Bi-ideal-simple semirings

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1 Comment.Math.Univ.Carolin. 46,3(2005) Bi-ideal-simple semirings VáclavFlaška,TomášKepka,JanŠaroch Abstract. Commutative congruence-simple semirings were studied in[2] and [7](but see also[1],[3] [6]). The non-commutative case almost(see[8]) escaped notice so far. Whatever, every congruence-simple semiring is bi-ideal-simple and the aim of this very short note is to collect several pieces of information on these semirings. Keywords: bi-ideal-simple, semiring, zeropotent Classification: 16Y60 1. Introduction A semiring is a non-empty set equipped with two binary operations, denoted as addition and multiplication, such that the addition makes a commutative semigroup, the multiplication is associative and distributes over the addition from both sides. The additive(multiplicative, resp.) semigroup of the semiring may, but need not, contain a neutral and/or an absorbing element. An element will be called bi-absorbing if it is absorbing for both the operations. If such an element exists,itwillbedenotedbythesymbol o(= o S ).Wethushave o+x=ox=xo=o forevery x S. Let Sbeasemiring. Weput A+B= {a+b; a A, b B}, AB= {ab; a A, b B}and2A={a+a; a A}foranytwosubsets Aand Bof S. A semiring S is called congruence-simple if it has just two congruence relations. 2. Bi-ideals Let Sbeasemiring. Anon-emptysubset Iof Siscalledabi-ideal of Sif (S+I) SI IS I(i.e., Iisanidealbothoftheadditiveandthemultiplicative semigroup of the semiring S). The following seven lemmas are easy. 2.1Lemma. Aone-elementsubset {w}of Sisabi-idealifandonlyif w=o S is a bi-absorbing element of S. This research has been supported by the Grant Agency of the Charles University, Grant #444/2004 and by the institutional grant MSM

2 392 V.Flaška,T.Kepka,J.Šaroch 2.2Lemma. Thesubsets S, S+ S, SS+ S, SSS+ S,...arebi-idealsof S. 2.3Lemma. SaS+ Sisabi-idealof Sforevery a S. Intheremaininglemmas,assumethat o S. 2.4Lemma. Theset {x S; xsx=o}isabi-ideal. 2.5Lemma. Theset {x S; x+xsx=o}isabi-ideal. 2.6Lemma. Theset {x S;2x=o}isabi-ideal. 2.7Lemma. Thesets(o:S) l = {x S; xs= o},(o:s) r = {x S; Sx=o} and(o:s) m = {x S; SxS= o}arebi-ideals. 3. Bi-ideal-free semirings 3.1 Proposition. A semiring S is bi-ideal-free(i.e., it has no proper bi-ideal) if andonlyif S= SaS+ Sforevery a S. Proof: The assertion follows easily from Bi-ideal-simple semirings introduction Asemiring Swillbecalledbi-ideal-simpleif S 2and I= Swhenever Iis abi-idealof Swith I Proposition. Asemiring Sisbi-ideal-simpleifandonlyifatleastone(and thenjustone)ofthefollowingfivecasestakesplace: (1) S =2; (2) S 3and S= SaS+ Sforevery a S; (3) S 3, o Sand S= SaS+ Sforevery a S, a o; (4) S 3, o S, S+ S= oand SaS= Sforevery a S, a o; (5) S 3, o S, SS= oand S+ a=sforevery a S, a o. Proof: We prove only the direct implication, the converse one being trivial. So, let Sbeabi-ideal-simplesemiring. Wewillassumethat S 3and,moreover, inviewof3.1,that Sisnotbi-ideal-free. Then o Sby2.1andwehaveto distinguish the following three cases. (i) Let S+ S= o.if SS= o,theneverysubsetof Scontaining oisabi-ideal andthen S =2,acontradiction.Thus SS o,andhence(o:s) l = oby 2.7. Similarly(o:S) r = oand,bycombination,weget(o:s) m = o(use 2.7again). Now,if a S, a o,then SaS oand,since S+ S= o,the set SaSisabi-ideal.Thus SaS= Sand(4)istrue. (ii) Let SS= o. Theneveryidealof S(+)isabi-ideal,andsothesemigroup S(+) is ideal-simple. Now,(5) is clear. (iii) Let S+ S o SS. Wehave(o : S) l = o = (o : S) r by2.7, and hence(o:s) m = o,too. Further, S+ S = S by2.2. Now,if a S,

3 Bi-ideal-simple semirings 393 a o,then bac oforsome b, c S,and b=d+e, d, e S. Then o bac=dac+eac SaS+ Sandconsequently, SaS+ S= Sby2.3. It meansthat(3)istrue. Intherestofthissection,weassumethat Sisabi-ideal-simplesemiringwith o S. 4.2Proposition. If Sisanil-semiring(i.e.,forevery x Sthereexists n 1 with x n = o),then SS= o. Proof:Theresultisclearfor S =2,andsolet S 3.If a S, a o,issuch thateither S= SaS+Sor S= SaS,then a=bac+w, b, c S, w S {0},and wehave a=b 2 ac 2 + bwc+w=b 3 ac 3 + b 2 wc 2 + bwc+w=...,acontradiction with b n = o.thus SaS+ S S SaSandtherestisclearfrom Lemma. If S 3and SS o,thenforevery a S, a o,thereexist elements b, c, d Ssuchthat bac o ada. Proof: Combine 2.4, 2.7 and Lemma. Justoneofthefollowingtwocasestakesplace: (1) x+xsx=oforevery x S; (2)forevery a S, a o,thereexistsatleastone b Swith a+aba o. Proof: Use Lemma. Justoneofthefollowingtwocasestakesplace: (1)2x=oforevery x S; (2)2a oforevery a S, a o. Proof: Use Congruence-simple semirings 5.1 Proposition. Every congruence-simple semiring is bi-ideal-simple. Proof:If Iisabi-ideal,thentherelation(I I) id S isacongruenceof S. 5.2Proposition. Let Sbeabi-ideal-simplesemiringwith o Sandlet a S, a o. If risacongruenceof Smaximalwithrespectto(a, o) / r,then S/ris a congruence-simple semiring. Proof: S/risnon-trivial. If sisacongruenceof Ssuchthat r sand r s, then(a, o) sand I= {x S;(x, o) s}isabi-idealof S. Since I 2,we have I= Sand s=s S. 5.3Corollary. Let Sbeabi-ideal-simplesemiringwith o S.Then Scanbe imbedded into the product of congruence-simple factors of S.

4 394 V.Flaška,T.Kepka,J.Šaroch 6. Bi-ideal-simple semirings of type 4.1(4) 6.1. If S is a bi-ideal-simple semiring of type 4.1(4), then the multiplicative semigroup of S is ideal-simple Let Sbeamultiplicativeideal-simplesemigroupwith S 3and o S. Setting S+ S= owegetabi-ideal-simplesemiringoftype4.1(4). 7. Bi-ideal-simple semirings of type 4.1(5) 7.1. If Sisabi-ideal-simplesemiringoftype4.1(5),then T(+)isan(abelian) subgroupof S(+),where T= S \ {o} Let T(+)beanabeliangroup, T 2, o / T and S= T {o}. Setting SS= S+ o=o+s= owegetabi-ideal-simplesemiringoftype4.1(5). 8. Additively zeropotent semirings In this section, let S be an additively zeropotent semiring(a zp-semiring for short). Thatis, o S and2s = o. Wedefinearelation on S by a b iff b (S+ a) {a}. Itiseasytocheckthat isarelationoforderwhichis compatiblewiththetwooperationsdefinedon S.Thatis, isanorderingofthe semiring S. Clearly, o is the greatest element of S. 8.1Lemma. If S 2,thenanelement a S, a o,ismaximalin S \ {o}if andonlyif S+ a=o. Proof: Obvious. Intherestofthissection,wewillassumethat S= S+ S. 8.2Lemma. If S 2,then Shasnominimalelements. Proof:If a S, a o,then a=b+c, b a.if b=a,then a=a+c=a+2c= a+o=o,acontradiction. 8.3Corollary. Either S =1or Sisinfinite. 8.4 Lemma. The only idempotent element of S is the bi-absorbing element o. Proof: Let b 2 = bforsome b S. Then b=c+dand b=b 3 = b(c+d)b= bcb+bdb.ofcourse, c b, d b,andhence cd bd, cd cb, o=2cd bd+cb and bd+cb=o.finally, o=bob=bdb+bcb=b. 8.5Corollary. If Scontainsaleft(orright)unit,then S =1. 8.6Lemma. If a k = a l forsome a Sand1 k < l,then a k = o. Proof: Therearepositiveintegers m, nsuchthat m(l k)=k+ n. Now,if b=a k+n,then b=a k a n = a l a n = a k a l k a n = a l a l k a n = a k a l k a l k a n = =a k a m(l k) a n = a 2k+2n = b 2.By8.4, b=o,andhence a k = a l = a k a l k = a k a l k a l k = =a k a m(l k) = a k a k+n = a k b=o.

5 Bi-ideal-simple semirings Lemma. Let a, b Sand k, l 1besuchthat a k = a l + b.then a 2k = o. Moreover,if 2k l,then a k = o. Proof:Wehave a 2l + a l b=a l (a l + b)=a k+l =(a l + b)a l = a 2l + ba l. Consequently, a 2k =(a l +b) 2 = a 2l +a l b+ba l +b 2 = a 2l +ba l +ba l +b 2 = o.if2k l, then a 2k = oimplies a l = oandhence a k = a l + b=o. 8.8Lemma. If a Sisanon-nilpotentelement,thenthepowers a 1, a 2, a 3,... are pair-wise incomparable. Proof: Combine 8.6 and Lemma. If a, b Saresuchthat aba a,then a=o. Proof: If aba=a,then(ab) 2 = ab, ab=oby8.4and a=aba=oa=o. If aba+c=a,then ab=abab+cb=(ab) 2 + cb, ab=oby8.7and a=o,too. 8.10Lemma. If a, b Saresuchthat ab=a o(ab=b o,resp.),then a b(b a,resp.). Proof: Firstly, a bby8.4. Now,if a b,then b=a+c, ac bc, c b, ac ab=a, o=2ac a+bcand a+bc=o. Thus o=a(a+bc)=a 2 + abc= a 2 + ac=a(a+c)=ab=a,acontradiction.similarlythesecondcase. 8.11Proposition. Let Sbeazp-semiringwith S+ S= Sand S 2.Then: (i) Sisinfinite; (ii) theorderedset(s, )hasnominimalelements; (iii) the bi-absorbing element o is the only idempotent element of S; (iv) Scontainsneitheraleftnorarightunit; (v)if a Sisnotnilpotent,thentheelements a i, i 1,arepair-wiseincomparablein(S, ); (vi)if a o,then aba aforevery b S; (vii) if o a b,then ab a; (viii) if o b a,then ab b. Proof:See8.2,8.3,8.4,8.5,8.8,8.9and Bi-ideal-simple zp-semirings 9.1 Proposition. Let S be a zp-semiring with S 3. Then S is bi-ideal-simple ifandonlyifatleastone(andthenjustone)ofthefollowingtwocasestakes place: (1) S+ S= oand SaS= Sforevery a S, a o; (2) S= S+ SaSforevery a S, a o. Proof: The result follows easily from 4.1. Intherestofthissection,let Sbeabi-ideal-simplesemiringsuchthat S+S o. Then S+ S= Sand Sisinfinite(see8.11).Moreover,by4.2, Sisnotnil.

6 396 V.Flaška,T.Kepka,J.Šaroch 9.2Lemma. Let V beafinitesubsetof S \ {o}.thenthereexistsatleastone element a Ssuchthat a oand a vforevery v V. Proof:Firstly,by9.1,wehave S= S+SbSforevery b S, b o.inparticular, SbS o, SS oand,by4.3,forevery w V thereisatleastone a w Swith wa w w o. Then a w oand wa w w wby8.9. Now,asequence v 1,...,v k, k 2,ofelementsfrom V willbecalledadmissibleinthesequeliftheseelements arepair-wisedistinctand v i a i v i v i+1 forsome a i S,1 i k 1. Ifthereisnoadmissiblesequence,then wa w w oand wa w w v forall w, v V. Theresultisprovedinthiscase, andhencewecanassumethat v 1,...,v k, k 2,isanadmissiblesequencewithmaximallength k. Let mbemaximalwithrespectto1 m kand v k bv k v m foratleastone b S.Then m < kby8.9and v m a m v m v m+1 implies v k bv k a m v k bv k v m+1, acontradictionwiththemaximalityof m. Wehavethusshownthat v k cv k v i forall1 i kand c S.Inparticular, o v k a k v k v i forsome a k Sand all i,1 i k.finally,itfollowsfromthemaximalityof kthat v k a k v k vfor every v V. 9.3Corollary. Denoteby Athesetofmaximalelementsof (S \ {o}, )(see 8.1)andassumethateveryelementfrom S \ {o}issmallerorequaltoanelement from A.Thentheset Aisinfinite. 9.4 Proposition. Just one of the following two cases takes place: (1) x+xsx=oand x m + x n = oforevery x Sandallpositiveintegers m, n; (2)forevery a S, a o,thereexistsatleastone b Ssuchthat a+aba o. Proof: Takingintoaccount4.4,wemayassumethat x+xsx=oforevery x S. Then x+x 3 = oandweput I= {a S; a+a 2 = o}. Clearly, Iisan idealof S(+). Moreover,if a Iand b S,then ab+abab=(a+aba)b=o. Thus ab I,similarly ba Iandweseethat Iisabi-idealof S. If I= {o},then a+a 2 oforevery a S, a o.but a 2 + a 4 = a(a+a 3 )= ao=oand a 2 I. Itfollowsthat a 2 = oforevery a S,acontradictionwith 4.2. Thus I {o}andweget I= Sand x+x 2 = oforevery x S. Further, x+x=obythezp-propertyand x+x n = oforevery n 3,since x+xsx=o. If2 n m,then x n + x m = x n 1 (x+x m n+1 )=x n 1 o=o. 9.5Proposition. Denoteby Athesetofmaximalelementsoftheorderedset (S \ {o}, ).If Aisnon-empty,then x+xsx=oforevery x S(i.e.,thecase 9.4(1) takes place). Proof: Combine 8.1 and An open problem Noexampleofanon-trivialzp-semiring Swith S+ S= S(see8.11)is known(atleasttotheauthorsofthepresentbriefnote).

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