrow x$, such that $x \leq z$. (3.18) yields $y = (y\rightarrow x) \rightarrow x = z \rightarrow x$. By (ⅰ), the uniqueness holds. □

Definition 3.5. ^[13] We define a semibrace $A$ to be a commutative monoid $(A, \; \wedge)$ with an additional binary operation $A\times A \dot{\rightarrow } A$ such that

Let $L$ be an $L$-algebra. We define a $\wedge$-closure of $L$ to be an $L$-algebra $\theta(L)$, which is a semibrace such that $L$ is an $L$-subalgebra of $\theta(L)$, and every $a\in \theta(L)$ is of the form $a = x_{1}\wedge \cdot\cdot\cdot \wedge x_{n}$ with $x_{i}\in L$.

Definition 3.6. ^[14] We call an $L$-algebra $L$ $\wedge$-closed if $L = \theta(L)$.

Proposition 3.2. Every $OM$-$L$-algebra is a $\wedge$-closed $L$-algebra.

Proof. It suffices to prove that the intersection in $L$ is consistent with the intersection in $S(L)$. For $\forall x, y\in L$, by (3.11), we have $0\rightarrow (x \rightarrow y)x$ = $((x \rightarrow 0)\rightarrow (x \rightarrow y))$$(0 \rightarrow x) = ((0 \rightarrow x)\rightarrow$ $(0 \rightarrow y)) = 1$, then $0 \leq (x\rightarrow y)x$. By Lemma 3.3 and (3.9), we have $x \wedge y = (x \rightarrow y)x\in L$. Thus, every $OM$-$L$-algebra is a $\wedge$-closed $L$-algebra. □

Lemma 3.7. ^[15] Let $L$ be a $\wedge$-closed $L$-algebra. $x, y\in L$. Then, the following are satisfied.

Corollary 3.1. Let $L$ be a $\wedge$-closed $L$-algebra. $a, b\in L$. If $a \wedge b = 0$ and $\langle a \rangle \subseteq \langle b \rangle$. Then, $\langle b \rangle = L$.

Proof. By (3.21), we have $\langle b \rangle = \langle a \rangle \vee \langle b \rangle = \langle a \wedge b \rangle = \langle 0 \rangle = L$. □

Proposition 3.3. Let $L$ be an $L$-algebra. For $\forall I \in I(L)$. Then, $I = \underset{i \in I_{0}}{\bigcup\limits}\{\langle x_{i} \rangle \mid \forall x_{i} \in I \}$, where $I_{0}$ represents the index set.

Proof. For $\forall t \in I$, we have $t \in \langle t \rangle \subseteq \underset{i \in I_{0}}{\bigcup\limits}\{ \langle x_{i} \rangle \mid \forall x_{i} \in I \}$. Then, $I \subseteq \underset{i \in I_{0}}{\bigcup\limits}\{ \langle x_{i} \rangle \mid \forall x_{i} \in I \}$. For $\forall t \in \underset{i \in I_{0}}{\bigcup\limits}\{ \langle x_{i} \rangle \mid \forall x_{i} \in I \}$, $\exists\; x_{0}\in I$, such that $t \in \langle x_{0}\rangle \subseteq I$, which yields $\underset{i \in I_{0}}{\bigcup\limits} \{ \langle x_{i} \rangle \mid \forall x_{i} \in I \} \subseteq I$. Thus, $I = \underset{i \in I_{0}}{\bigcup\limits}\{\langle x_{i} \rangle \mid \forall x_{i} \subseteq I \}$. □

Definition 3.7. Let $L$ be an $L$-algebra. $L$ is irreducible if $L$ is isomorphic to a product $L_{1}\times L_{2}$ of $L$-algebra implies that $|L_{1}| = 1$ or $| L_{2} | = 1$. $| L_{1} |$, $| L_{2}|$, respectively, the cardinality of $L$-algebras $L_{1}$ and $L_{2}$.

Proposition 3.4. Let $L$ be an irreducible $OM$-$L$-algebra. Assume that $a\in L$; $a$ is an atom. Then, $\langle a \rangle = L$.

Proof. Let $a$ be an atom, and $L$ is irreducible. Lemma 3.5 and Remark 3.1 imply that there exists $x_{0}\in L$ and $x_{0} \neq 0, 1$, such that $ax_{0} \neq x_{0}a \Leftrightarrow a \nleqslant x_{0}\rightarrow a \Leftrightarrow$ $(x_{0} \rightarrow a)\parallel a$. Since $a$ is an atom, then $a \wedge (x_{0} \rightarrow a) = 0$. Since $a, x_{0}\rightarrow a\in \langle a\rangle$, then $0 = a \wedge (x_{0} \rightarrow a) \in \langle a \rangle$, thus $\langle a \rangle = L$. □

Let $L$ be an $OM$-$L$-algebra, a subset $I\subseteq L$. By ^[16] and ^[11], we obtain that $I$ is an $L$-ideal if and only if $I$ is an invariant $L$-subalgebra satisfying $({I0})$. Next, we will prove the $p$-filters in $L$ ($L$ as an orthomodular lattice) are consistent with the $L$-ideals ($L$ as $OM$-$L$-algebra).

Proposition 3.5. Let $L$ be an $OM$-$L$-algebra. The $p$-filter of orthomodular lattice $L$ is the same as the $L$-ideals of $OM$-$L$-algebra $L$.

Proof. Let $F$ be a $p$-filter of $L$. We need to prove that $F$ satisfies $({I1})$ and is an invariant $L$-subalgebra of $L$. Let $x, x\rightarrow y\in F$. Since $x\rightarrow y = x' \vee (x \wedge y) \in F$ and $F$ is a $p$-filter, we have $x \wedge (x \rightarrow y)\in F$. Then, $x \wedge (x \rightarrow y) = (x\rightarrow (x\rightarrow y))x = (x\rightarrow y)x = x\wedge y\leq y \in F$. So $y \in F$. For $\forall y \in L, x \in F$, we get $y \rightarrow x = y' \vee (y \wedge x) \in F$, whence $F$ is an invariant $L$-subalgebra of $L$.

Conversely, let $I$ be an $L$-ideal of an $OM$-$L$-algebra $L$. Set $a, b\in I$. We have $a\rightarrow b \in I$. Then, $(a\rightarrow b)\rightarrow (a \wedge b)$ = $(a\rightarrow b)\rightarrow (a\rightarrow b)a$ = $((a\rightarrow (a\rightarrow b))\rightarrow (a\rightarrow b))((a\rightarrow b)\rightarrow a)$ = $((a\rightarrow b)\rightarrow (a\rightarrow b))a = 1a$ = $a\in I$. By $({I1})$, $a \wedge b\in I$. Let $a\in I, b\in L, a\leq b$. Then $a\rightarrow b = 1\in I$. By $({I1})$, we have $b \in I$. Then, $I$ is a filter. If $a\in I$, for $\forall x\in L$, by (3.15), we obtain $x \vee (x' \wedge a) = x'\rightarrow a \in I$. □

Proposition 3.6. Let $L$ be an $OM$-$L$-algebra. $a\in L$, $a \in \mathcal{C}(L)$ if and only if $a^{\uparrow} = \{x\rightarrow a\mid \forall x \in X \}$.

Proof. Let $a$ be the central element; $\forall y \in a^{\uparrow}$, $a \leq y$. By Proposition 3.1, there exists a unique element $z \in L$ such that $a\leq z$ and $y = z \rightarrow a$. Then, $y \in \{x\rightarrow a\mid \forall x \in L \}$. Hence, $a^{\uparrow}\subseteq \{x\rightarrow a\mid \forall x \in L \}$. Set $\forall y\in \{x\rightarrow a\mid \forall x \in L \}$, there exists $x_{0}\in L$, such that $y = x_{0} \rightarrow a$. Since $a$ is the central element, then Lemma 3.4 implies that $ax_{0} = x_{0}a$. Since $x_{0}a\leq a \Leftrightarrow ax_{0} \leq a \Leftrightarrow a\leq x_{0} \rightarrow a = y$, we have $y \in a^{\uparrow}$, then $\{x\rightarrow a\mid \forall x \in L \} \subseteq a^{\uparrow}$. This shows that $a^{\uparrow} = \{x\rightarrow a\mid \forall x \in L \}$.

Conversely, if $a^{\uparrow} = \{x\rightarrow a\mid \forall x \in L \}$, for $\forall x \in L$, we have $a \leq x\rightarrow a$. Lemma 3.5 implies $\forall x \in L$, $a \leq x\rightarrow a \Leftrightarrow ax = xa$. Thus, $a \in \mathcal{C}(L)$. □

Proposition 3.7. Let $L$ be an $OM$-$L$-algebra, $a\in L$. Then, $a \in \mathcal{C}(L)$ if and only if $\langle a \rangle = a^{\uparrow}$.

Proof. If $a \in \mathcal{C}(L)$, by Corollary 2.2 and Proposition 3.5, $a^{\uparrow} = [a, 1]$ is an $L$-ideal. Since $a \in a^{\uparrow}$, we have $\langle a \rangle \subseteq a^{\uparrow}$. For $\forall y \in a^{\uparrow}$, we obtain $a \leq y$. Since $\langle a \rangle$ is an upper set, it follows that $y \in \langle a \rangle$. Hence, $a^{\uparrow} \subseteq \langle a \rangle$. Thus, we have $\langle a \rangle = a^{\uparrow}$.

Conversely, if $a\notin \mathcal{C}(L)$, then $\exists\; x_{0} \in L$, such that $ax_{0} \neq x_{0}a \Leftrightarrow a \nleq x_{0}\rightarrow a$, which is impossible since $x_{0}\rightarrow a \in \langle a \rangle = a^{\uparrow}$. Thus, $a \in \mathcal{C}(L)$. □

Definition 3.8. ^[17] An element $p$ of an $L$-algebra $L$ is said to be prime. If $p < 1$ and $x\leq p$ or $x\rightarrow p \leq p$ for all $x\in L$.

Definition 3.9. ^[17] We call an $L$-algebra prime if all of its elements $p < 1$ are prime.

Remark 3.2. Let $\Omega$ be a partially ordered set with the greatest element 1. By ^[3],

makes $\Omega$ into an $L$-algebra which, satisfies the inequality (3.5). Thus, $\Omega$ is a prime $KL$-algebra. Note that for any $KL$-algebra, $x\rightarrow p\leq p$ implies that $x\rightarrow p = p$.

A prime $KL$-algebra $L$ is completely determined by its underlying partial order. So, the above example shows that there is a unique prime $KL$-algebra for each partially ordered set with greatest element 1.

Definition 3.10. ^[11] We call an element $x$ of an $L$-algebra $L$ spatial if $x = \bigwedge \{p \in P(L) \mid x \leq p\}$ holds in $\theta(L)$. If all $x\in L$ are spatial, we say that $L$ has enough primes.

Definition 3.11. ^[11] We define a closure $L$-algebra to be an $L$-algebra $L$ such that for each $x\in L$, the meet $\bar{x}: = \bigwedge \{ p \in P(L)\mid p \geq x\}$ exists in $\theta(L)$ and belongs to $L$. We call an $L$-algebra $L$ complete if $L = \theta(L)$ and its underlying $\wedge$-semilattice is a complete lattice.

Definition 3.12. ^[11] We call an $L$-algebra $L$ distributive if its lattice of ideals is distributed.

Lemma 3.8. ^[17] The lattice of ideals of an $L$-algebra $L$ is distributive.

Definition 3.13. ^[18] A complete lattice is said to be a $locale$ if it satisfies the infinite distributive law

Lemma 3.9. ^[11] For the distributive $L$-algebra $L$. The ideal lattice $I(L)$ is spatial locale, and the operation on $I(L)$ is

where $\langle x \rangle$ represents the ideal generated by $x$. Then, $\forall I, J, K \in I(L)$. We have

and

Theorem 3.1. Let $L$ be an $L$-algebra. Then, $I(L)$ is a chain under the set inclusion if and only if $I(L)$ is a bounded prime $KL$-algebra.

Proof. Let $I(L)$ be a chain, $I\in I(L)$, $J\in I(L)$. If $J\subseteq I$. Then, $J \rightarrow I = L$. If $J \nsubseteq I$, since $I(L)$ is a chain, then $I \subsetneqq J$. Let $\forall x \in J\rightarrow I$, we have $\langle x \rangle \subseteq J\rightarrow I$, by (3.26), $\langle x \rangle \subseteq J\rightarrow I\Leftrightarrow \langle x \rangle \cap J \subseteq I$. Then, by Lemma 3.8, we can obtain $I = I \vee (\langle x \rangle \cap J) = (I \vee \langle x \rangle)\cap (I\vee J) = (I \vee \langle x \rangle)\cap J$. If $J \subseteq J\rightarrow I$, by (3.26), we get $J = J\cap J \subseteq I$, which is a contradiction. Then, $x \in J \rightarrow I \subseteq J$. Obviously, $x \in \langle x \rangle \subseteq I \vee \langle x \rangle$, we have $x\in(I \vee \langle x \rangle)\cap J = I$, i.e., $J \rightarrow I\subseteq I$. Since $I(L)$ is a $KL$-algebra, we have $I \subseteq J\rightarrow I$, then $I = J\rightarrow I$. Remark $4$ gives $I(L)$ is a prime $KL$-algebra. Clearly, $I(L)$ is bounded. Thus, $I(L)$ is a bounded prime $KL$-algebra.

Conversely, assume that $I, J \in I(L)$ is incomparable. Since $I\parallel J$, we obtain $I\neq I \cap J$, i.e., $I\nsubseteq I \cap J$. Since $I(L)$ is a bounded prime $KL$-algebra, by Remark 3.2, we have $I\rightarrow J = J$ and $I\rightarrow (I \cap J) = I \cap J$. By (3.27), we have $J = L\rightarrow J = (I \rightarrow I)\rightarrow (I \rightarrow J) = I\rightarrow (I \cap J) = I\cap J \subseteq I$, which is a contradiction. Thus, $I(L)$ is a chain under the set inclusion. □

It immediately, we have:

Corollary 3.2. Let $L$ be an OML. Then, $I_p(L)$ is a chain under the set inclusion relationship if and only if $I_p(L)$ is a bounded prime $KL$-algebra.

Proof. By Lemma 3.2, Proposition 3.5, Remark 2.1, and Theorem 3.1, it clearly. □

4.   Conclusions

This paper aims to give an answer to the question raised in Kalmbach's book: When are the $p$-ideals of an irreducible orthomodular lattice well ordered under set inclusion? For irreducible orthomodular lattice $L$, we prove that if $L$ is Noetherian or completely with the relative central properties or complete atomic satisfying the exchange axiom, then the $p$-ideal sets of $L$ are a chain (or well ordered) under set inclusion. In Section 3, from the perspective of $L$-algebra, we obtain that $L$-ideals lattice is a chain under the set inclusion if and only if $L$-ideals lattice is a bounded prime $KL$-algebra. Remarkably, this paper only partially answers the issues mentioned above. Therefore, we look forward to a more in-depth study about it.

Author contributions

Z. T. Zhao and J. Wang are mainly responsible for writing the initial draft; Y. L. Wu is responsible for reviewing, commenting, and revising before and after publication. All authors have read and approved the final version of the manuscript for publication.

Acknowledgments

This research is supported by the following grant: Cultivation Fund of Qilu University of Technology (Shandong Academy of Sciences) (2023PX068).

Conflict of interest

All authors declare no conflicts of interest in this paper.