EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 4, 2017, 717-729 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Centre of Core Regular Double Stone Algebra A R J Srikanth1,2,∗, R V G Ravi Kumar3 1 G V P College of Engineering, Andhra Pradesh, India 2 Jawaharlal Nehru Technological University Kakinada, Andhra Pradesh, India 3 G V P College of Engineering, Andhra Pradesh, India Abstract. In literature there is an elegant characterization of factor congruences on a distributive lattice. In this paper, we make an attempt such type of characterization of factor congruences on a Core Regular Double Stone Algebra (CRDSA) and we identify that the factor congruences on a CRDSA A with certain elements of A and proved that set of all factor congruences forms a Boolean centre for A. Further Birkhoff centre is defined for CRDSA and finally it is shown that Birkhoff centre of CRDSA is isomorphic to its Boolean centre 2010 Mathematics Subject Classifications: 06D99, 03G10,06D15 Key Words and Phrases: Core Regular Double Stone Algebra, Boolean centre, Birkhoff centre 1. Introduction The concept of a core regular double Stone algebra was introduced by Ravi Kumar etal and obtained a decomposition theorem for a complete atomic core regular double Stone algebra [5]. In [6], U.M. Swamy and G.S. Murti introduced the concept of the Boolean center of an universal Algebra. In this paper we make an attempt to characterize the Boolean centre of a CRDSA A and the concept of Birkhoff’s ‘Centre’ of a bounded poset is extended to CRDSA A and referred to this, as ‘Birkhoff centre’ of A. It is also proved that Birkhoff centre BC(A) is isomorphic to Boolean centre of A. 2. Preliminaries In this section the concept of the isomorphism of RDSA is extended to CRDSA and a new characterization for centre of a CRDSA based on core element is done. We start with certain basic definitions and properties of RDSA. ∗Corresponding author. Email addresses: rvgravikumar@yahoo.com (R V G Ravi Kumar), seeku.ammu@gmail.com (A R J Srikanth) http://www.ejpam.com 717 c© 2017 EJPAM All rights reserved. A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 718 Definition 1. A Regular double Stone algebra (RDSA) < A,∧,∨, ∗,+, 0, 1 > is an algebra of type < 2, 2, 1, 1, 0, 0 > such that (i) < A,∧,∨, 0, 1 > is a bounded distributive lattice. (ii) ∗ is a pseudo complementation satisfying the Stone identity x∗ ∨ x∗∗ = 1 (iii) + is a dual pseudo complementation satisfying the dual Stone x+ ∧ x++ = 0 (iv) For any x, y ∈ A, x∗ = y∗ and x+ = y+ then x = y Example 1. Consider the hasse diagrams of lattices L1 and L2. 1 c d a b 0 Figure 1: L1 1 d b c a 0 Figure 2: L2 Clearly L1, L2 are bounded distributive lattices, pseudo complemented and dual pseudo complemented , and in L1, 0 ∗ = 1, a∗ = b, b∗ = c, c∗ = b, d∗ = 0, 1∗ = 0 and 0+ = 1, a+ = 1, b+ = c, c+ = b, d+ = c, 1+ = 0. Clearly L1 is a regular double Stone algebra where as in L2, a ∗ = b∗ = c∗ = d∗ = 1∗ = 0, 0∗ = 1 and a+ = b+ = c+ = d+ = 0+ = 1, 1+ = 0. Here a∗=b∗ and a+=b+but a 6= b therefore L2 is not a regular double Stone algebra. Definition 2. Let A be a regular double Stone algebra. An element a of A is called a central element of A if a∗ = a+. The set of all central elements of A is called the centre of A and is denoted by C(A); that is, C(A) = {a ∈ A|a∗ = a+} Note that C(A) can be described in various ways as follows; C(A) = {a ∈ A|a = a∗∗} = {a∗|a ∈ A} = {a ∈ A|a = a++} A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 719 = {a+|a ∈ A} = {a ∈ A|a ∨ a∗ = 1} = {a ∈ A | a ∧ a+ = 0} = {a ∈ A | a ∧ b = 0 and a ∨ b = 1 for some b ∈ A} Theorem 2.1. Let A be a regular double Stone algebra.Then C(A) is a Boolean sub algebra of A with respect to the induced operations ∧,∨ and ∗. Definition 3. Let A be a regular double Stone algebra. The set D(A) := {a ∈ A | a∗ = 0} is called the dense set of A and the elements of D(A) are called dense elements of A. The dual of D(A) := {a ∈ A | a+ = 1} is called dual dense set of A and denoted by D(A). The elements of D(A) are called dual dense elements of A. Note that D(A) = {a ∨ a∗ | a ∈ A} and D(A) = {a ∧ a+ | a ∈ A}. Theorem 2.2. Let A be a regular double Stone algebra. Then D(A) is a filter of A and D(A) is an ideal of A. Definition 4. The core of a double Stone algebra A is defined to be K(A) = D(A)∩D(A) K(A) is non empty if and only if A does not have 2 ={0, 1} as a factor. When it is non empty the behavior of K(A) in certain respects governs the behaviour of A. It is easy to prove that in any RDSA there exists at most one core element. We call a regular double Stone algebra with non empty core as Core Regular Double Stone Algebra(CRDSA). Note: In any CRDSA A, |K(A)| = 1. Example 2. Every three element chain is CRDSA. We call it as a discrete CRDSA Example 3. Consider the hasse diagrams of RDSAs L1 = (L1,∧,∨, ∗,+, 0, 1) and L3 = (L3,∧,∨, ∗,+, 0, 1). 1 c d a b 0 Figure 3: L1 1 e f c k d a b 0 Figure 4: L3 A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 720 Clearly L1, L3 are RDSAs, and it is seen that core of L1 is empty where as L3 has the core element k hence it a CRDSA Theorem 2.3. If A is CRDSA with core element k, then every element x of A can be written as x = x∗∗ ∧ (x++ ∨ k) and x = x++ ∨ (x∗∗ ∧ k) Proof. Let y = x∗∗ ∧ (x++ ∨ k). Then y∗∗ = (x∗∗ ∧ (x++ ∨ k))∗∗ = x∗∗ and y++ = (x++ ∧ (x++ ∨ k))++ = x++. Thus by regularity x = y. Other one follows from duality. Definition 5. Suppose that A and B are two CRDSAs with core elements k1, k2 respec- tively. A mapping f : A −→ B is called a homomorphism from A to B if (i) f is lattice homomorphism from A to B (ii) for a ∈ A, f(a∗) = f(a)∗ and f(a+) = f(a)+ (iii) f(k1) = k2 A necessary and sufficient condition for two CRDSAs is isomorphic is discussed in the following theorem. Theorem 2.4. Two CRDSAs are isomorphic if and only if their centers are isomorphic Proof. Let A1, A2 be CRDSAs with Core elements k1, k2 respectively. First suppose that f : C(A1) −→ C(A2) is an isomorphism. Define the map φ on A1 to A2 by φ(x) = f(x∗∗) ∧ (f(x++) ∨ k2). By using distributive property and the fact that f is a homomorphism it can be easily verify that φ(x) = f(x++) ∨ (f(x∗∗) ∧ k2). And also observe that, for x ∈ C(A1), φ(x) = f(x∗∗)∧ (f(x++) ∨ k2) = f(x)∨ (f(x) ∧ k2) = f(x), i.e. φ coincides with f on C(A1) To show that φ is one-one suppose that φ(x) = φ(y) for x, y in A1. Then (φ(x))∗ = (φ(y))∗ and (φ(x))+ = (φ(y))+, by using the definition of φ and the fact that f is one-to- one , it gives x∗ = y∗ and x+ = y+ and by regularity x = y. Hence φ is one-one. To show that φ is onto, y ∈ A2 and consider the following cases A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 721 Case (i) : y ∈ C(A2). Since f is onto from C(A1) to C(A2), there exists an element x ∈ C(A1) such that f(x) = y and φ(x) = f(x∗∗) ∧ ( f(x++) ∨ k2 ) = f(x) ∧ (f(x) ∨ k2) = f(x) = y Case (ii) : y = k2 Then φ(k1) = f(k∗∗1 ) ∧ ( f(k++ 1 ) ∨ k2 ) = 1 ∧ k2 = k2 Case (iii) : k2 6= y and y /∈ C(A2) Then y∗∗, y++ ∈ C(A2) and from the fact that f is onto there exists x1, x2 ∈ C(A1) such that φ (x1) = f (x1) = y ∗∗ and φ (x2) = f (x2) = y ++, now φ (x1 ∧ (x2 ∨ k1)) = f(x1) ∧ (f (x1 ∧ f (x2)) ∨ k2) = f (x1) ∧ (f (x2) ∨ k2) = y∗∗ ∧ ( y++ ∨ k2 ) = y Hence φ is onto. The remaining conditions which verifies that φ is homomorphism is straightforward. Conversely suppose that φ : A1 −→ A2 is an isomorphism. Let x ∈ C(A1) be any element, then (φ(x))∗∗ = φ(x∗∗) = φ(x). Therefore φ(x) ∈ C(A2), and hence φ(C(A1)) ⊆ C(A2). On the other hand, if y ∈ C(A2) then there exist x ∈ A1 and φ(x) = y. Now φ(x∗∗) = (φ(x))∗∗ = y∗∗ = y = φ(x) As φ is one-one, we get x∗∗ = x and hence x ∈ C(A1). Therefore φ(C(A1)) = C(A2) and hence they are isomorphic. Hence Boolean isomorphism between centre of a CRDSA can be extended to whole algebra so that core elements are mapped each other. LetA be a regular double Stone algebra. For a ∈ A the ∗− centralizer of a is denoted by A∗a and defined as A∗a = {x∗∗ | x ≤ a} = {x∗∗ ∧ a∗∗ | x ∈ A}. A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 722 Definition 7. LetA be a regular double Stone algebra. For a ∈ A the +− centralizer of a is denoted by A+ a and defined as A+ a = {x++ | x ≥ a} == {x++ ∨ a++ | x ∈ A}. Theorem 2.5. Let A be a Core Regular double Stone algebra. The relativized algebra A∗a =< A∗a,∧,∨,′ 0, a∗∗ > is a Boolean algebra Proof. Let x, y ∈ A∗a. Then x = p∗∗, y = q∗∗ for some p, q ∈ A and p, q ≤ a. Which gives p∨ q ≤ a, p∧ q ≤ a. Hence (p∨ q)∗∗ = p∗∗ ∨ q∗∗ ≤ a∗∗ and (p∧ q)∗∗ = p∗∗ ∧ q∗∗ ≤ a∗∗ . Therefore x ∨ y, x ∧ y ∈ A∗a. Therefore A∗a is closed with respect to ∨ and ∧. It is a routine verification that < A∗a,∧,∨ > is distributive lattice. Clearly 0∗∗ = 0 ≤ a, so 0 ∈ A∗a. Since a ≤ a we get a∗∗ ∈ A∗a. Let x ∈ A∗a be any element then x = p∗∗ for some p ∈ A and p ≤ a which gives p∗∗ = x ≤ a∗∗. Therefore a∗∗ is the greatest element of A∗a. Hence < A∗a,∧,∨, 0, a∗∗ > is a bounded distributive lattice. Finally for x = p∗∗ ∈ A∗a we have p ≤ a and x∗ ∧ a = p∗ ∧ a ≤ a which gives (x∗ ∧ a)∗∗ = x∗ ∧ a∗∗ ∈ A∗a and x ∧ (x∗ ∧ a∗∗) = 0 and x ∨ (x∗ ∧ a∗∗) = a∗∗. Therefore x∗ ∧ a∗∗ is the compliment of x in A∗a i.e. x′ = x∗ ∧ a∗∗. Hence A∗a =< A∗a,∧,∨,′ 0, a∗∗ > is a Boolean algebra Theorem 2.6. Let A be a Core Regular double Stone algebra and k is the core element of A. Then A∗k = A+ k Proof. Let x∗∗ ∈ A∗k and y = k ∨ x∗∗. Then y ≥ k and y++ = (k ∨ x∗∗)++ = x∗∗.Therefore x∗∗ ∈ A+ and hence A∗k ⊆ A+ k . Now take x++ ∈ A+ k put y = k ∧ x++ then y ≤ k and hence y∗∗ = (k ∧ x++)∗∗ = x++ ∈ A∗k. So A+ k ⊆ A ∗ k and hence A∗k = A+ k . In fact we have the stronger result in the following theorem. Theorem 2.7. Let A be a Core Regular double Stone algebra and k is the core element of A then A∗a = A+ a if and only if a = k Proof. First suppose that for some a ∈ A,A∗a = A+ a . Since 0 ∈ A∗a = A+ a there exists b ∈ A such that a ≤ b and 0 = b++. So b+ = 1 and b+ ≤ a+. Hence a+ = 1. As 1 ∈ A+ a = A∗a there exists c ∈ A such that c ≤ a and 1 = c∗∗. So c∗ = 0 and a∗ ≤ C∗. Hence a∗ = 0. Therefore a ∈ K(A) = k. Other part is clear from theorem 2.6. Throughout this paper we denote A∗k = A+ k by k(A). The following theorem discuss the relation between centralizer of core and centre of CRDSA . Theorem 2.8. Let A be a Core Regular double Stone algebra and k is the core element of A then k(A) = C(A) . Proof. Clearly k(A) ⊆ C(A). Let a be any element of C(A), then (a ∧ k)∗∗ = a∗∗ = a ∈ k(A). Hence k(A) = C(A). Theorem 2.8 gives another characterization for centre of a CRDSA based on the core element. A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 723 3. Boolean centre In [6], Swamy and Murthy introduced the concept of ‘Balanced congruence’ on any algebraA and showed the set B(A) of all balanced (direct) factor congruences which admit a balanced complement as a permutable Boolean sublattice of the lattice C(A) of all con- gruences on A. They referred to B(A) as the ‘Boolean centre’ of A. The main goal of this Section is to characterize the Boolean centre of a CRDSA A in terms of central elements. Let A be a Core Regular double Stone algebra. Let θx denote the equivalence relation associated to the function x→ x∧p from A to itself: θx = {(p, q) ∈ A×A |x ∧ p = x ∧ q} . We will write pθxq to indicate (p, q) ∈ θx. Theorem 3.1. Let A be a Core Regular double Stone algebra and x, y ∈ A Then (i) θy ⊆ θx if and only if x = x ∧ y. (ii) θy = θx if and only if x = y. (iii) θx is compatible with ∧,∨, ∗ (iv) θx is compatible with + if and only if x ∈ k(A) (v) θx congruence on A if and only if x ∈ k(A). (vi) θ0 = A×A (vii) θ1 = ∆A (viii) θx ∩ θy = θx∨y (ix) θx ◦ θy = θy ◦ θx (x) θx ◦ θy = θx∧y (xi) θx ◦ θx∗ = θx∗ ◦ θx = A×A (xii) θx∨x∗ = ∆A if and only if θx is a congruence relation (xiii) for x ∈ k(A), θx is the smallest congruence containing (1, x) Proof. (i) Let x, y ∈ A and suppose that θy ⊆ θx. Since y ∧ (x ∨ y) = y = y ∧ y we have (y, x∨ y) ∈ θy, by our supposition (y, x∨ y) ∈ θx ; that is x∧ y = x∧ (x∨ y) or x∧ y = x. Conversely suppose that x ∧ y = x. Let (p, q) ∈ θy. Then y ∧ p = y ∧ q. Now, x ∧ p = (x ∧ y) ∧ p = x ∧ (y ∧ p) A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 724 = x ∧ (y ∧ q) = (x ∧ y) ∧ q = x ∧ q Therefore, (p, q) ∈ θx and hence θy ⊆ θx. (ii) Clear from (i). (iii) If p, q, r, s ∈ A satisfy (p, q) ∈ θx and (r, s) ∈ θx. From associativity and distributivity in A it follows that ((p ∧ r), (q ∧ s)) ∈ θx and ((p ∨ r), (q ∨ s)) ∈ θx. Also if p, q ∈ A and (p, q) ∈ θx, it follows that (x∗ ∨ p∗) = (x∗ ∨ q∗), so that x∧ (x∗ ∨ p∗) = x∧ (x∗ ∨ q∗) using distributivity we conclude that (p∗, q∗) ∈ θx. (iv) Suppose that θx is compatible with +. Put y = x ∨ k then y++ = x++. As (1, x) and (k, k) ∈ θx which gives (1, x ∨ k) = (1, y) ∈ θx. Since θx is compatible with +, we get(1, y++) ∈ θx. Hence x = x ∧ y++ = y++. Therefor x ∈ k(A). Conversely suppose that x ∈ k(A) then x = y++ for some y ≥ k, if p, q ∈ A satisfy (p, q) ∈ θx = θy++ then y++ ∧ p = y++ ∧ q and hence y+ ∨ p+ = y+ ∨ q+ , it follows that y++ ∧ (y+ ∨ p+) = y++ ∧ (y+ ∨ q+) which gives (y++ ∧ p+) = (y++ ∧ q+) hence (p+, q+) ∈ θy++ = θx. (v) is clear from (iii) and (iv). (vi) and (vii) are clear from the definition of θx. (viii) Let (p, q) ∈ θx ∩ θy. Then x ∧ p = x ∧ q and y ∧ p = y ∧ q. Now, (x ∨ y) ∧ p = (x ∧ p) ∨ (y ∧ p) = (x ∧ q) ∨ (y ∧ q) = (x ∨ y) ∧ q Therefore (p, q) ∈ θx∨y. Hence θx ∩ θy ⊆ θx∨y. Conversely suppose that (p, q) ∈ θx∨y then (x ∨ y) ∧ p = (x ∨ y) ∧ q. Now, x ∧ ((x ∨ y) ∧ p) = x ∧ ((x ∨ y) ∧ q) (x ∧ (x ∨ y)) ∧ p = (x ∧ (x ∨ y)) ∧ q) x ∧ p = x ∧ q A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 725 Therefore (p, q) ∈ θx, similarly it can be shown that (p, q) ∈ θy. So (p, q) ∈ θx ∩ θy and hence θx∨y ⊆ θx ∩ θy. Therefore θx ∩ θy = θx∨y. (ix) Let (p, r) ∈ θx ◦ θy. Then there exists q ∈ A such that (p, q) ∈ θx and (q, r) ∈ θy. So x ∧ p = x ∧ q and y ∧ q = y ∧ r. Put t = (x ∧ r) ∨ (y ∧ p). Then, x ∧ t = x ∧ ((x ∧ r) ∨ (y ∧ p)) = (x ∧ r) ∨ (x ∧ (y ∧ p)) = (x ∧ r) ∨ x ∧ (y ∧ p) = (x ∧ r) ∨ (x ∧ p ∧ p) = (x ∧ r) ∨ (x ∧ q ∧ y), since x ∧ p = x ∧ q = (x ∧ r) ∨ (x ∧ y ∧ r), since q ∧ y = y ∧ q = y ∧ r = x ∧ r Hence (t, r) ∈ θx. Similarly it can be shown that y ∧ t = y ∧ p which gives (p, t) ∈ θy. Therefore (p, r) ∈ θy ◦ θx i.e. θx ◦ θy ⊆ θy ◦ θx. Conversely suppose that (p, r) ∈ θy ◦ θx. Now by setting t = (y ∧ r) ∨ (x ∧ p) and proceeding as above it can be shown that θy ◦ θx ⊆ θx ◦ θy. Finally it gives θx ◦ θy = θy ◦ θx. (x) Let (p, r) ∈ θx ◦ θy. Then there exists q ∈ A such that (p, q) ∈ θx and (q, r) ∈ θy. So x ∧ p = x ∧ q and y ∧ q = y ∧ r Now, (x ∧ y) ∧ p = (x ∧ p) ∧ y = (x ∧ q) ∧ y, since x ∧ p = x ∧ q = x ∧ (y ∧ q) = x ∧ (y ∧ r), since y ∧ q = y ∧ r = (x ∧ y) ∧ r. Therefore (p, r) ∈ θx∧y and hence θx ◦ θy ⊆ θx∧y. Conversely suppose that (p, r) ∈ θx∧y then (x∧y)∧p = (x∧y)∧r. Put q = (x∧p)∨(y∧r). Then, x ∧ q = x ∧ (x ∧ p) ∨ (y ∧ r) = (x ∧ p) ∨ (x ∧ y ∧ r) = (x ∧ p) ∨ (x ∧ y ∧ p), since (x ∧ y) ∧ p = (x ∧ y) ∧ r = (x ∧ p) A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 726 hence (p, q) ∈ θx. By considering y ∧ q and proceeding as above it can shown that y ∧ q = y ∧ r, so (q, r) ∈ θy and hence (p, r) ∈ θx ◦ θy. Therefore θx∧y ⊆ θx ◦ θy, which completes the proof. (xi) is clear from (x), (vi) and the fact that x ∧ x∗ = 0. (xii) follows from (v), (ii) and (vii). (xiii) Let x ∈ k(A). Then θx is congruence and (1, x) ∈ θx. Suppose that θ be any congruence containing (1, x) and (a, b) ∈ θx i.e. a ∧ x = b ∧ x. Since θ is reflexive and (1, x) ∈ θ we get (a, a) ∧ (1, x) = (a, a ∧ x) ∈ θ and (b, b) ∧ (1, x) = (b, b ∧ x) ∈ θ which in turn gives (a, b ∧ x) ∈ θ and (b, b ∧ x) ∈ θ. Hence (a, b) ∈ θ and θx ⊆ θ. Therefore θx is the smallest congruence containing (1, x) for x ∈ k(A). Recall that a congruence θ on an algebra A is said to be factor congruence if there is a congruence ψ on A such that θ ∧ ψ = ∆ θ ∨ ψ = A×A and θ permutes with ψ In the following theorem the factor congruences of core regular double stone algebras were characterized. Theorem 3.2. Let A be a Core Regular double Stone algebra and θ be congruence on A. Then θ is factor congruence on A if and only if θ = θx for some x ∈ k(A). Proof. Suppose that θ = θx for some x ∈ k(A), then from (vi) ,(vii), (viii) and (x) of theorem 3.1 and theorem 2.8, we have θx ∧ θx∗ = ∆ and θx ∨ θx∗ = A × A and hence θ = θx is factor congruence on A Conversely suppose that θ is factor congruence on A. Then there exists a congruence ψ on A such that θ∧ψ = ∆ and θ∨ψ = A×A. Since (1, 0) ∈ A×A = θ∨ψ, there exists x ∈ A such that (1, x) ∈ θ and (x, 0) ∈ ψ. Now put y = x ∧ k then y∗∗ ∈ k(A) and from the fact that (1, x), (k, k) ∈ θ it follows that (1, y) ∈ θ and hence (1, y∗∗) ∈ θ. Also observe that (x, 0), (k, k) ∈ ψ gives (y, 0) ∈ ψ and hence (y∗∗, 0) ∈ ψ. Now we show that θ = θy∗∗. Since (1, y∗∗) ∈ θ, by (xiii) of theorem 3.1, we have θy∗∗ ⊆ θ. Next suppose that (p, q) ∈ θ then (y∗∗ ∧ p, y∗∗ ∧ q) ∈ θ. Since (y∗∗, 0), (p, p) and (q, q) ∈ ψ, we have (y∗∗ ∧ p, 0 ∧ p) and (y∗∗ ∧ q, 0 ∧ q) ∈ ψ; that is (y∗∗ ∧ p, 0) and (0, y∗∗∧q) ∈ ψ which imply that (y∗∗∧p, y∗∗∧q) ∈ ψ. Therefore,(y∗∗∧p, y∗∗∧q) ∈ θ∩ψ = ∆ and hence y∗∗ ∧ p = y∗∗ ∧ q. Therefore (p, q) ∈ θy∗∗, hence θ ⊆ θy∗∗. Thus θ = θy∗∗. Recall that a congruence θ on any Universal algebra A of any type, is called balanced if (θ∨ψ)∩ (θ∨ψ′) = θ for all factor congruence ψ and its complements ψ′ and the set B(A) A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 727 of all balanced factor congruences which admit a balanced complement is called the Boolean centre of A. Now we conclude this section by proving that, if A is a Core regular double stone algebra, then the Boolean centre B(A) is precisely the set D = {θx | x ∈ k(A)} and that the map x 7−→ θx is an isomorphism of k(A) onto B(A). First we prove the following. Lemma 1. Let A be a CRDSA and x ∈ k(A). Then θx is balanced. Proof. Let ψ be a factor congruence on A and ψ′ be its complement. Then there exist y, z ∈ k(A) such that ψ = θy and ψ′ = θz. Now, (θx ∨ ψ) ∩ (θx ∨ ψ′) = (θx ∨ θy) ∩ (θx ∨ θz) = θx∧y ∩ θx∧z = θ(x∧y)∨(x∧z) = θx∧(y∨z) = θx ∨ θy∨z = θx ∨ (θy ∩ θz) = θx ∨ (ψ ∩ ψ′) = θx ∨∆A = θx Therefore, θx is balanced. Thus we have proved the following. Theorem 3.3. Let A be a CRDSA. Then the Boolean centre B(A) of A is precisely the set {θx | x ∈ k(A)}. The following theorem is a consequence of lemma 1 and above theorem 3.3 Theorem 3.4. Let A be a CRDSA. Then the Boolean centre B(A) = {θx | x ∈ k(A)} of A, is a Boolean algebra and the map x 7−→ θx is an isomorphism of k(A) onto B(A). 4. Birkhoff Centre An element a of a bounded poset P is called a ‘central element’ of P if there exist bounded posets P1 and P2 and an order isomorphism of P onto P1 × P2 such that a is mapped onto (1, 0). The set of all central elements of P are called the ’Birkhoff centre’ of P and is denoted by BC(P ). It is known that BC(P ) is a Boolean algebra in which the operations are g.l.b and l.u.b with respect to the partial order in P . In this section we extend the concept of Birkhoff centre to core regualr double Stone algebra. A R J Srikanth, R V G Ravi Kumar / Eur. J. Pure Appl. Math, 10 (4) (2017), 717-729 728 Definition 8. An element a of an RDSA A is called a Birkhoff central element if there exist RDSAs A1 and A2 and an isomorphism A onto A1 ×A2 such that a is mapped onto (1, 0). The set BC(A) of all central elements of P is called the Birkhoff centre. Recall that the ideal generated by an element x of A in a RDSA is called a relativized algebra and is denoted by (x]A. In [9] it is proved that if A in a CRDSA with core element k then for x ∈ At(C(A)),the relativized algebra (x]A is a three element chain i.e. a discrete CRDSA. In fact we have the following theorem. Theorem 4.1. Let A be a Core Regular double Stone algebra. The relativized algebra (a]A is a CRDSA if and only if a ∈ k(A). Proof. Assume that a ∈ k(A). Then a∗ ∨ a = 1 and a+ ∧ a = 0. It is a routine verifi- cation that (a]A = ((a],∧,∨, ∗a,+a, 0, a) is a double Stone algebra where a is the greatest element and for x ∈ (a], x∗a = x∗ ∧ a and x+a = x+ ∧ a. To prove that (a]A is regular consider x, y ∈ (a]A such that x∗a = y∗a and x+a = y+a, that is, x∗ ∧ a = y∗ ∧ a and x+ ∧ a = y+ ∧ a. Then (x∗ ∧ a) ∨ a∗ = (y∗ ∧ a) ∨ a∗ and (x+ ∧ a) ∨ a+ = y+ ∧ a ⇒ (x∗ ∨ a∗) ∧ (a ∨ a∗) = (y∗ ∨ a∗) ∧ (a ∨ a∗) and (x+ ∨ a+) ∧ (a ∨ a+) = (y+ ∨ a+) ∧ (a ∨ a+)− (∗) Since x, y ∈ (a] we have x, y ≤ a⇒ a∗ ≤ x∗, y∗ and a+ ≤ x+, y+. Also since a∗∨a = 1 Therefore (*) gives x∗ = y∗ and x+ = y+ and by regularity in A, x = y. Hence (a]A is a regular double Stone algebra. Moreover a∧ k ∈ (a] and (a∧ k)∗a = a∧ k∗ = 0, (a∧ k)+a = a ∧ k+ = a. Theretofore a ∧ k is the core element of (a]A. So (a]A is a CRDSA. Conversely suppose that for a ∈ A, (a]A = ((a],∧,∨, ∗a,+a, 0, a) is a CRDSA with the above defined operations. Since a is the greatest element of (a]A, we have a∗a = a+a and therefore a+ ∧ a = 0. Hence a is complimented element. So a ∈ C(A) = k(A) By the principle of duality and theorem 2.6 we have the following theorem. Theorem 4.2. Let A be a Core Regular double Stone algebra. Then relativized algebra [a)A = ([a),∧,∨, ∗a,+a, a, 1) is a CRDSA ifand only if a ∈ k(A). Theorem 4.3. Let A be a Core Regular double Stone algebra. a ∈ BC(A) if and only if a ∈ k(A) . Proof. Let a ∈ BC(A). Then there exist CRDSAs A1 and A2 and an isomorphism f from A onto A1 ×A2 such that a is mapped onto (1, 0). By theorem 2.4, C(A) is isomor- phic to C(A1) × C(A2) and (1, 0) ∈ C(A1) × C(A2) which in turn gives a ∈ C(A) and hence by theorem 2.8 a ∈ k(A). REFERENCES 729 Conversely suppose that a ∈ k(A). By theorems 4.1 and 4.2 (a]A and [a)A are CRDSAs. Now define a map f : A→ (a]A × [a)A by f(x) = (a∧ x, a∨ x). Then f is a isomorphism from A onto (a]× [a)A, such that f(a) = (a, a) = (1, 0). Hence a ∈ BC(A). Thus we have proved the following. Theorem 4.4. Let A be a CRDSA. Then the Birkhoff centre BC(A) of A is precisely the set {a | a ∈ k(A)} = {a | a ∈ C(A)}. The following theorem is a consequence of theorem 3.4 and above theorem 4.4. Theorem 4.5. Let A be a CRDSA. Then the Boolean centre B(A) of A, is isomorphic to Birkhoff centre BC(A) of A . References [1] Balbes R. and Dwinger PH., Distributive lattices, U.Missouri Press (1974). [2] Grätzer.George and Schmidt, E. T. On a problem of M. H. Stone, Acta Mathematica Academiae Scientiarum Hungaricae, 8 (1957), 455–460. [3] Grätzer.George, Lattice Theory. First concepts and distributive lattices, W. H. Freeman and Co. San Francisco, 1971. [4] G. C. MoislL, Sur les logiques de,Lukasiewicz a un nombre fini de valeurs, Rev. Roumaine Math. Pures Appl, 9 (1964), 905–920. [5] Ravi Kumar. R.V.G, M.P.K.Kishore and A.R.J.Srikanth, Core Regular double Stone algebra, Journal of Calcutta Mathematical Society, 11 (2015) 1–10. 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