SOME STRUCTURE THEOREMS FOR ATOMISTIC ALGEBRAIC …matradi/files/2000_1.pdf · 2 S. RADELECZKI This relation will prove a useful tool also in the proof of our Main Theorem and Theorem

Acta Math. Hungar.86 (1-2) (2000),1-15.

SOME STRUCTURE THEOREMSFOR ATOMISTIC ALGEBRAIC LATTICES

S. RADELECZKI (Miskolcr

IAbstract. We prove that any atomistic algebraic lattice is a direct product of

sub directly irreducible lattices iff its congruence lattice is an atomic Stone lattice.We define on the set A(L) of all atoms of an atomistic algebraic lattice L a relationR as follows: for a, bE A(L), (a, b) E R <=} IJ(O,a) /\ IJ(O,b) =f. L-con L. We provethat Con L is a Stone lattice iff R is transitive and we give a characterization ofCen (L) using R. We also give a characterization of weakly modular atomisticalgebraic lattices.

1. Introduction

1

It is a known fact that geometric lattices (i.e. the algebraic atomisticsemimodular lattices) are direct products of sub directly irreducible geomet-ric lattices. We generalize this structure theorem proving that this factor-ization is not related properly with semimodularity but it follows from theweak modularity (in the sense of [3]) of geometric lattices. In fact we provethe following:

MAIN THEOREM. Let L be an atomistic algebraic lattice. Then the fol-lowing statements are equivalent:

(i) ConL is an atomic Stone lattice.(ii) L is a direct product of subdirectly irreducible lattices.(iii) For any a E A(L), the congruence B(O,a) contains exactly one atom

of ConL.

We also show that an algebraic atomistic lattice is weakly modular if andonly if the sublattices of its standard and neutral elements coincide (Theorem3.4).

The starting point of this paper is a theorem of L. Libkin [7] assertingthat any atomistic algebraic lattice is a direct product of directly irreduciblelattices. To prove this result 11e showed that Cen (L), the lattice of centralelements of an atomistic algebraic lattice L, is a complete atomic Booleanlattice. Here we give an other proof of this fact, describing constructivelyCen (L) by the use of an equivalence R defined on the set A(L) of atoms of L.

* Research partially supported by Hungarian National Foundation for Scientific Re-search, Grant No. T16432.

023.6-5294/0/$ 5.00 © 2000 Akademiai Kiad6, Budapest

2 S. RADELECZKI

This relation will prove a useful tool also in the proof of our Main Theoremand Theorem 3.4.

Preliminaries

A lattice L with 0 is called atomic if for any a E L, a#-O the interval[0, a] contains at least one atom of L. If the join of atoms below a is equalto a, then the lattice is called atomistic. The set of all atoms under a isdenoted by A(a). It is clear that for any x, y E L we have A(x /\ y) = A(x)n A(y) and A(x) U A(y) ~ A(x V y). We note that the atoms of an atomisticalgebraic lattice L are compact elements in L.

An element a E L is called standard if

x/\(aVy)=(x/\a)V(x/\y), forall x,yEL,

and a is called neutral if for all x, y E L, the equality

(a /\ x) V (x /\ y) V (y /\ a) = (a V x) /\ (x V y) /\ (y V a)

holds (see [3]).The complemented neutral elements are called central elements of L. A

complement of an element a E L (if it exists) is denoted by a'. The stan-dard, neutral and central elements of a lattice L form sublattices denoted byS(L),N(L) and Cen(L) respectively. Obviously, we have Cen(L) ~ N(L)~ S(L).

REMARK 1.1. The following statements can be found e.g. in [3]:(i) s E L is a standard element of L iff the relation defined as B(s )

= {(x, y) I x V y = (x /\ y) Va, a ~ s} is a congruence on L.(ii) For any Sl, S2 E S(L) we have B(Sl V S2) = B(sd V B(S2) and B(Sl/\ S2)

= B(Sl) /\ B(S2). If Sl ~ S2, then B(Sl) ~ B(S2), if B(Sl) = B(S2), then Sl = S2·(iii) If B1 and B2 are factor congruences of a bounded lattice L, i.e.

L ~ L/B1 x L/B2' then there are C1, C2 E Cen (L) such that B1 = B(cd andB2 = B(C2)' j

The following results from [7] will be frequently used in our proofs:

LEMMA 1.2. For an algebraic atomistic lattice L we have(i) a E L is a standard element in L iff A(a V x) = A(a) U A(x), for all

x E L.(ii) Cen (L) = N(L).(iii) A(a') = A(L) \ A(a) for all a E Cen(L).

Set ~L = {A(s) I s E S(L)}, then Lemma 1.2(i) makes it easy to verifythat (~L' n, U) is a lattice. Now we can formulate

Acta Mathematica Hungarica 86, 2000

ATOMISTIC ALGEBRAIC LATTICES 3

COROLLARY 1.3. The map a : S(L) ---* ~L, a(s) = A(s) is a lattice iso-morphism.

PROOF. Since A(SI /\ S2) = A(sd n A(S2), and since by Lemma 1.2(i) wehave A(SI V S2) = A(SI) u A(S2), a is a lattice homomorphism which is ontoby definition. Since A(SI) = A(S2) obviously implies SI = S2, a is a latticeisomorphism. 0

REMARK 1.4. (i) According to Lemma 1.2(i), to prove that an elementa E L is standard in an atomistic algebraic lattice L, it is enough to showthat A(a V x) ~ A(a) U A(x), \Ix E L.

(ii) By Corollary 1.3 for any choice of SI, S2, ... , Sn E S(L) we have

For the identical and unit relation on L we use the notations !:::. and'V, respectively. As usual, Con L (Tol L) stands for the congruence lat-tice (tolerance lattice) of L. An ordered pair of elements a, bEL satisfyingb ~ a is denoted by alb and is called a quotient. The following definitionsare from [3]:

DEFINITION 1. 5. (i) Given two quotients a/band c/ d we write c/ d ,,"Walb iff b ~ d and c = a V d. Similarly, we write c/ d / w alb iff c ~ a and b /\ c= d. If c/d ,,"W alb or c/d / w alb, then we say that c/d is weakly perspectiveinto alb and we write c/d ""w a/b.

(ii) If for some sequence c/ d = xO/Yo, xI/Yl, ... ,xn/Yn = alb we have xi/Vi"'w Xi+I/Yi+l for all i = 0,1, ... ,n - 1, then we say that c/ d is weakly pro-jective into alb and we write c/ d ~w a/b.

The principal congruences of a lattice can be built up by using the rela-tion ~w. More precisely, we have the following theorem.

THEOREM 1.6 (R. P. Dilworth). Let L be an arbitrary lattice and a, b, c, dE L, b ~ a, d ~ c. Then we have (c, d) E B( a, b) if and only if there existsa sequence of elements c = eo ~ el ~ ... ~ en = d, such that for all i = 0,... ,n -1 the relation ei/ei+l ~w alb holds.

DEFINITION 1.7. The lattice L is called weakly modular (or weakly pro-jective) if for any a, b, c, dEL, b < a, d < c the relation c/ d ~w alb impliesthe existence of a#, b# E L SUell that b ~ b# < a# ~ a and a# /b# ~w c/ d.

It is well-known that any modular and any relatively complemented lat-tice is weakly modular.

In what follows let L denote an atomistic algebraic lattice.

PROPOSITION 1.8. S(L) is a complete sublattice of L and for any set{Si liE I, I i= 0} of standard elements of L the equality A( V Si) = U A(Si)

iEI iEIholds.


4 s. RADELECZKI

PROOF. First we prove the latter statement. The inclusion

A( V Si) ~ U A(Si)iEI iEI

is clear. In order to prove the converse inclusion take p E A(L) such thatp ~ V Si· Since p is compact in L there exist iI, ... ,in E I such that p ~ Sil

iEIV ... V Sin' By Remark 1.4 we get p E A(Sil) UA(Si2) u... UA(SiJ ~ U A(Si).

iEITo prove that V Si is standard in L, by Remark 1.4 it is enough to show

iEIthat

A( (V Si) V x) ~ A( V Si) U A(x).iEI iEI

Take apE A( ( V Si) V x) and suppose that p tf. A(x), then we only haveiEI

to prove p E A( V Si) .iEI

Now there exist il, ... ,in E I such that p ~ Sil V ... V Sin V x. SinceS = Sil V ... V Sin is a standard element of L, we have A(s V x) = A(s) U A(x).Thus p must belong to A( s), according to our assumption. Finally S ~ V Si

iEIimplies p E A( V Si) .

iEIThe fact that /\ Si is a standard element in L is equivalent to the in-

iEIclusion A( ( /\ Si) V x) ~ A( /\ Si) U A(x), Vx E L, which can be checked

iEI iEIimmediately, since for any a E A (( /\ Si) V x) we have a E n A( Si V x).

iEI iEIoCOROLLARY 1.9. (I;L, n,U) is a complete lattice.

Let T E Tol L be a tolerance of L. It is easy to check that the set T(O)= {x ELI (O,x) E T} 'is an ideal of L. We shall make use of the notationsA(T) = A(L) n T(O) = {a E A(L) I (0, a) E T} and 'WT = V {a Ia E A(T)}.

REMARK 1.10. Let S be an arbitrary standard element of L. Since forany a E A(L) we have (O,a) E B(s) {:} a ~ s, it is clear that A(B(s)) = A(s)and 'WO(s) = s.

The following two preliminary statements form the starting point of ourfurther results.



PROPOSITION 1.11. For any T E Tol L we have A(T) = A( WT) and WTis a standard element of L.1 For a fixed To E Tol L, B(WTo) is the greatestelement in {T E Tol L I A(T) = A(To)} .

PROOF. Since A(T) ~ A( WT) holds by definition, we have to show onlythe converse inclusion. Set p E A( WT) arbitrary. Since p ~ V { a Ia E A(T)}and since p is compact in L, there are al, ... , an E A(T) such that p ~ al

V ... Van. Since al, ... ,an belong to the ideal T(O), we get p E T(O), i.e.p E A(T).

To prove WT E S(L) it is enough to check A(WT V x) ~ A(WT) U A(x)for all x E L. For this purpose take any p E A( WT V x) and assume thatp r:f. A(x). We need p E A(WT):

Since p is compact in L from p ~ WT V x it follows that there are aI, ... , anE A(WT) = A(T) such that p ~ al V ... V an V x (n EN). Using aO = al V ...Van E T(O), we can write p ~ aO V x. Now (0, an) E T implies (x, aO V x) E Tand so we get (0,])) = (x /\]), (aO V x) /\ p) E T, i.e. p E A(T) = A(WT).

Let us prove the second part of the proposition. Now WTo E S(L) and byRemark 1.10 we have A(B(WTo)) = A(WTo) = A(To). Take aT E TolL withA(T) = A(To); we have to show T ~ B(WTo).

First observe that WT E S(L) and A(WT) = A(T) = A(To) = A(WTo),hence WT = WTo. On the other hand for any (x,y) E T, x,y E L, and forevery]) E A(y) \ A(x) we have (0,])) = (x /\ p, Y /\p) E T, i.e. p E A(T), whichimplies ]) ~ WT = WTo. Since

y ~ (x/\y) V (V {pip E A(y) \A(x)}),

we get y ~ (x /\ y) V WTo. Similarly, from (y, x) E T we obtain x ~ (x /\ y)V WTo. Thus x V y ~ (x 1\ y) V WTo. In consequence, we can write x /\ y ~ xV y ~ (x /\ y) V WTo. Using (x /\ y, (x /\ y) V WTo) E B(WTo) we obtain (x /\ y,xVy) E B(WTo), i.e.: (x,y) E B(WTo). 0

Let R denote the transitive hull of a binary relation R.LEMMA 1.12. (i) For any T E TolL, A(T) = A(T). If A(T) = 0, then

T = 6..(ii) For any ai E Con L, i E I (I =1= 0) we have

A( 1\ ai) = n A(adiEI iEI

and A( V ai) = U A(ai).iEI iEI

PROOF. (i) By definition TI ~ T2 implies A(Td ~ A(T2) for any TI, T2E Tol L. Thus A(T) ~ A (T). Since B(WT) is transitive and by Proposition

1 The proof of this assertion in the finite case can be found also in [4J.


6 S. RADELECZKI

1.11 T ~ e(wr), it follows that T ~ e(wr). Now we can write A(T) ~ A(T)~ A(e(wr)) = A(wr) = A(T), hence A(T) = A(T).

Let us assume now that A(T) = 0, but (a, b) E T for some a, bEL,ai-b. Now we have (al\b,aVb) ET, and since al\b<aVb, there isat least one atom p ~ a V b such that p rt A(a 1\ b). We can write (O,p)= ((al\b)l\p, (aVb)l\p) ET, thus we get pEA(T) - a contradiction.Thus T = 6.

(ii) The relations

A( /\ ai) = nA(ai) andiEI iEI iEI iEI

follow immediately from the definition of the sets A(ai). Since ai E TolL,i E I, according to Proposition 1.11 and Remark 1.1(ii) we have ai ~ e(woJ~ e ( V wQi) , thus we conclude V ai ~ e ( V wQi)· Since V WQi is a stan-

iEJ iEI iEI iEIdard element of L (according to Proposition 1.8), using Remark 1.10 andProposition 1.8 we obtain

A( V ai) ~ A(e( V wQi)) = A( V wQi) = U A(wQJ = U A(ai).iEI iEI iEI iEI iEI

Thus we get A( V ai) = U A(ai). 0iEI iEI

COROLLARY 1.13. The mapping <p: ConL ----t '2:L, <p(a) = A(a) is acomplete lattice homomorphism which separates 6.

PROOF. Since <p(a) = A(a) = A(wQ) E '2:L, the assertion is an immediateconsequence of Lemma 1.12.

2. The characterization of Cen (L) for atomistic algebraiclattices

In this section by means of principal congruences we define a relation R.This relation is used for the characterization Cen (L) of an atomistic algebraiclattice L which is equivalent to the characterization of direct decompositionsof L. This characterization is the subject of this section.

DEFINITION 2.1. On the set of atoms A(L) of the lattice L, we definethe relations f- and R as follows:

(i) a f- b ~ e(O,a) ~ e(O,b) (a,b E A(L)) ,(ii) aRb ~ there is acE A(L) such that a f- c and b f- c.

REMARK 2.2. The following properties of the above defined relationscan be easily obtained:

Acta Mathematica Hungal'ica 86, 2000


(i) a I- b ¢} A(e(O,a)) ~ A(e(O,b)) ¢} (O,b) E e(O,a).(ii) aRb ¢} e (0, a) 1\ e (0, b) # D ¢} A ( e (0, a)) n A ( e (0, b)) # 0.For an arbitrary relation p ~ X x X and for a set B ~ X we define p(B)

= {x E X 13b E B such that (b, x) E p}. We say that the set B is closedrelative to the relation p if p(B) ~ B, i.e. if for any b E B and x E X from(b,x) E p it follows x E B.

The following proposition will playa key role in the proofs of the mainresults:

PROPOSITION 2.3. (i) B ~ A(L) is closed relative to the relation I- iffSB = V{blb E B} is a standard element of Land A(SB) = B.

(ii) If B is closed relative to R, then ZB = V{b I bE B} is a central ele-ment of L. If z is a central element of L, then A(z) is closed relative toR.

PROOF. (i) Suppose that B ~ A(L), B # 0 is closed relative to 1-. Thenby Remark 2.2(i) for any b E B we have A( e(O, b)) ~ B. Now let us considerthe congruence TB = V e(O, b). It is clear that B ~ A(TB)' On the other

bEBhand, according to Lemma 1.12(ii) we have

A(TB) = U A(e(O,b)) ~ B.bEB

Hence B = A( TB), and now by Proposition 1.11 we get that SB = V {b IbE B} = V {b I bE A(TB)} is a standard element of L, and A(SB) = A(TB)=B.

Conversely, let us assume that SB = V {b I bE B} E S(L) and B = A(SB)'Set now x E Band y E A(L) such that x I- y. Then e(O,y) ~ e(O,x) andx E A(SB) implies e(O,x) ~ e(SB), thus we get (O,y) E e(SB)' But nowA( e(SB)) = A(SB) = B implies y E B, whence we get that the set B is closedrelative to 1-.

(ii) Set z E Cen(L), x E A(z) and assume that (x,y) E R. By the def-inition of R this implies the existence of acE A(L) such that x I- c andy I- c. Since z E S(L), according to (i) we have c E A(z). Now we prove thaty rt A(z) is impossible:

Indeed, by [7] we have A(z') = A(L) \ A(z). Since z' E S(L), assumingthat y rt A(z), from y E A(z') and y I- c we obtain c E A(z'), i.e. crt A(z)which is a contradiction. Thus we must have y E A(z) and we conclude thatA(z) is closed relative to R.

Conversely, assume that B ~ A is closed relative to R. Then by the defi-nition of R and I- the sets Band A(L) \ B are closed relative to I- also, thus weget SB = V{bl bE B} E S(L), s'B = V {bl bE A(L) \ B} E S(L) and A(SB)= B, A(s'B) = A(L) \ B. Since now we have A(SB 1\ s'B) = A(SB) n A(s'B) = 0


8 s. RADELECZKI

and A(SB V s'B) = A(SB) U A(s'B) = A(L), we obtain that SB 1\ s'B = 0 andSB V s'B = 1, i.e. that s'B = s~. Now by Remark 1.1(ii) we can write e(SB)1\ e(s~) = e(SB 1\ s~) = t::, and e(SB) V e(s~) = e( SB V s~) = \7, i.e. e(SB)and e( s~) are complements of each other in Con L. Since any two stan-dard congruences of a lattice are permutable (see [3]), e(SB) and e(s~) mustbe factor congruences of L and now in view of Remark 1.1(iii) and (ii) weconclude that SB, s~ E Cen (L). 0

THEOREM 2.4. (Cen(L),I\, V) ~ (P(A(L)/R),n,u).

PROOF. The set B ~ A(L) is closed relative to R iff it is closed rela-tive to R, which is an equivalence on A(L). Thus the subsets closed relativeto R of the set A(L) appear as unions of R equivalence classes and con-versely, the union of some R-equivalence classes is closed relative to R. Evi-dently, the sets of R-equivalence classes form a Boolean lattice with respect tothe operations nand u, which is isomorphic to (P(A(L)/R) ,n,u). Now,to finish our proof it is enough to prove the isomorphism (Cen (L), 1\, V)~ (ZR(A),n,U) where ZR(A) denotes the subsets of A(L) closed relativeto R.

In view of Proposition 2.3(ii) we can define the mapping

(; : Cen(L) ---7 ZR(A), (;(z) = A(z), \jz E Cen(L),

which is the restriction of the isomorphism a : (S(L),I\, V) ---7 (~L,n,U)introduced in Corollary 1.3 relative to the sub lattice Cen (L). In consequence(; is a one to one lattice homomorphism. Since by Proposition 2.3(ii) forany BE ZR(A) the element ZB = V{bl bE B} belongs to den(L) and (;(ZB)= A(ZB) = B, {;is onto. Hence the mapping (; is a lattice isomorphism, whichcompletes our proof. 0

COROLLARY2.5. (i) Cen (L) is a complete atomistic Boolean lattice andZo E L is an atom of the lattice Cen (L) iff A(zo) is an equivalence class ofR.

(ii) L is directly irreducible iff aRb holds for all a, bE A(L).

PROOF. The first part of the assertion (i) is a consequence of the isomor-phism Cen(L) ~ P(A(L)/R) proved in Theorem 2.4. Evidently, the atomsof Cen(L) correspond by isomorphism to the atoms of P(A(L)/R), whichare sets consisting of a single element, i.e. of a single R-equivalence class.Hence Zo E L is an atom of Cen(L) iff (;(zo) = A(zo) is a single R class.

(ii) By [3], L is directly irreducible if and only if Cen(L) = {OL, 1£}.If Cen (L) is a two element lattice then the above mentioned isomorphismimplies that A(L) / R is a one element set, i.e. the set A(L) itself is the onlyR-equivalence class.



3. Proof of the main results

It is well-known that for any lattice L both Tol L and Con L are pseu-docomplemented algebraic lattices (see for example [1]). We shall denote byT* (respectively by 8*) the pseudo complement of aTE Tol L (respectivelyof a 8 E ConL).

DEFINITION 3.1. A pseudo complemented lattice L (with o£ and 1d iscalled a Stone lattice if a* V a** = 1£ holds for all a E L (here a* stands forthe pseudo complement of a).

For an atomistic algebraic lattice L we can formulate:

PROPOSITION 3.2. For any T E Tol L, T* is a congruence and we haveT* = 8(s) for s = V {a Ia E A(T*)} E S(L) and A(T*) = A(L) \ R( A(T)).

PROOF. If T* = 6 or T* = V, then the assertion is satisfied trivially fors = o£ or s = 1£. Set T* E TolL \ {6, V} and let us consider the standardelement s = V { a Ia E A(T*)} and the congruence 8(s). Now by Proposition1.11, we have T* ~ 8(s) and A( 8(s)) = A(T*). On the other hand we canwrite .

A( 8(s) 1\ T) = A( 8(s)) n A(T) = A(T*) n A(T) = A(T* 1\ T) = 0.

Now Lemma 1.12(i) implies 8(s) 1\ T = 6, thus 8(s) ~ T* (according to thedefinition of T*). As a conclusion we get T* = 8( s) E Con L.

Take now a E A(L) \ R( A(T)) arbitrary; to prove a E A(T*) by defini-tion is enough to show 8(0, a) ~ T*, i.e. 8(0, a) 1\ T = 6. Indeed, if we assume8(0, a) 1\ T =I 6, then by Lemma 1.12(i) there is acE A(L) such that (0, c)E 8(0, a) 1\ T. But this fact means that c E A(T) and a f- c. Since the latterrelation implies (c,a) E R, we get a E R(A(T)) which is a contradiction.

Conversely, take a E A(T*), then 8(0, a) 1\ T ~ T* 1\ T = 6. Let us as-sume now that a E R( A(T)). In this case there are elements bE A(T) andc E A(L) such that a f- c and b f- c. Therefore we obtain

A(8(0,c)) ~ A(8(0,a)) nA(8(0,b)) ~ A(8(0,a)) nA(T)

=A(8(0,a)) nA(T)=A(8(0,a)I\T) =A(6)=0,

implying by Lemma 1.12(i) 8(0, c) = 6, which is a contradiction. Thus wecan conclude that a E A(L) \ R(A(T)). 0

Before proving our main theorem we establish the following:THEOREM 3.3. If L is an atomistic algebraic lattice, then Con L is a

Stone lattice if and only if R = R.PROOF. Let us assume that R = R. We have only to prove 8* V 8**= 'V

for arbitrary 8 E Con L.


10 s. RADELECZKI

For 8* = l:, the equality is immediate. For 8* i- l:, we can write

A(8*) = A(L) \ R(A(8)) = A(L) \ R(A(8)),

according to Proposition 3.2. Since the set R( A( 8)) is closed relative to R,its complement A(L) \R(A(8)) must be also closed. Now by Proposition2.3(ii) z = V{a Ia E A(L) \ R( A(8))} is a central element of L and we haveA(z) = A(L) \ R(A(8)) = A(8*). On applying Proposition 3.2 with T = 8we get 8* = 8(z).

Since 8(z) 1\ 8(z') = l:, implies 8** ;; 8(z'), we obtain 8* V 8** ;; 8(z) V8(z') = 8(z V z') = V. In consequence 8* V 8** = V holds.

Conversely, let us assume that Con L is a Stone lattice and let us provethat in this case the relation R is transitive:

First, notice that the relation 8* V 8** = V is equivalent to the fact thatthe congruences 8* and 8** are complements of each other in Con L. Since byProposition 3.2 there are 81, 82 E S(L) such that 8* = 8(81) and 8** = 8(82)and since any two congruences induced by standard elements permute, inthis case we conclude that 8* and 8** are factor congruences of L. Nowthe application of Remark l(iii) and (ii) gives 81,82 E Cen (L). Since A(8*)= A(8d and A(8**) = A(82), by Proposition 2.3(ii) we get that both A(8*)and A( 8**) are closed relative to the relation R.

Set now a, b, c E A(L) such that aRb and bRc. To show that R is tran-sitive we need only aRc. Let us consider R( {a}) = {x E A(L) I (a, x) E R}and assume that c rJ. R( {a}). In this case according to Remark 2.2(ii)we have 8(0, a) 1\ 8(0, c) = l:" implying 8*(0, a) i- l:, and 8(0, c) ~ 8*(0, a).Since A( 8*(0, a)) is closed relative to Rand c E A( 8*(0, a)), we get thatbE A(8*(0,a)) and a E A(8*(0,a)). But the latter relation implies (O,a)E 8(0, a) 1\ 8*(0, a), contrary to 8(0, a) 1\ 8*(0, a) = l:,. 0

Now we are ready to prove our Main Theorem.

THE PROOF OF MAIN THEOREM. (i) ~ (ii). According to [7], any atom-istic algebraic lattice L can be written as a direct product L ~ IILi, I i- 0,

iEIwhere any Li (i E I) is a directly irreducible atomistic algebraic lattice.

Now for any i E I we can write L ~ Bi X Li, where Bi = II Lj. SincejEI\{i}

in this case we also have ConL ~ ConBi x ConLi (see for example [2] or [3]),it is easy to show that from (i) follows that any Con Li is an atomic Stonelattice. Indeed, any Con Li is a pseudo complemented distributive lattice andthe fact that Con L is an atomic lattice immediately implies that any Con Lias a direct factor of it is an atomic lattice too. Thus we have only to verify 8*V 8** = Vi, for any 8 E ConLi (here Vi stands for the unit relation definedon Li).



Let us consider a E Con L, a = Lc,B; X e, where Lc,B; denotes the identicalrelation on Bi. (More precisely, a is that congruence of L, which correspondsby isomorphism to Lc,B; X e. )

It can be easily checked now that a* = VB; X e* and a** = Lc,Bi x e**,where VB; stands for the unit relation on Bi. Since a* Va** = Lc, by as-sumption, we obtain e* V e** = Vi.

Let us prove now that any Li (i E I) is a sub directly irreducible lattice.Let'Y be an atom of ConLi, we have to show that e ~ 'Y for all e E ConLi,e i- Lc,i. In contrary, let us assume that e /\ 'Y= Lc,i for some e E ConLi,e i- Lc,i. Then we have e* i- Vi and e* ~ 'Y> Lc,i. The latter relation im-plies e** i- Viand since e** ~ e, we have e** i- Lc,i. Thus we get e*, e**1. {Lc,i, vd· On the other hand, from the fact that ConLi is a Stone latticeit follows that e* and e** are complements of each other. By Proposition 3.2we have e* = e(81) and e** = e(82) for some 81,82 E S(L). Since e* and e**are congruences of Li induced by standard elements, they permute, thereforethey are factor congruences of Li, i.e. we have Li = Lde* x Lde**. Sincenone of this two factor lattices is trivial, we conclude that the lattice Li isdirectly reducible, contrary to pur assumption.

(ii) =? (iii). If L ~ IT Li (i E I), then it can be checked immediatelyiEI

that any atom p E A(L) is of the form p = (Pi)iEI' where Pk is an atomof Lk for some k E I and Pi = Oi for all i i- k. (Here 0i stands for theleast element of Ld Putting Bk = IT Li, we can write L ~ Bk X Lk

iEI\{k}and Con L ~ Con Bk x Con Lk. Form the congruence Lc,Bk x e(Ok,Pk) andtake a congruence e E ConL containing the pair (O,p). Since e = e1 x e2E ConBk x ConLk, it is easy to check now that e ~ Lc,Bk x e(Ok,Pk) holds.Since (0, p) = ((OBi' Ok), (OBi' Pk)) E Lc,Bk x e(Ok, Pk) we conclude that e(O, p)= Lc,Bk x e(Ok,]Jk)'

Since Lk is subdirectly irreducible, Con Lk has a unique atom, the leastnonzero element of Con Lk which we denote by Wk. It is easy to check nowthat W = Lc,Bk X Wk is an atom of ConL and W ~ Lc,Bk xe(Ok,pd = e(O,p).Take any atom a of ConL, with a ~ e(O,p). Then a is of the form a = a1x a2, and a ~ e(O,p) implies a1 = Lc,Bk' Since Wk ~ a2 we get W ~ a, W i- Lc"i.e. W = a. Therefore e(O, p) contains a unique atom of Con L (namely theatom w).

(iii) =? (i) Since by Lemma 1.12(i) for any e E ConL, e i- Lc, there existsat least one a E A(L) such that (0, p) E e, from (iii) evidently follows thatCon L is an atomic lattice. Thus we only have to check that Con L is a Stonelattice, which is equivalent (by Theorem 3.3) to the fact that the relation Rin Definition 2.1(ii) is transitive on A(L).

Set now a, b, c E A(L), (a, b) E Rand (b, c) E R arbitrary, and let 0 be theunique atom of Con L which is less than e(O, b). Since by Remark 2.2(ii) wehave A(e(O,a)) nA(e(O,b)) i- 0 and A(e(O,b)) nA(e(O,c)) i- 0, there are


12 S. RADELECZKI

x, Y E A(L) such that B(O, x) ~ B(O, a) 1\ B(O, b) and B(O, y) ~ B(O, b) 1\ B(O, c).By assumption there are atoms 51 and 52 of Con L satisfying the relations51 ~ B(O,x) and 52 ~ B(O,y). Since now 81,82 ~ B(O,b), by (iii) we must have81 = 82 = 8. Thus we get 5 ~ B(O, x) 1\ B(O, y) ~ B(O, a) 1\ B(O, c) hence B(O, a)1\ B(O, c) # 6. But according to Remark 2.2(ii) the last relation implies aRcwhich means that R is a transitive relation. D

In the following theorem we give a characterization of weakly modularatomistic algebraic lattices; this characterization can be considered as thesecond main result of this paper.

THEOREM 3.4. If L is an algebraic atomistic lattice, then the followingare equivalent:

(i) N(L) = S(L).(ii) L is weakly modular.(iii) For any a E A(L) the congruence B(O, a) is an atom of Con L.(iv) I- is an equivalence relation on A(L).PROOF. (i) ::::}(ii). Take a, b, c, dEL, b < a, d < c arbitrary. According

to Definition 1.7 we have to prove that the relation alb ~w c/ d implies theexistence of a pair c', d' E [d, c], d' < c', satisfying c' / d' ~w a/b.

Since by [7] in any atomistic algebraic lattice N(L) = Cen (L), the as-sumption (i) is equivalent to S(L) = Cen(L). Let now P = A(c) \A(d) andset So = V {p Ip E A( B(a, b)) }. By Proposition 1.11 we have So E S(L), andthus So E Cen(L). Therefore So has a complement s~ E Cen(L) and accord-ing to [7] we have A(s~) = A(L) \ A(so).

First let us show that P n A ( B( a, b)) = 0 is impossible. Indeed, ifwe assume that the latter equality holds, then A(so) = A( B(a, b)) impliesP ~ A(s~), i.e. V P = V{p IpEP} ~ s~. Since c = d V (V P), now we getd < c ~ d V s~, a fact which implies (d, c) E B(s~), i.e. B(d, c) ~ B(s~). On theother hand in view of Proposition 1.11, we have B(a, b) ~ B(so), thus B(a, b)I\B(d,c) ~ B(so)I\B(s~) = B(sol\s~) = 6, in contradiction with (a,b) E B(d,c)(a consequence of the assumption alb ~w c/d).

Since P nA( B(a, b)) # 0, there is an atom pEP such that (O,p) E B(a, b).In this case we have d < d V P and (d, d V p) E B(a, b), hence, according toTheorem 1.6, there are eo, e1, ... , et E [d, d V p] such that d = eo ~ e1 ~ ...~ et = d V P and ei+I!ei ~w alb for i E {O,... , t -I}. Since eo < et, we haveeio # eio+1 for some io E {O, ... , t - I}, and thus we get d ~ eio < eio+1 ~ c.If we substitute now d' = eio and c' = eio+1, we obtain c' / d' ~w a/b.

(ii) ::::}(iii). We have only to show that in the case when L is weaklymodular for any a E A(L) and x,y E L, x # y, the relation (x,y) E B(O,a)implies B(x, y) = B(O, a).

Let p be an atom of y such that p rJ. A(x). Then (O,p) = (p 1\ x, P 1\ y)E B(x, y) ~ B(O, a). Since the interval [O,p] does not contain elements differ-ent from 0 and p, by Theorem 1.6 (O,p) E B(O, a) implies p/O ~w a/O. SinceL is weakly modular there are c',d' E [0, a], d' < c' with c'/d' ~w p/O. Since

Acta Mathematica Hunga1'ica 86, 2000


a is an atom of L, we must have d' = ° and c' = a, whence we get thata/O ~w p/o, a fact which implies (0, a) E e(O,p). Thus we can write e(O, a)~ e(O,p) ~ e(x,y) ~ e(O,a), implying e(x,y) = e(O,a).

(iii) =} (iv). We have by definition a f- b {:> e(O, b) ~ e(O, a) for a, bE A(L). Since by assumption e(O,b) and e(O,a) are atoms in ConL, weobtain e(O, b) ~ e(O, a) {:> e(O, a) = e(O, b) and thus f- is evidently an equiva-lence relation on A(L).

(iv) =} (i). Let us check first that if f- is an equivalence relation, Randf- coincide. Indeed, f- is contained in R by definition. Conversely, if we have(a, b) E R for some a, bE A(L), then there is acE A(L) such that a f- c andb f- c. Now f- is an equivalence relation by assumption, whence we get a f- b,i.e. that R is contained in f-.

Let s E L be now an arbitrary standard element of L. By Proposi-~ tion 2.3(i) the set A( s) is closed relative to the relation f-, thus it must

be also closed relative to R. Applying now Proposition 2.3(ii) it follows thats = V {p Ip E A(s)} is a central element of L. Thus we get S(L) ~ Cen (L).Since the inclusions Cen (L) ~ N(L) ~ S(L) are straightforward, we concludeS(L) = N(L) = Cen (L). 0

COROLLARY 3.5. Any atomistic algebraic weakly modular lattice is a di-rect product of subdirectly irreducible lattices.

PROOF. By Theorem 3.4 in the present case e(O, a) is an atom of Con Lfor all a E A(L), thus the condition (iii) of Main Theorem is satisfied. Ap-plying Main Theorem (ii) we obtain the required statement.

REMARK 3.6. The geometric lattices are atomistic and algebraic by defi-nition. Since they are relatively complemented, they are also weakly modular(in the sense of Definition 1.7). Now we can obtain the well-known struc-ture theorem of geometric lattices (as indicated in the introduction) as aparticular case of Corollary 3.5.

Finally, we close our paper by presenting an interesting application ofTheorem 3.4 to another special class of atomistic lattices.

PROPOSITION 3.7. If L is an atomistic lattice, then any dltally standardelement of L is neutral.

PROOF. By [3] it is enough to show that any dually standard element ofL is standard. Let a E L be a dually standard element in L and set an x E Larbitrary. First of all, we prove that A(a V x) ~ A(a) U A(x).

Let us consider an atom p E A(a V x), p (j. A(a). To prove the aboveinclusion we have to show only p E A(x). Since x V p ~ x V a, we can writex = x V 0L = X V (a /\p) = (x V a) /\ (x Vp) = (x Vp). Thus we get p ~ x, i.e.P E A(x).


14 S. RADELECZKI

Applying now the above proven inclusion we can write

A( x 1\ (a V y)) = A(x) n A(a V y) ~ A(x) n (A(a) U A(y))

= A(x 1\ a) U A(y 1\ a) ~ A( (x 1\ a) V (y Va)),

i.e. we get x 1\ (a V y) ~ (x 1\ a) V (y Va). Since the inverse inequality holdsin any lattice, we conclude that a E L is a standard element of L. 0

COROLLARY 3.8. If L is an atomistic and dually atomistic algebraiclattice, then it is weakly modulaT.

PROOF. Since L is dually atomistic, applying the dual of Proposition 3.7we get that any standard element of L is neutral, i.e. that S(L) = N(L).Applying now Theorem 3.4 we obtain that L is weakly modular.

PROBLEM. Distributive pseudo complemented lattices satisfying theidentity

considered as universal algebras (L, V, 1\, *, 0, 1) form an equational class forany n E N. E.g. for n = 1 we obtain the class of Stone algebras. In [5] and [6]we can find a general characterization of those semi-discrete lattices whosecongruence lattices are of the class (Ln). According to Proposition 2.3, Stonealgebras which are the congruence lattice of an atomistic algebraic latticeL are characterized by the fact that the relation R (Definition 2.1(ii)) istransitive, i.e. R2 = R. Is it true that for an atomistic algebraic lattice L,ConL belongs to the class (Ln) iff the relation R on L satisfies an equalityof type RE(n)+1 = RE(n), where E(n) is a natural number depending on n?



References

[1J R.-J. Bandelt, Tolerance relations on lattices, Bull. Austral. Soc., 23 (1981), 367-38l.[2J S. Burris and R. P. Sankappanavar, A Course in Universal Algebra, Springer (New

York, 1981).[3J G. Gratzer, General Lattice Theory, Birkhauser Verlag (Basel-Stuttgart, 1978).[4J G. Gratzer, R. Lakser and E. T. Schmidt, Isotone maps as maps of congruences. I,

Acta Math. Hungar., 75 (1997), 105-135.[5J M. Raviar, Lattices whose congruence lattices satisfy Lee's identities, Demonstratio

Math., XXIV (1991), 247-26l.[6J M. Raviar and T. Katririak, Semi-discrete lattices with (Ln)-congruence lattice, Con-

tributions to Gen. Algebra, 7, Verlag Rolder-Pichler-Tempsky (Wien, 1991).[7J L. Libkin, Direct decompositions of atomistic algebraic lattices, Algebra Universalis,

33 (1995), 127-135.

(Received April 6, 1998; revised December 8, 1998)

c

INSTITUTE OF MATHEMATICSUNIVERSITY OF MISKOLCH-3515 MISKOLC-EGYETEMV AROSHUNGARYE-MAIL: [email protected]


SOME STRUCTURE THEOREMS FOR ATOMISTIC ALGEBRAIC …matradi/files/2000_1.pdf · 2 S. RADELECZKI This relation will prove a useful tool also in the proof of our Main Theorem and Theorem

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