Balanced factor congruences of double MS-algebras

Theorem 1

(Blyth and Varlet [6]) For any two elements a,b of an MS-algebra L, we have

(1) 0^∘∘=0 and 1^∘∘=1,

(2) a≤b⇒b^∘≤a^∘,

(3) a^∘∘∘=a^∘,

(4) a^∘∘∘∘=a^∘∘,

(5) (a∨b)^∘=a^∘∧b^∘,

(6) (a∨b)^∘∘=a^∘∘∨b^∘∘,

(7) (a∧b)^∘∘=a^∘∘∧b^∘∘.

Proposition 1

For any two elements a,b of a dual MS-algebra (L;⁺), we have

(1) 0⁺⁺=0 and 1⁺⁺=1,

(2) a≤b⇒b⁺≤a⁺,

(3) a⁺⁺⁺=a⁺,

(4) a⁺⁺⁺⁺=a⁺⁺,

(5) (a∨b)⁺=a⁺∧b⁺,

(6) (a∨b)⁺⁺=a⁺⁺∨b⁺⁺,

(7) (a∧b)⁺⁺=a⁺⁺∧b⁺⁺.

Theorem 2

(Blyth [12]) An equivalence relation on a lattice L is a lattice congruence on L if and only if (a,b)∈θ implies (a∨z,b∨z)∈θ and (a∧z,b∧z)∈θ for all z∈L.

Definition 1

An element x of a double MS-algebra L is called a Stone element of L if x^∘∨x^∘∘=1 and x⁺∧x⁺⁺=0. Let L_S denote the set of all Stone elements of L, that is, L_S={x∈L:x^∘∨x^∘∘=1,x⁺∧x⁺⁺=0}.

Definition 2

Let L₁ be a bounded sublattice of a double MS-algebra L. Then, L₁ is called a subalgebra of L if x^∘,x⁺∈L₁ for every x∈L₁.

Definition 3

A subalgebra L₁ of a double MS-algebra L is called a Stone subalgebra if x^∘∨x^∘∘=1 and x⁺∧x⁺⁺=0, for all x∈L₁.

Proposition 2

L_S is the greatest Stone subalgebra of a double MS-algebra L.

Proof

Hence, x⁺∈L_S. Similarly, we can prove that x^∘∈L_S for all x∈L_S. Therefore, L_S is a subalgebra of a double MS-algebra L. Since x^∘∨x^∘∘=1 and x⁺∧x⁺⁺=0 for every x∈L_S, then L_S is a Stone subalgebra of L. To prove that L_S is the greatest Stone subalgebra of L, let S be any Stone subalgebra of L. Let x∈S. Then, x is a Stone element of L, and hence, x∈L_S. So S⊆L_S as claimed. □

Definition 4

An element a of double MS-algebra L is called a central element if x∨x^∘=1 and x∧x⁺=0. The set of all central elements of L is called the center of L and is denoted by C(L), that is, C(L)={x∈L:x∨x^∘=1,x∧x⁺=0}.

Example 1

Consider the bounded distributive lattice L in Fig. 1. Define unary operations ^∘,⁺ on L by

$$ b^{\circ}=x^{\circ}=a, d^{\circ}=y^{\circ}=c, 1^{\circ}=z^{\circ}=0,0^{\circ}=1,c^{\circ}=d,a^{\circ}=b $$

(1)

It is clear that (L;^∘,⁺)is a double MS-algebra. Then, L^∘∘,L_S, and C(L) are given in Figs. 2, 3, and 4, respectively.

Theorem 3

Let L be a double MS-algebra. Then

(1) C(L)=L^∘∘∩L_S,

(2) C(L) is the greatest Boolean subalgebra of L, L_S, and L^∘∘,

(3) C(L)=C(L_S)=C(L^∘∘).

Proof

Thus, x∈C(L), and hence, L^∘∘∩L_S⊆C(L).

Since a^∘=a⁺ for all a∈C(L), then ^∘ coincide with ⁺ on C(L). Therefore, (C(L),∨,∧,^∘,0,1) is a subalgebra of L. Since a∨a^∘=1 and a∧a^∘=a^∘∘∧a^∘=(a^∘∨a)^∘=1^∘=0 for all a∈C(L), then (C(L),∨,∧,^∘,0,1) is a Boolean subalgebra of L. Suppose that B is any Boolean subalgebra of L and x∈B. Then, a∨a^∘=1 and a∧a⁺=a∧a^∘=0. Hence, a is a central element of L and a∈C(L). So, B⊆C(L) and C(L) is the greatest Boolean subalgebra of L. Using similar agrement, we can show that C(L) is also the greatest Boolean subalgebra of both L_S and L^∘∘.

(3) It follows (1) and (2). □

Theorem 4

If L and M are isomorphic double MS-algebras, then their centers are isomorphic.

Proof

Let h:L→M be an isomorphism and a∈C(L). Then, a∨a^∘=1 and a∧a⁺=0. Hence, h(a∨a^∘)=h(a)∨h(a^∘)=h(a)∨(h(a))^∘=h(1)=1 and h(a)∧(h(a))⁺=h(0)=0. Therefore, h_C(L)(a)=h(a)∈C(M). It is clear that h_C(L) is an injective (0,1) lattice homomorphism. Let b∈C(M). Then, there exists b∈L such that b=h(a)=h_C(L)(a) as h is onto. It follows that b^∘∘=(h(a))^∘∘=h(a^∘∘)=h(a)=h_C(L)(a). Thus, h_C(L) is onto. Obviously, h_C(L) preserves ^∘ and ⁺. Then, h_C(L) is an isomorphism, and hence, C(L)≅C(M). □

Theorem 5

Let L be a double MS-algebra. Suppose that a∈C(L), then the relativized algebra \(\phantom {\dot {i}\!}(a]_{L}=((a],\wedge,\vee,^{\circ _{a}},^{+_{a}},a,1)\) is a double MS-algebra, where \(\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\wedge a\) and \(\phantom {\dot {i}\!}x^{+_{a}}=x^{+} \wedge a\). Conversely, if \(\phantom {\dot {i}\!}(a]_{L}=((a],\wedge,\vee,^{\circ _{a}},^{+_{a}},a,1)\) is a double MS-algebra, then a∈L_S.

Proof

This deduce that \(\phantom {\dot {i}\!}x^{\circ _{a}+_{a}}=x^{\circ \circ }\). Also, we can get \(\phantom {\dot {i}\!}x^{+_{a}\circ _{a}}=x^{+_{a}+_{a}}\). Therefore, \(\phantom {\dot {i}\!}(a]_{L}=((a],\vee,\wedge,^{\circ _{a}},^{+_{a}},0,a)\) is a double MS-algebra.

Conversely, suppose that a∈L, \(\phantom {\dot {i}\!}(a]_{L}=((a],\vee,\wedge,^{\circ _{a}},^{+_{a}},0,a)\) is a double MS-algebra with \(\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\wedge a\) and \(\phantom {\dot {i}\!}x^{+_{a}}=x^{+}\wedge a\). Since a is the greatest element of (a]_L, then \(\phantom {\dot {i}\!}a^{+_{a}}=0\) and \(\phantom {\dot {i}\!}a^{\circ _{a}}=0\). This gives a⁺∧a=0 and a^∘∧a=0, respectively. Consequently, a⁺∧x⁺⁺=(a⁺∧a)⁺⁺=0⁺⁺=0 and a^∘∘∨a^∘=(a^∘∧a)^∘=0^∘=1. Therefore, a is a Stone element of L. □

Theorem 6

Let L be a double MS-algebra. If a∈C(L), then the relativized algebra \(\phantom {\dot {i}\!}[a)_{L}=([a),\wedge,\vee,^{\circ _{a}},^{+_{a}},a,1)\) is a double MS-algebra, where \(\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\vee a\) and \(\phantom {\dot {i}\!}x^{+_{a}}=x^{+}\vee a\). Conversely, if \(\phantom {\dot {i}\!}[a)_{L}=([a),\wedge,\vee,^{\circ _{a}},^{+_{a}},a,1)\) is a double MS-algebra, then a∈L_S.

Theorem 7

Let L be a double MS-algebra. If a∈C(L), then ((a]_L×[a)_L,^∘,⁺) is a double MS-algebra, where

(a]_L×[a)_L={(x,y):x∈(a]_L,y∈[a)_L},

and

(x,y)^∘=(x^∘∧a,y^∘∨a) and (x,y)⁺=(x⁺∧a,y⁺∨a) for all (x,y)∈(a]_L×[a)_L.

Definition 5

An element a of a double MS-algebra L is called a Birkhoff central element if there exist double MS-algebras L₁ and L₂ and an isomorphism from L to L₁×L₂ such that a is mapped to (1,0). The set BC(L) of all Birkhoff central elements of L is called the Birkhoff center.

Theorem 8

Let L be a double MS-algebra. Then, BC(L)=C(L).

Proof

Let a∈BC(L). Then, there exist double MS-algebras L₁ and L₂ and an isomorphism h from L to L₁×L₂ such that h(a)=(1,0). By Theorem 4, C(L) is isomorphic to C(L₁×L₁)=C(L₁)×C(L₂). Thus, (1,0)∈C(L₁)×C(L₂). Therefore, a=h⁻¹(1,0)∈C(L) and BC(L)⊆C(L).Conversely, let a∈C(L). Then, by Theorem 5 and Theorem 6, L₁=(a]_L and L₂=[a)_L are double MS-algebras, respectively. The direct product L₁×L₂=(a]_L×[a)_L is a double MS-algebra, by Theorem 7. Notice that \(1_{L_{1}}=a\) is the greatest element of L₁ and \(0_{L_{2}}=a\) is the smallest element of L₂. Define h:L→L₁×L₂ by h(x)=(a∧x,a∨x). It is already seen that h is an isomorphism of L onto L₁×L₂. Then, \(h(a)=(a,a)=(1_{L_{1}},0_{L_{2}})\) implies a∈BC(L). Therefore, C(L)⊆BC(L). □

Proposition 3

For any two elements a and b of a double MS-algebra L, we have

(1) θ_a is a lattice congruence on L with Ker θ_a=(a],

(2) a≤b iff θ_a⊆θ_b,

(3) a=b iff θ_a=θ_b,

(4) θ₀=Δ and θ₁=∇,

(5) θ_a is the smallest lattice congruence containing (0,a).

Proof

(2) Let a≤b and (x,y)∈θ_a. Hence, x∨a=y∨a. Then, x∨a∨b=y∨a∨b implies x∨b=y∨b. This gives (x,y)∈θ_a and θ_a⊆θ_b. Conversely, let θ_a⊆θ_b. Since (a∧b)∨a=a=a∨a, then (a∧b,a)∈θ_a. By hypotheses, (a∧b,a)∈θ_b. Then, (a∧b)∨b=a∨b implies b=a∨b. Therefore, a≤b.

(3) It is obvious.

(4) Since for any (x,y)∈θ₀, we have x=y. Then, θ₀=Δ. For all x,y∈L, we have x∨1=1=y∨1 and hence (x,y)∈θ₁. Hence, θ₁=∇.

(5) Let θ be a lattice congruence containing (0,a). Suppose that (x,y)∈θ_a. Then, x∨a=y∨a. Since (x,x),(0,a)∈θ, then (x,x∨a)∈θ. Also, (y,y),(0,a)∈θ give (y,y∨a)∈θ. Then, (x,x∨a),(x∨a,y)∈θ imply (x,y)∈θ. So, θ_a⊆θ. □

Proposition 4

For any two elements a and b of a double MS-algebra L, we have

(1) θ_a∧b=θ_a∩θ_b,

(2) θ_a∨b=θ_a∨θ_b,

(3) θ_a∘θ_b=θ_b∘θ_a,

(4) θ_a∘θ_b=θ_a∨θ_b,

Proof

Therefore, θ_a∩θ_b⊆θ_a∧b and θ_a∧b=θ_a∩θ_b.

(2) Since a,b≤a∨b, then θ_a,θ_b⊆θ_a∨b. Hence, θ_a∨b is an upper bound of θ_a and θ_b. Assume that θ_c is an upper bound of θ_a and θ_b. Then, by Proposition 3(2), θ_a,θ_b⊆θ_c imply that a,b≤c. We prove that θ_a∨b⊆θ_c. Let (x,y)∈θ_a∨b. Then, x∨a∨b=y∨a∨b. Hence, x∨a∨b∨c=y∨a∨b∨c implies x∨c=y∨c and (x,y)∈θ_c. This shows that θ_a∨b is the least upper bound of θ_a and θ_b, that is, θ_a∨b=θ_a∨θ_b.

Then, (x,q)∈θ_a. Also, we can get (q,y)∈θ_b. Therefore, (x,y)∈θ_a∘θ_b and θ_a∨b⊆θ_a∘θ_b. □

Theorem 9

For any two elements a and b of a double MS-algebra L, we have

(1) θ_a is compatible with ^∘ if and only if a∨a^∘=1,

(2) θ_a is compatible with ⁺ if and only if a⁺∧a⁺⁺=0,

(3) θ_a is a congruence on L if and only if a∈C(L).

Proof

Then, (x^∘,y^∘)∈θ_a. Conversely, let θ_a is compatible with ^∘. Since (0,a)∈θ_a by Proposition 3(5), then (1,a^∘))∈θ_a. Hence, (a,a),(1,a^∘))∈θ_a implies (1,a∨a^∘)∈θ_a. Therefore, 1=1∨a=a∨(a∨a^∘)=a∨b.

Then, (x⁺,y⁺)∈θ_a. Conversely, let θ_a is compatible with ⁺. Then, (0,a)∈θ_a implies (1,a⁺))∈θ_a. Since (a,a),(1,a⁺))∈θ_a, then (1,a∨a⁺)∈θ_a. Hence, 1=1∨a=a∨a⁺. It follows that a⁺∧a⁺⁺=(a∨a⁺)⁺=1⁺=0.

(3) As a∈C(L), then a∨a^∘=1, a∧a⁺=0, and a=a^∘∘, the proof follows (1) and (2). □

Definition 6

A congruence θ on a double MS-algebra L is called a factor congruence if there is a congruence ψ on L such that θ∧ψ=Δ, θ∨ψ=∇ and θ permutes with ψ.

Theorem 10

Let L be a double MS-algebra and θ a congruence on L. Then, θ is a factor congruence on L if and only if θ=θ_a for some a∈C(L).

Proof

Therefore, θ_a is a factor congruence on L, whenever a∈C(L). Conversely, let θ be a factor congruence on L. Then, there exists a congruence ψ on L such that θ∨ψ=∇ and θ∩ψ=Δ. Since (0,1)∈∇=θ∨ψ=θ∘ψ, then there exists x∈L such that (0,x)∈θ and (x,1)∈ψ. Thus, (0,x^∘∘)∈θ and (x^∘∘,1)∈ψ. We prove that \(\phantom {\dot {i}\!}\theta =\theta _{x^{\circ \circ }}\) such that x^∘∘∈C(L). Since (0,x^∘∘)∈θ, then by Proposition 3(5), \(\theta _{x^{\circ \circ }}\subseteq \theta \). Now, let (p,q)∈θ. Then, (p,q),(x^∘∘,x^∘∘)∈θ implies (p∨x^∘∘,q∨x^∘∘)∈θ. Since (x^∘∘,1),(p,p),(q,q)∈ψ, then (x^∘∘∨p,1),(x^∘∘∨q,1)∈ψ. Hence, (x^∘∘∨p,x^∘∘∨q)∈ψ. Therefore, (x^∘∘∨p,x^∘∘∨q)∈θ∩ψ=Δ. It follows that x^∘∘∨p=x^∘∘∨q and hence \(\phantom {\dot {i}\!}(p,q)\in \theta _{x^{\circ \circ }}\). So, θ⊆θ_∘∘ and \(\phantom {\dot {i}\!}\theta =\theta _{x^{\circ \circ }}\). This deduce that \(\phantom {\dot {i}\!}\theta _{x^{\circ \circ }}\) is a congruence on L. So, by Theorem 9(3), x^∘∘∈C(L). □

Definition 7

A congruence θ on a double MS-algebra L is called balanced if \((\theta \vee \alpha)\cap (\theta \vee \acute {\alpha })=\theta \) for all factor congruence α and its complement \(\acute {\alpha }\). The set B(L) of all balanced factor congruences which admit a balanced complement is called the Boolean center of L.

Example 2

It is observed that the Boolean lattice B(L), of all balanced factor congruences is B(L)={θ₀,θ_a,θ_b,θ₁} which is represented in Fig. 5. Clearly C(L) and B(L) are isomorphic Boolean lattices.

Lemma 1

Let L be a double MS-algebra and x∈C(L). Then, θ_x is balanced.

Proof

Then, θ_x is balanced. □

Theorem 11

Let L be a double MS-algebra. Then, the Boolean center B(L) of L is precisely the set {θ_a:a∈C(L)}.

Theorem 12

Let L be a double MS-algebra. Then, the Boolean center B(L) is a Boolean algebra and the mapping a↦θ_a is an isomorphism of C(L) onto B(L).

Proof

The set of all balanced factor congruences of L is B(L)={θ_a:a∈C(L)} by Theorem 11. It is clear that θ₁=∇ is the greatest element of B(L) and θ₀=Δ is the smallest element of B(L) by Proposition 3(4). Also, by Proposition 4(1),(2), respectively, we have θ_a∩θ_b=θ_a∧b and θ_a∨θ_b=θ_a∨b for all θ_a,θ_b∈B(L). Then, (B(L);∩,∨,θ₀,θ₁) is a bounded lattice. For all θ_a,θ_b,θ_c∈B(L), by distributivity of C(L), we get θ_a∩(θ_b∨θ_c)=θ_a∩θ_b∨c=θ_a∧(a∨b)=θ_{(a∨b)∧(a∨c)}=θ_a∨b∩θ_a∨c=(θ_a∨θ_b)∩(θ_a∨θ_c). Thus, B(L) is a distributive lattice. The complement of θ_a is \(\phantom {\dot {i}\!}\theta _{a^{\circ }}\). Then, B(L) is a Boolean algebra. The proof of the rest part of this theorem is straightforward. □

Authors’ contributions

The author read and approved the final manuscript.