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ANNALES SOC1ETATIS MATHEMAT1CAE POLONAE Series I: COMMENTATIONES MATHEMAT1CAE XXVI (1986) ROCZNIKl POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO

Séria I: PRACE MATEMATYCZNE XXVI (1986)

An d r z e j Wa l e n d z ia k

(Warszawa)

Solution of Gratzer’s problem

1. Abstract. In this paper, Problem IV. 15 of Gratzer [2] is solved. We find here a common generalization of the KuroS-Ore Theorem (cf. [2], p. 163) and the Schmidt-Ore Theorem (cf.

[1], p. 168) for modular lattices. For modular lattices of finite lenght this is done in E. T.

Schmidt [3].

2. Introduction. Let L be a modular lattice with 0 and 1. Denote by v or л the join or meet in L, respectively. Let = be a congruence relation of L, and let / be the ideal kernel of = , that is I = [x e L : x = 0}.

A finite subset \a l , a2, ..., an} of L is said to be l-independent if, for each j — 1, 2, ..., n,

(i) aj and ax v ... v a ^ x v aJ+l v ... v a„ are not comparable, and

(ü) dj

a

(ax v ... v ! v aj+1 v ... v a j e / .

The element a e L is called an I-direct join of the elements ax, a2, a„, written a = al + a2+ ... +an, if a = ax v a2 v ... v a„ and \al , a2, ..., a„] is an /-independent subset of L.

A nonzero element a of L that cannot be represented as an /-direct join of two elements of L, is said to be I-indecomposable.

An element b is called an I-direct summand of a if a = b + x for some element x.

Let

(1) 1 = ûi + a

2

+ ... •+• a„.

For je { 1, 2, ..., n}, üj = ax v ... v a ^ x v a

J+1

v ... v an. Denote by ccj the function of L defined by the formula

XOCj = Uj л {x

V

Щ).

The maps

olj

, j = 1, 2, ..., n, are called the decomposition functions related to decomposition (1), any

olj

shall be called the decomposition function of L.

Let F(L) denote the smallest set satisfying (i) and (ii):

(2)

350 A. W ale n d ziak

(i) if (p is a decomposition function of L, then (peF{L\

(ii) if (p, \j/eF(L), then (pif/eF{L) (*).

Let a e L . We denote by F {I, a) the set of all functions <peF{L) such that aq> = a and from c ^ a, c(pel follows c e l.

De f in it io n

1. Let an /-indecomposable element a e L be an /-direct summand of 1. Let b an element of L such that

(2) 1 = a + b,

and let

(3) 1 = d + e

be an arbitrary /-direct decomposition of 1 with two summands. Let a, fi and ô, e be the decomposition functions related to decompositions (2) and (3), respectively. If adaeF (I, a) or otm eF(I, a), then we say that a satisfies the B- condition.

In this paper we prove the following

Th e o r e m 1.

I f

1 = a± -H ... an = bi + ... + bm

are two decompositions o f 1 as an l-direct join of I-indecomposable elements satisfying the В-condition, then for every at there is a bj such that

1 = + ... + a^ _ j + bj + ci,- + J + ... + a„

and n = m.

The case / = {0} yields the theorem of Schmidt and Ore, while / = L produces the result of KuroS and Ore.

3. Properties of the /-direct join. Let M be a finite set of indices. If {ay. j e M } is an /-independent subset of L, then the /-direct join of elements aj, j e M we denote by ]T a}.

j e M

I. Let M be a finite set o f indices and Mj, j = 1, 2, ..., k, be nonvoid

к

subsets of M with (J Mj = M and Mjl n M

j 2

= 0 for any

j = i

j i J i e l i , 2, k] such that j i # ; 2. I f a = J] a{ and bj = £ a,-, then

i eM i eM :

к J

a = I bi-

j = 1

i 1) <рф is the шар of L defined by х(<рф) = (х(р)ф, x e L .

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Solution o f Gratzer's problem 351

P ro o f. Obviously, for each bj and bt v ... v v v bj+l v ... v bk are not comparable. Moreover,

bj л V

l ^ j

bi = ( V a<)л ( V

i e Mj i e M —Mj

ai) ^ ( V

i e Mj

ai )л Л ( V

i e Mj l e M —ÿ)

ai )

(observe af ^ V at f°r 1 ^ anc* aPPty modularity)

l e M - Г}

= V (ai Л V

i e Mj l e M —{i}

al)ç L

к

Therefore, the set {blt b2, ..., bk} is /-independent, and clearly, a = ]T ty.

Let decomposition (2) of 1 be given, and let a, f be the decomposition functions related to decomposition (2).

II. Let x e L . Then x ^ a implies xot = x.

P ro o f. Indeed,

xol

— a

a

(x v b) (by modularity and x ^ a)

= x v (а л b) = x v

0

— x.

III. Let x e L . I f

xol

e l and x

a

b e l, then x e L P ro o f. Compute:

x v b = (a v b)

a

(x v b) (by modularity and b ^ x v b)

= [a

a

(x v b)] v b — xot v b = 0 v b = b.

Hence, x л (x v b) = x л b, i.e., x = x л b. This implies that x = 0, since x

a

b = 0, and so x e L

IV. For x e L , x ^ xot v x f.

P ro o f. Compute:

xa v x f

[a л (x v b)] v [b

a

(

xv

a)]

(observe [a л (x v b)] < x v a and apply modularity)

= {[а л (x v b)] v b} л (x v a)

(observe b < x v b and apply modularity)

= (

û

v b) л (x v b) л (x v û) = (x v я) л (x v b) ^ x.

m m

V. For x u x 2, ..., xme L , ( \ / x,)a = V *;a-

i= 1 /==1

m P ro o f. By Property IV, x, ^ x,a v x,/? ^ x,a v b. Then \ / x,

i = 1

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352 A. W a le n d zia k

m m m m

^ ( \J x, a) v b, and hence ( \ / x() a ^ а л ( \ / xf a v b). Since \ / xf a ^ a,

i = 1 l'=l i = 1 i = 1

by modularity,

m m m

а л ( \J x{ a v b) = ( \ / x,- a) v (a л b) = \ / x, a .

i = 1 i = l i= 1

m m

Therefore, ( \ / x, )a ^ \ / xfa.

i = l i = 1

m

On the other hand, x,a ^

( V

x,)a, and hence

i = l

m m

V x«a ^ (V x.-)a-

i = 1 i = 1

VI. Let (peF(L), x, y e L , and let xq> = y. Then for any y 'e L with y' ^ y, there is an x 'e L satisfying x' ^ x, x' <p = ÿ , and y' ^ x' q>.

P roof. First we shall prove the statement for q> = a. Let x'

= {ÿ v b) л x. By modularity, [x л ( / v b)] v b = (x v b) л (y' v b), i.e., x' v b = (x v b) л (y' v b). Now compute:

x' a = a л (x' v b) = (y' v b)

a

xa

= (y' v b) л y (by modularity and y' ^ y)

= y' v (b

a

y) = y' v [b л a

a

(x v b)]

= y' v (a

a

b).

Hence x'a = y' and y' < x'a.

Now let us assume the statement to hold for ф and let (p — ф<х. Let {хф)ос = у and ÿ ^ y. Then there exist x

1

^ хф such that x l <x = y' and y' < x x a. By the induction hypothesis there exist x' ^ x such that х'ф = х х and х х ^ х 'ф . Since y ' ^ x ^ and х х ^ х 'ф , we obtain that y '^ x 'tp . Furthermore, x' q> = (х'ф)<х = x x a = y'. This finishes the induction.

Let {ax, a2, ..., a„] a L. If a = ax v a

2

v ... v and, for each i

= 1, 2, ..., и, а,- л («! v ... v a,--! v ai + 1 v ... v a„)el, then we shall write a = ax v a

2

v ... v a„.

VII. Let a — ax v ... v a„ and let a{ = aiX v ... v aim. for i — 1, 2, ..., n.

Then a = \ / a{j (i = 1, 2, ..., n; j = 1 ,2 ,..., m,).

i.j

P roof. Indeed, by modularity,

% л (V ak v V au) =

a

u(- л (V % v V a.-/)

= a,7 A [(«,- A \J ak) v \ J aa] = аи л \ / оц = 0.

l * j l * j

VIII. Let M be a finite set o f indices and Mj, j = 1, 2, ..., k, be nonvoid

(5)

Solution o f Gralzer's problem 353

к

subsets of M with у Mj = M and MJl n M

j 2

= 0 for any j

j = t

j 2e

[1, 2,

k} such that j i ^ j 2- I f a =

V

a{ and bj —

v

at, then

— b± v b

2

v ... v bk.

i eM ieM:

1>

a

P ro o f. This proof is obvious from Property I.

4. Preliminary lemmas. Let two decompositions (2) and (3) of the unit element of L be given. Let a, ft and Ô, e be the pairs of decomposition functions related to decompositions (2) and (3), respectively.

Lemma

1. The following conditions are equivalent:

(i) a<5aeF(/, a).

(ii) 1 = [d л (a v ef\ + b.

Proof. Let a b a e F (/, a). Then aixôot = a. Hence а л (ab v b) = a. Thus [с/ л {aS v b)] v b = (a v b) = 1. Since aô v b ^ b, by modularity,

[a л (aô v b)] v b = (a v b) л (ab v b), and so 1 = ub v b.

We will prove that x = aô л b e l. By Property VI, there exists x' ^ a such that x'b = x. Hence using Property II we obtain

x ’otôoL = x'ba = xa = а л (x v b) = а л b = 0.

By the fact that x'aba = 0 we infer that x 'e /. Hence x 'ô e l, and so x e l . Since aô and b are not comparable, and aô

a

b e l, the set {aô, bj is 1- independent. Therefore, l = ab + b. Now we shall prove that a

a

e e l.

Indeed, using Property II we obtain

(a

a

ejotôot = (a

a

e)ôot =

( d a

e)a

=

0.

Then a

a

e e l, by the definition of F (I, a).

Conversely, let 1 = aô + b and a

a

e e l. It is easy to see that actÔot = a.

Suppose now that x ^ a and xaba e I. By Property II, we have xa = x. Thus, xd a el. Since xô

a

b < aô

a

b, and aô A b e l, so xô

a

b e l. Hence, using Property III, we obtain x ô e l. Moreover, x л e e l, because x

a

e ^ а

а

ее I, and hence (by Property III) x e l .

Lemma 2.

I f

1

—a + b

=

d

+

e, where d is I-indecomposable, then a b a e F (/, a) if and only if 1 = d + b = a + e.

P ro o f. Necessity. From Lemma 1 we conclude that 1 — aô + b. First we shall prove that aô = d. Suppose on the contrary that d Ф aô. Since 1 = aÔ + b, so d = d

a

(aô v b), and hence, by modularity, d = aô v (d

a

b).

Moreover, aô

a

(d

a

h) < aÔ

a

be I, and therefore d — aô + (d

a

b), but then

d is not /-indecomposable. Therefore d = aô, and so 1 = d + b.

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354 A. W a le n d zia k

Moreover, 1 = ad + e = d

a

(a v e) + e = {a v e) л (d + e) = a v e, and since by Lemma 1, a

a

e e l, so 1 = a + e.

Sufficiency. Let 1 — d + b = a + e. Hence, 1 = d + b = [cl л (a + e)]+ 6, and by Lemma 1, we conclude that atôoteF(I, a). Thus the proof is completed.

L

emma

3. Let two decompositions (2) and (3) of 1 be given and let a satisfy the В-condition. I f ада ф F (I, a), then

1 = [b л (a v e)] + d.

P ro o f. By Definition 1, we conclude that ocôcteF {I, a). In view of Lemma 1, we have

(4) 1 = d

a

{a v e) + b.

Hence,

d

= [d

a( a

v e)] v

(b a d).

Therefore, 1 =

d

v

e

= [d л

( a

v e)]

v

( b a d )

v

e — \_{d

v

e) a ( a

v e)] v

{b a d) = ( a

v e) v

( b a d).

Then

b

=

b a

[(a v

e)

v

{b a

df] =

e f

v

{b a d),

and so

b

v

d

=

e p

v

d.

Since 1

=

a ô

v

b

^

d

v

b,

so

b

v

d

= 1, therefore 1 = e/1 v

d.

Clearly, the set

{ e f i , d }

is redundant (that is,

e f

and

d

are not comparable). Furthermore, by (4), e/? л

d e l .

Therefore, 1 = [6 л

( a

v

e)~]

+

d.

L

emma

4. Let two decompositions of 1: (2) and

(5) 1

= d i + d 2 + . . . + d n

be given. Let a, ft and Sh i = 1, 2, ..., n, be the decomposition functions related to decompositions (2) and (5), respectively. I f a satisfies the В-condition, then there exists an ie {1, 2, ..., n} such that а<5; а е Т ( /, a).

P ro o f. We prove Lemma 4 by induction on n. If n = 2, then Lemma 4 follows from Definition 1. Let us assume this lemma for n— 1 and let а<5„а^F {I,a ). By Lemma 3, we obtain

(6) 1 = (d i+ ... +d„_i) + b л (dH v a).

Obviously, there is an i, 1 < / < n — 1 such that dt and d\ v d

2

v ... v d,-_i v d

i+1

v ... v d

„ - 1

v dnf are incomparable. Let for example i = n — 1. From (6) we conclude that dn- x A d e l, where d

= d± v ... v d

n - 2

v d„f. Therefore

(7) 1 = d n_

1

+ d.

Let £ '_ г, S be the pair decomposition functions related to decomposition (7).

Since a satisfies the В-condition, so адщ-^аеF(I, a) or aSaeF(I, a).

We shall consider two cases.

C ase 1. Let aân- t a e F (I, a). By Lemma 1,

1 =

a

[a v d ' v dnfF\ +

b ,

where d' = dx + ... +dn- 2•

(7)

Solution o f Gratzer’s problem 355

Hence, since

a v

d „

[I — a v [b л

( d n

v a)] = (a v b) л (d„ v a) = a v

d n ,

we obtain 1 = d„_, л (a v d' v */n) + b.

From this we conclude, by Lemma 1, that oieF(I, a).

C ase 2. Suppose that a b ^ a £ F ( / , a). Applying Lemma 3 to (7), we conclude that 1

= d + b a

(d„_! v a). Without any loss of generality we can assume that the set

{ d t ,

..., d„_2, e'}, where ё =

dn- !

/? v

d n f i ,

is irredundant. Therefore

(8) 1 = d i + d 2 + • •. + d „ _ 2 + ё .

We denote by S-, ё, i = 1, 2, ..., n — 2, the decomposition functions related to decomposition (8). Observe that

t x ë c t $ F ( I , a).

Indeed, suppose on the contrary that

a

=

a a ë

a. Then

a

=

а

л (aae' v b) = a л b (since aae' ^

b).

Hence

a

^ b, a contradition.

By the induction hypothesis, there is an i, —2, such that ab('a e F ( / , a). Let for example, i = 1. Applying Lemma 1 to (8), we conclude that 1

= d x a

[a v

( d 2 + . . . + d„_ 2 +

e')]

+ b .

Since

a v

d n —

j /1 = a v [b л (d„_! v a)] = (a v

b )

л (a v d„_!) = a v </„_!

and similarly, a v dnfi = a v d„, so a v ë = a v */„_! v d„. Therefore, 1 = dx л [я v (d2+ . . . + d„)] + b. Hence, by Lemma 1, ab1a e F ( /,a ) . This ends the proof of Lemma 4.

In the proof of this lemma we applied Properties I, VII and VIII several times.

5. Proof of the theorem. Let a,, / = 1, 2, ..., n, pJt j = 1, 2, ..., m, be the decomposition functions related to decompositions (1) and

(9) 1 —

b t + b 2 +

• • .

+ b m ,

respectively. By Property I,

(10) 1 = al + al .

Applying Lemma 4 to (10) and (9), we conclude that there exists a je { 1, 2 „ ..., m} such that a 1^J a 1e F ( /, аД Let for example j = 1. Then, by

Lemma 2,

(11) 1 = bt = ax + b ,.

Observe that the set {bl5 a2, ..., an} is irredundant. Indeed, if for instance a

2

^ b l v a

3

v ... v an, then 1 = bi +(a

3

v ... v an). By Lemmas 4 and 2, there exists an гe {1, 2, ..., n} such that 1 = а г+ (а3+ ...+ а„). This means that the set {at , a2, a„] is redundant, contrary to our assumptions.

11 — Prace Matematyczne 26.2

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356 A. W a le n d zia k

Therefore the set {h1? a2, a„} is irredundant, and with (11) we have

1

= Ь j + d

2

~Ь Й

3

+ ... + an, proving the first statement.

Repeating this we eventually obtain l = b j + . . . + b j , and so

\jl , . . . , j n} = {1, 2, m}. This shows that n = m.

6. Modular finite-Ienght lattices. Suppose that the lattice L is of finite lenght. Let q>eF(L). We denote by k((p) the join of all x e L such that xq>el.

By Property V, we have

(12) k((p)(pel.

Note that

(13) k{(pn) ^ k ( ( p n+1) for all w = l ,2 , ...

Indeed, k((pn)(p

n+1

= (k(</>")(pn)(p e/, and by the definition of k(q>n+l) we get inequality (13).

Lf.m m a

5. Let n he a natural number. If k((pn) — k((pn+l), then 1 (pn л k((pn)e l.

Proof. We proof by induction on i that k{(pn + l) = k((pn). This is true for i = 1. We suppose that k{(pn+l) — k{(pn). By (12), we conclude that [k((pn+l+ *) qy] (pn+le /. Therefore, k((pn+l +1) q> ^k((pn+l) = k(cpn), and hence k((pn+l + l)(p

n+1

^ k((pn) (pnE I. Thus k((pn+, + l ) ^ k((pn+l). Moreover, by (13), k((pn+l) ^ k{(pn+, + i) and we conclude that k((pn+t+1) = k((pn+l) = k{(pn).

Thus, by induction, we obtain k{(pn+l) = k((pn) for all / = 1 ,2 ,... In particular,

(14) k((p2n) = k((pn).

By Property VI, we conclude that there exists a y e L such that y(pn

= 1 (pn л k((pn). Hence y((p2n) = [1 cpn л k((pn)'](pn ^ k(q>n)(pne l , and thus y(plne l, that is y ^ k ( ( p 2n). Therefore, using equality (14) we get y ^ k { ( p n).

Hence yep"el, and thus 1 (pn

a

k((pn)e l.

Le m m a 6 .

Let x x, x

2

eL, x

1

^k((p), x

2

^k((p). I f x 1cp = x

2

(p, then

* 1

= x 2.

P roof. We use induction on the lenght of q>. Let decomposition (2) be given and let a, /? be the pair of decomposition functions related to decomposition (2). Let *1 ^ k(ot) and x

2

^ k(a). Suppose that x

1

a = x

2

oc.

Then

[а л (*! v by] v b = [a л (x

2

v 6)] v b, and by modularity, we obtain

(15) Xj v b = x

2

v b.

(9)

Solution o f Gratzer s problem 357

Since bot = а л b e l, so b ^ к (a). Therefore, x x ^ b and x

2

^ b. Hence in view of (15) we obtain x x = x2. Thus, for <p = a, the proof of Lemma 6 is completed.

Now let us assume the statement holds for ф and let q> = фос. Let X! q>

= x2tp (xx ^ к (tp), x

2

^ k(<p)), that is а л (х1ф v b) = а л (х2ф v b). Then х хф v b = х 2ф v b, and hence

\ф л (

х

^ ф v b) =

1

ф л (x2 ф v b).

Since Xj ф < 1^ and x2^ ^ li/f, by modularity, we obtain (16) х хф v (

1

ф л b) = х 2ф v (

1

ф л b).

Observe that, by Property VI, there exists a z e L such that гф = \ф

a

b and \ф л b ^ zi/f. Therefore, we have

z<p = гфа = (It// л b)ot = а л Ь =

0

.

Hence ztpel, that is z < /c(<p). Thus 1ф

a

b ^ к((р)ф ^ х хф, and similarly,

a

b ^ х

2

ф. Hence in view of (16) we obtain х хф = х

2

ф. Obviously, we have к(ф) ^ k((p), and therefore к(ф) ^ x l5 к(ф) ^ x2. Applying the induction hypothesis we obtain x, = x 2. This ends the proof of Lemma 6.

L

emma

7. Let a be an I-direct summand of 1. I f a is 1-indecomposable, then a satisfies the B-condition.

P ro o f. Let two decompositions (2) and (3) of 1 be given. Let a, ft and

<5, e be the pairs of decomposition functions related to decompositions (2) and (3), respectively. Set tp = otôot, and cp = oteot. Since L is a lattice of finite lenght, so 1 (pn = 1 (pn+l and k((pn) — k((pn+x) for some integer n ^ 1. Observe that (17) 1<ри = [1<р" v k{<pnf\<pn.

Indeed, [1 <pn v /c(<p")] <p" = (by Property V) — (1 (pn)(pn v k((p")(pn = (since (1 tp”)<pn — 1 tpn) = 1 (pn v k{(pn)tpn = (sineek((pn)(pn ^ 1 (pn) = \tpn.

From (17) we conclude, by Lemma 6, that 1 = 1 cpn v k((pn). Hence, since Lpn ^ a, by modularity, a = l(pn v [a

a

/c(<p")]. Moreover, by Lemma 5, 1 cpn

a

a

a

k((p”)e l.

Suppose that а Ф l<p" and а Ф a

a

k{tpn). Then

й

= 1<рЧ[а л /с (</>")], and а is not /-indecomposable, a contradition. Therefore a = \(pn or a — a

a

k((pn).

We shall consider two cases.

C ase 1. Let a — l<pn. We shall prove that (p e F (I,a ). Since a ^ a tp

= Itp ^ l<p” = a, so atp — a. Suppose now that x ^ a and xtp el. Hence x ^ a

a

k((p) ^ \cpn

a

k{(pn). By Lemma 5, 1 q>n

a

k{(pn)e l, thus x e l .

C ase 2. Let a — a A k ((p n) and а Ф l(pn. We shall prove that cp e F (I ,a ). Observe that

(18) x ^ xcp v x<p for all x ^ a.

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358 A. W a le n d zia k

Indeed, by Property IV, xa ^ xa<5 v xae. Hence, by Property V, xaa ^ xaôcc v xaea.

But x ^ xaa, and (18) is proved.

Now we prove by induction on i that

(19) a < atp

1

v аф for j = 1, 2, ...

This is true for i = 1. Suppose that a < скр' v аф. From (18) we obtain aq

) 1

^ a<pl q> v acp

1

ф, and hence

aç>' v аф ^ atpl + i v a<p.

Applying the induction hypothesis we get a ^ atp

1

+ 1 v a<p. This finishes the proof of (19).

Thus, in particular, a ^ a(pn v аф. On the other hand, a(pn v аф < a.

Therefore, я = v аф. Moreover, а<рп л афе1. Indeed, a<p" л аф < л a (by assumption)

= a<pn д а л A; (</>") ^ l<p” л k((pn) e l,

by Lemma 5. Finally, a = аф, otherwise a = 1 (рп+аф, contradicting a is an /-indecomposable element of L.

Suppose now that x ^ a and хфе1. We shall prove that for every natural number i there exists an element c e l such that x ^ x<p‘ v c.

For i = 1, this proof is obvious from (18). Let (20) x ^ x(p‘ v c for some c e l.

From (18) we obtain that xq>‘ ^ x(p

i +1

v (хф)(р*. Therefore, by (20) we have x < xq>‘ v c ^ X(p, +1 v (хф) q

>1

v c,

that is x ^ x(pl + i v c\ where с' =(хф)(р

1

v c e l. Thus, in particular, there exists an element c e l such that x < x(pn v c. Finally,

x(pn ^ l(pn = l(pn л а л k((pn) ^ l(pn л k((pn) e l.

by Lemma 5. Hence x e l , and therefore ф е¥(1,а). Thus the proof of Lemma 7 is completed.

Combining this lemma and Theorem 1 we get at once

C

orollary

1 (E. T. Schmidt [3]). Let L be a modular lattice of finite length. I f

1 = c/j Ü

2

F ... -b am and 1 = b^ + A

>2

+ ... -j- bn

are two decompositions of the unit element of L as an 1-direct join of I- indecomposable elements, then n = m and for every at there is a bj such that

1 = a j -t- ... 4- üj _ i -j- b j -f- <zf

+ 1

-t- ... + an.

(11)

Solution o f Grdtzer's problem 359

R em ark. The case / = {0} yields the theorem of Schmidt and Ore (cf.

G. Birkhoff [1]).

Observe that if / = L, then a set [x{, x 2, ..., xk] cz L is /-independent if and only if it is irredundant. Furthermore, in this case, an element x e L — {0}

is /-indecomposable if and only if it is join-irreducible.

From Theorem 1 we obtain now

Corollary

2 (KuroS-Ore theorem). Let L be a modular lattice. I f 1 = fl1 v ... v an and 1 = hi v ... v bm are irredundant representations of 1 as joins of join-irreducible elements, then for every at there is a bj such that

l = at v ... v a,_! v bj v ai + l v ... v an and n = m.

P ro o f. We shall prove that, if a is a join-irreducible summand of 1, then a satisfies the B-condition.

Let two decompositions (2) and (3) be given, let a, /? and <5, e be the pairs of decomposition functions related to decompositions (2) and (3), respectively. From Property IV we have a ^ aô v as. Hence (b v aô) v (b v as) ^ b v a = 1, that is

1 = (b v aô) v (b v as).

By the Isomorphism Theorem (cf. Grà'tzer [2]) the lattices [h, 1] (2) and [u л b, a] are isomorphic. But a is join-irreducible in L, and, therefore, in [а л b, a], thus 1 is join-irreducible in [fe, 1]. Hence 1 = b v ad or 1 = b v as.

If 1 = b v aô, then axâcc = аба = а л {aô v b) = a, and therefore

<xô<xeF(L, a). Similarly/if 1 — b v as, then ccsoceF(L, a). Thus a satisfies the В-condition.

The statement now follows from Theorem I.

References

[1] G. Birkhoff, Lattice theory, Third Edition, Providence 1967.

[2] G. Grà'tzer, General lattice theory, Berlin 1978.

[3] E. T. Schmidt, Eine Verallgemernerung des Scitzes von Schmidt-Ore, Publ. Math. Debrecen 17 (1970), 283-287.

(2) (X v] = {ze L : x ^ z ^ y} (x, y e L, x < y).

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