The Uniqueness of the Decomposition of Distributive Lattices into Sums of Boolean Lattices

Let A =< A,≤A> and B =< B,≤B> be lattices such that A ∩ B is a filter in A and an ideal in B, and the orderings ≤A and ≤B coincide on A ∩ B. Then ≤A ∪ ≤B ∪ (≤A ◦ ≤B), is a lattice ordering on A ∪ B and the resulting lattice, called a sum of A and B, is denoted by A ⊕ B. The sum operation was introduced by Wroński [5], and its special case with A∩B = {1A} = {0B} by Troelstra [4]. In particular, if B is a two–element Boolean algebra then A⊕ B is the same as A⊕, where ⊕ is the Jaśkowski operation of adding to A the top element ( so called “mast” ), see [2]. Kotas, Wojtylak [3] proved that the closure of the class of all finite Received November 21, 1997; Revised June 23 1998 94 JOANNA GRYGIEL, PIOTR WOJTYLAK Boolean algebras with respect to the sum operation is the class of all finite distributive lattices. It means that for every finite distributive lattice D there is a finite family {Bi}i∈T of Boolean algebras such that D is a sum of that family. In this case we shall write D = ⊕{Bi}i∈T . As the sum operation is nonassociative and noncommutative, this notation does not give us any clue about the ordering in which the summation should be performed. As a matter of fact, there is no uniqueness of doing it. Despite this, we are going to prove that all maximal elements of the decomposition are uniquely determined. Let D be a lattice. A sublattice D1 of D is said to be a fragment of D, and we shall write D1 v D in this case, if a ≤ c ≤ b and a, b ∈ D1 ⇒ c ∈ D1, for every a, b, c ∈ D. If, additionally, D1 is a Boolean lattice, then D1 is said to be a Boolean fragment of D. Any filter ( ideal ) of D is its fragment and, if D is finite, then any fragment of D is its interval, that is a set of the form {x ∈ D : a ≤ x ≤ b} for some a and b. Clearly, v is transitive and Di v D1 ⊕ D2 for i = 1, 2. Theorem 1. Let D1 and D2 be distributive lattices and B be a Boolean algebra. If B v D1 ⊕D2 then B v D1 or B v D2. Proof. Let B v D1 ⊕ D2 and assume that there are elements a ∈ B \ D2 and b ∈ B \ D1. Let 0B be the zero of B and 1B be the unit of B. Since 0B ≤ a and a ∈ D1, then 0B ∈ D1 \ D2. Similarly, we conclude that 1B ∈ D2 \ D1. Since 0B ≤ 1B , there is an element x ∈ D1 ∩ D2 ∩ B by the definition of the ordering in D1 ⊕D2. Let y ∈ B be the complement of x in B. If y ∈ D1 then 1B = x ∨ y ∈ D1, which is impossible. Otherwise y ∈ D2, so 0B = x∧y ∈ D2, which also contradicts our assumptions. Thus, B ⊆ D1 or B ⊆ D2, and it yields B v D1 or B v D2 as B v D1 ⊕D2. Since every finite distributive lattice D is a sum ⊕ of Boolean lattices Bi, then D is also the set-theoretical sum ⋃ Bi of its Boolean fragments. UNIQUENESS OF DECOMPOSITIONS OF DISTRIBUTIVE LATTICES 95 The components of the sum are not uniquely determined as it is possible that Bi ⊆ Bj for some i, j. We call a family {Bi} of Boolean lattices a scarce decomposition of D iff D = ⊕ Bi and Bi ⊆ Bj does not hold for any i 6= j. As the immediate Corollaries of Theorem 1 we get: Corollary 1. If a lattice D is a sum ⊕ of a family {Bi} of Boolean lattices, then the family contains all maximal Boolean fragments of D. Corollary 2. There is at most one scarce decomposition of any finite distributive lattice D and the decomposition consists of all maximal Boolean fragments of D. It may happen, however, that a finite distributive lattice D does not have the scarce decomposition. More specifically, it is sometimes impossible to get D as a sum ⊕ of its maximal Boolean fragments without taking subalgebras of maximal fragments or repeating them in the sum operations. Example 1. Let FD(3) be a free distributive lattice on three generators. The lattice has five maximal Boolean fragments: but it is impossible to get FD(3) as their sum ⊕. It means that FD(3) is not a sum A1 ⊕ A2, where each of Ai is a sum of disjoint subsets of 96 JOANNA GRYGIEL, PIOTR WOJTYLAK {B1,B2,B3,B4,B5}. Indeed, if A1 = B1, then any sum of the rest does not form a distributive lattice. We deal with the other possibilities in a similar way. If we decompose FD(3) into a sum of proper fragments A1 ⊕A2, then either the lower lattice will contain a proper subalgebras of B1 or the upper lattice will contain a subalgebra of B5. Now, we are going to show a practical method of finding all maximal Boolean fragments of a given distributive lattice D. First let us identify the “top” maximal Boolean fragment of D. Theorem 2. The filter ∇ generated by all coatoms of any finite distributive lattice D is a maximal Boolean fragment of D. Proof. Let A be the set of all coatoms of D. It is clear that ∇ v D with the unit element 1 of D as the greatest and Θ = ∧ A as the least element. Let x ∈ ∇. Then x = ∧ {a ∈ A : x ≤ a}. Let −x = ∧ {a ∈ A : x ∧ a < x}. Clearly, −x ∈ ∇, and x ∧ −x = ∧ A = Θ, and x ∨ −x = ∧ {x ∨ a; a ∈ A and a < x ∨ a} = 1 Thus −x is the complement of x in ∇ and hence ∇ is a Boolean algebra. We need to prove that ∇ is a maximal Boolean fragment of D. Let ∇ v B, where B is a Boolean fragment of D. Then for every x ∈ B we have Θ ∨ x ≤ Θ ∨ ∧ {a ∈ A : x ≤ a} = ∧ {a ∈ A : x ≤ a}, so x∨Θ = 1 iff x = 1. Hence Θ is the zero of B which means that ∇ = B. Let D be a finite distributive lattice and b ∈ D. An element a ∈ D is said to be a b-coatom iff a is a coatom in the fragment {x ∈ D : x ≤ b} of D. We denote by ∇(b) the Boolean fragment of D determined by all b-coatoms in D. In other words, ∇(b) is the “top” Boolean fragment of {x ∈ D : x ≤ b}. Dually, the “bottom” Boolean fragment ∆ of D is the ideal generated by all atoms. By ∆(b) we denote the Boolean fragment of D determined by all b-atoms, that is atoms in the fragment {x ∈ D : b ≤ x} of D. UNIQUENESS OF DECOMPOSITIONS OF DISTRIBUTIVE LATTICES 97 Theorem 3. Let B be a maximal Boolean fragment of a finite distributive lattice D and let b be a maximal element of B such that the set of all b-coatoms is not contained in B. Then b < 1B and ∇(b) is a maximal Boolean fragment of D. Proof. First, note that if some 1B-coatoms were not elements of B (i.e. b = 1B), then B would be a proper sublattice of ∇(b). Let b ∈ B and Θ = ∧ {y : y is a b-coatom}. We can easily prove that ∇(b) is a maximal Boolean fragment of D iff all Θ-atoms are below b (i.e. ∇(b) = ∆(Θ)). One implication follows from the (dual counterpart) of the above theorem. If ∇(b) is a maximal Boolean fragment of D, then it is also a maximal Boolean fragment of {x ∈ D : x ≥ Θ} and hence ∇(b) = ∆(Θ). On the other hand, suppose that all Θ-atoms are below b and ∇(b) is a sublattice of some Boolean fragment B1 of D. Let −Θ be the complement of Θ in B1. If −Θ ∧ b < b, then −Θ ∧ b ≤ y0 for some b-coatom y0 and hence b = (Θ ∨ −Θ) ∧ (b ∨ Θ) = (−Θ ∧ b) ∨ Θ ≤ y0 ∨ Θ = y0 which is impossible. Hence −Θ ∧ b = b and it means that Θ ≤ b ≤ −Θ. Thus, Θ is the zero element in B1. Since b must be then the unit element there, we get ∇(b) = B1. Now, let us assume that B is a maximal Boolean fragment of D and b is a maximal element of B such that the set of all b-coatoms is not contained in B. We need to show that ∇(b) is a maximal Boolean fragment of D. Suppose, to the contrary, that a ∧ b = Θ, for some Θ-atom a. Let us prove that a ∧ 1B = Θ. Suppose that a ≤ 1B . According to the choice of b, there is a b-coatom y0 such that y0 6∈ B. Since 0B ≤ b ≤ a ∨ b ≤ 1B , 1 We wish to thank an anonymous refree for his suggestions concerning the presentation of our argument. 98 JOANNA GRYGIEL, PIOTR WOJTYLAK we get a ∨ b ∈ B. If a ∨ y0 ∈ B, then y0 = y0 ∨ Θ = (b ∧ y0) ∨ (a ∧ b) = (a ∨ y0) ∧ b ∈ B which is impossible. Thus, a ∨ y0 6∈ B which means, in particular, that a ∨ y0 < a ∨ b. But b < a∨ b and hence (by the choice of the element b) the element a∨ y0 cannot be a (a ∨ b)-coatom. So a ∨ y0 < x0 < a ∨ b, for some x0.