Finitely Presented Modules over Right Non-Singular Rings

Finitely Presented Modules over Right Non-Singular Rings.

ULRICHALBRECHT(*)

ABSTRACT - Thispaper characterizesthe right non-singular ringsR for which M=Z(M) isprojective wheneverM isa cyclically (finitely) presented module.

Several related results investigate right semi-hereditary rings.

1. Introduction.

The straightforward attempt to extend the notion of torsion-freeness from integral domainsto non-commutative ringsencountersimmediate difficulties. To overcome these, one can concentrate on either the compu- tational or the homological propertiesof torsion-free modules. Goodearl and otherstook the first approach when they introduced the notion of a non- singular module [8]. A rightR-moduleMisnon-singularifZ(M)ˆ0 where Z(M)ˆ fx2MjxIˆ0 for some essential right idealIofR} denotesthe singular submodule of M. On the other hand,MissingularifZ(M)ˆM.

Moreover, a submodule U of an R-module M isS-closed ifM=U isnon- singular. Finally,Risaright non-singular ringifR_Risnon-singular.

The right non-singular rings are precisely the rings which have a regular, right self-injective maximal right ring of quotients, which will be denoted byQ^r(see [8] and [11] for details). Following [11, Chapter XI],Q^r isa perfect left localization of RifQ^risflat asa rightR-module and the multiplication mapQ^r_RQ^r!Q^ris an isomorphism. In particular,Q^risa perfect left localization of R if and only if every finitely generated non- singular rightR-module can be embedded into a projective module ([8] and [11]). We call a right non-singular ring with this propertyright strongly non-singular.

(*) Indirizzo dell'A.: Department of Mathematics, Auburn University, Auburn, AL 36849, U.S.A.

E-mail: albreuf@mail.auburn.edu

Hattori took the second approach by definingM to be torsion-freeif Tor^R₁(M;R=Rr)ˆ0 for allr2R[9]. The classes of torsion-free and non- singular rightR-modulescoincide if and only ifRisa right Utumi p.p.-ring without an infinite set of orthogonal idempotents [3, Theorem 3.7]. Here,R isa rightp.p.-ringif all principal right idealsofRare projective. Moreover, a right non-singular ringRisaright Utumi-ringif every S-closed right ideal ofRisa right annihilator.

Closely related to the notion of torsion-freeness are those of purity and relative divisibility. A sequence of rightR-modulesispure-exact (RD-exact) if every finitely presented (cyclically presented) module is projective with respect to it. Investigating RD- and pure-projective modules leads to the investigation of the condition thatM=Z(M) isprojective. The dual question when Z(M) isinjective hasbeen addressed in [8, Page 48, Example 24].

Section 2 discusses the question for which ringsM=Z(M) isprojective for all RD-projective modulesM. Theorem 2.1 shows that, providedRhasno in- finite set of orthogonal idempotents, these are precisely the right Utumi p.p.- rings discussed in [3]. The structure of pure-projective rightR-moduleswas described in [4] in case thatRis a right strongly non-singular, right semi- hereditary ringRwithout an infinite set of orthogonal idempotents. Theo- rem 2.3 shows that these conditions onRare not only sufficient, but also necessary for the structure-theorem (part b) of Theorem 2.3 to hold.

PruÈfer domainscan be characterized asthe domainswith the property that, whenever a torsion-free moduleMcontainsa projective submoduleU with M=U finitely generated, then M isprojective and M=U isfinitely presented [7, Chapter VI]. We show that the right non-singular rings having the corresponding property for non-singular modules are precisely the right strongly non-singular, right semi-hereditary rings of finite right Goldie dimension (Theorem 3.1).

Section 4 investigates pure-projective modules over right hereditary rings. As part of our discussion, we obtain a characterization of the right Noetherian right hereditary ringswith the restricted right minimum condition which are right strongly non-singular. The last results of this paper demonstrate that right invertible submodules of Q^r_R which were introduced in [11, ChaptersII.4 and IX.5] may fail to share many of the important propertiesof invertible modulesover integral domains. For instance, the lattice of finitely generated right ideals over a PruÈfer domain isdistributive, i.e.I\(J‡K)ˆ(I\J)‡(I\K) for all finitely generated idealsI,J, andKofR[7, Theorem III.1.1]. Example 4.7 shows that there exists a right strongly non-singular, hereditary, right and left Noetherian ring whose finitely generated right ideals do not have this property.

2. RD-Projective Modules.

LetUbe a submodule of a non-singular moduleM. TheS-closure of U in Misthe submoduleVofMwhich containsUsuch thatV=UˆZ(M=U).

THEOREM2.1. The following are equivalent for a right non-singular ring R without an infinite set of orthogonal idempotents:

a) R is a right Utumi p.p.-ring.

b) M=Z(M)is projective for every RD-projective module M.

PROOF. a))b): Since every RD-projective module isa direct sum- mand of a direct sum of cyclically presented modules, it suffices to verify b) in case thatMR=aRfor somea2R. LetJbe the right ideal ofRwhich containsaR such thatJ=aRˆZ(R=aR). Then, R=JM=Z(M) isa non- singular cyclic module which is projective by [3, Corollary 3.4].

b))a): LetIbe theS-closure ofrRfor somer2R. SinceR=rRisRD- projective, andR=I(R=rR)=Z(R=rR), we obtain thatR=Iisprojective.

Thus,Iisgenerated by an idempotent.

By [3, Lemma 3.5], it suffices to show that every S-closed right idealJ of Risgenerated by an idempotent. For this, select 06ˆr₀2J.

Since J is S-closed in R, it contains the S-closure I₀ of r₀R. By what hasbeen shown so far, I0ˆe0R for some idempotent e0 of R. Hence, Jˆe0R[J\(1 e0)R]. If J\(1 e0)R6ˆ0, select a non-zero r₁ˆ(1 e₀)r₁2J; and observe that J\(1 e₀)R is S-closed in R.

Hence, it containsthe S-closure I₁ of r₁R in R. By the previous paragraph, I₁ˆfR for some idempotent f of R. Write f ˆ(1 e₀)s for some s2R, and s et e1ˆf(1 e0). Since e0f ˆ0, we have e1e0ˆe0e1ˆ0 ande²₁ˆf² f²e0 fe0f‡(fe0)²ˆf fe0ˆe1. Thus , e0

and e1 are non-zero orthogonal idempotentswith e1RfR. On the other hand, f ˆf(1 e₀)sˆe₁s yields fRˆe₁R. Consequently, Rˆe₀R e₁R[J\(1 e₀ e₁)R]. Continuing inductively, we can construct non-zero orthogonal idempotents e0;. . .;en‡12J aslong

as J\(1 e0 . . . en)R6ˆ0. Since R doesnot contain an infinite

family of orthogonal idempotent, thisprocesshasto stop, say

J\(1 e₀ . . . e_n)Rˆ0. Then, e₀‡. . .‡e_m isan idempotent with

Jˆ(e₀‡. . .‡e_m)R. p

We now investigate which conditions R has to satisfy to ensure the validity of the structure theorem for pure-projectives in [4]. We want

to remind the reader that a right R-module is essentially finitely generated if contains an essential, finitely generated submodule.

LEMMA 2.2. The following are equivalent for a right non-singular ring R:

a) R is right semi-hereditary and has finite right Goldie-di- mension.

b) A finitely generated right R-module M is finitely presented if and only if p:d:M1.

PROOF. a))b): SinceRisright semi-hereditary, every finitely pre- sented module has projective dimension at most 1. Conversely, whenever MRⁿ=U for some projective module U, then U isessentially finitely generated sinceR has finite right Goldie dimension. By Sandomierski's Theorem [5, Proposition 8.24], essentially finitely generated projective modulesare finitely generated.

b))a): Clearly,R hasto be right semi-hereditary. If R hasinfinite right Goldie-dimension, then it contains a family fIng_n<vof non-zero, fi- nitely generated right ideals whose sum is direct. SinceRisright semi- hereditary, eachI_n isprojective, and the same holdsfor_n<vI_n. By b), R=_n<vI_nisfinitely presented, a contradiction. p THEOREM2.3. The following conditions are equivalent for a right non- singular ring R without an infinite set of orthogonal idempotents:

a) R is a (right Utumi), right semi-hereditary ring such that Q^ris a perfect left localization of R.

b) A right R-module M is pure-projective if and only if

i) Z(M)is a direct summand of a direct sum of finitely gen- erated modules of projective dimension at most1.

ii) M=Z(M)is projective.

PROOF. A right strongly non-sigular, right semi-hereditary ring without an infinite set of orthogonal idempotents is a right Utumi ring [3, Theorems3.7 and 4.2].

a))b): By [3],Rhas finite right Goldie-dimension. Because of Lemma 2.2,Z(M) ispure-projective if and only if condition i) in b) holds. It remains to show that M=Z(M) isprojective whenever M isfinitely presented.

However, thisfollowsfrom [11] sinceM=Z(M) isa finitely generated non- singular module, andQ^r isa perfect left localization ofR.

b))a): To see that R is right semi-hereditary, consider a finitely generated right idealIifR. TheS-closureJ ofIsatisfiesJ=IˆZ(R=I).

Hence,R=Jisprojective by b), andJ=Ihas projective dimension at most 1.

SinceJisprojective, thisisonly possible ifIisprojective.

To show thatRhasfinite right Goldie-dimension, consider a right ideal of Rof the forma0R. . .anR. . .where eachan6ˆ0. Form<v, letImbe theS-closure ofa0R. . .amR. SinceR=Im isprojective by b),Imisgen- erated by an idempotent em of R. WriteIm‡1ˆIm[Im‡1\(1 em)R].

Observe that [I_m‡1\(1 e_m)R] isgenerated by an idempotentf ofRasa direct summand ofR. Settingd_mˆf(1 e_m) yieldsan idempotentd_mofR such thate_md_mˆd_me_mˆ0, andI_m‡1ˆI_md_mRasin the proof of The- orem 2.1. Inductively, one obtainsan infinite family of orthogonal idempo- tentsfdmjm<vgofR, which is not possible. Thus,Rhasfinite right Goldie- dimension; and everyS-closed right idealJofRistheS-closure of a finitely generated right ideal. By b),R=Jisprojective; andRisa right Utumi-ring sinceJˆeRfor some idempotenteofR.

To establish thatQ^risa perfect left localization ofR, it suffices to show that every finitely generated non-singular rightR-moduleMisprojective.

Write MRⁿ=U and observe that U is essentially finitely generated.

Select a finitely generated essential submodule V of U. Then,

U=VˆZ(Rⁿ=U), andMisprojective by b). p

In the following, the injective hull of a moduleMisdenoted byE(M).

COROLLARY2.4. LetRbe a right semi-hereditary ring of finite right Goldie-dimension such thatQ^ris a perfect left localization ofR. A rightR- module M is pure-projective if and only if M=Z(M) is projective and Z(M) is isomorphic to a direct summand of a direct sum of finitely generated submodules of(Q^r=R)ⁿ.

PROOF. We first show that a finitely generated singular moduleMhas projective dimension at most 1 if and only if it can be embedded into a finite direct sum of copies of Q^r=R. If p:d:M1, then there exist a finitely generated free moduleFand an essential projective submodulePofFwith MF=P. Since R hasfinite right Goldie-dimension,P is essentially fi- nitely generated, and hence itself finitely generated by Sandomierski's Theorem [5, Proposition 8.24]. We can find a finitely generated projective module U such that PU isa finitely generated free module, say PURⁿ. Since M issingular, PU is an essential submodule of FU. Therefore,FUE(PU)ˆ(Q^r)ⁿ, andM(Q^r=R)ⁿ.

On the other hand, ifMisa finitely generated submodule of (Q^r=R)ⁿ, then there isa finitely generated submoduleUof (Q^r)ⁿcontainingRⁿsuch thatMˆU=Rⁿ. SinceRis right semi-hereditary, and sinceQ^risa perfect left localization of R, U isprojective and p:d:M1. The corollary now

followsdirectly from Theorem 2.3. p

3. Essential Extensions of Projective Modules.

In the commutative setting, PruÈfer domainsare characterized by con- ditionsb) and c.ii) [7].

THEOREM3.1. The following are equivalent for a right non-singular ring R:

a) R is a right semi-hereditary ring of finite right Goldie-dimen- sion for which Q^ris a perfect left localization of R.

b) Whenever a non-singular module M contains a projective submodule U such that M=U is finitely generated, then M is projective and M=U is finitely presented.

c) i) R is a right p.p.-ring.

ii) If a finitely generated non-singular right R-module M con- tains an essential projective submodule U, then M is projective, and M=U is finitely presented.

PROOF. a))b): LetW be theS-closure ofUinM. SinceM=W isfi- nitely generated asan image ofM=Uand non-singular, it is projective by a).

Hence,MˆWPfor some finitely generated projective moduleP. We may thusassume thatM=Uissingular.

SinceRisright semi-hereditary,Uˆ _IU_iwhere eachU_i isfinitely generated [1]. Because M=U issingular and M isnon-singular, U is essential in M. Thus , E(M)ˆE(U)ˆ _IE(U_i) in view of the fact that direct sums of non-singular injectives are injective if R hasfinite right Goldie-dimension [11, Proposition XIII.3.3]. Choose a finitely generated submoduleV of Msuch that MˆU‡V. There is a finite subsetJ ofI such that V _JE(U_i). Then, W₁ˆV‡ _JU_i isa finitely generated submodule of _JE(U_i) such that V\(_InJU_i)ˆ0. Consequently, MˆW_{1 InJ}U_i. But W₁ isprojective by a) showing that M ispro- jective and thatM=UˆW1=U isfinitely presented.

b))c): Observe that every finitely generated non-singular module is projective by choosingUˆ0 in b).

c))a): Assume that R containsa right ideal U of the form Uˆ n<vanRwhere eachan6ˆ0. By part i) of c),Uisprojective. Choose a right ideal V of R which ismaximal with respect to the property U\Vˆ0. SinceRisright non-singular,VisanS-closed submodule ofR and [R=V]=[UV=V]R=(UV) issingular. Therefore, the projective moduleUV=Vis essential in the non-singular moduleR=V. By c),R=V isprojective; and R=(UV) isfinitely presented. Hence, U isfinitely generated which is not possible.

To see that Ris a right strongly non-singular, right semi-hereditary ring, it suffices to show that a finitely generated non-singular right R- module Misprojective. By [11, Proposition XII.7.2],M(Q^r)ⁿ for some n<v. Since R hasfinite right Goldie dimension andRⁿ is essential in (Q^r)ⁿ, M hasfinite Goldie-dimension. Therefore, M containsuniform submodules U₁;. . .;U_m such that U₁. . .U_m isessential in M. Fur- thermore, we may assume that eachU_iiscyclic, sayU_iˆb_iR. SinceMis non-singular, ann_r(b_i)ˆ fr2Rjb_irˆ0g is not essential in R. Select ci2R with ciR\annr(bi)ˆ0. Then, Uicontainsa submodule ViciR.

Since R isa right p.p.-ring, V_i isprojective. Hence, M containsthe es- sential projective submoduleV₁. . .V_m. By c),Misprojective. p A submoduleU of a moduleM istightif bothU andM=U have pro- jective dimension at most 1. A module iscoherentif all itsfinitely gener- ated submodules are finitely presented.

COROLLARY3.2. LetRbe a right semi-hereditary ring of finite right Goldie-dimension such thatQ^ris a perfect localization ofR.

a) A right R-module M of projective dimension at most 1 is co- herent. Moreover, all its finitely generated submodules are tight.

b) If M is singular and a direct sum of countably generated modules, then p:d:M1if and only if M(Q^r=R)^(I)for some index-set I.

PROOF. a) WriteMF=PwhereFand itssubmodule Pare projec- tive. IfUisa finitely generated submodule ofM, then there isa submodule WofFcontainingPwithW=PU. By Theorem 3.1,Wisprojective and W=Pisfinitely presented. Clearly,UandM=Uhave projective dimension at most 1.

b) Without loss of generality, we may assume that M iscountably generated. Ifp:d:M1, thenMˆF=PwhereFisprojective andPR^(I) for some index-setI. SincePis essential inF, we haveFE(P)(Q^r)^(I)

by [11, Proposition XIII.3.3] becauseRhasfinite right Goldie-dimension.

Hence,M(Q^r=R)^(I).

Conversely, suppose thatM(Q^r=R)^(v), and select a submodule U of (Q^r)^(v) containing R^(v) such that MˆU=R^(v). Choose fu_njn<vg U such thatUˆS_n<vu_nR‡R^(v) andu₀ˆ0. SetV<sub>`</sub>ˆR<sup>(v)</sup>‡S<sup>`</sup>_nˆ1u_nR. By Theorem 3.1, each V` isprojective. LetW`ˆV`=R<sup>(v)</sup>M. Then,W0ˆ0 and Mˆ [<sup>v</sup><sub>nˆ1</sub>Wn. Observe that W`‡1=W`V`‡1=V` hasprojective di- mension at most 1. By Auslander's Theorem,p:d:M1. p

4. Hereditary Rings.

The first result describes the right strongly non-singular, right Noe- therian, right hereditary rings.

PROPOSITION4.1. The following conditions are equivalent for a right non-singular ringRof finite right Goldie dimension:

a) R is a right strongly non-singular, right hereditary ring with- out an infinite set of orthogonal idempotents.

b) R is a right strongly non-singular, right Noetherian and right hereditary.

c) i) R has finite right Goldie dimension.

ii) M is pure projective if and only if M=Z(M)is projective, and Z(M)is a direct summand of a direct sum of finitely generated modules.

PROOF. a))b): By [3, Theorems3.7 and 4.2],Rhasfinite right Goldie dimension. However, essentially finitely generated projective modules are finitely generated [5, Proposition 8.24].b))c) isobviousin view of Theo- rem 2.3.

c))a): LetIbe a right ideal ofR, andJitsS-closure inR. SinceRhas finite right Goldie dimension,Icontainsa finitely generated right idealK as an essential submodule. Thus,J istheS-closure ofK, and J=K isthe singular submodule of the finitely presented module R=K. By c),R=J is projective, andRˆJJ₁. Then,R=IJ=IJ₁. In particular,J=Iisa finitely generated singular module, which is pure-projective by c). Hence, J=I is a direct summand of a direct sum of finitely presented modules.

Clearly, thissum can be chosen to be finite. Therefore, J=I isfinitely presented, andI isfinitely generated sinceJ isa direct summand ofR.

Once we have shown that every finitely generated non-singular right R-

moduleMisprojective, we will have established thatRisa right heredi- tary ring with the property thatQ^risa perfect left localization ofR.

There exists a finitely generated free moduleFand a submoduleUofF such thatMF=U. SinceRhasfinite right Goldie-dimension,Ucontains a finitely generated essential submoduleV. Because,F=Uisnon-singular, U=V is the singular submodule of the finitely presented moduleF=V. By

c),F=U(F=V)=(U=V) isprojective. p

COROLLARY4.2. The following are equivalent for a right non-singular ringRwith finite right Goldie-dimension:

a) R is a right Noetherian, right hereditary ring which satisfies the restricted right minimum condition such that Q^r is a perfect left lo- calization of R.

b) M is pure projective if and only if M=Z(M)is projective and Z(M)is a direct summand of a direct sum of finitely generated Artinian modules.

PROOF. a))b): SinceRhasthe restricted minimum condition, every finitely generated singular right module is Artinian.

b))a): LetIbe an essential right ideal ofR. SinceRhasfinite right Goldie-dimension,Icontains a finitely generated essential right idealJ.

By c), the finitely presented moduleR=Jisa direct summand of a (finite) direct sum of finitely generated Artinian modules. But this is only pos- sible ifR=JisArtinian. But then,R=IisArtinian too. Arguing asin the proof of Proposition 4.1, we obtain thatRisa right strongly non-singular,

right hereditary. p

By [8, Proposition 5.27], a right hereditary, right Noetherian ring which isthe product of prime ringsand ringsMorita equivalent to lower trian- gular matrix rings over a division algebra is right strongly non-singular and hasthe restricted right minimum condition, i.e. R=I isArtinian for every essential right idealIofR.

Let U be a subset of Q^r, and set (R:U)_rˆ fq2Q^rjUqRg and (R:U)_`ˆ fq2Q^rjqURg.

THEOREM4.3. The following are equivalent for a ring R:

a) R is a right Noetherian right hereditary ring satisfying the restricted right minimum condition such that Q^r is a perfect left locali- zation of R.

b) R is a left Noetherian left hereditary ring satisfying the re- stricted left minimum condition such that Q^`is a perfect right localization of R.

c) R is a right and left Noetherian, hereditary, right and left Utumi-ring.

d) RˆR1. . .Rnwhere each Riis either a prime right and left

Noetherian hereditary ring or Morita equivalent to a lower triangular matrix ring over a division algebra.

PROOF. a))c): By [3, Theorem 4.2],Q^rˆQ^`is a semi-simple Artinian ring, andRisright and left Utumi. Furthermore,Risleft semi-hereditary by [3, Theorem 5.2]. It remainsto show thatRisleft Noetherian.

Suppose that R containsa left ideal I which isnot finitely gen- erated. Without loss of generality, we may assume that I is essential in R. Since Q^rˆQ^` is semi-simple Artinian, R hasfinite left Goldie- dimension [11, Theorem XII.2.5], and I containsa finitely generated essential left idealJ0. Since Iisnot finitely generated, we can find an ascending chain J0. . .Jn. . . of finitely generated essential left idealsinside I with J_n6ˆJ_n‡1 for all n.

SinceQ^risan injective leftR-module being the maximal left ring of quotientsof R, every map f:J_i!Q^r isright multiplication by some q2Q^r, which isuniquely determined by f since Ji isessential.

Therefore, we can identify HomR(J_i;R) and J_i ˆ(R:J_i)_r. Moreover, J_i isa finitely generated projective right R-module because J_i is projective sinceR isleft semi-hereditary. Furthermore, J_iˆ(R:J_i)_` satisfies J_iˆJ_i. To see this, observe that J_iJ_i by definition.

Conversely,Jihasa finite projective basissince it isfinitely generated and projective. There are a1;. . .;ak 2Ji and q1;. . .;qk2J_i such that

yˆyq1a1‡. . .‡yq_ka_k for all y2J_i. Since Q^r also is the maximal left

ring of quotientsofR, it isnon-singular asa left R-module. BecauseJ_i is an essential left ideal, 1ˆq₁a₁‡. . .‡q_ka_k. If z2J_i , then zq_i2R, and zˆ(zq1)a1‡. . .(zqk)ak 2Ji.

We obtain a descending chainJ₀. . .J_n. . .R ˆRof finitely generated submodules of Q^r_R. Since Q^r=R issingular, J₀=R isa finitely generated singular right R-module. By the restricted right minimum condition for R, J₀=R isArtinian. Consequently, there ism with J_m ˆJ_m‡1 , from which one obtainsJ_mˆJ_m‡1.

c))d): By [3, Theorem 5.2], the classes of torsion-free, non-singular and flat rightR-modulescoincide. Because of [3, Theorem 5.5],Rhasthe desired form.

d))a): By [3, Theorem 5.5],Risa right and left Noetherian heredi- tary ring for which the classes of torsion-free, flat and non-singular mod- ulescoincide. Thus,Q^risa perfect left localization ofRby [3, Theorem 5.2].

Finally,Rsatisfies the restricted right minimum condition by [5].

Since condition c) isright-left symmetric, the equivalence of a) and b)

followsimmediately. p

A submoduleU of the rightR-moduleQ^r isright invertibleif there exist u1;. . .;un2U and q1;. . .;qn 2(R:U)_{`</sub> with u1q1‡. . .unqnˆ1 [11, Chaptersii.4 and IX.5]. Over an integral domain R, a right in- vertible moduleUhasthe additional property thatU(R:U)<sub>`}ˆRwhich may fail in the non-commutative setting:

EXAMPLE4.4. There exists a right and left Artinian, hereditary ringR such thatQ^risa perfect right and left localization ofR, which containsan essential right idealIwith (R:I)_{`</sub>ˆQ<sup>r</sup>and (R:I)<sub>`}IˆI. Moreover, there existr₁;r₂;s₁;s₂2Iandq₁;q₂;p₁;p₂2(R:I)_`satisfying 1ˆr₁p₁‡r₂p₂ ˆ

ˆs1q1‡s2q2 such that r1p1;r2p22R, but s1q1;s2q2 62R. Thus, I(R:I)_`6R.

PROOF. LetFbe a field of characteristic different from 2, andRbe the lower triangular matrix ring overF. Clearly,Risa right and left Artinian ring. According to [8, Theorem 4.7],Risa right hereditary ring, which is also left hereditary by [5, Corollary 8.18]. Finally, by [8, Proposition 2.28 and Theorem 2.30], the maximal right and left ring of quotientsof Ris QˆMat₂(F). Inside R, we consider the idempotents e₁ˆ 1 0

0 0

and e2ˆ 0 0

0 1

. LetIˆ F 0 F 0

, a two-sided ideal ofRwhich is essential as a right ideal because a 0

b c

0 01 0

ˆ 0 0 c 0

for alla;b;c2F. Since IˆQ^re1, we have (R:I)_{`</sub>ˆQ<sup>r</sup>and (R:I)<sub>`}IˆI.

Finally, consider the elements s₁ ˆ 1 0 1 0

and s₂ˆ 1 0 1 0

of I and q₁ˆ ^{12 12} c 0

and q₂ˆ ^{12 12} c 0

of Q. It iseasy to see that s1q1;s2q262R although s1q1‡s2q2ˆ1. On the other hand, setting r1ˆp1ˆe12I, r2ˆ 0 0

1 0

2I, and p2ˆ 0 1 0 0

yields r1p1;r2p22R and r1p1‡r2p2ˆ1. p

In view of the previousexample, we define a submodule U of Q^r_R to be strongly invertible if there isa submodule M of RQ^r such that MUˆUMˆR.

LEMMA 4.5. Let R be a right and left non-singular, right and left Utumi-ring with maximal right and left ring of quotients Q. A submodule U of Q^ris strongly invertible if and only if it satisfies the following con- ditions:

i) U is also a submodule ofRQ.

ii) URis a finitely generated projective generator ofMR. iii) _RU is a finitely generated projective generator of_RM.

PROOF. Suppose that U is strongly invertible, and choose a submodule X of RQ^r with XUˆUXˆR. Then, X(R:U)_{`</sub>\(R:U)<sub>r</sub> and (R:U)<sub>`}UˆRˆU(R:U)_r. Moreover, RU ˆ(UX)UˆU(XU)ˆ

ˆURˆU yieldsthat U isa submodule of _RQ^r too. By symmetry, X is also a submodule of Q^r_R. Becaus e of this , (R:U)_{`</sub> and (R:U)<sub>r</sub> are submodules of both Q<sup>r</sup><sub>R</sub> and RQ<sup>r</sup> too. Therefore, U(R:U)<sub>`}ˆ

ˆU(R:U)_{`</sub>RˆU(R:U)<sub>`}U(R:U)_rˆU(R:U)_rˆR. By s ymmetry, (R:U)_rUˆR.

By what has been shown in the last paragraph, we can write 1ˆu₁q₁‡. . .‡u_nq_n with u₁;. . .;u_n2U and q₁;. . .;q_n2(R:U)_{`</sub>. Let f<sub>i</sub>:U!Rbe left multiplication byqi. Asin [11, Proposition IX.5.2], the setf(u1;f<sub>1</sub>);. . .;(un;f<sub>n</sub>)gisa projective basisforU. Selectv1;. . .;vm2U andp1;. . .;pm2(R:U)<sub>`} with 1ˆp1v1‡. . .‡pmvm. Definec:U^m!R

byc(x₁;. . .;x_m)ˆS^m_iˆ1p_ix_i. Then,cisonto, andU^mˆRW, i.e.Uisa

generator of M_R. By symmetry, _RU isa finitely generated projective generator ofRM.

Conversely, assume that U satisfies the three conditions. Observe that (R:U)<sub>`</sub> and (R:U)<sub>r</sub> are submodules of bothQ<sup>r</sup><sub>R</sub> and<sub>R</sub>Q<sup>r</sup>. Since it isa projective generator of M<sub>R</sub>, there is ` <v such thatU^`ˆRW.

Let p:U^{`</sup>!R be a projection with kernel W, and d<sub>j</sub>:U!U<sup>`} be the embedding into the j^th-coordinate. The map pd_j:U!R isleft multi- plication by someqj2Q^rsince Q^risa right self-injective ring. Clearly, since p isonto, there are u1;. . .;u` 2U with q1u1‡. . .‡q`u`ˆ1.

Since q₁;. . .;q_{`</sub> 2(R:U)<sub>`}, we have (R:U)_{`</sub>UˆR. By s ymmetry, U(U:R)<sub>r</sub>ˆR. Now, U(R:U)<sub>`} ˆU(R:U)_{`</sub>U(R:U)<sub>r</sub>ˆR yields (R:U)<sub>`}(R:U)_r. In the s ame way, (R:U)_`ˆ(R:U)_r, and U is

strongly invertible. p

PROPOSITION4.6. LetRbe a right and left non-singular, right and left Utumi-ring. IfIis a two-sided ideal ofRsuch thatRIandIRare finitely generated projective generators ofRMandMRrespectively, thenR=I is projective with respect to all RD-exact sequences.

PROOF. The proof of [7, LemmasI.7.2 and I.7.4] can be adapted to show that R=Iisprojective with respect to all RD-exact sequencesof R-modulesprovided there are r₁;. . .;r_n 2R and q₁;. . .;q_n2(R:I)_` such that r₁q₁‡. . .‡r_nq_nˆ1 and r₁q₁;. . .;r_nq_n2R. However, thisis

guaranteed by Lemma 4.5. p

Finally, the lattice of finitely generated right idealsover right strongly nonsingular, hereditary right and left Noetherian rings may not be dis- tributive:

EXAMPLE4.7. There exists a right strongly non-singular, hereditary, right and left Noetherian ringRfor which the lattice of right idealsisnot distributive.

PROOF. LetRbe the ring considered in Example 4.4, whose notation will be used in the following. Consider the right idealsJˆe1Rˆ Q 0

0 0

andKˆe2Rˆ 0 0 Q Q

. IfIˆ 1 0 1 0

Q 0 Q Q

ˆ f a 0 a 0

ja2Fg, thenI\JˆI\Kˆ0, whileI\(J‡K)ˆI. p

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Amer. Math. Soc.,12(1961), pp. 638-639.

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[7] L. FUCHS- L. SALCE,Modules over Non-Noetherian Domains, Mathematical Surveysand Monographs84, Amer. Math. Soc. (2000).

[8] K. GOODEARL,Ring Theory, Marcel Dekker, New York, Basel (1976).

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[10] J. ROTMAN, An Introduction to Homological Algebra; Academic Press, London (1979).

[11] B. STENSTROÈM, Rings of Quotients, Lecture Notesin Math. 217, Springer Verlag, Berlin, Heidelberg, New York (1975).

Manoscritto pervenuto in redazione il 12 gennaio 2007.