Noetherian and Artinian ordered groupoids—semigroups

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GROUPOIDS—SEMIGROUPS

NIOVI KEHAYOPULU AND MICHAEL TSINGELIS Received 9 February 2005 and in revised form 16 June 2005

Chain conditions, finiteness conditions, growth conditions, and other forms of finiteness, Noetherian rings and Artinian rings have been systematically studied for commutative rings and algebras since 1959. In pursuit of the deeper results of ideal theory in ordered groupoids (semigroups), it is necessary to study special classes of ordered groupoids (semigroups). Noetherian ordered groupoids (semigroups) which are about to be in-troduced are particularly versatile. These satisfy a certain finiteness condition, namely, that every ideal of the ordered groupoid (semigroup) is finitely generated. Our purpose is to introduce the concepts of Noetherian and Artinian ordered groupoids. An ordered groupoid is said to be Noetherian if every ideal of it is finitely generated. In this paper, we prove that an equivalent formulation of the Noetherian requirement is that the ideals of the ordered groupoid satisfy the so-called ascending chain condition. From this idea, we are led in a natural way to consider a number of results relevant to ordered groupoids with descending chain condition for ideals. We moreover prove that an ordered groupoid is Noetherian if and only if it satisfies the maximum condition for ideals and it is Ar-tinian if and only if it satisfies the minimum condition for ideals. In addition, we prove that there is a homomorphismπof an ordered groupoid (semigroup)Shaving an ideal Ionto the Rees quotient ordered groupoid (semigroup)S/I. As a consequence, ifSis an ordered groupoid andIan ideal ofSsuch that bothIand the quotient groupoidS/Iare Noetherian (Artinian), then so isS. Finally, we give conditions under which the proper prime ideals of commutative Artinian ordered semigroups are maximal ideals.

1. Introduction and prerequisites

Notherian and Artinian rings have been extensively studied since 1959. For Noetherian and Artinian rings we refer, for example, in [1]. In pursuit of the deeper results of ideal theory in ordered groupoids (semigroups), it is necessary to study special kinds of or-dered groupoids (semigroups). Noetherian oror-dered groupoids are particularly versatile since they satisfy a certain finiteness condition for ideals. The fact that, when dealing with Noetherian ordered groupoids, we can restrict our attention to finitely generated ideals is of great advantage. We call them Noetherian ordered groupoids, in honor of Emmy

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Noether who first initiated their study for rings. An equivalent definition of the Noether-ian ordered groupoids is that they satisfy the ascending chain condition for ideals. It is natural then to study ordered groupoids which satisfy the descending chain condition. These are the Artinian ordered groupoids (after Emil Artin). The aim of this paper is to introduce the concepts of Noetherian and Artinian ordered groupoids. We call an or-dered groupoidSNoetherian if every ideal ofSis finitely generated. We call an ordered groupoid Artinian if it satisfies the descending chain condition for ideals. We prove the following: An ordered groupoid is Noetherian if and only if it satisfies the ascending chain condition for ideals, equivalently, if it satisfies the maximum condition for ideals. An or-dered groupoid is Artinian if and only if it satisfies the minimum condition for ideals. Any homomorphism mapping of a Noetherian (resp., Artinian) ordered groupoid onto an ordered groupoid is Noetherian (resp., Artinian). We prove that there is a homomor-phismπof an ordered groupoid (resp., ordered semigroup)Shaving an idealIonto the Rees quotient ordered groupoid (resp., ordered semigroup)S/I. For an idealIof an or-dered groupoidS, we defineS/I:=S\I∪ {0}, where 0 is the zero ofS/I. IfSis an ordered groupoid andIan ideal ofS, then for each idealAofScontainingI, we haveπ(A)=A/I, and the setA/Iis an ideal ofS/I. As a consequence, ifSis an ordered groupoid andIan ideal ofSsuch that bothIand the quotient groupoidS/Iare Noetherian (resp., Artinian), then so isS. We finally find conditions under which the proper prime ideals of commu-tative Artinian ordered semigroups are maximal. We prove that ifSis a commutative, Artinian ordered semigroup having an elementbsuch that the ideal generated bybisS, then each proper prime ideal ofS, having the property “ifa∈S\P,z∈S1_,_n_∈_{N, and} ban_≤_zan+1_{, then}_b_≤_za_{”, is a maximal ideal of}_S_{. This leads to some conditions under} which in some Artinian ordered semigroups, the proper prime ideals are maximal ideals. A basic one is that in commutative, cancellative, Artinian ordered semigroups having an identity element, each proper prime ideal is a maximal ideal. Our results, with the appro-priate modifications, hold for groupoids (semigroups) without order, as well.

LetSbe an ordered groupoid. A nonempty subsetAofSis called an ideal ofSif (1) AS⊆A,SA⊆A; (2) ifa∈AandSb≤a, thenb∈A[2]. ForH⊆S, we denote (H] := {t∈S|t≤hfor someh∈H}. For a nonempty subsetAofS, we denote byI(A) the ideal ofSgenerated by A. We haveI(A)=(A∪AS∪SA∪SAS]. ForA= {a1,a2,...,an}, we

writeI(a1,a2,...,an) instead ofI({a1,a2,...,an}).An idealPofSis called proper ifP=S.

An idealP ofSis called prime if fora,b∈Ssuch thatab∈P, we havea∈P orb∈P. Equivalently, ifA,B⊆S,AB⊆P, this implies thatA⊆PorB⊆P. An ordered groupoid (S,·,≤) is called right (resp., left) cancellative if for eacha,b,c∈Ssuch thatac≤bc(resp., ca≤cb), we havea≤b. It is called cancellative if it is both right and left cancellative [3]. For commutative ordered groupoids, we just use the term “cancellative”. An identity of an ordered groupoidSis an element ofS, denoted bye, such thatea=ae=afor everya∈S. 2. Main results

Definition 2.1. An ordered groupoidSsatisfies theascending chain conditionfor ideals if, for any sequence of idealsI1,I2,...,Ii,...ofSsuch that

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there exists an elementn∈N such thatIm=In for eachm∈N,m≥n.N= {1, 2,...} always denotes the set of natural numbers.

Definition 2.2. An ordered groupoidSsatisfies themaximum conditionfor ideals if each nonempty set of idealsᏭofS, partially ordered by inclusion, has a maximal element. That is, for each nonempty setᏭof ideals ofS, there is an elementM∈Ꮽsuch that there is no elementT∈Ꮽsuch thatT ⊃M. Equivalently, ifT∈Ꮽsuch thatT⊇M, then T=M.

Definition 2.3. An idealTof an ordered groupoidSis calledfinitely generatedif there are elementsa1,a2,...,aninSsuch thatT=I(a1,a2,...,an).

Definition 2.4. An ordered groupoidSis calledNoetherianif every ideal ofSis finitely generated.

Theorem2.5. LetSbe an ordered groupoid. The following are equivalent: (i)Sis Noetherian;

(ii)Ssatisfies the ascending chain condition for ideals; (iii)Ssatisfies the maximum condition for ideals.

Proof. (i)⇒(ii) Let{Ti|i∈N}be a sequence of ideals ofSsuch that

T1⊆T2⊆ ··· ⊆Ti..., (2.2)

and letT:=i∈NTi. One can easily see thatT is an ideal ofS. Then, by (i), there exist

a1,a2,...,at∈Ssuch thatT=I(a1,a2,...,at). Clearly,a1,a2,...,at∈T. Letik∈Nsuch

thatak∈Tikfor eachk=1, 2,...,t. We putn:=max{i1,i2,...,it}.

Sinceik≤nfor allk=1, 2,...,t, we haveTik⊆Tnfor allk=1, 2,...,t. Hence

ak∈Tn ∀k=1, 2,...,t,.... (2.3)

Since Tn is an ideal of S, by (2.3), we haveI(a1,a2,...,at)⊆Tn. Hence we have T=

I(a1,a2,...,at)⊆Tn⊆T. ThenTn=T. We haveTm=Tnfor allNm≥n. Indeed, let

Nm≥n. Then we haveT=Tn⊆Tm⊆T, soTm=Tn.

(ii)⇒(iii) LetᏭbe a nonempty set of ideals ofS. Suppose there is no maximal ideal of Ꮽ. Then for eachJ∈Ꮽ, there exists

I∈Ꮽsuch thatJ⊂I. (2.4)

Indeed, letJ∈Ꮽ. Suppose there is noI∈Ꮽsuch thatJ⊂I. ThenJis a maximal element ofᏭ, which is impossible.

Let nowI1∈Ꮽ(Ꮽ= ∅). By (2.4), there existsI2∈Ꮽsuch thatI1⊂I2. SinceI2∈Ꮽ, by (2.4), there existsI3∈Ꮽsuch thatI2⊂I3. Continuing this way, we get a sequence of idealsI1,I2,...,Ii,...ofSsuch that

I1⊂I2⊂ ··· ⊂Ii···. (2.5)

Then, clearly, I1⊆I2⊆ ··· ⊆Ii···. Since Ssatisfies the ascending chain condition for

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(iii)⇒(i) LetTbe an ideal ofS. Let

Ꮽ:=K⊆S| ∃n∈N,a1,a2,...,an∈Tsuch thatK=Ia1,a2,...,an. (2.6) Leta∈T (T= ∅). SinceI(a)∈Ꮽ,Ꮽis a nonempty set of ideals of S. By (iii), there exists an elementM∈Ꮽwhich is maximal inᏭ. SinceM∈Ꮽ, there exist n∈Nand a1,a2,...,an∈T such thatM=I(a1,a2,...,an). ThenT=I(a1,a2,...,an). In fact, since

a1,a2,...,an∈T, we have I(a1,a2,...,an)⊆T. Letb∈T such thatb /∈I(a1,a2,...,an).

Since

M=Ia1,a2,...,an⊆Ia1,a2,...,an,b∈Ꮽ (2.7) andM is maximal inᏭ, we haveI(a1,a2,...,an,b)=I(a1,a2,...,an). Thenb∈I(a1,a2,

...,an), which is impossible.

Definition 2.6. An ordered groupoidSsatisfies thedescending chain conditionfor ideals if, for any sequence of idealsI1,I2,...,Ii,...ofSsuch that

I1⊇I2⊇ ··· ⊇Ii···, (2.8)

there exists an elementn∈Nsuch thatIm=Infor eachm∈N,m≥n.

Definition 2.7. An ordered groupoidSsatisfies theminimum conditionfor ideals if each nonempty set of idealsᏭofS, partially ordered by inclusion, has a minimal element. That is, for each nonempty setᏭof ideals ofS, there is an elementM∈Ꮽsuch that there is no elementT∈Ꮽsuch thatT⊂M. Equivalently, ifT∈Ꮽsuch thatT⊆M, thenT=M. Definition 2.8. An ordered groupoidSis calledArtinianifSsatisfies the descending chain condition for ideals.

Theorem2.9. An ordered groupoidSis Artinian if and only if it satisfies the minimum conditions for ideals.

Proof. “If ” part. This is the dual of (ii)⇒(iii) ofTheorem 2.5.

“Only if ” part. Let{Ti|i∈N}be a sequence of ideals ofSsuch that

T1⊇T2⊇ ··· ⊇Ti···. (2.9)

We putᏭ= {Ti|i∈N}. SinceᏭis a nonempty set of ideals ofS, by hypothesis, there is

an elementTninᏭwhich is minimal element ofᏭ. We haveTm=Tnfor everyNm≥n.

Indeed, letNm≥n. SinceTn⊇Tm∈ᏭandTn is a minimal element ofᏭ, we have

Tn=Tm.

A mapping f :S→T of an ordered groupoidSinto an ordered groupoidT is called homomorphism if (1) f(xy)= f(x)f(y) for allx,y∈S; and (2) ifx≤y implies that

f(x)≤f(y).

Lemma2.10. IfS,Tare two ordered groupoids, f :S→Ta homomorphism and onto map-ping, andIan ideal ofT, then f−1₍_I₎_{is an ideal of}_S_.

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Proposition2.11. LetSbe a Noetherian (resp., Artinian) ordered groupoid,Tan ordered groupoid, and f :S→Ta homomorphism and onto mapping. Then T is Noetherian (resp., Artinian).

Proof. LetSbe Noetherian and{Ji|i∈I}a family of ideals ofSsuch that

J1⊆J2⊆ ··· ⊆Ji···. (2.10)

LetIi:= f−1(Ji),i∈N.

ByLemma 2.10,Iiis an ideal ofSfor everyi∈N. Moreover, by (2.10), we have

I1⊆I2⊆ ··· ⊆Ii···. (2.11)

SinceSis Noetherian, there existsn∈Nsuch thatIm=Infor eachm∈N,m≥n. Then

Jm=Jnfor everym∈N,m≥n. In fact, letm∈N,m≥n. SinceIm= f−1(Jm) and f is onto, we have

fIm= ff−1Jm=Jm. (2.12)

Similarly, f(In)=Jn. SinceIm=In, we have f(Im)= f(In). ThenJm=Jn.

Lemma2.12 (cf. [4, Lemma 2]). Let(S,·,≤S)be an ordered groupoid, I an ideal of S. Let S/I:=S\I∪ {0}, where 0 is an arbitrary element ofI. Define an operation “∗” and an order “≤” onS/Ias follows:

∗:S/I×S/I−→S/I|(x,y)−→x∗y, x∗y:=

 

xy₀ if_ifxy_xy∈_∈S_I\I,

≤:=≤S∩ (S\I)×(S\I)∪(0,x)|x∈S/I.

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Then(S/I,∗,≤)is an ordered groupoid (called the Rees quotient of S by I) and 0 is its zero. In particular, if the multiplication on S is associative, thenS/Iis an ordered semigroup.

Lemma2.13. Let(S,·,≤S)be an ordered groupoid,Ian ideal ofS. Then the mapping

π:S,·,≤S−→(S/I,∗,≤)|a−→

  a

ifa∈S\I,

0 ifa∈I (2.14)

is a homomorphism and onto mapping.

Proof. The mappingπ is clearly well defined. Leta,b∈S. Thenπ(ab)=π(a)∗π(b). In fact, we have the following.

(1) Letab∈S\I. Then ifa∈I, thenab∈IS⊆I. Ifb∈I, thenab∈SI⊆I, which is impossible. Thus we havea,b∈S\I. Sincea,b,ab∈S\I, we haveπ(ab) :=ab, π(a) :=a,π(b) :=b. Sinceab∈S\I, we havea∗b:=ab. Hence we have

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(2) Letab∈I. Thenπ(ab) :=0. Then

(a) ifa,b∈I, thenπ(a) :=0,π(b) :=0, andπ(ab)=0=0∗0=π(a)∗π(b); (b) ifa∈I,b∈S\I, thenπ(a) :=0,π(b) :=b, andπ(ab)=0=0∗b=π(a)∗

π(b) (since 0b∈IS⊆I);

(c) ifa∈S\I,b∈I, thenπ(a) :=a,π(b) :=0, andπ(ab)=0=a∗0=π(a)∗ π(b);

(d) leta,b∈S\I. Thenπ(a) :=a,π(b) :=b. Sinceab∈I, we havea∗b:=0 and π(ab) :=0. Thus, we have

π(ab)=0=a∗b=π(a)∗π(b). (2.16) Leta,b∈S,a≤Sb. Thenπ(a)≤π(b). Indeed, we have the following.

(1) Ifa∈I, thenπ(a) :=0. Sinceb∈S, we haveπ(b)∈S/I. Then

π(a),π(b)=0,π(b)∈(0,x)|x∈S/I⊆≤. (2.17) (2) Leta∈S\I. Thenπ(a) :=a.

Ifb∈I, then sinceSa≤Sb∈IandIis an ideal ofS, we havea∈I, which is impossible. Thusb∈S\Iandπ(b) :=b. Then we have

π(a),π(b)=(a,b)∈≤S∩ (S\I)×(S\I)⊆≤. (2.18)

The mappingπis clearly an onto mapping. This is because the setIis nonempty. Notation 2.14. LetSbe an ordered groupoid andIan ideal ofS. IfAis an ideal ofSsuch thatI⊆A, we denoteA/I:=A\I∪ {0}, where 0 is the zero ofS/I.

Lemma2.15. Let(S,·,≤S)be an ordered groupoid and I an ideal ofS. LetAbe an ideal ofS

such thatI⊆A. Then the setA/Iis an ideal ofS/I.

Proof. (A) We haveπ(A)=A/I. In fact, letx∈π(A). Thenx=π(y) for some y∈A. Then the following hold.

(1) Ify∈I, thenx=π(y) :=0∈A/I. (2) Ify∈A\Ithen, sinceA\I⊆S\I, we have

x=π(y) :=y∈A\I⊆A/I. (2.19) Letx∈A/I. Thenx∈π(A). Indeed, we have the following.

(1) Ifx∈A\I, thenx∈S\I, andπ(x) :=x. Sincex∈A, we haveπ(x)∈π(A), thus we havex∈π(A).

(2) Letx=0. Take an elementy∈I(I= ∅). Thenπ(y) :=0. Since y∈I⊆A, we haveπ(y)∈π(A), thus we havex∈π(A).

(B) The setπ(A) is an ideal ofS/I. Indeed,∅ =π(A)⊆S/I(sinceA= ∅).

Leta∈S/I,b∈π(A). Leta=π(x) for somex∈S(πis onto) andb=π(y) for some y∈A. Sinceπis a homomorphism andxy∈SA⊆A, we have

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LetS/Ia≤b∈π(A). Thena∈π(A). Indeed, leta=π(x) for somex∈S(πis onto) andb=π(y) for somey∈A. Then the following hold.

(1) Ifx∈I, thenx∈A, hencea=π(x)∈π(A).

(2) Letx∈S\I. Thena=π(x) :=x∈S\I, soa=0 (since 0∈I).

Since (a,b)∈≤=(≤S∩[(S\I)×(S\I)])∪ {(0,x)|x∈S/I}anda=0, we have

a≤Sb,

a,b∈S\I. (2.21)

Ify∈I, thenS\Ib=π(y) :=0, which is impossible. Thus we havey∈S\I. Thenb= π(y) :=y∈A. SinceSa≤Sb∈AandAis an ideal ofS, we havea∈A, soπ(a)∈π(A).

Sincea∈S\I, we haveπ(a) :=a. Hence we havea∈π(A). Remark 2.16. IfSis a set andT1,T2,Iare subsets ofSsuch thatT1⊆T2,T2∩I⊆T1∩I andT2\I⊆T1\I, thenT1=T2. Indeed, leta∈T2. Ifa∈I, thena∈T2∩I⊆T1∩I⊆T1. Ifa /∈I, then

a∈T2\I⊆T1\I⊆T1. (2.22)

Clearly, ifT1⊆T2, thenT2∩I⊆T1∩Iis equivalent toT1∩I=T2∩IandT2\I⊆T1\I is equivalent toT1\I=T2\I.

Remark 2.17. If (S,·,≤S) is an ordered groupoid, then each nonempty subsetAofSwith

the multiplication “◦” and the order “≤A” onAdefined by

◦:A×A−→A|(a,b)−→a.b

≤A:=≤S∩(A×A) (2.23)

is an ordered groupoid (a subgroupoid ofS). In particular, if the multiplication onSis associative, thenAis an ordered semigroup.

Theorem 2.18. Let S be an ordered groupoid andI an ideal of S. If bothI andS/I are Noetherian, then so isS.

Proof. Let{Ti|i∈N}be a sequence of ideals ofSsuch that

T1⊆T2⊆ ··· ⊆Ti⊆ ···=⇒ ∃n∈Nsuch thatTm=Tn∀m∈N,m≥n? (2.24)

SinceTk,Iare ideals ofS,Tk∩Iis an ideal ofSfor allk∈N. We putIk:=Tk∩I(k∈N),

and we have

I1⊆I2⊆ ··· ⊆Ii⊆ ···. (2.25)

SinceIkis an ideal ofSandIk⊆I,Ikis an ideal ofIfor eachk∈N. SinceIis Noetherian, there existsp∈Nsuch that

Ip=Im ∀Nm≥p. (2.26)

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We putMk:=Tk∪I(k∈N), and we have M1⊆M2⊆ ··· ⊆Mi⊆ ···

=⇒M1\I⊆M2\I⊆ ··· ⊆Mi\I⊆ ···

=⇒M1\I

∪ {0} ⊆M2\I

∪ {0} ⊆ ··· ⊆Mi\I∪ {0}

⊆ ··· (where 0 is the zero ofS/I)

=⇒M1/I⊆M2/I⊆ ··· ⊆Mi/I⊆ ···.

(2.27)

SinceMkis an ideal ofSandI⊆Mk, byLemma 2.15, the setMk/Iis an ideal ofS/Ifor eachk∈N. SinceS/Iis Noetherian, there existsq∈Nsuch that

Mq/I=Mm/I ∀Nm≥q. (2.28)

We putn:=max{p,q}.

Letm∈N,m≥n. ThenTm=Tn. Indeed, sincem≥n, we have

Tn⊆Tm. (2.29)

Sincem≥n≥p, by (2.26), we haveIp=Im=In, then

Tm∩I=Tn∩I. (2.30)

Sincem≥n≥q, by (2.28), we haveMq/I=Mm/I=Mn/I, then

Mm\I∪ {0} =Mn\I∪ {0}. (2.31)

SinceMm:=Tm∪I⊆S, we haveMm\I⊆S\I. Since 0∈/ S\I, 0∈/ Mm\I.

Similarly, 0∈/ Mn\I. By (2.31), we have Mm\I =Mn\I. Then, sinceMm\I=(Tm∪

I)\I=Tm\IandMn\I=Tn\I, we have

Tm\I=Tn\I. (2.32)

By (2.29), (2.30), (2.32), andRemark 2.16, we haveTn=Tm.

In a similar way, we prove the following.

Theorem 2.19. LetS be an ordered groupoid andI an ideal of S. If bothI andS/I are Artinian, then so isS.

In the following, we consider ordered semigroups.

Definition 2.20. An idealMof an ordered semigroupSis called maximal if the following hold.

(1)M=S.

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For convenience, we denoteS1_:₌_S_{∪ {}₁_}_{, where 1 is a symbol denoting that}_a₁₌₁_a₌ afor alla∈S. We have

I(a)=(a∪Sa∪aS∪SaS]=(a]∪(Sa]∪(aS]∪(SaS]

= {t∈S| t≤aort≤za;z∈Sort≤ah;h∈Sort≤xay;x,y∈S}. (2.33)

Using this notation, for an elementt∈I(a), we can writet≤xayfor somex,y∈S1_. Also, the relationban≤zan+1_;_z_∈_S1_in_{Theorem 2.21}_{, means that}_ban_≤_an+1_or_ban_≤

zan+1_{for some}_z_∈_S_.

Theorem2.21. LetSbe a commutative, Artinian ordered semigroup. Letb∈Ssuch that I(b)=SandPa proper prime ideal ofShaving the property:

Ifa∈S\P,z∈S1_,_n_∈_N,_ban_≤_zan+1_,_then_b_≤_za. _(2.34) ThenPis a maximal ideal ofS.

Proof. By hypothesis,P=S. Let nowT be an ideal ofS such thatT⊃P. ThenT=S. Indeed, leta∈T,a /∈P. Clearly, we have

I(a)⊇Ia2_⊇_I_a3_{⊇ ··· ⊇}_I_an_{⊇ ···} _(2.35)

whereI(ak) is an ideal ofSfor everyk∈N. SinceSis Artinian, there existsn∈Nsuch thatI(am)=I(an) for everyNm≥n. SoI(an+1₎₌_I₍_an_{). Since}

ban_∈_San_⊆_I_an₌_I_an+1_, _(2.36)

there existx,y∈S1 _{such that}_ban_≤_xan+1_y₌_xyan+1_{. Since}_x_,_y_∈_S1_{, we have}_xy_∈_S1_. Sincea∈S\P,xy∈S1_,_n_∈_{N, and}_ban_≤₍_xy₎_an+1_{, by hypothesis, we have}

b≤(xy)a∈Sa⊆I(a). (2.37)

Sincea∈T andT is an ideal ofS, we haveI(a)⊆T. SinceSb≤(xy)a∈T, we have b∈T. ThenI(b)⊆T. By hypothesis,I(b)=S. So we haveT=S.

Corollary2.22. LetS be a commutative, Artinian ordered semigroup with an identity element “e”. Then each proper prime idealP ofShaving the property “ifa∈S\P,z∈S1_, n∈N,an_≤_zan+1_{, then}_e_≤_za_{”, is a maximal ideal of}_S_.

Proof. We have

I(e)=(e∪Se∪eS∪SeS]=e∪S∪S2₌₍_e_∪_S_]₌_S. _(2.38)

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Theorem2.23. LetSbe a commutative, Artinian ordered semigroup having an elementb such thatI(b)=S. Then each proper prime idealPofSfor which the ordered semigroupS\P is cancellative is a maximal ideal ofS.

Proof. LetPbe a proper prime ideal ofSand letS\Pbe cancellative. It is enough to prove that condition (2.34) ofTheorem 2.21is satisfied.

Leta∈S\P,z∈S1_,_n_∈_{N, and}_ban_≤_zan+1_{. Then}_b_≤_za_{. In fact, we have the} follow-ing.

(A)b /∈P. Sinceb∈Pimplies thatS=I(b)⊆P, thenP=S, which is impossible. (B)an∈/ P. Indeed, ifan∈P then, sinceP is prime anda /∈P, we havean−1_∈_P_.

Continuing this way, we get thata∈P, which is impossible.

(C)za /∈P. Indeed, since z∈S1 _and_a_∈_S_{, we have}_za_∈_S_{. Let} _za_∈_P_{. Then} _S ban_≤_zan+1₌₍_za₎_an_∈_PS_⊆_P_{. Then}_ban_∈_P_{. Since}_P_{is prime, we have}_b_∈_P oran∈P, which is impossible.

Sinceban≤(za)an;b,an,za∈S\PandS\Pis cancellative, we get thatb≤za.

Corollary2.24. LetSbe a commutative, cancellative, Artinian ordered semigroup having an elementbsuch thatI(b)=S. Then each proper prime idealPofSis a maximal ideal ofS.

Corollary2.25. LetSbe a commutative, Artinian ordered semigroup having an identity element e. Then each proper prime ideal P of S for which the ordered semigroupS\P is cancellative is a maximal ideal ofS.

Corollary2.26. LetSbe a commutative, cancellative, Artinian ordered semigroup having an identity elemente. Then each proper prime idealPofSis a maximal ideal ofS.

Theorem2.27. LetSbe a commutative, Artinian ordered semigroup,ban element ofSsuch thatI(b)=S, andPa proper prime ideal ofShaving the property:

Ifa,c∈S\Psuch thatbc≤ac,thenb≤a. (2.39) ThenPis a maximal ideal ofS.

Proof. It is enough to prove that condition (2.34) ofTheorem 2.21is satisfied. Leta∈ S\P,z∈S1_,_n_∈_{N, and}_ban_≤_zan+1_{. Then}_b_≤_za_{. In fact, as in}_{Theorem 2.23}_{, we prove} thatb /∈P,an∈/ P,za /∈P. On the other hand,

ban_≤₍_za₎_an_; _za_,_an_∈_S_\_P. _(2.40)

By hypothesis, we haveb≤za.

Corollary2.28. LetSbe a commutative, Artinian ordered semigroup having an identity elemente. Then each proper prime idealPofShaving the property “ifa,c∈S\Psuch that c≤ac, thene≤a”, is a maximal ideal ofS.

Acknowledgments

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the paper. This research was supported by the Special Research Account of the University of Athens (Grant no. 70/4/5630).

References

[1] D. M. Burton,A First Course in Rings and Ideals, Addison-Wesley, Massachusetts, 1970. [2] N. Kehayopulu,On weakly prime ideals of ordered semigroups, Math. Japon.35(1990), no. 6,

1051–1056.

[3] N. Kehayopulu and M. Tsingelis, On separative ordered semigroups, Semigroup Forum 56

(1998), no. 2, 187–196.

[4] ,Ideal extensions of ordered semigroups, Comm. Algebra31(2003), no. 10, 4939–4969.

Niovi Kehayopulu: University of Athens, Department of Mathematics, 15784 Panepistimiopolis, Greece

E-mail address:[email protected]