Residuated Lattices with Noetherian Spectrum

Abstract

In this paper, we characterize residuated lattices for which the topological space of prime ideals is a Noetherian space. The notion of i-Noetherian residuated lattice is introduced and related properties are investigated. We proved that a residuated lattice is i-Noetherian iff every ideal is principal. Moreover, we show that a residuated lattice has the spectrum of a Noetherian space iff it is i-Noetherian.

1. Introduction

Residuated lattices play the role of semantics for residuated logic. In 1939, Ward and Dilworth introduced commutative residuated lattices in [1]. Residuated lattices are known under many names: BCK-lattices, full BCK-algebras, FL${}_{ew}$-algebras, or integral residuated commutative l-monoids.

These ordered structures have two historical sources: the study of residuation in the ideal lattices of rings and the algebrization of implication in intuitionistic logic.

The theory of residuated lattices was used to develop algebraic counterparts of fuzzy logics. Important examples of residuated lattices related to logic are Boolean algebras corresponding to basic logic, BL algebras corresponding to Hajek logic, and MV algebras corresponding to ukasiewicz many-valued logic.

Complete studies on residuated lattices or their subvarieties were developed in [1,2,3,4,5,6,7].

Filters are important concepts in studying residuated lattices and the completeness of the corresponding logic, see [4,7,8].

Many authors [8,9,10,11] remarked that the notion of ideals is missing in residuated lattices and this lack is associated with the fact that there is no suitable algebraic addition in these structures. Refs. [9,10] introduced some types of ideals in residuated lattices. In MV algebras, by definition, ideals are kernels of homomorphisms, see [12]. In residuated lattices, ideals (in the sense of [10]) generalize the existing notion in $MV$ algebras. However, ideals and the dual of filters are quite different in residuated lattices; the reason for this is the involution law.

The main scope of this paper is to characterize residuated lattices for which the the topological space of prime ideals is a Noetherian space.

The paper is organized as follows:

In Section 2, we review some results that we use in the sequel.

Section 3 contains new results about ideals in residuated lattices. For a residuated lattice $L,$ the lattice of ideals $\left(\mathbf{I}\right(\mathbf{L}),\subseteq )$ is a complete Brouwerian lattice. We show that in this lattice, every finite generated ideal is principal (Proposition 2). It is known that for a divisible residuated lattice, the quotient residuated lattice, via ideals, is an MV algebra, not just a divisible residuated lattice as in the case of filters (see [13]). Proposition 6 gives characterizations for ideals in this quotient $MV$ algebra.

Using an interesting construction of Turunen and Mertanen (see [14]), which associates an $MV$ algebra with any semidivisible residuated lattice, we prove the first theorem of isomorphism for residuated lattices via ideals (Theorem 1) and compare this result with the one obtained using filters.

In Section 4, using the model of $MV$ algebras (see [15]), we introduce the concept of i-Noetherian residuated lattice as lattices in which $\left(\mathbf{I}\right(\mathbf{L}),\subseteq )$ is a Noetherian poset. We study this notion and show that a residuated lattice is i-Noetherian iff every ideal is principal (Theorem 2). Further, we establish some connections between i-Noetherian residuated lattices and residuated lattices for which $\left(\mathbf{F}\right(\mathbf{L}),\subseteq )$ is a Noetherian poset (Corollary 1).

We recall that a proper ideal of a residuated lattice $L,$ is prime if it is a prime element in ($\mathbf{I}\left(\mathbf{L}\right),\subseteq ).$ For a residuated lattice L, $Spec\left(L\right),$ the set of all prime ideals of $L,$ can be endowed with the Zariski topology ${\tau }_L$ and $(Spec\left(L\right),{\tau }_L)$ becomes a compact topological space (see [16]).

A topological space is called Noetherian if it satisfies the descending chain condition on closed subsets (see [17]).

In Section 5, we prove that $Spec\left(L\right)$ is a Noetherian space iff L is an i-Noetherian residuated lattice (see Corollary 4).

In Section 6, we study certain connections between ideals in residuated lattices and ideals in certain types of unitary commutative rings, used in algebraic number theory.

2. Preliminaries

A residuated lattice ([1,3,5,6]) is an algebra $(L,\vee ,\wedge ,\odot ,\to ,0,1)$ satisfying the following axioms:

(RL₁)

$(L,\vee ,\wedge ,0,1)$ is a bounded lattice;

(RL₂)

$(L,\odot ,1)$ is a commutative monoid;

(RL₃)

$x\odot z\le y$ iff $z\le x\to y,$ for every $x,y,z\in L.$

The class $\mathcal{RL}$ of residuated lattices is equational; so, following Birkhoff’s theorem (see [18]), $\mathcal{RL}$ is a variety.

Example 1

([4,7]). The real unit interval $I=[0,1]$ becomes a residuated lattice $(I,\max ,\min ,\odot$$,\to ,0,1)$ called Godel structure. The operations of multiplication and implication are given for $x,y\in [0,1]$ by $x\odot y=min\{x,y\}$ and $x\to y=1$ if $x\le y$ and y otherwise.

We recall three important subclasses of residuated lattices (see [4,7,19]):

A residuated lattice L is called the following:

(i)

divisible if L verifies $\left(DIV\right):$$x\odot (x\to y)=x\wedge y;$

(ii)

involutive if L verifies $\left(DN\right):$$x^{**}=x;$

(iii)

$MV$algebra if L verifies $\left(MV\right):$$(x\to y)\to y=(y\to x)\to x.$

We denote by $\mathcal{DIV}$ the class of divisible residuated lattices. Obviously, $\mathcal{DIV}$ is a subcategory of $\mathcal{RL}$.

Let L be a residuated lattice. For $x,y\in L$, we denote $x^{*}=x\to 0$ and $x⊞y={(x^{*}\odot y^{*})}^{*}.$ For a natural number $n\ge 1,$ we will use the notation $(n+1)x:=nx⊞x.$

In a residuated lattice $L,$ the following properties hold, for every $x,y,z\in L$ (see [1,5,7,10,13,16,20,21]):

(c₁)

$x\le y$ iff $x\to y=1;$

(c₂)

$x\to (y\to z)=(x\odot y)\to z=y\to (x\to z);$

(c₃)

$x\le x^{**},$${(x\odot y)}^{*}=x\to y^{*}=y\to x^{*}=x^{**}\to y^{*};$

(c₄)

$x,y\le x⊞y,$$x⊞y=y⊞x,$$(x⊞y)⊞z=x⊞(y⊞z),$

(c₅)

${(x⊞y)}^{**}=x⊞y=x^{**}⊞y^{**};$

(c₆)

If L is divisible, ${(x\to y)}^{**}=x^{**}\to y^{**}.$

Definition 1

([1,5,7]). Let $L_1$ and $L_2$ be residuated lattices. A function $f:L_1\to L_2$ is a morphism of residuated lattices if f is a morphism of bounded lattices and $f(x\odot y)=f\left(x\right)\odot f\left(y\right),$$f(x\to y)=f\left(x\right)\to f\left(y\right),$ for every $x,y\in L_1.$

3. Filters and Ideals in Residuated Lattices

Definition 2

([5,7]). Let L be a residuated lattice. A filter of L is a subset $F\subseteq L$ such that

(f₁)

If $x\in F$ and $x\le$$y,$ then $y\in F;$

(f₂)

$x,$$y\in F$ implies $x\odot y\in F.$

Filters are also called congruence filters or deductive systems, see [7].

We denote by $\mathbf{F}\left(\mathbf{L}\right)$ the set of all filters of $L.$

In general, in residuated lattices, a dual operation to ⊙ does not exist; so, a dual notion for filter does not exist either. Refs. [9,10] introduced some kind of ideal in residuated lattices, not dual to a filter. This concept generalizes the existing notion in $MV$ algebras (see [2,12]).

Definition 3

([13]). A subset $I\subseteq L$ of a residuated lattice L is called an ideal of L if it satisfies

(i₁)

If $y\in I$ and $x\le y$, then $x\in I;$

(i₂)

$x,y\in I$ implies $x⊞y\in I.$

Trivial examples of ideals are $\left\{0\right\}$ and $L.$

We denote by $\mathbf{I}\left(\mathbf{L}\right)$ the set of all ideals of $L.$

We recall that if $f:L_1\to L_2$ is a morphism of residuated lattices, $i-Ker\left(f\right)=f^{-1}\left(0\right)=\{x\in L_1:f\left(x\right)=0\}$ is a proper ideal of $L_1$. Moreover, ideals are i-kernels of homomorphisms of residuated lattices (see [16]).

Obviously, if $I\in \mathbf{I}\left(\mathbf{L}\right),$ then $0\in I$ and $x\in I$ iff $x^{**}\in I,$ (see [10]).

Ref. [13] gives an equivalent condition for ideals in residuated lattices: $I\in \mathbf{I}\left(\mathbf{L}\right)$ iff $[0\in I$ and $x,$$x^{*}\odot y\in I$ implies $y\in I].$

Example 2.

Let $L=\{0,a,b,c,1\}$ be such that $0<a,b<c<1,$$a,b$ are incomparable. Define → and ⊙ as follows:

$$\begin{array}{cccccc}\to & 0& a& b& c& 1\\ 0& 1& 1& 1& 1& 1\\ a& b& 1& b& 1& 1\\ b& a& a& 1& 1& 1\\ c& 0& a& b& 1& 1\\ 1& 0& a& b& c& 1\end{array},\begin{array}{cccccc}\odot & 0& a& b& c& 1\\ 0& 0& 0& 0& 0& 0\\ a& 0& a& 0& a& a\\ b& 0& 0& b& b& b\\ c& 0& a& b& c& c\\ 1& 0& a& b& c& 1\end{array}.$$

Then, $(L,\vee ,\wedge ,\odot ,\to ,0,1)$ becomes a residuated lattice (see [19]). We remark that $\mathbf{I}\left(\mathbf{L}\right)=\left\{\right\{0\},\{0,a\},\{0,b\},L\}$ and $\mathbf{F}\left(\mathbf{L}\right)=\left\{\right\{1\},\{1,c\},\{1,a,c\},\{1,b,c\},L\}.$

Remark 1.

In a residuated lattice $L,$ if $I\in \mathbf{I}\left(\mathbf{L}\right)$ and $x\in I,y\in L$ such that ${(x⊞y^{*})}^{*}\in I,$ then $y\in I.$ Indeed, since $I\in \mathbf{I}\left(\mathbf{L}\right)$ and ${(x⊞y^{*})}^{*}={(x^{*}\odot y^{**})}^{**}\in I,$ we deduce that $x^{*}\odot y^{**}\in I.$ Then, from $x,x^{*}\odot y^{**}\in I$ we obtain $y^{*}{}^{*}\in I,$ so $y\in I.$

For a nonempty subset X of a residuated lattice L, we denote by $\left(X\right]$ the ideal of L generated by X and for $x\in L$ we denote $\left(\right\{x\left\}\right]$ by $\left(x\right],$ the principal ideal of L generated by $x.$

Proposition 1

([10,16]). Let L be a residuated lattice, $X\subseteq L$, and $x,y\in L.$ Then,

(i)

$\left(X\right]=\{a\in L:a\le x_1⊞...⊞x_n,$ for some $n\ge 1$ and $x_1,...,x_n\in X\};$

(ii)

$\left(x\right]=\{a\in L:a\le nx,$ for some $n\ge 1\}$ and $\left(x\right]\cap \left(y\right]=(x^{**}\wedge y^{**}],$$\left(x\right]\vee \left(y\right]=(x⊞y];$

(iii)

$\left(\mathbf{I}\right(\mathbf{L}),\subseteq )$ is a complete Brouwerian lattice in which for $I,J\in \mathbf{I}\left(\mathbf{L}\right),$$I\wedge J=I\cap J$ and $I\vee J$$=(I\cup J].$

Proposition 2.

In a residuated lattice, every finitely generated ideal is principal.

Proof. 

Let $I\in \mathbf{I}\left(\mathbf{L}\right)$ be a finitely generated ideal. Then, there are $n\ge 1$ and $x_1,...,x_n\in L$ such that $I=\left(\{x_1,...,x_n\}\right].$ We show that $I=(x_1⊞...⊞x_n].$ By definition, $I=\cap \{J\in \mathbf{I}\left(\mathbf{L}\right):\{x_1,...,x_n\}\subseteq J\}.$ Since, by $\left(c_4\right),$ $x_i\le x_1⊞...⊞x_n$ for every $i\in \{1,...,n\}$ we deduce that $x_i\in (x_1⊞...⊞x_n]$; so, $\{x_1,...,x_n\}\subseteq (x_1⊞...⊞x_n].$ Thus, $I\subseteq (x_1⊞...⊞x_n].$ Now, let $J\in \mathbf{I}\left(\mathbf{L}\right)$ such that $\{x_1,...,x_n\}\subseteq J.$ Then, $x_1⊞...⊞x_n\in J,$ so, $(x_1⊞...⊞x_n]\subseteq J.$ Therefore, $(x_1⊞...⊞x_n]\subseteq I.$ Hence, I is principal and $\left(\{x_1,...,x_n\}\right]=(x_1⊞...⊞x_n]$. □

We recall some relationships between ideals and filters in residuated lattices proved in [13] using the set of complemented elements.

We denote a subset X of a residuated lattice L by

$$N\left(X\right)=\{x\in L:x^{*}\in X\}.$$

Proposition 3

([13]). Let L be a residuated lattice, $I\in \mathbf{I}\left(\mathbf{L}\right)$, and $F\in \mathbf{F}\left(\mathbf{L}\right).$ Then, $N\left(I\right)\in \mathbf{F}\left(\mathbf{L}\right),$$N\left(F\right)\in \mathbf{I}\left(\mathbf{L}\right),$$I=N\left(N\right(I\left)\right)$, and $F\subseteq N\left(N\right(F\left)\right).$

In the following, we establish other properties of this operator:

Lemma 1.

Let L be a residuated lattice; $X_1,X_2\subseteq L$; and $I_1,I_2\in \mathbf{I}\left(\mathbf{L}\right).$ Then,

(i)

$X_1\subseteq X_2$ implies $N\left(X_1\right)\subseteq N\left(X_2\right);$

(ii)

$I_1\subseteq I_2$ iff $N\left(I_1\right)\subseteq N\left(I_2\right);$

(iii)

$I_1=I_2$ iff $N\left(I_1\right)=N\left(I_2\right).$

Proof. 

(i)

Suppose that $X_1\subseteq X_2$ and let $x\in$$N\left(X_1\right).$ Then, $x^{*}\in X_1\subseteq X_2,$ so $x\in$$N\left(X_2\right)$ and $N\left(X_1\right)\subseteq N\left(X_2\right).$

(ii)

Suppose that $N\left(I_1\right)\subseteq N\left(I_2\right)$ and let $x\in I_1.$ Then, $x^{**}\in I_1$, so $x^{*}\in N\left(I_1\right)$$\subseteq N\left(I_2\right).$ Then, $x^{**}\in I_2$; so, $x\in I_2$ and $I_1\subseteq I_2.$ Using $\left(i\right)$, we deduce that $I_1\subseteq I_2$ iff $N\left(I_1\right)\subseteq N\left(I_2\right).$

(iii)

Apply $\left(ii\right).$

□

In [10], a residuated lattice L and an ideal I of L a congruence relation ${\approx }_I$ on L is defined by $x{\approx }_{I}y$ iff ${(x\to y)}^{*},{(y\to x)}^{*}\in I.$ For $x\in L,$ the congruence class of x is denoted by $x/I$ and the quotient residuated lattice $L/{\approx }_I$ by $L/I.$ Obviously, in $L/I,$$\mathbf{0}=0/I=\{x\in L:x\in I\}=I,$$\mathbf{1}=1/I=\{x\in L:x^{*}\in I\}$ and for $x,y\in L,$$x/I\le y/I$ iff ${(x\to y)}^{*}\in I.$

We recall that (see [12]) for an $MV$ algebra A and an ideal I of $A,$ the binary relation ${\sim }_I$ on A defined by $x{\sim }_{I}y$ iff $x^{*}\odot y,x\odot y^{*}\in I$, for $x,y\in A$ is a congruence relation on $A.$ Unlike in $MV$ algebras, in a residuated lattice L, for $I\in \mathbf{I}\left(\mathbf{L}\right),$${\sim }_I$ is only an equivalence relation on L (see [13]).

Proposition 4.

In a divisible residuated lattice L, the relations ${\sim }_I$ and ${\approx }_I$ coincide for every $I\in \mathbf{I}\left(\mathbf{L}\right).$

Proof. 

In [13], it is proved that ${\sim }_I$ is a congruence relation on $L,$ if L is divisible.

We have $x{\sim }_{I}y$ iff $x^{*}\odot y,$ $x\odot y^{*}\in I$ iff ${(x^{*}\odot y)}^{**},$ ${(x\odot y^{*})}^{**}\in I$ iff ${(y\to x)}^{*},$ ${(x\to y)}^{*}\in I$ iff $x{\approx }_{I}y$, since ${(x^{*}\odot y)}^{**}\stackrel{\left(c_3\right)}{=}{(y^{**}\to x^{**})}^{*}\stackrel{\left(c_6\right)}{=}{(y\to x)}^{**}{}^{*}={(y\to x)}^{*},$ for every $x,y\in L.$ □

It is known that, for a divisible residuated lattice, the quotient residuated lattice, via ideals, is an MV algebra:

Proposition 5

([13]). If L is a divisible residuated lattice and $I\in \mathbf{I}\left(\mathbf{L}\right),$$(L/{\sim }_I=L/{\approx }_I\stackrel{not}{=}L/I,⊞{,}^{*},I)$ is an $MV$ algebra.

Let L be a residuated lattice, $X\subseteq L$ be a nonempty subset, and $I\in \mathbf{I}\left(\mathbf{L}\right).$ We denote

$$X/I=\{x/I:x\in X\}.$$

In the following, for a divisible residuated lattice L, we give a characterization for ideals in the quotient $MV$ algebra $L/I$.

Proposition 6.

Let L be a divisible residuated lattice.

(i)

If $I,J\in \mathbf{I}\left(\mathbf{L}\right)$ and $I\subseteq J,$ then $J/I\in \mathbf{I}(L/I);$

(ii)

If $I\in \mathbf{I}\left(\mathbf{L}\right)$ then in $MV$ algebra$L/I,$ the set of ideals is $\mathbf{I}(L/I)$$=\{J/I:J\in \mathbf{I}(\mathbf{L})$ and $I\subseteq J\};$

(iii)

If $I,I_1,I_2\in \mathbf{I}\left(\mathbf{L}\right)$ with $I\subseteq I_1,I_2$ and $I_1/I\subseteq I_2/I,$ then $I_1\subseteq I_2.$

Proof. 

(i)

Since L is divisible and $I\in \mathbf{I}\left(\mathbf{L}\right),$ using Proposition 5, $L/I$ is an $MV$ algebra.

To prove that $J/I\in \mathbf{I}(L/I),$ first, let $x/I,y/I\in J/I.$ Then, $x,y\in J$ and since $J\in \mathbf{I}\left(\mathbf{L}\right)$, we deduce that $x⊞y\in J$; so, $(x⊞y)/I\in J/I.$

If $x/I\in L/I,$$y/I\in J/I$ and $x/I\le y/I,$ then $x\in L,y\in J$ and ${(x/I)}^{*}⊞(y/I)=1/I.$ We deduce that $(x^{*}⊞y)/I=1/I$; so, ${(x^{*}⊞y)}^{*}\in I.$ Since $I\subseteq J$, we have ${(x^{*}⊞y)}^{*}\in J$. Thus, $y,{(x^{*}⊞y)}^{*}\in J$ and since $J\in \mathbf{I}\left(\mathbf{L}\right),$ using Remark 1, we conclude that $x\in J$; so, $x/I\in J/I$ and $J/I\in \mathbf{I}(L/I).$

(ii)

Using $\left(i\right)$, $J/I\in \mathbf{I}(L/I),$ for every $J\in \mathbf{I}\left(\mathbf{L}\right)$ with $I\subseteq J.$

Now, let $K/I\in \mathbf{I}(L/I).$ Since $I=0/I\in$$K/I,$ we deduce that $I\subseteq K.$ To prove that $K\in \mathbf{I}\left(\mathbf{L}\right),$ let $x,y\in K.$ Then, $x/I,y/I\in$$K/I$, which is an ideal of $L/I.$ We deduce that $(x/I)⊞(y/I)=(x⊞y)/I\in K/I$; so, $x⊞y\in K.$ If $x\in L$ such that $x\le y$ and $y\in K,$ then $x\to y=1$; so, ${(x\to y)}^{*}=0\in K.$

Since L is divisible, using $\left(c_6\right)$, we deduce that ${(x\to y)}^{*}={(x\to y)}^{***}={(x^{**}\to y^{**})}^{*}={(x^{**}\odot y^{*})}^{**}={(x^{*}⊞y)}^{*}.$ Since $y,{(x^{*}⊞y)}^{*}\in K,$ using Remark 1, we obtain $x\in K$; thus, $K\in \mathbf{I}\left(\mathbf{L}\right)$.

(iii)

Using $\left(i\right)$, $I_1/I,$$I_2/I\in \mathbf{I}(L/I).$ Now, let $x\in I_1.$ Then, $x/I\in I_2/I,$ so $x/I$$=y/I,$ for some $y\in I_2.$ Thus, ${(x\to y)}^{*},{(y\to x)}^{*}\in I.$ Since $I\subseteq I_2$ and ${(x\to y)}^{*}={[x^{*}⊞y]}^{*}$ we deduce that $x\in I_2$—that is, $I_1\subseteq I_2.$

□

In [14], Turunen and Mertanen defined the $MV$ center of a divisible residuated lattice L,

$$MV\left(L\right)=\{x^{*}:x\in L\}=\{x\in L:x^{**}=x\},$$

and proved that, in this case, $(MV\left(L\right),⊞{,}^{*},0)$ is an $MV$ algebra. Using this construction, which associates an $MV$ algebra to any divisible residuated lattice, in the following we prove the first theorem of isomorphism for residuated lattices via ideals:

Theorem 1.

Let $L_1$ and $L_2$ be residuated lattices such that $L_1$ is divisible. If $f:L_1\to L_2$ is a morphism of residuated lattices, $L_1/i-Ker\left(f\right)\approx MV\left(Imf\right)(\mathrm{as}MV\mathrm{algebras}).$

Proof. 

Since $\mathcal{DIV}$ is a subvariety of $\mathcal{RL}$ and i-$Ker\left(f\right)\in \mathbf{I}\left(\mathbf{L}\right)$, $L_1$/$i-Ker\left(f\right)$ and $Imf$ are divisible residuated lattices.

Moreover, $L_1$/i-$Ker\left(f\right)$ and $MV\left(Imf\right)$ are $MV$ algebras.

Now, we define $\varphi :L_1$/i-$Ker\left(f\right)\to MV\left(Imf\right)$ by $\varphi (x$/${i-Ker\left(f\right))=f\left(x\right)}^{**},$ for every $x\in L_1.$

Obviously, $\varphi$ is well-defined and a one-to-one map since for every $x,y\in L_1$ we have x/$i-Ker\left(f\right)=y$/$i-Ker\left(f\right)$ iff $x^{**}$/$i-Ker\left(f\right)=y^{**}$/$i-Ker\left(f\right)$ iff $x^{*}\odot y^{**},$ $x^{**}\odot y^{*}\in i-Ker\left(f\right)$ iff $f(x^{*}\odot y^{**})=f($$x^{**}\odot y^{*})=0$ iff $f{\left(x\right)}^{*}\odot f{\left(y\right)}^{**}=f($${x)}^{**}\odot f{\left(y\right)}^{*}=0$ iff $f{\left(y\right)}^{**}{\le }_{MV}f{\left(x\right)}^{**},$ $f{\left(x\right)}^{**}{\le }_{MV}f{\left(y\right)}^{**}$ iff $f{\left(x\right)}^{**}=f{\left(y\right)}^{**}$ iff $\varphi (x$/$i-Ker\left(f\right))=\varphi (y$/$i-Ker\left(f\right)).$

By definition, $\varphi$ is onto and clearly a morphism of MV algebra since f is a morphism of residuated lattices:

$$\varphi ((x/i-Ker\left(f\right)⊞(y/i-Ker\left(f\right)))=\varphi ((x⊞y)/i-Ker\left(f\right))=f{(x⊞y)}^{**}=$$

$$={[f\left(x\right)⊞f\left(y\right)]}^{**}\stackrel{\left(c_5\right)}{=}{\left[f\left(x\right)\right]}^{**}⊞{\left[f\left(y\right)\right]}^{**}=\varphi ((x/i-Ker\left(f\right))⊞\varphi (y/i-Ker\left(f\right)),$$

$$\varphi \left({(x/i-Ker\left(f\right))}^{*}\right)=\varphi (x^{*}/i-Ker\left(f\right))={\left[f\left(x^{*}\right)\right]}^{**}={\left[f{\left(x\right)}^{**}\right]}^{*}={\left[\varphi (x/i-Ker\left(f\right))\right]}^{*}$$

$$\mathrm{and}\varphi (0/i-Ker\left(f\right))={\left[f\left(0\right)\right]}^{**}=0^{**}=0,\mathrm{for}\mathrm{every}x,y\in L_1.$$

We conclude that f is an isomorphism of $MV$ algebras. □

Remark 2.

Unlike in $MV$ algebras, we remark that ideals and filters behave quite differently in residuated lattices and generate different constructions.

4. i-Noetherian Residuated Lattices

In the following, using the model of MV algebras, see [15], we introduce and characterize the concept of i-Noetherian residuated lattice.

We recall that a poset $(A,\le )$ is Noetherian if every increasing chain $a_1\le a_2\le ...$ of elements of A is stationary, i.e., there is a natural number $n\ge 1$ such that $a_i=a_n,$ for every $i\ge n$ (see [22]).

Definition 4.

A residuated lattice L is called i-Noetherian if the poset $\left(\mathbf{I}\right(\mathbf{L}),\subseteq )$ is Noetherian.

Example 3.

Let L be the Godel structure, see Example 1. Then, for every $x\ne 0,$$x^{*}=0$; so, $x^{**}=1.$ We deduce that Godel structure is an i-Noetherian residuated lattice since $\mathbf{I}\left(\mathbf{L}\right)=\left\{\right\{0\},L\}.$

We recall that [23] introduced the notion of Noetherian BL algebra, Ref. [20] generalized these results and studied the concept of Noetherian residuated lattice as a lattice with the property that every increasing chain of filters is stationary.

In this paper, we study some connections between i-Noetherian and Noetherian residuated lattices.

Proposition 7.

If L is a Noetherian residuated lattice (in the sense of [20]), the poset $\left(\mathbf{I}\right(\mathbf{L}),\subseteq )$ is Noetherian.

Proof. 

Let $I_1\subseteq I_2\subseteq ...$ be an increasing chain of ideals of $L.$ Using Proposition 3 and Lemma 1 $\left(ii\right)$, $N\left(I_1\right)\subseteq N\left(I_2\right)\subseteq ...$ is an increasing chain of filters of $L.$ Since L is Noetherian, there is a natural number $n\ge 1$ such that $N\left(I_i\right)=N\left(I_n\right),$ for every $i\ge n.$ Using Lemma 1 $\left(iii\right)$, $I_i=I_n,$ for every $i\ge n.$ □

Remark 3.

The converse implication in Proposition 7 is not true. For example, let L be the Godel structure, which is i-Noetherian, see Example 3. Then, for every $n\ge 1,$$F_n=[\frac{1}{n},1]$ is a filter of L and $F_1\subseteq F_2\subseteq ...$ is an increasing chain in $\left(\mathbf{F}\right(\mathbf{L}),\subseteq )$ that is not stationary; so, L is not Noetherian.

The next result is a consequence of Proposition 7 and Remark 3:

Corollary 1.

Every Noetherian residuated lattice is i-Noetherian.

We conclude that for a residuated lattice

$$Noetherian\Rightarrow i-Noetherian$$

$$i-Noetherian\nRightarrow Noetherian.$$

Corollary 2.

If L is an involutive residuated lattice, the notions of Noetherian and i-Noetherian coincide.

Proof. 

Suppose that L is i-Noetherian.

First, we prove that for $F_1,F_2\in \mathbf{F}\left(\mathbf{L}\right),$ $N\left(F_1\right)\subseteq N\left(F_2\right)$ implies $F_1$$\subseteq F_2.$ Let $x\in F_1.$ Since $F_1\in \mathbf{F}\left(\mathbf{L}\right)$ and $x\le x^{**}$, we have $x^{**}\in F_1$. Then, $x^{*}\in N\left(F_1\right)$$\subseteq N\left(F_2\right)$; so, $x^{**}\in F_2.$ Since L is involutive, $x=x^{**}\in F_2$ and $F_1\subseteq F_2.$

Using Lemma 1 $\left(i\right)$, we deduce that $F_1=F_2$ iff $N\left(F_1\right)=N\left(F_2\right).$

Now, let $F_1\subseteq F_2\subseteq ...$ be an increasing chain in $\mathbf{F}\left(\mathbf{L}\right).$ Using Proposition 3 and Lemma 1, $N\left(F_1\right)\subseteq N\left(F_2\right)\subseteq ...$ is an increasing chain in $\mathbf{I}\left(\mathbf{L}\right).$ Since L is i-Noetherian, there exists a natural number $n\ge 1$ such that $N\left(F_i\right)=N\left(F_n\right),$ for every $i\ge n.$ Thus, $F_i=F_n,$ for every $i\ge n.$ We deduce that L is Noetherian.

Using Corollary 1, we conclude that the notions of Noetherian and i-Noetherian coincide if L is involutive. □

Remark 4.

For $MV$ algebras, the notions of Noetherian and i-Noetherian coincide since $MV$ algebras are involutive residuated lattices and the notions of ideal and filter are dual.

We recall that a poset $(A,\le )$ is Noetherian iff every nonempty subset of A has a maximal element, see [22]. Using this result, we give a characterization for i-Noetherian residuated lattices:

Theorem 2.

A residuated lattice L is i-Noetherian iff every ideal of L is principal.

Proof. 

First, suppose that every ideal of L is principal and let $I_1\subseteq I_2\subseteq ...$ be an increasing chain in $\mathbf{I}\left(\mathbf{L}\right).$

Since $I=\underset{i\ge 1}{\cup }{I}_i\in \mathbf{I}\left(\mathbf{L}\right),$ there exists $x\in L$ such that $I=\left(x\right].$ Then, there is a natural number $n\ge 1$ such that $x\in I_n.$ Thus, $I=($$x]\subseteq I_n\subseteq I.$ Hence, $I_i=I_n,$ for every $i\ge n$ and L is i-Noetherian.

Conversely, suppose that L is i-Noetherian and let $I\in \mathbf{I}\left(\mathbf{L}\right)$ such that I is not principal. If we denote that ${\mathcal{S}}_I=\{K\in \mathbf{I}\left(\mathbf{L}\right):$K is principal and $K\subseteq I\},$ then we remark that $<0>=\left\{0\right\}\subseteq I$; so, ${\mathcal{S}}_I\ne \varnothing .$

Since L is i-Noetherian, $\left(\mathbf{I}\right(\mathbf{L}),\subseteq )$ is a Noetherian poset and ${\mathcal{S}}_I$⊆$\mathbf{I}\left(\mathbf{L}\right)$ has a maximal element $J.$ Thus, $J\subseteq I$ and J is principal—that is, $J=\left(j\right],$ for some $j\in L.$ Since I is not principal, $J\ne I$; so, there is $i\in I$∖$J.$ Hence, $J\subset (i⊞j]\in {\mathcal{S}}_I$ is a contradiction, since J is the maximal element of ${\mathcal{S}}_I$. □

Example 4.

Let L be the residuated lattice $L=\{0,a,b,c,1\}$ from Example 2. Then, $\mathbf{I}\left(\mathbf{L}\right)=\left\{\right\{0\},\{0,a\},\{0,b\},L\}$ and every ideal is principal $\left\{0\right\}=\left(0\right],\{0,a\}=\left(a\right],\{0,b\}=\left(b\right],L=\left(1\right].$ By Theorem 2, L is i-Noetherian.

Proposition 8.

Every subalgebra of an i-Noetherian residuated lattice is i-Noetherian.

Proof. 

Let L be an i-Noetherian residuated lattice and $L^{\prime }\subseteq L$ be a subalgebra of $L.$ Obviously, $\mathbf{I}\left({\mathbf{L}}^{\prime }\right)=\{I\cap L^{\prime }:I\in \mathbf{I}\left(\mathbf{L}\right)\}.$ We deduce that $L^{\prime }$ is also i-Noetherian. □

Proposition 9.

Let L be a divisible residuated lattice and $I\in \mathbf{I}\left(\mathbf{L}\right).$ If L is i-Noetherian, $L/I$ is an i-Noetherian MV algebra.

Proof. 

Let $I_1/I\subseteq I_2/I\subseteq ...$ be an increasing chain of ideals in $L/I,$ see Proposition 6$\left(ii\right).$ Using Proposition 6 $\left(iii\right),$ we obtain an increasing chain of ideals in $L:$$I\subseteq I_1\subseteq I_2\subseteq ....$ Since L is i-Noetherian, there is $n\ge 1$ such that $I_i=I_n,$ for every $i\ge n$. Thus, $I_i/I=I_n/I,$ for every $i\ge n$; so, $L/I$ is i-Noetherian. □

Theorem 3.

The MV center of any homomorphic image of a divisible and i-Noetherian residuated lattice is i-Noetherian.

Proof. 

Let $L_1$ be a divisible residuated lattice that is i-Noetherian, $L_2$ be a residuated lattice, and $f:L_1\to L_2$ be a morphism of residuated lattices. Since $\mathcal{DIV}$ is a subvariety of $\mathcal{RL}$, $f\left(L_1\right)$ is a divisible residuated lattice. Using Turunen and Mertanen’s result (see [14]), $MV\left(f\left(L_1\right)\right)$ is an $MV$ algebra. Since $i-Ker\left(f\right)\in \mathbf{I}\left({\mathbf{L}}_1\right)$ using Isomorphism Theorem 1, $L_1$/ i-$Ker\left(f\right)\approx MV\left(Imf\right),$ as $MV$ algebras. From Proposition 9, we deduce that $L_1$/ i-$Ker\left(f\right)$ is an i-Noetherian $MV$ algebra; so, $MV\left(Imf\right)$ is i-Noetherian. □

Remark 5.

Theorem 3 generalizes Dymek’s result ([15]): Any homomorphic image of an i-Noetherian (= Noetherian) MV algebra is Noetherian.

Using Theorem 3, we deduce the following:

Corollary 3.

Let $L_1,L_2$ be residuated lattices and $f:L_1\to L_2$ be an onto morphism of residuated lattices. If $L_1$ is divisible and i-Noetherian, $L_2$ is divisible and $MV\left(L_2\right)$ is i-Noetherian.

5. Noetherian Spectrum in Residuated Lattices

In the following, we establish connections between i-Noetherian residuated lattices and those residuated lattices for which their spectrum is a Noetherian space.

In this way, we translated some important results from theory of rings to the case of residuated lattices.

We recall that an ideal P of a residuated lattice L is called prime if $P\ne L$ and P is a prime element in $\left(\mathbf{I}\right(\mathbf{L}),\subseteq ),$ see [13].

For a residuated lattice $L,$ we denote by $Spec\left(L\right)$ the set of prime ideals. It is known that $Spec\left(L\right)$ can be endowed with the Zariski topology ${\tau }_L:$

${\left\{V\left(I\right)\right\}}_{I\in \mathbf{I}\left(\mathbf{L}\right)}$ is the family of closed subsets of $Spec\left(L\right)$ and ${\left\{D\left(I\right)\right\}}_{I\in \mathbf{I}\left(\mathbf{L}\right)}$ is the family of open subsets of $Spec\left(L\right),$ where for $I\in I\left(L\right)$ and $x\in L,$

$$V\left(I\right)=\{P\in Spec(L):I\subseteq P\},D\left(I\right)=\{P\in Spec(L):I⊈P\}$$

$$\mathrm{and}D\left(x\right)=D\left(\right(x\left]\right)=\{P\in Spec(L):x\notin P\}.$$

Thus, ${\tau }_L={\left\{D\left(I\right)\right\}}_{I\in \mathbf{I}\left(\mathbf{L}\right)}$ is a topology on $Spec\left(L\right)$ and the topological space $(Spec\left(L\right),{\tau }_L)$ is called the prime spectrum of $L.$ Moreover, the family ${\left\{D\left(x\right)\right\}}_{x\in L}$ is a basis for the topology ${\tau }_L$ on $Spec\left(L\right),$ see [16].

Proposition 10

([16]). Let L be a residuated lattice. Then,

(i)

$D\left(\right\{1\left\}\right)=D\left(L\right)=\mathcal{P}\left(L\right)$ and $D\left(\right\{0\left\}\right)=D(⌀)=⌀;$

(ii)

For every family ${\left\{I_k\right\}}_{k\in K}\in \mathbf{I}\left(\mathbf{L}\right),$$\underset{k\in K}{\cup }D\left(I_k\right)=D\left(\underset{k\in K}{\vee }{I}_k\right);$

(iii)

For every $I,J\in \mathbf{I}\left(\mathbf{L}\right),$$D\left(I\right)\cap D\left(J\right)=$$D(I\cap J)$ and $\left[D\right(I)=D(J)$ iff $I=J];$

(iv)

$D(x^{**}\wedge y^{**})=D\left(x\right)\cap D\left(y\right)$ and $D\left(x\right)\cup D\left(y\right)=D(x⊞y),$ for every $x,y\in L.$

Lemma 2.

If L is a residuated lattice and $I\in \mathbf{I}\left(\mathbf{L}\right),$ then $D\left(I\right)=\underset{x\in I}{\cup }D\left(x\right).$

Proof. 

Using Proposition 10, $D\left(x\right)\subseteq D\left(I\right),$ for every $x\in I,$ so $\underset{x\in I}{\cup }D\left(x\right)\subseteq D\left(I\right).$

Now, let $P\in D\left(I\right).$ Then, $I⊈P.$ Thus, there is $x_0\in I$ such that $x_0\notin P.$ We deduce that $P\in D\left(x_0\right)$; so, $P\in$$\underset{x\in I}{\cup }D\left(x\right).$ Then, $D\left(I\right)\subseteq \underset{x\in I}{\cup }D\left(x\right).$ □

We give a characterization for compact open subsets of $Spec\left(L\right):$

Theorem 4.

The compact open subsets of $Spec\left(L\right)$ are $D\left(x\right)$ with $x\in L.$

Proof. 

Obviously, for every $x\in L,$ $D\left(x\right)$ is an open subset of $Spec\left(L\right)$.

To prove that $D\left(x\right)$ is compact, let $D\left(x\right)=\underset{k\in K}{\cup }D\left(x_k\right).$

From Proposition 10, we deduce that $D\left(x\right)=D\left(\underset{k\in K}{\cup }\left\{x_k\right\}\right)$; so, $\left(x\right]=\left(\underset{k\in K}{\cup }\left\{x_k\right\}\right].$ Then, $x\in \left(\underset{k\in K}{\cup }\left\{x_k\right\}\right]$; so, there are $m\ge 1,$ $k_1,...,k_m\in K,$ such that $x\le x_{k_1}⊞...⊞x_{k_m}.$

It follows that $D\left(x\right)\subseteq D(x_{k_1}⊞...⊞x_{k_m})=D\left(x_{k_1}\right)\cup ...\cup D\left(x_{k_m}\right).$ Since $D\left(x_{k_1}\right)\cup ...\cup D\left(x_{k_m}\right)\subseteq D\left(x\right),$ we deduce that $D\left(x\right)=D\left(x_{k_1}\right)\cup ...\cup D\left(x_{k_m}\right)$—that is, $D\left(x\right)$ is compact.

Conversely, we will prove that for any open compact subset $D\left(I\right)$ of $Spec\left(L\right),$ with $I\in \mathbf{I}\left(\mathbf{L}\right),$ there is some $x\in L$ such that $D\left(I\right)=D\left(x\right).$

By Lemma 2, $D\left(I\right)=\underset{x\in I}{\cup }D\left(x\right).$ Since $D\left(I\right)$ is compact, there are $n\ge 1$ and $x_1,...,x_n\in I$ such that $D\left(x\right)=D\left(x_1\right)\cup ...\cup D\left(x_n\right)=D(x_1⊞...⊞x_n),$ by Proposition 10.   □

Example 5.

Let $L=\{0,a,b,c,1\}$ be the residuated lattice from Example 2. We remark that

$$\mathbf{I}\left(\mathbf{L}\right)=\left\{\right\{0\},\{0,a\},\{0,b\},L\}andSpec\left(L\right)=\left\{\right\{0,a\},\{0,b\left\}\right\}$$

$$D\left(\right\{0\left\}\right)=⌀,D\left(L\right)=Spec\left(L\right),D\left(\right\{0,a\left\}\right)=\{0,b\},D\left(\right\{0,b\left\}\right)=\{0,a\},$$

$$D\left(0\right)=⌀,D\left(a\right)=\{0,b\},D\left(b\right)=\{0,a\},D\left(c\right)=D\left(1\right)=Spec\left(L\right)$$

so, ${\tau }_L=\mathcal{P}\left(Spec\left(L\right)\right).$

We recall that a topological space is called Noetherian [17] if it satisfies the descending chain condition on closed subsets (that is, every decreasing chain of closed subsets is stationary).

Remark 6.

If L is a residuated lattice with $Spec\left(L\right)$ finite, $Spec\left(L\right)$ is a Noetherian space.

Example 6.

If L is the residuated lattice $L=\{0,a,b,c,1\}$ from Example 2, $Spec\left(L\right)$ is finite; so, $Spec\left(L\right)$ is a Noetherian space.

Another characterization for Noetherian spaces is the following:

Proposition 11

([17]). A topological space is Noetherian iff every open set is compact.

Using this result, we deduce the following:

Theorem 5.

Let L be a residuated lattice. Then, the following are equivalent:

(i)

$Spec\left(L\right)$ is Noetherian;

(ii)

Every ideal of L is principal.

Proof. 

Using Proposition 11 and Theorem 4, $Spec\left(L\right)$ is Noetherian iff for every $I\in \mathbf{I}\left(\mathbf{L}\right)$ there is $x\in L$ such that $D\left(I\right)=D\left(x\right).$ Since $D\left(x\right)=D\left(\right(x\left]\right),$ by Proposition 10, $D\left(I\right)=D\left(\right(x\left]\right)$ iff $I=\left(x\right].$ We conclude that $Spec\left(L\right)$ is Noetherian iff for every $I\in \mathbf{I}\left(\mathbf{L}\right)$ there is $x\in L$ such that $I=\left(x\right].$ □

By Theorems 2 and 5, we obtain the relationship between i-Noetherian residuated lattices and residuated lattices with Noetherian spectrum:

Corollary 4.

A residuated lattice L is i-Noetherian iff $Spec\left(L\right)$ is a Noetherian space.

6. Ideals in Residuated Lattices and Ideals in Unitary Commutative Rings (Similarities and Differences)

6.1. Differences

If in any residuated lattice, every finitely generated ideal is principal (according to Proposition 2), there are unitary commutative rings in which there are finitely generated ideals that are not principal. We give an example in this regard by considering the quadratic field $\mathbb{Q}\left(\sqrt{26}\right).$ Since $26\equiv 2$ (mod 4), the ring of algebraic integers of this quadratic field is $\mathbb{Z}\left[\sqrt{26}\right]=\left\{a+b\sqrt{26}|a,b\in \mathbb{Z}\right\}.$

In the proof of Proposition 2, we used the fact that if $x,y$∈I (where I is an ideal in a residuated lattice L), $x\le x⊞y$ and $y\le x⊞y.$ However, this thinking is generally not true in an ideal of unitary commutative ring. For example, in the ring $\left(\mathbb{Z}\left[\sqrt{26}\right],+,·\right),$ if we take the ideal $I=\mathbb{Z}\left[\sqrt{26}\right]$ and $x,y$∈$\mathbb{Z}\left[\sqrt{26}\right],$ $x=1+\sqrt{26},$ $y=-3-2\sqrt{26},$ it results $x+y=-2-\sqrt{26}$ and x $\nleqq$ $x+y$; therefore, for ideals in unitary rings, a proof similar to that in Proposition 2 does not work.

We recall the following results.

Proposition 12

([24]). For any algebraic number field $K,$ the ring of algebraic integers of K is a Dedeking ring.

Proposition 13

([24,25]). In a Dedekind ring, any ideal is finitely generated, with a maximum of 2 generators.

So, in the ring $\mathbb{Z}\left[\sqrt{26}\right]$, any ideal is finitely generated, with a maximum of 2 generators. However, we are showing that $\mathbb{Z}\left[\sqrt{26}\right]$ is not a principal ring.

It is easy to prove that $2,13,\sqrt{26}$ are irreducible elements of this ring; so, $26=2·13=\sqrt{26}·\sqrt{26}$ are two decompositions into irreducible elements of 26 in the ring $\mathbb{Z}\left[\sqrt{26}\right]$. It results that $\mathbb{Z}\left[\sqrt{26}\right]$ is not a factorial ring; so, it is not a principal ring.

In conclusion, $\mathbb{Z}\left[\sqrt{26}\right]$ is a Dedekind ring (so, it is a Noetherian ring), but it is not a principal ring.

6.2. Similarities

If $\left(R,+,·\right)$ is a unitary commutative ring and x∈$R,$ we denote by $\left(x\right)$ the principal ideal generated by $x,$ of the ring $R.$

We asked ourselves if there are unitary commutative rings in which every finitely generated ideal is principal and also if these rings can be endowed similar to a residuated lattices. The answer is yes:

Example 7.

Let M be a nonempty set and let $P\left(M\right)$ be the set of all subsets of the set $M.$ If we consider the composition laws “$+,$”, “⋂”: $P\left(M\right)\times P\left(M\right)\to P\left(M\right),$ $A+B=\left(A\setminus B\right)\bigcup \left(B\setminus A\right)=A\Delta B,$$\left(\forall \right)$$A,B$∈$P\left(M\right),$$A\odot B=A\bigcap B,$$\left(\forall \right)$$A,B$∈$P\left(M\right),$ it is easy to remark that $\left(P\left(M\right),+,\odot \right)$ is a unitary commutative boolean ring, with identity elements: ∅ for “+”, M for “⋂”. So, $A^2=A$ and $A+A=\varnothing ,$ $\left(\forall \right)$ A∈ $P\left(M\right).$

Moreover, $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right)$ is a residuated lattice, in which $A\vee B=A\bigcup B,$$A\wedge B=A\bigcap B$, and $A\to B=C_{M}A\bigcup B,$ $\left(\forall \right)$$A,B$∈$P\left(M\right).$ We remark that, in this residuated lattice, for $A,B$∈$P\left(M\right),$ we have

$$A⊞B={\left(A^{*}\odot B^{*}\right)}^{*}=C_M\left(\left(C_{M}A\right)\bigcap \left(C_{M}B\right)\right)=$$

$$=\left(C_M\left(C_{M}A\right)\right)\bigcup \left(C_M\left(C_{M}B\right)\right)=A\bigcup B.$$

We consider the case when M is a finite set, so all ideals of the ring $\left(P\left(M\right),+,\odot \right)$ are finite generated.

For example, if we take Card $M=3,$ let us to look at the ideals in the unitary commutative boolean ring $\left(P\left(M\right),+,\odot \right)$ and at the ideals in the residuated lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right).$

We denote $M=\left\{1,2,3\right\}.$ Let I be an ideal of the ring $\left(P\left(M\right),+,\odot \right).$ Since Card $\left(P\left(M\right)\right)=8$ and applying Lagrange’s theorem, it results that CardI $\in$ $\left\{1,2,4,8\right\}.$ Knowing that the ring $\left(P\left(M\right),+,\odot \right)$ is boolean, we obtain that the ideals of the ring $\left(P\left(M\right),+,\odot \right)$ are as follows:

$$I_1=\left\{\varnothing \right\}=\left(\varnothing \right),I_2=\left\{\varnothing ,\left\{1\right\}\right\}=\left(\left\{1\right\}\right),$$

$$I_3=\left\{\varnothing ,\left\{2\right\}\right\}=\left(\left\{2\right\}\right),I_4=\left\{\varnothing ,\left\{3\right\}\right\}=\left(\left\{3\right\}\right),$$

$$I_5=\left\{\varnothing ,\left\{1\right\},\left\{2\right\},\left\{1,2\right\}\right\}=\left\{\left\{1,2\right\}\bigcap X|X\in P\left(M\right)\right\}=\left(\left\{1,2\right\}\right),$$

$$I_6=\left\{\varnothing ,\left\{1\right\},\left\{3\right\},\left\{1,3\right\}\right\}=\left\{\left\{1,3\right\}\bigcap X|X\in P\left(M\right)\right\}=\left(\left\{1,3\right\}\right),$$

$$I_7=\left\{\varnothing ,\left\{2\right\},\left\{3\right\},\left\{2,3\right\}\right\}=\left\{\left\{2,3\right\}\bigcap X|X\in P\left(M\right)\right\}=\left(\left\{2,3\right\}\right),$$

$$I_8=P\left(M\right)=\left(\left\{1,2,3\right\}\right).$$

So, all the ideals of the ring $\left(P\left(M\right),+,\odot \right)$ are principal ideals.

We are finding all the ideals of the residuated lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right),$ which is a Boolean algebra. According to Proposition 1, if X⊆$P\left(M\right),$ then, the ideal generated by X in the residuated lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right)$ is $\left(X\right]=\{A\in P\left(M\right):A\subseteq X_1⊞...⊞X_n,$ for some $n\ge 1$ and $X_1,...,X_n\in X\}.$ It results that the ideals of the residuated lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right)$ are as follows:

$$J_1=\left\{\varnothing \right\}=\left(\varnothing \right],J_2=\left\{\varnothing ,\left\{1\right\}\right\}=\left(\left\{1\right\}\right],$$

$$J_3=\left\{\varnothing ,\left\{2\right\}\right\}=\left(\left\{2\right\}\right],J_4=\left\{\varnothing ,\left\{3\right\}\right\}=\left(\left\{3\right\}\right],$$

$$J_5=\left(\left\{1\right\}⊞\left\{2\right\}\right]=\left(\left\{1,2\right\}\right]=\left\{\varnothing ,\left\{1\right\},\left\{2\right\},\left\{1,2\right\}\right\},$$

$$J_6=\left(\left\{1\right\}⊞\left\{3\right\}\right]=\left(\left\{1,3\right\}\right]=\left\{\varnothing ,\left\{1\right\},\left\{3\right\},\left\{1,3\right\}\right\},$$

$$J_7=\left(\left\{2\right\}⊞\left\{3\right\}\right]=\left(\left\{2,3\right\}\right]=\left\{\varnothing ,\left\{2\right\},\left\{3\right\},\left\{2,3\right\}\right\},$$

$$J_8=\left(\left\{1\right\}⊞\left\{2\right\}⊞\left\{3\right\}\right]=\left(\left\{1,2,3\right\}\right]=P\left(M\right).$$

So, we obtain that all ideals of the residuated lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right)$ are principal ideals and $I_l=J_l,$ for $\left(\forall \right)$ $l=\overline{1,8}$—that is, the ideals of the boolean ring $\left(P\left(M\right),+,\odot \right)$ coincide with the ideals of the residuated lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right).$ Further, according to Theorem 2, it results that the lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right)$ is an i-Notherian residuated lattice. Moreover $\left(P\left(M\right),+,\odot \right)$ is a Notherian ring (in the sense of [24,25]).

We asked ourselves if this analogy between the ideals of the boolean ring $\left(P\left(M\right),+,\odot \right)$ and the ideals of the residuated lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right)$ is preserved for any finite set $M.$ The answer is affirmative. To obtain this, we need some results.

Proposition 14

([25]). Let R be a unitary commutative ring and let I be an idempotent ideal of the ring $R.$ If I is finitely generated, there is an idempotent element e∈I such that $I=eR.$

Definition 5

([24]). A Bezout domain is an integral domain in which every finitely generated ideal is principal.

There are rings in which every finitely generated ideal is principal, but they have divisors of zero.

We introduce the following definition.

Definition 6.

A unitary commutative ring with zero divisors, in which every finitely generated ideal is principal, is called a Bezout ring with zero divisors.

We obtain the following results.

Proposition 15.

If $\left(R,+.·\right)$ is a Boolean ring, any ideal of R is idempotent.

Proof. 

Let I be an ideal of $R.$

$I^2=\left\{x_1·y_1+x_2·y_2+....+x_n·y_n|n\in {\mathbb{N}}^{*},x_i,y_i\in I,i=\overline{1,n}\right\}.$ It is clear that $I^2\subseteq I.$ We prove that $I\subseteq I^2.$ Let $x\in I.$ Since $x=x^2,$ it results that $x\in I^2,$ so $I\subseteq I^2.$ We obtain that $I^2=I.$ □

Proposition 16.

Any Boolean ring is a Bezout ring with zero divisors.

Proof. 

It results immediately, using Proposition 15, Proposition 14, and Definition 6. □

Taking into account the results obtained, we deduce the following:

Corollary 5.

Let M be a nonempty set. Then,

$\left(i\right)$

In the residuated lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right)$ all finitely generated ideals are principal;

$\left(ii\right)$

The ring $\left(P\left(M\right),+,\odot \right)$ is a Bezout ring with zero divisors;

$\left(iii\right)$

If M is finite, the ideals of the boolean ring $\left(P\left(M\right),+,\odot \right)$ are principal and these ideals coincide with the ideals of the residuated lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right).$ Further, the lattice $\left(P\left(M\right),\subseteq ,\vee ,\wedge ,\odot ,\to ,\varnothing ,M\right)$ is an i-Notherian residuated lattice and the ring $\left(P\left(M\right),+,\odot \right)$ is a Notherian ring.