Ifa1/0 anda2/0 are strongly cofinitely⊕−supplemented, thenLis also strongly cofinitely⊕−supplemented

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Miskolc Mathematical Notes HU e-ISSN 1787-2413 Vol. 21 (2020), No. 1, pp. 81–89 DOI: 10.18514/MMN.2020.2904

COFINITELY⊕-SUPPLEMENTED LATTICES

C¸ I ˘GDEM BIC¸ ER AND CELIL NEBIYEV Received 20 March, 2019

Abstract. In this work, cofinitely⊕−supplemented and strongly cofinitely⊕−supplemented lattices are defined and investigated some properties of these lattices. LetL be a lattice and 1=⊕

i∈Iaiwithai∈L. Ifai/0 is cofinitely⊕−supplemented for everyi∈I, thenLis also cofinitely

⊕−supplemented. LetLbe a distributive lattice and 1 =a1⊕a2witha1,a2∈L. Ifa1/0 anda2/0 are strongly cofinitely⊕−supplemented, thenLis also strongly cofinitely⊕−supplemented. Let Lbe a lattice. If every cofinite element ofLlies above a direct summand inL, thenLis cofinitely

⊕−supplemented.

2010Mathematics Subject Classification: 06C05; 06C15

Keywords: lattices, compact elements, small elements, supplemented lattices

1. INTRODUCTION

Throughout this paper, all lattices are complete modular lattices with the smallest element 0 and the greatest element 1. LetLbe a lattice,a,b∈Landa≤b. A sublattice{x∈L|a≤x≤b}is called aquotient sublattice, denoted byb/a. An element a⁰ of a latticeL is called a complement ofa ifa∧a⁰ =0 and a∨a⁰=1, this case we denote 1=a⊕a⁰ (aanda⁰ also is called direct summandsofL). L is called a complemented latticeif each element has at least one complement inL. An element cofLis said to becompactif for every subsetX ofLsuch thatc≤ ∨X, there exists a finiteF ⊆X such that c≤ ∨F. A lattice L is said to be compactly generated if each of its elements is a join of compact elements. A latticeLis said to becompact if 1 is a compact element of L. An elementaof a latticeLis said to becofinite if 1/a is compact. An elementa ofLis said to besmall orsuperfluous and denoted byaLifb=1 for every elementbofLsuch thata∨b=1. The meet of all the maximal elements(6=1)of a latticeLis called theradicalofLand denoted byr(L).

An elementcofLis called asupplementofbinLif it is minimal forb∨c=1. ais a supplement of bin a latticeLif and only ifa∨b=1 anda∧ba/0. A lattice Lis said to besupplementedif every element ofLhas a supplement inL. Lis said to becofinitely supplementedif every cofinite element ofLhas a supplement inL. L is said to be⊕−supplementedif every element ofLhas a supplement that is a direct

c

2020 Miskolc University Press

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summand inL.We say that an elementbofL lies abovean element aofLifa≤b andb1/a.Lis said to behollowif every element(6=1)is superfluous inL, andL is said to belocalifLhas the greatest element(6=1). An elementaofLis called a weak supplementofbinLifa∨b=1 anda∧bL. A latticeLis said to beweakly supplemented, if every element ofLhas a weak supplement inL.Lis said to becofinitely weak supplemented, if every cofinite element ofLhas a weak supplement inL.

An elementa∈Lhasample supplementsinLif for everyb∈Lwitha∨b=1,ahas a supplementb⁰inLwithb⁰≤b. Lis called anamply supplemented lattice, if every element ofLhas ample supplements inL. It is clear that every supplemented lattice is weakly supplemented and every amply supplemented lattice is supplemented. A latticeLis said to bedistributiveifa∧(b∨c) = (a∧b)∨(a∧c)for everya,b,c∈L.

LetLbe a lattice. It is definedβ∗relation on the elements ofLbyaβ_∗bwitha,b∈L if and only if for eacht∈Lsuch thata∨t=1 thenb∨t=1 and for eachk∈Lsuch thatb∨k=1 thena∨k=1.

More details about (amply) supplemented lattices are in [1,2,7]. The definitions of cofinitely (weak) supplemented lattices and some properties of these lattices are in [1,2]. The definition of⊕−supplemented lattices and some properties of these lattices are in [5]. More results about (amply) supplemented modules are in [6,11].

Some important properties of⊕−supplemented modules are in [8,9]. The definition of⊕−cofinitely supplemented modules and some properties of these modules are in [4]. The definition ofβ∗ relation on lattices and some properties of this relation are in [10]. The definition ofβ^∗relation on modules and some properties of this relation are in [3].

Lemma 1. Let L be a lattice and a,b,c∈L with a≤b. If c is a supplement of b in L, then a∨c is a supplement of b in1/a.

Proof. Similar to proof of [7, Proposition12.2(7)].

Lemma 2([7, Lemma 7.4]). Let L be a lattice, a,b∈L and a≤b. If ab/0then aL.

Lemma 3 ([7, Lemma 7.5]). In a lattice L let c⁰ c/0 and d⁰ d/0. Then c⁰∨d⁰(c∨d)/0.

Lemma 4([7, Exercise 7.3]). If L is a lattice and a∈L, then r(a/0)≤r(L).

Lemma 5([7, Lemma 12.3]). In any modular lattice[(c∨d)∧b]≤[c∧(b∨d)]∨ [d∧(b∨c)]holds for every b,c,d∈L.

Lemma 6(See also [5]). Let L be a lattice, a,b∈L and a≤b. Then b lies above a if and only if aβ∗b.

Proof. (=⇒)See [10, Theorem 3].

(⇐=) Letb∨t=1 witht∈1/a. Sinceaβ∗b, a∨t=1 and since a≤t,t=1.

Henceb1/aandblies abovea.

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2. COFINITELY⊕−SUPPLEMENTED LATTICES

Definition 1. LetLbe a lattice. Lis called a cofinitely⊕−supplemented lattice, if every cofinite element ofLhas a supplement that is a direct summand ofL.

Clearly we can see that every⊕−supplemented lattice is cofinitely⊕−supplemented and every cofinitely⊕−supplemented lattice is cofinitely supplemented.

Proposition 1. Let L be a lattice. Then L is cofinitely⊕−supplemented if and only if for every cofinite b∈L, there exists a direct summand c of L such that b∨c=1 and b∧cc/0.

Proof. Clear from definition.

Proposition 2. Let L be a lattice. If every cofinite element of L has a weak supplement that is a direct summand of L, then L is cofinitely⊕−supplemented.

Proof. Letabe a cofinite element ofLandbbe a weak supplement ofainLthat is a direct summand ofL. Sincebis a weak supplement ofainL,a∧bLand since bis a direct summand ofL,a∧bb/0. Hencebis a supplement ofainLandLis

cofinitely⊕−supplemented.

Lemma 7(See also [5]). Let L be a lattice, and a,b∈L. If x is a supplement of a∨b in L and y is a supplement of a∧(x∨b)in a/0, then x∨y is a supplement of b in L. (See also [5]).

Proof. Sincexis a supplement ofa∨binLandyis a supplement ofa∧(x∨b)in a/0, then 1=a∨b∨x,(a∨b)∧xx/0,a= [a∧(x∨b)]∨yand(x∨b)∧y=a∧ (x∨b)∧yy/0. Here 1=a∨b∨x= [a∧(x∨b)]∨y∨b∨x=b∨x∨y. By Lemma 5,(x∨y)∧b≤[(y∨b)∧x]∨[(x∨b)∧y]≤[(a∨b)∧x]∨[(x∨b)∧y](x∨y)/0.

Hencex∨yis a supplement ofbinL.

Lemma 8. Let L be a lattice and 1= ⊕

i∈Iai with ai ∈L. If ai/0 is cofinitely

⊕−supplemented for every i∈I, then L is also cofinitely⊕−supplemented.

Proof. Let x be any cofinite element of L. Since 1/x is compact and 1 =

i∈I∨(x∨ai), there exists a finite subset F = {i₁,i₂, ...,in} of I such that 1 =

∨n

t=1(x∨ai_t) =x∨

n

t=1∨ai_t

. Since x is a cofinite element of L, x∨ _n−1

∨

t=1ai_t

is a

cofinite element of L. Then by ¹

x∨

n−1 t=1∨a_it

=

x∨

n−1 t=1∨a_it

∨a_in x∨

n−1 t∨=1a_it

∼= ^aⁱⁿ

a_in∧

x∨

n−1 t=1∨a_it

, a_i_n∧

x∨ n−1

t=1∨a_i_t

is a cofinite element ofa_i_n/0 and sincea_i_n/0 is cofinitely⊕−supplemented, ai_n∧

x∨

n−1 t=1∨ai_t

has a supplement xi_n that is a direct summand in

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a_i_n/0. Since 0 is a supplement ofx∨

n

∨

t=1a_i_t

inLandx_i_n is a supplement ofa_i_n∧

x∨ n−1

t=1∨ai_t

inai_n/0, by Lemma7,xi_n=xi_n∨0 is a supplement ofx∨ n−1

∨

t=1ai_t

inL. Sincexis a cofinite element ofL,x∨ n−2

t=1∨ai_t

∨xi_n is a cofinite element ofL.

Then by 1 x∨

_n−2

t=1∨a_i_t

∨x_i_n

= x∨

_n−2

t=1∨a_i_t

∨x_i_n∨a_i_n−1 x∨

_n−2

t=1∨a_i_t

∨x_i_n

∼= ain−1

a_i_n−1∧

x∨ _n−2

t=1∨a_i_t

∨x_i_n ,

ain−1∧

x∨ n−2

∨

t=1ai_t

∨xi_n

is a cofinite element ofain−1/0 and sinceain−1/0 is cofinitely⊕−supplemented,ain−1∧

x∨

_n−2

t=1∨ai_t

∨xi_n

has a supplementxin−1 that is a direct summand ina_i_n−1/0. Sincex_i_n is a supplement ofx∨

_n−1

t=1∨a_i_t

inLandx_i_n−1 is a supplement ofai_n−1∧

x∨

n−2 t=1∨ai_t

∨xi_n

inai_n−1/0, by Lemma7,xi_n−1∨xi_n is a supplement ofx∨

n−2

∨

t=1ai_t

inL. If so, xhas a supplement ∨ⁿ

t=1xi_t inLwherexi_t is a direct summand ofai_t/0 for everyt=1,2, ...,n. Sincexi_t is a direct summand of ai_t/0 for everyt=1,2, ...,nand 1=⊕

i∈Iai, ∨ⁿ

t=1xi_t is a direct summand ofL. HenceL

is cofinitely⊕−supplemented.

Corollary 1. Let L be a lattice, a₁,a₂, ...,an∈L and1=a₁⊕a₂⊕...⊕an. If ai/0 is cofinitely⊕−supplemented for every i=1,2, . . . .,n, then L is cofinitely⊕−supplemented.

Proof. Clear from Lemma8.

Lemma 9. Let L be a lattice, a∈L and a= (a∧a₁)⊕(a∧a₂)for every a₁,a₂∈ L with 1=a₁⊕a₂. If L is cofinitely ⊕−supplemented, then 1/a is also cofinitely

⊕−supplemented.

Proof. Letxbe a cofinite element of 1/a. Then 1/xis compact andxis a cofinite element ofL. SinceLis cofinitely ⊕−supplemented, there existy,z∈L such that 1=x∨y, x∧yy/0 and 1=y⊕z. Sincey is a supplement ofxinLanda≤x, by Lemma 1, a∨y is a supplement of x in 1/a. Since 1=y⊕z, by hypothesis, a= (a∧y)⊕(a∧z). Then(a∨y)∧(a∨z) = [(a∧y)∨(a∧z)∨y]∧[(a∧y)∨(a∧ z)∨z] = [y∨(a∧z)]∧[(a∧y)∨z] = (a∧y)∨[(y∨(a∧z))∧z] = (a∧y)∨[(y∧

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z)∨(a∧z)] = (a∧y)∨(0∨(a∧z)) = (a∧y)∨(a∧z) =a. Hence 1/ais cofinitely

⊕−supplemented.

Corollary 2. Let L be a distributive lattice. If L is cofinitely⊕−supplemented, then1/a is also cofinitely⊕−supplemented for every a∈L.

Proposition 3. Let L be a cofinitely⊕−supplemented lattice and r(L)be a cofinite element of L. Then there exist a₁,a2∈L such that1=a₁⊕a₂, r(a1/0)a₁/0and r(a₂/0) =a₂.

Proof. SinceLis cofinitely⊕−supplemented andr(L)is a cofinite element ofL, there exista₁,a₂∈Lsuch that 1=r(L)∨a₁=a₁⊕a₂ andr(L)∧a₁a₁/0. Then by Lemma4,r(a₁/0)≤r(L)∧a₁a₁/0.

Assumexbe a maximal(6=a₂)element ofa₂/0. Since 1/(a1∨x) = (a1⊕a₂)/(a1∨ x) = (a₁∨x∨a₂)/(a₁∨x)∼=a₂/[a₂∧(a₁∨x)] =a₂/[(a₂∧a₁)∨x] =a₂/x,a₁∨xis a maximal element(6=1)ofLand since 1=r(L)∨a₁≤a₁∨x, this is a contradiction.

Hencer(a₂/0) =a₂.

Definition 2. LetL be a cofinitely supplemented lattice. L is called a strongly cofinitely⊕−supplemented lattice if every supplement element of any cofinite element inLis a direct summand ofL.

Clearly we can see that every strongly cofinitely ⊕−supplemented lattice is cofinitely⊕−supplemented and every strongly ⊕−supplemented lattice is strongly cofinitely⊕−supplemented.

Lemma 10(See also [5]). Let a be a supplement of b in L and x,y∈a/0. Then y is a supplement of x in a/0if and only if y is a supplement of b∨x in L.

Proof. (=⇒) Let y be a supplement of x in a/0 and b∨x∨z=1 with z≤y.

Because ofx,y∈a/0 andz≤y,x∨z≤a. Sinceais a supplement ofbinL,a=x∨z.

Since andyis a supplement ofxina/0,z=y. Henceyis a supplement ofb∨xinL.

(⇐=)Letybe a supplement ofb∨xinL. So,b∨x∨y=1 and(b∨x)∧yy/0.

Sincex∨y≤aandais a supplement ofbinL,x∨y=aandx∧y≤(b∨x)∧yy/0.

Henceyis a supplement ofxina/0.

Proposition 4. Let L be a strongly cofinitely⊕−supplemented lattice. Then for every direct summand a of L, the quotient sublattice a/0is strongly cofinitely⊕−supplemented.

Proof. Sincea is a direct summand ofL, there existsb∈Lsuch that 1=a⊕b.

SinceL is cofinitely supplemented, we can see that 1/bis cofinitely supplemented.

Then by 1/b= (a∨b)/b∼=a/(a∧b) =a/0, a/0 is cofinitely supplemented. Let x be a cofinite element of a/0 and y be supplement of x in a/0. By Lemma 10, y is a supplement of b∨x in L. By _b∨x¹ = â∨b∨x_b∨x ∼= _a∧(b∨x)â = _(a∧b)∨xâ = â_x, b∨x

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is a cofinite element of L. Since L is strongly cofinitely ⊕−supplemented, y is a direct summand ofL. Here there exists z∈L such that 1=y⊕z. By modularity, a=a∧1=a∧(y⊕z) =y⊕(a∧z). Thusyis a direct summand ofa/0. Hencea/0

is strongly cofinitely⊕−supplemented.

Lemma 11. Let L be a distributive lattice and a₁,a₂∈L with1=a₁⊕a₂. If a₁/0 and a₂/0are strongly cofinitely⊕−supplemented, then L is also strongly cofinitely

⊕−supplemented.

Proof. Letb be a cofinite element ofL anda be a supplement ofb inL. Since L is distributive,a=a∧1=a∧(a1⊕a₂) = (a∧a₁)⊕(a∧a₂) holds. By Lemma 10, a∧a₁ is a supplement of (a∧a₂)∨b inL. Then we can see thata∧a₁ is a supplement of a₁∧((a∧a₂)∨b) in a₁/0. Since b is a cofinite element of L, we can see thata₁∧((a∧a₂)∨b)is a cofinite element ofa₁/0. Sincea₁/0 is strongly cofinitely⊕−supplemented,a∧a₁is a direct summand ofa₁/0. Similarly we can see thata∧a₂is a direct summand ofa₂/0. Since 1=a₁⊕a₂anda= (a∧a₁)⊕(a∧a₂), ais a direct summand ofL. HenceLis strongly cofinitely⊕−supplemented.

Corollary 3. Let L be a distributive lattice, a₁,a₂, ...,a_n ∈L and 1=a₁⊕a₂⊕ ...⊕an. If ai/0is strongly cofinitely⊕−supplemented for every i=1,2, . . . .,n, then L is strongly cofinitely⊕−supplemented.

Proposition 5. Let L be a cofinitely supplemented lattice. The following state- ments are equivalent.

(i)L is strongly cofinitely⊕−supplemented.

(ii) Every supplement element of a cofinite element of L lies above a direct summand in L.

(iii) (a)For every nonzero supplement element a which is a supplement of a cofinite element of L, a/0contains a nonzero direct summand of L.

(b)For every nonzero supplement element a which is a supplement of a cofinite element of L, a/0contains a maximal direct summand of L.

Proof. (i) =⇒(ii)Clear, since every element ofLlies above itself.

(ii) =⇒(iii) Letabe a nonzero supplement element which is a supplement of a cofinite element of L. Assumea is a supplement of a cofinite element bofL. By hypothesis, there exists a direct summandx ofLsuch that alies abovex inL. By Lemma6,aβ∗xand sincea∨b=1,x∨b=1. Sinceais a supplement ofbinLand x≤a,a=xandais a nonzero direct summand ofL.

(iii) =⇒(i)Letabe a supplement of a cofinite elementbofLandxbe a maximal direct summand ofLwith x≤a. Assume 1=x⊕y withy∈L. Thena=a∧1= a∧(x⊕y) =x⊕(a∧y)and by Lemma 10, a∧y is a supplement ofb∨xin L. If a∧yis not zero, then by hypothesis,(a∧y)/0 contains a nonzero direct summandc ofL. Herex⊕cis a direct summand ofLandx⊕c≤a. This contradicts the choice

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ofx. Hencea∧y=0 anda=x. Thusais a direct summand ofLandLis strongly

cofinitely⊕−supplemented.

Lemma 12. Let L be a lattice. If every cofinite element of L isβ∗equivalent to a direct summand in L, then L is cofinitely⊕−supplemented.

Proof. Letabe a cofinite element of L. By hypothesis, there existx,y∈L with x⊕y=1 andaβ∗x. Thena∨y=1. Leta∨t=1 witht≤y. Sinceaβ∗x,x∨t=1 and since x⊕y=1, t=y. Hence y is a supplement ofa in L andL is cofinitely

⊕−supplemented.

Corollary 4. Let L be a lattice. If every cofinite element of L lies above a direct summand in L, then L is cofinitely⊕−supplemented.

Proof. Clear from Lemma6and Lemma12.

Example1. Consider the latticeL={0,a,b,c,1}given by the following diagram.

a 1

0 c

b

ThenLis cofinitely supplemented but not cofinitely⊕−supplemented.

Example2. Consider the latticeL={0,a,b,c,d,e,1}given by the following diagram.

a 1

0 c

b

e d

ThenLis cofinitely supplemented but not cofinitely⊕−supplemented.

Example 3. Consider the interval [0,1]with natural topology. Let Pbe the set of all closed subsets of [0,1]. Pis complete modular lattice by the inclusion (See [1, Example 2.10]). Here ∧

i∈ICi = ∩

i∈ICi and ∨

i∈ICi = ∪

i∈ICi for every Ci ∈P (i∈I)

i∈I∪Ciis the closure of ∪

i∈ICi

. Let X ∈P and X∨Y = [0,1] withY ∈P. Then

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[0,1]−X⊂Y and sinceY is closed[0,1]−X⊂Y. LetX⁰= [0,1]−X. ThenX⁰∈P, X∨X⁰=X∪X⁰= [0,1]andX⁰⊂Y for everyY ∈PwithX∨Y = [0,1]. HenceXhas ample supplements in P (hereX⁰ = [0,1]−X is the only supplement ofX inP) and P is amply supplemented. Let A= [0,a]∈P with 0<a <1. Here A⁰ = [0,1]−A= [a,1]is the only supplement ofAinP. LetA⁰∨B=A⁰∪B= [0,1]

with B∈P. Since A⁰∪B= [0,1], [0,a) = [0,1]−A⁰ ⊂B and since B is closed, [0,a]⊂B. This casea∈Band sincea∈A⁰,A⁰∧B=A⁰∩B6=∅. HenceA⁰ is not a direct summand ofPandPis not⊕−supplemented (See also [5]). We can see that [0,1]is only a cofinite element ofL. HencePis strongly cofinitely⊕−supplemented.

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Authors’ addresses

C¸ i˘gdem Bic¸er

Ondokuz Mayıs University, Department of Mathematics, Kurupelit-Atakum, 55270 Samsun, Turkey E-mail address:cigdem bicer184@hotmail.com

Celil Nebiyev

Ondokuz Mayıs University, Department of Mathematics, Kurupelit-Atakum, 55270 Samsun, Turkey E-mail address:cnebiyev@omu.edu.tr