G. C. RAO AND P. SUNDARAYYA
Received 12 April 2005; Revised 12 December 2005; Accepted 18 December 2005
We prove that ifA is aC-algebra, then for eacha∈A, Aa= {x∈A/x≤a} is itself a C-algebra and is isomorphic to the quotient algebra A/θa of A where θa= {(x,y)∈ A×A/a∧x=a∧y}. IfAisC-algebra withT, we prove that for everya∈B(A), the centre ofA,Ais isomorphic toAa×Aa and that ifAis isomorphicA1×A2, then there existsa∈B(A) such thatA1is isomorphicAaandA2is isomorphic toAa. Using this de- composition theorem, we prove that ifa,b∈B(A) witha∧b=F, thenAais isomorphic toAbif and only if there exists an isomorphismφonAsuch thatφ(a)=b.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
Introduction
In [1], Guzm´an and Squier introduced the variety ofC-algebras as a class of algebras of type (2, 2, 1) satisfying certain identities and proved that this variety is generated by the 3-element algebraC= {T,F,U}which is the algebraic semantic of the three valued conditional logic. In [3] Swamy et al. introduced the concept of the centreB(A)= {x∈ A/x∨x=T}of aC-algebraAwithTand proved thatB(A) is a Boolean algebra with induced operations and is equivalent to the Boolean Centre ofA. In [2], Rao and Sun- darayya defined a partial ordering on aC-algebraAand the properties ofAas a poset are studied.
In this paper, we prove that ifAis aC-algebra, then for eachx∈A,Ax= {s∈A/s≤x} is itself aC-algebra and is isomorphic to the quotient algebraA/θx, whereθx= {(s,t)∈ A×A/x∧s=x∧t}. IfAis aC-algebra withTthen, for everya∈B(A),Ais isomorphic toAa×Aa and conversely ifAis isomorphic toA1×A2, then there exists an element a∈B(A) such thatA1is isomorphic toAaandA2is isomorphic toAa. Using the above decomposition theorem we prove that for anya,b∈B(A) witha∧b=F,Aais isomor- phic toAbif and only if there exists an isomorphism onAwhich sendsatob.
1. Preliminaries
First, we recall the definition of aC-algebra and some results, which will be used in the later text.
Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 78981, Pages1–8
DOI10.1155/IJMMS/2006/78981
By aC-algebra we mean an algebra of type (2, 2, 1) with operations∧,∨, andsatisfy- ing the following properties:
(a)x=x;
(b) (x∧y)=x∨y; (c) (x∧y)∧z=x∧(y∧z);
(d)x∧(y∨z)=(x∧y)∨(x∧z);
(e) (x∨y)∧z=(x∧z)∨(x∧y∧z);
(f)x∨(x∧y)=x;
(g) (x∧y)∨(y∧x)=(y∧x)∨(x∧y).
Clearly, every Boolean algebra is aC-algebra. The set{T,F,U}is aC-algebra with oper- ations∧,∨, andgiven by
∧ T F U
T T F U
F F F F
U U U U
∨ T F U
T T T T
F T F U
U U U U
X X
T F
F T
U U
We denote this three-elementC-algebra byCand the two-elementC-algebra (Boolean algebra){T,F}by B. It can be observed that the identities (a), (b) imply that the variety of allC-algebras satisfies the dual statements of (b) to (g). In general∧and∨are not commutative inCand the ordinary right distributive law of∧over∨fails inC.
The following properties of aC-algebra can be verified directly [1,3]:
(i)x∧x=x;
(ii)x∧y=x∧(x∨y)=(x∨y)∧x;
(iii)x∨(x∧x)=(x∧x)∨x=x; (iv) (x∨x)∧y=(x∧y)∨(x∧y);
(v)x∨x=x∨x;
(vi)x∨y∨x=x∨y; (vii)x∧x∧y=x∧x.
If aC-algebraAhas an identity for∧, then it is unique and we denote it byT. In this case, we say thatAis aC-algebra withT. If we write F forT, then F is the identity for∨. In aC-algebra, we have the following [1,3]:
(viii)x∨y=Fif and only ifx=y=F;
(ix) ifx∨y=T, thenx∨x=T; (x)x∨T=x∨x;
(xi)T∨x=TandF∧x=F;
(xii) fora∈A,a=aif and only ifais left zero of both∧and∨.
If there exists an elementxinAsuch thatx=x, then it is unique and we denote it by U(Uis called the uncertain element ofA). An elementx∈Ais called a central element ofAifx∨x=T. The set{x∈A/x∨x=T}of all central elements ofAis called the centre ofAand is denoted byB(A). The setB(A) of all central elements ofAis a Boolean algebra with respect to the operations∨,∧, and(ofA) restricted toB(A) [3].
Forx∈Adefine the relationθxonAbyθx= {(p,q)∈A×A/x∧p=x∧q}thenθxis a congruence relation onAandθx∩θx=θx∨x[1].
The relation≤defined on aC-algebraAbyx≤yify∧x=xis a partial ordering onA in which, for everyx∈A, the supremum of{x,x} =x∨x, and the infimum of{x,x} = x∧x[2]. IfAis aC-algebra withT,x∈B(A) and y∈Aare such that x∧y=y∧x, thenx∨yis the lub of{x,y}and in this case y∨xneed not be the lub ofxand y. For example, in the algebraC,T∈B(C) andT∧U=U∧TbutU∨T=Uis not the lub of {U,T}. Ifx≤y, theny∧x=xand hencex∧y=x∧y∧x=x∧x=x. Thereforex≤y if and only ify∧x=x=x∧y.
2. TheC-algebraAx
Recall that for every Boolean algebraBanda∈Bthe set (a]= {x∈B/x≤a}([a)= {x∈ B/a≤x}) is a Boolean algebra under the induced operations∧and∨where complemen- tation is defined byx∗=a∧x(x∗=a∨x).
In this section, we prove that ifAis aC-algebra andx∈A, thenAx= {s∈A/s≤x} is aC-algebra withT(=x) under the induced operations and thatAxis isomorphic to a quotient algebra ofA.
Theorem 2.1. LetAbe aC-algebra,x∈A, andAx= {s∈A/s≤x}. Then Ax,∧,∨,∗is aC-algebra withTwhere∧and∨are the operations inArestricted toAx,s∗is defined by x∧s, and “x” is the identity for∧.
Proof. ClearlyAx is closed under∧and∨. Ifs∈Ax, thenx∧s∗=x∧(x∧s)=(x∧ x)∧s=x∧s=s∗. So thats∗∈Ax ands∗∗=(s∗)∗=(x∧s)∗=x∧(x∧s)=x∧ (x∨s)=x∧s=s(sinces≤x).
Now, fors,t∈Ax, (s∧t)∗=x∧(s∧t)=x∧(s∨t)=(x∧s)∨(x∧t)=s∗∨t∗. Finally, fors,t,u∈Ax,
(s∨t)∧u=x∧
(s∨t)∧u=x∧
(s∧u)∨(s∧t∧u)
=
(x∧s)∧(x∧u)∨(x∧s∧t∧u)
=(s∧u)∨
s∗∧t∧u.
(2.1)
The remaining identities hold inAxsince they hold inA.
Hence Ax,∧,∨,∗is aC-algebra with “x” as the identity for∧. Observe thatAxis itself aC-algebra but it is not a subalgebra ofAbecause the unary operation∗is not the restriction oftoAx. Now, we give some properties ofAx.
Theorem 2.2. LetAbe aC-algebra. Then the following hold:
(i)Ax= {x∧s/s∈A};
(ii)Ax=Ayif and only ifx=y;
(iii)Ax∩Ay⊆Ax∧y; (iv)Ax∩Ax=Ax∧x;
(v) (Ax)x∧y=Ax∧y.
Proof. (i), (ii), and (iii) can be verified routinely. We prove (iv) as follows. Lets∈Ax∧x, then (x∧x)∧s=s and hence x∧s=x∧(x∧x∧s)=x∧x∧s=s. Also we have x∧s=x∧(x∧x∧s)=s, sincex∧x=x∧x. Now we prove (v),
Ax
x∧y=
x∧y∧t/t∈Ax by (i)
=
x∧y∧x∧s /s∈A
=
x∧y∧s /s∈A=Ax∧y.
(2.2)
LetA1,A2be twoC-algebras withT1andT2. Then a mapping f :A1→A2that pre- serves∧,∨,and carriesT1toT2is called aT-preservingC-algebra homomorphism. In future, we deal withC-algebras withT only and hence by aC-algebra homomorphism we mean aT-preservingC-algebra homomorphism. The following lemma can be verified routinely.
Lemma 2.3. Let f :A1→A2be aC-algebra homomorphism whereA1,A2 areC-algebras withT1andT2. Then
(i) ifA1has the uncertain elementU, then f (U) is the uncertain element ofA2; (ii) ifa∈B(A1), then f(a)∈B(A2). The converse holds if f is one-one.
Now we prove the following.
Theorem 2.4. Let A be a C-algebra withT and x∈A, then the mappingαx:A→Ax
defined byαx(s)=x∧sfor alls∈Ais a homomorphism ofAontoAxwith kernelθxand henceA/θx∼=Ax.
Proof. Fors∈A,x∧s≤xand hencex∧s∈Ax. Lets,t∈A, then αx(s∧t)=x∧s∧t=x∧s∧x∧t=αx(s)∧αx(t),
αx(s)=x∧s=x∧(x∨s) by (ii) in the preliminaries
=x∧(x∧s)=(x∧s)∗= αx(s)∗.
(2.3)
Clearly,αx(s∨t)=αx(s)∨αx(t) andαx(T)=a. Henceαxis aC-algebra homomorphism.
Now, fors∈Ax, we haveαx(s)=s. Thereforeαx is onto homomorphism. Hence by the fundamental theorem of homomorphismA/Kerαx∼=Axand Kerαx= {(s,t)∈A×A/αx(s)
=αx(t)} = {(s,t)∈A×A/x∧s=x∧t} =θx. ThusA/θx∼=Ax. 3. Decompositions ofA
IfBis a Boolean algebra anda∈B, then we know thatBis isomorphic to (a]×[a). In this section we prove similar decompositions for aC-algebra. IfAis aC-algebra withT and a∈B(A), then we prove thatAis isomorphic toAaxAa and conversely. We also prove that ifa,b∈B(A) anda∧b=F, thenAa is isomorphic toAb if and only if there is an automorphism onAthat carriesatob. First we prove the following.
Lemma 3.1. LetAbe aC-algebra withT,a∈B(A) andx,y∈A. Then
a∨x=a∨y, a∨x=a∨y⇐⇒x=y. (3.1)
Proof. Leta∈B(A) andx,y∈A. Assume thata∨x=a∨yanda∨x=a∨y. Then x=F∨x=(a∧a)∨x=(a∨x)∧(a∨x)
=(a∨y)∧(a∨y)=(a∧a)∨y=F∨y=y. (3.2)
The converse is trivial
Note thatLemma 3.1fails ifa /∈B(A). For example, in theC-algebraC, we haveU /∈ B(C),U∨T=U∨F=U, andU∨T=U∨F=U, butT=F.
Now we prove the following decomposition theorem.
Theorem 3.2. IfAis aC-algebra withTanda∈B(A), thenA∼=Aa×Aa. Proof. Defineα:A→Aa×Aaby
α(x)=
αa(x),αa(x) ∀x∈A. (3.3)
Then, byTheorem 2.4,αis well defined andαis a homomorphism.
Now, α(x)=α(y)⇒a∧x=a∧y and a∧x=a∧y. Hence x=y (by the dual ofLemma 3.1). Finally, we proveαis onto. Let (x,y)∈Aa×Aa. Thenx≤aandy≤a. So thata∧x=xanda∧y=y.
Thus,a∧x=a∧a∧x=Fanda∧y=a∧a∧y=F.
Now,
x∨y∈A, α(x∨y)=
αa(x∨y),αa(x∨y)
=
a∧(x∨y),a∧(x∨y)
=
(a∧x)∨(a∧y), (a∧x)∨(a∧y)
=(x∨F,F∨y)=(x,y).
(3.4)
Henceαis an isomorphism.
Now we prove the converse of the above theorem in the following sense.
Theorem 3.3. LetA,A1,A2beC-algebras withTsuch thatA∼=A1×A2. Then there exists an elementa∈B(A) such that
A1∼=Aa, A2∼=Aa. (3.5)
Proof. Letφ:A→A1×A2be an isomorphism anda=φ−1(T1,F2) (whenT1,T2denote the∧-identities ofA1,A2, resp.)
Now (T1,F2)∈B(A1)×B(A2)=B(A1×A2) and hencea∈B(A).
Define f :A1→Aaby f(x1)=φ−1(x1,F2) for allx1∈A1. Now
a∧φ−1x1,F2
=φ−1T1,F2
∧φ−1x1,F2
=φ−1x1,F2
(sinceφ−1is a homomorphism). (3.6)
Thereforeφ−1(x1,F2)∈Aa. Thus f is well defined.
It can be routinely verified that f preserves∧,∨and that f is one-one.
Now we prove that f preserves the unary operation. Letx1∈A1, then
fx1
=φ−1x1,F2
=φ−1T1∧x1,F2∧T2
=φ−1T1,F2
∧φ−1x1,T2 sinceφ−1is homomorphism
=a∧
φ−1x1,F2
=a∧fx1
= fx1
∗ .
(3.7)
Finally, we prove f is onto.
Letx∈Aa. Thenφ(x)=(x1,x2) for somex1∈A1,x2∈A2. Now
x1,x2
=φ(x)=φ(a∧x)=φ(a)∧φ(x)
= T1,F2
∧ x1,x2
= x1,F2
. (3.8)
Thusx2=F2and f(x1)=φ−1(x1,F2)=φ−1(x1,x2)=x.
Hence f is onto. ThusA1∼=Aa. SimilarlyA2∼=Aa. Finally, fora,b∈B(A) witha∧b=F, we derive a necessary and sufficient condition forAato be isomorphic toAb. First we prove the following lemmas.
Lemma 3.4. IfAis aC-algebra withT,a∈B(A),x∈Aa, andy∈Aa, thenx∨y=y∨x. Proof. Letx∈Aa,y∈Aa. Thenx≤aandy≤a. Hencea∧y=F=a∧x. Now
a∧(x∨y)=(a∧x)∨(a∧y)=x∨F=x,
a∧(y∨x)=(a∧y)∨(a∧x)=F∨x=x. (3.9)
Therefore,a∧(x∨y)=a∧(y∨x). Similarlya∧(x∨y)=a∧(y∨x).
By the dual ofLemma 3.1,
x∨y=y∨x. (3.10)
Lemma 3.5. LetAbe aC-algebra withT. Then, fora,b∈B(A),a∧b∈B(Aa).
Proof. Clearlya∧b≤a. Now
(a∧b)∨(a∧b)∗=(a∧b)∨
a∧(a∧b)
=(a∧b)∨
a∧(a∨b)=(a∧b)∨(a∧b)
=a∧(b∨b)=a∧T=a.
(3.11)
Hence,a∧b∈B(Aa).
Now, we prove the theorem.
Theorem 3.6. LetAbe aC-algebra withTanda,b∈B(A) such thata∧b=F. ThenAa
is isomorphic toAbif and only if there exists an isomorphismα:A→Asuch thatα(a)=b. Proof. Leta,b∈B(A) witha∧b=F. Letφ:Aa→Abbe an isomorphism.
Nowa∧b=(a∧b)∨F=(a∧b)∨(a∧b)=(a∨a)∧b=b becauseB(A) is a Boolean algebra. So thatb∈Aaandb∗=a∧b. Similarly,b∧a=a. Now by Theorems 2.2,3.2, andLemma 3.5, we have
(i)A∼= Aa×Aa∼=Aa×Aa∧b×A(a∧b)∗=Aa×Ab×Aa∧b
under the isomorphismx→β (a∧x,b∧x, (a∧b)∧x);
(ii)A∼=Ab×Ab∼=Ab×Ab∧a×A(b∧a)∗∼=Ab×Aa×Aa∧b
under the isomorphismx→γ (b∧x,a∧x, (a∧b)∧x);
(iii)Aa×Ab×Aa∧b∼=Ab×Aa×Aa∧b
under the isomorphism (x,y,z)→δ (φ(x),φ−1(y),z).
Now defineα:A→Abyα=γ−1◦δ◦β. Thenαis an isomorphism ofAontoAand α(a)=
γ−1◦δ◦β(a)=γ−1δ(a,F,F) (sinceb∧a=F=a∧a)
=γ−1(b,F,F) sinceφ(a)=b,φ(F)=F
=b sinceγ(b)=(b,F,F).
(3.12)
Henceαis an isomorphism ofAsuch thatα(a)=b.
Conversely, suppose thatα:A→Ais an isomorphism such thatα(a)=b.
Letλbe the restriction ofαtoAa. Now we prove thatλis an isomorphism ofAaonto Ab. Forx∈Aa,
b∧λ(x)=b∧α(x)=α(a)∧α(x)=α(a∧x)=α(x)=λ(x). (3.13)
So thatλ(x)∈Ab. Henceλis well defined. Clearlyλis a homomorphism and one-one.
Letx∈Ab. Sinceαis onto, there existsy∈Asuch thatα(y)=x. Nowa∧y∈Aa and λ(a∧y)=α(a∧y)=α(a)∧α(y)=b∧x=x(sincex≤b).
Henceλis an isomorphism ofAaontoAb.
Acknowledgments
The authors thank Professor U. M. Swamy for his support and guidance in the prepara- tion of this paper. The second author gratefully acknowledges University Grants Com- mission for their financial support in the form of UGC-JRF. We thank referees for their helpful suggestions.
References
[1] F. Guzm´an and C. C. Squier, The algebra of conditional logic, Algebra Universalis 27 (1990), no. 1, 88–110.
[2] G. C. Rao and P. Sundarayya,C-algebra as a poset, International Journal of Mathematical Sci- ences, 4 (2005), no. 2, 225–236.
[3] U. M. Swamy, G. C. Rao, and R. V. G. Ravi Kumar, Centre of aC-algebra, Southeast Asian Bulletin of Mathematics 27 (2003), no. 2, 357–368.
G. C. Rao: Department of Mathematics, Andhra University, Visakhapatnam 530 003, India E-mail address:[email protected]
P. Sundarayya: Department of Mathematics, Andhra University, Visakhapatnam 530 003, India E-mail address:[email protected]
