We introduce the concept of Stone elements of a double MS-algebra L. Then, we show that the set LS of all Stone elements of L is the greatest Stone subalgebra of 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 LS denote the set of all Stone elements of L, that is, LS={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
LS is the greatest Stone subalgebra of a double MS-algebra L.
Proof
It is clear that 0,1∈LS. Let x,y∈LS. Then, x∘∨x∘∘=1, x+∧x++=0, y∘∨y∘∘=1, and y+∧y++=0. Thus, we get
$$\begin{array}{@{}rcl@{}} (x\vee y)^{\circ}\vee(x\vee y)^{\circ\circ}&=&(x^{\circ}\wedge y^{\circ})\vee(x^{\circ\circ}\vee y^{\circ\circ})\text{ by Theorem 1(5),(6)}\\ &=&(x^{\circ}\vee x^{\circ\circ}\vee y^{\circ\circ})\wedge(y^{\circ}\vee x^{\circ\circ}\vee y^{\circ\circ})\text{ by distributivity of }L\\ &=&1\wedge 1=1\text{ as }x^{\circ}\vee x^{\circ\circ}=1,y^{\circ}\vee y^{\circ\circ}=1,\\ (x\vee y)^{+}\wedge(x\vee y)^{++}&=&(x^{+}\wedge y^{+})\wedge(x^{++}\vee y^{++})\text{ by Proposition 1(5),(6)}\\ &=&(x^{+}\wedge y^{+}\wedge x^{++})\vee(x^{+}\wedge y^{+}\wedge y^{++})\text{ by distributivity of }L\\ &=&0\vee 0=0\text{ as }x^{+}\wedge x^{++}=0,y^{+}\wedge y^{++}=0. \end{array} $$
Then, x∨y∈LS. Using a similar way, we get x∧y∈LS. Therefore, (LS,∨,∧,0,1) is a bounded distributive sublattice of L. Now, we prove that x+∈LS for all x∈LS.
$$\begin{array}{@{}rcl@{}} x^{+\circ}\vee x^{+\circ\circ}&=&x^{++}\vee x^{+++}\text{as }x^{+\circ}=x^{++}\\ &=&(x^{+}\wedge x^{++})^{+}\text{ by Proposition 1(5)}\\ &=&0^{+}=1\text{ as }x^{+}\wedge x^{++}=0,\\ x^{+\circ}\wedge x^{+\circ\circ}&=&x^{++}\wedge x^{+++}\text{ as }x^{+\circ}=x^{++}\\ &=&x^{++}\wedge x^{+}=0\text{ by Proposition 1(3)}. \end{array} $$
Hence, x+∈LS. Similarly, we can prove that x∘∈LS for all x∈LS. Therefore, LS is a subalgebra of a double MS-algebra L. Since x∘∨x∘∘=1 and x+∧x++=0 for every x∈LS, then LS is a Stone subalgebra of L. To prove that LS 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∈LS. So S⊆LS as claimed. □
On the following, we introduce the notion of central elements of a double MS-algebra L and prove that the set C(L) of all central elements of L is the greatest Boolean subalgebra of L. Also, the relationship among LS, C(L), and L∘∘ is investigated.
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)
and
$$ a^{+}=z^{+}=b, c^{+}=y^{+}=d, 0^{+}=x^{+}=1,b^{+}=a,d^{+}=c,1^{+}=0. $$
(2)
It is clear that (L;∘,+)is a double MS-algebra. Then, L∘∘,LS, 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∘∘∩LS,
(2) C(L) is the greatest Boolean subalgebra of L, LS, and L∘∘,
(3) C(L)=C(LS)=C(L∘∘).
Proof
(1). Let x∈C(L). Then, x∨x∘=1 and x∧x+=0. Then
$$\begin{array}{@{}rcl@{}} x^{++}&=&x^{++}\vee 0\\ &=&x^{++}\vee(x\wedge x^{+})\\&=&(x^{++}\vee x)\wedge(x^{++}\vee x^{+})\text{ by distributivity of }L\\&=&x\wedge(x^{+}\wedge x)^{+}\text{ as }x\geq x^{++}\\&=&x\wedge0^{+}=x\wedge 1=x. \end{array} $$
Thus, x∈L∘∘. Also,
$$\begin{array}{@{}rcl@{}} x^{++}\wedge x^{+}&=&x^{++}\wedge x^{+++}\text{ by Proposition 1(3)}\\ &=&(x\wedge x^{+})^{++}\text{ by Proposition 1(7)}\\&=&0^{++}=0\text{ by Proposition 1(1)},\\ x^{\circ\circ}\vee x^{\circ}&\geq& x\vee x^{\circ}\hbox { as }x^{\circ\circ}\geq x\\&=&1. \end{array} $$
Then, x++∧x+=0 and x∘∘∨x∘=1 imply x∈LS. Therefore, C(L)⊆L∘∘∩LS. Conversely, let x∈L∘∘∩LS. Then, x=x∘∘=x++, x∘∨x∘∘=1, and x+∧x++=0. Now,
$$\begin{array}{@{}rcl@{}} x^{\circ}\vee x&=&x^{\circ}\vee x^{\circ\circ} =1,\\ x\wedge x^{+} &=&x^{++}\wedge x^{+}=0. \end{array} $$
Thus, x∈C(L), and hence, L∘∘∩LS⊆C(L).
(2) Clearly 0,1∈C(L). Let a,b∈C(L). Then, we have
$$\begin{array}{@{}rcl@{}} (a\vee b)\vee(a\vee b)^{\circ}&=&a\vee b\vee(a^{\circ}\wedge b^{\circ})\text{ by Proposition 1(5)}\\ &=&(a\vee b\vee a^{\circ})\wedge(a\vee b\vee b^{\circ})\text{ by distributivity of }L\\ &=&1\wedge 1=1,\\ (a\vee b)\wedge(a\vee b)^{+}&=&(a\vee b)\wedge(a^{+}\wedge b^{+})\text{ by Theorem 1(5)}\\ &=&(a\wedge a^{+}\wedge a^{+})\vee(b\wedge a^{+}\wedge b^{+})\text{ by distributivity of }L\\ &=&0\vee 0=0. \end{array} $$
Then, a∨b∈C(L). Similarly a∧b∈C(L). Therefore, (C(L);∨,∧,0,1) is a bounded sublattice of L. Now, we observe that a∘∈C(L) for all a∈C(L),
$$\begin{array}{@{}rcl@{}} a^{\circ\circ}\vee a^{\circ\circ\circ}&=&a\vee a^{\circ}\text{ as }a^{\circ\circ}=a,\forall a\in C(L)\text{ and }a^{\circ\circ\circ}=a^{\circ}\\ &=&1,\\ a^{\circ+}\wedge a^{\circ++}&=&a^{\circ\circ}\wedge a^{\circ\circ\circ}\text{ as }a^{\circ+}=a^{\circ\circ}\\ &=&a^{\circ\circ}\wedge a^{\circ}\text{ as }a^{\circ\circ\circ}=a^{\circ}\\ &=&(a^{\circ}\vee a)^{\circ}=1^{\circ}=0\text{ by Theorem 1(5)}. \end{array} $$
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 LS and L∘∘.
(3) It follows (1) and (2). □
The following theorem shows that the centers of isomorphic double MS-algebras are isomorphic Boolean algebras.
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, hC(L)(a)=h(a)∈C(M). It is clear that hC(L) is an injective (0,1) lattice homomorphism. Let b∈C(M). Then, there exists b∈L such that b=h(a)=hC(L)(a) as h is onto. It follows that b∘∘=(h(a))∘∘=h(a∘∘)=h(a)=hC(L)(a). Thus, hC(L) is onto. Obviously, hC(L) preserves ∘ and +. Then, hC(L) is an isomorphism, and hence, C(L)≅C(M). □
For an MS-algebra (L,∘), it is proved in [3] that \(\phantom {\dot {i}\!}(a]_{L}=((a],^{\circ _{a}})\) is an MS-algebra if and only if a∘∈C(L), where (a]={x∈L:x≤a}=[0,a] is a principal ideal of L generated by the element a of L, a unary operation \(\phantom {\dot {i}\!}^{\circ _{a}}\) is defined on (a] by \(\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\wedge a\) for all x∈(a] and C(L)={x∈L:x∨x∘=1} is the center of L.
For a double MS-algebra (L;∘,+), the answer of the following question is given: Under what conditions a principal ideal (a],a∈L constructs a double MS-algebra?
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∈LS.
Proof
Assume that a∈C(L). Hence, a∘∈C(L). Then, by ([13], Theorem 3.5), \(\phantom {\dot {i}\!}((a],\vee,\wedge,^{\circ _{a}},0,a)\) is an MS-algebra, whenever \(\phantom {\dot {i}\!}x^{\circ _{a}}=x^{\circ }\wedge a\). Now, we prove that \(\phantom {\dot {i}\!}((a],\vee,\wedge,^{+_{a}},0,a)\) is a dual MS-algebra, where \(\phantom {\dot {i}\!}x^{+_{a}}=x^{+}\wedge a\) for any x∈(a]. Let x∈(a], we have
$$\begin{array}{@{}rcl@{}} x^{+_{a}\circ+_{a}}\vee x&=&(x^{+}\wedge a)^{+_{a}}\vee x\\ &=&((x^{+}\wedge a)^{+}\wedge a)\vee x\\ &=&((x^{++}\vee a^{+})\wedge a)\vee x \\ &=&(x^{++}\wedge a)\vee(a^{+}\wedge a)\vee x\text{ by distributivity of }L\\ &=&(x^{++}\wedge a)\vee x\text{ as }a^{+}\wedge a=0\\ &=&x\text{ as }x\geq x^{++}\geq x^{++}\wedge a. \end{array} $$
Then, \(\phantom {\dot {i}\!}x\geq x^{+_{a}+_{a}}\). Let x,y∈(a]
$$\begin{array}{@{}rcl@{}} (x\wedge y)^{+_{a}}&=&(x\wedge y)^{\circ}\wedge a\\ &=&(x^{+}\vee y^{\circ+})\wedge a\\ &=&(x^{+}\wedge a)\vee (y^{+}\wedge a)\text{ by distributivity of }L\\ &=&x^{+_{a}}\vee y^{+_{a}}, \end{array} $$
Also, \(\phantom {\dot {i}\!}0^{+_{a}}=a\). Now, for every x∈(a], we have
$$\begin{array}{@{}rcl@{}} x^{\circ_{a}+_{a}}&=&(x^{\circ}\wedge a)^{+_{a}}\\ &=&(x^{\circ}\wedge a)^{+}\wedge a\\ &=&(x^{\circ+}\vee a^{+})\wedge a\\ &=&(x^{\circ\circ}\vee a^{+})\wedge a\\ &=&(x^{\circ\circ}\wedge a)\vee (a^{+}\wedge a)\\ &=&x^{\circ\circ}\wedge a\text{ as }a^{+}\wedge a=0,\\ x^{\circ_{a}\circ_{a}}&=&(x^{\circ}\wedge a)^{\circ}\wedge a\\ &=&(x^{\circ\circ}\vee a^{\circ})\wedge a\\ &=&(x^{\circ\circ}\wedge a)\vee(a^{\circ}\wedge a)\text{ by distributivity}\\ &=&x^{\circ\circ}\wedge a\text{ as }a^{+}\wedge a=0. \end{array} $$
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. □
Similarly for the principal filter [a) of a double MS-algebra, we establish the following result, where [a)={x∈L:x≥a}=[a,1].
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∈LS.
Let L1,L2 are double MS-algebras. Then, L1×L2 is a double MS-algebra, where ∘ and + are defined by (x,y)∘=(x∘,y∘) and (x,y)+=(x+,y+). Moreover, \((L_{1}\times L_{1})^{\circ \circ }=L_{1}^{\circ \circ }\times L_{2}^{\circ \circ } \) and C(L1×L2)=C(L1)×C(L2).
As a consequence of Theorem 5 and Theorem 6, we have
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.
Now, we introduce the concept of Birkhoff center for a double MS-algebra.
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). □