Molecules and linearly ordered ideals of MV-algebras
Abstract
Hoo, C. S.
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Publicacions Matem`atiques, Vol 41 (1997), 455–465. MOLECULES AND LINEARLY ORDERED IDEALS OF MV-ALGEBRAS C. S. Hoo Abstract We show that an ideal Iof an MV -algebra Ais linearly ordered if and only if every non-zero element of Iis a molecule. The set of molecules of Ais contained in Inf(A)∪B2(A) where B2(A) is the set of all elements x∈Asuch that 2xis idempotent. It is shown that I={0}is weakly essential if and only if B⊥⊂ B(A).Connections are shown among the classes of ideals that have various combinations of the properties of being implicative, essential, weakly essential, maximal or prime. 1. Introduction In [8] and [15] we deduced various properties of MV-algebras on the assumption that certain ideals were linearly ordered. It was also shown in [8] that atoms generate linearly ordered ideals. In this paper, we characterize completely those elements that generate linearly ordered ideals. We also characterize linearly ordered ideals in terms of a property of their elements. This property is the concept of a molecule. Molecules were introduced by A. Abian in [1] as a generalization of atoms. Not much has been done using this concept although there was a paper [20] by Yaqub on the molecules of Post algebras. Recall that a non-zero element mof a partially ordered set Pwith zero is a molecule if every two non-zero elements of Pwhich are less than or equal to m have a non-zero lower bound. It was shown in [1] that an element of a Boolean algebra is a molecule if and only if it is an atom. In general, molecules exist even when atoms do not; and in a chain, there is at most one atom while every non-minimum element is a molecule. 1991 Mathematics subject classifications: 06F35, 03G25.
456 C. S. Hoo We shall characterize molecules in MV-algebras and deduce some of their properties. We shall also show when the orthogonal complement of the ideal generated by a molecule is implicative. Other properties of ideals such as “essential” and “weakly essential” are also studied. We shall show connections among the classes of ideals that have various combinations of the properties of being implicative, essential, weakly essential, maximal or prime. In order not to lengthen the paper, we shall not review the definitions and basic concepts of MV-algebras. Rather, we refer the reader to the references for the theory of BCK and MV-algebras, in particular to [2], [4], [8], [16], [17], [18] and [19]. We shall follow the notation and terminology of [8] and shall assume the results there without further reference, as well as the results in [2] and [4]. We shall state explicitly the results of [14] and [15] that we need, if and when we need them. We use freely the BCK-algebra operation found in [8] and [18] as we feel that computations involving the BCK-algebra operation are often more transparent than those using just the MV-algebra operations. In this paper, Ashall denote a general MV-algebra, B(A) its Boolean subalgebra of idempotents, At(A) its set of atoms, and Mol(A) its set of molecules. Ishall denote an ideal of A. In order to avoid trivialities, it shall always be assumed that A={0,1}. We shall also denote B(A)−{1} by B1(A). Then the only idempotent in A−B1(A)is1. Recall that Iis implicative if whenever xn∈Ifor some integer n≥1, then x∈I(see [8, Theorem 3.6] and [11, Theorem 2.4]). These are precisely the ideals which give quotients A/I which are Boolean algebras. Let Inf(A)={x∈A|x2=0}={x∧x|x∈A},N(A)={x∈A|xn= 0 for some integer n≥1}and Rad(A) = intersection of all maximal ideals of A. Observe that Iis implicative if and only if Inf(A)⊂I(see [11]). Also Rad A={x∈A|nx ≤xfor all integers n≥0}(see [6], [7] and [11]). Let IRad(A) be the intersection of all implicative ideals of A.It was observed in [11] that we have Rad(A)⊂Inf(A)⊂N(A)⊂IRad(A), and Inf(A)=N(A)=IRad(A). Recall that Iis essential if I⊥={0}. Also I={0}is weakly essential if for all ideals Jsuch that J∩{A−B1(A)}=∅we have I∩J={0} (see [10]). Observe that if Jis a proper ideal, then J∩{A−B1(A)}is idempotent free. Thus J∩{A−B1(A)}=∅means that it contains a non-zero non-idempotent. It was shown in [10, Theorem 3.15] that if Jis a proper ideal, then J∩{A−B1(A)}=∅if and only if J∩Inf(A)={0}. It is easily verified that essential ideals are weakly essential (see [10]). In [15, Theorem 2.13] we showed that if I={0}is implicative, then it is weakly essential, and in [15, Theorem 3.11], we showed that if Iis weakly essential, then I⊥⊂B(A). We also showed in [15, Theorem 2.3] that if I
Molecules and linearly ordered ideals 457 is prime then I⊥is linearly ordered. It was also shown in [8, Lemma 5.1] that if Iis linearly ordered and contains an idempotent x= 0, then xis the largest element of I. Finally, observe that if a∈At(A), then either a2=0ora2=a, that is, either a∈Inf(A)ora∈B(A). Thus we have At(A)⊂Inf(A)∪B(A). 2. Molecules Definition 2.1. A non-zero element mof a poset Pwith 0 is a molecule if whenever 0 <x,y≤m, then {x, y}a non-zero lower bound. Thus m∈Ais a molecule if and only if whenever x,y∈Asatisfy 0<x,y≤m, then x∧y>0. Let Mol(A) denote the set of all molecules of A. Observe that At(A)⊂Mol(A). Recall that if ∅=X⊂A, then X⊥={a∈A|a∧x= 0 for all x∈X}. Then X⊥is an ideal of A; it is proper if X={0}(see [2] and [8]). The ideal Xgenerated by Xis the set of all a∈Asuch that a≤k1x1+···+knxnfor some integer n≥1, integers k1,... ,k n≥0, and x1,... ,x n∈X. We shall denote {a} by a. Observe that {a}⊥= a⊥. Theorem 2.2. m∈Mol(A)if and only if m⊥is a prime ideal. Proof: Suppose that m∈Mol(A). Then m⊥is a proper ideal of A. Suppose that x∧y∈m⊥, that is, x∧y∧m=0. Ifx/∈m⊥and y/∈m⊥, then we have 0 <x∧m,y∧m≤m, and hence x∧y∧m>0, a contradiction. Thus either x∈m⊥or y∈m⊥, proving that m⊥is prime. Conversely, suppose that m⊥is a prime ideal of A. Suppose that x,y∈Asatisfy 0 <x,y≤m. Then x=x∧m= 0 and y=y∧m=0, that is x/∈m⊥and y/∈m⊥. This means that x∧y/∈m⊥, and hence x∧y∧m=0. Thusx∧y>0. In [8, Theorem 4.14], it was shown that I⊥is prime if and only if Iis linearly ordered and ={0}. Consequently, we have the following result. Corollary 2.3. m∈Mol(A)if and only if mis linearly ordered and ={0}. Theorem 2.4. Iis linearly ordered if and only if every non-zero element of Iis a molecule. Proof: Suppose that Iis linearly ordered and 0 =m∈I. Then m is linearly ordered and ={0}. Hence m∈Mol(A). Conversely suppose
458 C. S. Hoo that every non-zero element of Iis a molecule of A. Let m1,m 2∈I. Then (m1∗m2)∧(m2∗m1) = 0. If m1and m2are not comparable, then m1∗m2>0 and m2∗m1>0. Now 0 <m 1∗m2,m2∗m1≤m1+m2. Since m1+m2∈Mol(A), we have that (m1∗m2)∧(m2∗m1)>0, a contradiction. Hence either m1∗m2=0orm2∗m1= 0, that is, either m1≤m2or m2≤m1. Corollary 2.5. If m∈Mol(A), then km ∈Mol(A)for each integer k≥1. Theorem 2.6. Let a∈At(A)and m∈Mol(A).Ifa∧m=0, then a+m∈Mol(A). Proof: Suppose that 0 <x,y≤a+m. Then x∗m≤aand y∗m≤a. We have several possibilities: (i) x∗m= 0 and y∗m=0, (ii) x∗m=a, and y∗m=a, and (iii) x∗m=aand y∗m=0,ory∗m=aand x∗m=0. In case (i), we have 0 <x,y≤mand hence x∧y>0. In case (ii), we have (x∧y)∗m=(x∗m)∧(y∗m)=a>0 and hence x∧y>0. In case (iii) we need only consider the possibility x∗m=aand y∗m= 0, that is, y≤m. Then we also have x∗a≤m. We claim that x∗a>0. For if x∗a= 0, then 0 <x≤aand hence x=a. This means that a∗m=a, that is, a∧m= 0, contradicting the assumption that a∧m=0. Thus we have 0 <x∗a,y≤m. Hence x∧y≥(x∗a)∧y>0. Theorem 2.7. Let m∈Mol(A). Then for each e∈B(A), either m≤eor m≤e. Proof: We may assume that e=0,1. If both me > 0 and me>0, we have 0 <me,me≤mand hence (me)∧(me)>0, that is, m(e∧e)>0. But e∧e= 0, a contradiction. Hence either me =0orme = 0, that is, either m≤eor m≤e. Theorem 2.8. If B(A)={0,1}, then 1/∈Mol(A). Proof: By hypothesis, there exists e∈B(A) with e=0,1 and hence e=0,1. Then 0 <e,e<1, and hence if 1 ∈Mol(A), this is a contradiction because e∧e=0. Recall that the order of an element x∈A, ord x, was defined in [4] as the smallest positive integer nsuch that nx = 1. If no such integer exists, then ord x=∞.
Molecules and linearly ordered ideals 459 Corollary 2.9. If B(A)={0,1},then for each m∈Mol(A), ord m=∞. Proof: If ord m=nthen nm = 1. Hence 1 ∈Mol(A), a contradiction. Definition 2.10. Let B2(A)={a∈A|2a=3a}. Proposition 2.11. B2(A)={a∈A|2a∈B(A)}. Proof: If a∈B2(A), then 2a=3aand hence 2a∈B(A). Conversely if 2a∈B(A), then 2a=4a. Hence (3a)∗(2a)=(3a)∗(4a) = 0. This means that 2a=3a. Note. Of course B(A)⊂B2(A). Theorem 2.12. Mol(A)⊂Inf(A)∪B2(A). Proof: Let m∈Mol(A). If m/∈Inf(A) and m/∈B2(A), then m2>0 and 2m=3m, that is, m2=m3.Thusm2∧m=m2∗m3= 0. Hence we have 0 <m 2,m2∧m≤m. This means that m2∧m2∧m>0, that is, m2∧m2>0. But m2∧m2=(m∧m)2= 0, a contradiction. Remark. Theorem 2.12 means that Mol(A)∩B(A)=∅if and only if Mol(A)⊂Inf(A). Theorem 2.13. Let a∈At(A). Then a<mfor some m∈Mol(A) if and only if a∈Inf(A). Proof: Suppose that a<mfor some m∈Mol(A). Now At(A)⊂ Inf(A)∪B(A). If a/∈Inf(A), then a∈B(A), and by Theorem 2.7, either m≤aor m≤a. Hence we must have m≤a, which means that a<m≤a, giving a2= 0, a contradiction. Conversely, suppose that a∈Inf(A). Now, a<a+a∈Mol(A) by Corollary 2.5. Theorem 2.14. Mol(A)∩Inf(A)=∅if and only if Mol(A)⊂B(A). Proof: Clearly, if Mol(A)⊂B(A), then if there exists m∈Mol(A)∩ Inf(A), we would have m∈B(A)∩Inf(A)={0}, a contradiction. Conversely, suppose that Mol(A)∩Inf(A)=∅. Let m∈Mol(A). If m∩At(A)=∅, then we have an element a∈m∩At(A). Since mis linearly ordered, we must have a≤m.Now,At(A)∩Inf(A)⊂
460 C. S. Hoo Mol(A)∩Inf(A)=∅, that is, At(A)⊂B(A). Thus a∈B(A)∩mand hence ais the largest element of mby [8, Lemma 5.1]. This means that m≤a, that is, m=a∈B(A). On the other hand, if m∩At(A)=∅ then by [8, Theorem 5.21] there exists a non-zero element y∈msuch that 2y≤m. Then y∈Mol(A) and hence y∈B2(A) by Theorem 2.12. This means that 2y∈B(A), and hence 2yis the largest element of m. Thus m≤2y, that is, m=2y∈B(A). Lemma 2.15. If m∈Inf(A), then mis essential. Proof: We have m2= 0, that is, m≤m. Hence m⊂mand hence m⊥⊂m⊥. But if x∈m⊥, then x∧m= 0. Therefore, x=xm ≤ m, that is, m⊥⊂m. This means that m⊥⊂m⊥⊂mand hence m⊥={0}. Theorem 2.16. If Mol(A)⊂Inf(A), then At(A)⊂Mol(A)⊂Rad(A) and for each m∈Mol(A),mis essential. Proof: Let m∈Mol(A). Then m2= 0 and m≤m. In general, if n≥1 is an integer, we have nm ∈Mol(A) and hence nm ≤mn≤ m.Thusm∈Rad(A) (see [6], [7] and [11] for this characterization of Rad(A)). The rest of the theorem follows from Lemma 2.15. The hypothesis that Mol(A)⊂Inf(A) is extremely strong, as is evidenced by the next two results. Corollary 2.17. Suppose that Mol(A)⊂Inf(A)and Iis a linearly ordered ideal. Then I⊂Rad(A), and hence if I={0}, it cannot have a largest element. Proof: We have I−{0}⊂Mol(A)⊂Rad(A) and hence I⊂Rad(A). If I={0}and if it has a largest element e, then e∈B(A)∩Rad(A)= {0}. Corollary 2.18. Suppose that Mol(A)⊂Inf(A). If there exists an implicative linearly ordered ideal I, then Mol(A)⊂I= Rad(A)= Inf(A)=N(A)=IRad(A), and hence every maximal ideal of Ais implicative. Proof: Suppose that Iis linearly ordered and implicative. Then Inf(A)⊂I⊂Rad(A) and hence IRad(A)=Inf(A)⊂I⊂Rad(A)= Inf(A)⊂N(A)⊂IRad(A). Hence if Mis a maximal ideal of A, then Inf(A)=IRad(A) = Rad(A)⊂Mand hence Mis implicative.
Molecules and linearly ordered ideals 461 3. Essential and Implicative Ideals If m∈Mol(A), we have seen that m={0}and is linearly ordered, and hence m⊥is prime. We wish to consider when m⊥is implicative as well. It was shown in [8, Theorems 3.7 and 3.8] that Iis prime and implicative if and only if it is maximal and implicative. Generally, if I={0}is linearly ordered, then I⊥is prime. We shall answer our question for this general situation. To do so, we recall that in [15, Theorem 2.5] we proved the following result. Theorem 3.1. Suppose that Iis prime and implicative. If Iis not essential, then there exists x∈B(A)∩At(A)⊂AtB(A)such that I⊥={0,x}=xand I={y∈A|y≤x}=x. Theorem 3.2. Suppose that I={0}is linearly ordered. Then I⊥is implicative if and only if I={0,a}where a∈B(A)∩At(A). Proof: Suppose that I⊥is implicative. Since it is prime and 0 = I⊂I⊥⊥, it follows by Theorem 3.1 that I=I⊥⊥ ={0,a}=afor some a∈B(A)∩At(A). Conversely, suppose that I={0,a}where a∈B(A)∩At(A). Let x∈Inf(A). Then x∧a=0ora.Ifx∧a=a, then a≤xand hence a=a2≤0, a contradiction. Hence x∧a= 0, that is, x∈I⊥. Thus Inf(A)⊂I⊥. For each a∈A, let C(a)={x∈A|x=xa +xa}. It is easily verified that C(a)=Aif and only if a∈B(A). It was shown in [14, Theorem 2.3] that x∈C(a) if and only if x∧x∧a∧a= 0, and in [14, Theorem 2.10] that C(a) is a subalgebra of Acontaining B(A). If m∈Mol(A), we can identify the subalgebra C(m)ofAin terms of the element m. To do so, let us recall that an ideal Igenerates a subalgebra AI=I∪I, where x∈Iif and only if x∈I(see [2] and [8]). Theorem 3.3. Let m∈Mol(A). Then C(m)=Am∧m⊥. Proof: If m∈B(A), then C(m)=Aas observed above. Also m∧m= 0 and hence m∧m⊥=A.ThusAm∧m⊥=A. We need therefore only consider the case m∧m>0. Then m∧m∈Mol(A), and m∧m={0}and is linearly ordered. Hence m∧m⊥is prime. Let x∈C(m). Then x∧x∧m∧m= 0, that is, x∧x∈m∧m⊥, and hence x∈m∧m⊥or x∈m∧m⊥. This means that x∈Am∧m⊥. Conversely, if x∈Am∧m⊥, then x∈m∧m⊥or x∈m∧m⊥.Thus x∧x∈m∧m⊥, proving that x∧x∧m∧m= 0, and hence x∈C(m).
462 C. S. Hoo Remark. If m∈Mol(A), then mmay not be proper. For example, if m∈Inf(A) then m2= 0, and hence 2m=1.Thusm=A. However, if m/∈N(A), then m∈B2(A) and we have the following result. Theorem 3.4. Let m∈Mol(A).Ifmis not nilpotent, then mis a prime ideal. Proof: Since mis not nilpotent, then mis proper. Let x∈m⊥. Then x∧m= 0, that is, x=x∗m=xm≤m.Thusm⊥⊂m. Since m⊥is prime, it follows from the prime extension property for MV-algebras (see [10, Corollary 2.11]) that mis prime. We now characterize weakly essential ideals, Theorem 3.5. I={0}is weakly essential if and only if I⊥⊂B(A). Proof: If Iis weakly essential, we have I⊥⊂B(A)by[15, Theorem 3.11]. Conversely, suppose that I⊥⊂B(A). Let Jbe an ideal such that J∩{A−B1(A)}=∅, that is, there exists x∈Jsuch that x∈A−B1(A). If x= 1, then J=Aand hence I∩J=I={0}.We may therefore assume that x<1. Then x/∈B(A) and hence x/∈I⊥. This means that we can find y∈Isuch that x∧y>0. Thus we have 0=x∧y∈I∩J, proving that Iis weakly essential. We showed in [15, Corollary 2.9] the following result. Theorem 3.6. Suppose that Iis prime and Rad(A)⊂I. Then there exists a unique e∈B(A)such that I⊥=e. As a result we have the following. Theorem 3.7. Suppose that Mol(A)⊂Inf(A). Then every maximal ideal I={0}is essential. Proof: By Theorem 3.6, we have I⊥=efor some e∈B(A), and I⊥is linearly ordered since Iis prime. Hence e∈Mol(A), and hence e∈Mol(A)∩B(A)⊂Inf(A)∩B(A)={0}.ThusIis essential. Theorem 3.8. Suppose that I={0}is not essential. If Iis weakly essential and prime, then I⊥=eand I=efor some e∈B(A)∩ At(A)⊂AtB(A). Proof: We have I⊥⊂B(A). Since I⊥is linearly ordered and ={0}, we have I⊥={0,e}for some e∈B(A) with e= 0. Clearly e∈At(A)
Molecules and linearly ordered ideals 463 since if 0 <x∈Ais such that 0 <x≤e, then x∈I⊥and hence x=e.Thuse∈B(A)∩At(A)⊂AtB(A).Now,I⊂I⊥⊥ =e. But e∧e=0∈Iand since e= 0, we have e/∈I. Hence e∈I.Thuse⊂I, proving that I=e. Corollary 3.9. Suppose that At(A)∩B(A)=∅.IfI={0}is weakly essential and prime, then it is essential. Theorem 3.10. Suppose that I={0}is weakly essential and maximal. Then Iis either essential or implicative. Proof: We have that I⊥⊂B(A) and I⊥is linearly ordered. Hence I⊥∩Rad(A)={0}.In[15, Theorem 2.17], we showed the following result. Suppose that Jis a linearly ordered ideal such that J∩Rad(A)= {0}. Then every element x∈Acan be written as x=x1+x2for a unique x1∈Jand x2∈J⊥, and there exists a unique e∈B(A) such that J={x|x≤e}=eand J⊥={x|x≤e}=e.Thusin our case, we can find e∈B(A) such that I⊥=eand I⊥⊥ =e. Let x∈Inf(A). Then x=x1+x2where x1∈I⊥⊂B(A) and x2∈I⊥⊥. Thus 0 = x2=(x1+x2)2=(x1∨x2)2since x1∧x2= 0. Hence 0=x2 1∨x2 2=x2 1+x2 2=x1+x2 2using the fact that x2 1∧x2 2= 0. Hence x1= 0 and x2 2= 0, that is, x=x2∈I⊥⊥. Thus Inf(A)⊂I⊥⊥. This means that I⊥⊥ is implicative. Also since I⊂I⊥⊥ and Iis maximal, then either I=I⊥⊥ or I⊥⊥ =Awhich means that either Iis implicative or Iis essential. Using these results and the earlier mentioned result that the set of ideals that are both prime and implicative is precisely the set of ideals that are both maximal and implicative, we can state the following. Theorem 3.11. Suppose that I={0}is not essential. Then the following are equivalent: (1) Iis weakly essential and maximal (2) Iis implicative and maximal (3) Iis implicative and prime. We have seen that if Iis implicative, then I⊥⊂B(A). To get the converse, we need to consider only those maximal ideals which are not essential. Theorem 3.12. Suppose that Iis maximal but not essential. Then Iis implicative if and only if I⊥⊂B(A). Proof: If Iis implicative, we have already shown that I⊥⊂B(A)in [15, Lemma 2.1], as mentioned above. Conversely, suppose that I⊥⊂