The Unique Maximal GF-Regular Submodule of a Module

Abstract

An 𝑅-module 𝐴 is called 𝐺𝐹-regular if, for each 𝑎∈𝐴 and 𝑟∈𝑅, there exist 𝑡∈𝑅 and a positive integer 𝑛 such that 𝑟^𝑛𝑡𝑟^𝑛𝑎=𝑟^𝑛𝑎. We proved that each unitary 𝑅-module 𝐴 contains a unique maximal 𝐺𝐹-regular submodule, which we denoted by 𝑀𝐺𝐹(𝐴). Furthermore, the radical properties of 𝐴 are investigated; we proved that if 𝐴 is an 𝑅-module and 𝐾 is a submodule of 𝐴, then 𝑀𝐺𝐹(𝐾)=𝐾∩𝑀𝐺𝐹(𝐴). Moreover, if 𝐴 is projective, then 𝑀𝐺𝐹(𝐴) is a 𝐺-pure submodule of 𝐴 and 𝑀𝐺𝐹(𝐴)=𝑀(𝑅)⋅𝐴.

1. Introduction

Throughout this paper, 𝑅 is a commutative ring with identity and all modules are left unitary, unless otherwise stated. Recall that an element 𝑟∈𝑅 is said to be regular if there exists 𝑡∈𝑅 such that 𝑟𝑡𝑟=𝑟; a ring 𝑅 is called regular if and only if each element of 𝑅 is regular. An ideal 𝐼 of a ring 𝑅 is regular if each of its elements is regular in 𝑅; indeed, a regular ideal 𝐼 of 𝑅 is itself a regular ring [1]. Brown and McCoy proved in [1] that each ring 𝑅 contains a unique maximal regular ideal 𝑀(𝑅) which satisfies the well-known radical properties. The ideal 𝑀(𝑅) is called the regular radical of 𝑅.

The concept of regularity was extended to modules in several ways and in [2] the notion of 𝐹-regular modules (in the sense of Fieldhouse [3]) was generalized to 𝐺𝐹-regular modules. Let 𝐴 be an 𝑅-module; an element 𝑎∈𝐴 is said to be 𝐺𝐹-regular if for each 𝑟∈𝑅 there exist 𝑡∈𝑅 and a positive integer 𝑛 such that 𝑟^𝑛𝑡𝑟^𝑛𝑎=𝑟^𝑛𝑎. An 𝑅-module 𝐴 is called 𝐺𝐹-regular if and only if all its elements are 𝐺𝐹-regular; in particular, a ring 𝑅 is 𝐺𝐹-regular if and only if 𝑅 is 𝐺𝐹-regular as an 𝑅-module. On the other hand a ring 𝑅 is 𝜋-regular if and only if 𝑅 is a 𝐺𝐹-regular 𝑅-module; recall that a ring 𝑅 is 𝜋-regular if, for each 𝑟∈𝑅, there exist 𝑡∈𝑅 and a positive integer 𝑛 such that 𝑟^𝑛𝑡𝑟^𝑛=𝑟^𝑛. A submodule 𝑁 of an 𝑅-module 𝐴 is called 𝐺𝐹-regular if each element of 𝑁 is 𝐺𝐹-regular and every submodule of a 𝐺𝐹-regular module is a 𝐺𝐹-regular module. Also, in [2] the concept of 𝐺-pure submodules was introduced; a submodule 𝑃 of an 𝑅-module 𝐴 is called 𝐺-pure if, for each 𝑟∈𝑅, there exists a positive integer 𝑛 such that 𝑃∩𝑅𝑟^𝑛𝐴=𝑅𝑟^𝑛𝑃.

In this paper we show that each module contains a unique maximal 𝐺𝐹-regular submodule, which we denote by 𝑀𝐺𝐹(𝐴), and we show that 𝑀𝐺𝐹(𝐴) satisfies some but not all of the usual radical properties.

2. Main Results

Theorem 1. Let 𝑅 be any ring. Every 𝑅-module contains a unique maximal 𝐺𝐹-regular submodule.

Proof. Let 𝑅 be any ring, let 𝐴 be an 𝑅-module, and let 𝐺=𝑁∣𝑁isa𝐺𝐹-regularsubmoduleof𝐴, (1) where 𝐺≠𝜙 because (0) is a 𝐺𝐹-regular submodule of 𝐴. Let {𝑁_𝑖} be an ascending chain in 𝐺 and 𝐵=⋃_𝑖∈Λ𝑁_𝑖. Let 𝑏∈𝐵; there exists 𝑗∈Λ such that 𝑏∈𝑁_𝑗, but 𝑁_𝑗 is a 𝐺𝐹-regular submodule; then; for each 𝑟∈𝑅, there exist 𝑡∈𝑅 and a positive integer 𝑛 such that 𝑟^𝑛𝑡𝑟^𝑛𝑏=𝑟^𝑛𝑏; therefore 𝑏 is a 𝐺𝐹-regular element in 𝐵 which implies that 𝐵 is a 𝐺𝐹-regular 𝑅-module. Now, by Zorn’s lemma, 𝐺 contains a maximal element which we call 𝑀𝐺𝐹. To prove the uniqueness of 𝑀𝐺𝐹, assume that 𝑀𝐺𝐹1 and 𝑀𝐺𝐹2 be two maximal 𝐺𝐹-regular submodules in 𝐴; then for any maximal ideal 𝑃 of 𝑅 each of 𝑀𝐺𝐹1_𝑝 and 𝑀𝐺𝐹2_𝑝 is semisimple over 𝑅_𝑝 [2, Proposition 21]. Now, let 𝑀𝐺𝐹1_𝑝∩𝑀𝐺𝐹2_𝑝=𝐾_𝑝; then 𝐾_𝑝⊆𝑀𝐺𝐹1_𝑝 and 𝐾_𝑝⊆𝑀𝐺𝐹2_𝑝; thus 𝑀𝐺𝐹1_𝑝=𝐾_𝑝+𝐴1_𝑝 and 𝑀𝐺𝐹2_𝑝=𝐾_𝑝+𝐴2_𝑝, where 𝐴1_𝑝 and 𝐴2_𝑝 are two submodules of 𝐴_𝑝 [4]. Hence, 𝑀𝐺𝐹1_𝑝+𝑀𝐺𝐹2_𝑝=𝐴1_𝑝+𝐾_𝑝+𝐴2_𝑝, but each of 𝐴1_𝑝, 𝐴2_𝑝, and 𝐾_𝑝 is a semisimple submodule; thus 𝑀𝐺𝐹1_𝑝+𝑀𝐺𝐹2_𝑝 is a semisimple submodule which implies that 𝑀𝐺𝐹1_𝑝+𝑀𝐺𝐹2_𝑝 is 𝐺𝐹-regular [2]. So 𝑀𝐺𝐹1+𝑀𝐺𝐹2 is a 𝐺𝐹-regular submodule [2, Theorem 20]. Now, each of 𝑀𝐺𝐹1 and 𝑀𝐺𝐹2 is a maximal 𝐺𝐹-regular submodule and hence 𝑀𝐺𝐹1+𝑀𝐺𝐹2=𝑀𝐺𝐹2=𝑀𝐺𝐹1.

Remark 2. We denote the unique maximal 𝐺𝐹-regular submodule of an 𝑅-module 𝐴 by 𝑀𝐺𝐹(𝐴). It is obvious that 𝑀𝐺𝐹(𝐴) contains every 𝐺𝐹-regular submodule of 𝐴; this means that 𝑀𝐺𝐹(𝐴) is a 𝐺𝐹-regular submodule which is not contained properly in any other 𝐺𝐹-regular submodule. In fact, 𝑀𝐺𝐹(𝐴) is the sum of all 𝐺𝐹-regular submodules of 𝐴 and 𝑀𝐺𝐹(𝐴)=𝐴 if and only if 𝐴 is a 𝐺𝐹-regular module.

Example 3. (a) Since the 𝑍-module 𝑍_𝑛 is 𝐺𝐹-regular for each positive integer 𝑛 [2], then 𝑀𝐺𝐹(𝑍_𝑛)=𝑍_𝑛.
(b) Each element in the 𝑍-module 𝑄 is not 𝐺𝐹-regular [2]; hence 𝑀𝐺𝐹(𝑄)=(0).
(c) Let 𝑝 be a prime number and let 𝐴=𝑍_𝑝∞=⋃_∀𝑖𝑍_𝑝𝑖 be a 𝑍-module. Let 𝑎∈⋃_∀𝑖𝑍_𝑝𝑖; then there exists a positive integer 𝑚 such that 𝑎∈𝑍_𝑝𝑚, but 𝑍_𝑛 is a 𝐺𝐹-regular 𝑍-module for each positive integer 𝑛; hence 𝑎 is a 𝐺𝐹-regular element, so 𝑎∈𝑀𝐺𝐹(𝑍_𝑝∞) which implies that 𝑀𝐺𝐹(𝑍_𝑝∞)=𝑍_𝑝∞.
(d) Let 𝑄/𝑍 be a 𝑍-module; since 𝑄/𝑍 is a torsion 𝑍-module, then 𝑄/𝑍 is a 𝐺𝐹-regular 𝑍-module [2, Proposition 6]. Since 𝑄/𝑍=∑_𝑝𝑍_𝑝∞ for each prime number 𝑝, then 𝑀𝐺𝐹(𝑄/𝑍)=∑_𝑝𝑍_𝑝∞ for all primes 𝑝.

Proposition 4. Let 𝐴 and 𝐵 be 𝑅-modules, and let 𝐾 be a submodule of 𝐴; then (a)𝑀𝐺𝐹(𝐾)=𝐾∩𝑀𝐺𝐹(𝐴), (b)𝑀𝐺𝐹(𝐴⊕𝐵)⊆𝑀𝐺𝐹(𝐴)⊕𝑀𝐺𝐹(𝐵).

Proof. (a) Let 𝐾 be a submodule of 𝐴, and let 𝑘∈𝑀𝐺𝐹(𝐾); then 𝑘∈𝐾 and 𝑘 is 𝐺𝐹-regular in 𝐾 which implies that 𝑘 is 𝐺𝐹-regular in 𝐴, thus 𝑘∈𝐾∩𝑀𝐺𝐹(𝐴). Conversely, let 𝑘∈𝐾 and 𝑘∈𝑀𝐺𝐹(𝐴); therefore 𝑘 is 𝐺𝐹-regular in 𝐾 which means that 𝑘∈𝑀𝐺𝐹(𝐾) and hence 𝑀𝐺𝐹(𝐾)=𝐾∩𝑀𝐺𝐹(𝐴).
(b) Let 𝑐∈𝑀𝐺𝐹(𝐴⊕𝐵); then 𝑐=(𝑎,𝑏), where 𝑎∈𝐴 and 𝑏∈𝐵. Since 𝑐 is 𝐺𝐹-regular, then each of 𝑎 and 𝑏 is 𝐺𝐹-regular which means that 𝑎∈𝑀𝐺𝐹(𝐴) and 𝑏∈𝑀𝐺𝐹(𝐵); hence 𝑐∈𝑀𝐺𝐹(𝐴)⊕𝑀𝐺𝐹(𝐵).

Proposition 5. Let 𝐴 and 𝐴^′ be 𝑅-modules, and let 𝑓∶𝐴→𝐴^′ be an 𝑅-homomorphism; then 𝑓(𝑀𝐺𝐹(𝐴))⊆𝑀𝐺𝐹(𝑓(𝐴)).

Proof. If 𝑎∈𝑀𝐺𝐹(𝐴), then 𝑅/𝑎𝑛𝑛(𝑎) is a 𝜋-regular ring, but 𝑎𝑛𝑛(𝑎)⊆𝑎𝑛𝑛(𝑓(𝑎)); thus 𝑅/𝑎𝑛𝑛(𝑓(𝑎)) is an epimorphic image of 𝑅/𝑎𝑛𝑛(𝐴); hence it is a 𝜋-regular ring. Therefore, 𝑓(𝑎)∈𝑀𝐺𝐹(𝑓(𝑎)) and 𝑓(𝑀𝐺𝐹(𝐴))⊆𝑀𝐺𝐹(𝑓(𝐴)).

Remark 6. (a) If 𝑓∶𝐴→𝐴^′ is an 𝑅-epimorphism, then 𝑓(𝑀𝐺𝐹(𝐴))≠𝑀𝐺𝐹(𝐴^′) in general. In fact, let 𝜋∶𝑍→𝑍₄≃𝑍/4𝑍 be the natural map, where 𝑍 and 𝑍₄ are 𝑍-modules. It is easy to check that 𝑀𝐺𝐹(𝑍)=(0), 𝑓(𝑀𝐺𝐹(𝑍))=(0), but 𝑀𝐺𝐹(𝑍₄)≃𝑍₂.
(b) It is shown in [1] that, for a ring 𝑅, 𝑀(𝑅/𝑀(𝑅))=(0) which is not true in case of 𝐺𝐹-regular modules; this means that 𝑀𝐺𝐹(𝐴/𝑀𝐺𝐹(𝐴))≠(0) (as in (a)).

Corollary 7. For each 𝑅-module 𝐴, 𝑀(𝑅)⋅𝐴⊆𝑀𝐺𝐹(𝐴).

Proof. For each 𝑎∈𝐴, let 𝑓∶𝑅→𝐴 be an 𝑅-homomorphism defined by 𝑓(𝑟)=𝑟𝑎. Then 𝑓(𝑀(𝑅))⊆𝑀𝐺𝐹(𝐴) by Proposition 5, but 𝑀(𝑅)⋅𝐴=∑_𝑎𝑓(𝑀(𝑅)); hence 𝑀(𝑅)⋅𝐴⊆𝑀𝐺𝐹(𝐴).

Let 𝐽(𝑅) be the Jacobson radical of a ring 𝑅. Brown and McCoy proved in [1] that 𝑀(𝑅)∩𝐽(𝑅)=(0). However, this is not true for 𝐺𝐹-regular modules; for example, if 𝐴=𝑍₄ is a 𝑍-module, then 𝑀𝐺𝐹(𝐴)≃𝑍₂, 𝐽(𝐴)≃𝑍₂ and 𝑀𝐺𝐹(𝐴)∩𝐽(𝐴)≠(0).

Lemma 8. Let 𝐴 be an 𝑅-module and let 𝑃 be a 𝐺-pure submodule of 𝐴. For any 𝑟∈𝑅, there exists a positive integer 𝑛 such that 𝑃=𝑅𝑟^𝑛𝑃 if and only if 𝑃⊆𝑅𝑟^𝑛𝐴.

Proof. Since 𝑃 is 𝐺-pure in 𝐴, then, for each 𝑟∈𝑅, there exists a positive integer 𝑛 such that 𝑃∩𝑅𝑟^𝑛𝐴=𝑅𝑟^𝑛𝑃. If 𝑃=𝑅𝑟^𝑛𝑃, then 𝑃∩𝑅𝑟^𝑛𝐴=𝑃, and hence 𝑃⊆𝑅𝑟^𝑛𝐴. Conversely, if 𝑃⊆𝑅𝑟^𝑛𝐴, then 𝑃∩𝑅𝑟^𝑛𝐴=𝑃, but 𝑃∩𝑅𝑟^𝑛𝐴=𝑅𝑟^𝑛𝑃; therefore 𝑃=𝑅𝑟^𝑛𝑃.

Lemma 9. Let 𝑟∈𝐽(𝑅); if 𝑃 is a finitely generated 𝐺-pure submodule of an 𝑅-module 𝐴 such that 𝑃⊆𝑅𝑟^𝑛𝐴 for some positive integer 𝑛, then 𝑃=0.

Proof. By Lemma 8 we get that 𝑃=𝑅𝑟^𝑛𝑃 and by Nakayama’s lemma [5], 𝑃=0.

Theorem 10. Let 𝐴 be an 𝑅-module. If 𝑀𝐺𝐹(𝐴) is a 𝐺-pure submodule of 𝐴, then 𝑀𝐺𝐹(𝐴)∩𝐽(𝑅)⋅𝐴=(0).

Proof. Let 𝑟∈𝑀𝐺𝐹(𝐴)∩𝐽(𝑅)⋅𝐴, and let 𝑃=𝑅𝑟. It is clear that 𝑃⊆𝑀𝐺𝐹(𝐴). Since 𝑀𝐺𝐹(𝐴) is a 𝐺𝐹-regular module, then 𝑃 is a 𝐺-pure submodule in 𝑀𝐺𝐹(𝐴) [2, Theorem 11]. But 𝑀𝐺𝐹(𝐴) is 𝐺-pure in 𝐴; hence 𝑃 is 𝐺-pure in 𝐴. Now, 𝑃⊆𝐽(𝑅)⋅𝐴, so 𝑃=0 by Lemma 9. Therefore 𝑀𝐺𝐹(𝐴)∩𝐽(𝑅)⋅𝐴=(0).

Recall that 𝑀(𝑅) is always a pure ideal in 𝑅. Hence 𝑀(𝑅) is 𝐺-pure [2].

Theorem 11. Let 𝐴 be a projective 𝑅-module; then (a)𝑀𝐺𝐹(𝐴)=𝑀(𝑅)⋅𝐴, (b)𝑀𝐺𝐹(𝐴) is a 𝐺-pure submodule of 𝐴, (c)𝑀𝐺𝐹(𝐴)∩𝐽(𝐴)=(0).

Proof. (a) By the dual basis lemma [4], for each 𝑎∈𝐴 we have that 𝑎=∑_𝑖𝑓_𝑖(𝑎)𝑎_𝑖, where 𝑎,   𝑎_𝑖∈𝐴 for all 𝑖 and 𝑓_𝑖∈𝐴^∗∶=Hom_𝑅(𝐴,𝑅). If 𝑎∈𝑀𝐺𝐹(𝐴), then the submodule 𝑅𝑎 is 𝐺𝐹-regular and 𝑓_𝑖(𝑅𝑎) is a 𝐺𝐹-regular ideal in 𝑅 by Proposition 5, hence 𝑀(𝑅). Thus 𝑀𝐺𝐹(𝐴)⊆𝑀(𝑅)⋅𝐴. We get the other direction of the inclusion by Corollary 7.
(b) First we claim that; for any two ideals 𝐾 and 𝐿 of 𝑅, (𝐾∩𝐿)𝐴=𝐾𝐴∩𝐿𝐴; it is enough to show this locally; thus we may assume that 𝐴 is free. It is clear that (𝐾∩𝐿)𝐴⊆𝐾𝐴∩𝐿𝐴. On the other hand, let 𝑥∈𝐾𝐴∩𝐿𝐴; then 𝑥=𝑟_𝑖𝑥_𝑖=𝑠_𝑖𝑥_𝑖𝑟_𝑖∈𝐾,𝑠_𝑖∈𝐿. (2) By freeness, 𝑟_𝑖=𝑠_𝑖 and 𝑥∈(𝐾∩𝐿)𝐴. Now, let 𝐾 be any ideal in 𝑅; then by (a) we get 𝑀𝐺𝐹(𝐴)∩𝐾𝐴=𝑀(𝑅)∩𝐾𝐴=(𝑀(𝑅)∩𝐾)𝐴. But 𝑀(𝑅) is 𝐺-pure ideal, so, for each 𝑟∈𝑅, there exists a positive integer 𝑛 such that 𝑀(𝑅)∩𝑅𝑟^𝑛=𝑅𝑟^𝑛𝑀(𝑅); hence 𝑀𝐺𝐹(𝐴)∩𝑅𝑟^𝑛𝐴=𝑅𝑟^𝑛𝑀(𝑅)⋅𝐴=𝑅𝑟^𝑛𝑀𝐺𝐹(𝐴).
(c) Since 𝐴 is projective, then 𝐽(𝐴)=𝐽(𝑅)⋅𝐴 [4] which implies that 𝑀𝐺𝐹(𝐴)∩𝐽(𝐴)=(0) by Theorem 10.

Corollary 12. Let 𝑅 be any ring, and let 𝐴 be any 𝑅-module such that 𝑀𝐺𝐹(𝐴) is a 𝐺-pure submodule and 𝐽(𝐴)=𝐽(𝑅)⋅𝐴; then 𝑀𝐺𝐹(𝐴)∩𝐽(𝐴)=(0).

Proof. Since 𝐽(𝐴)=𝐽(𝑅)⋅𝐴, then by Theorem 10 we get that 𝑀𝐺𝐹(𝐴)∩𝐽(𝐴)=(0).

Remark 13. If 𝐴 is a 𝐺𝐹-regular 𝑅-module, then 𝑀𝐺𝐹(𝐴)=𝐴; hence 𝐽(𝑅)⋅𝐴=𝐽(𝑅)⋅𝐴∩𝐴=𝐽(𝑅)⋅𝐴∩𝑀𝐺𝐹(𝐴)=(0). In fact, this shows that Theorem 10 is a generalization of [2, Proposition 28].
In [2] we noticed that every module over 𝜋-regular ring is 𝐺𝐹-regular, but the converse need not be true in general. The next result shows how the converse may be true, but first we recall that if 𝐴 is an 𝑅-module, then the trace of 𝐴 is tr(𝐴)=∑_𝑓∈𝐴^∗𝑓(𝐴), where 𝐴^∗=Hom(𝐴,𝑅).

Proposition 14. Let 𝐴 be a 𝐺𝐹-regular 𝑅-module. If tr(𝐴)=𝑅, then 𝑅 is a 𝜋-regular ring.

Proof. For each 𝑎∈𝐴 and 𝑓∈𝐴^∗=Hom(𝐴,𝑅), since 𝑅𝑎 is a 𝐺𝐹-regular submodule of 𝐴, then by [2, Proposition 7] we get that 𝑓(𝑅𝑎) is a 𝜋-regular ideal. Thus 𝑓(𝑅𝑎)⊆𝑀(𝑅), but 𝑓(𝑅𝑎)⊆tr(𝐴); hence tr(𝐴)=𝑅⊆𝑀(𝑅), which implies that 𝑀(𝑅)=𝑅 and 𝑅 is 𝜋-regular.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

Sheng Chen was supported by the National Natural Science Foundation of China (Grants nos. 11001064 and 11101105), by the Fundamental Research Funds for the Central Universities (Grant no. HIT. NSRIF. 2014085), and by the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry, and Areej M. Abduldaim was supported by China Scholarship Council (CSC no. 2011368T09).