Sigma-Cotorsion Rings$

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Advances in Mathematics 191 (2005) 11–28 http://www.elsevier.com/locate/aim

Sigma-cotorsion rings$

Pedro A. GuilAsensio a, and Ivo Herzogb a Departamento de Matema´ticas Universidad de Murcia, Espinardo, Murcia 30100, Spain b The Ohio State University at Lima, Lima, OH 45804, USA

Received 18 August 2003; accepted 23 January 2004

Communicated by Mark Hovey

Abstract

It is proved that a ring R is right perfect if and only if it is S-cotorsion as a right module over itself. Several other conditions are shown to be equivalent. For example, that every pure submodule of a free right R-module is strongly pure-essential in a direct summand, or that the ð@0Þ countable direct sum CðRÞ of the cotorsion envelope of RR is cotorsion. If CR is a ﬂat S-cotorsion module, then CR admits a decomposition into a direct sum of indecomposable modules with a local endomorphism ring. The Jacobson radical JðSÞ of the endomorphism ring S ¼ EndRC is characterized as the maximum ideal that acts locally T-nilpotently on CR: If R is semilocal and C ¼ CðRÞ; then the radicalconsists of those endomorphisms f : C-C whose image is contained in CJ: r 2004 Elsevier Inc. All rights reserved.

MSC: 16D50; 16D90; 16L30

Keywords: Sigma-cotorsion ring; Perfect ring; Cotorsion envelope

Let R be an associative ring with identity. A right R-module MR is said to be 1 cotorsion if for every ﬂat right R-module F; ExtRðF; MÞ¼0: This concept, which generalizes the notion of a pure-injective module, was introduced by Harrison [13] in the context of abelian groups. Recently, Bican et al. [2] showed that every right R-module admits a cotorsion envelope. This notion of cotorsion envelope is also closely related to the study of decompositions of modules into direct summands.

$Pedro A. Guil Asensio was partially supported by the DGI (BFM2003-07569-C02-01, Spain) and by the Fundacio´ nSe´ neca (PI-76/00515/FS/01). Ivo Herzog was partially supported by NSF Grant DMS-02- 00698. Corresponding author. E-mail addresses: [email protected] (P.A. Guil Asensio), [email protected] (I. Herzog).

12 P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28

Namely, let M be a right R-module and let add½M denote the full subcategory of Mod-R consisting of direct summands of finite direct sums of copies of M: Then add½M is equivalent to the category proj-S; of finitely generated projective right S-modules, where S ¼ EndRðMÞ (see e.g. [7, Lemma 1.7]). Therefore, the category of flat functors Flatðadd½Mop; AbÞ is equivalent to the category Flatððproj-SÞop; AbÞD Flat-S; where Flat-S denotes the category of flat right modules over S: As observed in [14], the cotorsion envelope of S in Mod-S corresponds to the pure-injective envelope of S in Flat-S: On the other hand, the ring S is right perfect if and only if every object in Flat-S is a direct sum of finitely presented indecomposable modules and this decomposition complements direct summands ([1, Theorems 24.4 and 28.14]). Equivalently, the locally finitely presented additive category Flatðadd½Mop; AbÞ is pure-semisimple [3,p.1655]. Following these ideas, the authors have proved in [12] that if the ring R is cotorsion as a right module over itself, and J ¼ JðRÞ denotes the Jacobson radicalof R; then the quotient ring R=J is Von Neumann regular and right self-injective and idempotents lift modulo J: Furthermore, Example 1 of [12] shows that the endomorphism ring of a cotorsion flat right R-module is also right cotorsion. Let us point out that the category Flat-S is not abelian in general. This means that standard functor category techniques used in the study of pure-injectivity to transform pure-injective modules into injective objects in the functor category cannot be applied in this new framework. ðIÞ A right R-module MR is called S-cotorsion if the direct sum M is cotorsion for every index set I; the ring R is right S-cotorsion if it is S-cotorsion as a right module over itself. This article continues the investigation initiated in [12] by extending to flat S-cotorsion modules results obtained by Garavaglia [5,6], Gruson and Jensen [11] and Zimmermann [21] for S-pure-injective or, equivalently, totally transcendental, modules. If CR is a flat S-cotorsion module, we note (Proposition 7) that it admits a decomposition into indecomposable modules with a local endomorphism ring; this is a decomposition that complements direct summands. If the endomorphism ring is denoted by S ¼ EndRðCÞ; this may be used to show (Theorem 9) that the Jacobson radical JðSÞ acts locally T-nilpotently on C: Consequently, if the ring R is S-cotorsion as a right module over itself, then it is right perfect. We give several other conditions on R that are equivalent to R being right perfect.

Theorem. The following conditions are equivalent for a ring R:

(1) The ring R is right perfect. (2) The ring R is right S-cotorsion. (3) The cotorsion envelope CðRRÞ of R is S-cotorsion. ðIÞ (4) If MRDR is a pure submodule of a free right R-module, then MR is strongly pure-essential in a direct summand of RðIÞ: (5) The category of projective right R-modules is closed under strongly pure-essential extensions. The deﬁnition of strongly pure-essential submodules is given in Section 1. If R is right perfect, then every ﬂat right R-module is projective, and therefore every right ARTICLE IN PRESS

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R-module MR is cotorsion [20, Proposition 3.3.1]. Thus Condition (1) implies the others. The implications (4) or (5) ) (3) are proved (Proposition 4 and Theorem 6) as a consequence of Kaplansky’s Theorem on projectives and a variant (Theorem 2) of the Faith–Walker Theorem for S-cotorsion modules. In proving the implication (3) ) (1), we also characterize the Jacobson radical JðSÞ of the endomorphism ring of the cotorsion envelope C ¼ CðRRÞ when R is a semilocal ring—the endomorphisms that belong to JðSÞ are those whose image is contained in CJ: Conditions (4) and (5) in the above theorem can also be seen as a pure version of the concept of right S-extending rings [4]. Thus, this theorem also extends Oshiro’s results characterizing right S-extending rings [17]. In fact, our results establish a bridge between purity and the theory of (S)-extending rings that has been profusely developed in the last years (see e.g. [4,9,17]). A naturalquestion is whether the cardinalityof the set I that appears in Condition (4) of the above theorem can be bounded. We prove in Section 3 that it is possible to bound it by @¼maxf@1; jRjg (Theorem 22). We also give an example in Section 1 that shows that it is not possible to bound it by @0: Throughout the article, R will denote an associative ring with identity and J the Jacobson radicalof R: Unless otherwise stated, an R-module C will mean a right R- module CR with endomorphism ring S ¼ EndRðCÞ acting on the left; the Jacobson radicalof S will be denoted JðSÞ: The standard reference for cotorsion envelopes is Xu [20].

1. R-cotorsion modules and strong purity

In this section, we prove a variant of the Faith–Walker Theorem [1, Theorem 25.8] for S-cotorsion modules, and apply it to show that if every local direct summand of a free right R-module is strongly pure-essential in a direct summand, then the cotorsion envelope of R is right S-cotorsion.

Lemma 1. Let I be an index set and MR; a cotorsion module. If the cotorsion envelope ðIÞ ðIÞ CR ¼ CðM Þ of the direct sum M admits a decomposition

C ¼ C0"C00; where jIj4jC0jjMj; then M is isomorphic to a direct summand of C00:

ðIÞ Proof. For every iAI; denote by fi : M-M the structuralinjection associated to 0 the direct sum and let Mi ¼ Im fi be the image of fi: Let p : C-C be the projection determined by the decomposition C ¼ C0"C00: Consider the function 0 0 M 0 i/pfi : M-C from the index set I to the set ðC Þ of functions from M to C : From the hypothesis jIj4jC0jjMj ¼jðC0ÞM j; we deduce that this function cannot be one-to-one. Thus there are distinct i; jAI for which pfi ¼ pfj; that is, pð fi À fjÞ¼0and 00 so the image Im ð fi À fjÞDC : Now fi À fj : M-Mi"Mj is a split-monomorphism, ARTICLE IN PRESS

14 P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28 because it has the retraction given by the structuralprojection pi : Mi-M associated ðIÞ to the direct sum M : The image Im ð fi À fjÞ is thus a direct summand of Mi"Mj ðIÞ isomorphic to M: Since Mi"Mj is a direct summand of CR ¼ CðM Þ; the image 00 Im ð fi À fjÞ is a direct summand of C; and therefore of C : &

Theorem 2. Let MR be a nonzero cotorsion R-module. If there exists a cardinal @ such that for every k; the cotorsion envelope CðMðkÞÞ is a direct sum of modules of cardinality at most @; then MR is S-cotorsion.

Proof. Let @XjMjþ@0 be any cardinalthat satisﬁes the hypothesis of the theorem, @ ðkÞ and let k42 : Write C ¼ CðM Þ¼"iAI Ci; where jCijp@; for every iAI: Then kpjCjp@ÁjIj; so that jIjXk: For every ordinal ao@; we will deﬁne a subset IaDI; jIajp@; with the property that if aob; then Ia and Ib are disjoint, and the summand " Ca ¼ iAIa Ci of C has a direct summand isomorphic to M: Taking the direct sum of these @-many copies of MR; one obtains a direct summand of C isomorphic to Mð@Þ: Since we can replace @ by any bigger cardinal, M is S-cotorsion. We proceed by recursion on the ordinals ao@:

* a ¼ 0: Let MDC be a direct summand, and deﬁne I0 to be the support of M; I0 :¼fiAI : piðMÞa0g; where pi : C-Ci are the structuralprojections associated to the direct sum decomposition of C: Since @XjMj; we have that jI0jp@Á@0 ¼ D " @: Thus M C0 :¼ iAI0 Ci is a direct summand. * a40: Suppose that for every boa; the subsets Ib have been deﬁned. The cardinality of their union is bounded by the product jajÁ@¼@: Thus the direct 0 " oo sum C ¼ boa Cb has cardinality bounded by j@ j¼@ (see [15, Corollary M I.10.13]). Since jC0jj jp@@ ¼ 2@okpjIj; the lemma applies and there is a direct 00 " summand of C ¼ ie,Ib Ci isomorphic to M: As in the initialstep, let Ca be the support of M in C00: &

To obtain our ﬁrst characterization of right S-cotorsion rings, we will need the notion of a strongly pure-essential extension of ﬂat modules. Recall that a morphism m : MR-NR of right R-modules is a pure monomorphism if for every left R-module RX; the morphism of abelian groups m#RX : M#RX-N#RY is a monomorphism. Due to the existence of pure-injective envelopes, this is equivalent to the condition that every morphism f : MR-UR with UR pure-injective, may be extended to N: Since we now know that cotorsion envelopes exist [2], we call

m : MR-NR a strongly pure monomorphism if every f : MR-CR with CR cotorsion, extends to N: Because every pure-injective module is cotorsion, every strongly pure monomorphism is a pure monomorphism. Xu [20, p. 67] introduces terminology that suggests calling a monomorphism m : MR-NR an F-pure monomorphism if the cokernel N=M ¼ FR is a ﬂat module. Let us note that every F-pure monomorphism ARTICLE IN PRESS

P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28 15

- 1 m : MR NR is a strongly pure monomorphism, since ExtRðF; CÞ¼0 for any flat module F and any cotorsion module C: However, the class of strongly pure monomorphisms also includes the split monomorphisms. If m : FR-GR is a pure monomorphism of flat modules, then the cokernel Coker m is also flat, and so m is an F-pure monomorphism, and hence strongly pure. Thus when dealing with morphisms between flat modules, there is no difference between these three kinds of monomorphisms. A strongly pure monomorphism m : MR-NR is called strongly pure-essential if every morphism f : NR-XR whose restriction fm to MR is a strongly pure monomorphism is itself a strongly pure monomorphism; if MR is a submodule of NR and m : MR-NR is the inclusion map, then we say that NR is a strongly pure- essential extension of MR: It is clear that if MRDNR is a strongly pure-essential extension of M and NRDKR is a strongly pure-essential extension of N; then MRDKR is a strongly pure-essential extension of M: The assertions of the following proposition are easily derived using standard properties of the cotorsion envelope and the definition of a pure-essentialextension.

Proposition 2 (Xu [20, Section 3.4]). Let MR be a right R-module.

(1) (see [20, Lemma 2.1.2]). The cotorsion envelope MDCðMÞ is a maximal strongly pure-essential extension of MR: It is an F-pure extension and is unique up to isomorphism over M: (3) If MRDNR is a strongly pure-essential extension, then this inclusion is an F-pure monomorphism. If NDCðNÞ is a cotorsion envelope of N; then MDCðNÞ is a cotorsion envelope of M: (4) If MDNDCðMÞ is such that the inclusion MDNisanF-pure monomorphism, then the extension MDN is strongly pure-essential.

This proposition implies that a module CR is cotorsion if and only if there exist no strongly pure-essential extensions CDNR:

Proposition 4. Let R be a ring and assume that the category of projective right R-modules is closed under strongly pure-essential extensions. Then CðRRÞ is S-cotorsion.

Proof. As the category of projective modules is closed under strongly pure-essential extensions, the cotorsion envelope of CðRðIÞÞ is projective for every set I by Proposition 3(1). By Kaplansky’s Theorem [1, Theorem 26.1] on projective ðIÞ modules, it is a direct sum of countably generated modules. Clearly CðRR Þ is the ðIÞ cotorsion envelope of CðRRÞ : Now the result follows from Theorem 2, using @¼@0 þjRj: &

Our next lemma is a useful technical trick inspired by [10, Lemma 2.1]. ARTICLE IN PRESS

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Lemma 5. Let F be a free right R-module and let p : F-M be a pure epimorphism. Assume that:

(1) The module M contains a strongly pure-essential projective submodule P: (2) Every pure submodule of F is strongly pure-essential in a direct summand of F:

Then p is a splitting epimorphism.

Proof. Let e : P-M be the inclusion. Then e lifts to e0 : P-F; because P is projective. As M is ﬂat, e is a pure monomorphism. Therefore, the lifting e0 : P-F is likewise a pure monomorphism:

0 The image of e ; which we shall also denote by PDFR is thus a pure submodule whose image under p is strongly pure-essential in M: The kernel Y ¼ Ker p is a pure submodule of F; so by hypothesis, it is strongly pure-essentialin a direct summand YDL of F: The restriction of p to P is a monomorphism, hence Y-P ¼ 0: The direct sum P"YDF is a pure submodule, because F=ðP"YÞDM=P is a ﬂat module. The restriction of the quotient map p : F-F=P to Y is a morphism pY : Y-F=P with the same cokernel F=ðP"YÞ and is therefore also a pure monomorphism. As Y is strongly pure-essential in L; the restriction of p to L is also a pure monomorphism pL : L-F=P: This implies that P-L ¼ 0: Moreover, the direct sum P"LDF is a pure submodule, because the cokernelis the quotient module F=ðP"LÞ; which is the cokernelof the pure monomorphism pL: But

ðP"LÞ=YDP"ðL=YÞ is a strongly pure extension of P: It is therefore a strongly pure-essential extension, which forces L=Y ¼ 0 so that the kernel Y ¼ L is a direct summand of F; that is, p splits. Then M is also isomorphic to a direct summand of F: &

Theorem 6. If R is a ring with the property that every pure submodule of a free right R- module is strongly pure-essential in a direct summand, then the right R-module CðRRÞ is S-cotorsion.

Proof. Let PR be a projective right R-module. By (the proof of) Proposition 4, it sufﬁces to prove that CðPRÞ is also projective. Let p : F-CðPÞ be an epimorphism ARTICLE IN PRESS

P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28 17 from a free module FR: As CðPÞ is ﬂat, p is a pure epimorphism. By Lemma 5, it is a splitting epimorphism. &

Closer inspection of the proofs of Proposition 4 and Theorem 6 shows that we only used the fact that for every index set I; the cotorsion envelope of CðRðIÞÞ is projective. The following question naturally arises.

Question. Is it possible to bound the cardinality of the index set I?

An easy example shows that it is not possible to bound it by @0: Let V be an inﬁnite dimensionalvector space over a ﬁeld K and call R ¼ EndK ðVÞ: Then R is a right self-injective Von Neumann regular ring. It is shown in [4, Example

ð@0Þ ð@0Þ 12.20] that every submodule of RR is essentialin a direct summand of R : However, RR is not right S-cotorsion because in this case it would be S-injective and so, semisimple. We will prove in Section 3 that it is possible to bound the cardinality of I by @¼maxf@1; jRjg:

2. The radical of the endomorphism ring

If MR is a right R-module, a local direct summand LR ¼ "iAI LiDM is a direct sum such that for every ﬁnite subset KDI; the sum "iAK Li is a direct summand of MR:

Proposition 7. Let CR be a ﬂat S-cotorsion module. Then every local direct summand of CR is a summand, and CR admits a direct sum decomposition M CR ¼ Ci; i where each of the summands Ci has a local endomorphism ring.

Proof. Let LR ¼ "iAI LiDCR be a local direct summand. Each of the Li is a direct ðIÞ summand of CR: Embed LR into the direct sum C ; so that Li is a direct summand of the ith summand of CðIÞ: Thus, L is isomorphic to a direct summand of CðIÞ: As ðIÞ C is cotorsion, so is L: For each ﬁnite subset KDI; let SðKÞ¼"iAK Ci: Then C=LD lim C=SðKÞ; so C=L is ﬂat, and we get that L is a direct summand of C : - R R K If every local direct summand of CR is a summand, then [16, Theorem 2.17] implies that CR has a decomposition as a direct sum C ¼ "i Ci of indecomposable modules. As CR is cotorsion, so is each of the Ci: By [12, Example 1], the endomorphism ring of each Ci is a right cotorsion ring. Therefore, it is local by [12, Corollary 7]. & ARTICLE IN PRESS

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By [16, Theorem 2.25], the decomposition given in this proposition complements direct summands. Also, the family fCigi of indecomposable factors is locally semi-T- nilpotent. This means that for every subfamily fCiðnÞgnAN; and sequence of non- - isomorphisms fn : CiðnÞ Ciðnþ1Þ; and any xACið1Þ; there is a kAN (depending on x) such that fk fkÀ1?f2 f1ðxÞ¼0: If CR is a ﬂat cotorsion module, then S ¼ EndRðCÞ is a right cotorsion ring. An endomorphism f : C-C belongs to the Jacobson radical JðSÞ if for every nonzero direct summand C0 of C; the restriction of f to C0 is not a pure monomorphism. Because, if the restriction of f to C0 is pure, then it is a split monomorphism. If e0 is an idempotent in S such that e0C ¼ C0; then there is a gAS such that gfe0 ¼ e0: Thus f cannot belong to the Jacobson radical. On the other hand, if f eJðSÞ; then the left ideal Sf contains a nonzero idempotent e; gf ¼ e (see just before Theorem 6 in [12]). The restriction of f to the direct summand eC is then a pure monomorphism.

Lemma 8. Let CR be a ﬂat cotorsion module, S ¼ EndRðCÞ and let ei; ejAS be two primitive idempotents. If f : eiC-ejC is not an isomorphism, then the endomorphism ej fei of C belongs to the Jacobson radical of S:

Proof. Assume on the contrary that ej feieJðSÞ: There would be a gAS such that gej fei ¼ e is a nonzero idempotent endomorphism of CR: Then eei ¼ gej feiei ¼ gej fei ¼ e: Consider the direct sum decomposition Ci ¼ eiC ¼ eieC"eið1 À eÞC: As 2 e ¼ e ¼ eeie is nonzero, so is eie: Thus eieCa0; and hence eið1 À eÞ¼0: In other words, ei ¼ eie and hence eigej fei ¼ eie ¼ ei: But that means that f : Ci-Cj has a retraction given by eigej; and is therefore one-to-one. Moreover, f is a split monomorphism and, because Cj is indecomposable, it is also onto. This contradicts the assumption that f is not an isomorphism. &

Theorem 9. Let CR be a ﬂat S-cotorsion module and S ¼ EndRðCÞ; the endomorphism ring of C: The Jacobson radical JðSÞ is the maximum left (respectively, right) ideal of S that acts locally T-nilpotently on C:

Proof. Let us ﬁrst show that JðSÞ acts locally T-nilpotently on C: Decompose C ¼ "i Ci into a direct sum of indecomposable modules having a local endomorphism ring. Let eiAS be the idempotent corresponding to the summand Ci with respect to this decomposition. Suppose there is an element c0AC and a sequence f1; f2; f3; yAJðSÞ such that for every nAN;

cn ¼ fn fnÀ1?f2 f1ðc0Þa0:

0 Using the cn and fn; we will ﬁnd fi AJðSÞ and ciðnÞ satisfying: A 0? 0 0 (1) For every n N; ciðnÞ ¼ fn f2f1ðcið0ÞÞ is nonzero and belongs to an indecomposable CiðnÞ; iðnÞAI: a (2) For all k4n; fk fkÀ1?fnþ1ðciðnÞÞ 0: ARTICLE IN PRESS

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0 Suppose fi and ciðnÞ have already been constructed. Then we may take 0 0 0 fn ¼ eiðnÞ fneiðnÀ1Þ: Because the fn all belong to JðSÞ; this will give a sequence of 0 - morphisms fn : CiðnÀ1Þ CiðnÞ that are not pure monomorphisms. As the Ci are ﬂat 0 cotorsion indecomposables, the fn are all nonisomorphisms. This contradicts the fact ðoÞ " ðoÞ that the summands of the decomposition C ¼ i Ci form a locally semi-T- nilpotent family. 0 To deﬁne the ciðnÞ; nX0andfn; nX1; we proceed by recursion on n:

* n ¼ 0: Since c0 has only finitely many nonzero coordinates with respect to the " decomposition C ¼ i Ci; one of the nonzero coordinates cið0Þ satisfies a fn fnÀ1?f2 f1ðcið0ÞÞ 0; for every nAN: * 0 n ¼ k þ 1: Suppose that ciðkÞ and fk have been defined. We have that a fp fpÀ1?fkþ1ðciðkÞÞ 0 for all p4k: As above, there is nonzero coordinate of ciðnÞ (in some indecomposable summand CiðnÞ)offkþ1ðciðkÞÞ¼fnðciðkÞÞ such that a 0 0 fq fqÀ1?fnþ1ðciðnÞÞ 0 for all q4n: Let fn ¼ eiðnÞ fn: Clearly, fnAJðSÞ:

We have shown that JðSÞ acts locally T-nilpotently on C: Let us now prove that JðSÞ is the maximum left (respectively, right) ideal of S that acts locally T-nilpotently on CR: Let SK be a left ideal of S with the property that every element f of K acts locally T-nilpotently on C: Then for every gAS; gf AK and therefore acts locally N i nilpotently on C: It makes sense to deﬁne the power series Si¼0ðgf Þ ; which is an endomorphism of C and is the inverse of 1 À gf : Thus f AJðSÞ and hence SKDJðSÞ: &

Corollary 10. If R is a ring for which the module RR is S-cotorsion, then R is right perfect.

Proof. By Proposition 7, RR has a direct sum decomposition by indecomposable modules with a local endomorphism ring. The ring R is therefore semiperfect. As R ¼ EndRðRÞ; Theorem 12 implies that J ¼ JðRÞ is right T-nilpotent. &

Notation. From now on, let CR ¼ CðRRÞ denote the cotorsion envelope of the ring R considered as a right module over itself, and S ¼ EndRC the endomorphism ring of CR: The envelope CR is a flat cotorsion R-module. For each element aAC the morphism la : RR-C determined by lað1Þ¼a extends to an endomorphism fa : CR-CR of the cotorsion envelope. This extension is unique modulo the annihilator annSð1Þ¼annSðRÞ which, by the definition of envelope, is contained in the Jacobson radical JðSÞ of the endomorphism ring. Thus for every aAC; a ¼ fað1Þ; so that as a left S-module, SC ¼ S1 is the cyclic module isomorphic to S=annSðRÞ: The remainder of this section is devoted to showing that if f AS is an endomorphism of CR for which f ð1ÞACJ; then f belongs to the Jacobson radical of S: If MR and NR are flat R-modules, then a morphism m : MR-NR is a pure monomorphism if m#R1R=I : M#RR=I-N#RR=I is a monomorphism for all ARTICLE IN PRESS

20 P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28 cyclic left R-modules R=I (see [19, Section 2.1]). Let FR be a ﬂat right R-module and T ¼ EndRðFÞ the endomorphism ring. For every left ideal RI; the canonical morphism of T-modules F#RI-F determined by the rule f #i/fi; is a monomorphism [18, Proposition I.10.6] whose image is the T-submodule FI of F; generated by linear combinations of elements in F with scalars in the ideal I: We will routinely identify F#RI with FI; and F#RR=I with the T-module F=FI: Thus a morphism m : FR-GR of ﬂat right R-modules is a pure monomorphism if for every left ideal RI; the induced morphism of abelian groups m=I : F=FI-G=GI; given by a þ FI/mðaÞþGI is a monomorphism. In other words, whenever aAF is such that mðaÞAGI; then aAFI: If I is a left ideal and rAR; let ðr : IÞ :¼fsAR : srAIg: This is a left ideal, which is part of an exact sequence of left R -modules

r 0-ðr : IÞ-R !r R=I;

- / where rr : R R=I is the map s sr þ I: If FR is a ﬂat right R-module, apply the exact functor F#R- to get an exact sequence of T-modules, T ¼ EndRF:

r ðFÞ 0-Fðr:IÞ-F r ! F=FI;

- / where rrðFÞ : F F=FI is the map f fr þ FI: This shows that Fðr : IÞ¼ðr : FIÞ : ¼faAF : arAFIg:

Lemma 11. Let FR be a ﬂat right R-module, and PR a projective module. A morphism m : PR-FR is a pure monomorphism if for every simple left R-module RX; the morphism of abelian groups m#R1X : P#RX-F#RX is a monomorphism. In other words, for every maximal left ideal RK; the induced morphism m=K : P=PK-F=FK is a monomorphism.

Proof. We need to prove that given aAP and a left ideal RI; then mðaÞAFI; implies aAPI: Suppose there are aAP and RI for which this fails. Thus mðaÞAFI; but aePI: By Zorn’s Lemma, there is a left ideal I 0+I maximalwith respect to the property that aePI 0; it also satisfies mðaÞAFI 0: We may thus replace I by I 0 and assume that I is a left ideal maximal with respect to the property that aePI: ðaÞ ðaÞ As PR is projective, there is a free module R such that P"Q ¼ R : Then PI"QI ¼ RðaÞI ¼ I ðaÞ: As aAP has only finitely many nonzero coordinates in RðaÞ; we may represent it as a ¼ rað1Þ þ rað2Þ þ ? þ raðnÞ according to the direct sum ðaÞ decomposition R : There is an 1pipn such that r ¼ raðiÞeI: In fact, our choice of I ensures that I is a maximalleftidealwith respect to reI: Let us note that the left ideal ðr : IÞ¼fsAR : srAIg is maximal. For, if the inclusion I 0+ðr : IÞ were proper, then I 0r þ I would strictly contain I: We could find a tAI 0 and sAI such that tr þ s ¼ r: Consequently, ð1 À tÞrAI and therefore 1 À tAðr : IÞDI 0 and hence 1AI 0: ARTICLE IN PRESS

P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28 21

As reI; we get, by choice of I; that aAPI þ Pr: There is a bAP such that a À brAPI: Thus mðaÞÀmðbÞr ¼ mða À brÞAFI: Since mðaÞAFI; mðbÞrAFI; which we can express as

mðbÞAðr : FIÞ¼Fðr : IÞ:

Because ðr : IÞ is a maximalleftideal,the hypothesis impliesthat bAPðr : IÞ¼ ðr : PIÞ: But then brAPI; which forces the contradiction aAPI: &

The lemma may be applied to give the following sufﬁcient condition for an endomorphism f : CðRRÞ-CðRRÞ to belong to the Jacobson radical of S ¼ EndRðCðRÞÞ:

Proposition 12. Let R be a ring, RRDC the cotorsion envelope of RR; and S ¼ EndRðCÞ the endomorphism ring of the cotorsion envelope. If f AS satisﬁes f ð1ÞACJ; then f AJðSÞ; the Jacobson radical of S: In particular, if C0 is a direct 0 0 0 summand of CR such that C J ¼ C ; then C ¼ 0:

Proof. If not, then the left ideal Sf contains a nonzero idempotent e ¼ gf : We still 0 0 have that eð1Þ¼gf ð1ÞACJ: Denote by e the restriction of e to RR: Since e ð1ÞDCJ; 0 it follows that e ðRÞDCJ: If RK is a maximalleftideal,then K+J; so that e0ðRÞDCJDCK: Thus e0=K : R=K-C=CK is the zero morphism. It follows that for every simple left R-module RX; the morphism of abelian groups 0 e #1X : R#RX-C#RX is zero. The restriction of 1 À e to RR is thus a pure 0 0 monomorphism, because ð1 À e Þ#1X ¼ 1#1X À e #1X ¼ 1#1X ; and the embed- ding of RR into CR is a pure monomorphism. As RRDCR is a strongly pure-essential extension, the extension 1 À e : CR-CR is also necessarily a pure monomorphism. But then e ¼ 0; a contradiction. If C0 is a direct summand of C such that C0J ¼ C0; consider an idempotent eAS that projects onto C0: Then eð1ÞAC0DCJ; so eAJðSÞ: It follows that e ¼ 0 and hence that C0 ¼ 0: &

3. Semilocal rings

Recall that a ring R is called semilocal if the quotient ring R=J is semisimple artinian.

Proposition 13. Let R be a ring and C ¼ CðRRÞ the cotorsion envelope of RR: If the direct sum CðNÞ is cotorsion, then R is semilocal. In particular, if C is S-cotorsion, then R is semilocal.

Proof. We will prove that the endomorphism ring S ¼ EndRðCÞ is semilocal. Then it follows that R is also semilocal. For, suppose that n is the length of the S-module S=JðSÞ: As SC ¼ S1 is a cyclic left S-module (see remarks below Corollary 10), the ARTICLE IN PRESS

22 P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28 length of C=radSðCÞ is at most n; where radSðCÞ denotes the intersection of all maximalleft S-submodules of SC: Now if R were not semilocal, we could ﬁnd n þ 1 maximalleftideals I1; I2; y; Inþ1 such that the intersection

n\þ1 I ¼ Ii j¼1 is irredundant. Then R=I ¼ "j R=Ij: Applying the functor C#R- shows that C=CI ¼ "j C=CIj: Because RR is pure in C; all n þ 1 of the S-modules C=CIj are nonzero. Since SC is cyclic, so are all of the quotients C=CIj: This would imply that C=radðCÞ contained a direct sum of n þ 1 nonzero modules, a contradiction. By [12, Theorem 6] S=JðSÞ is Von Neumann regular, so if S is not semilocal, there exists a countably infinite family of orthogonal idempotents in S=JðSÞ: As idempotents lift modulo JðSÞ [12, Corollary 4], Lemma 13 of [22] yields a countably infinite family fengnAN of orthogonalidempotents in S: Thus L ¼ "n enC is a local direct summand of CR: As each enC is a direct summand of C; we can embed L into a direct sum "nAN C so that each enC is a direct summand of the nth summand of CðNÞ: Then L is isomorphic to a direct summand of ðNÞ C ; which is cotorsion by hypothesis. Thus LR is cotorsion and we may write 0 " " CR ¼ LR ð nAN enCÞ: The element 1ARDC has only finitely many nonzero coordinates with respect to this decomposition. But then R itself is contained in only finitely many of these summands. This contradicts that CR is the cotorsion envelope of RR: &

If R is semilocal, then Lemma 11 may be stated more succinctly. This will enable us to characterize the ﬂat cover of a simple module over a semiperfect ring.

Lemma 11, revisited. Let R be a semilocal ring, FR a ﬂat right R-module, and PR a projective module. A morphism m : PR-FR is a pure monomorphism if the induced morphism m=J : P=PJ-F=FJ of right R=J-modules is a monomorphism.

Proof. The condition is equivalent to the morphism m#1R=J : P#RR=J-F#RR=J being a monomorphism. But every simple left R-module RX occurs as a direct summand of R=J: Thus m#1X must also be a monomorphism, and Lemma 11 applies. &

Proposition 14. Let R be a semiperfect ring and XR a simple right R-module. If c : FðXÞ-X is the ﬂat cover of X; then the kernel K ¼ Ker c ¼ FðXÞJ:

Proof. Clearly FðXÞJDKR; so the quotient K=ðFðXÞJÞ is a semisimple right R-module. If it is nonzero, there is a simple module YRDK=FðXÞJ: The projective cover PðYÞ-Y lifts to a map g : PðYÞ-K: The following situation ARTICLE IN PRESS

P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28 23 arises:

The morphism h : PðYÞ-C½PðYÞ is the cotorsion envelope. Since K is the kernelof a ﬂat cover, it is a cotorsion module [20, Lemma 2.1.1]. The morphism g : PðYÞ-K thus factors through the cotorsion envelope as in the diagram. By the revisited Lemma 11, the composition kg : PðYÞ-FðXÞ is a pure monomorphism. For, consider the induced map kg=J : PðYÞ=PðYÞJ-FðXÞ=FðXÞJ: Its image is just the simple module YDK=FðXÞJ; and it is thus a monomorphism. Because C½PðYÞ is a strongly pure-essential extension of PðYÞ; the morphism kf : C½PðYÞ-FðXÞ is a pure monomorphism and therefore a split monomorphism. Since it factors through K; it produces a nonzero direct summand of FðXÞ contained in K: But the kernelof a ﬂat cover contains none of the cover’s nonzero direct summands [20, Corollary 1.2.8]. &

Theorem 15. Let R be a semiperfect ring and XR a simple right R-module. Let c : FðXÞ-X be the ﬂat cover. Then FðXÞ is an indecomposable cotorsion module. If g : PðXÞ-XR is the projective cover of XR; then any morphism e : PðXÞ-FðXÞ that makes the diagram

commute is a cotorsion envelope of PðXÞ:

Proof. Since R=J is semisimple artinian, the simple right R-module XR is ﬁnite- dimensionalwhen considered as a vector space over its endomorphism ring. It is therefore S-pure-injective and therefore cotorsion. As the outer terms of the short exact sequence

c 0-K-FðXÞ !X-0 are both cotorsion, so is the middle term FðXÞ: ARTICLE IN PRESS

24 P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28

The ﬂat cover FðXÞR is an indecomposable right R-module, because if FðXÞ¼ C1"C2; then XDFðXÞ=FðXÞJ ¼ C1=C1J"C2=C2J; so it must be that one of C1=C1J and C2=C2J is zero. By Proposition 14, the corresponding summand lies in the kernelof the ﬂat cover, and must itselfbe zero. To prove the last statement, we must show that the morphism e : PðXÞ-FðXÞ is a pure monomorphism. Because FðXÞ is an indecomposable cotorsion module, this morphism will be a cotorsion envelope. But if aAPðXÞ is such that eðaÞAFðXÞJ; then pðaÞ¼ceðaÞAXJ ¼ 0; so that aAPðXÞJ by the revisited Lemma 11. &

Corollary 16. If R is a local ring, then the cotorsion envelope CðRRÞ is an indecomposable right R-module.

Next, we will show that for a semilocal ring, the sufﬁcient condition given in Proposition 12 for an element of the endomorphism ring of CðRÞ to belong to the Jacobson radicalis alsonecessary.

Lemma 17. Let R be a semilocal ring, C ¼ CðRRÞ and S ¼ EndRC: Suppose that a; bAC=CJ and that there is an endomorphism f : C=CJ-C=CJ such that f ðaÞ¼b: Then there is an endomorphism g : CR-CR such that the induced endomorphism g=J : C=CJ-C=CJ also maps a to b:

Proof. Let a and b in C=CJ be given as in the lemma with f ðaÞ¼b and denote R=J % % % by R: Because R is semisimple artinian, and RR% DC=CJ; we know that we can ﬁnd a cyclic R%-module cR% containing aR% and isomorphic to R%: The naturalepimorphism of R-modules R-cR%; 1/c is lifts to a morphism m : R-C which is necessarily a pure monomorphism. This morphism m then extends to a pure monomorphism on the cotorsion envelope m0 : CðRÞ-C: Since the image of m0 is a direct summand of C; there is a morphism g : C-C such that m0ð1Þ/b0; where b0 þ CJ ¼ f ðcÞ: But m0ð1Þ¼c; so that g : C-C induces an endomorphism of C=CJ that maps c to f ðcÞ and therefore a to f ðaÞ¼b: &

Theorem 18. Let R be a semilocal ring, C the cotorsion envelope of RR; and S the endomorphism ring of C: Then f AJðSÞ if and only if f ð1ÞACJ: This is equivalent to the image of f being contained in CJ:

Proof. Because of Proposition 12, all we need to prove is that if f AJðSÞ; then Im f DCJ: We may infer from Lemma 17 that if f : CR-CR is an endomorphism whose image Im f is not contained in CJ; then f cannot belong to the Jacobson radicalof the endomorphism ring S of CR: For suppose that cAC is such that f ðcÞeCJ: As C=CJ is semisimple, there is an endomorphism h : C=CJ-C=CJ such that fhð f ðcÞþCJÞ¼f ðcÞþCJ: By the lemma, we may assume that h is induced by an endomorphism of CR: Thus 1 À fhinduces an endomorphism of C=CJ that is not a unit, because the kernelcontains f ðcÞþCJ: Then 1 À fhitself cannot be a unit in S; and so f does not belong to the Jacobson radical JðSÞ: & ARTICLE IN PRESS

P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28 25

This allows us to weaken the hypothesis of Corollary 10 as follows.

Theorem 19. Let R be a ring and C ¼ CðRRÞ the cotorsion envelope of RR: If the direct sum CðNÞ is cotorsion, then R is right perfect.

Proof. First we prove that CðNÞ has no nonzero direct summands C0 such that C0 ¼ C0J: The naturalembedding RðNÞDCðNÞ is strongly pure-essential, and because ðNÞ ðNÞ ðNÞ C is cotorsion, it is the cotorsion envelope of R : Let eAEndRC be an idempotent that projects onto the summand C0 ¼ C0J: As in the proof of Proposition 12, the restriction e : RðNÞ-CðNÞ has the property that for every simple left R -module RX; e#1X ¼ 0: Thus ð1 À eÞ#1X is a monomorphism for every ðNÞ simple left R-module RX: By Lemma 11, the restriction of 1 À e to R is a pure monomorphism. As the cotorsion envelope is a pure-essential extension, the endomorphism 1 À e of CðNÞ must also be a pure monomorphism. But then e ¼ 0 and hence C0 ¼ 0: By Proposition 13, R is semilocal, so it remains to show that the Jacobson radical J is right T-nilpotent. Let r1; r2; y be a sequence of elements of J: For every iAN; there is a morphism fi : CR-CR such that fið1Þ¼ri: By Theorem 18 each fiAJðSÞ: Consider the direct limit L lim C of the ordered system C AN; where C C ¼ - f igi i ¼ for every iAN; and the structuralmorphisms are given by the fi:

f f f C !1 C !2 C !3 C?:

As in [8, Lemma 2.1], the kernel K of the induced epimorphism CðNÞ- lim C is the - union of a chain of summands indexed by N: As such, it is a local direct summand of ðNÞ C with countably many components K ¼ "nAN Kn: Because each of the Kn is a direct summand of CðNÞ; K is isomorphic to a direct summand of the cotorsion ðNÞ ðNÞ ðNÞ module ðC Þ : Thus KR is itself cotorsion and the inclusion KDC is a split- monomorphism. It follows that the limit L lim C is also isomorphic to a direct ¼ - ðNÞ summand of C : But the image of every fi : C-C is contained in CJ; by Theorem 18. Thus L ¼ LJ: By the foregoing observations, L ¼ 0 and hence for some nAN; rnrnÀ1?r2r1 ¼ fn fnÀ1?f2 f1ð1Þ¼0: &

Corollary 20. If R is a ring with the property that the free right R-module RðNÞ is cotorsion, then R is right perfect.

We are going to ﬁnish this section by showing that Theorem 6 remains true when we bound the cardinality of the set I by maxf@1; jRjg: We ﬁrst need a technical lemma. We will say that a right R-module M is @-generated, for some cardinal number @; if there exists a generating set of M of cardinality at most @: ARTICLE IN PRESS

26 P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28

Lemma 21. Let N be a submodule of a right R-module M Then N is contained in a pure submodule of M of cardinality at most maxfjNj; jRj; @0g:

Proof. Adapt the arguments given in [2, Lemma 2.1]. &

ð@Þ Theorem 22. Let R be a ring and @¼maxf@1; jRjg: If every pure submodule of RR is strongly pure-essential in a direct summand, then R is right perfect.

Proof. Let C be the right cotorsion envelope of R; and let I be an inﬁnite countable set. By Theorem 19, we only need to show that CðIÞ is cotorsion. Assume on the ðIÞ ðIÞ contrary that C is not cotorsion. Let us write C ¼ "iAI Ci with each Ci ¼ C: By [9, Lemma 2.1], there exist subsets A; KD2I satisfying:

* A is a partition of I with jAj¼@0 and jAj¼@0 for every AAA: * ADK; jKj¼@1 and jKj¼@0 for each KAK: * K-K0 is a ﬁnite set for all K; K0AK such that KaK0:

0 0 Let C ¼ Cð"iAI CiÞ be the cotorsion envelope of "iAI Ci: Call S ¼ EndRðC Þ and let J ¼ JðSÞ be the Jacobson radicalof S: We know by [12] that S=J is Von Neumann regular and right self-injective and idempotents lift modulo J: Let us fix an 0 orthogonalset of idempotents feigiAI DS such that Ci ¼ eiC for each iAI: For any 0 0 KAK; let eK AS be an idempotent such that eK C and ð1 À eK ÞC are the cotorsion envelopes of "iAK Ci and "iAI\K Ci; respectively. Let us note that eK ei ¼ ei if iAI 0 but eK ei ¼ 0 otherwise. Moreover, eK C a"iAK Ci for any KAK; because otherwise CðKÞ would be cotorsion for some infinite set K: And this would mean that C is S- cotorsion by Theorem 19. 0 a 0 0 " 0 Let us choose K; K AK with K K : We claim that eK eK0 C ¼ iAK-K0 eiC : It is " 0 0 clear that iAK-K0 eiC is a direct summand of eK eK0 C ; because it is a direct 0 0 " 0 " summand of C : Say eK eK0 C ¼ð iAK-K0 eiC Þ X: We must show that X ¼ 0: 0- 0 PConsider the homomorphism F : C C defined as FðcÞ¼ð1 À eK eK0 Þc þ " 0 0 iAK-K0 eic: F is injective, because iAI eiC is strongly pure-essential in C and FðeiÞ¼ei for every iAI: Therefore X ¼ 0; because FðXÞ¼0: Reasoning by induction, we deduce from last claim that, if K1; y; KnAK are 0 ? 0 0 pairwise different, then eK1 C þ þ eKn C is a direct summand of C : Namely, let T be the union of the finite sets Ki-Kj for i; j ¼ 1; y; n with iaj: Call 0 0 e e S A e for l 1; y; n: e is an idempotent and it is straighforward to Kl ¼ Kl À i Kl -T i ¼ Kl check that ! ! M Mn e C0 ? e C0 e C0 " e0 C0 K1 þ þ Kn ¼ i Kl iAT l¼1 is a direct summandP of C0: 0 Let M ¼ KAKeK CK : M is a pure submodule of C because it is a directed union of direct summands. Let us choose, for any KAK; an element ARTICLE IN PRESS

P.A. Guil Asensio, I. Herzog / Advances in Mathematics 191 (2005) 11–28 27 P A 0\ " 0 ðIÞ xK eK C ð iAK eiC Þ: The module R þ KAKxK R is an @1-generated submodule of M: Thus, its cardinality is at most @¼maxf@1; jRjg: Lemma 21 above implies ðIÞ thatP there exists an @-generated ﬂat submodule N of M that contains R þ ðIÞ KAKxK R: In particular, N is a ﬂat module and R is strongly pure-essential in N: Using now Lemma 5, we deduce that N is projective. Therefore, N is a direct sum of countably generated modules, say N ¼ "H Ph: Moreover, H must be countable, because RðIÞ is a countably generated submodule of N that it is strongly pure- essential. Thus, N is countably generated. P 0D D 0 Let K K be a countable subset such that N KAK0 eK C : As K is not 0 countable, we can choose a K0AK\ K : Then: X X X M A 0 D 0C 0D 0 xK0 eK0 C -N eK0 eK C eiC eiC : 0 0 A KAK KAK iAK0-K i K0

A 0\ " 0 & But this is a contradiction because we have chosen xK0 eK0 C ð iAK0 eiC Þ:

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