INCIDENCE ALGEBRAS

INCIDENCE ALGEBRAS

SURJEET SINGH AND FAWZI AL-THUKAIR

Received 19 March 2005 and in revised form 12 July 2005

LetXbe any partially ordered set,Rany commutative ring, andT=I^∗(X,R) the weak incidence algebra ofX overR. LetZbe a finite nonempty subset ofX,L(Z)= {x∈X: xzfor somez∈Z}, andM=Te_Z. Various chain conditions onM are investigated.

The results so proved are used to construct some classes of right perfect rings that are not left perfect.

1. Introduction

Let R be a commutative ring andX a partially ordered set. Let T =I^∗(X,R) be the set of all functions f :X×X→R such that f(x,y)=0, whenever xy, and{(x,y) : f(x,y)=0 andx < y}is finite. ThenTis anR-algebra under the operations defined as follows. For any f,g∈T,x,y∈X, andr∈R, (f +g)(x,y)=f(x,y) +g(x,y), f g(x,y)=

x≤z≤yf(x,z)g(z,y), andr f(x,y)=r·f(x,y). The algebraTis calledweak incidence al- gebraofXoverR. For a locally finite partially ordered setY, the concept of incidence al- gebraI(Y,R) is well known [6]. It can be proved on similar lines as for incidence algebras that for any two partially ordered setsX,Zand any two indecomposable commutative ringsR,S,I^∗(X,R) andI^∗(Z,S) are isomorphic as rings if and only ifXandZare iso- morphic andRandSare isomorphic [5]. It has been seen in [1,5] that weak incidence algebras can be used to construct rings whose left and right maximal rings of quotients need not be isomorphic. Here we give some more such applications. IfXis infinite, ob- viouslyTis neither left nor right artinian or Noetherian. In the present paper we study chain conditions on a specific one-sided ideal ofT. LetZbe a finite nonempty subset of X,L(Z)= {x∈X:x≤zfor somez∈Z}, andM=TeZ, where for any subsetY ofX, e_Y∈Tis such thate_Y(x,x)=1 for everyx∈Y, ande_Y(x,y)=0 otherwise.Theorem 3.5 shows thatM is an artinian leftT-module if and only ifRis artinian andL(Z) satisfies dccand has no infinite antichain.Theorem 5.2gives a similar result forM to be Noe- therian. InSection 4, the construction of partially ordered sets satisfyingdccbut having no infinite antichains is studied. InSection 6, perfect rings are studied, as an application;

Theorem 3.5is used to construct a class of right perfect rings that are not left perfect.

Copyright©2005 Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences 2005:15 (2005) 2389–2397 DOI:10.1155/IJMMS.2005.2389

2. Preliminaries

Throughout, all rings have identity element 1=0. LetX be a partially ordered set and Ra commutative ring. A subsetSofXis called anantichaininXif no two members of Sare comparable [6]. We will apply the terminology for incidence algebras given in [6]

for weak incidence algebras. As usual, for anyx < yinX,exydenotes the corresponding matrix unit inT=I^∗(X,R). NowK^∗(X,R)= {f ∈I^∗(X,R) : f(x,x)=0 for eachx∈X} is an idealK^∗(X,R) contained in its lower nil radical, andT/K^∗(X,R)^∼₌Πx∈XRx, with eachRx=R. The following is immediate.

Lemma2.1. LetM be an artinian (Noetherian) left module overT=I^∗(X,R)such that K^∗(X,R)M=0, then for some finite subsetZ ofX,(1−e_Z)M=0. In particular, ifM is artinian, thenMhas finite composition length as anR-module.

3. Artinian modules

A partially ordered setX is said to satisfy strongdcc if it does not contain an infinite sequence x1,x2,. . .,xn,. . . such thatxjxi wheneveri < j. Let Z be a finite nonempty subset ofXandM=TeZ=

x∈ZTexx. NowTMis artinian if and only ifTexxis artinian for everyx∈Z. A finite union of subsets ofXsatisfies strongdccif and only if each of the subsets satisfies strongdcc. SupposeMis artinian. ThenRis artinian. SupposeL(Z) does not satisfy strongdcc. Then there exists anx0∈Zsuch thatL(x0) does not satisfy strong dcc. Therefore there exists an infinite sequence inL(x0) :x1,x2,. . .,x_n,. . .such thatx_jx_i wheneveri < j. For anyn≥1, letNn=

k≥nTexkx0. ThenNn+1⊂Nn⊆M, which contra- dicts the assumption thatMis artinian. HenceL(Z) satisfies strongdcc. We now discuss the converse of this result. Henceforth we assume thatR is artinian andL(Z) satisfies strongdcc. SupposeMis not artinian. Without loss of generality we takeZ= {x0}and M=Tex0x0. There exists an infinite properly descending chain ofT-submodules ofM: N1⊃N2⊃ ··· ⊃N_n⊃ ···.For eachi≥1 andx∈L(x0), letA⁽ⁱ⁾x = {a∈R:ae_xx0∈N_i}. Lemma3.1. (i)A⁽ⁱ⁾x ⊆A⁽ⁱ⁾y whenevery≤xinL(x0).

(ii)For anyx∈L(x0),A⁽ⁱ⁾x ⊆A^(j)x wheneverj≤i.

(iii)IfA⁽ⁱ⁾x ⊂A^(j)y , then eitherj≤iorxy.

(iv)IfA⁽ⁱ⁾x A^(j)y , then eitheryxori < j.

Proof. (i) and (ii) are obvious.

(iii) Suppose ji. Theni < j, andA^(j)x ⊆A⁽ⁱ⁾x ⊂A⁽y^j). Ifx≤y, thenA^(j)y ⊆A^(j)x ⊆A⁽ⁱ⁾x , which is a contradiction.

(iv) Suppose y≤x. ThenA^(j)x ⊆A⁽y^j). If j≤i, thenA⁽ⁱ⁾x ⊆A⁽x^j), thereforeA⁽ⁱ⁾x ⊆A^(j)y ,

which is a contradiction.

LetS be the set of allA⁽ⁱ⁾x withx∈L(x0) andi≥1. Let A∈S. For somex∈L(x0) and ani,A=A⁽ⁱ⁾x . AsL(x0) satisfiesdcc, by keepingifixed we can findxminimal with respect to the pair (A,i). If for some j > i,A=A^(j)x , then we can find minimalx≤xfor whichA=A^(j)_x. Hence we can find anx∈L(x0) and a positive integertsuch thatA=A^(t)x

such that if for someu≥tandy≤x,A=A^(u)y , thenx=y. Keeping this in mind, a triple

(A,t,x) is called acritical tripleifA∈S,A=A^(t)x , and if for someu≥t,y≤x,A=A^(u)y , theny=x. For any subsetVofS, the set of thosex∈L(x0) such that (A,t,x) is a critical triple for someA∈V andt1 is called theL(x0)-co-supportofV.

Lemma3.2. (a)LetA,B∈S. If for some positive integeri,(A,i,x)and(B,i,y)are critical triples andx=y, then one of the following holds:(i)x < yandB⊂A,(ii)y < xandA⊂B, and(iii)xandyare noncomparable.

(b)If(A,i,x)and(B,j,y)are two critical triples withAandBnoncomparable or equal, andx < y, then j < i.

Proof. (a) is immediate. (b) NowB=A^(j)y ⊆A^(j)x . Ifi≤j, thenA^(j)x ⊆A⁽ⁱ⁾x =A, therefore A=B=A⁽x^j) and (A,j,y) is a critical pair. But also A=A^(j)x , hencex=y, which is a

contradiction. Hence j < i.

Lemma3.3. LetY⊆Sbe an antichain. ThenYis finite.

Proof. LetZ be the co-support of Y. For anyi, let Y(i)= {A∈Y : (A,i,x) is a critical triple for somex∈Z}. LetZibe the set of thosex∈L(x0) such that (A,i,x) is a critical triple for someA∈Y(i). It follows fromLemma 3.2(a) thatZiis an antichain, soZi is finite. If for someA,B∈Y(i), (A,i,x) and (B,i,y) are critical triples andA=B, clearly x=y. HenceY(i) is finite. LetZ1 be the set of minimal members of Z. Fix anx∈Z1

and a critical triple (A,k,x). Consider any critical triple (B,i,y) withx < yandB∈Y. By Lemma 3.2(b),i < k. LetY_x= {B∈Y: there exists a critical triple (B,i,y) withx≤y}. It follows thatYx=_k

i=1(Yx∩Y(i)) is finite. AsZ1is finite andY =

x∈ZYx, we getY is

finite.

Lemma3.4. Sis finite.

Proof. For anyA∈S, the set S_A of all thoseB∈S which are minimal with respect to A < Bis finite byLemma 3.3. Also the setY1of minimal members ofSis finite. After this by using the fact thatRhas finite composition length, we getSis finite.

Theorem3.5. LetT=I^∗(X,R), whereXis any partially ordered set andRis a commutative ring. LetZ be a finite nonempty subset ofX andM=TeZ. ThenMis an artinian leftT- module if and only ifRis artinian andL(Z)satisfies strongdcc.

Proof. As remarked earlier it is enough to takeM=Te_x0x0. Suppose thatL(x0) satisfies strongdccandRis artinian. SupposeM is not artinian. SoM has an infinite properly descending chain ofT-submodules: N1⊃N2⊃ ··· ⊃Nn⊃ ···.We use the notations given above this result. LetA∈S. Fix anx∈L(x0). SupposeA=A⁽ⁱ⁾x for somei. Then either there exists a smallest positive integers_(x,A)such thatA₌A^(j)x for everyj_≥s_(x,A)or there exists a largest positive integerk_(x,A)such thatA=A^(kx^(x,A)). LetZ_Abe the set of those x∈Xfor whichAadmits the positive integerk(x,A). Suppose there is no upper bound on k(x,A)asxranges overZA. So there exists an infinite sequence:x1,x2,. . .,xn,. . .inZAsuch thatk_(xi,A)> k_(xj,A)wheneveri > j. Thenx_ix_jwheneveri < j. This contradicts the as- sumption thatL(x0) satisfies strongdcc. Hence there exists a positive integerkAsuch that k(x,A)< kAfor everyx∈ZA. AsSis finite, we can find a positive integerusuch that for

anyA∈S,x∈L(x0),s(x,A)< uandk_(x,A)< u, whenevers_(x,A)ork_(x,A)is defined. Con- siderNu. If for someA^(u)x ,A^(u)x ⊃A^(u+1)x , then forA=A^(u+1)x we havek(x,A)> kors(x,A)> u, which is a contradiction. HenceA^(u)x =A^(u+1)x . This proves thatNk=Nk+1, which is also a

contradiction. HenceMis artinian.

Remark 3.6. LetXbe a partially ordered set satisfying strongdcc, andRan artinian com- mutative ring. It follows from the above theorem that forT=I^∗(X,R), any finitely gen- erated left ideal contained inA=

x∈XTexx satisfiesdcc. As the idealK^∗(X,R)= {f ∈ T: f(x,x)=0 for everyx∈X} ⊆A, and it is nil,K^∗(X,R) is rightT-nilpotent. However this ideal need not be leftT-nilpotent. For example, letNbe the set of natural numbers with usual ordering. Then for any fieldF,K^∗(N,F) is not leftT-nilpotent.

4. Partially ordered sets

We now prove some results that can help in constructing partially ordered sets satisfying strongdcc.

Proposition4.1. A partially ordered setXsatisfies strongdccif and only if it satisfiesdcc and it has no infinite antichain.

Proof. IfXsatisfies strongdcc, obviously it cannot have an infinite antichain. Conversely, letX satisfy strongdccand have no infinite antichain. Suppose there exists an infinite sequence{xi}inXsuch thatxjxiwheneveri < j. Thesexiare distinct. LetAbe the set of thesexiandSthe set of minimal members ofA. ThenSis a finite nonempty set. So there exists anx_i∈Ssuch thatx_i< x_j for infinitely many values of j. As a consequence, we can find ak > isuch thatxi< xk, which is a contradiction. HenceX satisfies strong

dcc.

Theorem 4.2. Let X andY be two partially ordered sets satisfying strong dcc, then the partially ordered setZ=X×Ywith the ordering given by(a,b)≤(c,d)if and only ifa≤c andb≤dsatisfies strongdcc.

Proof. ThatZsatisfiesdccis obvious. SupposeZhas an infinite antichainS. LetA1and A2be sets ofX-components andY-components respectively of the members ofS. AsY does not contain an infinite antichain, for any fixedx∈A1, there are only finitely many y∈A2 such that (x,y)∈S. Also, the number of minimal members of A1 is finite. So there exists a minimal memberx1∈A1such thatT1= {(x,y)∈S:x1< x}is infinite. Fix an (x1,y1)∈S. If (x,y)∈T1, then eithery < y1oryandy1are noncomparable. ThusT1

satisfies one of the following conditions.

(i) There are infinitely many (x,y)∈T1such thaty < y1.

(ii) There are infinitely many (x,y)_∈T1such thatyandy1are noncomparable.

Suppose (i) is satisfied. ThenS1= {(x,y)∈T1:y < y1}is infinite. As forS, we can find an (x2,y2)∈S1such thatT2= {(x,y)∈S1:x2< x}is infinite. Nowy2< y1. Suppose T2 also satisfies (i), that gives rise to a subsetS2analogous toS1. Continue the process, and this gives a descending chain inY. AsY satisfiesdcc, this process must end after a finite number of steps. Thus we get a subset V1 of S1 and an element (u1,v1)∈V1

such that V2= {(x,y)∈V :u1< x, yandv1are not comparable} is infinite. Thus for

any infinite antichain S in Z, there exists a (u,v)∈S, such thatT = {(x,y)∈S:u <

x, yis not comparable withv}is infinite, soT satisfies (ii). Suppose for somen≥2, we have constructed infinite setsVi inS, for 1≤i≤n, (ui,vi)∈Vi, for 1≤i≤n−1 with Vi+1= {(x,y)∈Vi:ui< x,viandyare noncomparable}. NowVnhas an element (un,vn) such thatV_n+1= {(x,y)∈V_n:u_n< x,v_nandyare noncomparable}is infinite. This in- ductive process gives an infinite setL= {(ui,vi) :i≥1} ⊆Ssuch that ui< ui+1for any i≥1, butB= {vi:i≥1}is an infinite antichain inY. This is a contradiction. HenceZ

satisfies strongdcc.

Example 4.3. For any finite collection of well-ordered sets, their direct product as defined in the above theorem satisfies strongdcc.

Definition 4.4. LetXbe a partially ordered set satisfyingdcc. For any nonnegative integer, definesi(X) as follows. Firstly,s0(X) is the set of all minimal elements inX. For anyi≥0, anx∈s_i+1(X), if it is minimal with respect to the property that for somey∈s_i(X),y < x.

DefineB1(X)=

i≥0si(X).

Lemma4.5. LetXbe any partially ordered set satisfyingdcc.

(i)Everysi(X)is an antichain. In addition ifXsatisfies strongdcc, then everysi(X)is finite.

(ii)If for somei >0, anx∈si(X), then there exists a sequencex0< x1<···< xi=xsuch thatxj∈sj(X)for0≤j≤i.

(iii)Letx∈s_i(X)for somei,y∈s_j(X)for somej > i. Thenyx.

Proof. (i) is immediate from the definition andProposition 4.1.

(ii) follows by using induction oni.

(iii) Supposey≤x. By using (ii) we haveyi−1< y≤x. ByDefinition 4.4,y=x. At the same time, asj > i, by (ii), there existsz∈s_i(X) such thatz < y. This contradicts

(i). Hence the result follows.

Definition 4.6. Let X be a partially ordered set satisfying dcc. For any ordinal α, de- fine Bα(X) as follows.B0(X)= ∅, the empty set, if α=β+ 1, then Bα(X)=Bβ(X)∪ B1(X\Bβ(X)). Ifαis a limit ordinal, thenBα(X)=

β<αBβ(X).

Lemma4.7. LetXbe any partially ordered set satisfying strongdcc.

(i)B1(X)is countable.

(ii)For any two ordinalsβ < α, ifα=β+γ, thenBα(X)=Bβ(X)∪Bγ(X\Bβ(X)).

(iii)X=Bα(X)for some ordinalα.

(iv)SupposeX=B_α(X)for some smallest ordinalα. If for everyβ < α,B1(X\B_β(X))is linearly ordered, thenXis linearly ordered.

Proof. (i) is immediate fromLemma 4.5.

(ii) follows fromDefinition 4.4by using transfinite induction onγ.

(iii) If X=B1(X), there is nothing to prove. Suppose X=B1(X). Then B1(X) is countably infinite. It follows from the definition of aB_β(X) that ifX=B_β(X), then|Bβ(X)| ≥ |β|. Now there exists a smallest ordinal βsuch that|β|>|X|.

ThenX=Bβ(X). Finally (iv) is obvious.

Remark 4.8. LetX be a partially ordered set satisfying strongdcc. IfXis infinite, then eachs_i(X) is nonempty andB1(X) is countably infinite. So, the given ordering onB1(X) can be extended to a linear ordering such thatB1(X) becomes isomorphic to the set of natural numbers. Now extend the ordering onX as follows. Letx,y∈X. Ifx∈Bα(X) andy∈B_β(X) such thatα < βand y /∈B_α(X), then setx < y. For any ordinalα, extend the ordering onB1(X)\Bα(X), such that it embeds in the set of natural numbers. This makesX a linearly ordered set satisfyingdcc. The order on any partially ordered set can be extended to a linear order, [3, Chapter 1]. Here, we see thatXcan be made into a well- ordered set. Let (Y,) be any linearly ordered set with the following properties. (i) For some ordinalα,Yis a union of an ascending chain of subsets{Yβ}β≤α, with eachYβ+1\Yβ

embeddable in the set of natural numbers. (ii) For any limit ordinalβ≤α,Y_β=

γ<βY γ.

(iii) For anyy∈Y\Yβandx∈Yβ,x < y. ThenYsatisfiesdcc. For eachβ < α, consider any orderingβonYβ+1\Yβunder whichYβ+1\Yβsatisfies strongdccand the given ordering onY_β+1\Y_βextendsβ. This defines an orderingonY that for eachβ < αcoincides withβonYβ+1\Yβand equalsotherwise. Then (Y,) satisfies strongdcc.

5. Noetherian modules

LetX be a partially ordered set.Xis said to satisfy strongaccif it does not contain an infinite sequencex1,x2,. . .,xn,. . .such thatxjxiwheneverj > i.

As inSection 3, we considerM=TeZ, whereZis a finite nonempty subset ofX. IfTM is Noetherian, it follows on similar lines as inSection 3that R is Noetherian andL(Z) satisfies strongacc.

To prove the converse of the above remark, throughout we take R to be Noether- ian,Z= {x0}, and x0∈X such thatL(x0) satisfies strong acc. Let N be a submodule of M. For eachx∈L(x0), setA_x= {a∈R:ae_xx0∈N}. Each A_x is an ideal ofR and N=

x∈L(x0)Axexx0. Forx≤yin L(x0),Ay⊆Ax. LetSbe the set of allAx,x∈L(x0).

Consider any subsetK ofS. For anyA_∈K, asL(x0) satisfiesacc, we can findx_∈L(x0) maximal with respect to the property thatA=A_x. LetZ(K) be the set of all such maximal elements ofL(x0).

Lemma5.1. LetY⊆Sbe an antichain. ThenZ(Y)is an antichain andYis finite.

Proof. Letx,y∈Z(Y) such thatx≤y. For someA,B∈Y,A=AxandB=Ay. However Ay⊆Ax, soA=B. Asxis maximal with respect to A, we getx=y. HenceZ(Y) is an antichain, soZ(Y) is finite. For eachA∈Y, there exists anx∈Z(Y) such thatA=A_x. Thus there exists a mapping ofZ(Y) ontoY. HenceY is finite.

Theorem5.2. LetT=I^∗(X,R)whereXis a partially ordered set andZis a finite nonempty subset ofX. ThenM=TeZ is a NoetherianT-module if and only ifRis Noetherian and L(x0)satisfies strongacc.

Proof. Without loss of generality we takeZ= {x0}. We use notations given aboveLemma 5.1. LetRbe Noetherian andL(x0) satisfy strongacc. LetNbe aT-submodule ofM. AsR is Noetherian,Y1= {A∈S:Ais maximal inS}is nonempty and no two members ofY1

are comparable. SetZ1=Z(Y1). ConsiderN1=

x∈Z1TA_xe_xx0. Lety≤xwithx∈Z1, then Ax⊆Ay, thereforeAy=Ax andAyeyx0=eyx(Axexx0)⊆N1. HenceN1=

x∈L(Z1)Axexx0. Suppose, for somen≥1, we have already defined subsetsZ1,Z2,. . .,Zn,Vn=_n

i=1Zi, and

N_n=

x∈VnTA_xe_xx0 such that the following hold. (i)N_n=

x∈L(Vn)A_xe_xx0, (ii) for any y∈L(V_n), there exists anx inV_nsuch that y≤x andA_y=A_xi, and (iii) for any y∈ L(x0)\L(Vn), there existsx∈Znsuch thatAy< Ax. SetSn+1= {A∈S:A=Axfor some x∈L(x0)\L(Vn)}andYn+1the set of all maximal members ofSn+1. SetZn+1=Z(Yn+1), V_n+1=V_n∪Z_n+1, andN_n+1=

x∈VnTA_xe_xx0. The above three conditions are obviously satisfied byN1. Suppose they are satisfied byNnfor somen. Supposey∈Zn+1andx∈X such thatx < y. Then Ay⊆Ax. Ifx /∈L(Vn), Ax=Ay. If y∈L(Vn), by (ii) there ex- istsz∈V_n⊆V_n+1 such that y≤z and A_y=A_z. HenceN_n+1=

x∈L(Vn+1)A_xe_xx0. Thus Nn+1satisfies (i), (ii), and (iii). For eachifor whichZiis non-empty, fix anxi∈Zi. If anL(Zi)=S, obviouslyZi+1= ∅. Fori < j, asL(Vi)∩Zj= ∅,xjxi. AsL(x0) satisfies strongacc, it follows that there exists annsuch thatZ_n= ∅butZ_n+1= ∅. Consequently, L(x0)=L(Vn),Nn=N. AsVnis finite, eachAxis finitely generated as anR-module, and N=

x∈VnTAxexx0, it follows thatNis a finitely generatedT-module. HenceMis Noe-

therian.

Remark 5.3. LetX be the dual of a partially ordered setX. For any commutative ring R, setT=I^∗(X,R) andT=I^∗(X,R). These two algebras are naturally anti-isomorphic.

LetZbe a finite nonempty subset ofX,M=eZT, andU(Z)= {x∈X:x≥zfor somez∈ Z}. By using the anti-isomorphism betweenTandTand Theorems3.5and5.2, we get the following results:

(i)MT is artinian if and only ifRis artinian andU(Z) satisfies strongacc;

(ii)M_T is Noetherian if and only ifRis Noetherian andU(Z) satisfies strongdcc.

Remark 5.4. LetXbe a locally finite partially ordered set, andT=I(X,R) the incidence algebra of X over a commutative ringR. SupposeR is artinian and for somex0∈X, L(x0) satisfies strongdcc. AsL(x0) has finitely many minimal elements,L(x0) is a finite set, soM=Te_x0x0, being a finite direct sum of copies ofR, is trivially an artininian left T-module. HenceM is an artinian leftT-module if and only ifRis artinian andL(x0) satisfies strongdcc.

Now supposeRis Noetherian,L(x0) satisfies strongacc, andN is aT-submodule of M. As in the proof ofTheorem 5.2, we haveY1andZ1=Z(Y1). ConsiderN_nas defined in the proof ofTheorem 5.2. NowN1=

x∈Z1TAxexx0. For anyx∈Z1, letGx= {bx j: 1jnx}generateAxas anR-module. Consider any f ∈N1withDf = {z∈L(x0) : f(z,x0)=0} ⊆L(Z1). Letz∈L(Z1). Then for anyx∈Z1,A_z=A_x wheneverzx. So f(z,x0)=

zx_nx

j=1rzx jbx j, wherex∈Z1. Then the formal sumgx j=

zxrzx jezx∈T and f =

x∈Z1

_nx

j=1gx jbx jexx0∈N1. HenceN1= {f ∈N1:Df ⊆L(Z1)}. Inductively, one can prove that for anyn1,N_{n= {}f _∈N:D_{f ⊆}L(V_n)}. EachN_nis finitely generated.

Hence as inTheorem 5.2, we get thatN=N_nfor some n, henceNis finitely generated.

This proves thatMis a Noetherian leftT-module if and only ifRis Noetherian andL(x0) satisfies strongacc.

6. Perfect rings

A partially ordered set X is said to locally satisfy strong dcc, if for any finite subset S of X, L(S) satisfies strong dcc. Throughout, R is an artinian, commutative local

ring, X is a partially ordered set locally satisfying strong dcc, and T=I^∗(X,R). Let T=R+K^∗(X,R). ThenT is a local ring. We will prove that T is right perfect. We will writeKforK^∗(X,R).

Lemma6.1. Any finitely generated left ideal ofTcontained inK^∗is artinian.

Proof. LetTCbe any artinian module. By Lemma 2.1,C/K^∗Cis of finite composition length over R. Let A=_n

i=1Tbi be a finitely generated left ideal of T contained in K^∗. ThenB=_n

i=1Tb_iis a finitely generated left ideal ofT contained inK^∗. LetA= A1⊇A2⊇ ··· ⊇An⊇ ··· be a descending chain of left ideals ofT. AsTBis artinian, there exists a positive integermsuch thatK^∗Ai=K^∗Am for anyi≥m. LetBi=TAi. ThenK^∗B_i=K^∗A_i. NowB_i is an artinian leftT-module. It follows that for anyi≥m, Bi/K^∗Biis of finite composition length overR. Therefore there exists ann≥msuch that Aj/K^∗Aj=Aj/K^∗An=An/K^∗Anfor any j≥n. HenceAj=Anfor any j≥n.

Theorem6.2. Tis a local, right perfect ring.

Proof. It is enough to prove thatTsatisfiesdccon principal left ideals [2, Theorem 28.4].

Let A1⊇A2⊇ ··· ⊇A_n⊇ ··· be a descending chain of principal left ideals of T. In view ofLemma 6.1, we takeAi=T(αiI+bi) for someαi=0 inRandbi∈K^∗. Then αi+1I+bi+1=(βiI+ci)(αiI+bi) for someβi∈R, andci∈K^∗. This givesαi+1=βiαiand ann_R(α_i)⊆ann_R(α_i+1). AsR is Noetherian, there exists a positive integer msuch that annR(αi)=annR(αi+1) for anyi≥m. Thereforeβiis a unit for anyi≥mandβiI+ciis a

unit. HenceAi=Amfor anyi≥m.

The dualization of the above result gives the following.

Theorem6.3. LetXbe a partially ordered set such that for any finite nonempty subsetZ ofX,U(Z)satisfies strong acc,Ris an artinian commutative ring, andT=I^∗(X,R). Then T=R+K^∗(X,R)is left perfect.

Examples of rings that are right perfect but not left perfect are well known (one such example is the dual of example given in [2, Exercise 2, page 322]). By using the above theorem, we end this section by constructing a class of right perfect rings that are not left perfect.

Example 6.4. LetXbe any partially ordered set that locally satisfies strong dcc, but has a finite, nonempty subset Z such that L(Z) is not finite. As L(Z) satisfies strong dcc, L(Z) has a subset V isomorphic to the set of natural number. Any infinite well- ordered set not embeddable in the set of natural numbers is such a setX. ThusV is given by elements:x1< x2<···< xn<···.Let R be a local artinian ring, andT= R+K^∗(X,R). ByTheorem 6.2, T is right perfect, however{ex1xiT}i≥2 is an infinite, nonterminating descending sequence of principal right ideals inT. HenceTis not left perfect.

Acknowledgment

The authors are thankful to the referees for their comments and suggestions.

References

[1] F. Al-Thukair, S. Singh, and I. Zaguia,Maximal ring of quotients of an incidence algebra, Arch.

Math. (Basel)80(2003), no. 4, 358–362.

[2] F. W. Anderson and K. R. Fuller,Rings and Categories of Modules, Graduate Texts in Mathemat- ics, vol. 13, Springer, New York, 1974.

[3] P. C. Fishburn,Interval Orders and Interval Graphs, Wiley-Interscience Series in Discrete Math- ematics, John Wiley & Sons, New York, 1985.

[4] T. Y. Lam,Lectures on Modules and Rings, Graduate Texts in Mathematics, vol. 189, Springer, New York, 1999.

[5] S. Singh and F. Al-Thukair,Weak incidence algebra and maximal ring of quotients, Int. J. Math.

Math. Sci.2004(2004), no. 53, 2835–2845.

[6] E. Spiegel and C. J. O’Donnell,Incidence Algebras, Monographs and Textbooks in Pure and Applied Mathematics, vol. 206, Marcel Dekker, New York, 1997.

Surjeet Singh: Department of Mathematics, King Saud University, P.O. Box 2455, Riyadh 11451, Kingdom of Saudi Arabia

E-mail address:[email protected]

Fawzi Al-Thukair: Department of Mathematics, King Saud University, P.O. Box 2455, Riyadh 11451, Kingdom of Saudi Arabia

E-mail address:[email protected]

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