CLASSICAL QUOTIENT RINGS OF GENERALIZED MATRIX RINGS

An associative ring R with identity is a generalized matrix ring with idempotent set E if E is a finite set of orthogonal idempotents of R whose sum is 1. We show that, in the presence of certain annihilator conditions, such a ring is semiprime right Goldie if and only if eRe is semiprime right Goldie for all e∈E, and we calculate the classical right quotient ring of R.

In this paper, we show that, in the presence of certain non-degeneracy conditions, a generalized matrix ring R with idempotent set E is semiprime right Goldie if and only if eRe is semiprime right Goldie for all e E E, and we calculate the classical right quotient ring of R. Kerr's example [4] of a right Goldie ring whose matrix ring is not right Goldie shows that our semiprimeness condition cannot be omitted.
Examples of generalized matrix rings include incidence algebras of directed graphs with finite number of vertices (see [5] and [9]), structural matrix rings (see Van Wyk [13] and subsequent papers), endomorphism rings of finite direct sums of modules and Morita context rings.Sands [10] observed that if [S,V, W,T] is a Morita context, then is a ring.These Morita context rings are precisely generalized matrix rings with idempotent sets E such that EI 2, and they have been widely studied.In particular, we note Amitsur's paper [1], the survey paper [6], McConnell and Robson's treatment in ([7], .
A generalized matrix ring R with idempotent set E is called a piecewise domain if for all e,f, gEE, :rEeRf and yGfR9, we have,ry=0implies z=0or 9=0.These rings have been studied in sone detail-see, for instance, [2] and [3].
We denote the 1)rime radical of a ring R 1)y p(R) and if and f are idempotents of R, 7 f, p(eRf) denotes the set {a'e eRf:xfRe C_ p( Re)}.
e,fE PROOF.If EI there is nothing to prove and if EI 2 this is Theorem in Sands [10].Assume now that EI -n > 2 and that the theorem is true for generalized matrix rings with idempotent sets of cardinality less than n.Let e E and set E= {e, 1-e}.Then R is a generalized matrix ring with idempotent set E and so Sands' result implies that p(R) p(eRe)+ p(eR(1-e)) + p((1-e)Re)+ p((1-e)R(1 -e)).
Let R be a generalized matrix ring with idempotent set E. We say that the pair (R,E) satisfies the left (respectively, right) annihilator condition if for all e, f E, 0 x e err implies that xfRe 0 (respectively, fRex 7 0).This concept is defined in [12] where right and left are interchanged.
COROLLARY 1.2.(Wauters and Jepers [12]).The following conditions on a generalized matrix ring with idempotent set E are equivalent.
(b) (R,E) satisfies the left annihilator condition and ere is semiprime for all e E. (c) (R,E) satisfies the right annihilator condition and ere is semiprime for all e E. Ci 2. THE GOLDIE CONDITIONS.
The right singular ideal of a ring S will be denoted by Z(S), and the right singular submodule of a right S-module M will be denoted by Z(M).So, if R is a generalized matrix ring with idempotent set E and e, f e E with e f, then Z(eRe) is the right singular ideal of the ring eRe and Z(eRf)is the right singular submodule of the right fRf-module err.PROPOSITION 2.1.Let R be a generalized matrix ring with idempotent set E and suppose that (R, E) satisfies the left annihilator condition.Then Z(R) Z(Rf).
e,fE PROOF.Let e, f E E and suppose that x Z(R).Then ezf Z(R), so there is an essential right ideal I of R such that exfI 0. To show that ezf e Z(eRf) it suffices to sho, that fir is an essential right ideal of fRf.Let A be a nonzero right ideal of fRf.Because I is essential, IOAFtO.L('t OuEIClAR.SinccINARisarightidealthereisanidempotent 9EEsuch that 0 y ug E I NAR, and ug .lug 1)('cans(, ug E AR C_ fRfR.Since (R,E) satisfies the left annihilator condition, 0 # (.fug)gRf C_ (I AR)fl fRf C_ fRf C A. It follows that Z(R) C_ Z(cR.f).,.fEE Conversely, suppose that e,f E E and egf E Z(eRf).Then 9H 0 hn" some essential right dcal H of fRf.Let J {rE R:.f,'E HR}.Clearly, J is a right ideal of R and 9,!=egfJ=(eyf)fJC_(egf)HR=O, so to show that 9EZ(R)it is enough to show that J is essential in R. Let B be a nonzero right ideal of R. IffB=0, then B_CJ and so BClJ0.Now assume fB -O.Then fBg # 0 for som, 9 E E, and so the left annihilator condition implies that fBf 7 O.So we see that fBf is a nonzero right ideal of fRf.Thus fBf cl H # 0 and so B VI J # 0 because H C_ HR. 13 COROLLARY 2.2.If R is a generalized natrix ring with idempotent set E such that (R, E) satisfies the left annihilator condition, then R is nonsingulax if and only if eRe is nonsingular for all e E E.
PROOF.In view of the proposition, we need only show that Z(R) 0 implies that Z(eRe) 0 for some e E E. Suppose that 0 y x E Z(R).Then 0 7 exf E Z(R) for some e,f E. The right annihilator condition implies that (ezf)fRe 0 and so ere Cl Z(R) O.It now follows from the proposition that Z(eRe) O.

I-i
The right uniform dimension of a ring R (respectively, right R-module M) will be denoted by d(R) (respectively, d(i)).PROPOSITION 2.3.Let R be a generalized matrix ring with idempotent set E such that (R,E) satisfies the left annihilator condition.If d(R)< o0 then d(eRe)< e for all e E E.Moreover, if R is semiprime and d(eRe)< oc for all e E E, then d(eRf)< cx for all e,f E E and hence d(R) < cx.
PROOF.Assume that d(R) < cx,e E E and ]A, is a direct sum of nonzero right ideals of ere.To prove that d(eRe)< oc it is enough to show that ]A,R is direct, and to accomplish this we need only show that ] A,Rf is direct for each f E E. Suppose that f E E and b, E A,Rf are such that ] b, 0. Since b,fRe C_ A, and ]A, is direct, b,fRe 0 for all i.Thus the left annihilator condition implies that b, 0 for all and hence ] A,Rf is direct.Now assume that d(eRe) < 0 for all e E E and suppose that ] N, is a direct sum of nonzero fRf-submodules of eRf.Since 0 7 N, C_ err the left annihilator condition implies that N,fRe # 0, and each NfRe is a right ideal of eRe.Let K NffRefl 2 {N,fRe:i # j}.Then KeRr C_ N, Ci {N,:i 7 j} and so KeRr 0. Since If C_ KeRf, K 0 and so K 0 because eRe is semiprime by Corollary 2. Thus Z N,fRe is direct and so d(eRf)<_ d(eRe).It follows that R has finite right uniform dimension as a right ( 2 eRe-module and so certainly d(R) < odi3 From Corollary 2, Corollary 4 and Proposition 5 we otan the following theorem.
THEOIM 2.4.Let R be a generalized matrix ring with idempotent set E. If R is semiprime right Goldie, then so too axe the rings eRe, e E E. Conversely, if (R,E) satisfies the left annihilator condition and ere is semiprime right Goldie for all e E, then R is semiprime right Goldie.Let S and T bc rings and let M be an S-T-bimodule.We say that M satisfies the right D.G. t'OOLE AND P. N. STEWART bimodule Ore condt,on if for each , E :! an,l each regular 'lemcnt c S there is an m M and a regular c T such that PROPOSITION 3.1.If R is a scmiprime right Goldie generalized matrix ring with id'mpotent set E. tlwn R." satisfies the right ]fimodule Ore condition for all .." E.
PROOF.Let m R.f and suppose that c is regular in Rt.Define &cR.fccRf by O(x) cx for all .v6eR.f.Clearly is a mononaorphisn.Suppose that a'6 eRr and cx= 0. Then c.rfRc 0 which implies that xfRe 0 becausc c is regular.But then the left annihilator condition implies that .r 0as required.
From Theorem 6 and Proposition 5 we know that eRf has finite right uniform dimension as a right fRf-module.Since eRf and ceRf are isomorphic fRf-modules, d(ceRf)= d(eRf)and so ceRf is an essential fRf submodule of eR.f.Hence {y fRf:y ceR.f} is an essential right ideal of fRf which, since fRf is semiprime right Goldie, must contain the required regular element c.
If S is semiprime right Goldie, Q(S) denotes the clsical right quotient ring of S d if M is a right S-module, Q(M)= M sQ(S).Using the right common denominator property of Ore sets we see that every element of Q(M) is of the form m @ cwhere M and c is regul in S. In what follows we shall write THEOM 3.2.If R is a semiprime right Goldie generMized matrix ring with idempotent set E, then O(n)= O(nY).
For each eE, eRe embeds in Q(eRe) and we now check that for e,f E,e # f, eRf embeds in Q(eRf).Suppose that c is regular in fRf, err d zc=0.
Then fRezc 0 and so fRez 0 because c is regular in fRf.Thus ezfRezf 0 d hence 0 ezf z since R is semiprime.This shows that eRr is a torsion free fRf-module d so eRf embeds in Q(eRf).Let e, f, 9 E, err, fR9 and suppose that c is regular in fRf and d is regul in 9R9.Define (ze-)(yd -) zycTdwhere Ya and c are obtned from the right bimodule Ore condition" yc eye.It is straightforward to check that this multiplication is well-defined and hat a result Q Q(eRf) becomes a generalized matrix ring with idempotent set E. We now show that s semiprime.It follows from Threm 6 that eRe is semiprime right Goldie for all e E and hence Q(eRe) is semiprime for all e suces to show that (Q, E) satisfies the right nihilator condition.Let yd-Q(fRe) be such that (eRf)ud-' 0. Then (eRf)(yd-') 0 and so (eRf)u 0. From Corolly 1.2 we s that (R,E) satisfies the right annihilator condition and so y=0.Thus (Q,E) satisfies the right nihilator condition and hence Q is semiprime.
Let e, f E,e f.From Proposition 2.3 we see tha err has finite uniform dimension a right fRf-module d so Q(eRf) has finite uniform dimension as a right Q(fRf)-module.Since Q(fRf) is semisimple Artinian it follows that Q(eRf)is an artinian Q(fRf)-module, and hence Q is right Artinian by an argument similar to ([7], 1.1.7).Since we have already seen that Q is semiprime, Q is a semisimple Arfinian ring.
To complete the prf, we need only show that R is z Z z(e,f) where z(e,f)G Q(eRf)for all e,f E. Using the right common denominator bimodule Ore condition if for each m E M and each regular element c G 5' there is an m M and a regular c T such that me1 cm.idempotent set E, then eRr satisfies the right bimodule Ore condition for all e, f E. PROOF.Let rn eRr and suppose that c is regular in ere.Define O:eRf--,ceRf by O(x) cx for all x err.Clearly 0 is an fRf-module homomorphism and we now check that 0 is a monomorphism.Suppose that x eRr and cx O. Then cxfRe 0 which implies that xfRe 0 because c is regular.But then the left annihilator condition implies that x 0 as required.
From Theorem 6 and Proposition 5 we know that eRr has finite right uniform dimension as a right fRf-module.Since eRr and ceRf are isomorphic fRf-modules, d(ceRf)= d(eRf) and so ceRf is an essential fRf submodule of err.Hence {y fRf:rny e ceRf} is an essential right ideal of fRf which, since fRf is semiprime right Goldie, must contain the required regular element c.El If 5' is semiprime right Goldie, Q(S) denotes the classical right quotient ring of 5' and if M is a right 5,-module, Q(M)= M (R) sQ(5").Using the right common denominator property of Ore sets we see that every element of Q(M) is of the form m (R) c-1 where m M and c is regular in 5,.In what follows we shall write rnc-instead of m (R) c-1.THEOREM 3.2.If R is a semiprime right Goldie generalized matrix ring with idempotent set E, then Q(R)= y] Q(eRf).
e, f EE PROOF.
For each eE, eRe embeds in Q(eRe) and we now check that for e,f e E,e f, eRf embeds in Q(eRf).Suppose that c is regular in fRf, x err and xc O.
Then fRexc 0 and so fRex 0 because c is regular in fRf.Thus exfRexf 0 and hence 0 exf x since R is semiprime.This shows that err is a torsion free fRf-module and so eRf embeds in Q(eRf).
Let e,f,g E E, x _ eRf, y f Rg and suppose that c is regular in f R f and d is regular in gRg.Define (zc-1)(yd -x) xylcd -1 where yl and c are obtained from the right bimodule Ore condition: ycl cyl.It is straightforward to check that this multiplication is well-defined and that as a result Q Q(eRf) becomes a generalized matrix ring with idempotent set E.
We now show that /.E as semiprime.It follows from Theorem 6 that ere is semiprime right Goldie for all e E E and hence Q(eRe) is semiprime for all e E. In view of Corollary 1.2, it suffices to show that (Q,E) satisfies the right annihilator condition.Let yd-e Q(fRe) be such that Q(eRf)yd -1 0. Then (eRf)(yd-1) 0 and so (eRf)y 0. From Corollary 1.2 we see that (R,E) satisfies the right annihilator condition and so y 0. Thus (Q,E) satisfies the right annihilator condition and hence Q is semiprime.
Let e, f E, e f.From Proposition 2.3 we see that eRf has finite uniform dimension as a right fRf-module and so Q(eRf) has finite uniform dimension as a right Q(fRf)-module.Since Q(fRf) is semisimple Artinian it follows that Q(eRf)is an Artinian Q(fRf)-module, and hence Q is right Artinian by an argument similar to ([7], 1.1.7).Since we have already seen that Q is semiprime, Q is a semisimple Artinian ring.
To complete the proof, we need only show that R is a right order in Q.Let xQ, x x(e,f) where z(e,f) Q(eRf) for all e,f E. Using the right common denominator 3. THE QUOTIENT R/NG.

PROPOSITION 3 . 1 .
If R is a semiprime right Goldie generalized matrix ring with