New York Journal of Mathematics

New York J. Math. 14 (2008) 403–410.

A note on p.q.-Baer modules

Ebrahim Hashemi

Abstract. A module M_{R} is called right principally quasi-Baer (or simply right p.q.-Baer) if the right annihilator of a principal submodule of R is generated by an idempotent. Let R be a ring. Let α be an endomorphism of R and M_{R} be a α-compatible module and T = R[[x; α]]. It is shown that M[[x]]_{T} is right p.q.-Baer if and only if M_{R} is right p.q.-Baer and the right annihilator of any countably-generated submodule of M is generated by an idempotent. As a corollary we obtain a generalization of a result of Liu, 2002.

Contents

1. Introduction 403 2. Principally quasi-Baer modules 405 References 408

1. Introduction

Throughout the paper R always denotes an associative ring with unity and M_{R} will stand for a right R-module. Recall from [15] that R is a Baer ring if the right annihilator of every nonempty subset of R is generated by an idempotent. In [15] Kaplansky introduced Baer rings to abstract various properties of von Neumann algebras and complete ∗-regular rings. The class of Baer rings includes the von Neumann algebras. In [10] Clark defines a ring to be quasi-Baer if the left annihilator of every ideal is generated, as a left ideal, by an idempotent. Then he used the quasi-Baer concept to characterize when a finite-dimensional algebra with unity over an algebraically closed field is isomorphic to a twisted matrix units semigroup algebra. Every prime ring is a quasi-Baer ring. Another generalization

403^{ISSN 1076-9803/08}

Received February 26, 2008.

Mathematics Subject Classification. 16D80, 16S36.

Key words and phrases. Quasi-Baer modules, α-compatible modules, quasi-Armendariz modules.

This research is supported by Shahrood University of Technology of Iran.

of Baer rings are the p.p.-rings. A ring R is called right (resp. left) p.p. if the right (resp. left) annihilator of an element of R is generated by an idempotent. Birkenmeier et al. in [5] introduced the concept of principally quasi-Baer rings. A ring R is called right principally quasi-Baer (or simply right p.q.-Baer) if the right annihilator of a principal right ideal of R is generated by an idempotent.

In 1974, Armendariz considered the behavior of a polynomial ring over a Baer ring by obtaining the following result: Let R be a reduced ring (i.e., R has no nonzero nilpotent elements). Then R[x] is a Baer ring if and only if R is a Baer ring ([3], Theorem B). Armendariz provided an example to show that the reduced condition is not superfluous. In [6] Birkenmeier et al. showed that the quasi-Baer condition is preserved by many polynomial extensions. Also, Birkenmeier et al. [5] showed that a ring R is right p.q.- Baer if and only if R[x] is right p.q.-Baer. Recall from [7], that an idempotent e ∈ R is left semicentral in R if ere = er for all r ∈ R. Equivalently, e^{2} = e ∈ R is left semicentral if eR is an ideal of R. Since the right annihilator of a right ideal is a ideal, we see that the right annihilator of a principal right ideal is generated by a left semicentral idempotent in a right p.q.-Baer ring. In [21], Z. Liu showed that if all left semicentral idempotents of a ring R are central, then R[[x]] is right p.q.-Baer if and only if R is right p.q.-Baer and any countable family of idempotents in R has a generalized join in the set of idempotents of R.

From now on, we always denote the skew power series ring by T := R[[x; α]], where α : R → R is an endomorphism. The skew power series ring T is then the ring consisting of all power series of the form ∑_{i=0}^{∞} a_{i}x^{i} (a_{i} ∈ R), which are multiplied using the distributive law and the Ore commutation rule xa = α(a)x, for all a ∈ R.

Given a right R-module M_{R}, we can make M[[x]] into a right T -module by allowing power series from T to act on power series in M[[x]] in the obvious way, and applying the above “twist” whenever necessary. The verification that this defines a valid T -module structure on M[[x]] is almost identical to the verification that T is a ring, and it is straightforward.

For a nonempty subset X of M, put ann_{R}(X) = {a ∈ R | Xa = 0}. In [20], Lee–Zhou introduced Baer, quasi-Baer and p.p.-modules as follows:

(1) M_{R} is called Baer if, for any subset X of M, ann_{R}(X) = eR where e^{2} = e ∈ R.

(2) M_{R} is called quasi-Baer if, for any submodule X ⊆ M, ann_{R}(X) = eR where e^{2} = e ∈ R.

(3) M_{R} is called p.p. if, for any element m ∈ M, ann_{R}(m) = eR where

e^{2} = e ∈ R. Clearly, a ring R is Baer (resp. p.p. or quasi-Baer) if and only if R_{R} is Baer (resp. p.p. or quasi-Baer) module. If R is a Baer (resp. p.p. or quasi-Baer) ring, then for any right ideal I of R, I_{R} is Baer (resp. p.p. or quasi-Baer) module.

A module M_{R} is called principally quasi-Baer (or simply p.q.-Baer) if, for any m ∈ M, ann_{R}(mR) = eR where e^{2} = e ∈ R. It is clear that R is a right p.q.-Baer ring if and only if R_{R} is a p.q.-Baer module. Every submodule of a p.q.-Baer module is p.q.-Baer and every Baer module is quasi-Baer.

We use I(R), S_{ℓ}(R) and C(R) to denote the set of idempotents, the set of left semicentral idempotents, and the center of R, respectively.

In this note we show that, if M_{R} is α-compatible module, the M[[x]]_{T} is p.q.-Baer if and only if M_{R} is p.q.-Baer and the right annihilator of any countably-generated submodule of M_{R} is generated by an idempotent. As a corollary, we show that if R is α-compatible and S_{ℓ}(R) ⊆ C(R), then R[[x; α]] is p.q.-Baer if and only if R is p.q.-Baer and any countable family of idempotents in R has a generalized join in I(R). This result is a generalization of [21].

2. Principally quasi-Baer modules

According to Kim et al. [16], a ring R is called power-serieswise Armendariz if whenever f(x)g(x) = 0 where f(x) = ∑_{i=0}^{∞} a_{i}x^{i}, g(x) = ∑_{j=0}^{∞} b_{j}x^{j} ∈ R[[x]], we have a_{i}b_{j} = 0 for all i, j. Let α ∈ End(R) and M be an R-module. According to Lee and Zhou [20], a module M_{R} is called α-Armendariz of power series type if the following cinditions are satisfied:

(1) For m ∈ M and a ∈ R, ma = 0 if and only if mα(a) = 0.

(2) For any m(x) = ∑_{i=0}^{∞} m_{i}x^{i} ∈ M[[x]] and f(x) = ∑_{i=0}^{∞} a_{i}x^{i} ∈ R[[x; α]], m(x)f(x) = 0 implies m_{i}α^{i}(a_{j}) = 0 for all i, j.

Definition 2.1. Let M_{R} be an R-module and α be an endomorphism of R. We say M_{R} is power-serieswise α-quasi-Armendariz if whenever m(x) = ∑_{i=0}^{∞} m_{i}x^{i} ∈ M[[x]] and f(x) = ∑_{j=0}^{∞} b_{j}x^{j} ∈ R[[x; α]] satisfy

m(x)R[[x; α]]f(x) = 0,

we have m_{i}x^{i}Rb_{j}x^{j} = 0 for all i, j.

Definition 2.2 (Annin, [2]). Given a module M_{R}, an endomorphism α : R → R, we say that M_{R} is α-compatible if for each m ∈ M, r ∈ R, we have mr = 0 ⇔ mα(r) = 0.

Theorem 2.3. Let M_{R} be an α-compatible module and T = R[[x; α]]. (1) If M[[x]]_{T} is p.q.-Baer, then M_{R} is p.q.-Baer.

(2) If M_{R} is p.q.-Baer, then M is power-serieswise α-quasi-Armendariz.

Proof. (1) Let m ∈ M. Since M[[x]]_{T} is p.q.-Baer, there exists idempotent e(x) = e_{0} +e_{1}x+⋯ ∈ T, such that ann_{T} (mT) = e(x)T. Since mRe(x) = 0, so mRe_{0} = 0. Thus e_{0}R ⊆ ann_{R}(mR). Let b ∈ ann_{R}(mR). Then b ∈ ann_{T} (mT), since M is α-compatible. Thus b = e(x)b and b = e_{0}b ∈ e_{0}R. Therefore ann_{R}(mR) = e_{0}R and M_{R} is p.q.-Baer.

(2) Assume that (∑_{i=0}^{∞} m_{i}x^{i})T (∑_{j=0}^{∞} b_{j}x^{j}) = 0 with m_{i} ∈ M, b_{j} ∈ R. Let c be an arbitrary element of R. Then we have the following equation:

∑_{k=0}^{∞}( ∑_{i+j=k} m_{i}x^{i}cb_{j}x^{j}) = ∑_{k=0}^{∞}( ∑_{i+j=k} m_{i}α^{i}(cb_{j}))x^{k} = 0,

and hence

(2.1)  ∑_{i+j=k} m_{i}α^{i}(cb_{j}) = 0 for all k ≥ 0.

We show that m_{i}x^{i}Rb_{j}x^{j} = 0 for all i, j. We proceed by induction on i + j. From Equation (2.1), we obtain m_{0}Rb_{0} = 0. This proves the case i + j = 0. Now suppose that m_{i}x^{i}Rb_{j}x^{j} = 0 for i + j ≤ n − 1. Hence b_{j} ∈ ann_{R}(m_{i}R) for j = 0, …, n − 1 and i = 0, …, n − 1 − j. Now ann_{R}(m_{i}R) = e_{i}R for some idempotent e_{i} ∈ R. Thus, e_{i}b_{j} = b_{j} for j = 0, …, n − 1 and i = 0, …, n−1−j. If we put f_{j} = e_{0} … e_{n−1−j} for j = 0, …, n−1, then f_{j}b_{j} = b_{j} and f_{j} ∈ ann_{R}(m_{0}R) ∩ ⋯ ∩ ann_{R}(m_{n−1−j}R). For k = n replacing c by cf_{0} in (2.1) and using α-compatibility of M, we obtain m_{0}cb_{n} = m_{0}cf_{0}b_{n} = 0. Hence m_{0}Rb_{n} = 0. Continuing this process (replacing c by cf_{j} in (2.1), for j = 1, …, n − 1 and using α-compatibility of M), we obtain m_{i}Rb_{j} = 0 and so m_{i}x^{i}Rb_{j}x^{j} = 0 for i + j = n. Therefore M_{R} is power-serieswise α-quasi-Armendariz. □

Lemma 2.4. Let M_{R} be an α-compatible module and M_{R} be a p.q.-Baer module. Let ann_{T} (m(x)T) = e(x)T for some idempotent e(x) = e_{0} + e_{1}x + ⋯ ∈ T. Then ann_{T} (m(x)T) = e_{0}T and e_{0} is an idempotent of R.

Proof. Let m(x) = m_{0} +m_{1}x+· · ·. By Theorem 2.3, M is power-serieswise α-quasi-Armendariz. Since m(x)T e(x) = 0, so m_{i}Re_{0} = 0, for each i ≥ 0. Hence e_{0} ∈ ann_{T} (m(x)T), and e_{0}T ⊆ e(x)T. Now let f(x) = a_{0}+a_{1}x+⋯ ∈ ann_{T} (m(x)T). Then m_{i}Rb_{j} = 0, for all i, j, since M is power-serieswise α-quasi-Armendariz. Thus b_{j} ∈ ann_{T} (m(x)T) = e(x)T, since M_{R} is αcompatible. Hence b_{j} = e(x)b_{j} and b_{j} = e_{0}b_{j} for each j. Therefore f(x) = e_{0}f(x) ∈ e_{0}T. □

Theorem 2.5. Let M be an α-compatible module and T = R[[x; α]]. Then M[[x]]_{T} is p.q.-Baer if and only if M_{R} is p.q.-Baer and the right annihilator of any countably-generated submodule of M is generated by an idempotent.

Proof. If M[[x]]_{T} is p.q.-Baer, then by Theorem 2.3, M_{R} is p.q.-Baer. Let X = {a_{0}, a_{1}, … } be a countable subset of M and ⟨X⟩ be the right submodule of M generated by X. Let m(x) = a_{0} + a_{1}x + · · ·. M[[x]]_{T} is p.q.-Baer, so by Lemma 2.4, there exists an idempotent e ∈ R such that ann_{T} (m(x)T) = eT. Clearly, eR ⊆ ann_{R}(⟨X⟩). Let b ∈ ann_{R}(⟨X⟩). Then a_{i}Rb = 0 for each i. Hence m(x)T b = 0, since M_{R} is α-compatible. Thus b = eb ∈ eR. Consequently, ann_{R}(⟨X⟩) = eR.

Now assume M_{R} is p.q.-Baer and the right annihilator of any countablygenerated submodule of M is generated by an idempotent. Let m(x) = ∑_{i=0}^{∞} m_{i}x^{i} ∈ M[[x]]. Let N be the submodule of M generated by the coefficients {m_{0}, m_{1}, … }. Then ann_{R}(N) = eR for some idempotent e ∈ R. Since m_{i}Re = 0 for each i and M_{R} is α-compatible, so by α-compatibility of M_{R}, m(x)T e = 0 and that eT ⊆ ann_{T} (m(x)T). Now let f(x) = ∑_{j=0}^{∞} a_{i}x^{i} ∈ ann_{T} (m(x)T). Then m_{i}Ra_{j} = 0, for each i, j, since M is power-serieswise α-quasi-Armendariz. Then a_{j} ∈ eR, for each j, and a_{j} = ea_{j}. Therefore f(x) = ef(x) ∈ eT. Consequently, M[[x]]_{T} is p.q.-Baer. □

Corollary 2.6. Let M be a right R-module. Then M[[x]]_{R[[x]]} is right p.q.- Baer if and only if M_{R} is right p.q.-Baer and for any countably-generated submodule N of M, ann_{R}(N) = eR for an idempotent e ∈ R.

Remark 2.7. In [20], it was proved that, if M_{R} is α-Armendariz of power series type, then M[[x]]_{T} is p.p. if and only if for any countable subset X of M, ann_{R}(X) = eR where e^{2} = e ∈ R. By Zalesskii and Neroslavskii [9], there is a simple Noetherian ring R which is not a domain and in which 0 and 1 are the only idempotents. Thus R_{R} is p.q.-Baer ring which is not right p.p. Therefore our Corollary 2.6, is not implied from [20].

There is a p.q.-Baer module M_{R} such that M[[x]]_{R[[x]]} is not p.q.-Baer.

Example 2.8. Let M_{1} be a right p.q.-Baer R_{1}-module. Let

M = {(m_{n}) ∈ ∏_{n=1}^{∞} M_{n} | m_{n} is eventually constant},

where M_{n} = M_{1} for n > 1 and let

R = {(a_{n}) ∈ ∏_{n=1}^{∞} R_{n} | a_{n} is eventually constant},

where R_{n} = R_{1} for n > 1. Clearly M is a right R module. Clearly M is right p.q.-Baer. Let m be a nonzero element of M_{1}. Let m_{1} = (m, 0, 0, …), m_{2} = (m, 0, m, 0, 0, …), m_{3} = (m, 0, m, 0, m, 0, 0, …), . . .. Let ⟨X⟩ be the submodule of M generated by X = {m_{1}, m_{2}, … }. One can show that ann_{R}(⟨X⟩) is not generated by any idempotent, hence by Theorem 2.5, M[[x]]_{R[[x]]} is not right p.q.-Baer.

Definition 2.9 (Z. Liu [21]). Let {e_{0}, e_{1}, … } be a countable family of idempotents of R. We say {e_{0}, e_{1}, … } has a generalized join in I(R) if there exists an idempotent e ∈ I(R) such that:

(1) e_{i}R(1 − e) = 0.

(2) If f ∈ I(R) is such that e_{i}R(1 − f) = 0, then eR(1 − f) = 0.

Lemma 2.10. Let R be a ring and S_{ℓ}(R) ⊆ C(R). Then the following are equivalent:

(1) R is right p.q.-Baer and any countable family of idempotents in R has a generalized join in I(R).

(2) R is right p.q.-Baer and the right annihilator of any countably-gener-ated right ideal of R is generated by an idempotent.

Proof. (1)⇒(2) Let X = {a_{i}}_{i∈I} be a countable subset of R and ⟨X⟩ be the right ideal of R generated by X. Then for each a_{i} ∈ X, ann_{R}(a_{i}R) = e_{i}R for some idempotent e_{i} ∈ R. Let h be a generalized join of the set {1−e_{i} | i ∈ I}. Then (1 − e_{i})R(1 − h) = 0. Hence r(1 − h) = e_{i}r(1 − h) for all r ∈ R. Since e_{i} ∈ S_{ℓ}(R) ⊆ C(R), a_{i}r(1 − h) = a_{i}e_{i}r(1 − h) = 0 for all i and each r ∈ R. Hence (1 − h) ∈ ann_{R}(⟨X⟩) and (1 − h)R ⊆ ann_{R}(⟨X⟩). Suppose that b ∈ ann_{R}(⟨X⟩). Hence b = e_{i}b for each i. Since e_{i} ∈ S_{ℓ}(R) ⊆ C(R), bR(1 − e_{i}) = 0 for each i. Since R is right p.q.-Baer, so ann_{R}(bR) = fR, where f is a left semicentral idempotent of R. Thus (1 − e_{i}) ∈ ann_{R}(bR) = fR, so (1 − e_{i}) = f(1 − e_{i}) for each i. Hence from (1 − e_{i}) ∈ C(R), we have (1 − e_{i})R(1 − f) = 0. Since h is a generalized join of the set {1 − e_{i} | i ∈ I}, hR(1 − f) = 0. Hence b = b − bf = (1 − f)b = (1 − h)(1 − f)b ∈ (1 − h)R. Therefore ann_{R}(⟨X⟩) = (1 − h)R.

(2)⇒(1) Suppose that {e_{i} | i = 0, 1, … } is a countable family of idempotents of R. Let J be the right ideal of R generated by {e_{i} | i = 0, 1, … }. Then ann_{R}(J) = eR for some left semicentral idempotent e. Let h = 1 − e. Then e_{i}r(1 − h) = 0 for each r ∈ R. Suppose that f is an idempotent of R such e_{i}R(1 − f) = 0 for each i. Then r(1 − f) ∈ ann_{R}(J) for each r ∈ R. Thus r(1 − f) = er(1 − f) and hr(1 − f) = (1 − e)r(1 − f) = 0. Hence h is a generalized join of the set {e_{i} | i = 0, 1, … }. □

Theorem 2.11. Let R be a ring with S_{ℓ}(R) ⊆ C(R) and α be an endomorphism of R. Let R_{R} be an α-compatible module. Then the following are equivalent:

(1) R[[x; α]] is right p.q.-Baer.

(2) R is right p.q.-Baer and any countable family of idempotents of R has a generalized join in I(R).

Proof. This follows from Theorem 2.5 and Lemma 2.10. □

Corollary 2.12 (Z. Liu [21, Theorem 3]). Let R be a ring with S_{ℓ}(R) ⊆ C(R). Then the following conditions are equivalent:

(1) S = R[[x]] is right p.q.-Baer.

(2) R is right p.q.-Baer and any countable family of idempotents in R has a generalized join in I(R).

Acknowledgements. The author thanks the referee for his/her helpful suggestions.

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410 Ebrahim Hashemi

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Department of Mathematics, Shahrood University of Thechnology, Shahrood, Iran, P.O.Box: 316-3619995161 eb_{−}hashemi@yahoo.com or eb_{−}hashemi@shahroodut.ac.ir

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