Nilpotent Elements of Residuated Lattices

Abstract

Some properties of the nilpotent elements of a residuated lattice are studied. The concept of cyclic residuated lattices is introduced, and some related results are obtained. The relation between finite cyclic residuated lattices and simple MV-algebras is obtained. Finally, the notion of nilpotent elements is used to define the radical of a residuated lattice.

1. Introduction

Ward and Dilworth [1] introduced the concept of residuated lattices as generalization of ideal lattices of rings. The residuated lattice plays the role of semantics for a multiple-valued logic called residuated logic. Residuated logic is a generalization of intuitionistic logic. Therefore it is weaker than classical logic. Important examples of residuated lattices related to logic are Boolean algebras corresponding to basic logic, BL-algebras corresponding to Hajek’s basic logic, and MV-algebras corresponding to Lukasiewicz many valued logic. The residuated lattices have been widely studied (see [2–8]).

In this paper, we study the properties of nilpotent elements of residuated lattices. In Section 2, we recall some definitions and theorems which will be needed in this paper. In Section 3, we study the nilpotent elements of a residuated lattice and study its properties. In Section 4, we define the notion of cyclic residuated lattice and we obtain some related results. In particular, we will prove that a finite residuated lattice is cyclic if and only if it is a simple MV-algebra. In Section 5, we investigate the relation between nilpotent elements and the radical of a residuated lattice.

2. Preliminaries

In this section, we review some basic concepts and results which are needed in the later sections.

A residuated lattice is an algebraic structure (𝐴,∧,∨,→,∗,0,1) such that(1)(𝐴,∧,∨,0,1) is a bounded lattice with the least element 0 and the greatest element 1,(2)(𝐴,∗,1) is a commutative monoid where 1 is a unit element,(3)𝑥∗𝑦≤𝑧 if and only if 𝑥≤𝑦→𝑧, for all 𝑥,𝑦,𝑧∈𝐴. We denote the residuated lattice (𝐴,∧,∨,∗,→,0,1) by 𝐴. We use the notation 𝐿(𝐴) for the bounded lattice (𝐴,∧,∨,0,1).

Proposition 2.1 (see [5, 9]). Let 𝐴 be a residuated lattice. Then one has the following properties: for all 𝑥,𝑦,𝑧∈𝐴,(1)𝑥≤𝑦 if and only if 𝑥→𝑦=1,(2)1→𝑥=𝑥, 𝑥→1=1,(3)𝑥∗𝑦≤𝑥,𝑦, 𝑥∗(𝑥→𝑦)≤𝑥∧𝑦,(4)𝑥∗(𝑦∨𝑧)=(𝑥∗𝑦)∨(𝑥∗𝑧),(5)𝑥→(𝑦→𝑧)=(𝑥∗𝑦)→𝑧,(6)(𝑥→𝑦)∗(𝑦→𝑧)≤𝑥→𝑧,(7)if 𝑥≤𝑦, then 𝑧→𝑥≤𝑧→𝑦 and 𝑦→𝑧≤𝑥→𝑧.

An MV-algebra is an algebra (𝐴,⊕,¬,0) with one binary operation ⊕, one unary operation ¬, and one constant 0 such that (𝐴,⊕,0) is a commutative monoid and, for all 𝑥,𝑦∈𝐴, ¬¬𝑥=𝑥, 𝑥⊕¬0=¬0, ¬(¬𝑥⊕𝑦)⊕𝑦=¬(¬𝑦⊕𝑥)⊕𝑥. If 𝐴 is an MV-algebra, then the binary operations ∗, ∧, ∨, → and the constant 1 are defined by the following relations: for all 𝑥,𝑦∈𝐴, 𝑥∗𝑦=¬(¬𝑥⊕¬𝑦), 𝑥∧𝑦=(𝑥⊕¬𝑦)∗𝑦, 𝑥∨𝑦=(𝑥∗¬𝑦)⊕𝑦, 𝑥→𝑦=¬𝑥⊕𝑦, 1=¬0.

Remark 2.2. A residuated lattice 𝐴 is an MV-algebra if it satisfies the additional condition: (𝑥→𝑦)→𝑦=(𝑦→𝑥)→𝑥, for any 𝑥,𝑦∈𝐴.

Definition 2.3. A nonempty subset 𝐹 of 𝐴 is called a filter of 𝐴 if and only if it satisfies the following conditions:(i)for all 𝑥∈𝐹 and all 𝑦∈𝐴, if 𝑥≤𝑦 then 𝑦∈𝐹,(ii)for all 𝑥,𝑦∈𝐹, 𝑥∗𝑦∈𝐹.

Let 𝐹 be a filter of 𝐴. For all 𝑥,𝑦∈𝐴, we denote 𝑥≡𝑦 and say that 𝑥 and 𝑦 are congruent if and only if 𝑥→𝑦∈𝐹 and 𝑦→𝑥∈𝐹. ≡ is a congruence relation on 𝐴. The quotient residuated lattice with respect to the congruence relation ≡ is denoted by 𝐴/𝐹 and its elements are denoted by [𝑥], for 𝑥∈𝐴.

For all elements𝑥 of a residuated lattice 𝐴, define 𝑥⁰=1 and 𝑥^𝑛=𝑥^𝑛−1∗𝑥 for all 𝑛∈ℵ. The order of 𝑥∈𝐴, in symbols ord(𝑥), is the smallest positive integer 𝑛 such that 𝑥^𝑛=0. If such 𝑛 does not exist, we say 𝑥 has infinite order.

Definition 2.4. The residuated lattice 𝐴 is called simple if the only filters of 𝐴 are {1} and 𝐴.

Proposition 2.5 (see [10]). A residuated lattice 𝐴 is simple if and only if ord(𝑎)<∞, for every 𝑎∈𝐴 such that 𝑎≠1.

A filter 𝑀 of 𝐴 is called a maximal filter if and only if it is a maximal element of the set of all proper filters of 𝐴. The set of all maximal filters of 𝐴 is called the maximal spectrum of 𝐴 and is denoted by Max(𝐴). For any 𝑋⊆𝐴, we will denote 𝑆_Max(𝑋)={𝑀∈Max(𝐴)∣𝑋⊆𝑀̸}. For any 𝑥∈𝐴, 𝑆_Max({𝑥}) will be denoted by 𝑆_Max(𝑥).

Proposition 2.6 (see [10]). Let 𝐴 be a residuated lattice and 𝑀 a proper filter of 𝐴. Then the followings are equivalent:(i)𝐴/𝑀 is a simple residuated lattice,(ii)𝑀 is a maximal filter, (iii)for any 𝑥∈𝐴, 𝑥∉𝑀 if and only if 𝑥^𝑛→0∈𝑀, for some 𝑛≥1.

Definition 2.7. A residuated lattice is said to be local if and only if it has exactly one maximal filter.

Proposition 2.8 (see [10]). Any simple residuated lattice is local.

We denote by 𝐵(𝐴) the Boolean center of 𝐴, that is the set of all complemented elements of the lattice 𝐿(𝐴). The complements of the elements in the Boolean center of a residuated lattice are unique.

Theorem 2.9 (see [10]). If 𝐴 is a local residuated lattice, then 𝐵(𝐴)={0,1}.

3. Nilpotent Elements of Residuated Lattices

We recall that an element 𝑥∈𝐴 is called nilpotent if and only if ord(𝑥) is finite. We denote by 𝑁(𝐴) the set of the nilpotent elements of 𝐴.

Example 3.1 (see [5]). Consider the residuated lattice 𝐴 with the universe {0,𝑎,𝑏,𝑐,𝑑,𝑒,𝑓,1}. Lattice ordering is such that 0<𝑑<𝑐<𝑏<𝑎<1, 0<𝑑<𝑒<𝑓<𝑎<1, and elements from {𝑏,𝑐} and {𝑒,𝑓} are pairwise incomparable. The operations of ∗ and → are given by the following: ∗0𝑎𝑏𝑐𝑑𝑒𝑓10𝑎00000000𝑏0𝑐𝑐𝑐0𝑑𝑑𝑎𝑐0𝑐𝑐𝑐00𝑑𝑏𝑑0𝑐𝑐𝑐000𝑐𝑒0000000𝑑𝑓0𝑑000𝑑𝑑𝑒10𝑑𝑑00𝑑𝑑𝑓→0𝑎𝑏𝑐𝑑𝑒𝑓10𝑎𝑏𝑐𝑑𝑒𝑓10𝑎11111111𝑏𝑑1𝑎𝑎𝑓𝑓𝑓1𝑐𝑒11𝑎𝑓𝑓𝑓1𝑑𝑓111𝑓𝑓𝑓1𝑒𝑎1111111𝑓𝑏1𝑎𝑎𝑎1111𝑐1𝑎𝑎𝑎𝑎110𝑎𝑏𝑐𝑑𝑒𝑓1 (3.1) Then ord(0)=1, ord(𝑑)=2, ord(𝑒)=ord(𝑓)=3, and ord(𝑎)=ord(𝑏)=ord(𝑐)=ord(1)=∞. Hence 𝑁(𝐴)={0,𝑑,𝑒,𝑓}.

Remark 3.2. Let 𝐴 be a residuated lattice. Then 𝑥∈𝐴 has order 𝑛 if and only if 𝑥^𝑛→0=1, for some 𝑛∈ℵ.

Example 3.3. Let 𝐴_𝑛+1={𝑎₀,…,𝑎_𝑛}, for 𝑛∈ℵ and 𝑎_𝑖≤𝑎_𝑖+1 for all 0≤𝑖≤𝑛. Also, we have 𝑎₀=0 and 𝑎_𝑛=1. Define 𝑎_𝑖∗𝑎_𝑗=𝑎0if𝑖+𝑗≤𝑛,_{𝑖+𝑗−𝑛}𝑎if𝑛<𝑖+𝑗,_𝑖→𝑎_𝑗=𝑎_{𝑛−𝑖+𝑗}if𝑗<𝑖,1if𝑖≤𝑗. (3.2) Then 𝐴_𝑛+1 becomes a residuated lattice. Let 1≠𝑎_𝑖∈𝐴_𝑛+1. Then there exists 𝑘∈ℵ such that 𝑘𝑖≤(𝑘−1)𝑛. We get that (𝑎_𝑖)^𝑘=𝑎_𝑖∗(𝑎_𝑖)^𝑘−1=𝑎_𝑖∗𝑎_{(𝑘−1)𝑖−(𝑘−2)𝑛}=0, because 𝑖+(𝑘−1)𝑖−(𝑘−2)𝑛=𝑘𝑖−(𝑘−2)𝑛≤𝑛. Hence 𝐴_𝑛+1 is a simple residuated lattice.

Theorem 3.4. 𝑁(𝐴) is a lattice ideal of the residuated lattice 𝐴.

Proof. It is clear that 0∈𝑁(𝐴). Suppose that 𝑥≤𝑦 and 𝑦∈𝑁(𝐴). There exists 𝑚∈ℵ such that 𝑦^𝑚=0. We have 𝑥^𝑚≤𝑦^𝑚. Therefore 𝑥∈𝑁(𝐴).
Suppose that 𝑥,𝑦∈𝑁(𝐴). Then there exists 𝑚,𝑛∈ℵ such that 𝑥^𝑚=𝑦^𝑛=0. By Proposition 2.1(4), we have 𝑥∨𝑦^𝑚+𝑛=𝑥^𝑚+𝑛∨𝑥^{𝑚+𝑛−1}∗𝑦∨⋯∨𝑥^𝑚∗𝑦^𝑛∨⋯∨𝑦^𝑚+𝑛=0. (3.3) Hence 𝑥∨𝑦∈𝑁(𝐴), and then 𝑁(𝐴) is a lattice ideal of 𝐿.

Remark 3.5. An element 𝑥 of a residuated lattice 𝐴 is nilpotent if and only if there is no proper filter 𝐹 of 𝐴 such that 𝑥∈𝐹.

Theorem 3.6. Let 𝐴 be a residuated lattice and {𝐴_𝑖∶𝑖∈𝐼} a family of residuated lattices. Then(1)𝐵(𝐴)∩𝑁(𝐴)={0},(2)𝑁(Π_𝑖∈𝐼𝐴_𝑖)=Π_𝑖∈𝐼𝑁(𝐴_𝑖).

Proof. (1) Let 𝑥∈𝐵(𝐴)∩𝑁(𝐴). Since 𝑥∈𝐵(𝐴), then 𝑥=𝑥∗𝑥. Hence we get that 𝑥^𝑛=𝑥 for all 𝑛∈ℵ. Also, we have 𝑥∈𝑁(𝐴). So there exists 𝑚∈𝑁(𝐴) such that 𝑥^𝑚=0. Therefore, we obtain that 𝑥=0.
(2) Suppose that 𝑥∈Π_𝑖∈𝐼𝑁(𝐴_𝑖). Then 𝑥=(𝑥₁,…,𝑥_𝑛) where 𝑥_𝑖 is nilpotent in 𝐴_𝑖. Thus, for each 1≤𝑖≤𝑛, there exists 𝑚_𝑖 such that (𝑥_𝑖)^𝑚𝑖=0. Put 𝑚=Max{𝑚_𝑖∶1≤𝑖≤𝑛}. Then 𝑥^𝑚=0, that is 𝑥∈𝑁(Π_𝑖∈𝐼𝐴_𝑖). The proof of reverse inclusion is straightforward.

For a nonempty subset 𝑋⊆𝐴, the smallest filter of 𝐴 which contains 𝑋 is said to be the filter of 𝐴 generated by 𝑋 and will be denoted by ⟨𝑋⟩. If 𝑋={𝑥} with 𝑥∈𝐴, we denote by ⟨𝑥⟩ the filter generated by {𝑥}. Also, we have ⟨𝑥⟩={𝑦∈𝐴∶𝑦≥𝑥^𝑛 for some 𝑛∈ℵ}.

Theorem 3.7. Let 𝑥 be an element of a residuated lattice 𝐴. Then 𝑥 is nilpotent if and only if 𝐴=⟨𝑥⟩.

Proof. Let 𝑥 be a finite order. Then there exists integer 𝑛>0 such that 𝑥^𝑛=0, that is 0∈⟨𝑥⟩. Therefore 𝐴=⟨𝑥⟩.
Conversely, if 𝐴=⟨𝑥⟩, then 0∈⟨𝑥⟩. So there exists an integer 𝑛>0 such that 𝑥^𝑛≤0. Hence 𝑥 has finite order.

Theorem 3.8. Let 𝑥 be an element of order 𝑛 of a residuated lattice 𝐴. Then the elements 𝑥⁰→0=0, 𝑥^𝑛→0=1, 𝑥^𝑖→0, 𝑖=1,…,𝑛−1 of 𝐴 are pairwise distinct.

Proof. Suppose that 𝑥^𝑖→0=𝑥^𝑗→0, for some 0≤𝑖<𝑗≤𝑛. Then 1=𝑥^𝑛→0=(𝑥^𝑗∗𝑥^𝑛−𝑗)→0=𝑥^𝑛−𝑗→(𝑥^𝑗→0)=𝑥^𝑛−𝑗→(𝑥^𝑖→0)=𝑥^{𝑛−𝑗+𝑖}→0. But 𝑛−𝑗+𝑖<𝑛 which is a contradiction with the order of 𝑎. Hence 𝑥^𝑖→0≠𝑥^𝑗→0.

4. Cyclic Residuated Lattices

The order of a residuated lattice 𝐴 is the cardinality of 𝐴 and denoted by 𝑂(𝐴).

Definition 4.1. Let 𝐴 be a finite residuated lattice. 𝐴 is called cyclic, if there exists 𝑥∈𝐴 such that ord(𝑥)=𝑂(𝐴)−1. 𝑥 is called a generator of 𝐴.

Theorem 4.2. Let 𝐴 be a cyclic residuated lattice of order 𝑛+1. Then there exists an element 𝑥∈𝐿 of order 𝑛 such that 𝐴={𝑥^𝑖→0∶0≤𝑖≤𝑛} where 𝑥⁰→0=0, 𝑥^𝑛→0=1.

Proof. Since 𝐴 is cyclic of order 𝑛+1, then there exists an element 𝑥∈𝐿 of order 𝑛. By Theorem 3.8, the elements 𝑥⁰→0=0, 𝑥^𝑛→0=1 and 𝑥^𝑖→0, 𝑖=1,…,𝑛−1 of 𝐴 are pairwise distinct. Hence the cardinality of 𝑋={𝑥^𝑖→0∶𝑖=0,…,𝑛−1} is 𝑛+1. Since 𝑂(𝐴)=𝑛+1 and 𝑋⊆𝐴, we get that 𝐴={𝑥^𝑖→0∶𝑖=0,…,𝑛}.

An element 𝑎 of a residuated lattice 𝐴 is called a coatom if it is maximal among elements in 𝐿(𝐴)−{1}.

Theorem 4.3. Let 𝐴 be a cyclic residuated lattice of order 𝑛+1. Then 𝐴 is linearly ordered. Moreover the generator of 𝐴 is a unique coatom.

Proof. Since 𝐴 is a cyclic residuated lattice of order 𝑛+1, then there exists an element 𝑥∈𝐴 such that 𝐴={𝑥^𝑖→0∶𝑖=0,…,𝑛} by Theorem 4.2. If 0≤𝑗≤𝑖≤𝑛, then 𝑥^𝑖≤𝑥^𝑗. By Proposition 2.1(7),𝑥^𝑗→0≤𝑥^𝑖→0. Hence 𝐴 is linearly ordered.
It is clear that 𝑥^𝑛−1→0 is a coatom of 𝐴. We will show that 𝑥=𝑥^𝑛−1→0. Since 1≠𝑥∈{𝑥^𝑖→0∶𝑖=0,…,𝑛}, then there exists 0≤𝑖<𝑛 such that 𝑥=𝑥^𝑖→0. We have 𝑥^𝑖+1→0=𝑥→𝑥^𝑖→0=𝑥→𝑥=1. (4.1) Therefore 𝑥^𝑖+1→0=1. By definition order of 𝑥, we get that 𝑛≤𝑖+1. Hence 𝑛−1≤𝑖. Therefore 𝑥=𝑥^𝑛−1→0.

In the following example, we will show that the converse of the above theorem may not be true in general.

Example 4.4. Consider the residuated lattice 𝐴 with the universe {0,𝑎,𝑏,𝑐,1}. Lattice ordering is such that 0<𝑎<𝑏<𝑐<1. The operations of ∗ and → are given by the following: ∗0𝑎𝑏𝑐10𝑎00000𝑏0000𝑎𝑐0000𝑏1000𝑎𝑐→0𝑎𝑏𝑐10𝑎𝑏𝑐10𝑎11111𝑏𝑐1111𝑐𝑐𝑐1111𝑏𝑐𝑐110𝑎𝑏𝑐1 (4.2)𝐴 is a finite linearly residuated lattice of order 5 but it is not cyclic because we have ord(0)=1, ord(𝑎)=ord(𝑏)=2, ord(𝑑)=3, and ord(1)=∞.

In the following theorems, we characterize cyclic residuated lattices.

Theorem 4.5. Let 𝐴 be a cyclic residuated lattice of order 𝑛+1. Then 𝐴 is isomorphic to the simple residuated lattice 𝐴_𝑛+1 of Example 3.3.

Proof. By Theorems 4.2 and 4.3, there exists an element 𝑥∈𝐴 such that 𝐴={𝑥^𝑖→0∶𝑖=0,…,𝑛}, where 𝑥⁰→0=0<𝑥¹→0<⋯<𝑥^𝑛−1→0<𝑥^𝑛→0=1. We denote 𝑎₀=0, 𝑎_𝑛=1, and 𝑎_𝑖=𝑥^𝑖→0 for 1≤𝑖≤𝑛−1. By Theorem 4.3, 𝑥=𝑎_𝑛−1 is the generator of 𝐴 and it is a coatom. We will show that 𝑎_𝑖→𝑎_𝑗=𝑎_{𝑛−𝑖+𝑗}if𝑗<𝑖,1if𝑖≤𝑗. (4.3) It is clear that 𝑎_𝑖→𝑎_𝑗=1, if 𝑖≤𝑗. We will prove that 𝑎_𝑖→𝑎_𝑗=𝑎_{𝑛−𝑖+𝑗}, if 𝑗<𝑖. Suppose that 𝑖=𝑗+1. Then 𝑎_𝑛−1→𝑎_𝑖→𝑎_𝑗=𝑎_𝑖→𝑎_𝑛−1→𝑎_𝑗=𝑎_𝑖→𝑥→𝑥^𝑗→0=𝑎_𝑖→𝑥^𝑗+1→0=𝑎_𝑗+1→𝑎_𝑗+1=1. (4.4) Hence 𝑥=𝑎_𝑛−1≤𝑎_𝑖→𝑎_𝑗≤1. Since 𝑥 is a coatom, then 𝑎_𝑖→𝑎_𝑗=1 or 𝑥=𝑎_𝑖→𝑎_𝑗. If 𝑎_𝑖→𝑎_𝑗=1, then 𝑎_𝑖≤𝑎_𝑗. Since 𝑗<𝑖, then 𝑎_𝑗<𝑎_𝑖 which is a contradiction by Theorem 3.8. Hence 𝑎_𝑛−1=𝑥=𝑎_𝑖→𝑎_𝑗.
Now, suppose that 𝑖=𝑗+𝑘 where 𝑘>1. Since 𝑎_𝑖→𝑎_𝑗∈𝐴={𝑥^𝑖→0∶𝑖=0,…,𝑛}, then there exists 0≤𝑚≤𝑛 such that 𝑎_𝑖→𝑎_𝑗=𝑥^𝑚→0. We have 𝑥^𝑘−1→𝑎_𝑖→𝑎_𝑗=𝑎_𝑖→𝑥^𝑘−1→𝑎_𝑗=𝑎_𝑖→𝑥^𝑘−1→𝑥^𝑗→0=𝑎_𝑖→𝑥^𝑘−1∗𝑥^𝑗→0=𝑎_𝑖→𝑥^{𝑗+𝑘−1}→0=𝑎_𝑖→𝑎_𝑖−1=𝑥. (4.5) Therefore 𝑥=𝑥^𝑘−1→(𝑎_𝑖→𝑎_𝑗)=𝑥^𝑘−1→(𝑥^𝑚→0)=𝑥^{𝑚+𝑘−1}→0. We get that 1=𝑥→𝑥=𝑥^𝑚+𝑘→0. Thus, 𝑛≤𝑚+𝑘. On the other hand, since 1≠𝑥=𝑥^𝑛−1→0 and 𝑥^𝑡→0=1 for all 𝑡≥𝑛, then 𝑚+𝑘−1≤𝑛−1. Therefore 𝑚=𝑛−𝑘. We obtain that 𝑎_𝑖→𝑎_𝑗=𝑥^𝑛−𝑘→0=𝑎_𝑛−𝑘=𝑎_{𝑛−𝑖+𝑗}.
Now, we will show that 𝑎_𝑖∗𝑎_𝑗=𝑎0if𝑖+𝑗≤𝑛,_{𝑖+𝑗−𝑛}if𝑛<𝑖+𝑗. (4.6) Suppose that 𝑖+𝑗≤𝑛. Then 𝑎_𝑖≤𝑎_𝑛−𝑗. Since 𝑎_𝑛−𝑗=𝑎_𝑗→𝑎₀, we get that 𝑎_𝑖∗𝑎_𝑗≤𝑎₀=0. Thus 𝑎_𝑖∗𝑎_𝑗=0.
Suppose that 𝑛<𝑖+𝑗. Since 𝑖≤𝑛, we have 𝑖+𝑗−𝑛≤𝑗. Therefore 𝑎_𝑗→𝑎_{𝑖+𝑗−𝑛}=𝑎_𝑖. We get that 𝑎_𝑖∗𝑎_𝑗≤𝑎_{𝑖+𝑗−𝑛}. Let 𝑎_𝑖∗𝑎_𝑗=𝑎_𝑡. Then 𝑡≤𝑖+𝑗−𝑛. Consider the following cases.(1)If 𝑗≤𝑡<𝑖+𝑗−𝑛, then 0≤𝑡−𝑗<𝑖−𝑛≤0 which is a contradiction.(2)If 𝑖≤𝑡<𝑖+𝑗−𝑛, then 0≤𝑡−𝑖<𝑗−𝑛≤0 which is a contradiction.(3)𝑡≤𝑖,𝑗, then 𝑎_𝑗≤𝑎_𝑖→𝑎_𝑡=𝑎_{𝑛−𝑖+𝑡}. We get that 𝑗≤𝑛−𝑖+𝑡. Therefore 𝑗+𝑖−𝑛≤𝑡 which is a contradiction.We obtain that 𝑡=𝑖+𝑗−𝑛.
Hence 𝑎_𝑖↦𝑥_𝑖 is an isomorphism between 𝐴 and 𝐴_𝑛+1. Since 𝐴_𝑛+1 is a simple residuated lattice, then 𝐴 is a simple residuated lattice.

Remark 4.6. Consider the residuated lattice 𝐴_𝑛+1 in Example 3.3. Since (𝑎_𝑖→𝑎_𝑗)→𝑎_𝑗=(𝑎_𝑗→𝑎_𝑖)→𝑎_𝑖 for all 0≤𝑖, 𝑗≤𝑛, then 𝐴_𝑛+1 is an MV-algebra.

Corollary 4.7. Every cyclic residuated lattice is a finite simple MV-algebra.

Proof. It follows from Theorem 4.5 and Remark 4.6.

Corollary 4.8. Every cyclic residuated lattice 𝐴 is local and 𝐵(𝐴)={0,1}.

Proof. It follows from Theorem 4.5, Proposition 2.8, and Theorem 2.9.

Theorem 4.9. Every finite simple MV-algebra is a cyclic residuated lattice.

Proof. Any simple MV-algebra is isomorphic to a subalgebra of [0,1], and also 𝐿_𝑛={0,1/(𝑛−1),…,(𝑛−2)/(𝑛−1),1}, (𝑛≥2) is the only subalgebra of [0,1] with 𝑛 elements [11]. Since ((𝑛−2)/(𝑛−1))^𝑛−1=0, then ord((𝑛−1)/(𝑛−2))=𝑂(𝐿_𝑛)−1. Therefore it is cyclic.

Corollary 4.10. A finite residuated lattice is cyclic if and only if it is a simple MV-algebra.

Proof. It follows from Corollary 4.7 and Theorem 4.9.

Theorem 4.11. Every nonzero subalgebra of a cyclic residuated lattice is cyclic.

Proof. Suppose that 𝑆 is a nonzero subalgebra of a cyclic residuated lattice 𝐴. Then 𝐴 is a simple MV-algebra and 𝑆 is isomorphic to a subalgebra of [0,1]. Moreover 𝐿_𝑛={0,1/(𝑛−1),…,(𝑛−2)/(𝑛−1),1}, (𝑛≥2) is the only subalgebra of [0,1] with 𝑛 elements. Hence 𝑆 is a simple MV-algebra. By Theorem 4.9, 𝑆 is a cyclic MV-algebra.

Theorem 4.12. Every finite MV-algebra is a direct product of cyclic residuated lattices.

Proof. Every finite MV-algebra is isomorphic to a finite direct product of finite subalgebras of the standard MV-algebra [11]. Theorem 4.11 yields the theorem.

5. Semisimple Residuated Lattices

The intersection of the maximal filters of residuated lattice 𝐴 is called the radical of 𝐴 and will be denoted by Rad(𝐴).

Theorem 5.1. Let 𝐴 be a residuated lattice. Then Rad(𝐴)={𝑥∈𝐴∶𝑥^𝑛→0∈𝑁(𝐴) for all 𝑛∈ℵ}.

Proof. See [12].

Theorem 5.2. Let 𝐴 be a residuated lattice. Then(1)Rad(𝐿(𝐴)/𝑁(𝐴))=Rad(𝐿(𝐴))/𝑁(𝐴),(2)Max(𝐿(𝐴)) and Rad(𝐿(𝐴)/𝑁(𝐴)) are homomorphic topological spaces.

Proof. (1) It is easily seen that Max(𝐿(𝐴)/𝑁(𝐴))={𝑀/𝑁(𝐴)∶𝑀∈Max(𝐿(𝐴))}, thus 𝐿(𝐴)Rad𝑁(𝐴)=_{𝑀∈Max(𝐿(𝐴))}𝑀𝑁(𝐴)=⋂_{𝑀∈Max(𝐿(𝐴))}𝑀𝑁(𝐴)(𝐿(𝐴=Rad))𝑁(𝐴). (5.1)
(2) Define 𝜑∶Max(𝐿(𝐴))→Max(𝐿(𝐴)/𝑁(𝐴)), for all 𝑀∈Max(𝐿(𝐴)), 𝜑(𝑀)=𝑀/𝑁(𝐴). This function is well defined and surjective. For any 𝑀∈Max(𝐿(𝐴)) and any 𝑥∈𝐴, we have [𝑥]∈𝑀/𝑁(𝐴) if and only if 𝑥∈𝑀. We get that 𝜑 is injective.
Now, we will prove that 𝜑 is continuous and open. Let 𝑥∈𝐴. By using the above, we get 𝑆_Max𝑥=𝑀([])𝑁(𝐴)(𝐿(𝐴,[𝑥]∉𝑀∶𝑀∈Max))𝑁(𝐴)=𝑀𝑁(𝐴)(𝐴)=𝑀∶𝑀∈Max,𝑥∉𝑀𝑁(𝐴)∶𝑀∈𝑆_Max(𝑥=𝜑(𝑀))∶𝑀∈𝑆_Max(𝑥𝑆)=𝜑_Max(𝑥). (5.2) Thus 𝜑 is open. Since 𝜑 is injective and open, then 𝜑⁻¹(𝑆_Max([𝑥]))=𝑆_Max(𝑥). So 𝜑 is continuous.

Definition 5.3. A residuated lattice 𝐴 is called semisimple if the intersection of all congruences of 𝐴 is the congruence ▵_𝐴 (where, for all 𝑥,𝑦∈𝐴, 𝑥▵_𝐴𝑦 if and only if 𝑥=𝑦).

Remark 5.4 (see [5]). A residuated lattice 𝐴 is semisimple if and only if Rad(𝐴)={1}.

Lemma 5.5. Let 𝐴 be a residuated lattice, 𝑥∈𝐴, and 𝐹={𝑦∈𝐴∶𝑦→𝑥=𝑥}. Then 𝐹 is a filter of 𝐴.

Proof. (1) Suppose that 𝑦₁,𝑦₂∈𝐹. Then 𝑦₁→𝑥=𝑥=𝑦₂→𝑥. We have 𝑦₁∗𝑦₂→𝑥=𝑦₁→𝑦₂→𝑥=𝑦₁→𝑥=𝑥. (5.3) Therefore 𝑦₁∗𝑦₂∈𝐹.
(2) Let 𝑦≤𝑧, where 𝑦∈𝐹 and 𝑧∈𝐴. We have 𝑦→𝑥=𝑥. Since 𝑦≤𝑧, then 𝑥≤𝑧→𝑥≤𝑦→𝑥=𝑥. Therefore 𝑧→𝑥=𝑥 and 𝑧∈𝐹.
Hence 𝐹 is a filter of 𝐴.

Theorem 5.6. Let 𝐴 be a residuated lattice such that for all 𝑥∈𝐴 there exists 𝑛∈ℵ such that (𝑥^𝑛→0)→𝑥=𝑥. Then 𝐴 is semisimple.

Proof. We will show that Rad(𝐴)={1}. Suppose that 𝑥∈Rad(𝐴). By Theorem 5.1, we have that 𝑥^𝑛→0 is nilpotent. By assumption, there exists 𝑛∈ℵ such that (𝑥^𝑛→0)→𝑥=𝑥. We have 𝑥^𝑛→0∈𝐹={𝑦∈𝐴∶𝑦→𝑥=𝑥}. Since 𝐹 is a filter of 𝐴 by Lemma 5.5 and 𝑥^𝑛→0 is nilpotent, we get that 0∈𝐹. Thus 1=0→𝑥=𝑥. We obtain that Rad(𝐴)={1} and then 𝐴 is semisimple by Remark 5.4.