On EMV-algebras

TL;DR

This paper introduces EMV-algebras, a new algebraic structure extending MV-algebras without requiring a top element, and explores their properties, representations, and categorical equivalences.

Contribution

The paper defines EMV-algebras, investigates their properties, and establishes their relationship with MV-algebras, including embeddings, ideal structures, and categorical equivalences.

Findings

Every EMV-algebra can be embedded into an MV-algebra.

Semisimple EMV-algebras are isomorphic to EMV-clans of fuzzy functions.

The category of EMV-algebras is categorically equivalent to certain MV-algebra and $ ext{l}$-group categories.

Abstract

The paper deals with an algebraic extension of $MV$-algebras based on the definition of generalized Boolean algebras. We introduce a new algebraic structure, not necessarily with a top element, which is called an $EMV$-algebra and every $EMV$-algebra contains an $MV$-algebra. First, we present basic properties of $EMV$-algebras, give some examples, introduce and investigate congruence relations, ideals and filters on this algebra. We show that each $EMV$-algebra can be embedded into an $MV$-algebra and we characterize $EMV$-algebras either as $MV$-algebras or maximal ideals of $MV$-algebras. We study the lattice of ideals of an $EMV$-algebra and prove that any $EMV$-algebra has at least one maximal ideal. We define an $EMV$-clan of fuzzy sets as a special $EMV$-algebra. We show any semisimple $EMV$-algebra is isomorphic to an $EMV$-clan of fuzzy functions on a set. We consider the variety of $EMV$-algebra and we present an equational base for each proper subvariety of the variety of $EMV$-algebras. We establish a categorical equivalencies of the category of proper $EMV$-algebras, the category of $MV$-algebras with a fixed special maximal ideal, and a special category of Abelian unital $\ell$-groups.

Equations

$$x\odot y:=(x^{\prime }\oplus y^{\prime }{)}^{\prime },x\ominus y:=(x^{\prime }\oplus y{)}^{\prime }.$$

$$0.x=0,n.x=(n-1).x\oplus x,n\ge 1,$$

$$x^0=1,x^n=x^{n-1}\odot x,n\ge 1.$$

$$x\cdot y\le z\iff x\le z/y\iff y\le x\setminus z.$$

$$x/((x\vee y)\setminus x)=x\vee y=(x/(x\vee y))\setminus x.$$

$${\lambda }_{a,b}(x)=\min \{z\in [a,b]\mid x\oplus z=b\}$$

$$\overline{\overline{x}}$$

$$x{\odot }_{{}_a}y={\lambda }_a({\lambda }_a(x)\oplus {\lambda }_a(y)),x,y\in [0,a],$$

$$\forall b\in A\cap \mathcal{I}(M),\forall x\in [0,b{]}_A,\min \{z\in [0,b{]}_A\mid x\oplus z=b\}=\min \{z\in [0,b]\mid x\oplus z=b\},$$

$${\lambda }_b(y)={\lambda }_{f(u)}(y)\wedge b={\lambda }_{f(u)}(f(x))\wedge f(a)=f({\lambda }_u(x))\wedge f(a)=f({\lambda }_u(x)\wedge a).$$

$${\lambda }_a(x)\vee {\lambda }_b(a)$$

$$=({\lambda }_b(x)\vee {\lambda }_b(a))\wedge (a\vee {\lambda }_b(a))$$

$$=({\lambda }_b(x)\vee {\lambda }_b(a))\wedge b={\lambda }_b(x)\vee {\lambda }_b(a).$$

$$f({\lambda }_a(x))=f({\lambda }_u(x)\wedge a)={\lambda }_{f(u)}(f(x))\wedge f(a)={\lambda }_{f(a)}(f(x)).$$

$$[x]\vee [y]=[x\vee y],[x]\wedge [y]=[x\wedge y],[x]\oplus [y]=[x\oplus y],\forall x,y\in M.$$

$${\lambda }_{a/{\theta }_a}(x/{\theta }_a)=\min \{z/{\theta }_a\mid z\in [0,a],z/{\theta }_a\oplus x/{\theta }_a=a/{\theta }_a\}={\lambda }_a(x)/{\theta }_a,\forall x/{\theta }_a\in [0,a]/{\theta }_a.$$

$$\theta =\{(f,g)\mid f,g\in \sum _{i\in I}{M}_i,(f(i),g(i))\in {\theta }_i,\forall i\in I\}.$$

$$h(i)=\{\begin{array}{ll}1,& i\in \mathrm{Supp}(f)\cup \mathrm{Supp}(g),\\ 0,& \text{otherwise}.\end{array}$$

$${\lambda }_b(\alpha )(i)=\{\begin{array}{ll}{\lambda }_{b_i}(\alpha (i)),& i\in \mathrm{Supp}(f)\cup \mathrm{Supp}(g),\\ 0,& \text{otherwise. }\end{array}$$

$$\displaystyle(x,y)\in\theta_{I}\Longleftrightarrow\Big{(}\exists\,b\in\mathcal{I}(M):x,y\leq b\quad\&\quad\lambda_{b}(\lambda_{b}(x)\oplus y),\lambda_{b}(\lambda_{b}(y)\oplus x)\in I\Big{)}$$

$${\lambda }_u(s)={\lambda }_a(s)\oplus {\lambda }_u(a),{\lambda }_u(t)={\lambda }_b(t)\oplus {\lambda }_u(b),\forall s\in [0,a],\forall t\in [0,b].$$

$${\lambda }_u({\lambda }_u(x)\oplus y)\ominus {\lambda }_a({\lambda }_a(x)\oplus y)$$

$${\lambda }_u({\lambda }_u(y)\oplus x),{\lambda }_u({\lambda }_u(z)\oplus w),{\lambda }_u({\lambda }_u(w)\oplus z)\in I\cap [0,u].$$

$$\{z\in M\mid z\le x_1\oplus x_2\oplus \cdots \oplus x_n,\exists n\in \mathbb{N},\exists x_1,\dots ,x_n\in M\}.$$

$$⟨I\cup \{x\}⟩=\{z\in M\mid z\le a\oplus n.x,\exists a\in I,\exists n\in \mathbb{N}\}.$$

$$z=z\wedge b\le (a\oplus n.x)\wedge b\le (a\wedge b)\oplus (n.x\wedge b)\le (a\wedge b)\oplus n.x.$$

$$⟨I⟩=\downarrow I=\{x\in M\mid x\le z,\exists z\in I\}.$$

$${\lambda }_b({\lambda }_b(x)\oplus y)$$

$$\displaystyle=\lambda_{b}\big{(}(\lambda_{d}(x)\oplus y)\wedge(b\oplus y)\big{)}$$

$$\displaystyle=\lambda_{b}\big{(}(\lambda_{d}(x)\oplus y)\wedge b\big{)},\mbox{ since $y\leq b$ }$$

$$\displaystyle=\lambda_{d}\big{(}(\lambda_{d}(x)\oplus y)\wedge b\big{)}\wedge b$$

$$\displaystyle=\big{(}\lambda_{d}(\lambda_{d}(x)\oplus y)\vee\lambda_{d}(b)\big{)}\wedge b$$

$$\displaystyle=\big{(}\lambda_{d}(\lambda_{d}(x)\oplus y)\wedge b\big{)}\vee\big{(}\lambda_{d}(b)\wedge b\big{)}$$

$$={\lambda }_d({\lambda }_d(x)\oplus y)\wedge b\in I_b.$$

Full text

*AMS Mathematics Subject Classification (2010): 06C15, 06D35 *

*Keywords: Generalized Boolean algebra, MV-algebra, idempotent element, $qEMV$-algebra, $EMV$-algebra, $EMV$-clan, state-morphism, ideal, filter, variety, equational base, categorical equivalence *

*Acknowledgement: AD is thankful for the support by the grants VEGA No. 2/0069/16 SAV and GAČR 15-15286S *

1 Introduction

$MV$-algebras were defined by Chang [Cha] as an algebraic counterpart of many-valued reasoning. The principal result of the theory of $MV$-algebras is a representation theorem by Mundici [Mun1] saying that there is a categorical equivalence between the category of MV-algebras and the category of unital Abelian $\ell$-groups. Today the theory of $MV$-algebras is very deep and has many interesting connections with other parts of mathematics with many important applications to different areas. For more details on $MV$-algebras, we recommend the monographs [CDM, Mun3].

$GMV$-algebras, called also pseudo $MV$-algebras [GeIo] or non-commutative $MV$-algebras [Rac], are a non-commutative generalization of $MV$-algebras and the algebraic counterparts of non-commutative many valued logic. Moreover, Galatos and Tsinakis generalized the notion of an $MV$-algebra in the context of residuated lattices to include both commutative and unbounded structures in [GaTs] and introduced the notion of generalized $MV$-algebra. Indeed, a pseudo $MV$-algebra is a bounded integral generalized $MV$-algebra. They extended the relation between unital $\ell$-groups and pseudo $MV$-algebras established by Mundici [Mun1] for $MV$-algebras and by Dvurečenskij [Dvu2] for pseudo $MV$-algebras. Many other results in these structures can be find in [DDT, Dvu1, Dvu2, RaSa, ShLu]. We note that $MV$-algebras are studied in the last period also in the frames of involutive semirings, see [DiRu].

There is another way how to generalize the concept of $MV$-algebras considering the definition of generalized Boolean algebras. In the paper, first we use the definition of generalized Boolean algebras to extend the concept of $MV$-algebras. We call this structure an $EMV$-algebra. These algebras generalize $MV$-algebras, and in any $EMV$-algebra $M$, the interval $[0,a]$ forms an $MV$-algebra for each idempotent $a\in M$.

The paper is organized as follows. After Preliminaries, Section 2, we introduce $EMV$-algebras in Section 3. We present some examples and find relations between generalized Boolean algebras and $EMV$-algebras. We exhibit basic properties of $EMV$-algebras, and in particular, we show a one-to-one relationship between ideals and congruences of $EMV$-algebras. In Section 4, we introduce state-morphisms as analogues of finitely additive probability measures, and we show their intimate relationship with maximal ideals. We introduce $EMV$-clans as $EMV$-algebras of fuzzy sets where all algebraic operations are defined by points; they are exactly semisimple $EMV$-algebras up to isomorphism. We also show that any semisimple $EMV$-algebra can be embedded into an $MV$-algebra. In Section 5, we show a relationship between ideals and filters of $EMV$-algebras and we show that nevertheless an $EMV$-algebra has not necessarily a top element, it contains at least one maximal ideal. Finally, we prove that every $EMV$-algebra can be embedded into an $MV$-algebra. Moreover, every $EMV$-algebra is either an $MV$-algebra or it can be embedded into an $MV$-algebra as its maximal ideal. This allows us to study subvarieties of the variety of $EMV$-algebras and to present an equational base for each subvariety of $EMV$-algebras. In addition, in Section 6, we present mutual categorical equivalencies of the category of proper $EMV$-algebras with the special category of $MV$-algebras with a fixed maximal ideal having enough idempotents or with a special category of Abelian unital $\ell$-groups.

2 Preliminaries

In the section, we gather some basic notions relevant to $MV$-algebras which will be needed in the next sections. For more details, we recommend to consult [DiSe, CDM, Mun3] for $MV$-algebras.

An $MV$-algebra is an algebra $(M;\oplus,^{\prime},0,1)$ (henceforth write simply $M=(M;\oplus,^{\prime},0,1)$) of type $(2,1,0,0)$, where $(M;\oplus,0)$ is a commutative monoid with the neutral element [math] and for all $x,y\in M$, we have:

1. (i)

$x^{\prime\prime}=x$; 2. (ii)

$x\oplus 1=1$; 3. (iii)

$x\oplus(x\oplus y^{\prime})^{\prime}=y\oplus(y\oplus x^{\prime})^{\prime}$.

In any $MV$-algebra $(M;\oplus,^{\prime},0,1)$, we can define also the following operations:

[TABLE]

In addition, let $x\in M$. For any integer $n\geq 0$, we set

[TABLE]

and

[TABLE]

Moreover, the relation $x\leq y\Leftrightarrow x^{\prime}\oplus y=1$ is a partial order on $M$ and $(M;\leq)$ is a lattice, where $x\vee y=(x\ominus y)\oplus y$ and $x\wedge y=x\odot(x^{\prime}\oplus y)$. Note that, for each $x\in M$, $x^{\prime}$ is the least element of the set $\{y\in M\mid x\oplus y=1\}$. We use $\mathbb{MV}$ to denote the category of $MV$-algebras whose objects are $MV$-algebras and morphisms are $MV$-homomorphisms. A non-empty subset $I$ of an $MV$-algebra $(M;\oplus,^{\prime},0,1)$ is called an ideal of $M$ if $I$ is a down set which is closed under $\oplus$. The set of all ideals of $M$ is denoted by $I(M)$. It is well known that for each $x,y\in M$, if $I\in I(M)$ and $y,x\ominus y\in I$, then $x\in I$. For each ideal $I$ of $M$, the relation $\theta_{I}$ on $M$ defined by $(x,y)\in\theta_{I}$ if and only if $x\ominus y,y\ominus x\in I$ is a congruence relation on $M$, and $x/I$ and $M/I$ will denote $\{y\in M\mid(x,y)\in\theta_{I}\}$ and $\{x/I\mid x\in M\}$, respectively. A prime ideal is an ideal $I\neq M$ of $M$ such that $M/I$ is a linearly ordered $MV$-algebra, or equivalently, for all $x,y\in M$, $x\ominus y\in I$ or $y\ominus x\in I$. The set of all minimal prime ideals of $M$ is denoted by $Min(M)$. An element $a$ of an $MV$-algebra $(M;\oplus,^{\prime},0,1)$ is called a Boolean element if there is $b\in M$ such that $a\wedge b=0$ and $a\vee b=1$. The set of all Boolean elements of $M$ forms a Boolean algebra; it is denoted by $B(M)$.

Theorem 2.1**.**

[CDM, Thm. 1.5.3]* For every element $x$ in an $MV$-algebra $M$, the following conditions are equivalent:*

• (i)

$x\in B(M)$;

• (ii)

$x\vee x^{\prime}=1$;

• (iii)

$x\wedge x^{\prime}=0$;

• (iv)

$x\oplus x=x$;

• (v)

$x\odot x=x$;

• (vi)

$x\oplus y=x\vee y$* for all $y\in M$;*

• (vii)

$x\odot y=x\wedge y$* for all $y\in M$.*

Let $(M;+,0)$ be a monoid. An element $a\in M$ is called idempotent if $a+a=a$. The set of all idempotent elements of $M$ is denoted by $\mathcal{I}(M)$. A monoid $(G;+,0)$ is called partially ordered if it is equipped with a partial order relation $\leq$ that is compatible with $+$, that is, $a\leq b$ implies $x+a+y\leq x+b+y$ for all $x,y\in G$. A partially ordered monoid $(G;+,0)$ is called a lattice ordered monoid or simply an $\ell$-monoid if $G$ with its partially order relation is a lattice. In a similar way, a group $(G;+,0)$ is said to be a partially ordered group if it is a partially ordered monoid. A partially ordered group $(G;+,0)$ is called a lattice ordered group or simply an $\ell$-group if $G$ with its partially order relation is a lattice. An element $x\in G$ is called positive if $0\leq x$. An element $u$ of an $\ell$-group $(G;+,0)$ is called a strong unit of $G$ if, for each $g\in G$, there exists $n\in\mathbb{N}$ such that $g\leq nu$. A couple $(G,u)$, where $G$ is an $\ell$-group and $u$ is a fixed strong unit for $G$, is said to be a unital $\ell$-group.

If $(G;+,0)$ is an Abelian $\ell$-group with strong unit $u$, then the interval $[0,u]:=\{g\in G\mid 0\leq g\leq u\}$ with the operations $x\oplus y:=(x+y)\wedge u$ and $x^{\prime}:=u-x$ forms an $MV$-algebra, which is denoted by $\Gamma(G,u)=([0,u];\oplus,^{\prime},0,u)$. Moreover, if $(M;\oplus,0,1)$ is an $MV$-algebra, then according to the famous theorem by Mundici, [Mun1], there exists a unique (up to isomorphism) unital Abelian $\ell$-group $(G,u)$ with strong $u$ such that $\Gamma(G,u)$ and $(M;\oplus,0,1)$ are isomorphic (as $MV$-algebras). Let $\mathcal{A}$ be the category of unital Abelian $\ell$-groups whose objects are unital $\ell$-groups and morphisms are unital $\ell$-group morphisms (i.e. homomorphisms of $\ell$-groups preserving fixed strong units). It is important to note that $\mathbb{MV}$ is a variety whereas $\mathcal{A}$ not because it is not closed under infinite products. Then $\Gamma:\mathcal{A}\rightarrow\mathbb{MV}$ is a functor between these categories. Moreover, there is another functor from the category of $MV$-algebras to $\mathcal{A}$ sending $M$ to a Chang $\ell$-group induced by good sequences of the $MV$-algebra $M$, which is denoted by $\Xi:\mathbb{MV}\rightarrow\mathcal{A}$. For more details relevant to these functors, please see [CDM, Chaps 2 and 7].

Theorem 2.2**.**

[CDM, Thms 7.1.2, 7.1.7]* The composite functors $\Gamma\Xi$ and $\Xi\Gamma$ are naturally equivalent to the identity functors of $\mathbb{MV}$ and $\mathcal{A}$, respectively. Therefore, the categories $\mathcal{A}$ and $\mathbb{MV}$ are categorically equivalent.*

Recall that a residuated lattice is an algebra $(L;\vee,\wedge,\cdot,\setminus,/,e)$ of type $(2,2,2,2,2,0)$ such that $(L;\vee,\wedge)$ is a lattice, $(L;\cdot,e)$ is a monoid, and for all $x,y,z\in L$,

[TABLE]

A residuated lattice is called commutative if it satisfies the identity $x\cdot y=y\cdot x$ and is called integral if it satisfied the identity $x\wedge e=x$. Galatos and Tsinakis, [GaTs], introduced the concept of a generalized MV-algebra ($GMV$-algebra) which is a residuated lattice that satisfies the identities

[TABLE]

It is well known that bounded commutative integral $GMV$-algebras and $MV$-algebras coincide (see [JiTs, GaTs]).

3 $EMV$-algebras, Ideals, and Congruences

In the section, we define $qEMV$-algebras and $EMV$-algebras which form an important subclass of $qEMV$-algebras. We present some examples and we define subalgebras and homomorphisms. We show that $EMV$-algebras form a variety. Congruences on the class of $EMV$-algebras are in a one-to-one correspondence with the set of ideals. We show that every semisimple $EMV$-algebra can be embedded into an MV-algebra.

Definition 3.1**.**

An algebra $(M;\vee,\wedge,\oplus,0)$ of type $(2,2,2,0)$ is called a quasi extended $MV$-algebra ($qEMV$-algebra in short) if it satisfies the following conditions:

• ($\mathrm{EMV}1$) $(M;\vee,\wedge,0)$ is a distributive lattice with the least element [math];
• ($\mathrm{EMV}2$)  $(M;\oplus,0)$ is a commutative ordered monoid with neutral element [math];
• ($\mathrm{EMV}3$) for all $a,b\in\mathcal{I}(M)$ such that $a\leq b$, the element

[TABLE]

exists in $M$ for all $x\in[a,b]$, and the algebra $([a,b];\oplus,\lambda_{a,b},a,b)$ is an $MV$-algebra.

We say that an $qEMV$-algebra $(M;\vee,\wedge,\oplus,0)$ has enough idempotent elements if, for each $x\in M$, there is $a\in\mathcal{I}(M)$ such that $x\leq a$. An extended $MV$-algebra, an $EMV$-algebra in short, is a $qEMV$-algebra $(M;\vee,\wedge,\oplus,0)$ which has enough idempotent elements.

From now on, in this paper, we usually denote $\lambda_{0,b}$ by $\lambda_{b}$.

Now we present some examples of $qEMV$-algebras and $EMV$-algebras, respectively.

Example 3.2**.**

1. (1)

Any $MV$-algebra $(M;\oplus,^{\prime},0,1)$ is an $EMV$-algebra. Let $a,b\in B(M)$. By Proposition 2.1, for each $x,y\in[a,b]$, we have $x\oplus y\geq a\oplus y=a\vee y=y\geq a$ and $x\oplus y\leq x\oplus b=x\vee b=b$, thus $[a,b]$ is closed under $\oplus$. It can be easily seen that $\overline{x}:=(x^{\prime}\vee a)\wedge b$ is the least element of the set $\{z\in[a,b]\mid x\oplus z=b\}$. Moreover, for each $x\in[a,b]$,

[TABLE]

Therefore, $([a,b];\oplus,^{-},a,b)$ is an $MV$-algebra (for more details we refer to [DMN]), and so any $MV$-algebra is an $EMV$-algebra. 2. (2)

Any generalized Boolean algebra $(M;\vee,\wedge,0)$ (studied also as a Boolean ring, see [LuZa, Kel]) forms an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$, where $\oplus=\vee$ and if $a\leq b$, then $\lambda_{a,b}(x)$ is the unique relative complement of $x$ in the interval $[a,b]$. 3. (3)

Let $(B;\vee,\wedge)$ be a generalized Boolean algebra and $(M;\oplus,^{\prime},0,1)$ be an $MV$-algebra. Then it can be easily shown that $M\times B$ is an $EMV$-algebra. 4. (4)

Any bounded $qEMV$-algebra is an $MV$-algebra. Note that if $M$ is a $qEMV$-algebra with the greatest element $1$, then $M=[0,1]$ and $M$ is an $MV$-algebra. 5. (5)

Let $G$ be a non-trivial $\ell$-group. The set of positive elements $G^{+}$ of $G$ with the natural operation $+$ and natural ordering is a $qEMV$-algebra. Since [math] is the only idempotent element, so $G^{+}$ is not an $EMV$-algebra. 6. (6)

Let $\{(M_{i};\oplus,^{\prime},0,1)\}_{i\in I}$ be a family of $MV$-algebras and $S=\{f\in\prod_{i\in I}M_{i}\mid\mathrm{Supp}(f)\mbox{ is finite}\}$. Clearly, $S$ is closed under $\vee$, $\wedge$ and $\oplus$. Moreover, if $f\in S$, then $u=(u_{i})_{i\in I}$, where $u_{i}=1$ for all $i\in\mathrm{Supp}(f)$ and $u_{i}=0$ for all $i\in I\setminus\mathrm{Supp}(f)$, is an element of $S$ which is idempotent and $f\leq u$. It can be easily shown that $S$ is a $qEMV$-algebra and so $S$ is an $EMV$-algebra. We will denote this $qEMV$-algebra by $\sum_{i\in I}M_{i}$. 7. (7)

Let $(M;\oplus,^{\prime},0,1)$ be an $MV$-algebra and $A$ be any ideal of $M$. Then similarly to (1), we can see that $A$ is a $qEMV$-algebra. 8. (8)

Let $J$ be an ideal of an $MV$-algebra $(A;\oplus,^{\prime},0,1)$ and $B$ be a generalized Boolean algebra. Then $B\times J$ with the pointwise operations forms a $qEMV$-algebra. 9. (9)

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra. Then it is straightforward to show that $(\mathcal{I}(M);\vee,\wedge,0)$ is a generalized Boolean algebra. Moreover, $M$ is an $MV$-algebra if and only if $\mathcal{I}(M)$ is a Boolean algebra. 10. (10)

Every finite $EMV$-algebra is an $MV$-algebra. 11. (11)

Every $EMV$-clan of fuzzy sets on some $\Omega\neq\emptyset$ is an $EMV$-algebra, where all operations are defined by points, see Definition 4.9 and Proposition 4.10 below.

Remark 3.3**.**

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra.

(i) For each $x,y\in M$, there exist $a,b\in\mathcal{I}(M)$ such $a\leq x\oplus y\leq b$ and so $x,y\in[a,b]$. Since $([a,b];\oplus,\lambda_{a,b},a,b)$ is an $MV$-algebra, then the element $\lambda_{a,b}(\lambda_{a,b}(x)\oplus y)\oplus y$ is the supremum of $x$ and $y$ taken in the MV-algebra $[a,b]$ and it coincides with $x\vee y$. Similarly, $x\wedge y=\lambda_{a,b}(\lambda_{a,b}(x)\oplus\lambda_{a,b}(\lambda_{a,b}(x)\oplus y))$. In addition, if $a_{0}\leq x,y\leq b_{0}$ for some $a_{0},b_{0}\in\mathcal{I}(M)$, then $\lambda_{a,b}(\lambda_{a,b}(x)\oplus y)=x\vee y=\lambda_{a_{0},b_{0}}(\lambda_{a_{0},b_{0}}(x)\oplus y)$.

(ii) If $a,c,b$ are idempotents with $a\leq c\leq b$ and $x\in M$ such $x\in[a,b]$, then $x\oplus c=x\vee c$ and $x\wedge c=\lambda_{a,b}(\lambda_{a,b}(x)\oplus\lambda_{a,b}(\lambda_{a,b}(x)\oplus c))$.

(iii) The Riesz Decomposition Theorem holds: If $z\leq x\oplus y$, then there are $x_{z}\leq x$ and $y_{z}\leq y$ such that $x=x_{z}\oplus y_{z}$. Or if $x_{1}\oplus x_{2}=y_{1}\oplus y_{2}$, there are four elements $c_{11},c_{12},c_{21},c_{22}\in M$ such that $x_{1}=c_{11}\oplus c_{12}$, $x_{2}=c_{21}\oplus c_{22}$, $y_{1}=c_{11}\oplus c_{21}$ and $y_{2}=c_{21}\oplus c_{22}$. These facts follow from the analogous properties in the $MV$-algebra $[0,a]$, where $a\geq x,y,x_{1},x_{2},y_{1},y_{2}$.

Remark 3.4**.**

Let $(L;\vee,\wedge,\cdot,\setminus,/,e)$ be a residuated lattice such that $(L;\cdot,e)$ is commutative and $(L;\vee,\wedge)$ is a lattice with the least element [math]. Then $y/x=x\setminus y$ for all $x,y\in L$, and $x\setminus y$ is usually written $x\rightarrow y$ (see [GaTs, p. 12]). We claim that $L$ is not an $EMV$-algebra. Otherwise, since $0\rightarrow 0$ is the greatest element of $L$, then by Example 3.2(4), $L$ is an $MV$-algebra. Therefore, from Example 3.2 we get that there exists an $EMV$-algebra which is not a generalized $MV$-algebra in the sense of [GaTs].

The following proposition shows that the notions of a $qEMV$-algebra and of an $EMV$-algebra can be defined also in a simpler way.

Proposition 3.5**.**

Let $(M;\vee,\wedge,\oplus,0)$ be an algebra of type $(2,2,2,0)$. Then

• (i)

$M$* is a $qEMV$-algebra if and only if $([0,b];\oplus,\lambda_{b},0,b)$ is an $MV$-algebra for all $b\in\mathcal{I}(M)$. In such a case, $\lambda_{a,b}(x)=\lambda_{b}(x)\vee a$.*

• (ii)

$M$* is an $EMV$-algebra if and only if*

($\mathrm{EMV}_{I}1$)* $(M;\vee,\wedge,0)$ is a lattice with the least element 0;*

($\mathrm{EMV}_{I}2$)* $(M;\oplus,0)$ is a commutative monoid with neutral element [math];*

($\mathrm{EMV}_{I}3$)* for each $x\in M$, there is $b\in\mathcal{I}(M)$ with $x\leq b$ such that $([0,b];\oplus,\lambda_{b},0,b)$ is an $MV$-algebra.*

Proof.

(i) Let, for all $b\in\mathcal{I}(M)$, $([0,b];\oplus,\lambda_{b},0,b)$ be an $MV$-algebra. Take $a,b\in\mathcal{I}(M)$ such that $a\leq b$. We show that for each $x\in[a,b]$, $\lambda_{a,b}(x)$ exists and $\lambda_{a,b}(x)=\lambda_{b}(x)\vee a$. Indeed, $x\oplus(\lambda_{b}(x)\vee a)=(x\oplus\lambda_{b}(x))\vee(x\oplus a)=b$. Now, let $z\in[a,b]$ such that $x\oplus z=b$. Then by definition of $\lambda_{b}(x)$, we have $\lambda_{b}(x)\leq z$ and so $\lambda_{b}(x)\vee a\leq z\vee a=z$. Now, we can easily see that $([a,b];\oplus,\lambda_{a,b},a,b)$ is an $MV$-algebra. Therefore, $(M;\vee,\wedge,\oplus,0)$ is a $qEMV$-algebra. The proof of the converse is clear.

(ii) Let ($\mathrm{EMV}_{I}1$)–($\mathrm{EMV}_{I}3$) hold. First we show that $(M;\oplus,0)$ is an ordered monoid. Let $x,y\in M$ be such that $x\leq y$. For each $z\in M$, by the assumption, there exists $a\in\mathcal{I}(M)$ such that $y\vee z\leq a$ and $([0,a];\oplus,\lambda_{a},0,a)$ is an $MV$-algebra and so $x\oplus z\leq y\oplus z$ (since $x,y,z\in[0,a]$). That is, $(M;\oplus,0)$ is an ordered monoid. In a similar way, we can show that $(M;\vee,\wedge)$ is a distributive lattice. Now, by (i), it is enough to show that for all $b\in\mathcal{I}(M)$, $([0,b];\oplus,\lambda_{b},0,b)$ is an $MV$-algebra. Let $b$ be an arbitrary idempotent element of $M$. By the assumption, there is $u\in\mathcal{I}(M)$ such that $b\leq u$ and $([0,u];\oplus,\lambda_{u},0,u)$ is an $MV$-algebra. It can be easily seen that $\lambda_{b}(x)=b\wedge\lambda_{u}(x)$ for all $x\in[0,b]$, and similarly to Example 3.2(1), $([0,b];\oplus,\lambda_{b},0,b)$ is an $MV$-algebra. Therefore, $M$ is an $EMV$-algebra. Clearly, the converse holds. ∎

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra. Then for all $a\in\mathcal{I}(M)$, we have a well-known binary operation

[TABLE]

on the $MV$-algebra $([0,a];\oplus,\lambda_{a},0,a)$.

Inspired by the equivalence in Proposition 3.5(i), we can define the notion of a $qEMV$-subalgebra also in the following equivalent way.

Definition 3.6**.**

(i) Let $(M;\vee,\wedge,\oplus,0)$ be a $qEMV$-algebra. A subset $A\subseteq M$ is called a $qEMV$-subalgebra of $M$ if $A$ is closed under $\vee$, $\wedge$, $\oplus$ and [math] and for each $b\in\mathcal{I}(M)\cap A$ the set $[0,b]_{A}:=[0,b]\cap A$ is a subalgebra of the $MV$-algebra $([0,b];\oplus,\lambda_{b},0,b)$. Clearly, the last condition is equivalent to the following condition:

[TABLE]

and due to Proposition 3.5(i), this also means $\lambda_{a,b}(x)$ is defined in $[a,b]_{A}$ for all $a,b\in\mathcal{I}(M)\cap A$ with $a\leq b$ and $x\in[a,b]_{A}$. If $(M;\vee,\wedge,\oplus,0)$ is an $EMV$-algebra and $A$ is a $qEMV$-subalgebra of $M$ such that for all $x\in A$, there is $b\in A\cap\mathcal{I}(M)$ such that $x\leq b$, then $A$ is called an $EMV$-subalgebra of $M$.

(ii) Let $(M_{1};\vee,\wedge,\oplus,0)$ and $(M_{2};\vee,\wedge,\oplus,0)$ be $qEMV$-algebras. A map $f:M_{1}\rightarrow M_{2}$ is called a $qEMV$-homomorphism if $f$ preserves the operations $\vee$, $\wedge$, $\oplus$ and [math], and for each $b\in\mathcal{I}(M_{1})$ and for each $x\in[0,b]$, $f(\lambda_{b}(x))=\lambda_{f(b)}(f(x))$.

If $M_{1}$ and $M_{2}$ are two $EMV$-algebras, then each $qEMV$-homomorphism $f:M_{1}\rightarrow M_{2}$ is said to be an $EMV$-homomorphism.

Lemma 3.7**.**

Let $M_{1}$ and $M_{2}$ be two $qEMV$-algebras and $f:M_{1}\rightarrow M_{2}$ be a $qEMV$-homomorphism.

• (i)

If $B$ is a subalgebra of $M_{2}$, then $f^{-1}(B)$ is a subalgebra of $M_{1}$.

• (ii)

If $M_{1}$ and $M_{2}$ are $EMV$-algebras, $f:M_{1}\to M_{2}$ is an $EMV$-homomorphism, and $A$ is an $EMV$-subalgebra of $M_{1}$, then $f(A)$ is an $EMV$-subalgebra of $M_{2}$.

Proof.

(i) Clearly, $f^{-1}(B)$ is closed under the operations $\oplus$, $\vee$, $\wedge$ and [math]. Let $a\in f^{-1}(B)\cap\mathcal{I}(M_{1})$. We show that $[0,a]\cap f^{-1}(B)$ is a subalgebra of the $MV$-algebra $[0,a]$. Put $x,y\in[0,a]\cap f^{-1}(B)$. Then clearly, $x\oplus y\in[0,a]\cap f^{-1}(B)$. Also from $f(\lambda_{a}(x))=\lambda_{f(a)}(f(x))$, we get that $\lambda_{a}(x)\in[0,a]\cap f^{-1}(B)$, so $[0,a]\cap f^{-1}(B)$ is a subalgebra of $[0,a]$. Therefore, $f^{-1}(B)$ is a subalgebra of $M_{1}$.

(ii) Clearly, $f(A)$ is closed under $\oplus,\vee,\wedge$ and [math]. Now we have to show that, for each $b\in f(A)\cap\mathcal{I}(M_{2})$, the set $f(A)\cap[0,b]$ is a subalgebra of the $MV$-algebra $[0,b]$. By definition, since $A$ is an $EMV$-subalgebra of $M_{1}$, then for each $y\in f(A)$, there is an element $b\in f(A)\cap\mathcal{I}(M_{2})$ such that $y\leq b$. We only need to show that $f(A)\cap[0,b]$ is closed under $\lambda_{b}$. Put $y\in f(A)\cap[0,b]$. Then there exist $a,x\in A$ such that $f(a)=b$ and $f(x)=y$. Let $u\in\mathcal{I}(M_{1})\cap A$ such that $a,x\leq u$. Then $b,y\in[0,f(u)]$ and so

[TABLE]

Since $A$ is a subalgebra of $M_{1}$, then $[0,u]\cap A$ is a subalgebra of the $MV$-algebra $[0,u]$ and so $\lambda_{u}(x)\wedge a\in A$, which implies that $\lambda_{b}(y)\in f(A)$. Clearly, $\lambda_{b}(y)\in[0,b]$. Therefore, $\lambda_{b}(y)\in[0,b]\cap f(A)$. That is, $f(A)$ is an $EMV$-subalgebra of $M_{2}$. ∎

Remark 3.8**.**

(i) Definition 3.6 yields that each $MV$-homomorphism is an $EMV$-homomorphism, but the converse is not true in general case. Indeed, let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra and $a\in\mathcal{I}(M)$. Then $([0,a];\oplus,\lambda_{a},0,a)$ is an $MV$-algebra (and so an $EMV$-algebra). Clearly, the inclusion map $i:[0,a]\rightarrow M$ is an $EMV$-homomorphism. Now, if $(M;\oplus,^{\prime},0,1)$ is an $MV$-algebra and $a\in M\setminus\{0,1\}$ is its Boolean element, then the inclusion map $i:[0,a]\rightarrow M$ is not an $MV$-homomorphism (since $i(a)\neq 1$).

An $EMV$-homomorphism $f:M\rightarrow N$ is said to be strong if, for each $b\in\mathcal{I}(N)$, there exists $a\in\mathcal{I}(M)$ such that $b\leq f(a)$. Clearly, any $MV$-homomorphism is strong as an $EMV$-homomorphism. Moreover, if $(M;\oplus,^{\prime},0,1)$ is an $MV$-algebra and $f:M\rightarrow N$ is an $EMV$-homomorphism, then $N$ is an $MV$-algebra and $f$ is an $MV$-homomorphism, since $x\leq f(1)$ for all $x\in N$.

(ii) Let $f:M\rightarrow N$ be a strong $EMV$-homomorphism and $S$ be a full subset of $\mathcal{I}(M)$ (that is, for each $b\in\mathcal{I}(M)$, there exists $a\in S$ such that $b\leq a$). For each $a\in\mathcal{I}(M)$, set $f_{a}:=f\mid_{{}_{[0,a]}}$. Then we have

(F1) $\{f_{a}\mid a\in S\}$ is a family of $MV$-homomorphisms;

(F2) if $a,b\in S$ such that $a\leq b$, then $f_{a}=f_{b}\mid_{{}_{[0,a]}}$;

(F3) $\{f(a)\mid a\in S\}$ is a full subset of $\mathcal{I}(N)$.

Conversely, if $(M;\vee,\wedge,\oplus,0)$ and $(N;\vee,\wedge,\oplus,0)$ are two $EMV$-algebras, $S$ is a full subset of $M$ and $\{f_{a}:[0,a]\rightarrow N\mid a\in S\}$ is a family of maps satisfying conditions (F1)–(F3), then the map $f:M\rightarrow N$ defined by $f(x)=f_{a}(x)$, where $a\in S$ and $x\in[0,a]$, is a strong $EMV$-homomorphism.

Proposition 3.9**.**

Let $(M;\vee,\wedge,\oplus,0)$ be a $qEMV$-algebra, $a,b\in\mathcal{I}(M)$ such that $a\leq b$. Then for each $x\in[0,a]$, we have

• (i)

$\lambda_{a}(x)=\lambda_{b}(x)\wedge a$;

• (ii)

$\lambda_{b}(x)=\lambda_{a}(x)\oplus\lambda_{b}(a)$;

• (iii)

$\lambda_{a,b}(x)=\lambda_{b}(x)\vee a$;

• (iv)

$\lambda_{a}(x)\leq\lambda_{b}(x)$;

• (v)

$\lambda_{b}(a)$* is an idempotent, and $\lambda_{a}(a)=0$.*

Proof.

Since $([0,b];\oplus,\lambda_{b},0,b)$ is an $MV$-algebra, $a\in[0,b]$ and $a\oplus a=a$, we get that $a\vee x=a\oplus x$ for all $x\in[0,b]$. Let $x\in[0,a]$.

(i) From $\lambda_{b}(x)\wedge a\in[0,a]$ and $(\lambda_{b}(x)\wedge a)\oplus x=(\lambda_{b}(x)\oplus x)\wedge(a\oplus x)=b\wedge(a\vee x)=a$, it follows that $\lambda_{a}(x)\leq\lambda_{b}(x)\wedge a$. Also, $b=a\oplus\lambda_{b}(a)=(x\oplus\lambda_{a}(x))\oplus\lambda_{b}(a)=x\oplus(\lambda_{a}(x)\oplus\lambda_{b}(a))$, so $\lambda_{a}(x)\oplus\lambda_{b}(x)\geq\lambda_{b}(x)$. Hence $(\lambda_{a}(x)\oplus\lambda_{b}(a))\wedge a\geq\lambda_{b}(x)\wedge a$. Since $a$ is a Boolean element of the $MV$-algebra $[0,b]$, then so is $\lambda_{b}(a)$, which implies that $\lambda_{b}(x)\wedge a\leq(\lambda_{a}(x)\oplus\lambda_{b}(a))\wedge a=(\lambda_{a}(x)\vee\lambda_{b}(a))\wedge a=\lambda_{a}(x)\wedge a=\lambda_{a}(x)$. Summing up the above results, we get that $\lambda_{a}(x)=\lambda_{b}(x)\wedge a$.

(ii) By (i) we have

[TABLE]

On the other hand, $x,a\in[0,b]$ and $x\leq a$, it follows that $\lambda_{b}(a)\leq\lambda_{b}(x)$ and so $\lambda_{b}(x)\vee\lambda_{b}(a)=\lambda_{b}(x)$. Therefore, $\lambda_{b}(x)=\lambda_{a}(x)\vee\lambda_{b}(a)$.

(iii) It was proved in Proposition 3.5(i).

(iv) It follows from (i) or (ii).

(v) Since $[0,b]$ is an $MV$-algebra and $a\in[0,b]$ is an idempotent, $\lambda_{b}(a)$ is the relative complement of $a$ in $[0,b]$, so it is also an idempotent. The rest statement $\lambda_{a}(a)=0$ follows from definition of $\lambda_{a}$. ∎

Remark 3.10**.**

Let $(M;\vee,\wedge,\oplus,0)$ and $(N;\vee,\wedge,\oplus,0)$ be $EMV$-algebras and $f:M\rightarrow N$ be a map preserving $\oplus$ and [math]. If for each $x\in M$, there is $b\in\mathcal{I}(M)$ such that $x\in[0,b]$ and $f(\lambda_{b}(x))=\lambda_{f(b)}(f(x))$, then $f$ is an $EMV$-homomorphism. Indeed, if $x,y\in M$, there is $b\in\mathcal{I}(M)$ such that $x,y\in[0,b]$. Since $([0,b];\oplus,\lambda_{b},0,b)$ is an $MV$-algebra, $x\vee y=\lambda_{b}(\lambda_{b}(x)\oplus y)\oplus y$ and $x\wedge y=\lambda_{b}(\lambda_{b}(x)\vee\lambda_{b}(y))$. Hence, $f$ preserves $\vee$ and $\wedge$. It follows that for each $x\in M$, there is $a\in\mathcal{I}(M)$ such that $x\leq a$ and $f:[0,a]\rightarrow[0,f(a)]$ is a homomorphism of $MV$-algebras. Now, let $a$ be an arbitrary idempotent element of $M$. Then there exists $u\in\mathcal{I}(M)$ such that $a\leq u$ and $f:[0,u]\rightarrow[0,f(u)]$ is a homomorphism of $MV$-algebras. By Proposition 3.9(v), for each $x\in[0,a]$, we have

[TABLE]

It follows that $f$ is an $EMV$-homomorphism.

Theorem 3.11**.**

Let $\mathbb{EMV}$ be the class of $EMV$-algebras. Then $\mathbb{EMV}$ is a variety.

Proof.

The class $\mathbb{EMV}$ is closed under ${\rm HSP}$. ∎

Definition 3.12**.**

Let $(M;\vee,\wedge,\oplus,0)$ be a $qEMV$-algebra. An equivalence relation $\theta$ on $M$ is called a congruence relation or simply a congruence if it satisfies the following conditions:

• (i)

$\theta$ is compatible with $\vee$, $\wedge$ and $\oplus$;

• (ii)

for all $b\in\mathcal{I}(M)$, $\theta\cap([0,b]\times[0,b])$ is a congruence relation on the $MV$-algebra $([0,b];\oplus,\lambda_{b},0,b)$.

We denote by $\mathrm{Con}(M)$ the set of all congruences on $M$.

The next proposition makes our work easier when we want to verify that an equivalence relation on a $qEMV$-algebra is a congruence.

Proposition 3.13**.**

An equivalence relation $\theta$ on an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ is a congruence if it is compatible with $\vee$, $\wedge$ and $\oplus$, and for all $(x,y)\in\theta$, there exists $b\in\mathcal{I}(M)$ such that $x,y\leq b$ and $(\lambda_{b}(x),\lambda_{b}(y))\in\theta$.

Proof.

Suppose that for each $(x,y)\in\theta$ there exists $u\in\mathcal{I}(M)$ such that $x,y\leq u$ and $(\lambda_{u}(x),\lambda_{u}(y))\in\theta$. Put $b\in\mathcal{I}(M)$. We will show that $\theta\cap([0,b]\times[0,b])$ is a congruence on $([0,b];\oplus,\lambda_{b},0,b)$. Let $x,y\in[0,b]$ such that $(x,y)\in\theta$. Then by the assumption, there exists $u\in\mathcal{I}(M)$ such that $x,y\in[0,u]$ and $(\lambda_{u}(x),\lambda_{u}(y))\in\theta$. Suppose that $v\in\mathcal{I}(M)$ such that $u,b\leq v$ (for example, $v=u\oplus b$). By Proposition 3.9(ii), $\lambda_{v}(x)=\lambda_{u}(x)\oplus\lambda_{v}(u)$ and $\lambda_{v}(y)=\lambda_{u}(y)\oplus\lambda_{v}(u)$. Since $(\lambda_{u}(x),\lambda_{u}(y))\in\theta$ and $\theta$ is compatible with $\oplus$, we get that $(\lambda_{v}(x),\lambda_{v}(y))\in\theta$. It follows that $(\lambda_{v}(x)\wedge b,\lambda_{v}(y)\wedge b)\in\theta$ and so by Proposition 3.9(i), $(\lambda_{b}(x),\lambda_{b}(y))\in\theta$. The proof of the converse is clear. ∎

Let $\theta$ be a congruence relation on an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ and $M/\theta=\{[x]\mid x\in M\}$ (we usually use $x/\theta$ instead of $[x]$). Consider the induced operations $\vee$, $\wedge$ and $\oplus$ on $M/\theta$ defined by

[TABLE]

Clearly, $(\mathrm{EMV}_{I}1)$ and $(\mathrm{EMV}_{I}2)$ hold. By Proposition 3.5(ii), it suffices to show that for each $x/\theta\in M/\theta$, there exists an idempotent element $a/\theta\in M/\theta$ such that $x/\theta\leq a/\theta$ and $([0/\theta,a/\theta];\oplus,\lambda_{a/\theta},0/\theta,a/\theta)$ is an $MV$-algebra. Put $x/\theta\in M/\theta$. There is $a\in\mathcal{I}(M)$ such that $x\leq a$. Clearly, $x/\theta\leq a/\theta$ and $a/\theta$ is idempotent. Also, $([0,a];\oplus,\lambda_{a},0,a)$ is an $MV$ algebra and $\theta_{a}:=\theta\cap([0,a]\times[0,a])$ is its congruence relation, so $[0,a]/\theta_{a}$ (with the quotient operations) is an $MV$-algebra, where

[TABLE]

First, we show that for all $x/\theta\in[0/\theta,a/\theta]$, $\lambda_{a}(x)/\theta$ is the least element of the set $\{z/\theta\in[0/\theta,a/\theta]\mid z/\theta\oplus x/\theta=a/\theta\}$. For each $x/\theta\in[0/\theta,a/\theta]$, we have $x/\theta=x/\theta\wedge a/\theta=(x\wedge a)/\theta$ and $x\wedge a\in[0,a]$. So, we can assume that $x\in[0,a]$. If $y/\theta\in[0/\theta,a/\theta]$ such that $x/\theta\oplus y/\theta=a/\theta$, then $(x\oplus y,a)\in\theta$ and $x,y,a\in[0,a]$, thus $(x\oplus y,a)\in\theta_{a}$, that is $x/\theta_{a}\oplus y/\theta_{a}=a/\theta_{a}$ (which implies that $x/\theta\oplus y/\theta=a/\theta$). Hence by (3.1), $y/\theta_{a}\geq\lambda_{a}(x)/\theta_{a}$ and so $y/\theta\geq\lambda_{a}(x)/\theta$. Also, $\lambda_{a}(x)/\theta\oplus x/\theta=a/\theta$. Thus $\lambda_{a/\theta}(x/\theta)$ exists and is equal to $\lambda_{a}(x)/\theta$. Now, it is straightforward to check that $\lambda_{a/\theta}$ satisfies the conditions (1) and (3) in definition of $MV$-algebras. It follows that $([0/\theta,a/\theta];\oplus,\lambda_{a/\theta},0/\theta,a/\theta)$ is an $MV$-algebra. Therefore, by Proposition 3.5(ii), $(M/\theta;\vee,\wedge,\oplus,0/\theta)$ is an $EMV$-algebra, and the mapping $x\mapsto x/\theta$ is an $EMV$-homomorphism from $M$ onto $M/\theta$.

Example 3.14**.**

(i) Let $f:M\rightarrow N$ be a $qEMV$-homomorphism. Then $\ker(f)=\{(x,y)\in M\times N\mid f(x)=f(y)\}$ is a congruence on $M$.

(ii) Let $\{(M_{i};\oplus,^{\prime},0,1)\mid i\in I\}$ be a family of $MV$-algebras and $\theta_{i}$ be a congruence on $M_{i}$ for all $i\in I$. Set

[TABLE]

Then clearly, $\theta$ is an equivalence relation on $\sum_{i\in I}M_{i}$ which is compatible with $\vee$, $\wedge$ and $\oplus$. Let $(f,g)\in\theta$, for some $f,g\in\sum_{i\in I}M_{i}$. Define $h:I\rightarrow\bigcup_{i\in I}M_{i}$ by

[TABLE]

Clearly, $h$ is an idempotent element of $\sum_{i\in I}M_{i}$ and $f,g\leq h$. By Example 3.2(5), we know that $\sum_{i\in I}M_{i}$ is an $EMV$-algebra. Let $b\in\sum_{i\in I}M_{i}$ be an idempotent. Consider the $MV$-algebra $([0,b];\oplus,\lambda_{b},0,b)$. It can be easily seen that for all $\alpha\in[0,b]$,

[TABLE]

Since $\theta_{i}$ is a congruence on $M_{i}$ for all $i\in I$, then $(\lambda_{b}(f),\lambda_{b}(g))\in\theta$ and so by Proposition 3.13, $\theta$ is a congruence on the $qEMV$-algebra $\sum_{i\in I}M_{i}$.

Definition 3.15**.**

A non-empty subset $I$ of a $qEMV$-algebra $(M;\vee,\wedge,\oplus,0)$ is called an ideal if for each $x,y\in M$

• (i)

$x\oplus y\in I$ for all $x,y\in I$;

• (ii)

$x\leq y$ and $y\in I$ implies that $x\in I$.

The set of all ideals of $M$ is denoted by $\mathrm{Ideal}(M)$. Clearly $\{0\},M\in\mathrm{Ideal}(M)$. An ideal $I$ of $M$ is proper if $I\neq M$.

Similarly as for $MV$-algebras, see [CDM, Prop 1.2.6], we have a one-to-one relationship between the set of ideals and the set of congruences on a $qEMV$-algebra.

Theorem 3.16**.**

If $\theta$ is a congruence on an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$, then $I_{\theta}:=0/\theta$ is an ideal of $M$.

Conversely, let $I$ be an ideal of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$. Then the relation $\theta_{I}$ defined by

[TABLE]

is a congruence on $M$. In addition, the mapping $I\mapsto\theta_{I}$ is a bijection from the set $\mathrm{Ideal}(M)$ onto the set of congruences on $M$.

Proof.

Let $\theta$ be a congruence on an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$. Then it can be easily shown that $I_{\theta}:=0/\theta$ is an ideal of $M$.

Now, let $I$ be an ideal of $M$. Then for each $b\in\mathcal{I}(M)$, $I\cap[0,b]$ is an ideal of the $MV$-algebra $([0,b];\oplus,\lambda_{b},0,b)$. Define a relation $\theta_{I}$ on $M$ by (3.2). For each $(x,y),(z,w)\in\theta_{I}$, we can easily see that there exists $u\in\mathcal{I}(M)$ such that $x,y,z,w\in[0,u]$ and $(x,y),(z,w)\in\theta_{I_{u}}$, where $I_{u}=I\cap[0,u]$. Indeed, put $(x,y),(z,w)\in\theta_{I}$. Then there are $a,b\in\mathcal{I}(M)$ such that $x,y\in[0,a]$, $z,w\in[0,b]$ and $\lambda_{a}(\lambda_{a}(x)\oplus y),\lambda_{a}(\lambda_{a}(y)\oplus x)\in I$ and $\lambda_{b}(\lambda_{b}(z)\oplus w),\lambda_{b}(\lambda_{b}(w)\oplus z)\in I$. Since $a,b\leq a\oplus b$ and $M$ is an $EMV$-algebra, then there exists $u\in\mathcal{I}(M)$ such that $a\oplus b\leq u$. Thus $x,y,z,w\in[0,u]$. By Proposition 3.9,

[TABLE]

Since $\lambda_{u}(\lambda_{u}(x)\oplus y),\lambda_{a}(\lambda_{a}(x)\oplus y)\in[0,u]$, $\lambda_{a}(\lambda_{a}(x)\oplus y)\in I\cap[0,u]$ and $I\cap[0,u]$ is an ideal of the $MV$-algebra $([0,u];\oplus,\lambda_{u},0,u)$, then from

[TABLE]

It follows that $\lambda_{u}(\lambda_{u}(x)\oplus y)\in I\cap[0,u]$. In a similar way, we can show that

[TABLE]

Hence $(x,y),(y,z)\in\theta_{I_{u}}$. Since $\theta_{I_{u}}$ is a congruence on the $MV$-algebra $([0,u];\oplus,\lambda_{u},0,b)$, it proves that $\theta_{I}$ is a congruence on $M$.

In an analogous way, and using [CDM, Prop 1.2.6], we have that the mapping $I\mapsto\theta_{I}$ is a bijection in question. ∎

Definition 3.17**.**

Let $I$ be an ideal of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$. We denote the $EMV$-algebra $(M/\theta_{I};\vee,\wedge,\oplus,0/\theta_{I})$ simply by $M/I$, and $M/I$ is called the quotient $EMV$-algebra of $M$ induced by $I$.

In Example 3.2(2), we showed that any generalized Boolean algebra is an $EMV$-algebra. Now, let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra and $I$ be an ideal of $M$ such that for each $x\in M$, there is $a\in\mathcal{I}(M)$ such that $x\in[0,a]$ and $\lambda_{a}(x)\wedge x\in I$. Then $M/I$ is a generalized Boolean algebra. By definition, it suffices to show that for each $x\in M$, $x/I\oplus x/I=x/I$ (since in this case, the $MV$-algebra $[x/I,y/I]$ is a Boolean algebra). First, we note that for each $x\in M$, $x\in I$ if and only if $x/I=0/I$. Let $x\in M$. Then by the assumption, there exists $a\in\mathcal{I}(M)$ such that $x\wedge\lambda_{a}(x)\in I$, so in the $MV$-algebra $([0/I,a/I];\oplus,\lambda_{a/I},0/I)$ we have $x/I\wedge\lambda_{a/I}(x/I)=x/I\wedge\lambda_{a}(x)/I=0/I$. Hence, $x/I$ is a Boolean element of this $MV$-algebra and so $x/I\oplus x/I=x/I$.

We recall that a $qEMV$-algebra $M$ is simple if $M$ possesses only two congruences, and due to Theorem 3.16, this is equivalent to the condition $\mathrm{Ideal}(M)=\{\{0\},M\}$.

Theorem 3.18**.**

Any simple $EMV$-algebra is a simple $MV$-algebra.

Proof.

Let $(M;\vee,\wedge,\oplus,0)$ be a simple $EMV$-algebra. We claim that $([0,a];\oplus,\lambda_{a},0,a)$ is a simple $MV$-algebra for all $a\in\mathcal{I}(M)$. Otherwise, there are $a\in\mathcal{I}(M)\setminus\{0\}$ and an ideal $I$ of the $MV$-algebra $[0,a]$ such that $I\neq[0,a]$ and $\{0\}\neq I$. So, $I$ is an ideal of the $EMV$-algebra $M$ different from $\{0\}$ and $M$, which is a contradiction. Thus, $([0,a];\oplus,\lambda_{a},0,a)$ is a simple $MV$-algebra. We show that $a=\max M$. Put $x\in M$. Then there exists $b\in\mathcal{I}(M)$ such that $x,a\leq b$. Since $([0,b];\oplus,\lambda_{b},0,b)$ is simple and $a\in[0,b]$, then by [CDM, Thm. 3.5.1], there is $n\in\mathbb{N}$ such that $n.a=b$. From $a\in\mathcal{I}(M)$ it follows that $a=n.a$, hence $a=b$. That is $x\leq a$. Therefore, $M=[0,a]$ and so it is a simple $MV$-algebra. ∎

Example 3.19**.**

The $qEMV$-algebra in Example 3.2(5) is a simple $qEMV$-algebra if $G=\mathbb{R}$.

Proposition 3.20**.**

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra and $X\subseteq M$. Then $\langle X\rangle$, the least ideal of $M$ generated by $X$, is the set

[TABLE]

Proof.

The proof is straightforward. ∎

Corollary 3.21**.**

If $I$ is an ideal of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$, then for each $x\in M$,

[TABLE]

Definition 3.22**.**

An ideal $I$ of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ is maximal if, for all $x\in M\setminus I$, $\langle I\cup\{x\}\rangle=M$. The set of all maximal ideals of $M$ is denoted by $\mathrm{MaxI}(M)$. In Theorem 5.6, it will be proved that every proper $EMV$-algebra $M$ possesses at least one maximal ideal.

Proposition 3.23**.**

If $I$ is a maximal ideal of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$, then for each $b\in\mathcal{I}(M)$, $I_{b}:=I\cap[0,b]$ is equal to $[0,b]$ or $I_{b}$ is a maximal ideal of the $MV$-algebra $([0,b];\oplus,\lambda_{b},0,b)$.

In addition, if $B=\mathcal{I}(M)$, then $I\cap\mathcal{I}(M)$ is a maximal ideal of $B$.

Proof.

Let $b\in\mathcal{I}(M)$ and $I_{b}\neq[0,b]$. Let $x\in[0,b]\setminus I_{b}$. Then $\langle I\cup\{x\}\rangle=M$. By Proposition 3.21, for each $z\in[0,b]$, there exist $n\in\mathbb{N}$ and $a\in I$ such that $z\leq a\oplus n.x$ and so $z=z\wedge(a\oplus n.x)$. Since $([0,b];\oplus,\lambda_{b},0,b)$ is an $MV$-algebra, then by [GeIo, Prop. 1.17(1)], we have

[TABLE]

Hence $z\in\langle I_{b}\cup\{x\}\rangle$. Therefore, $I_{b}$ is a maximal ideal of the $MV$-algebra $([0,b];\oplus,\lambda_{b},0,b)$.

Since $I$ is a proper subset of $M$, there are $x\in M\setminus I$ and $a\in B$ such that $x\leq a$, whence, $a\notin I\cap\mathcal{I}(M)$, which says that $I\cap\mathcal{I}(M)$ is a proper ideal of $B$. Now let $b\in B\setminus I\cap\mathcal{I}(M)$. Then $b\notin I$. Hence, for each idempotent $a\in B$, there is an element $c\in I$ and an integer $n$ such that $a\leq c\oplus n.b=c\vee b$. Then $a=a\wedge(c\vee b)=(a\wedge c)\vee(a\wedge b)$ so that, $\lambda_{a}(a\wedge b)\leq(a\wedge c)\in I$. But according to Proposition 3.9(v), $\lambda_{a}(a\wedge b)$ is an idempotent of $M$, and thus $\lambda_{a}(a\wedge b)\in I\cap\mathcal{I}(M)$ which yields, $a\leq\lambda_{a}(a\wedge b)\vee b$, and finally, $I\cap B$ is a maximal ideal of $B$. ∎

The following theorem is in some sense a converse to Proposition 3.23.

Theorem 3.24**.**

Let $M$ be an $EMV$-algebra and let $B=\mathcal{I}(M)$ be the set of idempotents of $M$. Then $B$ is an $EMV$-subalgebra of $M$. For every maximal ideal $I$ of $B$, there is a maximal ideal $K$ of $M$ such that $I=K\cap B$. If $I_{1},I_{2}$ are two different maximal ideals of $B$ and if $K_{1},K_{2}$ are maximal ideals of $M$ such that $I_{1}=K_{1}\cap B$ and $I_{2}=K_{2}\cap B$, then $K_{1}\neq K_{2}$. Moreover, if $K$ is a maximal ideal of $M$ such that $K\supseteq\langle I\rangle$, where $\langle I\rangle$ is the ideal of $M$ generated by $I$, then $K\cap\mathcal{I}(M)=I$.

Proof.

Let $M$ be an $EMV$-algebra and $B$ be the set of all idempotent elements of $M$. Then $B$ is a subalgebra of $M$ (which is a generalized Boolean algebra).

By Proposition 3.21, for each ideal $I$ of $B$, the ideal of $M$ generated by $I$ is

[TABLE]

Clearly, $\langle I\rangle\cap B=I$. Now, let $I$ be a maximal ideal of $B$. Then $H=\downarrow I$ is an ideal of $M$. Put $x\in M\setminus H$. Since $M$ is an $EMV$-algebra, there exists $a\in B$ such that $x\leq a$. Clearly, $\uparrow a\cap H=\emptyset$ and so the set $S=\{J\in\mathrm{Ideal}(M)\mid H\subseteq J,\ J\cap\uparrow a=\emptyset\}$ is not empty. By Zorn’s lemma, $S$ has a maximal element, say $K$. If $Q$ is an ideal of $M$ such that $K\varsubsetneq Q$, then $I\varsubsetneq Q$ and $a\in Q$ which implies $B\subseteq Q$ and $\langle B\rangle\subseteq Q$. It entails $Q=M$. That is, $K$ is a maximal ideal of $M$ and, moreover, it contains $I$. Since $K\cap B$ is an ideal of $B$, from $I\subseteq K\cap B\subseteq B$, it follows that either $K\cap B=I$ or $K\cap B=B$. In the second case, $B\subseteq K$ which implies $K=M$. Therefore, $K\cap B=I$. If $I^{\prime}$ is another maximal ideal of $B$, $I^{\prime}\neq I$, the above construction gives a maximal ideal $K^{\prime}$ of $M$ such that $K^{\prime}\cap B=I^{\prime}$. Hence, $K^{\prime}\neq K$.

Now let $K$ be a maximal ideal of $M$ containing $\langle I\rangle$. Then $K\cap\langle I\rangle\supseteq I$ and the maximality of $I$ in $\mathcal{I}(M)$ guarantees $K\cap\langle I\rangle=I$. ∎

Theorem 3.24 will be strengthened for $EMV$-algebras satisfying the general comparability theorem in Theorem 4.4 below.

Using Theorem 3.24, Theorem 3.18 can be extended as follows. Another application of Theorem 3.24 will be done in Theorem 4.3 below.

Theorem 3.25**.**

If an $EMV$-algebra $M$ has finitely many maximal ideals, then $M$ is an $MV$-algebra.

In particular, if $M$ is linearly ordered, then $M$ has a unique maximal ideal, and $M$ is an $MV$-algebra.

Proof.

Let $M$ be an $EMV$-algebra and $B$ be the set of all idempotent element of $M$. Then $B$ is a subalgebra of $M$. Our aim is to show that $B$ is a finite set.

Theorem 3.24 implies that given a maximal ideal $I$ of $B$, there is a maximal ideal $K$ of $M$ such that $I=K\cap B$. In addition, if $I_{1}$ and $I_{2}$ are two different maximal ideals of $B$, and if $K_{1},K_{2}$ are maximal ideals of $M$ such that $K_{1}\cap B=I_{1}\neq I_{2}=K_{2}\cap I_{2}$, then $K_{1}\neq K_{2}$. This implies that if $M$ has finitely many maximal ideals, then $B$ has also finitely many ideals.

By the proof of [CoDa, Thm. 2.2], the generalized Boolean algebra can be embedded into a Boolean algebra of subsets of $\mathrm{MaxI}(B)$. Since $\mathrm{MaxI}(B)$ is finite, then $B$ is also a finite set. Therefore, the element $\bigvee\{a\mid a\in B\}$ is the top element of $B$ as well as of $M$ which implies $M$ is an $MV$-algebra.

Assume that $M$ is linearly ordered. If $I_{1}$ and $I_{2}$ are two different ideals, there are $x\in I_{1}\setminus I_{2}$ and $y\in I_{2}\setminus I_{1}$. If $x\leq y$, then $x\in I_{2}$, an absurd, and if $y\leq x$ again we get an absurd. Hence, $I_{1}=I_{2}$, $|\mathrm{MaxI}(M)|=1$, and by the first part, $M$ is an $MV$-algebra. ∎

Theorem 3.26**.**

Let $I$ be a maximal ideal of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$. Then $M/I$ is an $MV$-algebra.

Proof.

We claim that $M/I$ is a simple $EMV$-algebra. Let $B$ be an ideal of the $EMV$-algebra $M/I$. Set $A:=\{x\in M\mid x/I\in B\}$. Clearly, $0\in A$. If $x,y\in A$, then $x/I,y/I\in B$ and so $(x\oplus y)/I=x/I\oplus y/I\in B$. Also, if $x,y\in M$ such that $x\leq y\in A$, then clearly, $x/I\leq y/I\in B$ and so $x/I\in B$, that is $x\in A$. So, $A$ is an ideal of the $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ which clearly contains $I$. Hence $I=A$ or $A=M$. Therefore, $M/I$ is a simple $EMV$-algebra. Now, from Theorem 3.18, it follows that $M/I$ is an $MV$-algebra. ∎

Due to [CDM, Thm 3.5.1], if $A$ is a simple $MV$-algebra, then $A$ is isomorphic to a unique $MV$-subalgebra of the $MV$-algebra of the real interval $[0,1]$, hence in Theorem 3.26, $M/I$ can be embedded in a unique way into the real interval $[0,1]$ because there is an element $x\in M\setminus I$, so that $x/I>0/I$, and we can assume that the maximal value in $M/I$ is equal to the real number $1$.

4 State-morphisms, Maximal Ideals, and $EMV$-clans

In the section we introduce state-morphisms which in the case of $MV$-algebras are exactly extremal states. States are averaging of truth-values in Łukasiewicz logic and they correspond to an analogue of finitely additive measures in classical logic. We show that state-morphisms are in a one-to-one correspondence with maximal ideals. We present $EMV$-clans as $EMV$-algebras of fuzzy sets where all algebraic operations are defined by points. They are prototypes of semisimple $EMV$-algebras.

According to [Mun2], a mapping $s$ on an $MV$-algebra $M$ such that $s:M\to[0,1]$ is (i) a state if (a) $s(1)=1$ and (b) $s(a\oplus b)=s(a)+s(b)$ whenever $a\odot b=0$; (ii) a state-morphism if $s$ is an $MV$-homomorphism from $M$ into the $MV$-algebra of the real interval $[0,1]$; (iii) an extremal state if $s=\lambda s_{1}+(1-\lambda)s_{2}$, where $s_{1},s_{2}$ are states on $M$ and $\lambda$ is a real number such that $0<\lambda<1$, then $s_{1}=s_{2}=s$. Due to [Mun2] and [Dvu3], we have that (i) every non-degenerate $MV$-algebra possesses at leat one state; (ii) each state-morphism is a state, and it is an extremal state, and conversely, (iii) every extremal state is a state-morphism.

Inspired by the notion of a state-morphism on $MV$-algebras, we define a state-morphism on an $EMV$-algebra $M$ as follows: A mapping $s:M\to[0,1]$ is a state-morphism if $s$ is an $EMV$-homomorphism from $M$ into the $EMV$-algebra of the real interval $[0,1]$ such that there is an element $x\in M$ with $s(x)=1$. In the latter case, we can assume that there is an idempotent $a$ such that $s(a)=1$. We define the set $\mbox{\rm Ker}(s)=\{x\in M\mid s(x)=0\}$, the kernel of a state-morphism $s$.

The basic properties of state-morphisms are as follows.

Proposition 4.1**.**

Let $s$ be a state-morphism on an $EMV$-algebra $M$. Then

• (i)

$s(0)=0$;

• (ii)

$s(a)\in\{0,1\}$* for each idempotent $a\in M$;*

• (iii)

if $x\leq y$, then $s(x)\leq s(y)$;

• (iv)

$s(\lambda_{a}(x))=s(a)-s(x)$* for each $x\in[0,a]$, $a\in\mathcal{I}(M)$.*

• (v)

$\mbox{\rm Ker}(s)$* is a proper ideal of $M$.*

Proof.

(i) It is trivial.

(ii) Let $a\in\mathcal{I}(M)$ and assume $0<s(a)$. There is the least integer $n_{0}$ such that $n_{0}.s(a)=1$ in the $MV$-algebra of the real interval $[0,1]$. Then $s(a)=s(n_{0}.a)=1$.

(iii) Let $x\leq y$. There is an idempotent $a\in M$ such that $y\leq a$. If $s(a)=0$, then the restriction $s_{a}$ of $s$ onto the $MV$-algebra $[0,a]$ is the zero function, so that $s(x)=s(y)$. If $s(a)=1$, then the restriction $s_{a}$ is a state-morphism on the MV-algebra $[0,a]$, and the monotonicity of $s_{a}$ in $[0,a]$ implies $s(x)\leq s(y)$.

(iv) We have $\lambda_{a}(x)\oplus x=a$ for each $x\in[0,a]$. If $s(a)=0$, the statement follows from (iii). If $s(a)=1$, the restriction $s_{a}$ is a state-morphism on the MV-algebra $[0,a]$, and for $s_{a}$ we have $s_{a}(\lambda_{a}(x))=1-s(x)$ which proves (iv).

(v) It follows easily from (i) and (iii). ∎

Theorem 4.2**.**

(i)* If $I$ is a maximal ideal of an $EMV$-algebra $M$, then $M/I$ can be embedded in a unique way into the MV-algebra of the real interval $[0,1]$ such that the mapping $s_{I}:x\mapsto x/I$, $x\in M$, is a state-morphism.*

(ii)* If $s$ is a state-morphism, then $\mbox{\rm Ker}(s)$ is a maximal ideal of $M$. In addition, there is a unique maximal ideal $I$ of $M$ such that $s=s_{I}$.*

(iii)* If for state-morphisms $s_{1}$ and $s_{2}$ we have $\mbox{\rm Ker}(s_{1})=\mbox{\rm Ker}(s_{2})$, then $s_{1}=s_{2}$.*

Proof.

(i) Let $I$ be a maximal ideal. Due to Theorem 3.26 and Proposition 3.23, the mapping $s_{I}$ is in fact an $EMV$-homomorphism. Since $I$ is maximal, there is $x\in M\setminus I$, so that $x/I>0/I$. Without loss of generality, we can assume that the greatest value in $M/I$ is $1$. Hence, $s_{I}$ is a state-morphism on $M$.

(ii) Conversely, let $s$ be a state-morphism. Put $I=\mbox{\rm Ker}(s)$. Then $I$ is a proper ideal of $M$. Let $x\notin I$. There is an idempotent $a$ of $M$ such that $x\leq a\notin I$ and $s(a)=1$. Then the restriction $s_{a}$ of $s$ restricted to the $MV$-algebra $[0,a]$ is a state-morphism on $[0,a]$. By [Dvu3, Prop 4.3], $I\cap[0,a]=\{y\in[0,a]\mid s_{a}(y)=0\}$ is a maximal ideal of $[0,a]$. Consequently, there is an integer $n$ such that $\lambda_{a}(n.x)\in I\cap[0,a]\subseteq I$. Hence, $I$ is a maximal ideal of $I$. The uniqueness of $I$ follows from the [Mun1, Dvu3].

(iii) There is an idempotent $a$ such that $a\notin\{x\in M\mid s_{1}(x)=0\}$. Then for the restrictions of $s_{1}$ and $s_{2}$ onto the $MV$-algebra $[0,a]$, we have by [Dvu3, Prop 4.5] that $s_{1}(x)=s_{2}(x)$ for each $x\in[0,a]$, then $s_{1}(x)=s_{2}(x)$ for each $x\in M$. ∎

Let $\mathcal{SM}(M)$ denote the set of state-morphisms on an $EMV$-algebra $M$. In Theorem 5.6, it will be proved that every $M$ contains a maximal ideal, so that by Theorem 4.2(i), $\mathcal{SM}(M)$ is non-void whenever $M\neq\{0\}$. Using Theorem 3.24, we show that every state-morphism on $\mathcal{I}(M)$ can be extended to a state-morphism on an $EMV$-algebra $M$.

Theorem 4.3**.**

Every state-morphism $s$ on $\mathcal{I}(M)$ of an $EMV$-algebra $M$ can be extended to a state-morphism $\hat{s}$ on $M$.

Proof.

Let $s$ be a state-morphism on $B=\mathcal{I}(M)$. By Theorem 4.2, the set $I=\{x\in\mathcal{I}(M)\mid s(x)=0\}$ is a maximal ideal on $B$. Theorem 3.24 guarantees that there is a maximal ideal $K$ of $M$ such that $K\cap B=I$. Due to Theorem 4.2(i), there is a unique state-morphism $\hat{s}$ such that $K=\{x\in M\mid\hat{s}(x)=0\}$. Since by Proposition 4.1(ii), $s(a)\in\{0,1\}$, we see that $\hat{s}$ is an extension (not necessarily unique) of $s$ onto $M$. ∎

We note that a sufficient condition for the property “for every maximal ideal $I$ of the set $\mathcal{I}(M)$ of idempotent elements of an $MV$-algebra $M$, there is a unique maximal ideal $K$ of $M$ such that $K\cap\mathcal{I}(M)=I$” is a condition that the $MV$-algebra satisfies the general comparability property, see e.g. [Dvu4, Thm 4.4]. Therefore, we introduce this notion also for $EMV$-algebras as follows. We say that an $EMV$-algebra $M$ satisfies the general comparability property if it holds for every $MV$-algebra $([0,a];\oplus,\lambda_{a},0,a)$, i.e. if $a\in\mathcal{I}(M)$ and $x,y\in[0,a]$, there is an idempotent $e$, $e\in[0,a]$ such that $x\wedge e\leq y$ and $y\wedge\lambda_{a}(e)\leq x$. We note that every linearly ordered $EMV$-algebra satisfies the general comparability property, on the other hand, there are $MV$-algebras where the general comparability property fails, see e.g. [Dvu4, Ex 8.6].

As we promised above, now we strengthen Theorem 3.24 as follows.

Theorem 4.4**.**

Let $M$ be an $EMV$-algebra satisfying the general comparability property. If $I_{1}$ and $I_{2}$ are maximal ideals of $M$ such that $\mathcal{I}(M)\cap I_{1}=\mathcal{I}(M)\cap I_{2}$, then $I_{1}=I_{2}$.

Proof.

Suppose the converse, i.e. $I_{1}\neq I_{2}$. Since $I_{1}$ and $I_{2}$ are maximal ideals, there is $x\in I_{1}\setminus I_{2}$ and $y\in I_{2}\setminus I_{1}$. Choose an idempotent $a$ such that $x,y\leq a$. Then $a\notin I_{1}\cup I_{2}$. Then $I_{1}\cap[0,a]$ and $I_{2}\cap[0,a]$ are maximal ideals of $[0,a]$, see Proposition 3.23. In addition, $I_{1}\cap[0,a]\cap\mathcal{I}(M)=I_{2}\cap[0,a]\cap\mathcal{I}(M)$. Since the $MV$-algebra $[0,a]$ satisfies the general comparability property, using [Dvu4, Thm 4.4, Cor 4.5] and Theorem 4.2, we have $x\in I_{1}\cap[0,a]=I_{2}\cap[0,a]\ni y$ which gives $y\in I_{1}$ and $x\in I_{2}$, an absurd. Hence, $I_{1}=I_{2}$. ∎

Corollary 4.5**.**

Let $M$ be an $EMV$-algebra satisfying the general comparability property. Then every state-morphism on $\mathcal{I}(M)$ can be extended to a unique state-morphism on $M$.

Proof.

Let $s$ be a state-morphism on $\mathcal{I}(M)$. By Theorem 4.3, there is an extremal state $s_{1}$ on $M$ which is an extension of $s$. If there is another state-morphism $s_{2}$ on $M$ which extends $s$, then $I_{i}=\{x\in M\mid s_{i}(x)=0\}$ is by Theorem 4.2(iii) a maximal ideal of $M$ for $i=1,2$. Since $I_{1}\cap\mathcal{I}(M)=I_{2}\cap\mathcal{I}(M)$, Theorem 4.4 implies, $I_{1}=I_{2}$, which finally gives, Theorem 4.2, $s_{1}=s_{2}$. ∎

Now, there is a natural question “under which suitable condition on an ideal $I$ of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$, the quotient $EMV$-algebra induced by $I$, $M/I$, is an $MV$-algebra”?

Lemma 4.6**.**

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra, $I$ an ideal of $M$, $b,d\in\mathcal{I}(M)$ such that $b\leq d$, and let $I_{b}$ and $I_{d}$ be ideals of the $MV$-algebras $[0,b]$ and $[0,d]$, respectively. If $x,y\in[0,b]$ such that $x/I_{d}=y/I_{d}$, then $x/I_{b}=y/I_{b}$.

Proof.

Let $x,y\in[0,b]$ such that $x/I_{d}=y/I_{d}$. Then $\lambda_{d}(\lambda_{d}(x)\oplus y),\lambda_{d}(\lambda_{d}(y)\oplus x)\in I_{d}\subseteq I$. We will show that $\lambda_{b}(\lambda_{b}(x)\oplus y),\lambda_{b}(\lambda_{b}(y)\oplus x)\in I_{b}$.

[TABLE]

In a similar way, we can see that $\lambda_{b}(\lambda_{b}(y)\oplus x)\in I_{b}$. Therefore, $x/I_{b}=y/I_{b}$. ∎

The following equivalencies on the induced order for a quotient $EMV$-algebra are used in Theorem 4.7.

Let $I$ be an ideal of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ and $x,y\in M$. Then

[TABLE]

Theorem 4.7**.**

Let $I$ be an ideal of $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$. Then $M/I$ is an $MV$-algebra if and only if there exists $a\in\mathcal{I}(M)$ such that $\lambda_{b}(a)\in I$ for all $b\in\mathcal{I}(M)$ greater than $a$.

Proof.

Let $M/I$ be an $MV$-algebra. Then there exists $a\in M$ such that $x/I\leq a/I$ for all $x\in M$. Since $M$ is an $EMV$-algebra, there is $b\in\mathcal{I}(M)$ such that $a\leq b$ and so $a/I=b/I$. Thus, without loss of generality we can assume that $a\in\mathcal{I}(M)$. Let $b$ be an arbitrary element of $\mathcal{I}(M)$ greater than $a$. Since $a/I$ is the maximum of $M/I$, then $a/I=b/I$ and so there is $d\in\mathcal{I}(M)$ such that $a,b\leq d$ and $\lambda_{d}(\lambda_{d}(a)\oplus b),\lambda_{d}(\lambda_{d}(b)\oplus a)\in I$. Since $a,b\in[0,b]$, by Lemma 4.6, we get that $\lambda_{b}(a)=\lambda_{b}(\lambda_{b}(b)\oplus a)\in I$.

Conversely, let $x\in M$. Then there exists $b\in\mathcal{I}(M)$ such that $x,a\in[0,b]$. Since $\lambda_{b}(\lambda_{b}(x)\oplus a)\leq\lambda_{b}(a)\in I$, then $x/I\leq a/I$. Therefore, $M/I$ is an $MV$-algebra. ∎

From Theorem 3.26 and Theorem 4.7 we get that each maximal ideal satisfies the condition in Theorem 4.7.

We say that an $EMV$-algebra $M$ is semisimple if $\text{Rad}(M):=\bigcap\{I\mid I\in\mathrm{MaxI}(M)\}=\{0\}$; the set $\text{Rad}(M)$ is said to be the radical of $M$.

In what follows, we show that every generalized Boolean algebra is semisimple.

Lemma 4.8**.**

Let $M\neq\{0\}$ be a generalized Boolean algebra. Then:

• (i)

An ideal $I$ of $M$ is maximal if and only if, for each $a\notin I$ and each $b\in M$ with $a<b$, $\lambda_{b}(a)\in I$.

• (ii)

$M$* is a semisimple $EMV$-algebra.*

Proof.

If $M$ has the top element, $M$ is a Boolean algebra and the statement is well-known from the theory of Boolean algebras. Thus let us assume that $M$ has no top element. In generalized Boolean algebras we have $x\oplus y=x\vee y$.

(i) Let $I$ be a maximal ideal of $M$, and let $a,b\in M$ such that $a\notin I$ and $a<b$. Maximality of $I$ entails that $\lambda_{b}(a)\in\langle I\cup\{a\}\rangle$, by Corollary 3.21, there is an integer $n$ and $x\in I$ such that $\lambda_{b}(a)\leq x\oplus n.a=x\vee a\leq x$. Then $\lambda_{b}(a)=\lambda_{b}(a)\wedge(x\vee a)=(\lambda_{b}(a)\wedge x)\vee(\lambda_{b}(a)\wedge a)=\lambda_{b}(a)\wedge x\leq x$ which yields $\lambda_{b}(a)\in I$.

Conversely, let an ideal $I$ of $M$ satisfy conditions of (i). Choose $x\in\langle I\cup\{a\}\rangle\setminus I$ and let $b\in M$ be greater than $x$. Then $\lambda_{b}(x)\in I\subseteq\langle I\cup\{a\}\rangle$. Since $b=x\oplus\lambda_{b}(x)$, we have $b\in\langle I\cup\{a\}\rangle$. This is true for each $b\in M$ with $b>x$, therefore, we have $\langle I\cup\{a\}\rangle=M$, so that $I$ is maximal.

(ii) First we have to note that every generalized Boolean algebra $M\neq\{0\}$ possesses at least one maximal ideal, as it will be proved in Theorem 5.6 below.

Let $x\in\mathrm{Rad}(M)$. If $x>0$, using Zorn’s lemma, we have that there is a maximal filter $F$ of $M$ containing $x$. By Theorem 5.6 below, the set $I_{F}=\{\lambda_{a}(z)\mid z\in F,\,a\in\mathcal{I}(M),\,z\leq a\}$ is a maximal ideal of $M$. Let $b$ be an idempotent such that $b\notin I_{F}$ and $x\leq b$. Then $\lambda_{b}(x)\in I_{F}$ and $x\in I_{F}$. Hence, $b=x\oplus\lambda_{b}(x)\in I_{F}$ which is absurd. Consequently, $x=0$. ∎

An important family of $EMV$-algebras is a family of $EMV$-clans of fuzzy sets which as we show below are only semisimple $EMV$-algebras.

Definition 4.9**.**

A system $\mathcal{T}\subseteq[0,1]^{\Omega}$ of fuzzy sets of a set $\Omega\neq\emptyset$ is said to be an $EMV$-clan if

1. (i)

$0_{\Omega}\in\mathcal{T}$ where $0_{\Omega}(\omega)=0$ for each $\omega\in\Omega$; 2. (ii)

if $a\in\mathcal{T}$ is a characteristic function, then (a) $a-f\in\mathcal{T}$ for each $f\in\mathcal{T}$ with $f(\omega)\leq a(\omega)$ for each $\omega\in\Omega$, (b) if $f,g\in\mathcal{T}$ with $f(\omega),g(\omega)\leq a(\omega)$ for each $\omega\in\Omega$, then $f\oplus g\in\mathcal{T}$, where $(f\oplus g)(\omega)=\min\{f(\omega)+g(\omega),a(\omega)\}$, $\omega\in\Omega$, and $a$ is a characteristic function from $\mathcal{T}$; 3. (iii)

for each $f\in\mathcal{T}$, there is a characteristic function $a\in\mathcal{T}$ such that $f(\omega)\leq a(\omega)$ for each $\omega\in\Omega$; 4. (iv)

given $\omega\in\Omega$, there is $f\in\mathcal{T}$ such that $f(\omega)=1$.

Proposition 4.10**.**

Any $EMV$-clan $\mathcal{T}$ can be organized into an $EMV$-algebra of fuzzy sets where all operations are defined by points.

(i)* Let $f,g\in\mathcal{T}$ and $f,g\leq a,b$, where $a,b$ are characteristic functions from $\mathcal{T}$. Then $(f\odot g)(\omega)=\max\{f(\omega)+g(\omega)-a(\omega),0\}=\max\{f(\omega)+g(\omega)-b(\omega),0\}$ and $f\odot g\in\mathcal{T}$. Similarly, $f\ast g=f\odot(a-g)=f\odot(b-g)\in\mathcal{T}$.*

(ii)* If $f,g\in\mathcal{T}$ and $f(\omega)\leq g(\omega)$ for each $\omega\in\Omega$, then $g-f\in\mathcal{T}$.*

(iii)* If $f,g\in\mathcal{T}$, then $f\wedge g,f\vee g\in\mathcal{T}$, where $(f\wedge g)(\omega)=\min\{f(\omega),g(\omega)\}$ and $(f\vee g)(\omega)=\max\{f(\omega),g(\omega)\}$ for each $\omega\in\Omega$*

In addition, given $\omega\in\Omega$, the mapping $s_{\omega}(f):=f(\omega)$, $f\in\mathcal{T}$, is a state-morphism and if $I_{\omega}=\{f\in\mathcal{T}\mid f(\omega)=0\}$, then $I_{\omega}$ is a maximal ideal of $\mathcal{T}$.

Proof.

We assert that $\mathcal{T}$ is an $EMV$-algebra of fuzzy sets. By (iv) we have that if $f,g\in\mathcal{T}$, there is a characteristic function $a\in\mathcal{T}$ such that $f,g\leq a$. If $b\in\mathcal{T}$ is another characteristic function such that $a\leq b$, we have

[TABLE]

and

[TABLE]

If $a(\omega)=0$, then $f(\omega)=g(\omega)=0$ and $(f\oplus_{a}g)(\omega)=0=(f\oplus_{b}g)(\omega)$. If $a(\omega)=1$, then $b(\omega)=1$ and $(f\oplus_{a}g)(\omega)=(f\oplus_{b}g)(\omega)$. Hence, if $f,g\leq u,v$, where $u,v$ are characteristic functions from $\mathcal{T}$, there is a characteristic function $c\in\mathcal{T}$ such that $u,v\leq c$. Then $f\oplus_{u}g=f\oplus_{c}g=f\oplus_{v}g$, and the binary operation $\oplus$ does not depend on chosen characteristic functions $a,b,u,v,c\in\mathcal{T}$ dominating $f,g$, and $\oplus$ is a total binary operation such that $(\mathcal{T};\oplus,0_{\Omega})$ is a commutative ordered monoid. It is easy to see that, for $f\in\mathcal{T}$, we have $f\oplus f=f$ iff $f$ is a characteristic function. Finally $\lambda_{a}(f)=a-f$ whenever $f\leq a$ and $a\in\mathcal{T}$ is a characteristic function. So that $([0,a];\oplus,\lambda_{a},0,a)$ is an $MV$-algebra of fuzzy sets. Consequently, $(\mathcal{T};\vee,\wedge,\oplus,0_{\Omega})$ is an $EMV$-algebra.

(i) We can define $f\odot_{a}g$ in the similar but dual way as we defined already $f\oplus_{a}g$ if $f,g\in[0,a]$, and if $f,g\in[0,b]$, then $f\odot_{a}g=f\odot_{b}g$.

(ii) Let $f,g\in\mathcal{T}$ and $f\leq g$. There is a characteristic function $a\in\mathcal{T}$ such that $f,g$ belong to the $MV$-algebra $[0,a]$. Then $g-f=g\odot(a-f)\in[0,a]$ and similarly $f\odot_{a}(a-g)=f\odot_{b}(b-g)$, so we can define $f\ast g:=f\odot_{a}(a-g)$.

(iii) Let $f,g\in\mathcal{T}$, and let $a\in\mathcal{T}$ be a characteristic function such that $f(\omega),g(\omega)\leq a(\omega)$, $\omega\in\Omega$. Then $f\vee g=(f\ast g)\oplus g\in\mathcal{T}$, $f\wedge g=f\odot((a-f)\oplus g)\in\mathcal{T}$, and $(f\wedge g)(\omega)=\min\{f(\omega),g(\omega)\}$ and $(f\vee g)(\omega)=\max\{f(\omega),g(\omega)\}$ for each $\omega\in\Omega$.

The mapping $s_{\omega}$ is a homomorphism from $\mathcal{T}$ into the $MV$-algebra $[0,1]$. Since, for $\omega\in\Omega$, there is an $f\in\mathcal{T}$ such that $f(\omega)=1$, $s_{\omega}$ is a state-morphism.

In view of $I_{\omega}=\{f\in\mathcal{T}\mid s_{\omega}(f)=0\}$, due to Theorem 4.2(ii), $I_{\omega}$ is a maximal ideal of $\mathcal{T}$ for each $\omega\in\Omega$. ∎

Theorem 4.11**.**

An $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ is semisimple if and only if $M$ is isomorphic to an $EMV$-clan of fuzzy functions on some $\Omega\neq\emptyset$.

Proof.

Put $\Omega=\mathcal{SM}(M)$, and for each $x\in M$, let $\hat{x}:\mathcal{SM}(M)\to[0,1]$ be a mapping such that $\hat{x}(s):=s(x)$, $s\in\mathcal{SM}(M)$. Put $\widehat{M}:=\{\hat{x}\mid x\in M\}$. We assert that $\widehat{M}$ is an $EMV$-clan. Clearly $\hat{0}$ is the zero function. If $a\in\mathcal{I}(M)$, then by Proposition 4.1(ii), $\hat{a}$ is a characteristic function.

We have $\hat{x}\leq\hat{y}$ iff $x\leq y$. Indeed, there is an idempotent $a\in M$ such that $x,y\leq a$. By Proposition 3.23, for each maximal ideal $I$ on $M$, the set $I\cap[0,a]$ is either $[0,a]$ or a maximal ideal of $[0,a]$. Then $\bigcap\{I\cap[0,a]\mid I\in\mathrm{MaxI}(M)\}=\{0\}$, so that each $[0,a]$ is a semisimple $MV$-algebra. Then $\hat{x}\leq\hat{y}$ iff $s(x)\leq s(y)$ for each state-morphism $s$ on the semisimple $MV$-algebra $[0,a]$, consequently, $x\leq y$.

Then the mapping $x\mapsto\hat{x}$ preserves $0,\vee,\wedge,\oplus$, if $\hat{x}\leq\hat{a}$, then $x\leq a$, so that $\hat{\lambda}_{a}(x)=\hat{a}-\hat{x}\in\widehat{M}$, and $[\hat{0},\hat{a}]$ is an $MV$-algebra. In other words, $\widehat{M}$ is an $EMV$-clan of fuzzy sets, and the mapping $x\mapsto\hat{x}$ is an $EMV$-isomorphism from $M$ onto $\widehat{M}$.

Given a state-morphism $s$ on $M$, there is an element $x\in M$ such that $s(x)=1$. Then $\hat{x}(s)=1$.

Conversely, let $\phi:M\to\mathcal{T}$ be an $EMV$-isomorphism from $M$ onto an $EMV$-clan $\mathcal{T}$. If we set $I_{\omega}:=\{f\in\mathcal{T}\mid f(\omega)=0\}$, then by Proposition 4.10, $I_{\omega}$ is a maximal ideal of $\mathcal{T}$. Then $\bigcap\{I\mid I\in\mathrm{MaxI}(M)\}\subseteq\bigcap\{I_{\omega}\mid\omega\in\Omega\}=\{0_{\Omega}\}$, so that $\mathcal{T}$ is semisimple, and consequently so is $M$. ∎

A special type of an $EMV$-clan is a clan of fuzzy sets: We say that a system $\mathcal{C}$ of fuzzy sets of a set $\Omega\neq\emptyset$ is a clan if (i) $1_{\Omega}\in\mathcal{C}$, where $1_{\Omega}(\omega)=1$ for each $\omega\in\Omega$, (ii) if $f\in\mathcal{C}$, then $f^{\prime}:=1-f\in\mathcal{C}$, and (iii) if $f,g\in\mathcal{C}$, then $f\oplus g\in\mathcal{C}$, where $(f\oplus g)(\omega)=\min\{f(\omega)+g(\omega),1\}$, $\omega\in\Omega$. Then $(\mathcal{C};\oplus,^{\prime},0_{\Omega},1_{\Omega})$ is an $MV$-algebra where all $MV$-operations are defined by points. In addition, $\min\{f,g\},\max\{f,g\}\in\mathcal{C}$ whenever $f,g\in\mathcal{C}$. It is clear, that any clan can be understood as a bounded $EMV$-clan. In addition, an $EMV$-clan $\mathcal{T}$ is a clan iff $1_{\Omega}\in\mathcal{T}$.

We note that if $\mathcal{T}$ is any system of fuzzy sets of $\Omega\neq\emptyset$, then there is a minimal clan $\mathcal{C}_{0}(\mathcal{T})$ containing $\mathcal{T}$. In particular, if $\mathcal{T}$ is an $EMV$-clan, then $\mathcal{C}_{0}(\mathcal{T})$ is the least clan of fuzzy sets on $\Omega$ containing $\mathcal{T}$.

Corollary 4.12**.**

Any semisimple $EMV$-algebra can be embedded into an $MV$-algebra.

Proof.

Let $(M;\vee,\wedge,\oplus,0)$ be a semisimple $EMV$-algebra. We present two types of the proofs.

(i) By the proof of Theorem 4.11, the natural mapping

[TABLE]

is an embedding. By Theorem 3.26, $M/I$ is an $MV$-algebra for all $I\in\mathrm{MaxI}(M)$, so $\prod_{I\in\mathrm{MaxI}(M)}M/I$ is an $MV$-algebra.

(ii) By Theorem 4.11, there is an $EMV$-clan $\mathcal{T}$ of fuzzy sets on $\Omega\neq\emptyset$ such that $M$ and $\mathcal{T}$ are isomorphic. Then $\mathcal{C}_{0}(\mathcal{T})$ is the least clan containing $\mathcal{T}$. Then $M$ can be embedded into the $MV$-algebra $\mathcal{C}_{0}(\mathcal{T})$. ∎

Remark 4.13**.**

As we have already said, if $I$ is a maximal ideal of $M$, $M/I$ can be understood as an $MV$-subalgebra of the $MV$-algebra of the real interval $[0,1]$. If $\{I_{\alpha}\}_{\alpha}$ is a set of maximal ideals of $M$ such that $\bigcap_{\alpha}I_{\alpha}=\{0\}$, the embedding mapping $\phi:M\to\prod_{\alpha}M/I_{\alpha}$ defined by $\phi(x)=(x/I_{\alpha})_{\alpha}$ gets an $EMV$-clan $\phi(M)$ of fuzzy sets on $\Omega=\{I_{\alpha}\}_{\alpha}$, and the direct product $\prod_{\alpha}M/I_{\alpha}$ defines a clan of fuzzy sets on $\Omega$ such that $\mathcal{C}_{0}(\phi(M))\subseteq\prod_{\alpha}M/I_{\alpha}$. Question: Is $\mathcal{C}_{0}(\phi(M))$ equal to $\prod_{\alpha}M/I_{\alpha}$?

In the next example we show that the answer to the question posed in Remark 4.13 could be negative.

Example 4.14**.**

Let $\Omega$ be an infinite set and $\mathcal{T}$ be the set of characteristic functions of all finite subsets of $\Omega$. Then $\mathcal{T}$ is a generalized Boolean algebra that is not a Boolean algebra, more precisely $\mathcal{T}$ is an $EMV$-clan that is not a clan. It contains a system of maximal ideals $\{I_{\omega}\mid\omega\in\Omega\}$ such that $\bigcap_{\omega}I_{\omega}=\{0\}$ and $\mathcal{C}_{0}(\mathcal{T})\varsubsetneq\prod_{\omega}\mathcal{T}/I_{\omega}$.

Let $\mathcal{T}^{\prime}$ be the system of fuzzy sets on $\Omega$ such that $f\in\mathcal{T}^{\prime}$ if and only if there is a finite subset $A$ such that $f\leq\chi_{A}$. Then $\mathcal{T}^{\prime}$ is an $EMV$-clan of fuzzy sets, and $\mathcal{C}_{0}(\mathcal{T}^{\prime})$ consists of fuzzy sets $f$ on $\Omega$ such that either there is a finite subset $A$ of $\Omega$ such that $f\leq\chi_{A}$ or there is a co-finite subset $A$ such that $f\geq\chi_{A}$. In addition, $\mathcal{T}^{\prime}$ is a maximal ideal of $\mathcal{C}_{0}(\mathcal{T}^{\prime})$.

Proof.

It is evident that $\mathcal{C}_{0}(\mathcal{T})$ consists of all characteristic functions of all finite or co-finite subsets of $\Omega$. Given $\omega\in\Omega$, let $I_{\omega}=\{\chi_{A}\mid A\subseteq\Omega,\,\omega\notin A\}$. Lemma 4.8(i) implies that $I_{\omega}$ is a maximal ideal of $\mathcal{T}$ and $\bigcap\{I_{\omega}\mid\omega\in\Omega\}=\{\chi_{\emptyset}\}$. The mapping $\chi_{A}\mapsto\{\chi_{A}/I_{\omega}\mid\omega\in\Omega\}\in\prod_{\omega\in\Omega}\mathcal{T}/I_{\omega}=2^{\Omega}$ is an embedding of $\mathcal{T}$ into the clan $2^{\Omega}$. Clearly, $\mathcal{C}_{0}(\mathcal{T})\varsubsetneq 2^{\Omega}$.

It is easy to verify that $\mathcal{T}^{\prime}$ is an $EMV$-clan. Let $f\in\mathcal{C}_{0}(\mathcal{T}^{\prime})\setminus\mathcal{T}^{\prime}$. Then there is a co-finite subset $A$ of $\Omega$ such that $f\leq\chi_{A}$, which yields, $1-f\geq 1_{\Omega}-\chi_{A}=\chi_{\Omega\setminus A}$, i.e. $1_{\Omega}-f\in\mathcal{T}^{\prime}$, proving $\mathcal{T}^{\prime}$ is a maximal ideal of $\mathcal{C}_{0}(\mathcal{T}^{\prime})$. ∎

Now we generalize the latter example and [CoDa, Thm. 2.2].

Theorem 4.15**.**

Let $\mathcal{T}$ be an $EMV$-clan of fuzzy sets of $\Omega\neq\emptyset$ and let $1_{\Omega}\notin\mathcal{T}$. Then the minimal clan $\mathcal{C}_{0}(\mathcal{T})$ generated by $\mathcal{T}$ is the set

[TABLE]

In addition, $\mathcal{T}$ is a maximal ideal of $\mathcal{C}_{0}(\mathcal{T})$.

Proof.

Clearly, $\mathcal{T}\subseteq\mathcal{T}_{0}$, $1_{\Omega}\in\mathcal{T}_{0}$, and if $f\in\mathcal{T}_{0}$, then $1-f\in\mathcal{T}_{0}$. Now we show that $\mathcal{T}_{0}$ is closed under $\oplus$. Let $f,g\in\mathcal{T}_{0}$. We have the following three cases: (i) $f=f_{0}\in\mathcal{T}$, $g=g_{0}\in\mathcal{T}$. Then trivially $f\oplus g\in\mathcal{\mathcal{missing}}T_{0}$. (ii) Let $f=1-f_{0}$ and $g=1-g_{0}$ for some $f_{0},g_{0}\in\mathcal{T}$. Then $f\oplus g=(1-f_{0})\oplus(1-g_{0})=1-(f_{0}\odot g_{0})$ and $f_{0}\odot g_{0}\in\mathcal{T}$ by Proposition 4.10(i). (iii) $f=f_{0}\in\mathcal{T}$ and $g=1-g_{0}$, where $g_{0}\in\mathcal{T}$. Then

[TABLE]

Since $f_{0}\wedge g_{0}\in\mathcal{T}$, $g_{0}-(f_{0}\wedge g_{0})\in\mathcal{T}$ by Proposition 4.10(ii), so that $f\oplus g\in\mathcal{T}_{0}$. Then $\mathcal{C}_{0}(\mathcal{T})=\mathcal{T}_{0}$.

Now we show $\mathcal{T}$ is a maximal ideal of $\mathcal{C}_{0}(\mathcal{T})$. Let $g\leq f\in\mathcal{T}$ for $g\in\mathcal{C}_{0}(\mathcal{T})$. If $g\notin\mathcal{T}$, then $f\geq g=1-g_{0}$ for some $g_{0}\in\mathcal{T}$ which implies $f_{0}\odot g_{0}=1_{\Omega}\in\mathcal{T}$, absurd. Hence, $g\in\mathcal{T}$, and $\mathcal{T}$ is an ideal of $\mathcal{C}_{0}(\mathcal{T})$.

Let $g\in\mathcal{C}_{0}(\mathcal{T})\setminus\mathcal{T}$. Then $g=1-g_{0}$ for some $g_{0}\in\mathcal{T}$. Whence, $g_{0}=1-(1-g_{0})\in\mathcal{T}$ and $\mathcal{T}$ is a maximal ideal of $\mathcal{C}_{0}(\mathcal{T})$. ∎

Corollary 4.16**.**

Every proper semisimple $EMV$-algebra can be embedded into an $MV$-algebra as its maximal ideal.

Proof.

It follows from Theorem 4.15 and Theorem 4.11. ∎

5 Filters, Ideals and Representation of $EMV$-algebras

One of the main purposes of this part is to show that any $EMV$-algebra has at least one maximal ideal. For this reason, first we define the notion of a filter of an $EMV$-algebra, showing that for each filter $F$ there is an ideal related to it. Since any bounded $EMV$-algebra with top element $1$ is an $MV$-algebra, the existence of a maximal ideal is an easy application of Zorn’s lemma if $0\neq 1$. Therefore, we will prove the existence of a maximal ideal in any proper $EMV$-algebra $M$, that is, $M$ has no maximal element. Therefore, in a proper $EMV$-algebra $M$, for each $x\in M$, we can find an idempotent element $a$ such that $x<a$. In particular, we show that every $EMV$-algebra can be embedded into an MV-algebra, and we show a basic result saying that every $EMV$-algebra is either an $MV$-algebra or it can be embedded into an $MV$-algebra as its maximal ideal.

Lemma 5.1**.**

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra. For all $x,y\in M$, we define

[TABLE]

where $a\in\mathcal{I}(M)$ and $x,y\leq a$. Then $\odot:M\times M\rightarrow M$ is an order preserving, associative well-defined binary operation on $M$ which does not depend on $a\in\mathcal{I}(M)$ with $x,y\leq a$.

In addition, if $x,y\in M$, $x\leq y$, then

[TABLE]

for all idempotents $a,b$ of $M$ with $x,y\leq a,b$, and

[TABLE]

If $x,y\in[0,a]$ for some idempotent $a\in M$, then

[TABLE]

Moreover, a binary operation $\ast$ on $M$ defined by $x\ast y=x\odot\lambda_{a}(y)$ is correctly defined for all $x,y\in M$.

An element $x\in M$ is idempotent if and only if $x\odot x=x$.

Proof.

It suffices to show that $\odot$ is well defined. Put $x,y\in M$. We show that for all $a,b\in\mathcal{I}(M)$ such that $x,y\leq a,b$, we have $\lambda_{a}(\lambda_{a}(x)\oplus\lambda_{a}(y))=\lambda_{b}(\lambda_{b}(x)\oplus\lambda_{b}(y))$. That is $x\odot_{{}_{a}}y=x\odot_{{}_{b}}y$. Indeed, take $c\in\mathcal{I}(M)$ such that $a,b\leq c$. Then by Proposition 3.9, we have

[TABLE]

In a similar way, we can show that $\lambda_{c}(\lambda_{c}(x)\oplus\lambda_{c}(y))=\lambda_{b}(\lambda_{b}(x)\oplus\lambda_{b}(y))$.

To prove associativity, let $x$, $y$ and $z$ be elements of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$. Put $c\in\mathcal{I}(M)$ such that $x,y,z\leq c$. Then by definition of $\odot$, we have $x\odot y=\lambda_{c}(\lambda_{c}(x)\oplus\lambda_{c}(y))$, $y\odot z=\lambda_{c}(\lambda_{c}(y)\oplus\lambda_{c}(z))$ and both belong to $[0,c]$. It follows that $(x\odot y)\odot z=\lambda_{c}(\lambda_{c}(x\odot y)\oplus\lambda_{c}(z))$ and $x\odot(y\odot z)=\lambda_{c}(\lambda_{c}(x)\oplus\lambda_{c}(y\odot z))$, which implies that $(x\odot y)\odot z=x\odot(y\odot z)$. Therefore, in any $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ the binary operation $\odot$ is associative.

In a similar way, we can see that $\odot$ is order preserving.

Now let $x\leq y$, $x,y\leq a,b$ for some $a,b\in\mathcal{I}(M)$. There is an idempotent $c$ such that $a,b\leq c$. Check and use Proposition 3.9(ii)

[TABLE]

and

[TABLE]

because for $y\leq a\leq c$ we have $\lambda_{c}(a)\leq\lambda_{c}(y)$. This implies $y\odot\lambda_{a}(x)=y\odot\lambda_{c}(x)$. In the same way we have $y\odot\lambda_{b}(x)=y\odot\lambda_{c}(x)$ establishing $y\odot\lambda_{a}(x)=y\odot\lambda_{b}(x)$.

To prove (5.2), it is enough to calculate it in the $MV$-algebra $[0,a]$.

Now let $x,y\leq a$ for some $a\in\mathcal{I}(M)$. Then $x\odot\lambda_{a}(x\wedge y)=x\odot(\lambda_{a}(x)\vee\lambda_{a}(y))=(x\odot\lambda_{a}(x))\vee(x\odot\lambda_{a}(y))=x\odot\lambda_{a}(y)$.

Using (5.2), we can establish (5.3).

The property $x$ is an idempotent of $M$ iff $x\odot x$ follows from definition of the operation $\odot$. ∎

For any integer $n\geq 1$ and any $x$ of an $EMV$-algebra $M$, we can define

[TABLE]

and if $M$ has a top element $1$, we define also $x^{0}=1$.

Definition 5.2**.**

A non-empty subset $F$ of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ is called a filter if it satisfies the following conditions:

• (i)

for each $x,y\in M$, if $x\leq y$ and $x\in F$, then $y\in F$ (formally $F$ is an upset);

• (ii)

for each $x,y\in F$, $x\odot y\in F$.

The set of all filters of $M$ is denoted by $\mathrm{Fil}(M)$. Clearly, $M\in\mathrm{Fil}(M)$, and a filter $F$ is proper if $F\neq M$. A proper filter which cannot be a proper subset of another proper filter of $M$ is said to be maximal, and we denote by $\mathrm{MaxF}(M)$ the set of maximal filters of $M$. By Zorn’s lemma, $\mathrm{MaxF}(M)\neq\emptyset$.

Let $(M;\vee,\wedge,\oplus,0)$ be a proper $EMV$-algebra. Then there is a non-zero idempotent element $a\in M$. We can easily see that $\uparrow a$ is a filter of the $EMV$-algebra $M$, which is clearly a proper subset of $M$. In a similar way, we can see that $\uparrow a\setminus\{a\}$ is also a proper filter of $M$.

Proposition 5.3**.**

Let $F$ be a filter of a proper $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$. Then the set

[TABLE]

is an ideal of $M$.

Proof.

First, we note that for each $x\in M$, we have

[TABLE]

Let $x,y\in M$ such that $x\in I_{F}$ and $y\leq x$. Then there exists $a\in\mathcal{I}(M)$ such that $x\leq a$ and $\lambda_{a}(x)\in F$. Since $x,y\in[0,a]$, then $\lambda_{a}(x)\leq\lambda_{a}(y)$ and so by the assumption, $\lambda_{a}(y)\in F$. It follows that $y\in I_{F}$. Now, suppose that $x,y\in I_{F}$. Then there exist $a,b\in\mathcal{I}(M)$ such that $x\leq a$ and $y\leq b$ and $\lambda_{a}(x)\in F$ and $\lambda_{b}(y)\in F$. Put $c\in\mathcal{I}(M)$ such that $a,b\leq c$. Then by Proposition 3.9, $\lambda_{c}(x),\lambda_{c}(y)\in F$ and so $\lambda_{c}(x)\odot\lambda_{c}(y)\in F$. Since $\lambda_{c}(x),\lambda_{c}(y)\leq c$, $\lambda_{c}(x)\odot\lambda_{c}(y)=\lambda_{c}(x\oplus y)$, hence $x\oplus y\in I_{F}$. Therefore, $I_{F}$ is an ideal of $M$. ∎

Proposition 5.4**.**

Let $F$ be a proper filter of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$.

(i)* For each $x\in M$, the least filter $\lfloor F\cup\{x\}\rfloor$ of $M$ containing $F\cup\{x\}$ is the set $\{z\in M\mid z\geq y\odot x^{n},\ \exists\,n\in\mathbb{N},\exists\,y\in F\}$.*

(ii)* $F$ is a maximal filter if and only if, for each $x\notin F$, there are an integer $n$ and an idempotent $b$ with $x\leq b$ such that $\lambda_{b}(x^{n})\in F$.*

Proof.

The proof of the first part is straightforward.

For the second one, let $F$ be a maximal filter and $x\notin F$. By (i), there are an integer $n$ and an element $c\in F$ such that $0=c\odot x^{n}$. There is an idempotent $b\geq x,c$, so that $c,x$ are in the MV-algebra $[0,b]$. Then $c\odot x^{n}$ can be calculated in $[0,b]$, so that $c\leq\lambda_{b}(x^{n})$ and $\lambda_{b}(x^{n})\in F$.

The converse follows easily from (i). ∎

Lemma 5.5**.**

Let $F$ be a proper filter of a proper $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$.

• (i)

For each $a\in F\cap\mathcal{I}(M)$, $a\notin I_{F}$.

• (ii)

If $a\in\mathcal{I}(M)\cap F$, then for all $b\in\mathcal{I}(M)$ such that $a\leq b$, $\lambda_{b}(a)\in I_{F}$.

• (iii)

If $F$ is a maximal filter of $M$, then for each $a\in\mathcal{I}(M)$, $a\notin I_{F}$ implies $a\in F$.

• (iv)

If $J$ is a maximal ideal of $M$, then

[TABLE]

• (v)

If $J$ is an ideal of $M$ satisfying (5.4), then

[TABLE]

is a filter of $M$.

Proof.

(i) Otherwise, $a\in I_{F}$ implies that there exists $b\in\mathcal{I}(M)$ such that $a\leq b$ and $\lambda_{b}(a)\in F$ and so $\lambda_{b}(a),a\leq b$ and $\lambda_{b}(a),a\in F$. Thus $0=a\odot\lambda_{b}(a)\in F$, which is a contradiction.

(ii) It follows from definition of $I_{F}$.

(iii) Let $a\in\mathcal{I}(M)$ such that $a\notin I_{F}$. Then by definition, for all $b\in\mathcal{I}(M)$ with $a\leq b$, $\lambda_{b}(a)\notin F$. If $a\notin F$, then $\lfloor F\cup\{a\}\rfloor=M$ (since $F$ is maximal) and so there exist $n\in\mathbb{N}$ and $x\in F$ such that $0\geq x\odot a^{n}=x\odot_{u}a^{n}$ for some $u\in\mathcal{I}(M)$ such that $x,a\leq u$. Also, $([0,u];\oplus,\lambda_{u},0,u)$ is an $MV$-algebra and $a$ is a Boolean element of it, so $a^{n}=a$ and $\lambda_{u}(a)$ is the greatest element of $[0,u]$ satisfying the equation $z\odot a=0$. It follows that $x\leq\lambda_{u}(a)$ and $\lambda_{u}(a)\in F$, which is a contradiction.

(iv) Let $a$ be an idempotent element of $M$ such that $a\notin J$. For each $b\in\mathcal{I}(M)$ with $a\leq b$, we have $\lambda_{b}(a)\in\langle J\cup\{a\}\rangle$, so by Corollary 3.21, there exist $n\in\mathbb{N}$ and $x\in J$ such that $\lambda_{b}(a)\leq x\oplus n.a$. Hence we get

[TABLE]

whence $\lambda_{b}(a)\leq x\in J$. Therefore, $\lambda_{b}(a)\in J$ (since $J$ is an ideal).

(v) Let $x,y\in M$. If $y\geq x\in F_{J}$, then there exists $a\in\mathcal{I}(M)\setminus J$ such that $x<a$ and $\lambda_{a}(x)\in J$. Let $b\in\mathcal{I}(M)$ such that $a,y<b$. Then $\lambda_{b}(y)\leq\lambda_{b}(x)=\lambda_{a}(x)\oplus\lambda_{b}(a)$. By the assumption, $\lambda_{b}(a)\in J$, so $\lambda_{a}(x)\oplus\lambda_{b}(a)\in J$, which implies that $\lambda_{b}(y)\in J$. Thus $y\in F_{J}$ and $F_{J}$ is an upset. Moreover, if $x,y\in F_{J}$, then there exist $a,b\in\mathcal{I}(M)$ such that $x<a$ and $y<b$ and $\lambda_{a}(x),\lambda_{b}(y)\in J$. Let $c\in\mathcal{I}(M)$ such that $a,b<c$. Then by the assumption, $\lambda_{c}(a),\lambda_{c}(b)\in J$ and hence by Proposition 3.9, we have $\lambda_{c}(x)=\lambda_{a}(x)\oplus\lambda_{c}(a)\in J$ and $\lambda_{c}(y)=\lambda_{b}(y)\oplus\lambda_{c}(b)\in J$. It follows that $\lambda_{c}(x)\oplus\lambda_{c}(y)\in J$. Now, from definition of $F_{J}$, we have \lambda_{c}\big{(}\lambda_{c}(x)\oplus\lambda_{c}(y)\big{)}\in F_{J}. That is, $x\odot y=x\odot_{{}_{c}}y\in F_{J}$. Therefore, $F_{J}$ is a filter of $M$. ∎

Theorem 5.6**.**

Any proper $EMV$-algebra has at least one maximal ideal. In addition, if $F$ is a maximal filter of $M$, then

[TABLE]

is a maximal ideal of $M$.

Proof.

Let $(M;\vee,\wedge,\oplus,0)$ be a proper $EMV$-algebra. Then $M$ has a non-zero idempotent $a$ and $\uparrow a$ is a proper filter of $M$. By Zorn’s lemma, we can easily see that, $S$, the set of filters of $M$ not containing [math], has at least one maximal element which is clearly a maximal filter of $M$, $F$ say. Set

[TABLE]

By Proposition 5.3, $I_{F}$ is an ideal of $M$. Since $F\neq\emptyset$, there exists $a\in\mathcal{I}(M)\cap F$. By Lemma 5.5(i), $a\notin I_{F}$ and so $I_{F}\neq M$. We claim that $I_{F}$ is a maximal ideal. Let $J$ be an ideal of $M$ containing $I_{F}$. For each $a\in\mathcal{I}(M)$, if $a\notin J$, then $a\notin I_{F}$, so by Lemma 5.5(iii), $a\in F$. It follows that $\lambda_{b}(a)\in I_{F}\subseteq J$ for all $b\in\mathcal{I}(M)$ greater than $a$. By Lemma 5.5(v), $F_{J}:=\{\lambda_{a}(x)\mid x\in J,\ a\in\mathcal{I}(M)\setminus J,\ x<a\}$ is a filter of $M$. Let $x$ be an arbitrary element of $F$. Since $J$ is a proper ideal, then there is an idempotent element $v\in M$ which is not in $J$ (otherwise, $J=M$). Put $w\in\mathcal{I}(M)$ such that $x,v<w$. Then $w\notin J$ and by definition, $\lambda_{w}(x)\in I_{F}\subseteq J$, hence $x\in F_{J}$. That is $F\subseteq F_{J}$. Since $F$ is a maximal filter, then $F_{J}=F$ or $F_{J}=M$. From $F_{J}=M$, we get that $0\in F_{J}$ and so there are $x\in J$ and $a\in\mathcal{I}(M)$ such that $x<a$ and $0=\lambda_{a}(x)$ which is a contradiction (since $[0,a]$ is an $MV$-algebra). So, $F_{J}=F$. Let $x\in J$. Then there is $a\in\mathcal{I}(M)$ such that $x<a$ and $\lambda_{a}(x)\in F_{J}=F$. It follows that $x=\lambda_{a}(\lambda_{a}(x))\in I_{F}$, which simply shows that $I_{F}=J$. Therefore, $I_{F}$ is a maximal ideal of $M$. ∎

Definition 5.7**.**

A proper ideal $I$ of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ is called prime if, for each $x,y\in M$, $x\wedge y\in M$ implies that $x\in M$ or $y\in M$. We denote by $\mathcal{P}(M)$ the set of prime ideals of $M$.

We note that (i) in the next statement was already proved in Proposition 3.23, here we proved it in a different way using e.g. the Riesz Decomposition Property.

Proposition 5.8**.**

Let $I$ be a maximal ideal of an $EMV$-algebra $M$.

(i)* For each idempotent $a\in M$, either $I\cap[0,a]=[0,a]$ or $I\cap[0,a]$ is a maximal ideal of the $MV$-algebra $[0,a]$.*

(ii)* $I$ is a prime ideal.*

Proof.

(i) If $a\in I$, then $I\cap[0,a]=[0,a]$. If $a\notin I$, then $I\cap[0,a]$ is a proper ideal of the $MV$-algebra $[0,a]$. Let $x\in[0,a]\setminus(I\cap[0,a])$. Since $I$ is maximal, there are an integer $n$ and an element $c\in I$ such that $a\leq c\oplus n.x$. Applying the Riesz Decomposition Property, Remark 3.3(iii), there is $c_{1}\leq c$ and $x_{1},\ldots,x_{n}\leq x$ such that $a=c_{1}\oplus x_{1}\oplus\cdots\oplus x_{n}\leq c_{1}\oplus n.x\leq a$ because $c_{1}\in I\cap[0,a]$. Then $\lambda_{a}(n.x)\in I\cap[0,a]$ which proves by Proposition 5.4 means that $I\cap[0,a]$ is a maximal ideal of the $MV$-algebra $[0,a]$.

(ii) To prove $I$ is prime, let $x\wedge y\in I$ for some $x,y\in M$. If $M$ has the greatest idempotent, $M$ is an $MV$-algebra and the statement is well known. Thus assume $M$ is proper. Since $I$ is a proper ideal of $M$, there is an element $x_{0}\notin I$ and there is an idempotent $a$ of $M$ such that $x_{0},x,y\leq a$ and of course, $a\notin I$. Then by (i), $I\cap[0,a]$ is a maximal ideal of the $MV$-algebra $[0,a]$ and $x,y\in[0,a]$. Then $x\wedge y\in I\cap[0,a]$ which proves $x\in I\cap[0,a]$ or $y\in I\cap[0,a]$. Consequently, $I$ is prime. ∎

Proposition 5.9**.**

Every prime ideal of a proper $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ is contained in a unique maximal ideal of $M$.

Proof.

Let $I$ be a prime ideal of an $EMV$-algebra $M$. Put $a\in M\setminus I$. For each $b\in\mathcal{I}(M)$ with $a<b$, we have $\lambda_{b}(a)\wedge a=0\in I$ and so $\lambda_{b}(a)\in I$. Hence, $I$ satisfies condition (5.4) and so by Lemma 5.5(v), $F_{I}$ is a filter of $M$. It can be easily seen that $X=\{F\in\mathrm{Fil}(M)\mid F_{I}\subseteq F,\ 0\notin F\}$ has a maximal element, say $H$, (note that by definition [math] does not belong to $F_{I}$) which is clearly a maximal filter of $M$. Now, by Lemma 5.5(iii) and the proof of Theorem 5.6, $I_{H}$ is a maximal ideal of $M$ containing $I$.

Now let $J_{1},J_{2}$ be two different maximal ideals of $M$ containing $I$. Then there are $x\in J_{1}\setminus J_{2}$ and $y\in J_{2}\setminus J_{1}$. There is an idempotent $a\in\mathcal{I}(M)$ such that $x,y\in[0,a]$. Then $a\notin J_{1}\cup J_{2}$ and by Proposition 5.8(i), $J_{1}\cap[0,a]$ and $J_{2}\cap[0,a]$ are maximal ideals of the $MV$-algebra $([0,a];\oplus,\lambda_{a},0,a)$. It is easy to verify that $I\cap[0,a]$ is a proper ideal of $[0,a]$ which is also prime. Using [CDM, Cor 1.2.12], $J_{1}\cap[0,a]=J_{2}\cap[0,a]$ which implies $x,y\in J_{1},J_{2}$, an absurd. ∎

Proposition 5.10**.**

(i)* Let $P$ be a prime ideal of an $EMV$-algebra and let $I$ be a proper ideal of $M$ containing $P$. Then $I$ is a prime ideal of $M$.*

(ii)* For each prime ideal $J$ of $M$, the set $\mathcal{S}(J)=\{I\in\mathrm{Ideal}(M)\mid J\subseteq I\neq M\}$ is a linearly ordered set of prime ideals with respect to the set theoretical inclusion with a top element.*

Proof.

(i) Let $x\wedge y\in I$ for some $x,y\in M$. Since $I$ is a proper ideal of $M$, there is an idempotent $a$ of $M$ such that $x,y\in[0,a]$ and $a\notin I$. Then $I\cap[0,a]$ is a prime ideal of the $MV$-algebra $[0,a]$, $x\wedge y\in I\cap[0,a]$ and $I\cap[0,a]$ is an ideal of $[0,a]$. Applying [CDM, Thm 1.2.11(i)], we have $I\cap[0,a]$ is a prime ideal of $[0,a]$. Then $x\in I\cap[0,a]\subseteq$ or $y\in I\cap[0,a]\subseteq I$ proving $I$ is prime.

(ii) Let $I_{1},I_{2}$ be two proper ideals of $M$ containing $J$ such that $I_{1}$ and $I_{2}$ are not comparable. Then there are $x\in I_{1}\setminus I_{2}$ and $y\in I_{2}\setminus I_{1}$. We choose an idempotent $a\in\mathcal{I}(M)$ such that $x,y\in[0,a]$ and $a\notin I_{1}\cup I_{2}$. Then $J\cap[0,a]$ is a prime ideal of the $MV$-algebra $[0,a]$ which is contained in both $I_{1}\cap[0,a]$ and $I_{2}\cap[0,a]$. By [CDM, Thm 1.2.11(ii)], $I_{1}\cap[0,a]$ and $I_{2}\cap[0,a]$ are comparable ideals of $[0,a]$. If $I_{1}\cap[0,a]\subseteq I_{2}\cap[0,a]$, then $x\in I_{2}$, an absurd, and dually if $I_{2}\cap[0,a]\subseteq I_{1}\cap[0,a]$, then $y\in I_{1}$ also an absurd. Then $I_{1}$ and $I_{2}$ are comparable. The top element of $\mathcal{S}(J)$ is a unique maximal ideal of $M$ containing $J$ which is guaranteed by Proposition 5.9. ∎

Remark 5.11**.**

In the proof of Theorem 5.6 we showed that if $F$ is a maximal filter of a proper $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$, then $I_{F}$ is a maximal ideal of $M$. Now, let $I$ be a maximal ideal of $M$. By Lemma 5.5(v), $F_{I}$ is a filter of $M$ and so $F_{I}$ is contained in a maximal filter $H$.

• (i)

Since $F_{I}\subseteq H$, then from definition it follows that $I_{F_{I}}\subseteq I_{H}$.

• (ii)

$I_{H}\neq M$. Otherwise, if $I_{H}=M$, then $\mathcal{I}(M)\subseteq I_{H}$ and so by Lemma 5.5(i), $\mathcal{I}(M)\cap H=\emptyset$. That is, $H=\emptyset$.

• (iii)

Let $x\in I$. Put $a\in M\setminus I$ such that $x<a$. Then $\lambda_{a}(x)\in F_{I}$ and so $a\in F_{I}$. By definition, $x=\lambda_{a}(\lambda_{a}(x))\in I_{F_{I}}$. Therefore, $I\subseteq I_{F_{I}}$.

From (i), (ii) and (iii) it follows that $I\subseteq I_{I_{F}}\subseteq I_{H}\varsubsetneq M$ and so $I=I_{H}$. Therefore, any maximal ideal $I$ of $M$ is of the form $I_{H}$ for some maximal filter $H$ of $M$.

Theorem 5.12**.**

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra and let $I$ be a proper ideal of $M$, $a\in M\setminus I$. Then there exists an ideal $P$ of $M$ which is maximal with respect to the property $I\subseteq P$ and $a\in M\setminus P$. In addition, $P$ is prime.

Proof.

If $M$ is not proper and $M\neq\{0\}$, $M$ is a non-degenerated $MV$-algebra, and the statement is well-known [CDM, Prop 1.2.13]. So let $M$ be proper. By Zorn’s lemma, $X=\{J\in\mathrm{Ideal}(M)\mid I\subseteq J,\,a\notin J\}$ has a maximal element, $P$ say. Thus $a\notin P$. We will show that $P$ is prime. Let $x\wedge y\in P$ for some $x,y\in M$. If $x,y\notin P$, then $a\in\langle P\cup\{x\}\rangle$ and $a\in\langle P\cup\{y\}\rangle$, so by Corollary 3.21, $a\leq z_{1}\oplus m.x$ and $a\leq z_{2}\oplus m.y$ for some $m\in\mathbb{N}$ and $z_{1},z_{2}\in P$. It follows that

[TABLE]

Let $b\in\mathcal{I}(M)$ such that $x,y,z_{1},z_{2}\leq b$. Since $([0,b];\oplus,\lambda_{b},0,b)$ is an $MV$-algebra, then by [GeIo, Prop 1.17(i)], we have

[TABLE]

which is a contradiction. Therefore, $P$ is prime. ∎

Corollary 5.13**.**

Every proper ideal of an $EMV$-algebra $M$ can be embedded into a maximal ideal of $M$.

Proof.

Let $I$ be a proper ideal of $M$. By Theorem 5.12, there is a prime ideal $P$ of $M$ containing $I$. Applying Proposition 5.9, we have the assertion in question. ∎

Note that, if $P$ is a prime ideal of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$ and $x,y\in M$, then there exists $a\in\mathcal{I}(M)$ such that $x,y\leq a$. Since $([0,a];\oplus,\lambda_{a},0,a)$ is an $MV$-algebra, then by [CDM, Prop 1.1.7],

[TABLE]

which implies that $\lambda_{a}(\lambda_{a}(x)\oplus y)\in P$ or $\lambda_{a}(\lambda_{a}(y)\oplus x)\in P$.

Theorem 5.14**.**

Let $(M;\vee,\wedge,\oplus,0)$ be a proper $EMV$-algebra. Then the radical $\mbox{\rm Rad}(M)$ of $M$, the intersection of all maximal ideals of $M$, is the set

[TABLE]

Proof.

Let $x\in M\setminus\bigcap\{I\mid I\in\mathrm{MaxI}(M)\}$. Then there exists a maximal ideal $I$ such that $x\notin I$. By Remark 5.11, there exists a maximal filter $H$ such that $I=I_{H}$ and so $x\notin I_{H}$.

[TABLE]

If for all $n\in\mathbb{N}$, $n.x\leq\lambda_{a}(x)$, then $\lambda_{a}(n.x)\geq x$ and so $\lambda_{a}(x)^{n}\geq x$ (note that $([0,a];\oplus,\lambda_{a},0,a)$ is an $MV$-algebra). It follows that $x\in I_{H}=I$ which is a contradiction. Hence

[TABLE]

Now, let $x\in M\setminus\mathrm{Rad}(M)$. Then for all $a\in\mathcal{I}(M)$ with $x\leq a$, there exists $n\in\mathbb{N}$ that $n.x\nleq\lambda_{a}(x)$. It follows that $\lambda_{a}(\lambda_{a}(n.x)\oplus\lambda_{a}(x))>0$. By Theorem 5.12, there is a prime ideal $P$ of $M$ such that $\lambda_{a}(\lambda_{a}(n.x)\oplus\lambda_{a}(x))\notin P$ and so $\lambda_{a}(x\oplus n.x)\in P$ and $a\notin P$ (otherwise, from $\lambda_{a}(\lambda_{a}(n.x)\oplus\lambda_{a}(x))\leq a$ we have $\lambda_{a}(\lambda_{a}(n.x)\oplus\lambda_{a}(x))\in P$). By Proposition 5.9, there is a maximal ideal $J$ of $M$ containing $P$. We claim that the maximal filter $J$ induced from Proposition 5.9 does not contain $a$. Recall that $J=F_{H}$, where $H$ is a maximal filter of $M$ containing $F_{P}$. Check

[TABLE]

So our claim is true. From $\lambda_{a}((n+1).x)\in J$ it follows that $(n+1).x\notin J$ and so $x\notin J$. Hence $x\notin\bigcap\{I\mid I\in\mathrm{MaxI}(M)\}$. Therefore, $\bigcap\{I\mid I\in\mathrm{MaxI}(M)\}\subseteq\mathrm{Rad}(M)$. ∎

Remark 5.15**.**

Let $(B;\vee,\wedge)$ be a generalized Boolean algebra that is not a Boolean algebra and $(M;\oplus,^{\prime},0,1)$ be an $MV$-algebra. In Example 3.2(3), we showed that $M\times B$ is an $EMV$-algebra. By [CoDa, Thm. 2.2], there exists a Boolean algebra $\overline{B}$ such that $B$ is a maximal ideal of $\overline{B}$. Clearly, $M\times\overline{B}$ is an $MV$-algebra containing $M\times B$. It is straightforward to prove that $M\times B$ is a maximal ideal of $M\times\overline{B}$. Therefore, any $EMV$-algebra of the form $M\times B$, where $M$ is an $MV$-algebra and $B$ is a generalized Boolean algebra is a maximal ideal of an $MV$-algebra.

Proposition 5.16**.**

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra. If there exists $\{a_{i}\mid i\in\mathbb{N}\}\subseteq\mathcal{I}(M)$ such that $\{a_{1}\vee a_{2}\vee\cdots\vee a_{n}\mid n\in\mathbb{N}\}$ is a full subset of $\mathcal{I}(M)$ and $a_{i}\wedge a_{j}=0$ for all distinct elements $i,j\in\mathbb{N}$, then the $EMV$-algebra $M$ can be embedded into an $MV$-algebra.

Proof.

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra with the mentioned properties. For each $n\in\mathbb{N}$, $v_{n}:=a_{1}\vee\cdots\vee_{n}$ is an idempotent element of $M$ and so $[0,v_{n}]$ is an $MV$-algebra. Define a map $f:M\rightarrow\prod_{n\in\mathbb{N}}[0,v_{n}]$ by $f(x)=(x\wedge v_{n})_{n\in\mathbb{N}}$ for all $x\in M$. Clearly, $f$ is a one-to-one map which preserves $\vee$, $\wedge$ and [math]. Now, we show that $f$ preserves $\oplus$. Let $n$ be an arbitrary positive integer. We will show that $\pi_{n}\circ f:M\rightarrow[0,v_{n}]$ is a homomorphism of $EMV$-algebras, where $\pi_{n}$ is the $n$-th canonical projection map. Put $x,y\in M$. Then there exists $a\in\mathcal{I}(M)$ such that $x,y,v_{n}\in[0,a]$. Since $([0,a];\oplus,\lambda_{a},0,a)$ is an $MV$-algebra, then from

[TABLE]

it follows that $f$ preserves $\oplus$. From definition of the unary operation ′ in the $MV$-algebra $\prod_{n\in\mathbb{N}}[0,v_{n}]$ and Remark 3.10, it can be easily seen that $f$ is an $EMV$-algebra homomorphism. Therefore, $M$ can be embedded into an $MV$-algebra. ∎

Theorem 5.17**.**

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra. Then $(\mathrm{Ideal}(M);\subseteq)$ is a complete Brouwer lattice. Consequently, $\mathbb{EMV}$ is a congruence distributive variety.

Proof.

Indeed, we have:

(i) Clearly, for each family $\{J_{i}\mid i\in T\}\subseteq\mathrm{Ideal}(M)$, we have

[TABLE]

(ii) Let $I$ be an ideal of $M$. Then for each $x\in I\bigwedge(\bigvee_{i\in T}J_{i})$ by Proposition 3.20, there exist $n\in\mathbb{N}$ and $c_{i_{1}},\ldots,c_{i_{n}}\in\bigcup_{i\in T}J_{i}$ such that $x\leq c_{i_{1}}\oplus\cdots\oplus c_{i_{n}}$. Put $b\in M$ such that $x\leq b$. Since $([0,b];\oplus,\lambda_{b},0,b)$ is an $MV$-algebra, then by [GeIo, Prop 1.17],

[TABLE]

Hence, $\bigwedge$ distributes over arbitrary $\bigvee$.

(iii) For each ideal $I$ of $M$, from (i) and (ii) it follows that $\max\{J\in\mathrm{Ideal}(M)\mid I\bigcap J=\{0\}\}$ exists and it is denoted by $I^{\bot}$. In addition, $I^{\bot}=\{x\in M\mid x\wedge y=0\mbox{ for all }y\in I\}$.

(iv) Since $(\mathrm{Ideal}(M);\subseteq)$ is a Brouwer lattice, it is distributive see e.g. [Bly, p. 151]. Due to Theorem 3.2, the lattice of congruences on $M$, $(\mathrm{Con}(M);\subseteq)$ is also a Brouwer lattice. Therefore, $\mathbb{EMV}$ is a congruence distributive variety. ∎

Proposition 5.18**.**

Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra and $I$ be an ideal of $M$. Then $I^{\bot}=\bigvee_{a\in\mathcal{I}(M)}I_{a}^{\bot_{a}}$, where $I_{a}=I\cap[0,a]$ and $I^{\bot_{a}}$ is the pseudo complement of $I_{a}$ in the lattice of ideals of the $MV$-algebra $[0,a]$.

Proof.

Let $a\in\mathcal{I}(M)$. Then

[TABLE]

On the other hand, if $J$ is an ideal of $M$ such that $J\cap I=\{0\}$, then $J_{a}\cap I_{a}=\{0\}$ for all $a\in\mathcal{I}(M)$, which implies that $J_{a}\subseteq I_{a}^{\bot_{a}}$. Hence

[TABLE]

and so $I^{\bot}\subseteq\bigvee_{a\in\mathcal{I}(M)}I_{a}^{\bot_{a}}$. ∎

Now we show that every subdirectly irreducible $EMV$-algebra is linearly ordered similarly as does every subdirectly irreducible MV-algebra, see [CDM, Thm 1.3.3].

Proposition 5.19**.**

Every subdirectly irreducible $EMV$-algebra $M$ is linearly ordered.

Proof.

If $M=\{0\}$, the statement is clear. Let $M\neq\{0\}$ be a subdirectly irreducible $EMV$-algebra. Due to Theorem 3.16, this means that $M$ has the least non-trivial ideal $I$. Let $a>0$ be any idempotent of $M$. Then $[0,a]$ is an ideal of $M$ as well as of the $MV$-algebra $[0,a]$ which yields $I\subseteq[0,a]$ for each $a\in\mathcal{I}(M)$. It is clear that $I$ is also the least ideal of every $MV$-algebra $[0,a]$. Therefore, every $[0,a]$ is a linearly ordered $MV$-algebra. Let $x,y\in M$. There is an idempotent $a$ of $M$ such that $x,y\leq a$. Then $x\leq y$ or $y\leq x$ as was claimed. ∎

In Theorem 3.18, we showed that any simple $EMV$-algebra is an $MV$-algebra and by Theorem 3.26, we proved that for each maximal ideal $I$ of an $EMV$-algebra $(M;\vee,\wedge,\oplus,0)$, $M/I$ is an $MV$-algebra. Now, we want to generalize this result.

Theorem 5.20**.**

Any $EMV$-algebra can be embedded into an $MV$-algebra.

Proof.

We present two proofs.

(1) Let $(M;\vee,\wedge,\oplus,0)$ be an $EMV$-algebra. By Theorem 5.12, if $\mathcal{P}(M)$ is the set of all prime ideals of $M$, then $\cap\mathcal{P}(M)=\{0\}$. So, the natural map $\phi:M\rightarrow\prod_{I\in\mathcal{P}(M)}M/I$ defined by $\phi(x)=(x/I)_{I\in S}$ is a one-to-one $EMV$-homomorphism. It is easy to see that $M/I$ is a chain (since $I_{a}=I\cap[0,a]$ is a prime ideal of the $MV$-algebra $[0,a]$ for all $a\in\mathcal{I}(M)$ and all prime ideal $I$ of $M$). By Theorem 3.25, $M/I$ is an $MV$-algebra. Therefore, $\prod_{I\in\mathcal{P}(M)}M/I$ is an $MV$-algebra.

(2) By Theorem 3.11, the class of $EMV$-algebras is a variety. Therefore, due to the Birkhoff Subdirect Representation Theorem, see [BuSa, Thm 8.6], $M$ is a subdirect product of subdirectly irreducible $EMV$-algebras which are in view of Proposition 5.19 linearly ordered $EMV$-algebras. By Theorem 3.25, every linearly ordered $EMV$-algebra is an $MV$-algebra which gives the result. ∎

From [CoDa, Thm. 2.2], Theorem 4.15 and Corollary 4.16 we conclude that every proper generalized Boolean algebra, every proper $EMV$-clan and every proper semisimple $EMV$-algebra can be embedded into an $MV$-algebra as a maximal ideal of the $MV$-algebra. In the following, we present a basic result saying that this is true for each proper $EMV$-algebra.

Theorem 5.21**.**

[Basic Representation Theorem]* Every $EMV$-algebra $M$ is either an $MV$-algebra or $M$ can be embedded into an $MV$-algebra $N$ as a maximal ideal of $N$.*

Proof.

If $M$ possesses a top element, then $M$ is an $MV$-algebra, see Example 3.2(4). If $M$ has no top element, $M$ is proper and according to Theorem 5.20, there is an $MV$-algebra $N$ such that $M$ can be embedded into $N$ as an $EMV$-subalgebra of the $EMV$-algebra $N$. Let $1$ be the top element of $N$ and without loss of generality, we can assume that $M$ is an $EMV$-subalgebra of $N$. Let $N_{0}(M)$ be the least $EMV$-subalgebra of $N$ containing both $M$ and the element $1$. In what follows, we will use ideas of the proof of Theorem 4.15 to describe $N_{0}(M)$.

For each $x\in N$, let $x^{*}=\lambda_{1}(x)$.

Set

[TABLE]

We assert $N_{0}(M)=N_{0}$.

Clearly $N_{0}$ contains $M$ and $1$. Let $x,y\in N_{0}$. We have three cases: (i) $x=x_{0},y=y_{0}\in M$. Then $x\vee y,x\wedge y,x\oplus y\in N_{0}$. (ii) $x=x^{*}_{0}$, $y=y_{0}^{*}$ for some $x_{0},y_{0}\in M$. Then $x\vee y=x^{*}_{0}\vee y^{*}_{0}=(x_{0}\wedge y_{0})^{*}$, $x\wedge y=(x_{0}\vee y_{0})^{*}$ and $x\oplus y=x^{*}\oplus y^{*}=(x_{0}\odot y_{0})^{*}\in N_{0}$. (iii) $x=x_{0}$ and $y=y_{0}^{*}$ for some $x_{0},y_{0}\in M$. Then

[TABLE]

where $b$ is an idempotent of $M$ such that $x_{0},y_{0}\leq b$; for the last equality we use equality (5.1) of Lemma 5.1. Using again Lemma 5.1, we have $y_{0}\odot\lambda_{b}(x_{0}\wedge y_{0})\in M$ so that $x\oplus y\in N_{0}$.

In addition, if we apply equality (5.1), we have

[TABLE]

where $a$ is an idempotent of $M$ such that $x,y\leq a$. So that $x\wedge y\in N_{0}$. Using $x\vee y=(x^{*}\wedge y^{*})^{*}$, we have, $x\vee y\in N_{0}$.

We have just proved that $N_{0}$ is an $EMV$-algebra containing $M$ and $1$, so that $N_{0}$ is an $MV$-algebra contained in $N_{0}(M)$. Therefore, $N_{0}=N_{0}(M)$ and $N_{0}$ contains $M$ properly.

Now we prove that $M$ is a maximal ideal of $N_{0}$. Since $M$ is a proper $EMV$-algebra, $M$ is a proper subset of $N_{0}$. To show that $M$ is an ideal it is sufficient assume $y\leq x\in M$. If $y=y_{0}^{*}$, this is impossible while $1\notin M$. Therefore, $M$ is a proper ideal of $N_{0}=N_{0}(M)$. Now let $y\in N_{0}\setminus M$, then $y=y_{0}^{*}$ for some $y_{0}\in M$. Then $y^{*}=y_{0}\in M$ showing $M$ is a maximal ideal of the $MV$-algebra $N_{0}$. ∎

It is important to note that the converse to Theorem 5.21, i.e. whether a maximal ideal of an $MV$-algebra is an $EMV$-algebra, is not true, in general. Indeed, if we take the Chang $MV$-algebra $N=\Gamma(\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z},(0,1))$, where $\mathbb{Z}\,\overrightarrow{\times}\,\mathbb{Z}$ denotes the lexicographic product of the group of natural numbers $\mathbb{Z}$ with itself, then the set $I=\{(0,n)\mid n\geq 0\}$ is a unique maximal ideal of $N$, but $I$ is only a $qEMV$-algebra but not an $EMV$-algebra because $I$ has only one idempotent, namely $0=(0,0)$. However, if $M$ is an $MV$-algebra and $I$ is a maximal ideal of $I$ having enough idempotent elements, i.e., for each $x\in I$, there is an idempotent element $a$ of $M$ belonging to $I$ such that $x\leq a$, then $I$ is an $EMV$-algebra. It is well known that if $a$ is a Boolean element of $M$, then $[0,a]\subseteq I$ and $([0,a];\oplus_{a},^{{}^{\prime}a},0,a)$ is an $MV$-algebra, where $x\oplus_{a}y=(x+y)\wedge a=x\oplus y$, $x^{{}^{\prime}a}=a-x=\lambda_{a}(x)$, $x,y\in[0,a]$. Then due to Theorem 5.21, we have that the set $I\cup I^{\prime}$, where $I^{\prime}=\{x^{\prime}\mid x\in I\}$, is the least $MV$-subalgebra of $M$ containing $I$ and $1$.

More about $MV$-algebras for which a proper $EMV$-algebra can be embedded as their maximal ideal will be done at the end of this section, see Theorem 6.4.

Theorem 5.20 allows us to show that the lattice of all subvarieties of the variety $\mathbb{EMV}$ of $EMV$-algebras is countably infinite similarly as in the case of the lattice of subvarieties of the variety $\mathbb{MV}$ of MV-algebras.

Theorem 5.22**.**

The lattice of subvarieties of the variety $\mathbb{EMV}$ of $EMV$-algebras is countably infinite.

Proof.

According to Komori [Kom], the lattice of subvarieties of the variety $\mathbb{MV}$ of MV-algebras is countably infinite. Di Nola and Lettieri presented in [DiLe] an equational base of any subvariety of the variety $\mathbb{MV}$ which consists of finitely many MV-equations using only $\oplus$ and $\odot$. Hence, let $\mathbb{V}$ be any subvariety of MV-algebras with a finite equational base $\{f_{i}(x_{1},\ldots,x_{n})=g_{i}(y_{1},\ldots,y_{m})\mid i=1,\ldots,n\}$, where $f_{i},g_{i}$ are finite MV-terms using only $\oplus$ and $\odot$. Let $\mathcal{E}(\mathbb{V})$ be the subvariety of $EMV$-algebras satisfying equations $f_{i}(x_{1},\ldots,x_{n})=g_{i}(y_{1},\ldots,y_{m})$ for $i=1,\ldots,n$.

Now let $M$ be any $EMV$-algebra. It generates the subvariety $Var(M)$ of $EMV$-algebras. According to Theorem 5.20, there is an $MV$-algebra $N$ such that $M$ can be embedded into $N$. The MV-algebra $N$ generates the subvariety $\mathbb{V}(N)$ of MV-algebras, hence, $M$ belongs to the variety $\mathcal{E}(\mathbb{V}(N))$ which proves $Var(M)\subseteq\mathcal{E}(\mathbb{V}(N))$.

On the other hand, by the proof of Theorem 5.20, we know that $N$ can be chosen in such a way that $N$ is the direct product of the family $\{M/I\mid I\in\mathcal{P}(M)\}$. Clearly, $M/I$ as a homomorphic image of $M$ belongs to $Var(M)$ for all $I\in\mathcal{P}(M)$, and so the direct product $\prod_{I\in\mathcal{P}(M)}M/I$ also belongs to $Var(M)$, which implies that $N\in Var(M)$. Since any $MV$-algebra is an $EMV$-algebra, then $\mathbb{V}(N)\subseteq Var(N)\subseteq Var(M)$, where $Var(N)$ is the variety of $EMV$-algebras generated by $N$. Then $\mathcal{E}(\mathbb{V}(N))\subseteq Var(M)$ and finally, $\mathcal{E}(\mathbb{V}(N))=Var(M)$.

Now let $\{M_{\alpha}\mid\alpha\in A\}$ be any system of $EMV$-algebras. For every $M_{\alpha}$, there is an $MV$-algebra $N_{\alpha}$ such that $M_{\alpha}$ can be embedded into $N_{\alpha}$. Then $Var(M_{\alpha})=\mathcal{E}(\mathbb{V}(N_{\alpha}))$ for each $\alpha\in A$ which entails $Var(\{M_{\alpha}\mid\alpha\in A\})=\mathcal{E}(\mathbb{V}(\{N_{\alpha}\mid\alpha\in A\}))$. Hence, the cardinality of the set of subvarieties of $\mathbb{EMV}$ is $\aleph_{0}$. ∎

Corollary 5.23**.**

For every subvariety $\mathbb{V}_{E}$ of the variety $\mathbb{EMV}$, there is a subvariety $\mathbb{V}$ of $MV$-algebras such that $\mathbb{V}_{E}=\mathcal{E}(\mathbb{V})$, and the equational base from [DiLe] for $\mathbb{V}$ is also an equational base for $\mathbb{V}_{E}$.

Proof.

The statement follows directly from the proof of Theorem 5.22 and [DiLe]. ∎

For example, (i) the subvariety satisfying the equation $x=0$ is a singleton containing the one-element $EMV$ algebra $\{0\}$. (ii) The equation $x\oplus x=x$ defines the subvariety of generalized Boolean algebras, which is contained in any non-trivial subvariety of $EMV$-algebras. Indeed, if $\mathbb{B}$ is the variety of Boolean algebras, or equivalently, $\mathbb{B}$ is the subvariety $MV$-algebras that satisfy equation $x\oplus x=x$, then $\mathcal{E}(\mathbb{B})$ is the subvariety of generalized Boolean algebras. Then $\mathbb{B}\subseteq\mathbb{V}$ for any non-trivial variety $\mathbb{V}$ of $MV$-algebras, and $\mathcal{(}\mathbb{B})\subseteq\mathcal{E}(\mathbb{V})$. (iii) The equation $x=x$ determines the whole variety $\mathbb{EMV}$.

6 Categorical Equivalencies

In what follows, we present a categorical equivalence of the category of proper $EMV$-algebras with the special category of $MV$-algebras $N$ with a fixed maximal ideal $I$ having enough idempotents and $N=I\cup I^{\prime}$.

Let $\mathcal{PEMV}$ be the category of proper $EMV$-algebra whose objects are proper $EMV$-algebras and morphisms are homomorphisms of $EMV$-algebras. Now let $\mathcal{PMV}$ be the category whose objects are couples $(N,I)$, where $N$ is an $MV$-algebra and $I$ is a fixed maximal ideal of $N$ having enough idempotent elements such that $N=I\cup I^{\prime}$. If $(N_{1},I_{1})$ and $(N_{2},I_{2})$ are two objects of $\mathcal{PMV}$, then a morphism in $\mathcal{PMV}$ from $(N_{1},I_{1})$ into $(N_{2},I_{2})$ is a homomorphism of $MV$-algebras $\phi:N_{1}\to N_{2}$ such that $\phi(I_{1})\subseteq I_{2}$.

If $G$ is an arbitrary Abelian $\ell$-group, then the $MV$-algebra $N=\Gamma(G\,\overrightarrow{\times}\,\mathbb{Z},(0,1))$ (perfect $MV$-algebras, see [CDM, Sec 7.4]) has a unique maximal ideal $I=\{(g,0)\mid g\in G^{+}\}$ and for it we have $I\cup I^{\prime}=N$. However, $I$ is not an $EMV$ algebra because $(0,0)$ is a unique idempotent of $I$.

We note that if $\phi:(N_{1},I_{1})\to(N_{2},I_{2})$ is a morphisms, then $(\phi(N_{1}),\phi(I_{1}))$ is an object of $\mathcal{PMV}$, and it is easy to verify that $\mathcal{PEMV}$ and $\mathcal{PMV}$ are indeed categories. In addition, we underline that $\mathcal{PEMV}$ is not a variety, since due to Theorem 3.26, if $I$ is a maximal ideal of a proper $EMV$-algebra $M$, then $M/I$ is an $MV$-algebra and thus $M/I$ does not belong to $\mathcal{PEMV}$.

Define a mapping $\Phi:\mathcal{PMV}\to\mathcal{PEMV}$ as follows: For any object $(N,I)\in\mathcal{PMV}$, let

[TABLE]

and if $(N_{1},I_{1})$ and $(N_{2},I_{2})$ are objects of $\mathcal{PMV}$ and $\phi:(N_{1},I_{1})\to(N_{2},I_{2})$ is a morphism, then

[TABLE]

Proposition 6.1**.**

$\Phi$* is a well-defined functor that is faithful and full from the category $\mathcal{PMV}$ into the category $\mathcal{PEMV}$.*

Proof.

First, we show that $\Phi$ is a well-defined functor. In other words, we have to establish that if $\phi:(N_{1},I_{1})\to(N_{2},I_{2})$ is a morphism of proper $EMV$-algebras, then $\Phi(\phi)$ is a morphism in $\mathcal{PEMV}$. Indeed, the mapping $\Phi(\phi)$ is in fact an $EMV$-homomorphism from the $EMV$-algebra $I_{1}$ into the $EMV$-algebra $I_{2}$.

Let $\phi_{1}$ and $\phi_{2}$ be two morphisms from $(N_{1},I_{1})$ into $(N_{2},I_{2})$ such that $\Phi(\phi_{1})=\Phi(\phi_{2})$. Then $\phi_{1}(x)=\phi_{2}(x)$ for each $x\in I_{1}$. If $x\in N_{1}\setminus I_{1}$, then there is an element $x_{0}\in I_{1}$ such that $x=x_{0}^{\prime}$. Then $\phi_{1}(x)=\phi_{1}(x^{\prime}_{0})=(\phi_{1}(x_{0}))^{\prime}=(\phi_{2}(x_{0}))^{\prime}=\phi_{2}(x)$ which entails $\phi_{1}=\phi_{2}$, i.e. $\Phi$ is a faithful functor.

To prove that $\Phi$ is a full functor, let $h:I_{1}\to I_{2}$ be a morphism from $\mathcal{PMV}$, i.e. $h$ is a homomorphism of $EMV$-algebras. By Theorem 5.21, there are MV-algebras $N_{1}$ and $N_{2}$ such that $I_{1}$ and $I_{2}$ can be embedded into $N_{1}$ and $N_{2}$, respectively, as their maximal ideals. Without loss of generality, we can assume that $I_{i}$ is a subalgebra of $N_{i}$ for $i=1,2$. We assert that there is a morphism $\phi:(N_{1},I_{1})\to(N_{2},I_{2})$ such that $\Phi(\phi)=h$. In other words $h$ can be extended to an $MV$-homomorphism $\phi$ from $N_{1}$ into $N_{2}$ for some objects $(N_{1},I_{1})$ and $(N_{2},I_{2})$ from $\mathcal{PMV}$. By (5.6), $N_{1}=N_{0}(I_{1})$. So let $x\in N_{1}\setminus I_{1}$. There is a unique element $x_{0}\in I_{1}$ such that $x=x_{0}^{\prime}$. Then we set $\phi(x)=h(x_{0})^{\prime}$. Clearly $\phi(1)=1$, $\phi(x)=h(x)$ if $x\in I_{1}$, and $\phi(x^{\prime})=(\phi(x))^{\prime}$, $x\in N_{1}$. Now let $x,y\in N_{1}$. There are three cases: (1) $x,y\in I_{1}$, then clearly $\phi(x\oplus y)=\phi(x)\oplus\phi(y)$. (2) $x=x_{0}^{\prime}$ and $y=y_{0}^{\prime}$ for some $x_{0},y_{0}\in I_{1}$. Then $\phi(x\oplus y)=\phi(x_{0}^{\prime}\oplus y_{0}^{\prime})=(\phi(x_{0}\odot y_{0}))^{\prime}=(\phi(x_{0})\odot\phi(y_{0}))^{\prime}=(\phi(x)^{\prime}\odot\phi(y)^{\prime})^{\prime}=\phi(x)\oplus\phi(y)$. (3) $x=x^{\prime}_{0}$ and $y=y_{0}$ for some $x_{0},y_{0}\in I_{1}$. There is an idempotent $a\in I_{1}$ such that $x,y\leq a$. Applying (5.1) of Lemma 5.1, we get

[TABLE]

Therefore, $\phi$ is a homomorphism of $MV$-algebras which is an extension of $h$. Whence, $\Phi(\phi)=h$ and $\Phi$ is a full functor. ∎

Proposition 6.2**.**

Let $M$ be a proper $EMV$-algebra and $h_{i}:M\to N_{i}$ be an embedding of $M$ into an $MV$-algebra $N_{i}$ for $i=1,2$. Then $N_{0}^{i}:=N_{0}(h_{i}(M_{i}))$ are isomorphic $MV$-algebras and $(N_{i}^{0},h_{i}(M))\in\mathcal{PMV}$ for $i=1,2$.

Proof.

Let $h_{i}:M\to N_{i}$ be an embedding for $i=1,2$. By (5.6) of Theorem 5.21, $N_{i}^{0}=N_{0}(h_{i}(M))$ for $i=1,2$. Let us define $\psi:N_{1}^{0}\to N_{2}^{0}$ such that $\phi(x)=h_{2}(x_{0})$ if $x=h_{1}(x_{0})$ for $x_{0}\in M_{1}$ and $\phi(x)=(h_{2}(x_{0}))^{\prime}$ if $x=h_{1}(x_{0})^{\prime}$ for $x_{0}\in M_{1}$. Then, similarly as in the proof of the Proposition 6.1 that $\Phi$ is a full functor, we can prove that $\phi$ is a homomorphism of $MV$-algebras. In addition, $\phi$ is a bijection, so that it is an isomorphism. Clearly, $(N_{i}^{0},h_{i}(M))\in\mathcal{PMV}$ for $i=1,2$. ∎

Let $\mathcal{A}$ and $\mathcal{B}$ be two categories and let $f:\mathcal{A}\to\mathcal{B}$ be a functor. Suppose that $g,h$ are functors from $\mathcal{B}$ to $\mathcal{A}$ such that $g\circ f=id_{\mathcal{A}}$ and $f\circ h=id_{\mathcal{B}}$; then $g$ is a left-adjoint of $f$ and $h$ is a right-adjoint of $f$.

Proposition 6.3**.**

The functor $\Phi$ from the category $\mathcal{PMV}$ into the category $\mathcal{PEMV}$ has a left-adjoint.

Proof.

We claim, for a proper $EMV$-algebra $M$, there is a universal arrow $((N,I),f)$ i.e., $(N,I)$ is an object in $\mathcal{PMV}$ and $f$ is a morphism from $M$ into $\Phi(N,I)=I$ such that if $(N^{\prime},I^{\prime})$ is an object from $\mathcal{PEMV}$ and $f^{\prime}$ is a morphism from $M$ into $\Phi(N^{\prime},I^{\prime})$, then there exists a unique morphism $f^{*}:(N,I)\to(N^{\prime},I^{\prime})$ such that $\Phi(f^{*})\circ f=f^{\prime}$.

Indeed, by Theorem 5.21 and Proposition 6.2, there is a unique (up to isomorphism of $MV$-algebras) $MV$-algebra $N$ and an injective $EMV$-homomorphism $f:M\to N$ such that $f(N)$ is a maximal ideal of $N$. We assert that $((N,I),f)$ is universal arrow for $M$. Let $(N^{\prime},I^{\prime})$ be an object from $\mathcal{PEMV}$ and let $f^{\prime}$ be a morphism from $M$ into $\Phi(N^{\prime},I^{\prime})$. We can define a mapping $f^{*}:N\to N^{\prime}$ such that $f^{*}(f(x)):=f^{\prime}(x)$ if $x\in M$ and if $y\in N\setminus f(M)$, there is $y_{0}\in M$ such that $y=(f(y_{0}))^{\prime}$, and we set $f^{*}(y)=(f^{\prime}(y_{0}))^{\prime}$. Then $f^{*}:N\to N^{\prime}$ is a unique $MV$-homomorphism such that $\Phi(f^{*})\circ f=f^{\prime}$.

Define a mapping $\Psi:\mathcal{PEMV}\to\mathcal{PMV}$ by $\Psi(M):=(N,I)$ whenever $((N,I),f)$ is a universal arrow for $M$ and if $f^{\prime}:M\to M^{\prime}$ is an $EMV$-homomorphism, there is a unique morphism $f^{*}:(N,I)\to(N^{\prime},I^{\prime})$, where $\Phi(N^{\prime},I^{\prime})=M^{\prime}$, then we set $\Psi(f^{\prime}):=f^{*}$. Using Theorem 5.21, we have that $\Psi$ is a left-adjoint functor of the functor $\Phi$. ∎

Theorem 6.4**.**

The functor $\Phi$ defines a categorical equivalence of the category $\mathcal{PMV}$ and the category of proper $EMV$-algebras $\mathcal{PEMV}$.

In addition, if $h:\Phi(N,I)\to\Phi(N^{\prime},I^{\prime})$ is a morphism of proper $EMV$-algebras, then there is a unique homomorphism $\phi:(N,I)\to(N^{\prime},I^{\prime})$ of $MV$-algebras such that we have $h=\Psi(\phi)$, and

1. (i)

if $h$ is surjective, so is $\phi$; 2. (ii)

if $h$ is injective, so is $\phi$.

Proof.

According to [MaL, Thm IV.4.1 (i),(iii)], since $\Psi$ is faithful and full, it is necessary to show that, for any proper $EMV$-algebra $M$ there is an object $(N,I)$ in $\mathcal{PMV}$ such that $\Phi(N,I)$ is isomorphic to $M$. To show that it is sufficient to take any universal arrow $((N,I),f)$ of $M$. ∎

Let $(G,u)$ be an Abelian unital $\ell$-group. An $\ell$-ideal is a convex $\ell$-subgroup $I$ of $G$. An $\ell$-ideal $I$ is maximal if it is a value of the strong unit $u$, i.e. a maximal proper $\ell$-ideal of $(G,u)$ not containing $u$. Using categorical equivalence between the category of $MV$-algebras and the category of Abelian unital $\ell$-groups, Theorem 2.2, we have by [CDM, Thm 7.2.2] or [Dvu3, Thm 6.1]: (i) If $I$ is a (maximal) $\ell$-ideal of $(G,u)$, then $I_{0}=I\cap[0,u]$ is a (maximal) ideal of the $MV$-algebra $N=\Gamma(G,u)$; (ii) If $I_{0}$ is a (maximal) ideal of $\Gamma(G,u)$, then $I=\{x\in G\mid|x|\wedge u\in I_{0}\}$ is a (maximal) $\ell$-ideal of $(G,u)$ such that $I\cap[0,u]=I_{0}$. In addition,

[TABLE]

Proposition 6.5**.**

Let $I_{0}$ be a maximal ideal of $N=\Gamma(G,u)$ and let $I$ be a unique maximal $\ell$-ideal of $(G,u)$ generated by $I_{0}$. We define $I^{u}=\{nu-y\mid n\geq 1,\,y\in I,\,0\leq y<nu\}$. Then $I_{0}\cup I_{0}^{\prime}=N$ if and only if $G^{+}=(I^{+})\cup I^{u}$.

Proof.

It is clear that $I^{u}=\{x\in G\mid\exists\,y_{1},\ldots,y_{n}\in I_{0},\,x=y_{1}^{\prime}+\cdots+y_{n}^{\prime}\}$. In addition, $I\cap I^{u}=\emptyset$.

Let $I_{0}\cup I_{0}^{\prime}=\Gamma(G,u)$ and choose $x\in G^{+}$. Then $x=x_{1}+\cdots+x_{m}+(u-y_{1})+\cdots+(u-y_{n})$, where $x_{i},y_{j}\in I_{0}$, so that $x=x_{0}+nu-y_{0}$, where $x_{0}=x_{1}+\cdots+x_{m}\in I^{+}$ and $y_{0}=y_{1}+\cdots+y_{n}\in I^{+}$. If $n=0$, then $x=x_{0}\in I^{+}$. If $n>0$, then $x\wedge u=(x_{1}+\cdots+x_{m}+(u-y_{1})+\cdots+(u-y_{n}))\wedge u=x_{1}\oplus\cdots\oplus x_{m}\oplus(u-y_{1})\oplus\cdots\oplus(u-y_{n})\in\Gamma(G,u)$. Then $m=0$, $x_{0}$ and $x=nu-y_{0}\in I^{u}$.

Conversely, let $G^{+}=I^{+}\cup I^{n}$. Then $N=N\cap G^{+}=(N\cap I^{+})\cup(N\cap I^{u})=N\cup(N\cap I^{u})$. Let $x\in N\cap I^{u}$, then $x=nu-y$ for some $0\leq y<nu$, $y\in I$. Using the Riesz Decomposition Property for $\ell$-groups, we have $y=y_{1}+\cdots+y_{n}$, where $y_{i}\in I_{0}$. Then $x=(nu-y)=((u-y_{1})+\cdots+(u-y_{n}))\wedge u=y_{1}^{\prime}\oplus\cdots\oplus y_{n}^{\prime}=(y_{1}\odot\cdots\odot y_{n})^{\prime}$, but $y_{0}:=y_{1}\odot\cdots\odot y_{n}\in I_{0}$ so that $x=y_{0}\in I_{0}^{\prime}$ which yields $N\cap I^{u}=I_{0}^{\prime}$. Then $N=I_{0}\cup I_{0}^{\prime}$. ∎

Inspired by the previous categorical equivalence, let $\mathcal{PUALG}$ be the category of unital Abelian $\ell$-groups with a fixed maximal $\ell$-ideal with a special property. Namely, the objects are triples $(G,u,I)$ such that $(G,u)$ is an Abelian unital $\ell$-group and $I$ is a fixed maximal $\ell$-ideal $I$ of $(G,u)$ such that $G^{+}=I^{+}\cup I^{u}$ and the ideal $I_{0}=I\cap[0,u]$ of $\Gamma(G,u)$ has enough idempotent elements. If $(G_{1},u_{1},I_{1})$ and $(G_{2},u_{2},I_{2})$ are two objects of $\mathcal{PUALG}$, then a mapping $f:(G_{1},u_{1},I_{1})\to(G_{2},u_{2},I_{2})$ is a morphism if $f$ is a homomorphism of unital $\ell$-groups such that $f(I_{1})\subseteq I_{2}$. Our aim is to show that $\mathcal{PUALG}$ is categorically equivalent to the category $\mathcal{PMV}$. We will follows techniques used in the previous categorical equivalence.

Let us define a functor $\Gamma_{I}:\mathcal{PUALG}\to\mathcal{PMV}$ as follows: if $(G,u,I)$ is an object of $\mathcal{PUALG}$, then

[TABLE]

and if $f$ is a morphism from an object $(G_{1},u_{1},I_{1})$ into another one $(G_{2},u_{2},I_{2})$, then

[TABLE]

Proposition 6.6**.**

$\Gamma_{I}$* is a well-defined functor that is faithful and full.*

Proof.

Clearly $\Gamma_{I}(G,u,I)=(\Gamma(G,u),I\cap[0,u])\in\mathcal{PMV}$. If $f:(G_{1},u_{1},I_{1})\to(G_{2},u_{2},I_{2})$ is a morphism, then the restriction of $f$ onto $\Gamma(G_{1},u_{1})$ is in fact a homomorphism of $MV$-algebras with $f(I_{1})\subseteq I_{2}$, so that $\Gamma_{I}(f)(I_{1}\cap[0,u_{1}])\subseteq I_{2}\cap[0,u_{2}]$, and $\Gamma_{I}$ is a correctly defined functor.

Let $f_{1}$ and $f_{2}$ be two morphisms from $(G_{1},u_{1},I_{1})$ into $(G_{2},u_{2},I_{2})$ such that $\Gamma(f_{1})=\Gamma(f_{2})$. Then $f_{1}(x)=f_{2}(x)$ for each $x\in\Gamma(G_{1},u_{1})$. Since $f_{i}$ for $i=1,2$ is a homomorphism of unital $\ell$-groups, it is easy to see that $f_{1}(x)=f_{2}(x)$ for each $x\in G_{1}$ and $f_{1}=f_{2}$.

Now let $\mu:\Gamma_{I}(G_{1},u_{1},I_{1})=(\Gamma(G_{1},u_{1}),I_{1}\cap[0,u_{1}])\to\Gamma_{I}(G_{2},u_{2},I_{2})=(\Gamma(G_{2},u_{2}),I_{2}\cap[0,u_{2}])$ be a morphism, i.e. $\mu$ is an $MV$-homomorphism from $\Gamma(G_{1},u_{1})$ into $\Gamma(G_{2},u_{2})$ such that $\mu(I_{1}\cap[0,u_{1}])\subseteq I_{2}\cap[0,u_{2}]$. Using methods of the proof of [Dvu2, Prop 6.1], we can uniquely extend $\mu$ to a homomorphism of unital $\ell$-groups $f:G_{1}\to G_{2}$. Since $I_{i}\cap[0,u_{i}]$ can be uniquely extended to the $\ell$-ideal $I_{i}$, $i=1,2$, we have that $f$ is a morphism from $(G_{1},u_{1},I_{1})$ into $(G_{2},u_{2},I_{2})$, which proves $\Gamma_{I}$ is a full functor because $\Gamma_{I}(f)=\mu$. ∎

Now we introduce the following notions. On every $MV$-algebra $N$ we can define a partial addition $+$ such that $x+y$ is defined iff $x\odot y=0$, and in such a case, $x+y:=x\oplus y$; if $N=\Gamma(G,u)$, then the partial addition coincides with the group addition related to $[0,u]$. We say that a couple $(G,f)$ is a universal group for an $MV$-algebra $N$ if (i) $f$ is a mapping from $M$ into a po-group $G$ which preserves partial addition $+$ on $N$ such that $G=G^{+}-G^{+}$, $f(M)$ generates $G^{+}$ as a semigroup, (ii) for any group $K$ and any $+$-preserving mapping $h:N\to K$, there is a group homomorphism $\phi:G\to K$ such that $h=\phi\circ f$. Due to [Dvu2, Thm 5.3] if $N\cong\Gamma(G,u)$, then $(G,f)$ is a universal group for $N$, where $f$ is an isomorphism $f:N\to\Gamma(G,u)$.

Proposition 6.7**.**

The functor $\Gamma_{I}$ from the category $\mathcal{PUALG}$ into the category $\mathcal{PMV}$ has a left-adjoint.

Proof.

We assert that for an object $(N,I_{0})$, there is a universal arrow $((G,u,I),f)$, i.e. $(G,u,I)$ is an object from $\mathcal{PUALG}$ and $f$ is a morphism from $(N,I_{0})$ into $\Gamma_{I}(G,u,I)=(\Gamma(G,u),I\cap[0,u])=(\Gamma(G,u),I_{0})$ such that if $(G^{\prime},u^{\prime},I^{\prime})$ is an object from $\mathcal{PUALG}$ and $f^{\prime}$ is a morphism from $(N,I_{0})$ into $\Gamma_{I}(G^{\prime},u^{\prime},I^{\prime})=(\Gamma(G^{\prime},u^{\prime}),I^{\prime}\cap[0,u^{\prime}])$, then there is a unique morphism $f^{*}:(G,u,I)\to(G^{\prime},u,I^{\prime})$ such that $\Gamma_{I}(f^{*})\circ f=f^{\prime}$.

Take the universal group $(G,f)$ for the $MV$-algebra $N$. Then $f$ is an $MV$-bijection from $M$ onto $\Gamma(G,u)$. We assert $((G,u,I),f)$ is a universal arrow for $(N,I_{0})$, where $I$ is an $\ell$-ideal of $G$ generated by $f(I_{0})$. Indeed, take an object $(G^{\prime},u^{\prime},I^{\prime})$ from $\mathcal{PUALG}$ and let $f^{\prime}$ be a morphism from $(N,I_{0})\to\Gamma_{I}(G^{\prime},u^{\prime},I^{\prime})=(\Gamma(G^{\prime},u^{\prime}),I^{\prime}_{0})$, where $I^{\prime}_{0}=I^{\prime}\cap[0,u^{\prime}]$. Since $f:M\to\Gamma(G,u)\subset G^{+}$ is a $+$-preserving mapping and $f^{\prime}:M\to\Gamma(G^{\prime},u^{\prime})\subseteq G^{\prime}$ is also a $+$-preserving mapping, then there is a unique homomorphism of unital $\ell$-groups $f^{*}:G\to G^{\prime}$ such that $f^{*}\circ f=f^{\prime}$. First $x\in I_{0}$, then $f(x)\in I$ and thus $f^{*}(f(x))=f^{\prime}(x)\in I^{\prime}_{0}$. If we take a general $x\in I$, see (6.5), then $f^{\prime}(x)\in I^{\prime}$, so that $f^{*}$ is also a morphism from $(G,u,I)$ to $(G^{\prime},u,^{\prime}I^{\prime})$, i.e. $((G,u,I),f)$ is a universal arrow in question.

Define a mapping $\Xi_{I}:\mathcal{PMV}\to\mathcal{PUALG}$ by $\Xi_{I}(N,I_{0})=(G,u,I)$ if $((G,u,I),f)$ is a universal arrow for $(N,I_{0})$ and $I$ is a maximal $\ell$-ideal of $G$ generated by $f(I_{0})$. If $f^{\prime}$ is a morphism from $(N,I_{0})$ into $(N^{\prime},I^{\prime}_{0})$, there is a unique morphism $f^{*}:(G,u,I)\to(G^{\prime},u^{\prime},I^{\prime})$, where $N^{\prime}\cong\Gamma(G^{\prime},u^{\prime})$ and $I^{\prime}$ is a maximal $\ell$-ideal of $G^{\prime}$ generated by $f^{\prime}(I_{0}^{\prime})$, then $\Xi_{I}(f^{\prime}):=f^{*}$. Therefore, $\Xi_{I}$ is a left-adjoint of $\Gamma_{I}$. ∎

Theorem 6.8**.**

The functor $\Gamma_{I}$ defines a categorical equivalence of the category $\mathcal{PUALG}$ and the category $\mathcal{PMV}$.

Proof.

The statement follows from [MaL, Thm IV.4.1(i),(iii)] and Propositions 6.6–6.7. ∎

Corollary 6.9**.**

The categories $\mathcal{PUALG}$, $\mathcal{PMV}$ and $\mathcal{PEMV}$ are mutually categorically equivalent.

Proof.

It follows from Theorem 6.4 and Theorem 6.8. ∎

7 Conclusion

We have introduced the notion of an $EMV$-algebra, Definition 3.1, which generalizes the notion of an $MV$-algebra and of a generalized Boolean algebra. We have exhibited its basic properties and notions as ideals, congruences, filters, and their mutual relationship, Theorem 3.16. Nevertheless an $EMV$-algebra $M$ has not necessarily a top element, $M$ has a maximal ideal, Theorem 5.6. We have defined an $EMV$-clan as an $EMV$-algebra of fuzzy sets. We have shown that every $EMV$-algebra is semisimple iff it is isomorphic to some $EMV$-clan of fuzzy sets, Theorem 4.11. A state-morphism is any $EMV$-homomorphism from $M$ into the $MV$-algebra of the real interval $[0,1]$ which attains the value 1. State-morphisms are in a one-to-one relationship with maximal ideals of $M$, Theorem 4.2.

We have shown that every $EMV$-algebra can be embedded into an $MV$-algebra, Theorem 5.20. Theorem 5.21 characterizes any $EMV$-algebra saying that either it is an $MV$-algebra or it can be embedded into an $MV$-algebra as its maximal ideal.

The class of $EMV$-algebras forms a variety, Theorem 3.11. Using the equational base of any subvariety of the variety of $MV$-algebras, [DiLe], we describe a functional base of any subvariety of the variety $\mathbb{EMV}$ of $EMV$-algebras, Corollary 5.23, and the cardinality of all subvarieties of the variety $\mathbb{EMV}$ is $\aleph_{0}$, Theorem 5.22. Finally, we presented mutually categorical equivalencies of the category of proper $EMV$-algebras, a special category of $MV$-algebras $N$ with a fixed maximal ideal $I$ having enough idempotents, and a special categories of Abelian unital $\ell$-groups, Theorem 6.4, Theorem 6.8 and Corollary 6.9.

With the present paper we have opened a new and interesting window into the realm of unbounded generalizations of $MV$-algebras and generalized Boolean algebra, and we hope to continue in this research, for example with a variant of the Loomis–Sikorski theorem for $\sigma$-complete $EMV$-algebras.