# On Subdirect Decompositions of Finite Distributive Lattices

Authors: Yizhi Chen; Jing Tian; Zhongzhu Liu
Journal: Discrete Dynamics in Nature and Society
 (2017)

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## Abstract

Subdirect decomposition of algebra is one of its quite general and important constructions. In this paper, some subdirect decompositions (including subdirect irreducible decompositions) of finite distributive lattices and finite chains are studied, and some general results are obtained.


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## Body

## 1. Introduction and Preliminaries

A semiringR is an algebraic structure (R,+,·) consisting of a nonempty set R together with two binary operations + and · on R such that (R,+) and (R,·) are semigroups connected by distributivity, that is, a(b+c)=ab+ac and (b+c)a=ba+ca, for all a,b,c∈R [1, 2]. A semiring R is called a partially ordered semiring if it admits a compatible ordering ≤, that is, ≤ is a partial order on R satisfying the following condition: for any a,b,c,d∈R, if a≤b and c≤d, then a+c≤b+d and ac≤bd. A partially ordered semiring R is said to be a totally ordered semiring if the imposed partial order is a total order [1, 2].A distributive lattice is a lattice which satisfies the distributive laws [3]. In the following, we will denote L=〈L,+,·,0,1〉 as a finite distributive lattice, where 0 and 1 are the least and the greatest elements of L, respectively, and the addition and the multiplication on L are defined as follows:(1)a+b=a∨b=max⁡a,b,a·b=a∧b=min⁡a,b∀a,b∈L.Also, we denote L={0,1,…,m} as a finite chain with usual ordering [4]. Clearly, both the finite distributive lattices and the finite chains are partially ordered semirings.A semiringR is said to be a subdirect product of an indexed family Ri(i∈I) of semirings if it satisfies R≤Πi∈IRi and Rπi=Ri for each i∈I.An embeddingϕ:R→Πi∈IRi is subdirect if Rϕ is a subdirect product of Ri. At this time, we also say that R has subdirect decomposition of Ri or R is isomorphic to the subdirect product of {Ri}i∈I.A semiringR is called subdirectly irreducible if for every subdirect embedding ϕ:R→Πi∈IRi, there is an i∈I such that ϕ∘πi:R→Ri is an isomorphism. From the above definition, it is easy to see that any two-element semiring is subdirectly irreducible.From [4–7] we know that the subdirect product is a quite general construction. As for as semirings concerned, there are several ways of approaching subdirect decompositions of semirings. In most cases they can be obtained from various semirings theoretical constructions. Another way is based on the famous Birkhoff representation theorem. Formulated in terms of semirings, it asserts that every semiring can be represented as a subdirect product of subdirectly irreducible semirings, and it can often reduce studying the structure of semirings from a given class to studying subdirectly irreducible members of this class. Also, there is a third way of approaching subdirect decompositions which is based on another Birkhoff theorem verified in [4], which, in terms of semirings, says that a semiring R is a subdirect product of a family of semirings {Ri}i∈I if and only if there exists a family of factor congruences {ρi}i∈I on R such that ⋂i∈Iρi=εR and R/ρi=Ri for each i∈I; here, εR is the identity congruence on R.The main aim of this paper is to investigate the subdirect decompositions of a special class of semirings called finite distributive lattices. Although some subdirect decompositions of a finite distributive lattice are discussed in [8], and the subdirect decompositions of a finite chain are studied in [2], the results in this paper will be more general. We will investigate some subdirect decompositions (including subdirect irreducible decompositions) of finite distributive lattices and finite chains, whose proofs are also different from [8].For notations and terminologies occurred but not mentioned in this paper, the readers are referred to [1, 4].


## 2. Subdirect Decompositions of Finite Distributive Lattices

To obtain our main results in this section, we will also need the following lemmas and concepts.Lemma 1 (see [4]).
In the equational class of distributive lattices the only nontrivial subdirectly irreducible algebra is the two-element chain.LetR be a semiring and ConR be the set of all congruences on R. By Lemma  8.2 in [4], we have the following lemma.Lemma 2.
Ifρi∈ConR for i∈I and ∩i=1mρi=εR, then the natural homomorphism (2)α:R→Πi∈IRidefined by (3)iaα=aρiis a subdirect embedding.In the following, we will discuss the subdirect decompositions of a finite distributive lattice.Definition 3 (see [4]).
LetL be a lattice. The element a∈L is called join irreducible of L if for x,y∈L, a=x∨y implies x=a or y=a.Example 4.
LetL36 denote the divisible lattice which is generated by all the positive factor of 36; then the set J of all the join irreducible of L36 is {1,2,3,4,9}.Definition 5 (see [4]).
LetL be a finite distributive lattice and a∈L. If there exist join irreducible elements xi(i=1,2,…,s) such that a=⋁i=1sxi, where xi⩽xj for any 1≤i≠j≤s, then we call ⋁i=1sxi the join irreducible decomposition of a.Lemma 6 (see [4]).
LetL be a finite distributive lattice and J be the set of all the join irreducible elements of L. Then for any a∈L,a has a unique join irreducible decomposition and a=⋁p∈J,p≤ap.Lemma 7 (see [4]).
LetL be a finite distributive lattice and J be the set of all the join irreducible elements of L. If p∈J and p≤⋁i=1mxi(xi∈J,1≤i≤m), then there exists i0, 1≤i0≤m, such that p≤xi0.By Lemmas6 and 7, we immediately obtain the following corollary.Corollary 8.
LetL be a finite distributive lattice and J be the set of all the join irreducible elements of L. If p∈J and p≤⋁i=1kxi(xi∈L,1≤i≤k), then there exists i0, 1≤i0≤k, such that p≤xi0.Now, we can give some subdirect irreducible decompositions of a finite distributive latticeL.Theorem 9.
LetL be a finite distributive lattice with m+1 elements where m≥1 and J={0,p1,p2,…,ph}(m≥h≥1) are the set of all the join irreducible elements of L. Then L is isomorphic to a subdirect product of subdirect irreducible elements Li={0,pi}(i=1,2,…,h).Proof.
Fori=1,2,…,h, define ϕi:L→Li by(4)aϕi=pi,a≥pi,0,otherwise.Then, it is a routine way to verify that ϕi(i=1,2,…,h) is a homomorphism.
Firstly,ϕi(i=1,2,…,h) is clearly a mapping.
Secondly, for anya,b∈L, we will show that (a∨b)ϕi=aϕi∨bϕi(i=1,2,…,h).(i)
Ifa∨b≥pi, then, by Corollary 8, a≥pi or b≥pi, and then we get (a∨b)ϕi=pi=aϕi∨bϕi.(ii)
Ifa∨b≱pi, then a≱pi and b≱pi, and then (a∨b)ϕi=0=0∨0=aϕi∨bϕi.
Thirdly, we show that(a∧b)ϕi=aϕi∧bϕi(i=1,2,…,h) for any a,b∈L.(i)
Ifa∧b≥pi, then a≥pi and b≥pi. Thus, we have (a∧b)ϕi=pi=pi∧pi=aϕi∧bϕi.(ii)
Ifa∧b≱pi, then a≱pi or b≥pi, and then (a∧b)ϕi=0=aϕi∧bϕi.
Summing up all the discussions above, we have shown thatϕi(i=1,2,…,h) is a homomorphism.
Now, letρi=Kerϕi. Next, we show that ρ=∩i=1hρi=εL.
Assume thata,b∈L and (a,b)∈ρ, then we have (a,b)∈ρi(i=1,2,…,h). By Lemma 6, there exist the join irreducible decompositions of a and b. Assume that a=pj1⋁pj2⋁⋯⋁pjs and b=pi1⋁pi2⋁⋯⋁pit, where pj1,pj2,…,pjs,pi1,pi2,…,pit∈J. Then we get a≥pj1,pj2,…,pjs and b≥pi1,pi2,…,pit. Since (a,b)∈ρi(i=1,2,…,h), we have b≥pj1,pj2,…,pjs, a≥pi1,pi2,…,pit, and then a=pj1⋁pj2⋁⋯⋁pjs≤b, b=pi1⋁pi2⋁⋯⋁pit≤a. Hence, we obtain a=b.
Now, by Lemmas1 and 2, we immediately verify that L is isomorphic to a subdirect product of subdirect irreducible elements Li(i=1,2,…,h).Example 10.
LetL={0,a,b,c,1} be a finite distributive lattice as the Hasse diagram shown in Figure 1.
Clearly,J={0,a,b,1}. Now, we can take L1={0,a}, L2={0,b}, and L3={0,1} and define (5)ϕ:L⟶∏i=13Li,0⟼0,0,0,a⟼a,0,0,b⟼0,b,0,c⟼a,b,0,1⟼a,b,1.Then it is a routine way to check that ϕ is a subdirect embedding homomorphism from L to ∏i=13‍Li. Hence, L is isomorphic to a subdirect product of subdirect irreducible elements Li(i=1,2,3).Figure 1In general, if we replace the finite distributive latticeL with a finite lattice, we cannot get the corresponding subdirect irreducible decomposition of L.Example 11.
LetL={0,a,b,c,1} be a finite lattice as the Hasse diagram shown in Figure 2.
Clearly,J={0,a,b,c}. Now, if we take L1={0,a}, L2={0,b}, and L3={0,c}, then there is not any existing subdirect embedding homomorphism ϕ from L to ∏i=13‍Li. Hence, L is not isomorphic to a subdirect product of subdirect irreducible elements Li(i=1,2,3).Figure 2Next, we will discuss more general subdirect decompositions of a finite distributive lattice.LetL be a finite distributive lattice with m+1 elements where m≥1 and J={0,p1,p2,…,ph}(m≥h≥1) are the set of all the join irreducible elements of L. Then J can be expressed as ⋃i=lrL¯i; here L¯i={0,pi1,…,pisi} is a more than 2 elements maximal subchain of J (and also a subchain of L) satisfying 0=pi0<pi1<⋯<pisi, s1+s2+⋯+sr=h, and L¯i⋂L¯k={0} if i≠k. Note that the maximal chain from 0 to a maximal element of J may be not unique, and the expression ⋃i=lrL¯i may be not unique in general.Further, we can takeL¯i=⋃j=llLij (1≤i≤r), where Lij={0,pij1,…,pijtj} is a more than 2 elements subchain of L¯i (and also a subchain of L) and satisfying 0=pij0<pij1<⋯<pijtj, t1+t2+⋯+tl=si, and Lij⋂Lik={0} if j≠k. Clearly, we have J=⋃i=lrL¯i=⋃i=lr⋃j=llLij, where Lij={0,pij1,…,pijtj} is a subchain of L and constructed as above.Example 12.
LetL be a finite distributive lattice whose Hasse diagram given as shown in Figure 3. Clearly, J={0,a,b,c,1}, and we can take J=L¯1⋃L¯2, where L¯1={0,a,b,1} and L¯2={0,c} (also, we can take L¯1={0,a,c,1} and L¯2={0,b}). If we take L11={0,a,b}, L12={0,1}, and L21={0,c}, then J=L11⋃L12⋃L21. If we take L11={0,a}, L12=0,b, L13={0,1}, and L21={0,c}, then we can get J=L11⋃L12⋃L13⋃L21.Figure 3Theorem 13.
LetL be a finite distributive lattice with m+1 elements where m≥1 and J={0,p1,p2,…,ph}(m≥h≥1) are the set of all the join irreducible elements of L. Then L is isomorphic to a subdirect product of Lij(1≤i≤r,1≤j≤l) which is constructed as above.Proof.
For1≤i≤r, 1≤j≤l, define ϕij:L→Lij by(6)aϕij=pijtj,a≥pijtj,pijtk,a≥pijtk,a≱pijtk+10≤tk≤tj-1.
Then, it is a routine way to verify thatϕij(1≤i≤r,1≤j≤l) is a homomorphism.
Firstly,ϕij(1≤i≤r,1≤j≤l) is clearly a mapping.
Secondly, for anya,b∈L, we will show that (a∨b)ϕij=aϕi∨bϕij(1≤i≤r,1≤j≤l).(i)
Ifa∨b≥pijtj, then, by Corollary 8, a≥pijtj or b≥pijtj, and then we get (a∨b)ϕij=pijtj=aϕij∨bϕij.(ii)
Ifa∨b≥pijtka∨b≱pijtk+1(0≤tk≤tj-1), then, by Corollary 8, a≥pijtk or b≥pijtk, and also we have a≱pijtk+1 and b≱pijtk+1. Thus, (a∨b)ϕi=pijtk=aϕij∨bϕij.
Thirdly, we show that(a∧b)ϕij=aϕij∧bϕij(1≤i≤r,1≤j≤l) for any a,b∈L.(i)
Ifa∧b≥pijtj, then a≥pijtj and b≥pijtj. Thus, we have (a∧b)ϕij=pijtj=pijtj∧pijtj=aϕij∧bϕij.(ii)
Ifa∧b≥pijtk, a∧b≱pijtk+1(0≤tk≤tj-1), then we have a≥pijtk and b≥pijtk, and also a≱pijtk or b≥pijtk. Hence, (a∧b)ϕij=pijtk=aϕij∧bϕij.
Summing up all the discussions above, we have shown thatϕij(1≤i≤r,1≤j≤l) is a homomorphism.
Now, letρij=Kerϕij. Next, we show that ρ=∩1≤i≤r,1≤j≤l=εL.
Assume thata,b∈L and (a,b)∈ρ; then we have (a,b)∈ϕij(1≤i≤r,1≤j≤l). By Lemma 6, there exist the join irreducible decompositions of a and b. Assume that a=pλ1⋁pλ2⋁⋯⋁pλs′ and b=pμ1⋁pμ2⋁⋯⋁pμt′, where pλ1,pλ2,…,pλs′,pμ1,pμ2,…,pμt′∈J. Then we get a≥pλ1,pλ2,…,pλs′ and b≥pμ1,pμ2,…,pμt′. Since (a,b)∈ρij(1≤i≤r,1≤j≤l), we have b≥pλ1,pλ2,…,pλs′ and a≥pμ1,pμ2,…,pμt′ and then a=pλ1⋁pλ2⋁⋯⋁pλs′≤b and b=pμ1⋁pμ2⋁⋯⋁pμt′≤a. Hence, we obtain a=b.
Now, by Lemma2, we immediately verify that L is isomorphic to a subdirect product of Lij(1≤i≤r,1≤j≤l).In the following, we will discuss the subdirect product decomposition of the finite chainL={0,1,…,m}.Since for a finite chainL={0,1,…,m}, the set J of all the join irreducible elements of L is just equal to L, and also note that a finite chain must be a finite distributive lattice, then we immediately obtain the following corollary by Theorem 9.Corollary 14.
LetL={0,1,…,m} be a finite chain and Li={0,i}(i=1,2,…,m). Then L is isomorphic to a subdirect product of subdirect irreducible elements Li(i=1,2,…,m).In general, the subdirect decomposition manners of a finite chain into the subdirect irreducible elements can be various.Example 15.
LetL={0,1,…,m} be a finite chain. By the above corollary, we have shown that L is isomorphic to a subdirect product of subdirect irreducible elements Li, where Li={0,i}(i=1,2,…,m). On the other hand, we can also take Li′={i-1,i}(i=1,2,…,m) and define (7)ϕ′:L→∏i=1mLi′,0⟼0,1,2,…,m-2,m-1,1⟼1,1,2,…,m-2,m-1,2⟼1,2,2,…,m-2,m-1,3⟼1,2,3,…,m-2,m-1,⋮m-1⟼1,2,3,…,m-1,m-1,m⟼1,2,3,…,m-1,m;then it is not hard to check that ϕ′ is also a subdirect embedding homomorphism from L to ∏i=1m‍Li′. And so L is isomorphic to a subdirect product of subdirect irreducible elements Li′(i=1,2,…,m). Obviously, the two manners of subdirect decomposition of L into the subdirect irreducible elements are different.Next, we will denoteh=⌈m/2⌉, which is the least integer greater than or equal to m/2, and take Li={0,2i-1,2i}(i=1,2,…,h-1) and Lh={0,2h-1,2h}, when m=2h, Lh={0,2h-1} when m=2h-1. Also, we have the following theorem.Theorem 16.
LetL={0,1,…,m} be a finite chain. Then L is isomorphic to a subdirect product of Li(i=1,2,…,h) constructed above.Proof.
Fori=1,2,…,h, define ϕi:L→Li by(8)aϕi=2i,a≥2i,2i-1,2i>a≥2i-1,0,otherwise.Then, we can check that ϕi(i=1,2,…,h) is a homomorphism by a routine way. Now, let ρi=Kerϕi. Also, we can show that ∩i=1hρi=εL. By Lemma 2, L is isomorphic to a subdirect product of Li(i=1,2,…,h).Further, denoteh=⌈m/k-1⌉, which is the least integer greater than or equal to m/k-1, and take Li={0,…,(k-1)i-1,(k-1)i}(i=1,2,…,h-1) and Lh={0,…,(k-1)h-1,(k-1)h} when h=⌈m/k-1⌉=m/k-1; Lh=0,k-1h-1,…,m when ⌈m/k-1⌉>m/k-1. Generally, we will have the following theorem.Theorem 17.
LetL={0,1,…,m} be a finite chain. Then L is isomorphic to a subdirect product of Li(i=1,2,…,h) constructed as above.Proof.
Fori=1,2,…,h, define ϕi:L→Li by(9)aϕi=k-1i,a≥k-1i,a,k-1i>a≥k-1i-1,0,k-1i-1>a≥0.Then, it is a routine way to check that ϕi(i=1,2,…,h) is a homomorphism. Now, let ρi=Kerϕi. Also, we can show that ∩i=1hρi=εL. By Lemma 2, we can complete our proof.Finally, we will give a more general subdirect decomposition of a finite chain.LetL=⋃i=lhLi be a finite chain with m+1 elements where m≥1. Here Li={0,pi1,…,pisi} is a more than 2 elements subchain of L satisfying 0=pi0<pi1<⋯<pisi, s1+s2+⋯+sh=m, and also Li⋂Lj={0} if i≠j.Theorem 18.
LetL=⋃i=lhLi be a chain with m+1 elements where m≥1 and Li={0,pi1,…,pisi} are constructed as above. Then L is isomorphic to a subdirect product of Li(i=1,2,…,h).Proof.
Fori=1,2,…,h, define ϕi:L→Li by(10)aϕi=pisi,a≥pi,pisj,a≥pisj,a≱pisj+10≤sj≤si-1.
Then, it is a routine way to verify thatϕi(i=1,2,…,h) is a homomorphism.
Firstly,ϕi(i=1,2,…,h) is clearly a mapping.
Secondly, for anya,b∈L, we will show that (a∨b)ϕi=aϕi∨bϕi(i=1,2,…,h).(i)
Ifa∨b≥pisi, then, since L is a finite chain, we have a≥pisi or b≥pisi, and then we get (a∨b)ϕi=pisi=aϕi∨bϕi.(ii)
Ifa∨b≥pisj, a∨b≱pisj+1(0≤sj≤si-1), then a≥pisj or b≥pisj, a≱pisj+1 and b≱pisj+1, and then (a∨b)ϕi=pisj=aϕi∨bϕi.
Thirdly, we show that(a∧b)ϕi=aϕi∧bϕi(i=1,2,…,h) for any a,b∈L.(i)
Ifa∧b≥pisi, then a≥pisi and b≥pisi, and then we get (a∧b)ϕi=pisi=aϕi∧bϕi.(ii)
Ifa∧b≥pisj, a∧b≱pisj+1(0≤sj≤si-1), then a≥pisj and b≥pisj, a≱pisj+1 or b≱pisj+1, and then (a∧b)ϕi=pisj=aϕi∧bϕi.
Summing up all the discussions above, we have shown thatϕi(i=1,2,…,h) is a homomorphism.
Now, letρi=Kerϕi. Next, we show that ρ=∩i=1hρi=εL.
Assume thata=plsj, b=pkst∈L, and (a,b)∈ρ; then we have (a,b)∈ρi(i=1,2,…,h), clearly, (a,b)∈ρl and (a,b)∈ρk. By (a,b)∈ρl, we have b≥plsj=a. Similarly, by (a,b)∈ρk, we have a≥pkst=b. Hence, we obtain a=b.
Now, by Lemma2, we immediately verify that L is isomorphic to a subdirect product of Li(i=1,2,…,h).


## 3. Conclusion

Subdirect decomposition of algebra is one of its quite general and important constructions. In this paper, we investigate some subdirect decompositions (including subdirect irreducible decompositions) of finite distributive lattices and finite chains, and give a lot of concrete examples. Actually, the main results in this paper are good complements of the corresponding ones in [8].



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