Quasicomplemented residuated lattices

TL;DR

This paper introduces and studies quasicomplemented residuated lattices, a new subclass characterized by properties of prime filters and their relation to dense elements, expanding the understanding of residuated lattice structures.

Contribution

It defines quasicomplemented residuated lattices, introduces disjunctive residuated lattices, and provides characterizations using $eta$-filters, advancing lattice theory.

Findings

A residuated lattice is Boolean iff it is disjunctive and quasicomplemented.

Prime filters not containing dense elements are minimal prime filters.

Characterizations of quasicomplemented residuated lattices via $eta$-filters.

Abstract

In this paper, the class of quasicomplemented residuated lattices is introduced and investigated, as a subclass of residuated lattices in which any prime filter not containing any dense element is a minimal prime filter. The notion of disjunctive residuated lattices is introduced and it is observed that a residuated lattice is Boolean if and only if it is disjunctive and quasicomplemented. Finally, some characterizations for quasicomplemented residuated lattices are given by means of the new notion of $\alpha$-filters.

Figures

Equations

$$\displaystyle\begin{array}[]{l|lllllll}\odot&0&a&b&c&d&e&1\\ \hline\cr 0&0&0&0&0&0&0&0\\ a&0&a&a&a&a&a&a\\ b&0&a&b&a&b&a&b\\ c&0&a&a&a&a&c&c\\ d&0&a&b&a&b&c&d\\ e&0&a&a&c&c&e&e\\ 1&0&a&b&c&d&e&1\end{array}$$

$$X^{\perp }=\bigcap \{\mathfrak{m}\in Min(\mathfrak{A})|X⊈\mathfrak{m}\}.$$

$$\begin{array}[]{ll}F\cap(\vee^{\alpha}\mathcal{G})&=\alpha(F\cap(\veebar\mathcal{G}))\\ &=\alpha(\veebar_{G\in\mathcal{G}}(F\cap G))\\ &=\vee^{\alpha}_{G\in\mathcal{G}}(F\cap G).\end{array}$$

$$\begin{array}[]{ll}\alpha(F,x)\cap\alpha(F,y)&=\alpha(\mathscr{F}(F,x))\cap\alpha(\mathscr{F}(F,y))\\ &=\alpha(\mathscr{F}(F,x)\cap\mathscr{F}(F,y))\\ &=\alpha(\mathscr{F}(F,x\vee y)\\ &=\alpha(F,x\vee y).\end{array}$$

$$\begin{array}[]{ll}\alpha(F,x)\vee^{\alpha}\alpha(F,y)&=\alpha(\mathscr{F}(F,x))\vee^{\alpha}\alpha(\mathscr{F}(F,y))\\ &=\alpha(\mathscr{F}(F,x)\veebar\mathscr{F}(F,y))\\ &=\alpha(\mathscr{F}(F,x\odot y))\\ &=\alpha(F,x\odot y).\end{array}$$

$$\begin{array}[]{ll}\begin{array}[]{cclc}\Phi:&\mathscr{F}(\mathfrak{A})&\longrightarrow&\mathscr{F}(\gamma(\mathfrak{A}))\\ &F&\longmapsto&\{x^{\perp}|x\in F\},\end{array}&\begin{array}[]{cclc}\Psi:&\mathscr{F}(\gamma(\mathfrak{A}))&\longrightarrow&\mathscr{F}(\mathfrak{A})\\ &F&\longmapsto&\{x\in A|x^{\perp}\in F\}.\end{array}\end{array}$$

$$\begin{array}[]{ll}\Psi\Phi(F)&=\Psi(\Phi(F))\\ &=\{a\in A|a^{\perp}\in\Phi(F)\}\\ &=\{a\in A|a^{\perp}=x^{\perp},\exists x\in F\}\\ &=\{a\in A|a\in x^{\perp\perp},\exists x\in F\}=\cup_{x\in F}x^{\perp\perp}=\alpha(F).\\ \end{array}$$

$$\begin{array}[]{ll}a_{x}\vee a_{y}&\in(p_{x}\odot x^{n})^{\perp\perp}\cap(p_{y}\odot y^{m})^{\perp\perp}\\ &=((p_{x}\odot x^{n})\vee(p_{y}\odot y^{m}))^{\perp\perp}\\ &\subseteq((p_{x}\vee p_{y})\odot(p_{x}\vee y^{m}))\odot(x^{n}\vee p_{y})\odot(x\vee y)^{nm})^{\perp\perp}\\ &\subseteq P.\end{array}$$

Full text

Key words and phrases: quasicomplemented residuated lattice; disjunctive residuated lattice; $\alpha$-filter.

1 Introduction

As a generalization of distributive pseudo-complemented lattices, Varlet (1963) studied lattices which are just pseudo-complemented. He observed that a distributive lattice is pseudo-complemented if and only if each of its annulet is principal. By this motivation, Varlet (1968) introduced the notion of quasi-complemented lattices as a generalization of distributive pseudo-complemented lattices. Also, Speed (1969a) introduced the class of $\star$-lattices as a subclass of distributive lattices. Speed (1969b, Proposition 3.4) proved that these two classes are equivalent. Quasi-complemented lattices are studied extensively by Cornish (1972, 1973), Jayaram (1986), Speed (1974). Also, this notion is discussed for rings in Knox, et al. (2009). In this paper, we introduce the notion of quasicomplemented residuated lattice and generalize some results of Cornish (1973) and Speed (1969b) in this class of algebras.

This paper is organized in five sections as follow: In Sec. 2, some definitions and facts about residuated lattices are recalled and some of their propositions are proved. In Sec. 3, the notion of quasicomplemented residuated lattices, as a subclass of residuated lattices, is introduced and some of their properties are investigated. In Sec. 4, notions of disjunctive and weakly disjunctive residuated lattices are introduced and some of their characterizations are derived. It is proved that the lattice of a residuated lattice principal filters is Boolean if and only if it is weakly disjunctive and quasicomplemented. In Sec. 5, the notion of $\alpha$-filters is introduced and some of their properties are studied. Weakly disjunctive residuated lattices are characterized in terms of $\alpha$-filters and it is shown that a residuated lattice is quasicomplemented if and only if any its prime $\alpha$-filter is a minimal prime filter. We end this paper by deriving a set of equivalent conditions for any $\alpha$-filter to be principal.

2 Definitions and first properties

In this section, we recall some definitions, properties and results relative to residuated lattices, which will be used in the following.

An algebra $\mathfrak{A}=(A;\vee,\wedge,\odot,\rightarrow,0,1)$ is called a residuated lattice if $\ell(\mathfrak{A})=(A;\vee,\wedge,0,1)$ is a bounded lattice, $(A;\odot,1)$ is a commutative monoid and $(\odot,\rightarrow)$ is an adjoint pair. In a residuated lattice $\mathfrak{A}$, for any $a\in A$, we put $\neg a:=a\rightarrow 0$ and for any integer $n$ we right $x^{n}$ instead of $x\odot\cdots\odot x$ ($n$ times). An element $\mathfrak{A}$ is called nilpotent if $x^{n}=0$ for an integer $n$. The set of nilpotent elements of $\mathfrak{A}$ shall be denoted by $N(\mathfrak{A})$. It is well-known that $N(\mathfrak{A})$ is an ideal of $\ell(\mathfrak{A})$. Idziak (1984) showed that the class of residuated lattices is equational, and so it forms a variety. The properties of residuated lattices were presented in Galatos et al. (2007). For a survey of residuated lattices we refer to Jipsen and Tsinakis (2002).

Remark 1*.*

(Jipsen and Tsinakis, 2002, Proposition 2.2) Let $\mathfrak{A}$ be a residuated lattice. The following conditions are satisfied for any $x,y,z\in A$:

1. $r_{1}$

$x\odot(y\vee z)=(x\odot y)\vee(x\odot z)$; 2. $r_{2}$

$x\vee(y\odot z)\geq(x\vee y)\odot(x\vee z)$.

Let $\mathfrak{A}$ be a residuated lattice. The set of all complemented elements in $\ell(\mathfrak{A})$ is denoted by $B(\mathfrak{A})$ and it is called the Boolean center of $\mathfrak{A}$. For a survey of residuated lattices we refer to Ciungu (2009).

Proposition 2.1**.**

Ciungu (2009)** Let $\mathfrak{A}$ be a residuated lattice. The following assertions hold for any $e\in B(\mathfrak{A})$ and $a\in A$:

$e^{c}=\neg e$; 2. 2.

$e^{n}=e$, for each integer $n$; 3. 3.

$e\odot a=a\odot e=e\wedge a$; 4. 4.

$\neg\neg e=e$.

Example 2.2**.**

Let $A_{7}=\{0,a,b,c,d,e,1\}$ be a lattice whose Hasse diagram is below (see Figure 1). Define $\odot$ and $\rightarrow$ on $A_{7}$ as follows:

[TABLE]

Routine calculation shows that $\mathfrak{A}_{7}=(A_{7};\vee,\wedge,\odot,\rightarrow,0,1)$ is a residuated lattice.

Let $\mathfrak{A}$ be a residuated lattice. A non-void subset $F$ of $A$ is called a filter of $\mathfrak{A}$ if $x,y\in F$ implies $x\odot y\in F$ and $x\vee y\in F$ for any $x\in F$ and $y\in A$. The set of filters of $\mathfrak{A}$ is denoted by $\mathscr{F}(\mathfrak{A})$. A filter $F$ of $\mathfrak{A}$ is called proper if $F\neq A$. Clearly, $F$ is a proper filter if and only if $0\notin F$. For any subset $X$ of $A$ the filter of $\mathfrak{A}$ generated by $X$ is denoted by $\mathscr{F}(X)$. For each $x\in A$, the filter generated by $\{x\}$ is denoted by $\mathscr{F}(x)$ and called principal filter. The set of principal filters is denoted by $\mathscr{PF}(\mathfrak{A})$. Let $\mathcal{F}$ be a collection of filters of $\mathfrak{A}$. Set $\veebar\mathcal{F}=\mathscr{F}(\cup\mathcal{F})$. It is well-known that $(\mathscr{F}(\mathfrak{A});\cap,\veebar,\textbf{1},A)$ is a frame and so it is a complete Heyting algebra.

Example 2.3**.**

Consider the residuated lattice $\mathfrak{A}_{7}$ from Example 2.2. Then $\mathscr{F}(\mathfrak{A}_{7})=\{F_{1}=\{1\},F_{2}=\{b,d,1\},F_{3}=\{e,1\},F_{4}=\{a,b,c,d,e,1\},F_{5}=A_{7}\}$.

The following remark has a routine verification.

Remark 2*.*

Let $\mathfrak{A}$ be a residuated lattice and $F$ be a filter of $\mathfrak{A}$. The following assertions hold for any $x,y\in A$:

1. $(1)$

$\mathscr{F}(F,x):=F\veebar\mathscr{F}(x)=\{a\in A|f\odot x^{n}\leq a,~{}f\in F\}$; 2. $(2)$

$x\leq y$ implies $\mathscr{F}(F,y)\subseteq\mathscr{F}(F,x)$. 3. $(3)$

$\mathscr{F}(F,x)\cap\mathscr{F}(F,y)=\mathscr{F}(F,x\vee y)$; 4. $(4)$

$\mathscr{F}(F,x)\veebar\mathscr{F}(F,y)=\mathscr{F}(F,x\odot y)$; 5. $(5)$

$\mathscr{PF}(\mathfrak{A})$ is a sublattice of $\mathscr{F}(\mathfrak{A})$; 6. $(6)$

$\mathscr{F}(e)=\{a\in A|e\leq a\}$, for any $e\in B(\mathfrak{A})$.

A proper filter of a residuated lattice $\mathfrak{A}$ is called maximal if it is a maximal element in the set of all proper filters. The set of all maximal filters of $\mathfrak{A}$ is denoted by $Max(\mathfrak{A})$. A proper filter $P$ of $\mathfrak{A}$ is called prime, if for any $x,y\in A$, $x\vee y\in P$ implies $x\in P$ or $y\in P$. The set of all prime filters of $\mathfrak{A}$ is denoted by $Spec(\mathfrak{A})$. Since $\mathscr{F}(\mathfrak{A})$ is a distributive lattice, so $Max(\mathfrak{A})\subseteq Spec(\mathfrak{A})$. By Zorn’s lemma follows that any proper filter is contained in a maximal filter and so in a prime filter. A non-empty subset $\mathscr{C}$ of $\mathfrak{A}$ is called $\vee$-closed if it is closed under the join operation, i.e $x,y\in\mathscr{C}$ implies $x\vee y\in\mathscr{C}$.

Theorem 2.4**.**

(Rasouli and Kondo, 2018, Theorem 2.4)* If $\mathscr{C}$ is a $\vee$-closed subset of $\mathfrak{A}$ which does not meet the filter $F$, then $F$ is contained in a filter $P$ which is maximal with respect to the property of not meeting $\mathscr{C}$; furthermore $P$ is prime.*

Let $\mathfrak{A}$ be a residuated lattice and $X$ be a subset of $A$. A prime filter $P$ is called a minimal prime filter belonging to $X$ or $X$-minimal prime filter if $P$ is a minimal element in the set of prime filters containing $X$. The set of $X$-minimal prime filters of $\mathfrak{A}$ is denoted by $Min_{X}(\mathfrak{A})$. A prime filter $P$ is called a minimal prime if $P\in Min_{\{1\}}(\mathfrak{A})$. The set of minimal prime filters of $\mathfrak{A}$ is denoted by $Min(\mathfrak{A})$. For the basic facts concerning minimal prime filters of a residuated lattice belonging to a filter we refer to Rasouli and Kondo (2018).

Theorem 2.5**.**

(Rasouli and Kondo, 2018, Theorem 2.6)* Let $\mathfrak{A}$ be a residuated lattice. A subset $P$ of $A$ is a minimal prime filter if and only if $P^{c}$ is a $\vee$-closed subset of $\mathfrak{A}$ which it is maximal with respect to the property of not containing $1$.*

Let $\mathfrak{A}$ be a residuated lattice. For a subset $\mathfrak{M}$ of $Min(\mathfrak{A})$ we write $k(\mathfrak{M})=\bigcap\{\mathfrak{m}|\mathfrak{m}\in\mathfrak{M}\}$ and for a subset $X$ of $A$ we write $h(X)=\{\mathfrak{m}|X\subseteq\mathfrak{m}\}$. Also, let $d(X)=Min(\mathfrak{A})\setminus h(X)$. If the collection $\{h(x)|x\in A\}$ is taken as a closed basis, the resulting topology is called the hull-kernel topology which is denoted by $\tau_{h}$, and if the collection $\{h(x)|x\in A\}$ is taken as an open basis, the resulting topology is called the dual hull-kernel topology which is denoted by $\tau_{d}$. For a detailed discussion of spaces of minimal prime filters in residuated lattices we refer to Rasouli and Dehghani (2018).

Corollary 2.6**.**

(Rasouli and Dehghani, 2018, Corollary 4.4) Let $\mathfrak{A}$ be a residuated lattice. The following assertions hold:

1. (1)

$(Min(\mathfrak{A});\tau_{h})$ is zero-dimensional and consequently totally disconnected; 2. (2)

$\tau_{d}$ is finer than $\tau_{h}$.

Let $\mathfrak{A}$ be a residuated lattice. For any subset $X$ of $A$ we write $X^{\perp}=\{a\in A|a\vee x=1,\forall x\in X\}$. We set $\Gamma(\mathfrak{A})=\{X^{\perp}|X\subseteq A\}$ and $\gamma(\mathfrak{A})=\{x^{\perp}|x\in A\}$. Elements of $\Gamma(\mathfrak{A})$ and $\gamma(\mathfrak{A})$ are called coannihilators and coannulets, respectively. By Rasouli (2018, Proposition 3.13) follows that $(\Gamma(\mathfrak{A});\cap,\vee^{\Gamma},\{1\},A)$ is a complete Boolean lattice, where for any $\mathscr{G}\subseteq\Gamma(\mathfrak{A})$ we have $\vee^{\Gamma}\mathscr{G}=(F:(F:\cup\mathscr{G}))$, and by Rasouli (2018, Corollary 3.14) follows that $\gamma(\mathfrak{A})$ is a sublattice of $\Gamma(\mathfrak{A})$. A subset $X$ of $A$ is called dense if $X^{\perp}=\{1\}$. The set of all dense elements of $\mathfrak{A}$ shall be denoted by $\mathfrak{d}(\mathfrak{A})$. It is well-known that $\mathfrak{d}(\mathfrak{A})$ is an ideal of $\ell(\mathfrak{A})$.

Proposition 2.7**.**

(Rasouli, 2018)* Let $\mathfrak{A}$ be a residuated lattice. The following assertions hold for any $X,Y\subseteq A$ and $F,G\in\mathscr{F}(\mathfrak{A})$:*

1. $(1)$

$X\subseteq Y^{\perp}$* implies $Y\subseteq X^{\perp}$;* 2. $(2)$

$X^{\perp}\cap X^{\perp\perp}=\{1\}$; 3. $(3)$

$X\subseteq X^{\perp\perp}$; 4. $(4)$

$(\mathscr{F}(X))^{\perp}=X^{\perp}$;

Proposition 2.8**.**

Let $\mathfrak{A}$ be a residuated lattice and $F$ be a filter of $\mathfrak{A}$. Then $F^{\perp}$ is the pseudocomplement of $F$.

Proof.

By Proposition 2.7, it follows that $F\cap F^{\perp}=\{1\}$. Assume that $F\cap G=\{1\}$ for some filter $G$. Let $a\in G$ and $b\in F$. Since $a,b\leq a\vee b$ so $a\vee b\in F\cap G=\{1\}$. Thus $a\in F^{\perp}$ and it shows that $G\subseteq F^{\perp}$. ∎

Corollary 2.9**.**

(Rasouli and Dehghani, 2018, Corollary 2.11) Let $\mathfrak{A}$ be a residuated lattice. Then, for any subset $X$ of $A$, we have

[TABLE]

Proposition 2.10**.**

(Rasouli, 2018, Proposition 3.15)* Let $\mathfrak{A}$ be a residuated lattice. The following assertions hold for any $x,y\in A$:*

1. $(1)$

$x\leq y$* implies $x^{\perp}\subseteq y^{\perp}$;* 2. $(2)$

$x^{\perp}\cap y^{\perp}=(x\odot y)^{\perp}$; 3. $(3)$

$x^{\perp\perp}\cap y^{\perp\perp}=(x\vee y)^{\perp\perp}$; 4. $(4)$

$x^{\perp}\veebar y^{\perp}\subseteq x^{\perp}\vee^{\Gamma}y^{\perp}=(x\vee y)^{\perp}$; 5. $(5)$

$e^{\perp}=\mathscr{F}(\neg e)$, for any $e\in B(\mathfrak{A})$.

Proposition 2.11**.**

Let $\mathfrak{A}$ be a residuated lattice. Then any non-dense prime filter of $\mathfrak{A}$ is a coannulet.

Proof.

Let $P$ be a non-dense prime filter of $\mathfrak{A}$. So $P^{\perp}\neq\textbf{1}$. So there exists $1\neq x\in P^{\perp}$. Thus $P\subseteq P^{\perp\perp}\subseteq x^{\perp}$. Otherwise, $y\in x^{\perp}$ implies that $x\vee y=1\in P$. But $x\notin P$ since $x\in P$ states that $x\in P\cap P^{\perp}=\{1\}$ and it means that $x=1$ which it is a contradiction. So $y\in P$ and it shows that $P=x^{\perp}$. ∎

Let $\mathfrak{A}$ be a residuated lattice. For any ideal $I$ of $\ell(\mathfrak{A})$ we write $\omega(I)=\{a\in A|a\vee x=1,\exists x\in I\}$. We set $\Omega(\mathfrak{A})=\{\omega(I)|I\textrm{~{}is~{}an~{}ideal~{}of~{}}\ell(\mathfrak{A})\}$. By Rasouli and Kondo (2018, Proposition 3.3) follows that $(\Omega(\mathfrak{A});\cap,\vee^{\omega},\{1\},A)$ is a bounded distributive lattice, where $F\vee^{\omega}G=\omega(I_{F}\curlyvee I_{G})$, for any $F,G\in\Omega(\mathfrak{A})$ (by $\curlyvee$, we mean the join operation in the lattice of ideals of $\ell(\mathfrak{A})$). Also, by Rasouli and Kondo (2018, Proposition 3.8) follows that $\gamma(\mathfrak{A})$ is a sublattice of $\Omega(\mathfrak{A})$. For a prime filter $P$ of $\mathfrak{A}$, we write $D(P)=\omega(P^{c})$. The following corollary is a characterization for minimal prime filters.

Proposition 2.12**.**

A proper $\omega$-filter in a residuated lattice $\mathfrak{A}$ contains no dense elements.

Proof.

Let $F$ be a proper $\omega$-filter. Assume that $F$ contains a dense element as $d$. Hence $d\in x^{\perp}$ for some $x\in I_{F}$. It implies that $x\in d^{\perp}$ and it means $1\in I_{F}$. So $F=A$; a contradiction. ∎

Theorem 2.13**.**

(Rasouli and Kondo, 2018, Proposition 3.14)* Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:*

1. $(1)$

$P$* is a minimal prime filter;* 2. $(2)$

$P=D(P)$; 3. $(3)$

for any $x\in A$, $P$ contains precisely one of $x$ or $x^{\perp}$.

Corollary 2.14**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent for any $x,y\in A$:

1. $(1)$

$x^{\perp}=y^{\perp}$; 2. $(2)$

$h(x)=h(y)$; 3. $(3)$

$d(x)=d(y)$.

Proof.

Let $\mathfrak{m}\in h(x)$. By Theorem 2.13$(3)$ follows that $y^{\perp}\nsubseteq\mathfrak{m}$ and so $y\in\mathfrak{m}$. It implies that $\mathfrak{m}\in h(y)$. Thus we have $h(x)\subseteq h(y)$. The other inclusion is analogous by symmetry.

It is evident.

by Corollary 2.9 follows that $x^{\perp}=\bigcap d(x)=\bigcap d(y)=y^{\perp}$. ∎

Corollary 2.15**.**

(Rasouli and Kondo, 2018, Proposition 3.21) Let $\mathfrak{A}$ be a residuated lattice. We have $D(P)=\bigcap\{\mathfrak{m}\in Min(\mathfrak{A})|\mathfrak{m}\subseteq P\}$.

3 Quasicomplemented residuated lattices

In this section we introduce and study the notion of quasicomplemented residuated lattices.

Definition 3.1**.**

Let $\mathfrak{A}$ be a residuated lattice. $\mathfrak{A}$ is called quasicomplemented provided that for any $x\in A$, there exists $y\in A$ such that $x^{\perp\perp}=y^{\perp}$.

Proposition 3.2**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. $(1)$

$\mathfrak{A}$* is quasicomplemented;* 2. $(2)$

for any $x\in A$, there exists $y\in A$ such that $x\odot y\in\mathfrak{d(A)}$ and $x\vee y=1$; 3. $(3)$

$\gamma(\mathfrak{A})$* is a Boolean lattice.*

Proof.

1. $(1)$$\Rightarrow$$(2)$: Consider $x\in A$. So there exists $y\in A$ such that $x^{\perp\perp}=y^{\perp}$. Applying Proposition 2.7$(2)$ and 2.10$(2)$, it follows that $x\odot y$ is a dense element and Proposition 2.7$(2)$ shows that $x\in x^{\perp\perp}=y^{\perp}$ and so $x\vee y=1$.
2. $(2)$$\Rightarrow$$(3)$: By applying Proposition 2.10($(2)$ and $(4)$), it is evident.
3. $(3)$$\Rightarrow$$(1)$: Let $x\in A$. So there exists some $y\in A$ such that $x^{\perp}\cap y^{\perp}=\{1\}$ and $x^{\perp}\vee^{\Gamma}y^{\perp}=A$. Applying Proposition 2.8, the former states $y^{\perp}\subseteq x^{\perp\perp}$, and the latter states the reverse inclusion.

∎

In the following, we derive a sufficient condition for a residuated lattice to become quasicomplemented.

Proposition 3.3**.**

Let $\mathfrak{A}$ be a residuated lattice. $\mathfrak{A}$ is quasicomplemented provided that in which any coannulet is principal.

Proof.

Consider $x\in A$. So there exist $a\in A$ such that $x^{\perp}=\mathscr{F}(y)$. Using Proposition 2.7$(4)$, it follows that $x^{\perp\perp}=y^{\perp}$ and so $\mathfrak{A}$ is quasicomplemented. ∎

In the following proposition, we derive a necessary and sufficient condition for any residuated lattice to become quasicomplemented.

Proposition 3.4**.**

Let $\mathfrak{A}$ be a residuated lattice and $F$ be a filter of $\mathfrak{A}$. The following assertions are equivalent:

1. $(1)$

$\mathfrak{A}$* is quasicomplemented;* 2. $(2)$

any prime filter not containing any dense element is minimal prime; 3. $(3)$

any filter not containing any dense element is contained in a minimal prime filter.

Proof.

1. $(1)$$\Rightarrow$$(2)$: Let $P$ be a prime filter such that $P\cap\mathfrak{d}(\mathfrak{A})=\emptyset$. Consider $x\in P$. Applying Proposition 3.2$(2)$, there exists $y\in A$ such that $x\odot y$ is dense and $x\vee y=1$. It shows that $x\odot y\notin P$ and so $y\notin P$. Hence, $x\in D(P)$ and it states that $P=D(P)$. So the result holds by Theorem 2.13.
2. $(2)$$\Rightarrow$$(3)$: It follows by Theorem 2.4.
3. $(3)$$\Rightarrow$$(1)$: Let $x\in A$. By Theorem 2.13$(3)$ follows that $\mathscr{F}(x)\veebar x^{\perp}$ cannot be contained in any minimal prime filter and so it contains a dense element like $d$. Hence, there are $a\in\mathscr{F}(x)$ and $b\in x^{\perp}$ such that $a\odot b\leq d$. So for some integer $n$ follows that $x^{n}\odot b$ is dense. Let $u\in b^{\perp}$ and $v\in x^{\perp}$. Thus we have $(u\vee v)\vee b=1$ and $(u\vee v)\vee x^{n}=1$, and by using $r_{2}$ we deduce that $(u\vee v)\vee(x^{n}\odot b)=1$. It shows that $u\vee v=1$ and it means that $b^{\perp}\subseteq x^{\perp\perp}$. The other inclusion is evident by Proposition 2.7$(1)$, and so the result holds.

∎

Quasicomplemented residuated lattices are characterized under the name of $\star$-residuated lattices in Rasouli and Dehghani (2018). In the following theorem, we give a topological characterization for quasicomplemented residuated lattice.

Theorem 3.5**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. (1)

$\mathfrak{A}$* is quasicomplemented;* 2. (2)

$\tau_{h}$* and $\tau_{d}$ coincide;* 3. (3)

$(Min(\mathfrak{A});\tau_{h})$* is compact.*

Proof.

It follows by Rasouli and Dehghani (2018, Theorem 4.6). ∎

4 Disjunctive residuated lattices

Speed (1969b) introduced a certain class of distributive lattices with zero named disjunctive lattices. This notion has been discussed in semilattices by Büchi (1948) and in commutative semigroups by Kist (1963). Cornish (1972, Theorem 7.6) proved that if $\mathfrak{A}$ is a disjunctive normal lattice and $Max(\mathfrak{A})$, the space of maximal filters of $\mathfrak{A}$ with the hull-kernel topology, is a compact Hausdorff totally disconnected space, then $\mathfrak{A}$ is complementedly normal. Also, Cornish (1973, Proposition 2.3) showed that a disjunctive normal lattice is dual isomorphic to its lattice of annulets. Actually, disjunctive lattices are themselves important in the study of annulets; information can be obtained by dualizing Banaschewski’s results in (Banaschewski, 1964, Section 4). In this section, we introduce and study notions of disjunctive and weakly disjunctive residuated lattice.

Let $\mathfrak{A}$ be a residuated lattice. We set $D(\mathfrak{A})=\{d(x)|x\in A\}$ and $H(\mathfrak{A})=\{h(x)|x\in A\}$. By Rasouli and Dehghani (2018, Proposition 3.6 and 3.10), it follows that $(D(\mathfrak{A});\cap,\cup,d(1)=\emptyset,d(0)=Min(\mathfrak{A}))$ and $(H(\mathfrak{A});\cap,\cup,h(0)=\emptyset,h(1)=Min(\mathfrak{A}))$ are bounded lattices. Consider the following diagram;

Recalling that, if $\mathfrak{A}$ and $\mathfrak{B}$ are two algebras of a same type and $f:\mathfrak{A}\longrightarrow\mathfrak{B}$ is a homomorphism, then $\kappa(f)=\{(a_{1},a_{2})\in A^{2}|f(a_{1})=f(a_{2})\}$ is a congruence relation on $\mathfrak{A}$. The following remark has a routine verification.

Remark 3*.*

1. (1)

By Proposition 2.7$(4)$ and Corollary 2.14 follows that $f_{3},f_{4},f_{5}$ and $f_{6}$ are well-define. 2. (2)

$f_{2}=f_{4}f_{1}$ and $f_{5}f_{3}=f_{6}f_{4}$. 3. (3)

$f_{2},f_{3}$ are lattice epimorphisms and $f_{1},f_{4}$ are dual lattice epimorphisms. 4. (4)

$f_{5}$ is a dual lattice isomorphism and $f_{6}$ is a lattice isomorphism. 5. (5)

$\Re:=\kappa(f_{3})=\kappa(f_{4})$, so $f_{3}$ is injective if and only if $f_{4}$ is injective. 6. (6)

$f_{2}$ is injective if and only if $f_{1},f_{4}$ are injective. 7. (7)

$\mathscr{F}(1)/\Re=\{\mathscr{F}(1)\}$ and $\mathscr{F}(0)/\Re=\{\mathscr{F}(x)|x\in\mathfrak{d(A)}\}$. 8. (8)

$1/\kappa(f_{2})=\{1\}$ and $0/\kappa(f_{2})=\mathfrak{d(A)}$. 9. (9)

$1/\kappa(f_{1})=\{1\}$ and $0/\kappa(f_{1})=N(\mathfrak{A})$.

Proposition 4.1**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. (1)

$\mathfrak{A}$* is quasicomplemented;* 2. (2)

$\ell(\mathfrak{A})/\kappa(f_{2})$* is a Boolean lattice;* 3. (3)

$\mathscr{PF}(\mathfrak{A})/\Re$* is a Boolean lattice.*

Proof.

It is an immediate consequence of Proposition 3.2 and Remark 3. ∎

Definition 4.2**.**

Let $\mathfrak{A}$ be a residuated lattice. With notations of Figure 2, $\mathfrak{A}$ is called disjunctive provided that $f_{2}$ is injective and weakly disjunctive provided that $f_{3}$ (or, equivalently $f_{4}$) is injective.

By Remark 3(6), it is evident that if a residuated lattice is disjunctive, then it is weakly disjunctive. In the following proposition the interrelation between the subclasses of quasicomplemented and disjunctive residuated lattices is given (See Fig. 3).

Proposition 4.3**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. (1)

$\mathfrak{A}$* is quasicomplemented and disjunctive;* 2. (2)

$\ell(\mathfrak{A})$* is a Boolean lattice;* 3. (3)

$\mathfrak{A}$* is quasicomplemented, $\mathfrak{d(A)}=\{0\}$ and the operation $\neg$ is injective as a function.*

Proof.

It follows by Proposition 4.1.

Applying Proposition 2.1, it follows that the operation $\neg$ is injective as a function. By Remark 2$(6)$ and Proposition 2.10$(5)$, it follows that $\mathfrak{d(A)}=\{0\}$. Also, by Proposition 3.2 follows that $\mathfrak{A}$ is quasicomplemented.

Let $x^{\perp}=y^{\perp}$. So we have $(x\odot\neg y)^{\perp}=\{1\}$ and it implies that $\neg y\leq\neg x$. Analogously, we can conclude that $\neg x\leq\neg y$ and it implies that $\neg x=\neg y$. Since $\neg$ is an injective operation so the result holds. ∎

Remark 4*.*

According to (Ciungu, 2009, Corollary 3.2) follows that a residuated lattice $\mathfrak{A}$ is Boolean if and only if for any $a\in A$ we have $a\wedge\neg a=0$ and $a\vee\neg a=1$. It gives a new characterisation for quasicomplemented and disjunctive residuated lattices.

Proposition 4.4**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. (1)

$\mathfrak{A}$* is quasicomplemented and weakly disjunctive;* 2. (2)

$\mathscr{PF}(\mathfrak{A})$* is a Boolean lattice.*

Proof.

It follows by Proposition 4.1.

Let $(\mathscr{F}(x))^{c}=\mathscr{F}(y)$. So we have $\mathscr{F}(x)\cap\mathscr{F}(y)=1$ and $\mathscr{F}(x)\veebar\mathscr{F}(y)=A$. The former implies $\mathscr{F}(y)\subseteq x^{\perp}$ and so the latter implies $\mathscr{F}(x)\veebar x^{\perp}=A$. It shows that $\mathscr{F}(y)=x^{\perp}$. So $\mathfrak{A}$ is quasicomplemented since $x^{\perp\perp}=y^{\perp}$ and $\mathfrak{A}$ is weakly disjunctive since $x^{\perp}=y^{\perp}$ implies $(\mathscr{F}(x))^{c}=(\mathscr{F}(y))^{c}$ and so $\mathscr{F}(x)=\mathscr{F}(y)$. ∎

Remark 5*.*

Let $\mathfrak{A}$ be a residuated lattice. Applying Proposition 2, it is easy to see that $\mathscr{PF}(\mathfrak{A})$ is a Boolean lattice if and only if for any $a\in A$ there exists $b\in a^{\perp}$ such that $a\odot b\in N(\mathfrak{A})$. It gives a new characterisation for quasicomplemented and weakly disjunctive residuated lattices.

5 $\alpha$-filters

The notion of $\alpha$-ideals introduced by Cornish (1973) in distributive lattice with [math]. Jayaram (1986) generalized the concept of $\alpha$-ideals to [math]-distributive lattices. Some further properties of $\alpha$-ideals for [math]-distributive lattices were obtained by Pawar and Mane (1993), Pawar and Khopade (2010). Haveshki and Mohamadhasani (2015) proposed the concept of $\alpha$-filters in BL-algebras as the dual notion of $\alpha$-ideals. Recently, Dong and Xin (2018) extend the concept of $\alpha$-filters to residuated lattices. In this section we introduce and study the notion of $\alpha$-filter in residuated lattices.

Definition 5.1**.**

Let $\mathfrak{A}$ be a residuated lattice. A filter $F$ of $\mathfrak{A}$ is called an $\alpha$-filter if for any $x\in F$ we have $x^{\perp\perp}\subseteq F$. The set of $\alpha$-filters of $\mathfrak{A}$ is denoted by $\alpha(\mathfrak{A})$. It is obvious that $\{1\},A\in\alpha(\mathfrak{A})$.

Example 5.2**.**

Consider the residuated lattice $\mathfrak{A}_{7}$ from Example 2.2. With notations of Example 2.3, $F_{2}$ and $F_{3}$ are $\alpha$-filters of $\mathfrak{A}_{7}$.

Let $\mathfrak{A}$ be a residuated lattice. It is obvious that $\alpha(\mathfrak{A})$ is an algebraic closed set system on $\mathfrak{A}$. The closure operator associated with this closed set system is denoted by $\alpha^{\mathfrak{A}}:\mathcal{P}(A)\longrightarrow\mathcal{P}(A)$. Thus for any subset $X$ of $A$, $\alpha^{\mathfrak{A}}(X)=\cap\{F\in\alpha(\mathfrak{A})|X\subseteq F\}$ is the smallest $\alpha$-filter of $\mathfrak{A}$ contains $X$ which it is called the $\alpha$-filter of $\mathfrak{A}$ generated by $X$. When there is no ambiguity we will drop the superscript $\mathfrak{A}$. Hence $\alpha(\mathfrak{A})$ is a complete compactly generated lattice where the infimum is the set-theoretic intersection and the supremum of $\mathcal{F}\subseteq\alpha(\mathfrak{A})$ is $\vee^{\alpha}\mathcal{F}=\alpha(\cup\mathcal{F})$. It is obvious that $\alpha(X)=\alpha(\mathscr{F}(X))$ for any $X\subseteq A$ and so we have $\vee^{\alpha}\mathcal{F}=\alpha(\veebar\mathcal{F})$.

Proposition 5.3**.**

For any residuated lattice $\mathfrak{A}$, $(\alpha(\mathfrak{A});\cap,\vee^{\alpha})$ is a frame.

Proof.

We know that $\alpha(\mathfrak{A})$ is a complete lattice. Let $\{F\}\cup\mathcal{G}$ be a family of $\alpha$-filters. We have the following sequence of formulas:

[TABLE]

It shows that $\alpha(\mathfrak{A})$ is a frame. ∎

It is well-known that a lattice is a frame if and only if it is a complete Heyting algebra. So due to Proposition 5.3, we deduce that for any residuated lattice $\mathfrak{A}$, $(\alpha(\mathfrak{A});\cap,\vee^{\alpha},\hookrightarrow,\textbf{1},A)$ is a Heyting algebra, where $F\hookrightarrow G=\veebar\{H\in\alpha(\mathfrak{A})|F\cap H\subseteq G\}$ for any $F,G\in\alpha(\mathfrak{A})$.

Proposition 5.4**.**

Let $\{F\}\cup\{F_{i}\}_{i\in I}$ be a family of filters in a residuated lattice $\mathfrak{A}$ and $x,y\in A$. The following assertions hold:

1. $(1)$

$\alpha(F)=\cup_{x\in F}x^{\perp\perp}$; 2. $(2)$

$\vee^{\alpha}_{i\in I}F_{i}=\{a\in A|a\in(f_{i_{1}}\odot\cdots\odot f_{i_{n}})^{\perp\perp},~{}\exists n\in\mathbb{N},~{}i^{n}_{1}\in I,~{}f_{i_{j}}\in F_{i_{j}}\}$; 3. $(3)$

$\alpha(F,x):=\alpha(F\cup\{x\})=\cup_{f\in F,n\in\mathbb{N}}(f\odot x^{n})^{\perp\perp}$; 4. $(4)$

$\alpha(F,x)\cap\alpha(F,y)=\alpha(F,x\vee y)$; 5. $(5)$

$\alpha(F,x)\vee^{\alpha}\alpha(F,y)=\alpha(F,x\odot y)$.

Proof.

It proves quite in a routine way.

We have $\vee^{\alpha}_{i\in I}F_{i}=\alpha(\veebar_{i\in I}F_{i})=\{a\in A|a\in x^{\perp\perp},~{}for~{}some~{}x\in\veebar_{i\in I}F_{i}\}$. Let $\Sigma=\{a\in A|a\in(\odot_{t=i_{1}}^{i_{n}}f_{t})^{\perp\perp},\exists n,i_{1},\cdots,i_{n}\in\mathbb{N}~{}f_{i_{j}}\in F_{i_{j}}\}$. It is obvious that $\Sigma\subseteq\vee^{\alpha}_{i\in I}F_{i}$. Suppose that $a\in\vee^{\alpha}_{i\in I}F_{i}$. Thus $a\in x^{\perp\perp}$ for some $x\in\veebar_{i\in I}F_{i}$. So there exist an integer $n$, $i_{1},\cdots,i_{n}\in I$ and $f_{i_{j}}\in F_{i_{j}}$ such that $f_{i_{1}}\odot\cdots\odot f_{i_{n}}\leq x$. By Proposition 2.10$(1)$ follows that $x^{\perp\perp}\subseteq(f_{i_{1}}\odot\cdots\odot f_{i_{n}})^{\perp\perp}$ and it means $a\in\Sigma$.

It follows by $(1)$.

By Remark 2$(3)$, we have the following formulas:

[TABLE]

By Remark 2$(4)$, we have the following formulas:

[TABLE]

∎

In the following proposition, we give some equivalent assertions for a filter to be an $\alpha$-filter.

Proposition 5.5**.**

Let $\mathfrak{A}$ be a residuated lattice and $F$ be a filter of $\mathfrak{A}$. The following assertions are equivalent:

1. $(1)$

$F$* is an $\alpha$-filter;* 2. $(2)$

if $x^{\perp}=y^{\perp}$ and $x\in F$, then $y\in F$ for any $x,y\in A$; 3. $(3)$

if $d(x)=d(y)$ and $x\in F$, then $y\in F$ for any $x,y\in A$; 4. $(4)$

if $h(x)=h(y)$ and $x\in F$, then $y\in F$ for any $x,y\in A$.

Proof.

By Corollary 2.14, it follows that $(1)$, $(2)$ and $(3)$ are equivalent. So we only prove the other cases.

$(1)$$\Rightarrow$$(2)$: Let $x^{\perp}=y^{\perp}$ for some $y\in A$ and $x\in F$. Then $y\in y^{\perp\perp}=x^{\perp\perp}\subseteq F$.

$(2)$$\Rightarrow$$(1)$: Let $x\in F$ and $y\in x^{\perp\perp}$. So $x^{\perp}\subseteq y^{\perp}$. By Proposition 2.10$(4)$ follows that $y^{\perp}=x^{\perp}\vee^{\Gamma}y^{\perp}=(x\vee y)^{\perp}$ and it states that $y\in F$ since $x\vee y\in F$. ∎

Let $\mathscr{A}=(A;\leq)$ and $\mathscr{B}=(B;\preccurlyeq)$ be two posts. We recall that a pair $(f,g)$ is called a an Adjunction (or isotone Galois connection) between posets $\mathscr{A}$ and $\mathscr{B}$, where $f:A\longrightarrow B$ and $g:B\longrightarrow A$ are two functions such that for all $a\in A$ and $b\in B$, $f(a)\leq b$ if and only if $a\preccurlyeq g(b)$. It is well known that $(f,g)$ is an adjunction connection if and only if $gf$ is inflationary, $fg$ is deflationary and $f,g$ are isotone (García-Pardo et al., 2013, Theorem 2). It is well-known $\mathscr{C}_{gf}=g(B)$, where $\mathscr{C}_{gf}$ is the set of fixed point of the closure operator $gf$.

Theorem 5.6**.**

Let $\mathfrak{A}$ be a residuated lattice. We define

[TABLE]

Then the pair $(\Phi,\Psi)$ is an adjunction and we have $\Psi\Phi(F)=\alpha(F)$ for any $F\in\mathscr{F}(\mathfrak{A})$. In particular we have $\mathscr{C}_{\Psi\Phi}=\alpha(\mathfrak{A})$.

Proof.

Quite in a routine way we can show that the pair $(\Phi,\Psi)$ forms an adjunction. Let $F$ be a filter of $\mathfrak{A}$. We have the following formulas:

[TABLE]

The rest is evident. ∎

The next theorem should be compared with Theorem 2.4.

Theorem 5.7**.**

Let $\mathscr{C}$ be a $\vee$-closed subset of $\mathfrak{A}$ which does not meet the $\alpha$-filter $F$. Then $F$ is contained in an $\alpha$-filter $P$ which is maximal with respect to the property of not meeting $\mathscr{C}$; furthermore $P$ is prime.

Proof.

Let $\Sigma=\{G\in\alpha(\mathfrak{A})|F\subseteq G,~{}G\cap\mathscr{C}=\emptyset\}$. It is easy to see that $\Sigma$ satisfies the conditions of Zorn’s lemma. Let $P$ be a maximal element of $\Sigma$. Assume that $x\vee y\in P$ and neither $x\notin P$ nor $y\notin P$. By maximality of $P$ we have $\alpha(P,x)\cap\mathscr{C}\neq\emptyset$ and $\alpha(P,y)\cap\mathscr{C}\neq\emptyset$. Suppose $a_{x}\in\alpha(P,x)\cap\mathscr{C}$ and $a_{y}\in\alpha(P,y)\cap\mathscr{C}$. By Proposition 5.4$(3)$, there exist $p_{x},p_{y}\in P$ and integers $n,m$ such that $a_{x}\in(p_{x}\odot x^{n})^{\perp\perp}$ and $a_{y}\in(p_{y}\odot y^{m})^{\perp\perp}$. It follows that

[TABLE]

It is a contradiction. So $x\in P$ or $y\in P$ and it shows that $P$ is a prime filter. ∎

In the sequel for any residuated lattice $\mathfrak{A}$ we set $Spec_{\alpha}(\mathfrak{A})=Spec(\mathfrak{A})\cap\alpha(\mathfrak{A})$.

Corollary 5.8**.**

Let $F$ be an $\alpha$-filter of $\mathfrak{A}$ and $X$ be a non-empty subset of $A$. The following assertions hold:

1. (1)

If $X\nsubseteq F$, then there exists $P\in Spec_{\alpha}(\mathfrak{A})$ such that $F\subseteq P$ and $P$ is maximal with respect to the property $X\nsubseteq P$; 2. (2)

$\alpha(X)=\bigcap\{P\in Spec_{\alpha}(\mathfrak{A})|X\subseteq P\}$.

Proof.

1. (1):

Let $x\in X-F$. By taking $\mathscr{C}=\{x\}$ it follows by Theorem 5.7. 2. (2):

Set $\sigma_{X}=\{P\in Spec_{\alpha}(\mathfrak{A})|X\subseteq P\}$. Obviously, we have $\alpha(X)\subseteq\bigcap\sigma_{X}$. Now, let $a\notin\alpha(X)$. By (1) follows that there exits an $\alpha$-prime filter $P$ containing $\alpha(X)$ such that $a\notin P$. It shows that $a\notin\bigcap\sigma_{X}$.

∎

In the following proposition we characterize weakly disjunctive residuated lattices by means of $\alpha$ filters.

Proposition 5.9**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. $(1)$

$\mathfrak{A}$* is weakly disjunctive;* 2. $(2)$

any filter of $\mathfrak{A}$ is an $\alpha$-filter; 3. $(3)$

any prime filter of $\mathfrak{A}$ is an $\alpha$-filter.

Proof.

It follows by Proposition 5.5$(2)$.

It is trivial.

Let $x^{\perp}=y^{\perp}$ and $\mathscr{F}(x)\neq\mathscr{F}(y)$ for some $x,y\in A$. Without loss of generality suppose that $\mathscr{F}(x)\nsubseteq\mathscr{F}(y)$. Let $\Sigma=\{F\in\mathscr{F}(\mathfrak{A})|y\in F,~{}x\notin F\}$. It is obvious that $\mathscr{F}(y)\in\Sigma$ and $\Sigma$ satisfies the conditions of Zorn’s lemma. Let $P$ be the maximal element of $\Sigma$. Let $a\vee b\in P$ and $a,b\notin P$. So $\mathscr{F}(P,a),\mathscr{F}(P,b)\notin\Sigma$ and it states that $x\in\mathscr{F}(P,a)\cap\mathscr{F}(P,b)$. By Remark 2$(3)$ follows that $x\in\mathscr{F}(P,a)\cap\mathscr{F}(P,b)=\mathscr{F}(P,a\vee b)=P$ and it leads us to a contradiction. Thus $P$ is a prime filter and it implies $x\in x^{\perp\perp}=y^{\perp\perp}\subseteq P$; a contradiction. ∎

In the following proposition we show that any coannihilator filter and any $\omega$-filter is an $\alpha$-filter.

Proposition 5.10**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions hold:

1. $(1)$

$\Gamma(\mathfrak{A})\subseteq\alpha(\mathfrak{A})$; 2. $(2)$

$\Omega(\mathfrak{A})\subseteq\alpha(\mathfrak{A})$.

Proof.

Let $F$ be a coannihilator filter and $x\in F$. So $x^{\perp\perp}\subseteq F^{\perp\perp}=F$ and it shows that $F$ is an $\alpha$-filter.

Let $F$ be an $\omega$-filter of $\mathfrak{A}$. So there exists a lattice ideal $I$ of $\mathfrak{A}$ such that $F=\omega(I)$. Consider $y\in F$. So there exists $x\in I$ such that $y\in x^{\perp}$ and it follows that $x\in y^{\perp}$. Assume $z\in y^{\perp\perp}$. Thus $y^{\perp}\subseteq z^{\perp}$ and it follows that $x\in z^{\perp}$. So we observe that $z\in x^{\perp}\subseteq F$. ∎

Corollary 5.11**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions hold:

1. $(1)$

Any non-dense prime filter is an $\alpha$-filter; 2. $(2)$

any minimal prime filter is an $\alpha$-filter.

Proof.

: It is an immediate consequence of Proposition 2.11 and Proposition 5.10$(1)$.

: It is an immediate consequence of Theorem 2.13 and Proposition 5.10$(2)$. ∎

In the following proposition we give some equivalent conditions for each $\alpha$-filter to be a coannihilator.

Proposition 5.12**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. $(1)$

Any proper $\alpha$-filter is non-dense; 2. $(2)$

any dense filter contains a dense element; 3. $(3)$

any $\alpha$-filter is a coannihilator; 4. $(4)$

for any proper $\alpha$-filter $F$ there exists a proper $\alpha$-filter $G$ such that $F\cap G=\{1\}$ ($\alpha(\mathfrak{A})$* is semi-complemented);* 5. $(5)$

$\alpha(\mathfrak{A})$* has a unique dense element.*

Moreover, any of the above assertions implies that $\mathfrak{A}$ is quasicomplemented.

Proof.

1. $(1)$$\Rightarrow$$(2)$: Let $F$ be a dense filter of $\mathfrak{A}$. It implies that $\alpha(F)$ is a dense filter and it means that $\alpha(F)=A$. Therefore $0\in\alpha(F)$ and it means that $x^{\perp}=1$ for some element $x\in F$.
2. $(2)$$\Rightarrow$$(3)$: Let $F$ be an $\alpha$-filter. $F,F^{\perp}\subseteq F\veebar F^{\perp}$ and so $(F\veebar F^{\perp})^{\perp}\subseteq F^{\perp}\cap F^{\perp\perp}=\textbf{1}$. So $F\veebar F^{\perp}$ is a dense filter and so it contains a dense element. Assume that $x$ is a dense element in $F\veebar F^{\perp}$. So there exist $a\in F$ and $b\in F^{\perp}$ such that $a\odot b\leq x$. It implies that $a^{\perp}\cap b^{\perp}\subseteq x^{\perp}=\textbf{1}$ and since $a^{\perp\perp}$ is the pseudo-complement of $a^{\perp}$ it follows that $b^{\perp}\subseteq a^{\perp\perp}\subseteq F$ as $F$ is an $\alpha$-filter. On the other hand $b\in F^{\perp}$ implies that $F^{\perp\perp}\subseteq b^{\perp}$ and it states that $F^{\perp\perp}\subseteq F$. The other inclusion is evident.
3. $(3)$$\Rightarrow$$(4)$: Since the set of all coannihilators forms a Boolean lattice, it follows that $\alpha(\mathfrak{A})$ is semi-complemented.
4. $(4)$$\Rightarrow$$(5)$: It is obvious that $A$ is a dense element of $\alpha(\mathfrak{A})$. Now, if $F$ is a proper dense $\alpha$-filter, then there exists a filter $G\neq\textbf{1}$ such that $F\cap G=\textbf{1}$ and it implies that $G\subseteq F^{\perp}=\textbf{1}$. So $G=\textbf{1}$ and it is a contradiction.
5. $(5)$$\Rightarrow$$(1)$: It is trivial.

Now, let $\mathfrak{A}$ satisfies $(1)$ and $x\in A$. Set $F=x^{\perp}\veebar x^{\perp\perp}$. It implies that $\alpha(F)^{\perp}\subseteq x^{\perp}\cap x^{\perp\perp}=\textbf{1}$ and it means that $\alpha(F)$ is a dense filter. So $\alpha(F)=A$ and it states that $0\in\alpha(F)$. It follows that $y^{\perp}=1$ for some element $y\in F$. Hence there exist $a\in x^{\perp}$ and $b\in x^{\perp\perp}$ such that $a\odot b\leq y$. Therefore $a^{\perp}\cap b^{\perp}=\textbf{1}$ and it shows that $a^{\perp}\subseteq b^{\perp\perp}\subseteq x^{\perp\perp}$. On the other hand $a\in x^{\perp}$ gives $x^{\perp\perp}\subseteq a^{\perp}$. Combining both the inclusions follows that $x^{\perp\perp}=a^{\perp}$. Hence $\mathfrak{A}$ is quasicomplemented.

∎

In the following proposition we give some equivalent conditions for each $\alpha$-filter to be an $\omega$-filter.

Theorem 5.13**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. $(1)$

$\mathfrak{A}$* is quasicomplemented;* 2. $(2)$

every $\alpha$-filter is an $\omega$-filter; 3. $(3)$

every coannihilator is an $\omega$-filter; 4. $(4)$

For any $x\in A$, $x^{\perp\perp}$ is an $\omega$-filter.

Proof.

1. $(1)$$\Rightarrow$$(2)$: Let $F$ be an $\alpha$-filter and set $I_{F}=\{a\in A|f^{\perp}\subseteq a^{\perp\perp},for~{}some~{}f\in F\}$. It is obvious that $0\in I_{F}$. Let $a_{1},a_{2}\in I_{F}$. So $f_{1}^{\perp}\subseteq a_{1}^{\perp\perp}$ and $f_{2}^{\perp}\subseteq a_{2}^{\perp\perp}$ for some $f_{1},f_{2}\in F$. By Remark 2.10 follows that $(f_{1}\odot f_{2})^{\perp}\subseteq(a_{1}\vee a_{2})^{\perp\perp}$ and it states that $a_{1}\vee a_{2}\in I_{F}$. Now let $a_{1}\leq a_{2}$ and $a_{2}\in I_{F}$. So there exists $f\in F$ such that $f^{\perp}\subseteq a^{\perp\perp}_{2}$ and on the other hand we have $a^{\perp\perp}_{2}\leq a^{\perp\perp}_{1}$. Hence we have $f^{\perp}\subseteq a^{\perp\perp}_{1}$ and it means that $a_{1}\in I_{F}$. Thus $I_{F}$ is a lattice ideal of $\mathfrak{A}$. Now, let $x\in\omega(I_{F})$. Consequently, $x\in a^{\perp}$ for some $a\in I_{F}$ and so $x\in f^{\perp\perp}$ for some $f\in F$. Since $F$ is an $\alpha$-filter follows that $x\in F$. Otherwise, let $x\in F$. Since $\mathfrak{A}$ quasicomplemented follows that $x^{\perp\perp}=a^{\perp}$ for some $a\in A$. It means that $a\in I_{F}$. Therefore $x\in x^{\perp\perp}=a^{\perp}\subseteq\omega(I_{F})$ and it shows that $F\subseteq\omega(I_{F})$.
2. $(2)$$\Rightarrow$$(3)$: It is obvious, since any coannihilator is an $\alpha$-filter.
3. $(3)$$\Rightarrow$$(4)$: It is obvious, since for any $x\in A$, $x^{\perp\perp}$ is a coannihilator.
4. $(4)$$\Rightarrow$$(1)$: Let $x\in A$. So there exists a lattice ideal $I$ such that $x^{\perp\perp}=\omega(I)$. So $x\in\omega(I)$ and it implies that $x\in y^{\perp}$ for some $y\in I$. Thus $x^{\perp\perp}\subseteq y^{\perp}\subseteq\omega(I)=x^{\perp\perp}$.

∎

Lemma 5.14**.**

Let $F$ be a filter of a residuated lattice $\mathfrak{A}$. Then $\alpha(F)$ is proper if and only if $F$ contains no dense element.

Proof.

Let $F\cap\mathfrak{d(A)}\neq\emptyset$. So there exists a dense element $d$ in $F$. By Proposition 5.4$(1)$ we get that $A=d^{\perp\perp}\subseteq\alpha(F)$ and it implies $\alpha(F)=A$. Otherwise, if $\alpha(F)=A$ then $0\in\alpha(F)$. So there exists $x\in F$ such that $0\in x^{\perp\perp}$. Therefore $x^{\perp}\subseteq 0^{\perp}=\{1\}$ and it means that $x$ is a dense element. ∎

Lemma 5.15**.**

Let $\mathfrak{A}$ be a quasicomplemented residuated lattice and $P$ be a prime filter. The following assertions are equivalent:

1. $(1)$

$P$* is an $\alpha$-filter;* 2. $(2)$

$P$* contains no dense element;* 3. $(3)$

$P$* is minimal prime;* 4. $(4)$

$P$* contains precisely one of $x,y\in A$ such that $x^{\perp\perp}=y^{\perp}$.*

Proof.

$(1)$$\Rightarrow$$(2)$ follows by Lemma 5.14 and $(2)$$\Rightarrow$$(3)$ follows by Proposition 3.4$(2)$.

$(3)$$\Rightarrow$$(4)$: Let $x$ and $y$ be any pair of elements for which $x^{\perp\perp}=y^{\perp}$. Since $x\vee y=1$ and $P$ is prime so either $x\in P$ or $y\in P$. Also, $x^{\perp}\cap y^{\perp}=\{1\}$ implies that $x^{\perp}\subseteq P$ or $y^{\perp}\subseteq P$. Now we can deduce the result by Theorem 2.13$(3)$.

$(4)$$\Rightarrow$$(1)$: Let $x\in P$ and $y$ be an element such that $x^{\perp\perp}=y^{\perp}$. Since $y\notin P$ so $y^{\perp}\subseteq P$ and it shows that $P$ is an $\alpha$-filter. ∎

In the next corollary for which we characterize quasicomplemented residuated lattice in terms of $\alpha$-filters should be compared with Proposition 3.4.

Corollary 5.16**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. $(1)$

$\mathfrak{A}$ is quasicomplemented; 2. $(2)$

any prime $\alpha$-filter is minimal prime; 3. $(3)$

any proper $\alpha$-filter is the intersection of minimal prime filters; 4. $(4)$

any proper $\alpha$-filter is contained in a minimal prime filter.

Proof.

$(1)$$\Rightarrow$$(2)$ is followed by Proposition 3.4 and Lemma 5.15, $(2)$$\Rightarrow$$(3)$ is followed by Corollary 5.8(2) and $(3)$$\Rightarrow$$(4)$ is obvious.

$(4)$$\Rightarrow$$(1)$: Let $F$ be a filter such that $F\cap\mathfrak{d(A)}=\emptyset$. By Lemma 5.14, $\alpha(F)$ is proper and so it is contained in a minimal prime filter. It means that $F$ is contained in a minimal prime filter. Hence, the result is followed by Proposition 3.4$(3)$. ∎

We end this paper by deriving a set of equivalent assertions for any $\alpha$-filter of a residuated lattice to become principal.

Lemma 5.17**.**

Let $\mathfrak{A}$ be a residuated lattice, $F$ be an $\alpha$-filter and $a\in F$. If for any prime $\alpha$-filter $P$ which contains $a$ we have $F\subseteq P$, then $F=\alpha(a)$.

Proof.

It is a direct consequence of Proposition 5.8(2). ∎

Proposition 5.18**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. (1)

For any $\mathscr{P}\subseteq Spec_{\alpha}(\mathfrak{A})$ and any $\alpha$-filter $F$, $F\subseteq\bigcup\mathscr{P}$ implies $F\subseteq P$ for some $P\in\mathscr{P}$; 2. (2)

for any $\mathscr{P}\subseteq Spec_{\alpha}(\mathfrak{A})$ and any prime $\alpha$-filter $Q$, $Q\subseteq\bigcup\mathscr{P}$ implies $Q\subseteq P$ for some $P\in\mathscr{P}$; 3. (3)

any prime $\alpha$-filter is principal; 4. (4)

any $\alpha$-filter is principal.

Proof.

It is obvious.

Let $Q$ be a prime $\alpha$-filter which is not principal. By Lemma 5.17 for any $a\in Q$ there exists a prime $\alpha$-filter $P_{a}$ such that $Q\nsubseteq P_{a}$. We have $Q\subseteq\bigcup\{P_{a}|a\in Q\}$ and so $Q\subseteq P_{a}$ for some $a\in Q$; a contradiction.

It is obvious that $A$ is a principal $\alpha$-filter. Set $\Psi$ be the set of all proper non-principal $\alpha$-filters of $\mathfrak{A}$. Let $\Psi$ be not empty. In a routine way we can show that $\Psi$ satisfies the conditions of Zorn’s lemma. Let $M$ be a maximal element of $\Psi$. Let $a\vee b\in M$ and $a,b\notin M$. Thus there exist $x,y\in A$ such that $\alpha(M,a)=\alpha(x)$ and $\alpha(M,b)=\alpha(y)$. Applying Proposition 5.4, it follows that $M=\alpha(x\vee y)$; a contradiction. Thus $M$ is a prime $\alpha$-filter and so it is principal; a contradiction. Hence, $\Psi=\emptyset$ and it gets the result.

Let $F$ be an $\alpha$-filter such that $F\subseteq\cup\mathscr{P}$ for some $\mathscr{P}\subseteq Spec_{\alpha}(\mathfrak{A})$. By hypothesis, $F=\alpha(a)$ for some $a\in A$. So there exists $P\in\mathscr{P}$ such that $a\in P$. Hence, $F\subseteq P$ and it proves the implication. ∎

Lemma 5.19**.**

If any coannulet of a residuated lattice is principal, then any its prime $\alpha$-filter is a minimal prime filter.

Proof.

Let $P$ be a prime $\alpha$-filter of $\mathfrak{A}$. By Proposition 5.5$(2)$ it is obvious that $P$ has no any dense element. Let $x\in P$. By hypothesis $x^{\perp}=\mathscr{F}(y)$ for some $y\in A$. So we have $(x\odot y)^{\perp}=x^{\perp}\cap y^{\perp}=x^{\perp}\cap x^{\perp\perp}=\{1\}$ and so $x\odot y\notin P$. Hence $y\notin P$ and so $x\in D(P)$. By Theorem 2.13 the result holds. ∎

Proposition 5.20**.**

Let $\mathfrak{A}$ be a residuated lattice. The following assertions are equivalent:

1. $(1)$

Any $\alpha$-filter is principal; 2. $(2)$

any $\omega$-filter is principal; 3. $(3)$

any coannulet is principal and any minimal prime filter is non-dense; 4. $(4)$

any prime $\alpha$-filter is principal.

Further, any of the above assertions implies that $\mathfrak{A}$ is quasicomplemented.

Proof.

It is obvious, by Proposition 5.10$(2)$.

By $\gamma(\mathfrak{A})\subseteq\Omega(\mathfrak{A})$ follows that any coannulet is principal. Let $\mathfrak{m}$ be a minimal prime filter. So $\mathfrak{m}$ is an $\omega$-filter and it means that $\mathfrak{m}=\mathscr{F}(x)$ for some $x\in A$. If $\mathfrak{m}$ is dense, follows that $x^{\perp}=1$ and it contradicts with Proposition 2.12.

Let $P$ be a prime $\alpha$-filter. By Lemma 5.19, $P$ is a minimal prime filter and so it is non-dense. By proposition 2.11, $P$ is a coannulet and this means that $P$ is principal.

It follows by Proposition 5.18.

The rest is evident by Proposition 3.3. ∎