In this work,˚ supplemented and strongly˚ supplemented lattices are defined and investigated some properties of these lattices

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Vol. 20 (2019), No. 2, pp. 773–780 DOI: 10.18514/MMN.2019.2806

˚ SUPPLEMENTED LATTICES

C¸ I ˘GDEM BIC¸ ER AND CELIL NEBIYEV Received 15 January, 2019

Abstract. In this work,˚ supplemented and strongly˚ supplemented lattices are defined and investigated some properties of these lattices. LetLbe a lattice and1Da1˚a2˚:::˚an

with a1; a2; :::; an2L. If ai=0 is ˚ supplemented for each iD1; 2; :::; n, thenL is also

˚ supplemented. LetLbe a distributive lattice and1Da1˚a2˚:::˚anwitha1; a2; :::; an2 L. Ifai=0is strongly˚ supplemented for eachiD1; 2; :::; n, thenLis also strongly˚ supplemented. A latticeLhas.D1/property if and only if Lis amply supplemented and strongly

˚ supplemented.

2010Mathematics Subject Classification: 06C05; 06C15

Keywords: lattices, small elements, supplemented lattices, complemented lattices

1. INTRODUCTION

Throughout this paper, all lattices are complete modular lattices with the smallest element 0 and the greatest element 1. Let L be a lattice, a; b2L and ab. A sublattice fx2Ljaxbg is called a quotient sublattice, denoted by b=a. An elementa⁰of a latticeLis called acomplementofainLifa^a⁰D0anda_a⁰D1, this case we denote1Da˚a⁰ (a anda⁰ also is calleddirect summands ofL). L is called a complemented latticeif each element has at least one complement in L.

An elementaofLis said to besmallorsuperfluousand denoted byaLifbD1 for every elementbofLsuch thata_bD1. The meet of all the maximal elements .¤1/of a latticeLis called theradicalofLand denoted byr.L/. An elementcof Lis called asupplementofb inLif it is minimal forb_cD1. ais a supplement ofbin a latticeLif and only ifa_bD1anda^ba=0. A latticeLis said to be supplementedif every element ofLhas a supplement inL. We say that an element bofLlies abovean elementaofLifabandb1=a. Lis said to behollowif every element.¤1/is superfluous inL, andLis said to belocalifLhas the greatest element.¤1/. An elementaofLis called aweak supplementofbinLifa_bD1 anda^bL. A latticeLis said to beweakly supplemented, if every element ofL has a weak supplement inL. We say that an elementa2Lhasample supplements inLif for everyb2Lwitha_bD1,ahas a supplementb⁰inLwithb⁰b. Lis called anamply supplemented lattice, if every element ofLhas ample supplements

c 2019 Miskolc University Press

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inL. It is clear that every supplemented lattice is weakly supplemented and every amply supplemented lattice is supplemented. A latticeLis said to bedistributiveif a^.b_c/D.a^b/_.a^c/for everya; b; c2L. LetLbe a lattice. It is definedˇ relation on the elements ofLbyaˇbwitha; b2Lif and only if for eacht2Lsuch thata_tD1thenb_tD1and for eachk2Lsuch thatb_kD1thena_kD1.

LetLbe a lattice. Consider the following conditions.

.D1/ For every element a of L, there exist a1; a22 L such that1Da1˚a2, a12a=0anda2^aa2=0.

.D3/Ifa1anda2are direct summands ofLand1Da1_a2, thena1^a2is also a direct summand ofL.

More details about (amply) supplemented lattices are in [1,2] and [5]. More results about (amply) supplemented modules are in [4] and [9]. Some important properties of ˚ supplemented modules are in [6] and [7]. The definition of ˇ relation on lattices and some properties of this relation are in [8]. The definition ofˇrelation on modules and some properties of this relation are in [3].

In this paper, we generalize some properties of˚ supplemented modules to lattices. We constitute relationships between˚ supplemented quotient sublattices and

˚ supplemented lattices by Lemma11 and Corollary 2. We also constitute relationships between lattices which has.D1/ property and strongly˚ supplemented lattices by Proposition4. We give some examples at the end of this paper.

Lemma 1. LetLbe a lattice anda; b; c2Lwithab. Ifcis a supplement ofb inL, thena_c is a supplement ofbin1=a.

Proof. Similar to proof of [5, Proposition 12.2(7)].

Lemma 2([5, Lemma 7.4]). LetLbe a lattice,a; b2Landab. Ifab=0 thenaL.

Lemma 3([5, Lemma 7.5]). In a latticeL let c⁰c=0andd⁰d=0. Then c⁰_d⁰.c_d /=0.

Lemma 4([5, Lemma 7.6]). IfaL, thenar.L/.

Lemma 5([5, Exercise 7.3]). IfLis a lattice anda2L, thenr.a=0/r.L/.

Lemma 6([5, Lemma 12.3]). In any modular latticeŒ.c_d /^bŒc^.b_d /_ Œd^.b_c/holds for everyb; c; d2L.

Lemma 7. LetLbe a lattice,a; b2Landab. Thenblies aboveaif and only ifaˇb.

Proof. .H)/See [8, Theorem 3].

.(H/Letb_tD1witht 21=a. Sinceaˇb,a_tD1and sinceat,tD1.

Henceb1=aandblies abovea.

Lemma 8([8, Lemma 2]). LetLbe a lattice and a; b; c2L. Ifa_b D1and .a^b/_cD1, thena_.b^c/Db_.a^c/D1.

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2. ˚ SUPPLEMENTED LATTICES

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

Clearly we see that every˚ supplemented lattice is supplemented and every complemented lattice is ˚ supplemented. We also clearly see that hollow and local lattices are˚ supplemented.

Proposition 1. LetLbe a lattice. ThenLis˚ supplemented if and only if for everyb2L, there exists a direct summandcofLsuch thatb_cD1andb^cc=0.

Proof. Clear from definition.

Proposition 2. LetLbe a lattice. If every element ofLhas a weak supplement that is a direct summand ofL, thenLis˚ supplemented.

Proof. Letabe a weak supplement ofb inLandabe a direct summand of L.

Sinceais a weak supplement ofbinL,a^bLand sinceais a direct summand of L,a^ba=0. Henceais a supplement ofbinLandLis˚ supplemented.

Lemma 9. LetLbe a lattice, anda; b2L. Ifxis a supplement ofa_binLand yis a supplement ofa^.x_b/ina=0thenx_yis a supplement ofb inL.

Proof. Sincexis a supplement ofa_binLandyis a supplement ofa^.x_b/

ina=0, then1Da_b_x,.a_b/^xx=0,aDŒa^.x_b/_yand.x_b/^yD a^.x_b/^yy=0. Here1Da_b_xDŒa^.x_b/_y_b_xDb_x_y. By Lemma6,.x_y/^bŒ.y_b/^x_Œ.x_b/^yŒ.a_b/^x_Œ.x_b/^y

.x_y/=0. Hencex_y is a supplement ofbinL.

Lemma 10. Let L be a lattice and a1; a22 L where a1=0 and a2=0 are ˚ supplemented and1Da1˚a2. ThenLis˚ supplemented.

Proof. Letx be any element ofL. Then1Da1_a2_x anda1_a2_x has a supplement0inL. Sincea2=0is˚ supplemented,a2^.a1_x/has a supplement y that is a direct summand ina2=0. By Lemma 9,y is a supplement of a1_x in L. Sincea1=0is˚ supplemented,a1^.x_y/has a supplement´that is a direct summand ina1=0. By Lemma9,y_´is a supplement ofxinL. Sinceyis a direct summand ofa2=0and´is a direct summand ofa1=0, by1Da1˚a2,y_´Dy˚´ is a direct summand ofL. Finally,Lis˚ supplemented.

Corollary 1. LetLbe a lattice,a1; a2; :::; an2Land1Da1˚a2˚:::˚an. If ai=0is˚ supplemented for every iD1; 2; : : : :; n, thenLis˚ supplemented.

Proof. Clear from Lemma10.

Lemma 11. LetLbe a lattice,a2LandaD.a^a1/˚.a^a2/for everya1; a22 Lwith1Da1˚a2. IfLis˚ supplemented, then1=ais also˚ supplemented.

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Proof. Letx21=a. SinceLis˚ supplemented, there existy; ´2Lsuch that 1Dx_y,x^yy=0and1Dy˚´. Sinceyis a supplement ofxinLandax, by Lemma1, a_y is a supplement ofx in1=a. Since 1Dy˚´, by hypothesis, aD.a^y/˚.a^´/. Then.a_y/^.a_´/DŒ.a^y/_.a^´/_y^Œ.a^y/_ .a^´/_´DŒy_.a^´/^Œ.a^y/_´D.a^y/_Œ.y_.a^´//^´D.a^y/_ Œ.y^´/_.a^´/D.a^y/_.0_.a^´//D.a^y/_.a^´/Da. Hence1=ais

˚ supplemented.

Corollary 2. LetLbe a distributive lattice. IfLis˚ supplemented, then1=ais

˚ supplemented for everya2L.

Lemma 12. LetLbe a supplemented lattice anda=0is a quotient sublattice such thata^r.L/D0. Then every element ofa=0is a direct summand ofa=0.

Proof. Let x2a=0. Since L is supplemented, there exists an element y of L with1Dx_yandx^yy=0. Since1Dx_yandxa,aDx_.a^y/. Since x^yy=0, by Lemma 4,x^yr.L/. Thenx^.a^y/Da^x^ya^r.L/D 0anda^x^yD0. HenceaDx˚.a^y/ina=0 andx is a direct summand of

a=0.

Corollary 3. LetLbe a supplemented lattice anda=0is a quotient sublattice such thata^r.L/D0. Thena=0is complemented.

Proposition 3. LetLbe a˚ supplemented lattice. Then there exista1; a22L such that1Da1˚a2,r.a1=0/a1=0andr.a2=0/Da2.

Proof. SinceLis˚ supplemented, there exista1; a22Lsuch that1Dr.L/_ a1Da1˚a2 andr.L/^a1a1=0. Then by Lemma5,r.a1=0/r.L/^a1 a1=0.

Assume x be a maximal .¤a2/ element of a2=0. Since 1=.a1_x/D.a1˚ a2/=.a1_x/D.a1_x_a2/=.a1_x/Ša2=Œa2^.a1_x/Da2=Œ.a2^a1/_xD a2=x,a1_xis a maximal element.¤1/ofLand since1Dr.L/_a1a1_x, this

is a contradiction. Hencer.a2=0/Da2.

Definition 2. Let L be a lattice. L is called a completely ˚ supplemented lattice, if every quotient sublattice a=0 such that a is a direct summand of L is

˚ supplemented.

Theorem 1. LetLbe a˚ supplemented lattice with.D3/. ThenLis completely

˚ supplemented.

Proof. Letube a direct summand ofLandx2u=0. SinceLis˚ supplemented, then there exists a direct summandy ofL such that1Dx_y andx^y y=0.

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Because of 1 Dx_y, u_y D1 and because of L has .D3/, u^y is a direct summand of L and hence u^y is a direct summand of u=0. Since 1Dx_y and xu, uDx_.u^y/. Byxû^y Dx^yy=0, xû^y L. By xû^yu^yandu^yis a direct summand ofL,xû^yu^y. Thusu=0

is˚ supplemented.

Definition 3. LetL be a supplemented lattice. Lis called a strongly ˚ supplemented lattice if every supplement element inLis a direct summand ofL.

Clearly we see that every strongly ˚ supplemented lattice is˚ supplemented and every complemented lattice is strongly˚ supplemented. Hollow and local lattices are strongly˚ supplemented.

Lemma 13. Letabe a supplement ofbinLandx; y2a=0. Thenyis a supplement ofxina=0if and only ify is a supplement ofb_xinL.

Proof. .H)/Lety be a supplement of x ina=0andb_x_´D1 with´y.

Because of x; y2a=0 and ´y, x_´a. Since a is a supplement of b in L, aDx_´. Sincey is a supplement ofxina=0,´Dy. Henceyis a supplement of b_xinL.

.(H/Letybe a supplement ofb_x inL. So,b_x_yD1and.b_x/^y y=0. Sincex_y a and a is a supplement of b in L, x_y Da andx^y .b_x/^yy=0. Henceyis a supplement ofxina=0.

Lemma 14. LetLbe a strongly˚ supplemented lattice. Then for every direct summandaofL, the quotient sublatticea=0is strongly˚ supplemented.

Proof. Let1Da˚bwithb2L,x; y2a=0andybe supplement ofxina=0. By Lemma13,y is a supplement ofb_x inL. SinceLis strongly˚ supplemented, every supplement element is a direct summand ofLandy is a direct summand ofL.

Here there exists´2Lsuch that1Dy˚´. By modularity,aDa^1Da^.y˚´/D y˚.a^´/. Thusyis a direct summand ofa=0.

Corollary 4. Every strongly ˚ supplemented lattice is completely ˚ supplemented.

Lemma 15. LetLbe a distributive lattice anda1; a22Lwith1Da1˚a2. Ifa1=0 anda2=0are strongly˚ supplemented, thenLis also strongly˚ supplemented.

Proof. Let a be a supplement of b in L. Since L is distributive, aDa^1D a^.a₁˚a₂/D.aâ₁/˚.aâ₂/holds. By Lemma13,aâ₁is a supplement of .aâ2/_binL. We can also see thataâ1is a supplement ofa1^..aâ2/_b/

in a1=0. Since a1=0 is strongly ˚ supplemented, aâ1 is a direct summand of a1=0. Similarly we can see that aâ2 is a direct summand of a2=0. Since 1D a1˚a2andaD.aâ1/˚.aâ2/,ais a direct summand ofL. HenceLis strongly

˚ supplemented.

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Corollary 5. LetLbe a distributive lattice,a1; a2; :::; an2Land1Da1˚a2˚ :::˚an. If ai=0 is strongly ˚ supplemented for every iD1; 2; : : : :; n, then L is strongly˚ supplemented.

Lemma 16. LetLbe a supplemented lattice. The following statements are equivalent.

.i / Lis strongly˚ supplemented.

.i i /Every supplement element ofLlies above a direct summand inL.

.i i i / .a/For every nonzero supplement element ainL, a=0 contains a nonzero direct summand ofL.

.b/ For every nonzero supplement elementain L, a=0 contains a maximal direct summand ofL.

Proof. .i /H).i i /Clear, since every element ofLlies above itself.

.i i /H).i i i /Letabe a nonzero supplement element inL. Assumeais a supplement ofb inL. By hypothesis, there exists a direct summandx ofLsuch thata lies abovexinL. By Lemma7,aˇx and sincea_bD1,x_bD1. Sinceais a supplement ofbinLandxa,aDxandais a nonzero direct summand ofL.

.i i i /H).i /Letabe a supplement ofbinLandxbe a maximal direct summand ofLwithxa. Assume1Dx˚y withy 2L. ThenaDa^1Da^.x˚y/D x˚.a^y/and by Lemma13,a^y is a supplement ofb_x inL. Ifa^y is not zero, then by hypothesis,.a^y/ =0contains a nonzero direct summandcofL. Here x˚c is a direct summand ofL andx˚ca. This contradicts the choice of x.

Hencea^yD0 andaDx. Thus a is a direct summand ofL andL is strongly

˚ supplemented.

Proposition 4. LetLbe a lattice. The following statements are equivalent.

.i / Lhas.D1/property.

.i i /Every element ofLlies above a direct summand inL.

.i i i / Lis amply supplemented and strongly˚ supplemented.

Proof. .i /H).i i /Leta2L. SinceLhas.D1/property, there exista1; a22L such that 1Da1˚a2, a1a anda2âa2=0. Let a_t D1with t 21=a1. Since a1aand1Da1˚a2, aDa^1Da^.a1˚a2/Da1˚.aâ2/. Then 1Da_t Da1_.aâ2/_t D.aâ2/_t and sinceaâ2L, tD1. Hence a1=a1andalies abovea1.

.i i /H).i i i / Let a_b D1 with a; b2L. By hypothesis, a^b lies above a direct summand inL. Here there existx; y 2Lsuch that1Dx˚y anda^b lies abovex. Since1Dx˚y andxb,bDb^1Db^.x_y/Dx_.b^y/. Then 1Da_bDa_x_.b^y/Da_.b^y/. By hypothesis,b^y lies above a direct summand in L. Here there existx1; y1 2Lsuch that 1Dx1˚y1 andb^y lies abovex1. By Lemma 7,.b^y/ ˇx1 and since1Da_.b^y/,1Da_x1holds.

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Let.a^x1/_tD1 witht 2L. By a^x1a^b^y,.a^b^y/_t D1holds.

HereyDy^1Dy^..a^b^y/_t /D.a^b^y/_.y^t /and1Dx_yDx_ .a^b^y/_.y^t /Dx_.a^b/_.y^t /. Sincea^blies abovex, by Lemma7, .a^b/ ˇx. Then1Dx_.a^b/_.y^t /Dx_.y^t /and sinceyis a supplement ofx inLandy^t y, y^t Dy andyt. Hence 1D.a^b^y/_t Dt and a^x1L. Sincex1a direct summand ofL,a^x1x1=0andx1is a supplement ofainL. Moreover,x1b. HenceLis amply supplemented. By Lemma16,Lis strongly˚ supplemented.

.i i i /H).i /Leta be any element ofL. By hypothesis, ahas a supplement b in L. Here 1Da_b and a^bb=0. Since L is amply supplemented, b has a supplement x inLwith xa. By hypothesis, x is a direct summand ofLand there exists an elementyofLsuch that1Dx˚y. Let.a^y/_tD1witht 2L.

Since1Dx_yDa_y, by Lemma8,a_.y^t /D1. Since1Dx_bandxa, aDa^1Da^.x_b/Dx_.a^b/. Then1Da_.y^t /Dx_.a^b/_.y^t / and sincea^bL,1Dx_.y^t /. Since1Dx_.y^t /andyis a supplement of xinL,y^tDyandyt. Then1D.a^y/_tDt anda^yL. Sinceyis a direct summand ofL,a^yy=0. HenceLhas.D1/property.

Corollary 6. LetLbe a lattice with.D1/property. ThenLis˚ supplemented.

Proof. Clear from Proposition4and Corollary4.

Example1. Consider the latticeLD f0; a; b; c; 1ggiven by the following diagram.

a 1

0 c

b

ThenLis supplemented but not˚ supplemented.

Example2. Consider the latticeLD f0; a; b; c; d; e; 1ggiven by the following diagram.

a 1

0 c

b

e d

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ThenLis supplemented but not˚ supplemented.

Example 3. Consider the intervalŒ0; 1 with natural topology. Let P be the set of all closed subsets of Œ0; 1. P is complete modular lattice by the inclusion (See [1, Example 2.10]). Here ^

i2ICiD \

i2ICi and _

i2ICiD [

i2ICifor everyCi2P .i 2I /

i[2ICi is the closure of [

i2ICi

. LetX2P andX_Y DŒ0; 1withY 2P. Then Œ0; 1 X Y and sinceY is closedŒ0; 1 X Y. LetX⁰DŒ0; 1 X. ThenX⁰ 2 P,X_X⁰DX[X⁰DŒ0; 1andX⁰Y for everyY 2P withX_Y DŒ0; 1. Hence X has ample supplements in P .hereX⁰ D Œ0; 1 X is the only supplement of X inP /andP is amply supplemented. LetADŒ0; a2P with 0 < a < 1. Here A⁰DŒ0; 1 ADŒa; 1is the only supplement ofAinP. LetA⁰_BDA⁰[BDŒ0; 1

withB 2P. SinceA⁰[BDŒ0; 1, Œ0; a/DŒ0; 1 A⁰ B and since B is closed, Œ0; aB. This casea2Band sincea2A⁰,A⁰^BDA⁰\B¤¿. HenceA⁰is not a direct summand ofP andP is not˚ supplemented.

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Authors’ addresses

C¸ i˘gdem Bic¸er

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

Celil Nebiyev

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