In this work, cofinitely radical supplemented and cofinitely weak radical supplemented lattices are defined and some properties of them are investigated

Vol. 21 (2020), No. 2, pp. 993–999 DOI: 10.18514/MMN.2020.3219

COFINITELY RADICAL SUPPLEMENTED AND COFINITELY WEAK RADICAL SUPPLEMENTED LATTICES

CELIL NEBIYEV AND HASAN H ¨USEYIN ¨OKTEN Received 04 February, 2020

Abstract. In this work, cofinitely radical supplemented and cofinitely weak radical supplemented lattices are defined and some properties of them are investigated. LetL be a lattice, I be a nonempty index set andai∈Lfor everyi∈I. If 1= ∨

i∈Iaiandai/0 is cofinitely (weak) radical supplemented for everyi∈I, thenLis also cofinitely (weak) radical supplemented. LetLbe a cofinitely (weak) radical supplemented lattice anda∈L. Then 1/ais also cofinitely (weak) radical supplemented. LetLbe a lattice. ThenLis cofinitely weak radical supplemented if and only if every cofinite element of 1/r(L)is a direct summand of 1/r(L).

2010Mathematics Subject Classification: 06C05; 06C15

Keywords: lattices, small elements, supplemented lattices, generalized (radical) 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 sub- lattice{x∈L|a≤x≤b}is called aquotient sublattice, denoted byb/a. An element a⁰of a latticeLis called acomplementofaifa∧a⁰=0 anda∨a⁰=1 (in this casea anda⁰are said to bedirect summandsofLand denoted by 1=a⊕a⁰). A latticeLis said to becomplementedif each element ofLhas at least one complement inL. An elementcofLis said to becompactif for every subsetXofLsuch thatc≤ ∨X there exists a finiteF⊂X such thatc≤ ∨F. A latticeLis said to becompactly generated if each of its elements is a join of compact elements. A latticeLis said to becompact if 1 is a compact element ofL. An elementaof a latticeLis said to becofinite if 1/a is compact. An elementaofLis said to besmall orsuperfluous ifa∨b6=1 holds for everyb6=1 and denoted byaL. The meet of all the maximal(6=1)elements of a latticeL is called theradicalofL and denoted byr(L). An element cofL is called asupplementofbinLif it is minimal forb∨c=1.ais a supplement ofbin a latticeLif and only ifa∨b=1 anda∧ba/0. Lis called asupplemented lattice if every element ofLhas a supplement inL. We say that an elementbofL lies above an elementaofLifa≤bandb1/a.Lis said to behollowif every element(6=1)

c

2020 Miskolc University Press

elementaofLis called aweak supplementofbinLifa∨b=1 anda∧bL. Lis called aweakly supplemented lattice, if every element ofLhas a weak supplement in L. It is clear that every supplemented lattice is weakly supplemented. An elementa ofLis called ageneralized(radical)supplement(or briefly,Rad-supplement) ofbin Lifa∨b=1 anda∧b≤r(a/0). Lis said o beradical(generalized)supplemented if every element ofLhas a Rad-supplement inL.

More information about supplemented lattices are in [1,2] and [5]. More res- ults about supplemented modules are in [9] and [10]. The definitions of generalized supplemented modules and some properties of them are in [8]. More information about cofinitely Rad-supplemented modules are in [7]. We generalize cofinitely Rad- supplemented modules to lattices.

2. COFINITELYRADICAL SUPPLEMENTEDLATTICES

In this part, cofinitely radical supplemented lattices are defined and some proper- ties of them are given.

Definition 1. LetLbe a lattice. If every cofinite element ofLhas a Rad-supplement in L, then L is called a cofinitely radical supplemented (or cofinitely Rad-supple- mented) lattice.

Clearly we can see that every cofinitely supplemented lattice is cofinitely Rad- supplemented. Hollow and local lattices are cofinitely Rad-supplemented.

Proposition 1. Let L be a compact lattice. Then L is cofinitely Rad-supplemented if and only if L is Rad-supplemented.

Proof. Clear, since every element ofLis cofinite.

Lemma 1. Let L be a lattice, a∈L and x be a cofinite element of L. If x∨a has a Rad-supplement in L and a/0cofinitely Rad-supplemented, then x has a Rad- supplement in L.

Proof. Letbbe a Rad-supplement ofx∨ainL. Thenx∨a∨b=1 and(x∨a)∧b≤ r(b/0). Since x is a cofinite element of L, we clearly see that x∨b is a cofinite element of L. Then by _x∨b¹ = ^x∨a∨b_x∨b ∼= _a∧(x∨b)^a , a∧(x∨b) is a cofinite element of a/0. Since a/0 is cofinitely Rad-supplemented,a∧(x∨b)has a Rad-supplement c ina/0. Here(a∧(x∨b))∨c=aandc∧(x∨b) =c∧a∧(x∨b)≤r(c/0). Hence 1=x∨a∨b=x∨b∨(a∧(x∨b))∨c=x∨b∨candx∧(b∨c)≤(b∧(x∨c))∨ (c∧(x∨b))≤(b∧(x∨a))∨r(c/0)≤r(b/0)∨r(c/0)≤r((b∨c)/0). Thusb∨c

is a Rad-supplement ofxinL.

Corollary 1. Let L be a lattice, a₁,a₂, . . . ,an∈L and x be a cofinite element of L. If x∨a₁∨a₂∨ · · · ∨an has a Rad-supplement in L and ai/0 is cofinitely Rad- supplemented for every i=1,2, ...,n, then x has a Rad-supplement in L.

Proof. Clear from Lemma1.

Lemma 2. Let L be a lattice, I be a nonempty index set and ai∈L for every i∈I.

If 1= ∨

i∈Iai and ai/0 is cofinitely Rad-supplemented for every i∈I, then L is also cofinitely Rad-supplemented.

Proof. Letabe any cofinite element ofL. By hypothesis, 1= ∨

i∈Ia_i= ∨

i∈I(a∨a_i).

Since 1/ais compact anda∨ai∈1/afor everyi∈I, there existi₁,i₂, ...,in∈I such that 1=a∨a_i1∨a_i2∨...∨a_in. Since 0 is a Rad-supplement of 1=a∨a_i1∨a_i2∨...∨a_in andai_t/0 is cofinitely Rad-supplemented for everyt=1,2, ...,n, by Corollary1,ahas a Rad-supplement inL. HenceLis cofinitely Rad-supplemented.

Corollary 2. Let L be a lattice and1=a₁∨a₂∨...∨anin L. If ai/0is cofinitely Rad-supplemented for every i=1,2, ...,n, then L is also cofinitely Rad-supplemented.

Proof. Clear from Lemma2.

Proposition 2. Let L be a cofinitely Rad-supplemented lattice and a∈L. Then 1/a is also cofinitely Rad-supplemented.

Proof. Letxbe any cofinite element of 1/a. Then 1/xis compact andxis a cofinite element ofL. SinceLis cofinitely Rad-supplemented,xhas a Rad-supplementyin L. Sincea≤x, by [3, Lemma 5],a∨yis a Rad-supplement ofxin 1/a. Hence 1/a

is cofinitely Rad-supplemented.

Proposition 3. Let L be a cofinitely Rad-supplemented lattice. Then every cofinite element of1/r(L)is a direct summand of1/r(L).

Proof. Let x be any cofinite element of 1/r(L). Then 1/x is compact and x is a cofinite element of L. Since L is cofinitely Rad-supplemented, x has a Rad- supplementyinL. Here 1=x∨yandx∧y≤r(y/0)≤r(L). Then 1=x∨y∨r(L) and sincer(L)≤x,x∧(y∨r(L)) = (x∧y)∨r(L) =r(L). Hence 1=x⊕(y∨r(L))

in 1/r(L)andxis a direct summand of 1/r(L).

3. COFINITELYWEAKRADICAL SUPPLEMENTEDLATTICES

In this part, cofinitely weak radical supplemented lattices are defined and some properties of them are given.

Definition 2. LetLbe a lattice anda,b∈L. Ifa∨b=1 anda∧b≤r(L), thenb is called a weak radical supplement(or briefly, weak Rad-supplement)ofainL.

Definition 3. LetLbe a lattice. If every element ofLhas a weak Rad-supplement inL, thenLis called a weakly radical supplemented (or weakly Rad-supplemented) lattice. If every cofinite element of L has a weak Rad-supplement in L, then L is called a cofinitely weak radical supplemented (or cofinitely weak Rad-supplemented) lattice.

supplemented. It is also clear that every cofinitely Rad-supplemented lattice is cofinitely weak Rad-supplemented.

Proposition 4. Let L be a cofinitely weak Rad-supplemented lattice. If r(L)L, then L is cofinitely weak supplemented.

Proof. Clear from definitions.

Proposition 5. Let L be a compact lattice. Then L is cofinitely weak Rad- supplemented if and only if L is weakly Rad-supplemented.

Proof. Clear, since every element ofLis cofinite.

Lemma 3. Let L be a lattice, a∈L and x be a cofinite element of L. If x∨a has a weak Rad-supplement in L and a/0is cofinitely weak Rad-supplemented, then x has a weak Rad-supplement in L.

Proof. Let b be a weak Rad-supplement of x∨a inL. Then x∨a∨b=1 and (x∨a)∧b≤r(L). Sincex is a cofinite element of L, we clearly see that x∨b is a cofinite element of L. Then by _x∨b¹ = ^x∨a∨b_x∨b ∼= _a∧(x∨b)^a , a∧(x∨b) is a cofinite element of a/0. Since a/0 is cofinitely weak Rad-supplemented, a∧(x∨b) has a weak Rad-supplementcina/0. Here(a∧(x∨b))∨c=a andc∧(x∨b) =c∧a∧ (x∨b)≤r(a/0)≤r(L). Hence 1=x∨a∨b=x∨b∨(a∧(x∨b))∨c=x∨b∨cand x∧(b∨c)≤(b∧(x∨c))∨(c∧(x∨b))≤(b∧(x∨a))∨r(L)≤r(L)∨r(L) =r(L).

Thusb∨cis a weak Rad-supplement ofxinL.

Corollary 3. Let L be a lattice, a₁,a2, . . . ,a_n∈L and x be a cofinite element of L.

If x∨a₁∨a₂∨ · · · ∨anhas a weak Rad-supplement in L and ai/0is cofinitely weak Rad-supplemented for every i=1,2, ...,n, then x has a weak Rad-supplement in L.

Proof. Clear from Lemma3.

Lemma 4. Let L be a lattice, I be a nonempty index set and a_i∈L for every i∈I.

If 1= ∨

i∈Iai and ai/0is cofinitely weak Rad-supplemented for every i∈I, then L is also cofinitely weak Rad-supplemented.

Proof. Letabe any cofinite element ofL. By hypothesis, 1= ∨

i∈Ia_i= ∨

i∈I(a∨a_i).

Since 1/ais compact anda∨ai∈1/afor everyi∈I, there existi₁,i₂, ...,in∈I such that 1=a∨a_i1∨a_i2∨...∨a_in. Since 0 is a weak Rad-supplement of 1=a∨a_i1∨ ai₂∨...∨ai_n andai_t/0 is cofinitely weak Rad-supplemented for everyt=1,2, ...,n, by Corollary3,ahas a weak Rad-supplement inL. HenceLis cofinitely weak Rad-

supplemented.

Corollary 4. Let L be a lattice and1=a₁∨a₂∨...∨anin L. If ai/0is cofinitely weak Rad-supplemented for every i=1,2, ...,n, then L is also cofinitely weak Rad- supplemented.

Proof. Clear from Lemma4.

Lemma 5. Let L be a lattice, a,b,x∈L and x≤a. If b is a weak Rad-supplement of a in L, then x∨b is a weak Rad-supplement of a in1/x.

Proof. Sinceb is a weak Rad-supplement ofainL,a∨b=1 anda∧b≤r(L).

Thena∨x∨b=1 anda∧(x∨b)≤x∨(a∧b)≤x∨r(L)≤r(1/x). Hencex∨bis

a weak Rad-supplement ofain 1/x.

Proposition 6. Let L be a cofinitely weak Rad-supplemented lattice and a∈L.

Then1/a is also cofinitely weak Rad-supplemented.

Proof. Let x be any cofinite element of 1/a. Then 1/x is compact and x is a cofinite element of L. SinceL is cofinitely weak Rad-supplemented, xhas a weak Rad-supplementyinL. Sincea≤x, by Lemma5,a∨yis a weak Rad-supplement of xin 1/a. Hence 1/ais cofinitely weak Rad-supplemented.

Proposition 7. Let L be a lattice. Then L is cofinitely weak Rad-supplemented if and only if every cofinite element of1/r(L)is a direct summand of1/r(L).

Proof. (=⇒)Letxbe any cofinite element of 1/r(L). Then 1/xis compact andx is a cofinite element ofL. SinceLis cofinitely weak Rad-supplemented,xhas a weak Rad-supplementyinL. Here 1=x∨yandx∧y≤r(L). Then 1=x∨y∨r(L)and sincer(L)≤x,x∧(y∨r(L)) = (x∧y)∨r(L) =r(L). Hence 1=x⊕(y∨r(L))in 1/r(L)andxis a direct summand of 1/r(L).

(⇐=)Letxbe any cofinite element ofL. Here clearly we can see thatx∨r(L)is a cofinite element of 1/r(L). By hypothesis,x∨r(L)is a direct summand of 1/r(L).

Then there existsy∈1/r(L)such that 1=x∨r(L)∨y=x∨yand(x∨r(L))∧y= r(L). Since r(L)≤y, by modularity, r(L) = (x∨r(L))∧y = (x∧y)∨r(L) and x∧y≤r(L). Henceyis a weak Rad-supplement ofxinL. Therefore,Lis cofinitely

weak Rad-supplemented.

Proposition 8. Let L be a lattice and a L. If 1/a is cofinitely weak Rad- supplemented, then L is also cofinitely weak Rad-supplemented.

Proof. Letxbe any cofinite element ofL. Clearly we see that x∨ais a cofinite element of 1/a. Since 1/ais cofinitely weak Rad-supplemented, x∨ahas a weak Rad-supplement y in 1/a. Herex∨a∨y=1 and (x∨a)∧y≤r(1/a). Since x∨ a∨y=1 and a≤y, x∨y=1. Since aL, clearly we see that r(1/a) =r(L).

Hencex∨y=1 andx∧y≤(x∨a)∧y≤r(1/a) =r(L). ThusLis cofinitely weak

Rad-supplemented.

Letx,y∈L. It is defined a relationβ∗on the elements ofLbyxβ∗yif and only if for everyt∈Lwithx∨t=1 theny∨t=1 and for everyk∈Lwithy∨k=1 then x∨k=1. The definition ofβ∗relation and some properties of this relation are in [6].

The definition ofβ^∗ relation on modules and some properties of this relation are in [4].

weak Rad-supplement element in L, then L is cofinitely weak Rad-supplemented.

Proof. Letxbe a cofinite element ofL. By hypothesis, there exists a weak Rad- supplement elementy inLsuch thatxβ∗y. Letyis a weak Rad-supplement ofain L. Here y∨a=1 andy∧a≤r(L). Sincexβ_∗y andy∨a=1,x∨a=1. Assume x∧ar(L). Then there exists a maximal(6=1)elementtofLwithx∧at. Here (x∧a)∨t=1. By [6, Lemma 2],x∨(a∧t) =1 and sincexβ_∗y,y∨(a∧t) =1. Since a∨t=1, by [6, Lemma 2],(y∧a)∨t=1. Sincey∧a≤r(L)≤t,t= (y∧a)∨t= 1. This contradicts witht6=1. Hence x∧a≤r(L). Therefore,a is a weak Rad- supplement ofxinLandLis cofinitely weak Rad-supplemented.

Corollary 5. Let L be a lattice. If every cofinite element of L lies above a weak Rad-supplement element in L, then L is cofinitely weak Rad-supplemented.

Proof. Clear from Lemma6.

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Authors’ addresses

Celıl Nebıyev

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

Hasan H ¨useyın ¨Okten

Amasya University, Technical Sciences Vocational School, Amasya, Turkey E-mail address:hokten@gmail.com