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+1v ... v an. Denote by ccj the function of L defined by the formula
XOCj = Uj л {x
VЩ).
The maps
olj, j = 1, 2, ..., n, are called the decomposition functions related to decomposition (1), any
oljshall be called the decomposition function of L.
Let F(L) denote the smallest set satisfying (i) and (ii):
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 .
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 )
ai) ^ ( V i e Mj ai )л Л ( V i e Mj l e M —ÿ)ai )
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
xole l and x
ab 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
ab = 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(
xva)]
(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)
= (
ûv 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
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 ^
( Vx,)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)
axa
= (y' v b) л y (by modularity and y' ^ y)
= y' v (b
ay) = y' v [b л a
a(x v b)]
= y' v (a
ab).
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
2v ... 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
2v ... 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) =
au(- 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
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 =
Va{ and bj —
vat, then
— b± v b
2v ... 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
ab e l, the set {aô, bj is 1- independent. Therefore, l = ab + b. Now we shall prove that a
ae e l.
Indeed, using Property II we obtain
(a
aejotôot = (a
ae)ôot =
( d ae)a
=0.
Then a
ae e l, by the definition of F (I, a).
Conversely, let 1 = aô + b and a
ae 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
ab < aô
ab, and aô A b e l, so xô
ab e l. Hence, using Property III, we obtain x ô e l. Moreover, x л e e l, because x
ae ^ а
аее 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
ab).
Moreover, aô
a(d
ah) < aÔ
abe I, and therefore d — aô + (d
ab), but then
d is not /-indecomposable. Therefore d = aô, and so 1 = d + b.
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
ae 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
emma3. 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( av e)] v
(b a d).Therefore, 1 =
dv
e= [d л
( av e)]
v
( b a d )v
e — \_{dv
e) a ( av e)] v
{b a d) = ( av e) v
( b a d).Then
b=
b a[(a v
e)v
{b adf] =
e fv
{b a d),and so
bv
d=
e pv
d.Since 1
=
a ôv
b^
dv
b,so
bv
d= 1, therefore 1 = e/1 v
d.Clearly, the set
{ e f i , d }is redundant (that is,
e fand
dare not comparable). Furthermore, by (4), e/? л
d e l .Therefore, 1 = [6 л
( av
e)~]+
d.L
emma4. Let two decompositions of 1: (2) and
(5) 1
= d i + d 2 + . . . + d nbe 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
2v ... v d,-_i v d
i+1v ... v d
„ - 1v 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 - 2v 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•
Solution o f Gratzer’s problem 355
Hence, since
a v
d „[I — a v [b л
( d nv 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 ëa. Then
a=
ал (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
3v ... v an, then 1 = bi +(a
3v ... 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
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
ak((pn)e l.
Le m m a 6 .
Let x x, x
2eL, 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
1a = x
2oc.
Then
[а л (*! v by] v b = [a л (x
2v 6)] v b, and by modularity, we obtain
(15) Xj v b = x
2v b.
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) =
1ф л (x2 ф v b).
Since Xj ф < 1^ and x2^ ^ li/f, by modularity, we obtain (16) х хф v (
1ф л b) = х 2ф v (
1ф л b).
Observe that, by Property VI, there exists a z e L such that гф = \ф
ab 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
ab ^ к((р)ф ^ х хф, and similarly,
\ф
ab ^ х
2ф. Hence in view of (16) we obtain х хф = х
2ф. 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
emma7. 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
aa
ak((p”)e l.
Suppose that а Ф l<p" and а Ф a
ak{tpn). Then
й= 1<рЧ[а л /с (</>")], and а is not /-indecomposable, a contradition. Therefore a = \(pn or a — a
ak((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
ak((p) ^ \cpn
ak{(pn). By Lemma 5, 1 q>n
ak{(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.
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
1v 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
1ф, 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 +1v (хф)(р*. Therefore, by (20) we have x < xq>‘ v c ^ X(p, +1 v (хф) q
>1v c,
that is x ^ x(pl + i v c\ where с' =(хф)(р
1v 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
orollary1 (E. T. Schmidt [3]). Let L be a modular lattice of finite length. I f
1 = c/j Ü
2F ... -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.
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).