# Nilpotent Elements of Residuated Lattices

Authors: Shokoofeh Ghorbani; Lida Torkzadeh
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## 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.


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## Body

## 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 Section2, 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]).
LetA be a residuated lattice. Then one has the following properties: for all x,y,z∈A,(1)
x≤yif and only ifx→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)
ifx≤y, thenz→x≤z→yandy→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 latticeA 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 subsetF of A is called a filter of A if and only if it satisfies the following conditions:(i)
for allx∈F and all y∈A, if x≤y then y∈F,(ii)
for allx,y∈F, x*y∈F.LetF 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 xn=xn-1*x for all n∈ℵ. The order of x∈A, in symbols ord(x), is the smallest positive integer n such that xn=0. If such n does not exist, we say x has infinite order.Definition 2.4.
The residuated latticeA is called simple if the only filters of A are {1} and A.Proposition 2.5 (see [10]).
A residuated latticeA is simple if and only if ord(a)<∞, for every a∈A such that a≠1.A filterM 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 SMax(X)={M∈Max(A)∣X⊈M}. For any x∈A, SMax({x}) will be denoted by SMax(x).Proposition 2.6 (see [10]).
LetA be a residuated lattice and M a proper filter of A. Then the followings are equivalent:(i)
A/Mis a simple residuated lattice,(ii)
Mis a maximal filter,(iii)
for anyx∈A, x∉Mif and only ifxn→0∈M, for somen≥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 byB(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]).
IfA is a local residuated lattice, then B(A)={0,1}.


## 3. Nilpotent Elements of Residuated Lattices

We recall that an elementx∈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 latticeA 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:
(3.1)*0abcdef1000000000a0ccc0ddab0ccc00dbc0ccc000cd0000000de0d000ddef0dd00ddf10abcdef1→0abcdef1011111111ad1aafff1be11afff1cf111fff1da1111111eb1aaa111fc1aaaa1110abcdef1
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.
LetA be a residuated lattice. Then x∈A has order n if and only if xn→0=1, for some n∈ℵ.Example 3.3.
LetAn+1={a0,…,an}, for n∈ℵ and ai≤ai+1 for all 0≤i≤n. Also, we have a0=0 and an=1. Define
(3.2)ai*aj={0ifi+j≤n,ai+j-nifn<i+j,ai→aj={an-i+jifj<i,1ifi≤j.
Then An+1 becomes a residuated lattice. Let 1≠ai∈An+1. Then there exists k∈ℵ such that ki≤(k-1)n. We get that (ai)k=ai*(ai)k-1=ai*a(k-1)i-(k-2)n=0, because i+(k-1)i-(k-2)n=ki-(k-2)n≤n. Hence An+1 is a simple residuated lattice.Theorem 3.4.
N
(
A
) is a lattice ideal of the residuated lattice A.Proof.
It is clear that0∈N(A). Suppose that x≤y and y∈N(A). There exists m∈ℵ such that ym=0. We have xm≤ym. Therefore x∈N(A).
Suppose thatx,y∈N(A). Then there exists m,n∈ℵ such that xm=yn=0. By Proposition 2.1(4), we have
(3.3)(x∨y)m+n=xm+n∨(xm+n-1*y)∨⋯∨(xm*yn)∨⋯∨ym+n=0.
Hence x∨y∈N(A), and then N(A) is a lattice ideal of L.Remark 3.5.
An elementx 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.
LetA be a residuated lattice and {Ai:i∈I} a family of residuated lattices. Then(1)
B(A)∩N(A)={0},(2)
N(Πi∈IAi)=Πi∈IN(Ai).Proof.
(1) Letx∈B(A)∩N(A). Since x∈B(A), then x=x*x. Hence we get that xn=x for all n∈ℵ. Also, we have x∈N(A). So there exists m∈N(A) such that xm=0. Therefore, we obtain that x=0.
(2) Suppose thatx∈Πi∈IN(Ai). Then x=(x1,…,xn) where xi is nilpotent in Ai. Thus, for each 1≤i≤n, there exists mi such that (xi)mi=0. Put m=Max{mi:1≤i≤n}. Then xm=0, that is x∈N(Πi∈IAi). The proof of reverse inclusion is straightforward.For a nonempty subsetX⊆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≥xn for some n∈ℵ}.Theorem 3.7.
Letx be an element of a residuated lattice A. Then x is nilpotent if and only if A=〈x〉.Proof.
Letx be a finite order. Then there exists integer n>0 such that xn=0, that is 0∈〈x〉. Therefore A=〈x〉.
Conversely, ifA=〈x〉, then 0∈〈x〉. So there exists an integer n>0 such that xn≤0. Hence x has finite order.Theorem 3.8.
Letx be an element of order n of a residuated lattice A. Then the elements x0→0=0, xn→0=1, xi→0, i=1,…,n-1 of A are pairwise distinct.Proof.
Suppose thatxi→0=xj→0, for some 0≤i<j≤n. Then 1=xn→0=(xj*xn-j)→0=xn-j→(xj→0)=xn-j→(xi→0)=xn-j+i→0. But n-j+i<n which is a contradiction with the order of a. Hence xi→0≠xj→0.


## 4. Cyclic Residuated Lattices

The order of a residuated latticeA is the cardinality of A and denoted by O(A).Definition 4.1.
LetA 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.
LetA be a cyclic residuated lattice of order n+1. Then there exists an element x∈L of order n such that A={xi→0:0≤i≤n} where x0→0=0, xn→0=1.Proof.
SinceA 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, xn→0=1 and xi→0, i=1,…,n-1 of A are pairwise distinct. Hence the cardinality of X={xi→0:i=0,…,n-1} is n+1. Since O(A)=n+1 and X⊆A, we get that A={xi→0:i=0,…,n}.An elementa of a residuated lattice A is called a coatom if it is maximal among elements in L(A)-{1}.Theorem 4.3.
LetA be a cyclic residuated lattice of order n+1. Then A is linearly ordered. Moreover the generator of A is a unique coatom.Proof.
SinceA is a cyclic residuated lattice of order n+1, then there exists an element x∈A such that A={xi→0:i=0,…,n} by Theorem 4.2. If 0≤j≤i≤n, then xi≤xj. By Proposition 2.1(7),xj→0≤xi→0. Hence A is linearly ordered.
It is clear thatxn-1→0 is a coatom of A. We will show that x=xn-1→0. Since 1≠x∈{xi→0:i=0,…,n}, then there exists 0≤i<n such that x=xi→0. We have
(4.1)xi+1→0=x→(xi→0)=x→x=1.
Therefore xi+1→0=1. By definition order of x, we get that n≤i+1. Hence n-1≤i. Therefore x=xn-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 latticeA 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:
(4.2)*0abc1000000a0000ab0000bc000ac10abc1→0abc1011111ac1111bcc111cbcc1110abc1A 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.
LetA be a cyclic residuated lattice of order n+1. Then A is isomorphic to the simple residuated lattice An+1 of Example 3.3.Proof.
By Theorems4.2 and 4.3, there exists an element x∈A such that A={xi→0:i=0,…,n}, where x0→0=0<x1→0<⋯<xn-1→0<xn→0=1. We denote a0=0, an=1, and ai=xi→0 for 1≤i≤n-1. By Theorem 4.3, x=an-1 is the generator of A and it is a coatom. We will show that
(4.3)ai→aj={an-i+jifj<i,1ifi≤j.
It is clear that ai→aj=1, if i≤j. We will prove that ai→aj=an-i+j, if j<i. Suppose that i=j+1. Then
(4.4)an-1→(ai→aj)=ai→(an-1→aj)=ai→(x→(xj→0))=ai→(xj+1→0)=aj+1→aj+1=1.
Hence x=an-1≤ai→aj≤1. Since x is a coatom, then ai→aj=1 or x=ai→aj. If ai→aj=1, then ai≤aj. Since j<i, then aj<ai which is a contradiction by Theorem 3.8. Hence an-1=x=ai→aj.
Now, suppose thati=j+k where k>1. Since ai→aj∈A={xi→0:i=0,…,n}, then there exists 0≤m≤n such that ai→aj=xm→0. We have
(4.5)xk-1→(ai→aj)=ai→(xk-1→aj)=ai→(xk-1→(xj→0))=ai→((xk-1*xj)→0)=ai→(xj+k-1→0)=ai→ai-1=x.
Therefore x=xk-1→(ai→aj)=xk-1→(xm→0)=xm+k-1→0. We get that 1=x→x=xm+k→0. Thus, n≤m+k. On the other hand, since 1≠x=xn-1→0 and xt→0=1 for all t≥n, then m+k-1≤n-1. Therefore m=n-k. We obtain that ai→aj=xn-k→0=an-k=an-i+j.
Now, we will show that(4.6)ai*aj={0ifi+j≤n,ai+j-nifn<i+j.
Suppose that i+j≤n. Then ai≤an-j. Since an-j=aj→a0, we get that ai*aj≤a0=0. Thus ai*aj=0.
Suppose thatn<i+j. Since i≤n, we have i+j-n≤j. Therefore aj→ai+j-n=ai. We get that ai*aj≤ai+j-n. Let ai*aj=at. Then t≤i+j-n. Consider the following cases.(1)
Ifj≤t<i+j-n, then 0≤t-j<i-n≤0 which is a contradiction.(2)
Ifi≤t<i+j-n, then 0≤t-i<j-n≤0 which is a contradiction.(3)
t≤i,j, then aj≤ai→at=an-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.
Henceai↦xi is an isomorphism between A and An+1. Since An+1 is a simple residuated lattice, then A is a simple residuated lattice.Remark 4.6.
Consider the residuated latticeAn+1 in Example 3.3. Since (ai→aj)→aj=(aj→ai)→ai for all 0≤i, j≤n, then An+1 is an MV-algebra.Corollary 4.7.
Every cyclic residuated lattice is a finite simple MV-algebra.Proof.
It follows from Theorem4.5 and Remark 4.6.Corollary 4.8.
Every cyclic residuated latticeA is local and B(A)={0,1}.Proof.
It follows from Theorem4.5, 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 Ln={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(Ln)-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 Corollary4.7 and Theorem 4.9.Theorem 4.11.
Every nonzero subalgebra of a cyclic residuated lattice is cyclic.Proof.
Suppose thatS 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 Ln={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 latticeA is called the radical of A and will be denoted by Rad(A).Theorem 5.1.
LetA be a residuated lattice. Then Rad(A)={x∈A:xn→0∈N(A) for all n∈ℵ}.Proof.
See [12].Theorem 5.2.
LetA 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 thatMax(L(A)/N(A))={M/N(A):M∈Max(L(A))}, thus
(5.1)Rad(L(A)N(A))=⋂M∈Max(L(A))‍MN(A)=⋂M∈Max(L(A))‍MN(A)=Rad(L(A))N(A).
(2) Defineφ: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 thatφ is continuous and open. Let x∈A. By using the above, we get
(5.2)SMax([x])={MN(A):M∈Max(L(A)),[x]∉MN(A)}={MN(A):M∈Max(A),x∉M}={MN(A):M∈SMax(x)}={φ(M):M∈SMax(x)}=φ(SMax(x)).
Thus φ is open. Since φ is injective and open, then φ-1(SMax([x]))=SMax(x). So φ is continuous.Definition 5.3.
A residuated latticeA 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 latticeA is semisimple if and only if Rad(A)={1}.Lemma 5.5.
LetA be a residuated lattice, x∈A, and F={y∈A:y→x=x}. Then F is a filter of A.Proof.
(1) Suppose thaty1,y2∈F. Then y1→x=x=y2→x. We have
(5.3)y1*y2→x=y1→(y2→x)=y1→x=x.
Therefore y1*y2∈F.
(2) Lety≤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.
HenceF is a filter of A.Theorem 5.6.
LetA be a residuated lattice such that for all x∈A there exists n∈ℵ such that (xn→0)→x=x. Then A is semisimple.Proof.
We will show thatRad(A)={1}. Suppose that x∈Rad(A). By Theorem 5.1, we have that xn→0 is nilpotent. By assumption, there exists n∈ℵ such that (xn→0)→x=x. We have xn→0∈F={y∈A:y→x=x}. Since F is a filter of A by Lemma 5.5 and xn→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.



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