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RELATIONSHIP BETWEEN IDEALS OF BCI-ALGEBRAS AND ORDER IDEALS OF ITS ADJOINT SEMIGROUP
MICHIRO KONDO
(Received 4 October 2000 and in revised form 6 February 2001)
Abstract.We consider the relationship between ideals of a BCI-algebra and order ideals of its adjoint semigroup. We show that (1) ifIis an ideal, thenI=M−1(M(I)), (2)M(M−1(J)) is the order ideal generated byJ∩R(X), (3) ifXis a BCK-algebra, thenJ=M(M−1(J))for any order idealJofX, thus, for each BCK-algebraXthere is a one-to-one correspondence between the setᏵ(X)of all ideals ofXand the setᏻ(X)of all order ideals of it, and (4) the orderM(M−1(J))is an order ideal if and only ifM−1(J)is an ideal. These results are the generalization of those denoted by Huang and Wang (1995) and Li (1999). We can answer the open problem of Li affirmatively.
2000 Mathematics Subject Classification. 06F35, 03G25.
1. Introduction. In [1,2,4], the relationship betweenfiltersof a BCK-algebra and order idealsof itsadjoint semigroupis considered. There is a gap in the proofs of the results obtained there. For example, [2, Theorem 2.8] and [4, Theorem 2.3], where it is proved for a mapI→M(I)to be onto, is not correct. They only show that ifI is a filter thenM(I)is an order ideal and that ifJis an order ideal thenM−1(J)is a filter.
In order to fill the gap of the proof and to develop the theory of adjoint semigroups of BCI-algebras, we consideridealsinstead of filters and show more results about order ideals. As a particular case, we give an affirmative solution to the problem left open in [4].
First of all, we defineBCI-algebras.
LetX;∗,0be an algebraic structure of type2,0. We call it aBCI-algebrawhen it satisfies the conditions: for anyx, y, z∈X,
(1) (x∗y)∗(x∗z)≤z∗y, (2) x∗(x∗y)≤y,
(3) x∗x=0,
(4) ifx∗y=y∗x=0 thenx=y.
The relation “≤” is defined as follows:
x≤yx∗y=0. (1.1)
It is easy to show thatX;≤is a partially-ordered set and the following proposition holds.
Proposition1.1. LetXbe a BCI-algebra. For anyx, y, z∈X, (a) x∗0=x,
(b) 0∗(x∗y)=(0∗x)∗(0∗y),
(c) (x∗y)∗z=(x∗z)∗y, (d) (x∗z)∗(y∗z)≤x∗y, (e) x≤yimpliesx∗z≤y∗z,
(f) x≤yimpliesz∗x≤z∗y.
Moreover, a BCI-algebra is called a BCK-algebra if it satisfies the condition (5) 0∗x=0.
Any subsetIofXis said to be anidealif (I1) 0∈I,
(I2) x∗y,y∈I⇒x∈I.
LetX be any BCI-algebra. For an element a∈X, we define a map a−1 fromX into itself by
(x)a−1=x∗a. (1.2)
The mapa−1is called aright map in [3]. We also defineM(X)andR(X) for any BCI-algebraXas follows:
M(X)=
a−11 a−12 ···a−1n |a1, a2, . . . , an∈X
, R(X)=
a−1|a∈X
, (1.3) wherea−11 a−12 ···a−n1is a map defined by
(x)a−11a−21···a−1n =
···
x∗a1
∗a2
···
∗an. (1.4)
We call M(X)an adjoint semigroup of X. It follows from the properties of BCI- algebras thatM(X)is acommutative monoidwith unit 0−1. We introduce a relation onM(X):
b−11 b−12 ···bm−1 a−11 a−12 ···a−1n
⇐⇒
···
u∗a1
∗a2
···
∗an≤
···
u∗b1
∗b2
···
∗bm (u∈X).
(1.5)
It is obvious thatM(X), is a partially ordered set. We note that for any element a, b∈X,
a−1 b−1⇐⇒a≤b. (1.6)
In fact, ifa−1 b−1thenu∗b≤u∗afor everyu∈X. If we takeu=a, thena∗b=0 and thusa≤b. Conversely, if we supposea≤bthen, since(u∗b)∗(u∗a)≤a∗b=0 for anyu∈X, we haveu∗b≤u∗a. This means thata−1 b−1.
For any subsetSofX, we define M(S)=
a−11a−21···a−n1|ai∈S (1≤i≤n)
. (1.7)
A subsemigroupJofM(X)is called anorder idealif it satisfies the condition
∀x∈Jandy∈R(X), ify xtheny∈J. (1.8) Proposition1.2. LetXbe a BCI-algebra. Then the following statements are equiv- alent:
(a)X:BCK-algebra.
(b) 0−1is the smallest element inM(X).
(c)For any elementx−1, y−1∈R(X),x−1 x−1y−1.
Proof. (a)⇒(b). For every elementa−11 a−12 ···a−n1∈M(X)(ai∈X) andu∈X, since Xis a BCK-algebra, we have((···(u∗a1)∗···)∗an)∗(u∗0)=((···(0∗a1)∗···)∗ an)=0. This implies(···(u∗a1)∗···)∗an≤u∗0 for any elementu∈X. Thus we get 0−1 a−11 ···a−n1and 0−1is the smallest element ofM(X).
(b)⇒(c). Suppose that 0−1is the smallest element ofM(X). Since 0−1 y−1and the operation ofM(X)preserves the order , we havex−1=x−10−1 x−1y−1.
(c)⇒(a). If we takex−1=0−1then we have 0−1 y−1and hence 0≤yfor anyy∈X.
This means thatXis a BCK-algebra.
2. Basic properties of ideals and order ideals. In this section, we consider the basic properties of ideals and order ideals. Moreover, we investigate relations between ideals and order ideals. First of all we have the following result.
Proposition2.1. IfIis an order ideal, thenM(I)is an order ideal.
Proof. It is obvious thatM(I)is a subsemigroup ofM(X). Supposex∈M(I),y∈ R(X), andy x. There are elementsa1, . . . , an∈Iandb∈Xsuch thatx=a−11 ···a−1n andy=b−1. Sincey x, we have, for anyu∈X,
···
u∗a1
∗···
∗an≤u∗b. (2.1)
Especially, if we takeu=b, then ···
b∗a1
∗···
∗an=0∈I. (2.2)
SinceIis the ideal andai∈I, we haveb∈Iand hencey=b−1∈M(I).
Proposition2.2. For any subsetsS, S⊆X, ifS≠SthenM(S)≠M(S), that is,M is an injection.
Proof. Suppose thatS≠Sfor subsetsS, SofX. There exists an elementa∈X such thata∈Sbuta∉S. Sincea−1∈M(S)anda−1∉M(S), we getM(S)≠M(S).
Thus the mapMis an injection.
We define a mapM−1by
M−1(T )=
x|x−1∈T
(2.3) for everyT⊆M(X). It is clear thatM−1(T )=M−1(T∩R(X)).
Proposition2.3. IfJis an order ideal, thenM−1(J)is an ideal.
Proof. SinceJis an order ideal, it is of course a subsemigroup ofM(X)and hence 0−1∈J, that is, 0∈M−1(J).
We assume thatx, y∗x∈M−1(J)forx, y∈X. It follows from the definition ofM thatx−1(y∗x)−1∈J. On the other hand, we havey−1 x−1(y∗x)−1by the property of BCI-algebras. This yields thaty−1∈Jand hence thaty∈M−1(J).
From the above we have the following relations about ideals and order ideals:
I: ideal ⇒M(I): order ideal ⇒M−1 M(I)
: ideal, J: order ideal ⇒M−1(I): ideal ⇒M
M−1(J)
: order ideal. (2.4) It is natural to ask whetherIis identical withM−1(M(I))or so isJwithM(M−1(J)).
By simple calculation, we seeI⊆M−1(M(I))andM(M−1(J))⊆J. We answer the ques- tion for the case of ideals.
In general, we have the following result.
Theorem2.4. IfSis a subset ofXwith0, thenS=M−1(M(S)).
Proof. Letxbe any element ofS. It follows from the definition ofMthatx−1∈ M(S) and that x∈M−1(M(S)), which implies thatS ⊆M−1(M(S)). Thus we have M(S)⊆M(M−1(M(S))).
Suppose thatSis a subset ofXwith 0. It follows thatM(S)is a subsemigroup. In fact, we get 0−1∈M(S)by 0∈S. For any elementα=a−11···a−1n andβ=b1−1···b−1m (ai, bj∈S (1≤i≤n, 1≤j≤m)), we haveα·β=a−11 ···a−1n ·b−11 ···bm−1∈M(S).
This means thatM(S)is the subsemigroup. Ifx∈M(M−1(M(S))), then there exist ai∈M−1(M(S))(1≤i≤n) such thatx=a−11···a−1n . Sincea−i1∈M(S) andM(S) is the subsemigroup, it follows fromx=a−11 ···a−1n ∈M(S)thatM(M−1(M(S)))⊆ M(S). These imply thatM(S)=M(M−1(M(S)))andS=M−1(M(S))byProposition 2.2.
Corollary2.5. IfIis an ideal, thenI=M−1(M(I)).
3. Relations between order idealsJ andM(M−1(J)). We proceed to investigate relations between order idealsJandM(M−1(J)). In general, we haveM(M−1(J))⊆J for any order idealJ. Now the following question arises:
Are alwaysJ=M(M−1(J))? or Under what condition do we getJ=M(M−1(J))?
Theorem 3.1. For any order idealJ, M(M−1(J))is the order ideal generated by J∩R(X).
Proof. It is clear thatM(M−1(J))is an order ideal andJ∩R(X)⊆M(M−1(J)).
To establish a proof of the theorem, we only have to show thatM(M−1(J))⊆K for any order idealKcontainingJ∩R(X). Ifx∈M(M−1(J))then there are elements ai∈M−1(J)such thatx=a−11 ···a−n1. Froma−1i ∈J, we havea−1i ∈J∩R(X)⊆K.
SinceKis an order ideal, we also havex=a−11···a−1n ∈Kand henceM(M−1(J))⊆K.
We denote by(T ]the order ideal generated byT⊆M(X).
Corollary3.2. For any BCI-algebraX,M(X)is the order ideal generated byR(X), that is,M(X)=(R(X)].
To ask whetherJ=M(M−1(J))is equivalent to doJ=(J∩R(X)]for every order idealJ. As the next example shows, we cannot in general concludeJ=M(M−1(J)).
Example3.3. LetX= {0,1,2, . . . , a}and let∗ be an operation defined as in the following table:
0 0 1 2 3 ··· a
0 0 0 0 0 ··· a
1 1 0 0 0 ··· a
2 2 1 0 0 ··· a
3 3 2 1 0 ··· a
.. .
.. .
.. .
.. .
..
. . .. ...
a a a a a ··· 0
It turns out thatXis a BCI-algebra and bothJ= {0−1,1−1,2−1, . . . , a−1a−1}andJ∩ R(X)= {0−1,1−1,2−1, . . .}are order ideals. This means that the order idealM(M−1(J)) generated byJ∩R(X)is identical withJ∩R(X)itself. Hence we haveJ≠M(M−1(J)).
In spite of the above, in case of BCK-algebras, we can showJ=M(M−1(J)). It is proved in [4] thatM−1is aninjectionin case ofXbeing afinite BCK-algebra. We can prove the same result for any BCK-algebra without the assumption being finite.
Theorem3.4. LetXbe aBCK-algebra andJan order ideal. ThenJ=M(M−1(J)).
Proof. It suffices to prove thatJ ⊆M(M−1(J)). Suppose that x ∈J(⊆M(X)).
There existai∈Xsuch thatx=a−11···a−1n . SinceXis a BCK-algebra, we havea−i1 a−11 ···a−1n =x∈J for eachi. Froma−1i ∈Jandai∈M−1(J), we conclude thatx= a−1i ···a−n1∈M(M−1(J)). This implies thatJ⊆M(M−1(J)), that is,J=M(M−1(J)).
Corollary3.5. For any BCK-algebraX,Iis an ideal if and only ifM(I)is an order ideal.
Proof. It follows fromProposition 2.2that ifIis an ideal thenM(I)is an order ideal. Conversely, letM(I)be any order ideal. SinceXis aBCK-algebra, we haveM(I)= M(M−1(M(I)))fromTheorem 3.4. This impliesI=M−1(M(I))because the operator Mis injective. It follows thatIis the ideal ofX.
Theorem3.6. LetXbe a BCI-algebra. ThenM(X)=R(X)impliesJ=M(M−1(J)) for any order idealJ.
Proof. We only showJ⊆M(M−1(J)). If x∈Jthen there existai∈Xsuch that x=a−11 ···a−n1. It follows from the assumption thatx=a−11 ···a−n1=b−1for some b∈X. Sincex∈J, we haveb−1∈Jand hencex=b−1∈M(M−1(J))byb∈M−1(J).
This means thatJ⊆M(M−1(J)).
Remark3.7. The converses of Theorems3.4and 3.6do not hold as in the next example.
Example3.8. ForX= {0, a, b, c}, we define an operation∗as in the following table:
∗ 0 a b c
0 0 0 0 c
a a 0 a c
b b b 0 c
c c c c 0
It is clear thatXis not a BCK-algebra but a BCI-algebra. In this case we also have M(X)= {0−1, a−1, b−1, c−1, c−1c−1} and R(X)= {0−1, a−1, b−1, c−1}, that is,M(X)≠ R(X). But it turns out thatJ=M(M−1(J))for every order idealJofX.
4. Relations betweenI andM(I),JandM−1(J). Two mapsM andM−1are both monotone. WhileI=M−1(M(I))for any idealI, we have in generalJ≠M(M−1(J)).
This means thatMandM−1are not converse maps to each other. We also see that a mapM whose domain is restricted to the set of all ideals is an injection, but we do not know aboutM−1until now. Moreover there is a following problem left open in [4]:
letXbe a BCI-algebra andM(X)be an adjoint semigroup.
(a) Is a mapI→M(I)an injection?
(b) Is there a one-to-one correspondence between the setᏵ(X)of all ideals and the setᏻ(X)of all order ideals?
In the following, we give affirmative answers to the questions above.
Proposition4.1. For everyS⊆X, ifM−1(M(S))is an ideal thenM−1(M(S))is the ideal generated byS.
Proof. We only show thatM−1(M(S))⊆Ifor any idealIsuch thatS⊆I. Suppose thatx∈M−1(M(S)). Sincex−1∈M(S), there existsi∈S such thatx−1=s1−1···s−n1. It follows that(···(x∗s1)∗ ···)∗sn=0∈I and hencex∈Ifromsi∈S⊆I. This yields the desired result.
Proposition4.2. IfI is an ideal thenM(I)is an order ideal generated by {a−1| a∈I}.
Proof. It is clear fromProposition 2.2thatM(I)is an order ideal and{a−1|a∈ I} ⊆M(I). LetJ be an order ideal ofM(X)and{a−1|a∈I} ⊆J. Ifx∈M(I)then there existai∈I such thatx=a−11···a−1n . It follows fromai∈Ithata−i1∈ {a−1| a∈I} ⊆J. From supposition,Jis the order ideal and of course it is a subsemigroup.
Sincea−1i ∈J, we get thatx=a−11 ···a−n1∈J. This impliesM(I)⊆Jand henceM(I) is the order ideal generated by{a−1|a∈I}.
Proposition4.3. For any subsetS⊆X,M−1(M(S))is an ideal if and only ifM(S) is an order ideal.
Proof. Suppose thatM−1(M(S))is an ideal. It is obvious from the definition that M(M−1(M(S)))⊆M(S). SinceM−1(M(S))is an ideal, it is generated byS from the above, so we haveS ⊆M−1(M(S)). This means M(S)⊆M(M−1(M(S))) and conse- quentlyM(S)=M(M−1(M(S))), that is,M(S)is the order ideal byProposition 2.2.
IfM(S)is an order ideal, thenM−1(M(S))is an ideal fromProposition 2.3.
Theorem4.4. IfJis a subsemigroup ofM(X), then (1) M−1(M(M−1(J)))=M−1(J),
(2) M(M−1(J))is an order ideal if and only ifM−1(J)is an ideal.
Proof. (1) It is clear from assumption of the theorem thatM(M−1(J))⊆J and henceM−1(M(M−1(J)))⊆M−1(J). On the other hand, it follows from M−1(J)⊆X thatx∈M−1(J)impliesx−1∈M(M−1(J))andx∈M−1(M(M−1(J))). Thus we have M−1(J)⊆M−1(M(M−1(J)))andM−1(M(M−1(J)))=M−1(J).
(2) In case ofM(M−1(J))being an order ideal, sinceM−1(M(M−1(J)))is an ideal, M−1(J)is the ideal by the above. The converse is clear.
Proposition4.5. IfIis an ideal, thenx∈Iif and only ifx−1∈M(I).
Proof. Suppose thatx−1∈M(I). There existai∈I such thatx−1=a−11···a−1n . Since(···(x∗a1)∗···)∗an=0∈I, we getx∈I. The reverse case is clear.
A subsetSofXis calledclosed ifS=M−1(M(S)). It is clear from the definition of closedness thatSis a closed set if and only if the condition holds:x−1∈M(S)if and only ifx∈S.
From the above we have a relationship between an idealIand an order idealM(I).
Proposition4.6. For any subsetI,I is an ideal if and only ifIis a closed set and M(I)is an order ideal.
ByProposition 2.2,Mis aninjection. ForM−1, we have the following proposition.
Proposition4.7. For any mapM,M−1is an injection if and only ifJ=M(M−1(J)) for any order idealJ.
Proof. FromTheorem 4.4, we haveM−1(M(M−1(J)))=M−1(J)for any order ideal J. SinceM−1is injective, we haveJ=M(M−1(J)).
Suppose thatM−1(J)=M−1(K)for two order idealsJ, K. We obtainJ=M(M−1(J))
=M(M−1(K))=Kand henceM−1is injective.
Thus, fromTheorem 3.4, we have
X: BCK-algebra ⇒M−1: injection. (4.1) This generalizes the following result in [2] (which unfortunately has a gap in its proof):
X: finite BCK-algebra ⇒M−1: injection. (4.2) Also we have fromProposition 2.2 and Theorem 3.6that forX being a BCI-algebra and each idealI, I,
I≠I ⇒M(I)≠M(I),
M(X)=R(X) ⇒M−1: injection. (4.3) This gives affirmative answers to the Open Problem in [4] which says that: letXbe a BCI-algebra andM(X)an adjoint semigroup of it. Then
(a) Is a mapI→M(I)injective?
(b) IfM(X)=R(X), then is there a one-to-one correspondence from the set of all ideals ofXto the set of all order ideals ofM(X)?
Because as to (b) ifM(X)=R(X)thenM−1is injective and it follows fromTheorem 3.6 that J=M(M−1(J))for every order idealJ. This implies that M and M−1 are converse maps to each other and hence there is a one-to-one correspondence from Ᏽ(X)toᏻ(X).
5. Characterization of some BCI-algebras byR(X)andM(X)
Proposition5.1. For any BCI-algebraX,R(X)satisfies the condition that for any a−1∈R(X)there isb−1∈R(X)such thata−1b−1=0−1if and only ifXisp-semisimple, that is,0∗(0∗a)=afor everya∈X.
Proof. Suppose thatR(X)satisfies the condition that for for eacha−1there exists b−1∈R(X)(a−1b−1=0−1). This implies (u∗a)∗b=0 for everyu∈Xand hence 0∗a=b by takingu=b. We thus have(u∗a)∗(0∗a)=uand 0∗(0∗a)=aby takingu=a. This means thatXis ap-semisimple BCI-algebra.
It is easy to prove the converse by takingb=0∗a. For, if we takeb=0∗athen it suffices to show that(x∗a)∗b=xfor everyx∈X. It is clear from the following
(x∗a)∗(0∗a)=0∗ 0∗
(x∗a)∗(0∗a)
=0∗
0∗(x∗a)
∗
0∗(0∗a)
=0∗
(0∗x)∗(0∗a)
∗a
=0∗
0∗(0∗a)
∗x
∗a
=0∗
(a∗x)∗a
=0∗
(a∗a)∗x
=0∗(0∗x)
=x.
(5.1)
IfR(X)satisfies the condition that for anya−1∈R(X)there existsb−1∈R(X)such thata−1b−1=0−1, thenM(X)becomes anabelian group.
Corollary5.2. For any BCI-algebraX, ifXisp-semisimple thenM(X)is an abelian group.
In [2],positive implicative BCK-algebrasare characterized byR(X). We also give a characterization of those as follows.
Proposition5.3. The following conditions are equivalent to each other:
(1) for every BCI-algebraX,R(X)is idempotent (i.e, for alla−1∈R(X) a−1a−1= a−1),
(2) every BCI-algebra X is a positive implicative BCK-algebra (i.e, a BCK-algebra satisfying the condition: for allu, a∈X (u∗a)∗a=u∗a),
(3) {u∈M(X)|u a−1}is an order ideal for anya∈X.
Proof. It is straightforward from the definition of positive implicativeness that (1)(2).
(1)⇒(3). LetΓa= {u∈M(X)|u a−1}andu=x1−1···xn−1,v=y1−1···ym−1be inΓa. Sinceuv=x−11 ···xn−1·y1−1···ym−1 a−1a−1=a−1,Γa is a subsemigroup ofM(X).
Ifu=x1−1···xn−1∈Γaand b−1 u, then we haveb−1 a−1fromb−1 u a−1. It follows fromb−1∈ΓathatΓais an order ideal.
(3)⇒(1). IfΓais an order ideal, then it is also a subsemigroup. Thus, we havea−1a−1∈ Γa froma−1∈Γa, that is,a−1a−1 a−1. This yieldsu∗a≤(u∗a)∗afor anyu∈X.
If we takeu=a, then we have 0≤0∗a. It follows from 0∗(0∗a)≤athat 0≤a.
In this case we also have((u∗a)∗a)∗(u∗a)=0∗a=0,(u∗a)∗a≤u∗aand (u∗a)∗a=u∗a. This meansa−1a−1=a−1and henceR(X)is idempotent.
References
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[2] W. Huang and D. Wang,Adjoint semigroups of BCI-algebras, Southeast Asian Bull. Math.
19(1995), no. 3, 95–98.MR 96h:06039. Zbl 0859.06016.
[3] M. Kondo,Some properties of left maps in BCK-algebras, Math. Japon.36(1991), no. 1, 173–174.CMP 1 093 369. Zbl 0717.06011.
[4] J. Li,Relationship between filters of BCK-algebra and order ideals of its adjoint semigroup, Math. Japon.49(1999), no. 3, 447–450.MR 2000d:06028. Zbl 0932.06016.
Michiro Kondo: Department of Mathematics and Computer Science, Shimane University, Matsue,690-8504, Japan
E-mail address:[email protected]
