We consider  Baer rings and Baer semisimple R-modules which are generalizations of semisimple modules. Several characterization theorems of Baer semisimple modules are obtained. In particular, we prove that a ring R is a Baer ring if and only if R itself, regarded as a regular R-module, is Baer semisimple

Shum, K.P.

Xiaojiang Guo

Algebra and Discrete Mathematics

Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)

Journal Algebra Discrete Math.Algebra and Discrete Mathematics RESEARCH ARTICLENumber 2. (2008). pp. 42 – 49c© Journal “Algebra and Discrete Mathematics”Baer semisimple modules and Baer ringsXiaojiang Guo and K. P. ShumCommunicated by A. V. MikhalevAbstract. We consider Baer rings and Baer semisimpleR-modules which are generalizations of semisimple modules. Sev-eral characterization theorems of Baer semisimple modules are ob-tained. In particular, we prove that a ring R is a Baer ring if andonly if R itself, regarded as a regular R-module, is Baer semisimple.Throughout this paper, R is an associative ring with identity 1 and allR-modules are unital. Denote the set of idempotents of R by E(R).Let M be a left R-module and a right S-module. Also, let X be asubset of M , R or S, respectively. Then we denote the left [resp. right]annihilator of X by annℓ(X) [resp. annr(X)]. We also write annℓ({m})[resp. annr({m})] by annℓ(m) [resp. annr(m)].We call a ring R a Baer ring if the left annihilator of any subset ofR is generated by an idempotent. The properties of Baer rings and itsgeneralizations have been studied by many authors, for example, see ([3],[4], [11] and [13]). We observe that Baer rings can be generalized intoother forms, for example, rpp rings, etc. The rpp-rings and their gen-eralizations have been extensively studied in the literature after Hattori(see, [2]-[15]). Recently, the authors have introduced the concept of rightperpetual ideals and consequently, reduced pp rings are characterized byusing right perpetual submodules (see [8]).The research of the first author is jointly supported by the Natural Science Foun-dation of Jiangxi Province; the Science Foundation of the Education Department ofJiangxi Province and the Foundation of Jiangxi Normal University, China. The re-search of the second author is partially supported by a UGC (HK) grant (#2160297/05-06)2000 Mathematics Subject Classification: 16W60.Key words and phrases: Baer module; Baer semisimple module; perpetualsubmodule; Baer ring.Journal Algebra Discrete Math.X. J. Guo, K. P. Shum 43Recall that a right ideal I of R is a right perpetual ideal of R if forevery x ∈ I and y ∈ R, annℓ(x) ⊆ annℓ(y) implies that y ∈ I (see [8]).Clearly, for any X ⊆ R, there exists the smallest right perpetual ideal ofR containing X. We usually call this smallest right perpetual ideal of Rcontaining X the right perpetual ideal generated by X and is denoted byR∗(X). If X = {a}, then we write R∗(X) = R∗(a).The following results are known.Lemma 1. [8] The following statements hold in a ring R:(1) If e ∈ E(R), then R∗(e) = eR.(2) For all X ⊆ R, annr(X) is a right perpetual ideal of R.(3) A ring R is lpp if and only if for any a ∈ R, R∗(a) is generatedby an idempotent.Let M be a right R-module. Denote the ring of R-endomorphisms ofM by End(MR). If End(MR) is regarded as a set of left operations, innotation, Endℓ(MR), then M can be regarded as a left Endℓ(MR)- rightR-module. Inspiring by the definition of right perpetual ideals, we nowdefine the perpetual submodules.Definition 1. Let M be a right R-module. Then, we call a (right R-)submodule N of M a perpetual submodule of M if for all x ∈ N andy ∈M , annℓ(x) ⊆ annℓ(y) implies y ∈ N .It is clear that M and (0) are both trivial perpetual submodulesof M . Also, the intersection of perpetual submodules of M is still aperpetual submodule ofM and hence, there exists the smallest perpetualsubmodule ofM containingX forX ⊆M . Denote the smallest perpetualsubmodule of M containing X by SM∗(X). On the other hand, if R isregarded as a regular right R-module RR, then the left End(RR)-rightR-module R becomes a regular bimodule End(RR)RR. Thus in this case,every perpetual submodule of R is a right perpetual ideal of R (same asin rings).The following lemma can be easily proved.Lemma 2. Let M be a right R-module and X ⊆ Endℓ(MR). Then(1) annr(X) is a perpetual submodule of M .(2) If ϕ2 = ϕ ∈ Endℓ(MR), then ϕM is a perpetual submodule of M .The proof of the following lemma is straightforward.Lemma 3. Let M be a right R-module and x ∈ M . Then SM∗(x) =annr(annℓ(x)).The following result lemma is crucial in this paper but the proof canbe found in [1].Journal Algebra Discrete Math.44 Baer semisimple modules and Baer ringsLemma 4. [1] A R-submodule K of the right R-module M is a di-rect summand of M if and only if K = eM for some idempotent e ∈Endℓ(MR).Now, we formulate the following definition.Definition 2. Let M be a right R-module. Then(1) M is called a Baer simple R-module if M 6= 0, and M containsno perpetual submodules of M other than M itself and (0).(2) M is called a Baer semismiple R-module if every perpetualsubmodule of M is a direct summand of M .Evidently, a Baer simple R-module is itself Baer semisimple and theusual semisimple R-module is also Baer semisimple. Indeed, if M isa semisimple R-module, then every R-submodule N of M is a directsummand of M . By Lemma 4, N = eM, for some e2 = e ∈ Endℓ(MR).This implies that every R-submodule of M is a perpetual submodule ofM . Thus M is Baer semisimple.Proposition 1. LetM be a Baer semisimple R-module and N a perpetualsubmodule of M . Then the following statements hold:(i) N = eM for some idempotent e ∈ Endℓ(MR).(ii) N is Baer semisimple.Proof. (i) By our hypothesis, M is Baer semisimple and hence, N is adirect summand ofM . Now, by Lemma 4, N = eM, for some idempotente ∈ Endℓ(MR).(ii) It suffices to verify that any perpetual submodule of N is stilla perpetual submodule of M . In other words, we only need to provethat the smallest perpetual submodule SM∗M(x) ofM containing x is thesmallest perpetual submodule SM∗N(x) of N containing x, for all x ∈ N .By Lemma 5, we have N = eM, for some idempotent e ∈ Endℓ(MR).Denote the left annihilator of K related to the R-module M and relatedto the R-module N by annMℓ(K) and annNℓ(K), respectively. Now, byLemma 3, SM∗M(x) ⊆ N . Let f be an idempotent endomorphism inEndℓ(MR) such that SM∗M(x) = fM . Then, fM ⊆ eM . Thus, for anyx ∈M ,we havefx = ey = eey = efx (y ∈M),and thereby, f = ef . Hence, fe is an idempotent endomorphism inEndℓ(MR) and alsofM = ffM ⊆ fefM ⊆ feM ⊆ fM,that is, fM = feM . On the other hand, since the restriction fe|eM offe(= efe) to eM is an idempotent R-endomorphism which maps eMJournal Algebra Discrete Math.X. J. Guo, K. P. Shum 45into itself, we have fM = feM = fe(eM) and hence, by Lemma 2, fMis a perpetual submodule of N . Now, by the minimality of SM∗N(x), wehave SM∗N(x) ⊆ SM∗M(x).Now let ϕ ∈ annNℓ(x). Then, it can be easily observed that M =eM ⊕ (1− e)M . Hence, we can define a mappingϕ : M →M ; y 7→ ϕ(ey),which is a R-homomorphism of M into itself with ϕ|N = ϕ. Clearly,ϕ ∈ annMℓ(x). If y ∈ SM∗M(x), then by Lemma 3, ψ(y) = 0 for allψ ∈ annMℓ(x), and furthermore, ϕy = 0, for all ϕ ∈ annNℓ(x). Notethat SM∗M(x) ⊆ N and ϕ|N = ϕ. Thus ϕy = 0 implies that ϕy = 0.This shows that annNℓ(x) ⊆ annNℓ(y). Consequently, we can deducey ∈ SM∗N(x), by Lemma 3. This leads to SM∗M(x) ⊆ SM∗N(x). Thus,SM∗N(x) = SM∗M(x), as required.The following is a characterization theorem for the Baer simple R-modules.Theorem 1. Let M be a Baer semisimple R-module and N a perpetualR-submodule of M . Then N is Baer simple R-module if and only ifN = eM, for some primitive idempotent e ∈ Endℓ(MR).Proof. Suppose that N is a Baer simple R-submodule of M . Then, byLemma 4, N = eM for some idempotent e ∈ Endℓ(MR). Now let f2 =f ∈ Endℓ(MR) such that f ≤ e, i.e., f = ef = fe. Then fM ⊆ eM .Since N is Baer simple, fM = (0) or fM = eM .• If fM = (0), then f = 0.• If fM = eM, then for all x ∈M ,e(x) = f(y) = ff(y) = fe(x) = f(x) (y ∈M),and whence e = f .This shows that e is a primitive idempotent of Endℓ(MR). Conversely, weassume that N = eM , where e is a primitive idempotent of Endℓ(MR).Then N is a perpetual submodule ofM . LetK be a perpetual submoduleof N . Now, by using the proof of Proposition 1, we can show that K isstill a perpetual submodule of M , and by Lemma 4, K = fM for someidempotent f ∈ Endℓ(MR). Now, fM ⊆ eM implies that for all x ∈M ,we havef(x) = e(y) = ee(y) = ef(x) (y ∈M),and thereby, f = ef . By routine verification, fe is an idempotent ofEndℓ(MR), and fe ≤ e. But since e is primitive, fe = e or fe = 0.Journal Algebra Discrete Math.46 Baer semisimple modules and Baer rings• If fe = e, thenfM ⊆ eM = feM ⊆ fM,that is, K = N .• If fe = 0, thenK = fM = f(fM) ⊆ f(eM) = (0).This shows that the submodule N is indeed Baer simple.We next establish a “Schur Lemma " for Baer simple modules.Theorem 2. (Schur Lemma) If M is a Baer simple R-module, thenEndℓ(MR) is a domain (such a ring satisfies the cancellative law).Proof. It suffices to show that any ϕ ∈ Endℓ(MR)\{0} is injective. Forthis purpose, we only need to prove that annr(ϕ) = (0). By Lemma2, annr(ϕ) is a perpetual submodule of M and, since M is Baer sim-ple, annr(ϕ) = M or annr(ϕ) = (0). But since ϕ 6= 0, it is clear thatannr(ϕ) 6= M . Thus annr(ϕ) = (0) and hence ϕ is injective.Lemma 5. Any nonzero Baer semisimple R-module M contains a Baersimple R-module.Proof. Without loss of generality, we may assume that M is not a Baersimple R-module. Then we can pick a nonzero element x of M suchthat SM∗(x) ⊂ M . By Lemma 4, SM∗(x) = eM for some idempotentendomorphism e ∈ Endℓ(MR). By Lemma 1,K = (1−e)M is a perpetualsubmodule of M not containing x. Now, by Zorn’s lemma, there existsa perpetual submodule N of M which is maximal with respect to theproperty that x /∈ N . Choose a perpetual submodule N ′ of M such thatM = N ⊕N ′ (by Lemma 4). Then, we can finish our proof by showingthat N ′ is Baer simple. Indeed, if N ′′ is a nonzero perpetual submoduleof N ′, then by Proposition 1, N ′ is Baer semisimple and N ′ = N ′′⊕N ′′′,where N ′′′ is a submodule of N ′. Thus N⊕N ′′ is a direct summand ofM .Again by Lemma 4, N ⊕N ′′ = fM for some idempotent f ∈ Endℓ(MR)and by Lemma 2, N ⊕N ′′ is a perpetual submodule of M containing x(by the maximality of N) and N⊕N ′′ = M , which implies that N ′′ = N ′,as desired.Proposition 2. A Baer semisimple module is the direct sum of a familyof Baer simple submodules.Journal Algebra Discrete Math.X. J. Guo, K. P. Shum 47Proof. Assume thatM is a Baer semisimple module. Denote by A the setof Baer simple submodules of M . Then, we consider the subset B ⊂ Awith the following conditions:•∑J∈BJ is a direct sum.•∑J∈BJ is a perpetual submodule of M .By Lemma 5, A 6= ∅. Now, by Zorn’s lemma, we can consider the familyof all the above B’s with respect to the set inclusion. Thus we can pick aB to be the maximal element. For such a B, we can construct a perpetualsubmoduleM1 := ⊕J∈BJ . Now, by our hypothesis,M = M1⊕M2, whereM2 is a submodule of M . By Lemma 4, M2 is a perpetual submodule ofM and by Proposition 1, M2 is a Baer semisimple, and hence by Lemma5 again, M2 = K ⊕ Q, where K is a Baer simple submodule of M2 andQ a submodule of M2. Thus M1 ⊕K is a direct summand of M and ofcourse, M1 ⊕ K is a perpetual submodule of M, by Lemma 4. On theother hand, by using the proof of Proposition 1(ii), we can show that Kis Baer simple in M . This contradicts the maximality of B. ThereforeM = M1 = ⊕J∈BJ .Theorem 3. Let M be a R-module and P the set of submodules of theform eM , with e ∈ E(Endℓ(M)). Order the set P by set inclusion. Thenthe following statements are equivalent:(i) M is Baer semisimple.(ii) The following two conditions hold:(a) For any x ∈M , SM∗(x) is a direct summand of M .(b) P forms a complete lattice.Proof. (i) ⇒ (ii) Since condition (a) holds trivially, we need only toshow that condition (b) holds. Let T ⊆ P . Since every element of P isa perpetual submodule of M,⋂J∈TJ is a perpetual submodule of M .By our hypothesis,⋂J∈TJ is a direct summand of M . By Lemma 4,we have⋂J∈TJ ∈ P . Consider the smallest perpetual submodule K ofM containing J with J ∈ T . It is clear that K is a direct summand ofM , and whence K ∈ P . Thus K can be viewed as supJ∈T (J). Thus, Pindeed forms a complete lattice.(ii) ⇒ (i) Assume that (ii) holds. Let I be a perpetual submodule ofM . Consider I =⋃x∈ISM∗(x). Then by condition (a), I =⋃x∈IexM,where ex is the idempotent of Endℓ(M) such that SM∗(x) = exM , forany x ∈ I. By condition (b), I = eM for some e ∈ E(Endℓ(M)), thatis, I is a direct summand of M . Consequently, M is a Baer semisimplemodule.Journal Algebra Discrete Math.48 Baer semisimple modules and Baer ringsRecall that [14, Lemma 2.3] a ring R is Baer if and only if R is lppand under set inclusion, the set of all idempotent-generated principalright ideals forms a complete lattice. By using Lemma 1 and Theorem 3,we deduce the following characterization theorem of Baer rings.Theorem 4. A ring R is a Baer ring if and only if R itself, regarded asa regular R-module, is a Baer semisimple module.References[1] F.W. Anderson and K.R. Fuller, Rings and categories of modules (2nd edition),Graduate Texts in Mathematics, Springer-Verlag, Berlin, Heidelberg, New York,2004.[2] E.P. Armendariz, A note on extensions of Baer and PP rings, J. Austral. Math.Soc., Vol.18, 1974, pp.470-473.[3] G.F. Birkenmeier, Decompositions of Baer-like rings, Acta Math. Hungary,Vol.59, N.3-4, 1992, pp.319-326.[4] G.F. Birkenmeier, Baer rings and quasicontinuous rings having a MDSN, PacificJ. Math., Vol.97, N.2, 1981, pp.283-292.[5] S. Endo, Note on P.P. rings (a supplement to Hattori’s paper), Nagoga Math. J.,Vol.17, 1960, pp.167-170.[6] J.A. Fraser and W.K. Nicholson, Reduced pp rings, Math. Japonica, Vol.34, 1989,pp.715-725.[7] X.J. Guo and K.P. Shum, l.p.p.-rings which are left *-semisimple, InternationalMathematical Forum, Vol.2, N.56, 2007, pp.2787-2798.[8] X.J. Guo and K.P. Shum, Reduced p.p.-rings without identity, International Jour-nal of Mathematics and Mathematical Science, Vol.2006, 2006, Art. ID 93890, 7pp.[9] X.J. Guo and K.P. Shum, On p.p. rings which are reduced, International Journalof Mathematics and Mathematical Science, Vol.2006, 2006, Art. ID 34694, 5 pp.[10] Y. Hirano, M. Hongan and M. Oˆhori, On right PP rings, Math. J. OkayamaUniv., Vol.24, 1982, pp.99-109.[11] C.Y. Hong, N.K. Kim and T.K. Kwak, Ore extensions of Baer and P.P.-rings, J.Pure and Appl. Algebra, Vol.151, 2000, pp.215-226.[12] S. J∅ndrup, P.P. rings and finitely generated flat ideals, Proc. Amer. Math. Soc.,Vol.28, 1971, pp.431-435.[13] Z.K. Liu and J. Ahsan, P.P.-rings of generalized power series, Acta Math. Sinica(new ser.), Vol.16, 2000, pp.573-578.[14] S. Maeda, On a ring whose principal right ideals generated by idempotents, J. Sci.Hiroshima Univ. (Ser. A), Vol.24, 1960, pp.509-525.[15] L.W. Small, Semiherditary rings, Bull. Amer. Math. Soc., Vol.73, 1967, pp.656-658.Journal Algebra Discrete Math.X. J. Guo, K. P. Shum 49Contact informationXiaojiang Guo Department of Mathematics, Jiangxi Nor-mal University, Nanchang, Jiangxi 330022,P.R. ChinaE-Mail: xjguo@jxnu.edu.cnK. P. Shum Department of Mathematics, The Univer-sity of Hong Kong, Pokfulam Road, HongKong, P.R. China (SAR)E-Mail: kpshum@maths.hku.hkReceived by the editors: 01.05.2008and in final form 01.05.2008.

Baer semisimple modules and Baer rings

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Baer semisimple modules and Baer rings

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Наукова електронна бібліотека періодичних видань НАН України (Vernadsky National Library of Ukraine)