WEAK INCIDENCE ALGEBRA AND MAXIMAL RING OF QUOTIENTS

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PII. S0161171204311130 http://ijmms.hindawi.com

WEAK INCIDENCE ALGEBRA AND MAXIMAL RING OF QUOTIENTS

SURJEET SINGH and FAWZI AL-THUKAIR Received 12 November 2003

LetX,Xbe two locally finite, preordered sets and letRbe any indecomposable commutative ring. The incidence algebra I(X, R), in a sense, representsX, because of the well- known result that if the ringsI(X, R)andI(X, R)are isomorphic, thenXandXare isomorphic. In this paper, we consider a preordered setXthat need not be locally finite but has the property that each of its equivalence classes of equivalent elements is finite. Define I^∗(X, R)to be the set of all those functionsf:X×X→R such thatf (x, y)=0, when- everxyand the setSf of ordered pairs(x, y)withx < yandf (x, y)≠0 is finite. For anyf , g∈I^∗(X, R),r∈R, definef+g,f g, andr f inI^∗(X, R)such that(f+g)(x, y)= f (x, y)+g(x, y),f g(x, y)=

xzyf (x, z)g(z, y),r f (x, y)=r·f (x, y). This makes I^∗(X, R)anR-algebra, called theweak incidence algebraofXoverR. In the first part of the paper it is shown that indeedI^∗(X, R)representsX. After this all the essential one- sided ideals ofI^∗(X, R)are determined and the maximal right (left) ring of quotients of I^∗(X, R)is discussed. It is shown that the results proved can give a large class of rings whose maximal right ring of quotients need not be isomorphic to its maximal left ring of quotients.

2000 Mathematics Subject Classification: 16S60, 16S90, 16W20.

1. Introduction. Let X and X be two locally finite, preordered sets, and let R be a commutative ring. Under what conditions are incidence rings I(X, R) and I(X, R) isomorphic? In particular, under what conditions on Rcan one conclude that X and Xare isomorphic, when the two incidence ringsI(X, R)andI(X, R)are isomorphic?

The latter question has been discussed by many authors. One of the earliest results in this direction is by Stanley [9], who proved that ifRis a field, then the two incidence rings are isomorphic if and only ifXandXare isomorphic. Froelich [4] extended this result to the case of an indecomposable ringR. Similar questions have been examined in [1,3,10] in caseRneed not be commutative.

Now consider any preordered set X that need not be locally finite. Two elements x, y∈Xare said to beequivalent,x y, ifx≤y≤x. InSection 3, the isomorphism problem for weak incidence algebras is discussed. LetXandXbe two preordered sets in each of which every equivalence class is finite, and letR,Rbe two commutative rings such that the weak incidence algebrasI^∗(X, R)andI^∗(X, R)are isomorphic as rings.

In caseRandRare indecomposable,Theorem 3.10shows thatX,Xare isomorphic and R, R are isomorphic. The main aim of Section 4 is to prove some results that can help in studying the maximal ring of quotients of an I^∗(X, R). Similar work has been done in a recent paper [2] for certain classes of incidence algebras. In [7], Spiegel determines some essential ideals of an incidence algebra of a locally finite, partially

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ordered set. Here we are in a position to determine all the essential one-sided ideals of anS=I^∗(X, R)wheneverRis indecomposable. A particular essential right idealT is isolated and the ringQ= HomS(T , T )is discussed in Theorems4.8,4.9, and4.10. This ringQis used to give some results on maximal right (left) ring of quotients ofS.

2. Preliminaries. All rings considered here are with identity 1≠0. As the various concepts discussed here for weak incidence algebras are similar to those for incidence algebras, for details on incidence algebras one may consult [8]. We now collect some results on rings and modules.

Lemma2.1. For any commutative ringRand any positive integern, ifMR=R⁽ⁿ⁾ is isomorphic to its summandN, thenM=N.

Proof. NowM=N⊕K. For any maximal idealPofR, the localizationMP=NP⊕KP. As the ranks of the freeRP-modulesMP andNP are the same and finite,KP=0. Hence K=0.

Lemma2.2. LetR be a commutative ring and letKbe any ring such thatMn(R) Mm(K). Thenmdividesn. Ifn=m, thenRK.

Proof. The first part follows from Wedderburn’s structure theorem for simple ar- tinian algebras, and the second part is in [6].

Lemma2.3. LetT be any ring and lete,e,f,fbe any four idempotents inT such thateTeT,f TfT. TheneT f≠0if and only ifeT f≠0.

Proof. The hypothesis gives that HomT(f T , eT )HomT(fT , eT ), eT feT f, as abelian groups. This proves the result.

3. Isomorphism. LetXbe any preordered set (i.e.,Xis a set with a relationthat is reflexive and transitive). For anyx, y ∈X, setx y, ifx y x. Then is an equivalence relation. A preordered setX is said to be aclass finite, preordered set if, for anyx∈X, the equivalence class[x]= {y∈X:x≤y≤x}is finite. Henceforth we takeXto be a class finite, preordered set andRa commutative ring. The setK^∗(X, R)= {f∈I^∗(X, R):f (x, y)=0 wheneverx y}is a nil ideal. Indeed, givenf∈K^∗(X, R), f^m+1=0, form= |Sf|. Indeed, one can see that each member ofK^∗(X, R)is strongly nilpotent, as defined in [8, page 176], soK^∗(X, R)is contained in the lower nil radical ofI^∗(X, R). LetY be a representative partially ordered subset ofX. For anyx∈X, let

|[x]| =nx. For eachx∈X, the setBx= {f∈I^∗(X, R):f (u, v)=0 wheneveruxor vx}, is a ring with δx as identity, whereδx(u, v)=0, wheneverux,vx, or u≠v, andδx(u, u)=1 wheneveru x. Letδdenote the identity element ofI^∗(X, R).

For anyx, y∈X, withx y, letexy∈I^∗(X, R)be such thatexy(u, v)=0, for(u, v)≠ (x, y), andexy(x, y)=1. Each ofexyis called amatrix unit ofI^∗(X, R). We writeex= exx. ThenBxis thenx×nxfull matrix ring overRwith{euv:u x, v x}as its set of matrix units. LetMn(R)denote then×nfull matrix ring overR. Further,D^∗(X, R)= {f ∈I^∗(X, R):f (u, v)=0 wheneveruv} is a subring ofI^∗(X, R), eachBx is an ideal ofD^∗(X, R). SetS =I^∗(X, R), K=K^∗(X, R),D=D^∗(X, R). For any subsetZ of X, letEZ∈S be such thatEZ(u, u)=1 foru∈Z, andEZ(x, y)=0 otherwise. For any

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f∈S,support off, denoted by suppt(f ), equals{(x, y):f (x, y)≠0}, the cardinality of suppt(f )is called theweightoffand we denote it bywt(f ). LetXbe another class finite, preordered set. LetRbe another commutative ring. We use the same symbols for the matrix units ofI^∗(X, R)orI^∗(X, R)and so on, butS=I^∗(X, R),K=K^∗(X, R), andD=D^∗(X, R). LetY andYbe fixedrepresentative partially ordered subsets of XandX, respectively. For any two distinct membersy,zofY,δy,δzare orthogonal idempotents. Anyf∈Swill be sometimes denoted by the formal sum

x,yf (x, y)exy

(or by the matrix[f (x, y)]indexed byX). The following is obvious.

Lemma3.1. (i)I^∗(X, R)=D^∗(X, R)⊕K^∗(X, R)as abelian groups.

(ii)D^∗(X, R)=Πy∈YBy, whereY is any representative partially ordered subset ofX.

(iii)I^∗(X, R)/K^∗(X, R)Πy∈YMny(R)D^∗(X, R), whereY is any representative par- tially ordered subset ofX.

(iv)For anyf , exy∈I^∗(X, R),wt(f exy)is finite, that is,f exy=

u≤yauyeuy, with finitely manyaux≠0.

It follows from (ii) thatK^∗(X, R)does not equal the Jacobson radical ofS, unless the Jacobson radical ofR is zero. For anyf ∈S, we writef=fD+fK withfD∈D and fK∈K;fDis called thediagonaloff. The following is obvious.

Lemma3.2. For any nonempty subsetZofX,EZSEZI^∗(Z, R).

Lemma3.3. For any two idempotentsf , g∈S,f Sg≠0if and only iffDSgD≠0.

Proof. InS=S/K,f+K=fD+K. AsKis nil, we getf SfDS. After this,Lemma 2.3 completes the proof.

Lemma3.4. Let0≠e=e²∈S.

(i) eDis a nonzero idempotent andeDδy=δyeDfor anyy∈Y. (ii) There existsy∈Y such thateDδy=δyeD≠0.

(iii) For anyy∈Y,e=eeDδyeis an idempotent such thate(u, v)=

e(u, w1)e(w1, w2)e(w2, v), where the summation runs over w1, w2 in[y]∩[u, v]. Further, e−e,eare orthogonal idempotents. IfeDδy≠0, thene≠0.

Proof. (i) is obvious. NowS/K =D=Πy∈KBy D, δ=Πδy, and e=eD. It follows that for somey∈Y, eδy=eDδy≠0. This proves (ii). Consider anyy∈Y and e=eeDδye. The definition of the product of two members ofS gives thate(u, v)= e(u, w1)e(w1, w2)e(w2, v), where the summation runs over allw1, w2in[y]∩[u, v].

Then we have(e)²(u, v)=

uwve(u, w)e(w, v)=

e(u, w1)e(w1, w2)e(w2, w)e(w, w3)e(w3, w4)e(w4, v), where summation runs over allwi, w in[y]∩[u, v]such that w2 w w3. Thus(e)²(u, v)=

e(u, w1)e(w1, w4)e(w4, v)=e(u, v). Henceeis an idempotent. Asee=e=ee, it follows thate−eis an idempotent orthogonal toe. If eDδy≠0, as obviouslye=eDδy inS/K, we gete≠0.

Lemma3.5. (i)Ife∈S is an indecomposable idempotent, then there exists a unique y∈Y such thate=eeDδye.

(ii)Lete∈S be a nonzero idempotent such thateD∈By for somey∈Y. Thene= eeDδye; thisyis uniquely determined bye.

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Proof. (i) InS=S/K, e=eDis an indecomposable idempotent. So there exists a uniquey∈Y such thate=eDδy. ByLemma 3.4(iii),e=eeDδyeis a nonzero idempotent. Ase−eis orthogonal toeandeis indecomposable,e=e.

(ii) The hypothesis givese=eeDδye. ThenLemma 3.4(iii) givese=eeDδye.

Theorem3.6. LetRbe any indecomposable commutative ring andXany class finite, preordered set. Then for any automorphismσ ofS=I^∗(X, R),σ (K)=K.

Proof. Consider any f ∈ S\K. For some x y, f (x, y)≠0. Then g=exf eyx

is such thatg(x, x)≠0 and g=exgex. So σ (g)=eσ (g)e, where e=σ (ex) is an indecomposable idempotent. LetY be a representative partially ordered subset ofX.

ByLemma 3.5, there exists uniquez∈Y such thate=eeDδze,eD∈Bz. Thusσ (g)= eeDδzeσ (g)eeDδze≠0, δzeσ (g)eeDδz≠0, so for someu, v∈[z], σ (g)(u, v)≠0.

Henceσ (g)∉K. Consequently,σ (f )∉K. This proves the result.

Lemma3.7. For somey, y∈Y, let there exist idempotentse∈By,f∈By such that eSf≠0. TheneySey≠0.

Proof. The hypothesis gives thatδySδy≠0, so there existu∈[y],v∈[y]such thateuSev≠0. After this,Lemma 2.3completes the proof.

Lemma3.8. If for some idempotentf∈S,f SδyS for somey∈Y, thenfD=δy. Proof. We havefDSδyS. InS=S/K,fDSδyS, sofD∈ByandfDByBy. By Lemma 2.1,fD=δy.

Lemma3.9. LetR,Rbe indecomposable andσ:S→San isomorphism.

There exists a one-to-one mappingηofY ontoYsuch thatσ (δy)=δη(y)+gη(y)for somegy∈K,|[y]| = |[η(y)]|, andRR.

Proof. The hypothesis gives that for anyx∈X,exis an indecomposable idempotent inS. Nowσ (δy)S= ⊕

uyσ (eu)S. As theseσ (eu)Sare indecomposable and isomorphic right ideals, there exist uniqueη(y)∈Ysuch that eachσ (eu)D∈B_δ

η(y). Consequently, σ (δy)D ∈ B_η(y) and σ (δy)Dδη(y) = δη(y)σ (δy)D. By Lemma 3.5(ii), σ (δy)=σ (δy)σ (δy)Dδη(y)σ (δy). Similarly,

σ⁻¹ δη(y)

=σ⁻¹ δη(y)

σ⁻¹ δη(y)

Dδzσ⁻¹ δη(y)

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for somez∈Y. So,δη(y)=δη(y)σ ((σ⁻¹(δη(y)))D)σ (δz)δη(y). Thus, inS=S/K,

σ δy

=σ δy

σ δy

Dδη(y)σ σ⁻¹

δη(y)

D

σ δz

δη(y)σ δy

. (3.2)

InS,δη(y)is a central idempotent. Thus

σ δy

=σ δy

σ

σ⁻¹ δη(y)

D

σ δzδy

δη(y), (3.3)

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which equals zero, if z≠y. Hence z=y and ηis a bijection from Y ontoY. We get σ (δy)=δη(y)σ ((σ⁻¹(δη(y)))Dδy) and δη(y) =δη(y)σ ((σ⁻¹(δη(y)))Dδy). Hence σ (δy)=δη(y). This shows thatσ (δy)=δη(y)+gη(y)for somegη(y)∈K. NowδySδy= By. Asσ (δy)Sδη(y)S, it follows thatByB_η(y). ByLemma 2.2,|[y]| = |[η(y)]|and RR.

Theorem3.10. LetXandXbe two class finite, preordered sets. LetRandRbe any two indecomposable commutative rings. If there exists an isomorphism ofI^∗(X, R)onto I^∗(X, R), thenX,Xare isomorphic and the ringsR,Rare isomorphic.

Proof. We use the terminology developed beforeTheorem 3.10. Consider anyu, v∈Y such thatu v. TheneuSev≠0,σ (eu)Sσ (ev)≠0. It follows fromLemma 3.9 thatσ (eu)D∈B_η(u) ,σ (ev)D∈B_η(v) . ByLemma 3.3,σ (eu)DSσ (ev)D≠0,eη(u)Seη(v)≠ 0, hence η(u) η(v). Thusηis an isomorphism ofY ontoY. Also byLemma 3.9,

|[y]| = |[η(y)]|, hence it follows thatXandXare isomorphic. ByLemma 3.9,Rand Rare isomorphic.

Lemma3.11. For any commutative ringT and any class finite, preordered setX, the following hold.

(i)A central idempotente∈I^∗(X, T )is centrally indecomposable if and only ife= gEZ for some indecomposable idempotent g ∈ T and a connected component Z of X.

(ii)Letgandhbe two indecomposable idempotents inTand letZ,Zbe two connected components ofX; the ringsgEZI^∗(X, T ),hEZI^∗(X, T )are isomorphic if and only if the ringsgT,hT are isomorphic andZ,Zare isomorphic.

Proof. (i) Consider any central idempotente∈I^∗(X, T ). On the same lines as for incidence algebras, it can be easily seen thate(x, y)=0, wheneverx≠y. For any con- nected componentZofX, there exists an idempotentgZ∈Tsuch thate(x, x)=gZfor everyx∈X. Using this, (i) follows. (ii) AsgEZI^∗(X, T )I^∗(Z, gT )andhEZI^∗(X, T ) I^∗(Z·hT ), the result follows fromTheorem 3.10.

LetT be any ring. Let In(T )be the set of allcentrally indecomposable central idem- potentsofT. Two central idempotentsg,hofT are said to beequivalent if the rings gT andhT are isomorphic. For any central idempotentg∈T,[g]denotes the set of central idempotents inT equivalent tog.

Theorem3.12. LetRandRbe any two commutative rings and letX,Xbe two class finite, preordered sets. Letσ:I^∗(X, R)→I^∗(X, R)be a ring isomorphism. Letg∈In(R) and letZbe a connected component ofX.

(i)There exist uniqueg∈In(R)and unique connected componentZofXsuch that σ (gEZ)=gEZ; further,ZZ,|[g]||[Z]| = |[gEZ]| = |[gEZ]| = |[g]||[Z]|.

(ii)If the cardinalities of[g]and[g]are finite and equal, thenXandXare isomor- phic.

Proof. (i) The first part follows fromLemma 3.11(i); the second part follows from Lemma 3.11(ii). (ii) If|[g]| = |[g]|and they are finite, if follows from (i) that, given any

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connected componentZofX, there exists a connected componentZofXisomorphic toZ, and[Z],[Z]have the same cardinalities. Consequently,XandXare isomorphic.

The following is immediate fromTheorem 3.12.

Corollary3.13. LetRbe any commutative ring such thatR admits an indecom- posable idempotentgfor which the equivalence class[g]is finite. LetXandXbe any two class finite, preordered sets. If the ringsI^∗(X, R)andI^∗(Y , R)are isomorphic, then XandXare isomorphic.

4. Essential right ideals and maximal ring of quotients. ThroughoutS=I^∗(X, R), whereX is a class finite, preordered set andR is a commutative ring in which 1 is indecomposable. Anyx∈Xis said to be amaximal element if the equivalence class [x]is maximal in the partially ordered set of the equivalence classes in X. For any x, y ∈X, we sayx < y, if x ≤y but [x]≠[y]. Set X0= {x∈X :xis maximal}, Y0= {(x, y)∈X×X0:x≤y},Y1= {(x, y):x < y and there does not exist anyz∈ X0such thaty≤z}, Y2= {(x, y):x < yand there exists az∈X0such thaty < z}, andY3= {(x, y)∈X0×X0:[x]=[y]}. Further,K=K^∗(X, R). NowL=

(x,y)∈Y₃exyR is a right ideal ofS. In [2], maximal rings of quotients of certain incidence algebras have been discussed. Here we intend to prove some results that can help in studying the maximal rings of quotients ofS. Spiegel [7] has determined certain classes of essential ideals of an incidence algebra of a locally finite, preordered set. Here we determine all essential one-sided ideals ofS. For the definitions of anessential submodule, dense submodule, andsingular submoduleof a module, one may refer to [5]. LetM be any module, thenN⊂eM(N⊂dM) denotes thatNis anessential(dense) submodule ofM, andZ(M)denotes the singular submodule ofM. The concept of themaximal right ring of quotientsof a ring is discussed in [5, Section 13].

Lemma4.1. LetK1=K+L. ThenK1is an essential right ideal ofSandl·ann(K1)=0.

Indeed for any0≠f∈S, there existsexw∈K1such that0≠f exw∈K1.

Proof. Let 0≠f∈S. Thenf (u, v)≠0 for someu≤v. Supposef K1=0. Ifvis not maximal inX, there existsevz∈K, andf evz≠0, which is a contradiction. Hencev is maximal. Thenev∈K1withf ev≠0, which is again a contradiction. Hencel·ann(K1)= 0. In any case there exists exy ∈K1 such that f exy ≠0. By applying induction on wt(f exy), we prove that for someg∈S, 0≠f exyg∈K1, which will prove thatK1⊂eSS. Supposewt(f exy)=1. Thenf exy=aeuy, for some 0≠a∈R. Ifyis not maximal, for anyz > y,f exyeyz=aeuz∈K1. Ifyis maximal, theney∈K1, sof exyey=aeuy∈K1. To apply induction, suppose thatwt(f exy)=n >1, and for anyh∈S, if for some euv ∈K1,wt(heuv) < nandheuv ≠0, then for someevz∈S, 0≠heuvevz∈K1. We can writef exy=aeuy+h, wherewt(h)=n−1 andh(u, y)=0. For someeys∈K1, aeuyeys=aeus∈K1. Thenf exs=aeus+heys withwt(heys)=n−1. By the induction hypothesis, there existsesw ∈K1 such that 0≠heysesw ∈K1. Then 0≠f exw ∈K1. HenceK1⊂eSS.

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We call a subsetBofRanessential subsetofRif, for each 0≠r∈R, there exists an s∈Rsuch that 0≠r s∈B. Clearly the ideal ofRgenerated by an essential subset is an essential ideal.

Lemma4.2. LetE⊂eSS. For anyx≤y inX, letAxy= {r ∈R:r exy ∈E},Bxy =

∪y≤zAxz.

(i) Axy⊆Axw wheneverx≤y≤w.

(ii) Bxy is an essential subset ofR.

Proof. (i) is trivial. Let 0≠r∈R. Then for someg∈S, 0≠r exyg∈E. For some y≤w,r g(y, w)≠0. This givesr exygew =r g(y, w)exw ∈E,r g(y, w)∈Bxy. This proves thatBis an essential subset ofR.

Lemma4.3. Let{Axy: eitherx < y,orx≤yandyis maximal inX}be a family of ideals inR such that (i) Axy ⊆Axz whenever y≤z, and (ii) for any x≤y inX, Bxy= ∪y≤zAxzis an essential subset ofR. ThenE=

x,yAxyexy is an essential right ideal ofSandE⊆K1.

Proof. It is easy to verify thatEis a right ideal ofScontained inK1. Let 0≠f∈K1. By induction onwt(f ), we prove that 0≠f r exy∈Efor someexy∈K1,r∈R, which will prove thatE⊂eSS. Supposef=aexy. Asa≠0, there exists az≥yand anr∈R such that 0≠ar∈Axz. Then 0≠f r eyz=ar exz∈E. Here, ifyis not maximal, choose z > y; ify is maximal, choosey=z; in any caseexz∈K1. Thus the result holds for wt(f )=1. To apply induction, letwt(f )=n >1, and let the result hold for any positive integer less thann. We writef=aexy+h, with 0≠a∈R,exy∈K1,wt(h)=n−1, andh(x, y)=0. There exists anr eyz∈K1such that 0≠aexyr eyz=ar exz∈E. Then 0≠f r exz=ar exz+hr eyz. Ifhr eyz=0,f r exz=ar exz∈E and we finish. Suppose hr eyz≠0. By the induction hypothesis, there existsbezw∈K1, withb∈R, such that 0≠hr eyzbezw∈E. Then 0≠f r bexw∈E.

Let Minness(S)be the set of all essential right ideals of the form given inLemma 4.3.

Lemma4.4. Z(S)= {f∈S:f E=0for someE∈Minness(S)}.

Proof. Letf∈Z(S). For someE⊂eSS,f E=0. By Lemmas4.2and4.3, there exists anE∈Minness(S)such thatE⊆E. Thenf E=0. This proves the result.

Theorem4.5. Z(SS)=0if and only ifZ(R)=0.

Proof. Let Z(R)≠0. For somer ≠0 and an essential idealA of R, r A=0. In Lemma 4.3, by taking everyAxy=A, we get anE⊂eSSsuch thatr IE=0. ThusZ(S)≠0.

Conversely, letZ(S)≠0. Consider any 0≠f∈Z(S). For someE∈Minness(S),f E=0.

Nowf (u, v)≠0 for someu≤v. Then 0≠euf∈Z(S). Suppose there exists a maximal z≥v. Aszis maximal, it follows fromLemma 4.3(i) thatBvz=Avz, soevf evzAvz=0, f (u, v)Avz=0,f (u, v)∈Z(R). HenceZ(R)≠0.

Proposition4.6. For any(x, y)∈Y0, setAxy=R, for(x, y)∈Y1, setAxy=R, and for(x, y)∈Y2, setAxy=0. LetT=

x,yexyAxy.

(i) ThenT is an ideal ofS,T⊂eSS, andl·ann(T )=0.

(ii) S embeds in the ringQ=Hom(TS, TS)such thatSS is dense inQS.

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Proof. ThatTis an essential right ideal inSfollows fromLemma 4.3. Suppose that 0≠f∈l·ann(T ). Thenf (u, v)≠0 for someu≤v. Suppose there exists no maximal z≥v. Choose anyw > v. Thenevw∈T butf evw≠0, which is a contradiction. Hence there exists a maximalz≥v. Thenevz∈Tandf evz≠0, which is also a contradiction.

Hencel·ann(T )=0. Consider anyexy∈T. ByLemma 3.1,wt(f exy)is finite, sof exy=

u≤yauyeuy, a finite sum. By definition, the following two cases arise.

Case 1. yis maximal. Then everyeuy∈T, sof exy∈T.

Case 2. There does not exist any maximalz≥y. Thenu < y, Auy=R, euy∈T, hencef exy∈T.

This proves thatT is an ideal inS. For eachf∈S, letλ(f )be the left multiplication onT byf. Thenλis an embedding ofSinQ. Consider anyσ , η∈Q, withσ≠0. Then for somef∈T,σ (f )≠0. We see thatσ·λ(f )=λ(σ (f ))≠0 andη·λ(f )=λ(η(f ))∈ λ(S). HenceSS is dense inQS.

For eachx0∈X0, set T[x₀]=

{exyR:(x, y)∈Y3and[x0]=[y]}, and set T= {exyR:(x, y)∈Y1}. Observe that T[x₀] =T[x₁] if and only if[x0]=[x1]. Each of T[x₀],Tis a right ideal ofS contained inT, andT is a direct sum of these right ideals.

LetZ0be the set of equivalence classes inXgiven by the members ofX0. For any ring P, letPbe themaximal right ring of quotientsofP[5, Section 13]. The following result can be easily deduced from various results and exercises given in [5, Sections 8 and 13].

Theorem4.7. (I)For any family of rings{Pα:α∈Λ},P=

α∈ΛPα,P=

α∈ΛPα. (II)For any two subringsA,Bof a ringP, ifAA⊂dBA,BB⊂dPB, thenA=P.

(III)For any positive integernand any ringP,_M_n_{(P )}₌_M_n₍_{P ).}

Theorem4.8. (i)Q=Hom(TS, TS)(Π{HomS(T[x₀], T[x₀]):[x0]∈Z0})×Homs(T, T).

(ii)Maximal right rings of quotients ofS andQare the same.

(iii)LetP[x₀]=HomS(T[x₀], T[x₀])andP=Hom(T, T). ThenS(Π{_P_[x

0]:[x0]∈ Z0})×P.

Proof. To prove (i) it is enough to prove that HomS(T[x₀], T[x₁])=0 whenever[x0]≠ [x1], HomS(T[x₀], T)=0=HomS(T, T[x₀]). Considerσ ∈HomS(T[x₀], T[x₁]). For any exy ∈T[x₀], [x0]=[y], so euy ∈T[x₁], but σ (exy)=

u≤yauyeuy, auy ∈R. Thus σ (exy)=0,σ =0. Similarly, we can prove that the others are also zero. AsSSis dense inQS,S=Q. Because of (i) and Theorem 4.7, we getS(Π{P[x₀]:[x0]∈Z0})×P.

We now discuss matrix representations of HomS(T[x₀], T[x₀])and Homs(T, T).

Theorem 4.9. Let x0 be a maximal member ofX, Ux₀ = {x∈X:x ≤x0}. Then HomS(T[x₀], T[x₀])is isomorphic to the ring of column-finite matrices overRindexed by Ux₀.

Proof. Letσ∈HomS(T[x₀], T[x₀]). Forexy∈T[x₀],y x0. Ifσ (exy)=

u≤yauyeuy, then for any other exz ∈ T[x₀], σ (exz)=

u≤zauzeuz = σ (exy)eyz =

u≤yauyeuz, auy=auz. Conversely, anyσ∈HomR(T[x₀], T[x₀]), such that ifσ (exy)=

u≤yauyeuy,

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thenσ (exz)=

u≤yauyeuz fory z, is in HomS(T[x₀], T[x₀]). NowVx₀= {exy :x∈ Ux₀, y x0}is anR-basis ofT[x₀]. We writeσ (exy)=

u,vauvxyeuv, euv∈Vx₀. Then auvxy =0, for v≠y, auyxy =auzxz, whenever y z. We write bux=auyxy and bux=0 otherwise. We get matrix[bux]overRindexed byUx₀. This matrix is column finite;σ↔[bux]gives the desired isomorphism.

Theorem4.10. LetX= {y∈X:there exist no maximalz≥y}. LetGbe the set of arrays[avxy]overRindexed byX×X×Xsuch that it has following properties:

(i) avxy=0, wheneverx≮y,v≮y, orx < v < y,

(ii) for any fixed pair(x, y)withx < y, the number ofvfor whichavxy≠0is finite, (iii) fory≤z,avxy=avxzifv < y, andavxz=0ifv≮yandv < z.

InG, define addition componentwise and the product by[avxy][bvxy]=[cvxy]such thatcvxy=

wavwybwxy. ThenHomS(T, T)G.

In caseXhas the property that for every pair of elements u, v inXthere exists a w∈Xsuch thatu≤w,v≤w, then any array[avxy]∈Ghas the following additional properties:

(iv) ifu, v∈Xare not comparable, thenauxv=0, (v) forx < y,x < z,avxy=avxz.

Putbvx=avxy. Then[bvx]is a column finite matrix indexed byXwith the property thatbvx=0ifv > x, or there existsy > x such thatv≮y. Setbvx=0in all other cases. LetBbe the set of all such matrices. ThenBis a ring isomorphic toHomS(T, T).

Proof. Let σ ∈ HomS(T, T). For any x < y ≤ z ∈ X, we have σ (exy) = cuvxyeuv,cuvxy∈R,(u, v)∈Y1, withcuvxy=0, wheneverv≠y. So we can write σ (exy)=

v<yevyavxy, a finite sum. Fory≤z,σ (exz)=σ (exy)eyzgivesavxy=avxz

forv < y andavxz=0 wheneverv≮y,v < z. Suppose we have somex < v < y, by considering σ (exy)=σ (exv)evy it follows thatavxy =0. For any other(v, x, y)∈ X×X×X, setavxy =0. We get an array [avxy]with the desired properties. Con- versely, any such array gives anS-endomorphism ofT. This gives the desired isomorphism.

Suppose every pair of elements inX have a common upper bound. Consider any v, w∈Xthat are not comparable. By (i), avxw =0 for any x; this proves (iv). Sup- posex < y, x < z. There existsw∈Xsuch thaty < w,z < w. Thenσ (exz)ezw = σ (exy)eyw =σ (exw)gives (v). Setbvx=avxy. Because of (v),bvx is well defined. It gives a matrix[bvx]indexed byX, which is column finite and has the property that bvx=0 if eitherv > x, or there exists y > xsuch thatv≮y. LetBbe the set of all column-finite matrices[bvx]overRindexed byX×Xwithbvx=0, whenever either v > xor there exists ay > xsuch thatv≮y. Then HomS(T, T)is isomorphic to the ringB.

Remark4.11. LetXbe any locally finite, preordered set and letRbe any indecomposable commutative ring. Obviously,S=I^∗(X, R)is a subring ofS=I(X, R). ButSS

need not be dense or essential inS_S. So the maximal right rings of quotients ofS and Sneed not be the same; in fact, they need not be isomorphic (see the example given below). In caseSSis dense inS, the two rings will have the same maximal right ring of quotients. In that case,Scan help in studyingS.

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Theorem4.12[2]. LetXbe any partially ordered set such that for anyx∈X, there exists a maximal elementz≥xandLz= {y∈X:y≤z}is finite. LetX0be the set of max- imal elements ofX. For eachz∈X0, letnzbe the number of elementsy≤z. For the ring S=I(X, R),SΠ{Mn_z(R) :wherezruns over representatives of equivalence classes in X0}.

Proof. Letf , g ∈S=I(X, R) with g≠0. For someu, v ∈X, g(u, v)≠0. Then gev≠0. At the same time the hypothesis onXgives that the support off ev is finite, sof ev ∈S =I(X, R). HenceSS is dense inS. After this, Theorems4.7, 4.8, and4.9 complete the proof.

Example4.13. LetX=Nbe the set of natural numbers and letRbe any indecomposable commutative ring. ConsiderS=I^∗(N, R)andS=I(N, R). Let 0≠f∈S. For some r∈N,f er≠0. Clearly, the support off eris finite. HenceSSis dense inS. So the maximal right quotient rings ofSandSare the same. Considerg∈Sfor whichg(1, n)=1 for everyn, andg(n, m)=0 otherwise. Then for any h∈S,hg=0 or hg=kgfor some 0≠k∈N, soSSis not dense inS. Thus maximal left rings of quotients ofS and Sare not the same. We now show that they need not be isomorphic. ConsiderR=F a countable field. As Nhas no maximal element, K=T =T, Q=HomS(T, T). By Theorem 4.10,Qis isomorphic toS. ButS, as a rightS-module, is dense in the ring Lof all column-finite matrices overF, indexed byN. It is well known that the ring of all column-finite matrices over a field, indexed by any set, is right self-injective. HenceL is the maximal right ring of quotients ofSandS. LetNbe the set of natural numbers with reverse ordering. AsNhas unique maximal element 0,N=T0, byTheorem 4.9, the corresponding Q is isomorphic to the ring of all column-finite matrices over F, indexed byN. SoQis right self-injective. HoweverS=I^∗(N, F )is anti-isomorphic to S1=I^∗(N, F ). SoQ, the maximal left ring of quotients ofS, is isomorphic to the ring of all row-finite matrices overF, indexed byN. Nowe00Q is a countable set, and any minimal left ideal ofQis generated by an element ofe00Q, so the left socle ofe00Q is of countable rank. ForS, the left socle ise00S, which is of uncountable rank. Also Sis left nonsingular. SoL, the maximal left ring of quotients ofS, is such that its left socle is of uncountable rank. This proves that the maximal left rings of quotients ofS andSare not isomorphic.

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

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[7] E. Spiegel,Essential ideals of incidence algebras, J. Austral. Math. Soc. Ser. A68(2000), no. 2, 252–260.

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

E-mail address:[email protected]

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