EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 9, No. 3, 2016, 277-291
ISSN 1307-5543 – www.ejpam.com
On Generalized Ideals of Left Almost Semigroups
Waqar Khan
1,2,3, Faisal Yousafzai
2,∗, and Madad Khan
31School of Mathematics and Statistics, Southwest University, Beibei, Chongqing, China
2School of Mathematical Sciences, University of Science and Technology of China, Hefei, China
3Department of Mathematics, COMSATS Institute of Information Technology, Abbottabad, Pakistan
Abstract. In this paper, we study(m,n)-ideals of anL A-semigroup in detail. We characterize(0, 2) -ideals of anL A-semigroupSand prove thatAis a(0, 2)-ideal ofSif and only ifAis a left ideal of some left ideal ofS. We also show that anL A-semigroupSis 0−(0, 2)-bisimple if and only ifSis right 0-simple. Furthermore we study 0-minimal(m,n)-ideals in anL A-semigroupSand prove that ifR,
(L)is a 0-minimal right(l e f t)ideal ofS, then eitherRmLn={0}orRmLnis a 0-minimal(m,n)-ideal ofSform,n≥3. Finally we discuss(m,n)-ideals in an(m,n)-regularL A-semigroupSand show that
Sis(0, 1)-regular if and only if L=S LwhereLis a(0, 1)-ideal ofS. 2010 Mathematics Subject Classifications: 20M10, 20N99
Key Words and Phrases:L A-semigroups, left invertive law, left identity and(m,n)-ideals.
1. Introduction
A left almost semigroup (L A-semigroup) is a groupoidSsatisfying the left invertive law (a b)c = (c b)a for all a,b,c ∈ S. This left invertive law has been obtained by introducing braces on the left of ternary commutative lawa bc=c ba. The concept of anL A-semigroup was first given by Kazim and Naseeruddin in 1972[3]. AnL A-semigroup satisfies the medial law(a b)(cd) = (ac)(bd)for alla,b,c,d ∈S. SinceL A-semigroups satisfy medial law, they belong to the class of entropic groupoids which are also called abelian quasigroups[12]. If an L A-semigroup S contains a left identity (unitary L A-semigroup), then it satisfies the paramedial law(a b)(cd) = (d c)(ba)and the identitya(bc) =b(ac)for alla,b,c,d∈S[7].
An L A-semigroup is a useful algebraic structure, midway between a groupoid and a commutative semigroup. AnL A-semigroup is non-associative and non-commutative in gen-eral, however, there is a close relationship with semigroup as well as with commutative struc-tures. It has been investigated in[7]that if anL A-semigroup contains a right identity, then it becomes a commutative semigroup. The connection of a commutative inverse semigroup ∗Corresponding author.
Email addresses:wkyousafzai@163.com(W. Khan),yousafzaimath@gmail.com(F. Yousafzai), madadmath@yahoo.com(M. Khan)
with anL A-semigroup has been given by Yousafzaiet al. in[16]as, a commutative inverse semigroup(S,·) becomes anL A-semigroup(S,∗)under a∗b = ba−1r−1,∀a,b,r ∈S. An L A-semigroup S with left identity becomes a semigroup under the binary operation "◦e", defined as x◦e y = (x e)y for all x,y ∈S [17]. AnL A-semigroup is the generalization of a semigroup theory[7]and has vast applications in collaboration with semigroups like other branches of mathematics. Khan et al. studied an intra-regular class of an L A-semigroup in[4] and proved some interesting problems by using different ideals. They proved that the set of all two-sided ideals of intra-regularL A-semigroup forms a semilattice structure. They characterized an intra-regularL A-semigroup by using left, right, two-sided and bi-ideals. An L A-semigroup is the generalization of a semigroup theory[7]. Many interesting results on L A-semigroups have been investigated in[5, 9–11, 15].
Yaqoob, Corsini and Yousafzai[13]extended the concept ofLA-semigroups and introduced a new structure called left almost semihypergroup. Further Yaqoob and Gulistan [14] de-fined partial ordering on left almost semihypergroups. Gulistanet al. [2]defined Hv-L A
-semigroups which is a new generalization ofL A-semigroups andL A-semihypergroups.
2. Preliminaries and Examples
IfS is anL A-semigroup with product·:S×S −→S, thena b·c and(a b)cboth denote the product(a·b)·c.
If there is an element 0 of anL A-semigroup(S,·)such thatx·0=0·x= x ∀x ∈S, we call 0 a zero element ofS.
Example 1. Let S={a,b,c,d,e}with a left identity d. Then the following multiplication table shows that(S,·)is a unitaryL A-semigroup with a zero element a.
· a b c d e
a a a a a a
b a e e c e
c a e e b e
d a b c d e
e a e e e e
Example 2. Let S ={a,b,c,d}. Then the following multiplication table shows that(S,·) is an L A-semigroup with a zero element a.
· a b c d
a a a a a
b a d d c
c a c c c
d a c c c
Assume thatS is anL A-semigroup. Let us definea1=a,am+1=amaand am= ((((aa)a)a). . .a)a=am−1afor alla∈Swhere m≥1. It is easy to see that
am = am−1a = aam−1 for alla∈S and m≥ 3 ifS has a left identity. Also, we can show by induction,(a b)m=ambm andaman=am+nhold for alla,b∈S andm,n≥3.
A subsetAof anL A-semigroupSis called aright(left) ideal ofS ifAS⊆A(SA⊆A), and is called anidealofS if it is both left and right ideal ofS.
A subsetAof anL A-semigroupSis called anL A-subsemigroupofS ifA2⊆A.
The concept of(m,n)-ideals of a semigroup and anL A-semigroup was given in[6]and
[1]respectively.
AnL A-subsemigroupAof anL A-semigroupSis said to be an(m,n)-idealofS ifAmS· An⊆Awherem,nare non-negative integers such thatm=n6=0. HereAmorAnare suppressed ifm=0 orn=0, that isA0S=SorSA0=S. Note that ifm=n=1, then an(m,n)-idealAof anL A-semigroupS is called abi-idealofS. If we takem=0 orn=0, then an(m,n)-ideal Aof anL A-semigroupS becomes a left or a right ideal ofS.
An(m,n)-idealAof anL A-semigroupS with zero is said to be 0-minimalifA6={0}and {0}is the only(m,n)-ideal ofS properly contained inA.
An L A-semigroupS with zero is said to be 0-(0, 2)-bisimpleif S2 6= {0}and{0} is the only proper(0, 2)-bi-ideal ofS.
AnL A-semigroupSwith zero is said to benilpotentifSl ={0}for some positive integer l.
Let m,nbe non-negative integers andS be an L A-semigroup. We say thatS is(m,n) -regularif for every elementa∈S there exists some x ∈S such that a= (amx)an. Note that a0 is defined as an operator element such thata0y= y andza0=zfor any y,z∈S.
3.
0
-Minimal
(
0, 2
)
-Bi-Ideals in Unitary
L A
-Semigroups
IfS is a unitaryL A-semigroup, then it is easy to see thatS2=S,SA2=A2S andA⊆SA ∀A⊆S. Note that every right ideal of a unitaryL A-semigroupS is a left ideal ofS but the converse is not true in general. Example 1 shows that there exists a subset{a,b,e}ofS which is a left ideal ofS but not a right ideal ofS. It is easy to see thatSAandSA2 are the left and right ideals of a unitaryL A-semigroupS. ThusSA2 is an ideal of a unitaryL A-semigroup S.
Lemma 1. Let S be a unitaryL A-semigroup. Then A is a(0, 2)-ideal of S if and only if A is an ideal of some left ideal of S.
Proof. LetAbe a(0, 2)-ideal ofS, thenSA·A=AA·S=SA2⊆Aand A·SA=S·AA=SS·AA=SA2⊆A. HenceAis an ideal of a left idealSAofS.
Conversely, assume thatAis a left ideal of a left ideal LofS, then SA2=AA·S=SA·A⊆S L·A⊆LA⊆A,
Corollary 1. Let S be a unitaryL A-semigroup. Then A is a(0, 2)-ideal of S if and only if A is a left ideal of some left ideal of S.
Lemma 2. Let S be a unitaryL A-semigroup. Then A is a(0, 2)-bi-ideal of S if and only if A is an ideal of some right ideal of S.
Proof. Let Abe a (0, 2)-bi-ideal of S, then SA2·A= A2S ·A = AS·A2 ⊆ SA2 ⊆ Aand A·SA2=SS·AA2=A2A·SS=SA·A2⊆SA2⊆A. HenceAis an ideal of some right idealSA2 ofS.
Conversely, assume thatAis an ideal of a right idealRofS, then SA2=A·SA=A·(SS)A=A·(AS)S⊆A·(RS)R⊆AR⊆A, and(AS)A⊆(RS)A⊆RA⊆A, which shows thatAis a(0, 2)-ideal ofS.
Theorem 1. Let S be a unitaryL A-semigroup. Then the following statements are equivalent. (i) A is a(1, 2)-ideal of S;
(ii) A is a left ideal of some bi-ideal of S;
(iii) A is a bi-ideal of some ideal of S;
(iv) A is a(0, 2)-ideal of some right ideal of S; (v) A is a left ideal of some(0, 2)-ideal of S.
Proof. (i) =⇒(ii). It is easy to see thatSA2·S is a bi-ideal ofS. LetAbe a(1, 2)-ideal of S, then
(SA2·S)A=(SA2·SS)A= (SS·A2S)A= (S·A2S)A=A2S·A
=AS·A2⊆A,
which shows thatAis a left ideal of a bi-idealSA2·SofS. (ii) =⇒(iii). LetAbe a left ideal of a bi-idealB ofS, then
(A·SA2)A=(S·AA2)A⊆[S(SA·AA)]A= [S(AA·AS)]A
=[AA·S(AS)]A= [{S(AS)·A}A]A= [(AS·A)A]A ⊆[(BS·B)A]A⊆BA·A⊆A,
which shows thatAis a bi-ideal of an idealSA2 ofS. (iii) =⇒(i v). LetAbe a bi-ideal of an idealI ofS, then
SA2·A2=(A2·AA)S= (A·A2A)S⊆[A·(AI)A]S=AA·S
=SA·A⊆S I·S⊆I,
(i v) =⇒(v). It is easy to see thatSA3is a(0, 2)-ideal ofS. LetAbe a(0, 2)-ideal of a right idealRofS, then
A·SA3=A(SS·A2A) =A(AA2·S)⊆A[(SA·AA)S] =A[(AA·AS)S] =(AA)[(A·AS)S] = [S·A(AS)]A2= [A·S(AS)]A2
⊆RS·A2⊆RA2⊆A,
which shows thatAis a left ideal of a(0, 2)-idealSA3ofS. (v) =⇒(i). LetAbe a left ideal of a(0, 2)-idealOofS, then
AS·A2= (AA·SS)A=SA2·A⊆SO2·A⊆OA⊆A, which shows thatAis a(1, 2)-ideal ofS.
Lemma 3. Let S be a unitary L A-semigroup and A be an idempotent subset of S. Then A is a (1, 2)-ideal of S if and only if there exist a left ideal L and a right ideal R of S such that RL⊆A⊆R∩L.
Proof. Assume thatAis a(1, 2)-ideal ofS such thatAis idempotent. SettingL =SAand R=SA2, then
RL=SA2·SA=A2S·SA= (SA·SS)A2= (SS·AS)A2
=[S(AA·SS)]A2= [S(SS·AA)]A2= [S{A(SS·A)}]A2
=[A(S·SA)]A2⊆AS·A2⊆A. It is clear thatA⊆R∩L.
Conversely, letRbe a right ideal and Lbe a left ideal ofS such thatRL⊆A⊆R∩L, then AS·A2=AS·AA⊆RS·S L⊆RL⊆A.
Assume thatSis a unitaryL A-semigroup with zero. Then it is easy to see that every left (right) ideal ofSis a(0, 2)-ideal ofS. Hence ifOis a 0-minimal(0, 2)-ideal ofSandAis a left (right) ideal ofScontained inO, then eitherA={0}orA=O.
Lemma 4. Let S be a unitaryL A-semigroup with zero. Assume that A is a0-minimal ideal of S and O is anL A-subsemigroup of A. Then O is a(0, 2)-ideal of S contained in A if and only if O2={0}or O=A.
Proof. LetObe a(0, 2)-ideal ofScontained in a 0-minimal idealAofS. ThenSO2⊆O⊆A. SinceSO2is an ideal ofS, therefore by minimality ofA,SO2={0}orSO2=A. IfSO2=A, then A=SO2⊆Oand thereforeO=A. LetSO2={0}, thenO2S=SO2={0} ⊆O2, which shows thatO2is a right ideal ofS, and hence an ideal ofScontained inA, therefore by minimality of A, we haveO2={0}orO2=A. Now ifO2=A, thenO=A.
Corollary 2. Let S be a unitary L A-semigroup with zero. Assume that A is a0-minimal left ideal of S and O is anL A-subsemigroup of A. Then O is a(0, 2)-ideal of S contained in A if and only if O2={0}or O=A.
Lemma 5. Let S be a unitaryL A-semigroup with zero and O be a0-minimal(0, 2)-ideal of S. Then O2={0}or O is a0-minimal right(l e f t)ideal of S.
Proof. LetObe a 0-minimal(0, 2)-ideal ofS, then
S(O2)2=SS·O2O2=O2O2·S=SO2·O2⊆OO2⊆O2,
which shows thatO2is a(0, 2)-ideal ofScontained inO, therefore by minimality ofO,O2={0} orO2=O. Suppose thatO2=O, thenOS=OO·SS=SO2⊆O, which shows thatOis a right ideal ofS. LetRbe a right ideal ofS contained inO, thenR2S=RR·S⊆RS·S⊆R. ThusRis a(0, 2)-ideal ofS contained inO, and again by minimality ofO,R={0}orR=O.
The following Corollary follows from Lemma 4 and Corollary 2.
Corollary 3. Let S be a unitaryL A-semigroup. Then O is a minimal(0, 2)-ideal of S if and only if O is a minimal left ideal of S.
Theorem 2. Let S be a unitaryL A-semigroup. Then A is a minimal(2, 1)-ideal of S if and only if A is a minimal bi-ideal of S.
Proof. LetAbe a minimal(2, 1)-ideal ofS. Then
[(A2S·A)2S](A2S·A) =[{(A2S·A)(A2S·A)}S](A2S·A) ⊆[{(AS·A)(AS·A)}S](AS·A) =[{(AS·AS)(AA)}S](AS·A)
=[(A2S·AA)S](AS·A) ⊆[(AS·AS)S](AS·A) =(A2S·S)(AS·A)
⊆(AS·S)(AS·A) = (AS·AS)(SA) =A2S·SA=AS·SA2= (SA2·S)A
=(A2S·S)A= (SS·AA)A=A2S·A,
and similarly we can show that(A2S·A)2⊆A2S·A. ThusA2S·Ais a(2, 1)-ideal ofScontained inA, therefore by minimality ofA,A2S·A=A. Now
AS·A=(AS)(A2S·A) = [(A2S·A)S]A= (SA·A2S)A
=[A2(SA·S)]A⊆A2S·A=A,
Conversely, assume thatAis a minimal bi-ideal ofS, then it is easy to see thatAis a(2, 1) -ideal ofS. LetC be a(2, 1)-ideal ofS contained inA, then
[(C2S·C)S](C2S·C) =(SC·C2S)(C2S·C) = (SC2·C S)(C2S·C) =[C(SC2·S)](C2S·C) = [(C2S·C)(SC2·SS)]C
=[(C2S·C)(S·C2S)]C = [(C2S·C)(C2S)]C
=[C2{(C2S·C)S}]C ⊆C2S·C.
This shows that C2S·C is a bi-ideal of S, and by minimality of A, C2S ·C = A. Thus A=
C2S·C ⊆C, and thereforeAis a minimal(2, 1)-ideal ofS.
Theorem 3. Let A be a0-minimal(0, 2)-bi-ideal of a unitaryL A-semigroup S with zero. Then exactly one of the following cases occurs:
(i) A={0,a}, a2=0; (ii) ∀a∈A\{0}, Sa2=A.
Proof. Assume thatAis a 0-minimal(0, 2)-bi-ideal ofS. Leta∈A\{0}, thenSa2⊆A. Also Sa2 is a(0, 2)-bi-ideal ofS, thereforeSa2={0}orSa2=A.
LetSa2 ={0}. Sincea2∈A, we have eithera2=a ora2=0 ora2 ∈A\{0,a}. Ifa2=a, thena3 =a2a=a, which is impossible becausea3∈a2S=Sa2={0}. Leta2∈A\{0,a}, we have
S· {0,a2}{0,a2}=SS·a2a2=Sa2·Sa2={0} ⊆ {0,a2}, and
[{0,a2}S]{0,a2}={0,a2S}{0,a2}=a2S·a2⊆Sa2={0} ⊆ {0,a2}.
Therefore {0,a2}is a (0, 2)-bi-ideal of S contained inA. We observe that {0,a2} 6= {0} and{0,a2} 6=A. This is a contradiction to the fact thatAis a 0-minimal (0, 2)-bi-ideal ofS. Thereforea2=0 andA={0,a}.
IfSa26={0}, thenSa2=A.
Corollary 4. Let A be a0-minimal(0, 2)-bi-ideal of a unitaryL A-semigroup S with zero such that A26=0. Then A=Sa2 for every a∈A\{0}.
Lemma 6. Let S be a unitaryL A-semigroup. Then every right ideal of S is a(0, 2)-bi-ideal of S.
Proof. Assume thatAis a right ideal ofS, then
SA2=AA·SS=AS·AS⊆AA⊆AS⊆A, AS·A⊆A, and clearlyA2⊆A, thereforeAis a(0, 2)-bi-ideal ofS.
Theorem 4. Let S be a unitaryL A-semigroup with zero. Then Sa2=S∀a∈S\{0}if and only if S is0-(0, 2)-bisimple if and only if S is right0-simple.
Proof. Assume thatSa2 =S for every a ∈S\{0}. LetAbe a(0, 2)-bi-ideal ofS such that A6={0}. Leta∈A\{0}, thenS=Sa2⊆SA2⊆A. ThereforeS=A. SinceS=Sa2⊆SS=S2, we haveS2 = S 6= {0}. ThusS is 0−(0, 2)-bisimple. The converse statement follows from Corollary 4.
Let Rbe a right ideal of 0-(0, 2)-bisimpleS. Then by Lemma 6,R is a(0, 2)-bi-ideal ofS and soR={0}orR=S.
Conversely, assume that S is right 0-simple. Let a ∈S\{0}, then Sa2 = S. Hence S is 0-(0, 2)-bisimple.
Theorem 5. Let A be a0-minimal(0, 2)-bi-ideal of a unitaryL A-semigroup S with zero. Then either A2={0}or A is right0-simple.
Proof. Assume thatAis 0-minimal(0, 2)-bi-ideal ofS such thatA2 6={0}. Then by using Corollary 4, Sa2 = Afor every a ∈A\{0}. Since a2 ∈ A\{0} for every a ∈A\{0}, we have a4= (a2)2∈A\{0}for everya∈A\{0}. Leta∈A\{0}, then
(Aa2)S·Aa2=a2A·S(Aa2) = [(S·Aa2)A]a2⊆[(S·A)A]a2
=(AA·SS)a2=SA2·a2⊆Aa2, and
S(Aa2)2=S(Aa2·Aa2) =S(a2A·a2A) =S[a2(a2A·A)] =(aa)[S(a2A·A)] = [(a2A·A)S]a2
⊆(AA·SS)a2=SA2·a2⊆Aa2,
which shows thatAa2is a(0, 2)-bi-ideal ofScontained inA. HenceAa2={0}orAa2=A. Since a4∈Aa2anda4∈A\{0}, we getAa2=A. Thus by using Theorem 4,Ais right 0-simple.
4.
(
m
,
n
)
-Ideals in Unitary
L A
-Semigroups
In this section, we characterize a unitaryL A-semigroup in terms of(m,n)-ideals with the assumption thatm,n≥3. If we takem,n≥2, then all the results of this section can be trivially followed for a locally associative unitary L A-semigroup. IfS is a unitary L A-semigroup, then it is easy to see thatSAm=AmSandAmAn=AnAmform,n≥3 such thatA0=eif occurs, whereeis a left identity ofS.
Proof. LetRandLbe the right and left ideals ofSrespectively, then (RL)mS·(RL)n=(RmLm·S)(RnLn) = (RmLm·Rn)(S Ln)
=(LmRm·Rn)(S Ln) = (RnRm·Lm)(S Ln) =(RmRn·Lm)(S Ln) = (Rm+nLm)(S Ln)
=S(Rm+nLm·Ln) =S(LnLm·Rm+n)
=SS·Lm+nRm+n=S Lm+n·SRm+n
=Rm+nS·Lm+nS=SRm+n·S Lm+n, and
SRm+n·S Lm+n=(S·Rm+n−1R)(S·Lm+n−1L) =[S(Rm+n−2R·R)][S(Lm+n−2L·L)] =[S(RR·Rm+n−2)][S(L L·Lm+n−2)] ⊆(SS·RRm+n−2)(SS·L Lm+n−2) ⊆(SR·SRm+n−2)(S L·S Lm+n−2) ⊆(Rm+n−2S·RS)(L·S Lm+n−2) ⊆(Rm+n−2S·R)(S·L Lm+n−2) =(RS·Rm+n−2)(S Lm+n−1) ⊆RRm+n−2·S Lm+n−1 ⊆SRm+n−1·S Lm+n−1, therefore
(RL)mS·(RL)n⊆SRm+n·S Lm+n⊆SRm+n−1·S Lm+n−1⊆. . .⊆SR·S L ⊆(SS·R)L= (RS·S)L⊆RL,
and also
RL·RL=LR·LR= (LR·R)L= (RR·L)L⊆(RS·S)L⊆RL. This shows thatRLis an(m,n)-ideal ofS.
Theorem 6. Let S be a unitaryL A-semigroup with zero. If S has the property that it contains no non-zero nilpotent(m,n)-ideals and R(L)is a0-minimal right(l e f t)ideal of S, then either RL={0}or RL is a0-minimal(m,n)-ideal of S.
Proof. Assume thatR(L)is a 0-minimal right(l e f t)ideal ofSsuch thatRL6={0}, then by Lemma 7,RL is an(m,n)-ideal ofS. Now we show thatRL is a 0-minimal(m,n)-ideal ofS. Let{0} 6= M ⊆RL be an(m,n)-ideal ofS. Note that since RL ⊆R∩L, we have M ⊆ R∩L. HenceM⊆RandM⊆ L. By hypothesis,Mm6={0}andMn6={0}. Since{0} 6=S Mm=MmS, therefore
=RRm−2·RS⊆RRm−2·R=Rm, and
Rm⊆SRm=SS·RRm−1=Rm−1R·S= (Rm−2R·R)S
=(RR·Rm−2)S=SRm−2·RR⊆SRm−2·R
=(SS·Rm−3R)R= (RRm−3·SS)R= (RS·Rm−3S)R ⊆(R·Rm−3S)R= (Rm−3·RS)R⊆Rm−3R·R=Rm−1,
therefore{0} 6=MmS ⊆Rm ⊆Rm−1 ⊆. . .⊆R. It is easy to see that MmS is a right ideal ofS. Thus MmS=RsinceRis 0-minimal. Also
{0} 6=S Mn⊆ {0} 6=S Ln=S·Ln−1L=Ln−1·S L⊆Ln−1L= Ln, and
Ln⊆S Ln=SS·L Ln−1=Ln−1L·S= (Ln−2L·L)S=S L·Ln−2L ⊆L·Ln−2L=Ln−2·L L⊆ Ln−2L=Ln−1⊆. . .⊆L,
therefore{0} 6=S Mn⊆Ln⊆Ln−1⊆. . .⊆ L. It is easy to see thatS Mnis a left ideal ofS. Thus S Mn=Lsince Lis 0-minimal. Therefore
M⊆RL=MmS·S Mn=MnS·S Mm= (S Mm·S)Mn
=(S Mm·SS)Mn= (S·MmS)Mn= (Mm·SS)Mn
=MmS·Mn⊆M.
Thus M=RL, which means thatRL is a 0-minimal(m,n)-ideal ofS.
Theorem 7. Let S be a unitaryL A-semigroup. If R(L)is a0-minimal right(l e f t)ideal of S, then either RmLn={0}or RmLn is a0-minimal(m,n)-ideal of S.
Proof. Assume that R(L) is a 0-minimal right (l e f t) ideal of S such that RmLn 6= {0}, then Rm 6= {0} and Ln 6= {0}. Hence{0} 6= Rm ⊆ R and {0} 6= Ln ⊆ L, which shows that Rm = R and Ln = L since R (L) is a 0-minimal right (l e f t) ideal of S. Thus by Lemma 7, RmLn=RL is an(m,n)-ideal ofS. Now we show thatRmLn is a 0-minimal(m,n)-ideal ofS. Let{0} 6=M⊆RmLn=RL⊆R∩Lbe an(m,n)-ideal ofS. Hence
{0} 6=S M2=M M·SS=M S·M S⊆RS·RS⊆R
and{0} 6=S M ⊆S L⊆L. ThusR=S M2=M M·SS=S M·M⊆S M andS M =LsinceR(L) is a 0-minimal right(l e f t)ideal ofS. Therefore
M⊆RmLn⊆(S M)m(S M)n=SmMm·SnMn=SS·MmMn
=MnMm·S=S Mm·Mn=MmS·Mn⊆M,
Theorem 8. Let S be a unitaryL A-semigroup with zero. Assume that A is an(m,n)-ideal of S and B is an(m,n)-ideal of A such that B is idempotent. Then B is an(m,n)-ideal of S.
Proof. It is trivial that B is an L A-subsemigroupS. Secondly, since AmS·An ⊆ Aand BmA·Bn⊆B, then
BmS·Bn=(BmBm·S)(BnBn) = (BnBn)(S·BmBm)
=[(S·BmBm)Bn]Bn= [(Bn·BmBm)(SS)]Bn
=[(Bm·BnBm)(SS)]Bn= [S(BnBm·Bm)]Bn
=[S(BnBm·Bm−1B)]Bn= [S(BBm−1·BmBn)]Bn
=[S(Bm·BmBn)]Bn= [Bm(SS·BmBn)]Bn
=[Bm(BnBm·SS)]Bn= [Bm(SBm·Bn)]Bn
=[Bm{(SS·Bm−1B)Bn}]Bn= [Bm(BmS·Bn)]Bn ⊆[Bm(AmS·An)]Bn⊆BmA·Bn⊆B,
which shows that Bis an(m,n)-ideal ofS.
Lemma 8. Let〈a〉(m,n)=amS·an, then〈a〉(m,n) is an(m,n)-ideal of a unitaryL A-semigroup
S.
Proof. Assume thatSis a unitaryL A-semigroup andm,nare non-negative integers, then {〈a〉(m,n)}mH
{〈a〉(m,n)}n= [{((amH)an)}mH]{(amH)an}n = [{(ammHm)amn}H]{(amnHn)ann}
=[ann(amnHn)][H{(ammHm)amn}]
= [[H{(ammHm)amn}](amnHn)]ann
= [amn[[H{(ammHm)amn}]Hn]]ann ⊆(amnH)ann= (amnHn)ann
={(amH)an}n⊆ 〈a〉(m,n) n
⊆ 〈a〉(m,n),
and similarly we can show that 〈a〉(m,n) 2
⊆ 〈a〉(m,n).
Theorem 9. Let S be a unitaryL A-semigroup and〈a〉(m,n) be an(m,n)-ideal of S. Then the
following statements hold:
(i) 〈a〉(1,0) m
S=amS;
(ii) S 〈a〉(0,1) n
=San;
(iii) 〈a〉(1,0) m
S· 〈a〉(0,1) n
Proof. (i). As〈a〉(1,0)=aS, we have 〈a〉(1,0)
m
S=(aS)mS= (aS)m−1(aS)·S=S(aS)·(aS)m−1 =(aS)(aS)m−1= (aS)[(aS)m−2(aS)]
=(aS)m−2(aS·aS) = (aS)m−2(a2S)
=. . .= ¨
(aS)m−(m−1)(am−1S) ifmis odd (am−1S)(aS)m−(m−1) ifmis even =amS.
Analogously, we can prove(ii)and(iii)is simple.
Corollary 5. Let S be a unitaryL A-semigroup and let〈a〉(m,n) be an(m,n)-ideal of S. Then
the following statements hold:
(i) 〈a〉(1,0)mS=Sam;
(ii) S 〈a〉(0,1) n
=anS;
(iii) 〈a〉(1,0) m
S· 〈a〉(0,1) n
= (Sam)(anS).
LetL(0,
n),R(m,0)andA(m,n) denote the sets of(0,n)-ideals,(m, 0)-ideals and(m,n)-ideals of an L A-semigroup S respectively.
Theorem 10. If S is a unitaryL A-semigroup, then the following statements hold: (i) S is(0, 1)-regular if and only if∀L∈L(0,1), L=S L;
(ii) S is(2, 0)-regular if and only if∀R∈R(2,0), R=R2S such that every R is semiprime; (iii) S is(0, 2)-regular if and only if∀U ∈A(0,2), U=U2S such that every U is semiprime.
Proof. (i). LetSbe(0, 1)-regular, then fora∈Sthere existsx ∈Ssuch thata=x a. Since Lis(0, 1)-ideal, thereforeS L⊆L. Leta∈L, thena= x a∈S L⊆ L. Hence L=S L. Converse is simple.
(ii). LetS be(2, 0)-regular andRbe(2, 0)-ideal ofS, then it is easy to see thatR=R2S. Now fora∈S there existsx∈Ssuch thata=a2x. Leta2∈R, then
a=a2x ∈RS=R2S·S=SS·R2=R2S=R, which shows that every(2, 0)-ideal is semiprime.
Conversely, let R = R2S for every R ∈ R(2,0). Since Sa2 is a (2, 0)-ideal of S such that a2∈Sa2, thereforea∈Sa2. Thus
a∈Sa2= (Sa2)2S= (Sa2·Sa2)S= (a2S·a2S)S= [a2(a2S·S)]S
=(a2·Sa2)S= (S·Sa2)a2⊆Sa2=a2S, which implies thatSis(2, 0)-regular.
Lemma 9. If S is a unitaryL A-semigroup, then the following statements hold: (i) If S is(0,n)-regular, then∀L∈L(0,n), L=S Ln;
(ii) If S is(m, 0)-regular, then∀R∈R(m,0), R=RmS; (iii) If S is(m,n)-regular, then∀U∈A(m,n), U= (UmS)Un.
Proof. It is simple.
Corollary 6. If S is a unitaryL A-semigroup, then the following statements hold: (i) If S is(0,n)-regular, then∀L∈L(0,n), L=LnS;
(ii) If S is(m, 0)-regular, then∀R∈R(m,0), R=SRm; (iii) If S is(m,n)-regular, then∀U∈A(
m,n), U=Um+nS=SUm+n.
Theorem 11. Let S be a unitary(m,n)-regularL A-semigroup such that m=n. Then for every R∈R(m,0) and L∈L(0,n), R∩L=RmL∩RLn.
Proof. It is simple.
Theorem 12. Let S be a unitary(m,n)-regularL A-semigroup. If M(N)is a0-minimal(m, 0) -ideal((0,n)-ideal) of S such that M N ⊆ M∩N , then either M N ={0}or M N is a0-minimal
(m,n)-ideal of S.
Proof. LetM (N)be a 0-minimal(m, 0)-ideal ((0,n)-ideal) ofS. LetO=M N, then clearly O2⊆O. Moreover
OmS·On=(M N)mS·(M N)n= (MmNm)S·MnNn⊆(MmS)S·SNn
=S Mm·SNn=MmS·SNn⊆M N=O,
which shows thatO is an(m,n)-ideal ofS. Let{0} 6= P⊆Obe a non-zero(m,n)-ideal ofS. SinceS is(m,n)-regular, therefore by using Lemma 9, we have
{0} 6=P=PmS·Pn= (Pm·SS)Pn= (S·PmS)Pn= (Pn·PmS)(SS) =(PnS)(PmS·S) =PnS·S Pm=PmS·S Pn.
Hence PmS6={0}andPmS6={0}. FurtherP⊆O=M N⊆M∩N implies thatP⊆M and P⊆N. Therefore{0} 6= PmS⊆MmS ⊆M which shows that PmS=M sinceM is 0-minimal. Likewise, we can show thatS Pn=N. Thus we have
P⊆O=M N =PmS·S Pn=PnS·S Pm= (S Pm·SS)Pn
=(S·PmS)Pn=PmS·Pn⊆P.
Theorem 13. Let S be a unitary (m,n)-regular L A -semigroup. If M (N) is a 0-minimal
(m, 0)-ideal ((0,n)-ideal) of S, then either M∩N={0}or M∩N is a0-minimal(m,n)-ideal of S.
Proof. Once we prove thatM∩N is an(m,n)-ideal ofS, the rest of the proof is same as in Theorem 11. LetO=M∩N, then it is easy to see thatO2⊆O. Moreover
OmS·On⊆MmS·Nn⊆M Nn⊆SNn⊆N. But, we also have
OmS·On⊆MmS·Nn= (Mm·SS)Nn= (S·MmS)Nn= (Nn·MmS)S
=(Mm·NnS)(SS) = (MmS)(NnS·S) =MmS·SNn
=MmS·NnS=Nn(MmS·S) =Nn·S Mm=Nn·MmS
=Mm·NnS=Mm·SNn⊆MmN ⊆MmS⊆M. ThusOmS·On⊆M∩N=Oand thereforeOis an(m,n)-ideal ofS.
ACKNOWLEDGEMENTS The work of first author is supported by the NNSF (Grant No. 11371335) grant of China. The second author is highly thankful to CAS-TWAS President’s Fellowship.
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