AGQP Injective Modules

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Volume 2008, Article ID 469725,7pages doi:10.1155/2008/469725

Research Article

AGQP-Injective Modules

Zhanmin Zhu1 _{and Xiaoxiang Zhang}2

1_{Department of Mathematics, Jiaxing University, Jiaxing, Zhejiang 314001, China}

2_{Department of Mathematics, Southeast University, Nanjing 210096, China}

Correspondence should be addressed to Zhanmin Zhu,zhanmin [email protected]

Received 23 December 2007; Revised 20 April 2008; Accepted 20 June 2008

Recommended by Robert Lowen

LetRbe a ring and letMbe a rightR-module withSEndMR.Mis called almost general

quasi-principally injectiveor AGQP-injective for shortif, for any 0/s∈S, there exist a positive integern and a left idealXsnofSsuch thatsn/0 andlSKersn Ssn⊕X_sn. Some characterizations and

properties of AGQP-injective modules are given, and some properties of AGQP-injective modules with additional conditions are studied.

Copyrightq2008 Z. Zhu and X. Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

Throughout R is an associative ring with identity, and all modules are unitary. Recall that a ring R is called right principally injective 1 or right P-injective for short if, every homomorphism from a principal right ideal ofRtoRcan be extended to an endomorphism ofR, or equivalently,lra Rafor alla∈R. The concept of right P-injective rings has been generalized by many authors. For example, in2,3, right P-injective rings are generalized in two directions, respectively. Following2, a ringRis called right GP-injective if, for any 0/a ∈ R, there exists a positive integernsuch thatan/0 and any rightR-homomorphism fromanR toR can be extended to an endomorphism ofR. Note that GP-injective rings are also called YJ-injective in 4. From 5, we know that Ginjective rings need not to be P-injective. Following3, a rightR-moduleMR withS EndMRis called quasiprincipally

injective or QP-injective for short if, every homomorphism from anM-cyclic submodule of Mto M can be extended to an endomorphism ofM, or equivalently, lSKers Ss

for alls ∈ S. In 1998, Page and Zhou 6 generalized the concept of GP-injective rings to that of AGP-injective rings. According to 6, a ring R is called right AGP-injective if, for any 0/a ∈ R, there exist a positive integer n and a left ideal Xan such that an/0 and

lran _Ran_⊕_X_a_n_{. In}₇_{, the first author introduced the notion of GQP-injective modules}

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0/s ∈ S, there exists a positive integer nsuch thatsn/0 and any rightR-homomorphism from snM to M can be extended to an endomorphism of M, or equivalently, for any 0/s ∈ S, there exists a positive integer nsuch thatsn/0 andlSKersn Ssn. The nice

structure of AGP-injective rings and GQP-injective modules draws our attention to define almost GQP-injective modules, in a similar way to AGP-injective rings, and to investigate their properties.

2. Results

Definition 2.1. LetMRbe a rightR-module withSEndMR. Then,Mis said to be almost

general quasiprincipally injectivebriefly, AGQP-injectiveif, for any 0/s∈S, there exist a positive integernand a left idealXsn ofSsuch thatsn/0 andl_SKersn Ssn⊕X_sn.

Clearly, a ring R is right AGP-injective if and only if RR is Ainjective,

GQP-injective modules are AGQP-GQP-injective.

Our next result gives the relationship between the AGQP-injectivity of a module and the AGP-injectivity of its endomorphism ring.

Theorem 2.2. LetMRbe a rightR-module withSEndMR. Then,

1ifSis right AGP-injective, thenMRis AGQP-injective;

2ifMR is AGQP-injective andMgenerates Kersfor eachs ∈ S, thenSis right

AGP-injective.

Proof. 1Suppose thatSis right AGP-injective then for any 0/s ∈ S, there exist a positive integernand a left idealIsn ofSsuch thatsn/0 andl_Sr_Ssn Ssn⊕I_sn. Ifa∈ l_SKersn

andb∈rSsn, thensnb0, that is,bM⊆Kersn. Hence,abM0, that is,ab0. This

shows thatlSKersn ⊆ lSrSsn. Therefore, we haveSsn ⊆lSKersn⊆ Ssn⊕Isn, which

guarantees that

lSKersn Ssn⊕lSKersn∩Isn. 2.1

Thus,1is proved.

2Suppose thatMR is AGQP-injective then for any 0/s ∈ S, there exist a positive

integernand a left idealXsn ofSsuch thatsn/0 andl_SKersn Ssn⊕X_sn. Assume that a∈lSrSsnand Kersn t∈TtMfor some subsetT ofS. It is easy to see thatat0 for

eacht∈T, so we haveax 0 for eachx∈Kersn. This implies thatlSrSsn⊆lSKersn,

from which we have

Ssn_⊆_l

SrSsn⊆lSKersn Ssn⊕Xsn, 2.2

and hence

lSrSsn Ssn⊕lSrSsn∩Xsn. 2.3

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Recall that a moduleNis calledM-cyclic3, if it is a homomorphic image ofM. Let

SEndMR, following8, we writeWS {s∈S |Kers⊆essM}.

Theorem 2.3. LetMRbe an AGQP-injective module withSEndMR. Then,

1WS⊆JS,

2if every nonzero submodule ofMcontains a nonzeroM-cyclic submodule, thenWS

JS.

Proof. 1Lets∈WS. Then, for eacht∈S,ts∈WSand so 1−ts /0. SinceMRis

AGQP-injective, there exist a positive integernand a left ideal X₁₋tsn _{such that}₁−tsn/_{0 and}

lSKer1−tsn S1−tsn⊕X1−tsn_{. Note that}₁−tsn ₁−u_{for some}u ∈WS_{. Since}

Keru∩Ker1−u 0, we have Ker1−u 0, and thenSS1−u⊕X1−u. So 1exfor

somee∈S1−uandx∈X1−u, it follows thate2eandS1−u Se⊕S1−e∩S1−u Se.

Therefore, 1−uvefor somev∈S, since Keruis essential inMR, ife /1, then there exists

a nonzero element1−em∈1−eM∩Keru, and hence1−u1−em 1−em. But

1−u1−emve1−em0, a contradiction. Soe1,and hence 1−uis left invertible, which impliess∈JS.

2We need only to prove that JS ⊆ WS. Let s ∈ JS. Ifs /∈WS, then there exists 0/t∈Ssuch that Kers∩tM 0 by hypothesis. Clearly,st /0 and Kerst Kert. SinceMR is AGQP-injective, there exist a positive integernand a left idealXstn _{such that}

stn/0 and

lSKerstn Sstn⊕Xstn. _2.4

If m ∈ Kerstn, then stn−1m ∈ Kerst Kert, and som ∈ Kertstn−1. This shows that Kerstn Kertstn−1. Hence, tstn−1 ∈ Sstn⊕Xstn_{. Write} tstn−1 ustnv_,

whereu ∈ S, v ∈ Xstn_{. Then}₁−uststn−1 v_{, which gives that}stn s₁−us−1v ∈ Sstn∩Xstn _{0, a contradiction.}

Corollary 2.4see6, Corollary 2.3. IfRis a right AGP-injective ring, thenJR ZRR.

Following9, for a setX⊆HomNR, MR, the submodule

KerX∩{Kerg | g ∈X} 2.5

ofN is called anM-annihilator submodule ofN. By7, Lemma 9andTheorem 2.3, we have the following corollary.

Corollary 2.5. Let MR be an AGQP-injective module with S EndMR. If every nonzero

submodule ofM contains a nonzero M-cyclic submodule, andM/SocM satisfies ACC on M -annihilator submodules, thenJSis nilpotent.

Recall that a moduleMR is said to be a GC2 module10if every submoduleN ≤ M

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Theorem 2.6. LetMRbe an AGQP-injective module. Then,

1if M1 and M2 are submodules ofM such that M1 ⊆ M2 and M1 ∼ M2 | M, then

M1|M. In particularMis a GC2 module;

2ifM1andM2are simple submodules ofMsuch thatM1∼M2|M, thenM1 |M.

Proof. 1LetSEndMR. It is trivial in caseM1 0. Now suppose thatM1/0 andM2

f ∼

M1. ThenM1aMandM2 eM, wheree2e∈Sandafe. SinceMRis AGQP-injective,

there exist a positive integernand a left idealXansuch thatan/0 andl_SKeran San⊕X_an.

Leta0 e, thenf−1ai1M aiMi0,1, . . . , n−1sinceM1⊆M2eM. So we have

ai_M_|_ai−1_M_⇐⇒_f−1_ai1_M_|_f−1_ai_M_⇐⇒_ai1_M_|_ai_M _i₁_{, . . . , n}₋₁_. _2.6

Consequently,aM | eM ⇔ a2_M _| _aM _{⇔ · · · ⇔} _an_M _| _an−1_M_{. Thus, to show}_aM _| _M_{, it}

suﬃces to show thatanM | M. Note thata|eM : eM → eM is monic andanm anem

for everym ∈M,eM a

n

∼

an_M_{and hence Ker}_an _Ker_e_{. It follows that}_e _∈_l_S_Ker_e

lSKeran San ⊕Xan. Now, lete banxwith b ∈ S andx ∈ X_an, then an ane an_ban_an_x_an_ban_{. Finally, let}_g_an_b_{, then}_g2 _g_and_an_M_gM_{as required.}

2Let M2 e1M, wheree21 e1 ∈ S, and letM2

f1 ∼

M1. ThenM1 a1M, where a1 f1e1. SinceMRis AGQP-injective, there exist a positive integern1and a left idealXan1 1

such thatan1₁ /0 andlSKeran1₁ San1₁ ⊕Xan1

1 . Note that 0/a n1

1 M⊆a1M,anda1Mis simple.

We havean1₁ Ma1M. Clearly, Kere1 Kera1becausef1is a monomorphism. Sincea1M is simple, Kera1is a maximal submodule ofM. But Kera1⊆ Keran1₁ /M, so Kera1

Keran1₁ and then Kere1 Keran11 . It follows thate1∈lSKere1 lSKeran11 Sa

n1

1 ⊕ X_an₁

1 . Now, lete1 b1a n1

1 ywithb1 ∈Sandy∈Xan1

1 , thena n1

1 a

n1

1 e1 a

n1

1 b1a

n1

1 a

n1

1 y an1₁ b1an1₁ . Finally, letg1an1₁ b1, theng2

1 g1andM1 a1Ma

n1

1 Mg1Mas required.

Recall that a moduleMis said to be weakly injective11if, for any finitely generated submoduleN≤EM, there existsX≤EMsuch thatN⊆X∼M.

Corollary 2.7. LetMbe a finitely generated module. Then,Mis injective if and only ifMis weakly injective and AGQP-injective. In particular, a ringRis right self-injective if and only ifRRis weakly

injective and AGP-injective.

Proof. We need only to prove the suﬃciency. Letx∈EM. Then, there existsX⊆EMsuch thatMxR⊆X∼M. Hence,Xis AGQP-injective andM|Xfollows fromTheorem 2.61. ButMis essential inEM, soMXand hencex∈M.

Corollary 2.8. LetMRbe an AGQP-injective module withSEndMR.

1IfMRis of finite Goldie dimension, then S is semilocal. 2IfMRis a noetherian self-generator, thenSis semiprimary.

Proof. 1SinceMR is AGQP-injective, it satisfies the GC2-condition byTheorem 2.61and

then1follows immediately by12, Lemma 1.1.

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Recall that ifMandUare two rightR-modules, thenUis called M-projective in case for each epimorphismg : MR → NR and each homomorphism γ : UR → NR, there is an

R-homomorphismγ :UR →MRsuch thatγ gγ. A moduleMRis called quasiprojective if it

isM-projective.

LetRbe a ring. Recall that an elementa∈Ris calledπ-regular if there exists a positive integermsuch thatam ambam 13for someb∈R. An elementx∈Ris called generalized

π-regular if there exists a positive integer n such thatxn xnyx for some y ∈ R. A ring

R is calledπ-regularresp., generalizedπ-regularif every element in Risπ-regularresp., generalizedπ-regular. IfAis a subset ofR, then we say thatAis regular if every element in

Ais regular.

Proposition 2.9. LetMR be quasiprojective withS EndMR. Then,Sis regular if and only if

MRis AGQP-injective andsMisM-projective for everys∈S.

Proof. Assume that S is regular. Then, every right ideal of S is a direct summand of SS,

and so every homomorphism from a principal right ideal of S to S can be extended to an endomorphism of S. Hence, S is right P-injective and then right AGP-injective. By

Theorem 2.2,MR is AGQP-injective. The regularity of Salso implies that sMis a direct

summand ofMby14, Theorem 37.7. ButMis quasiprojective, sosMisM-projective for everys∈S.

Conversely, supposeMRis AGQP-injective andsMisM-projective for everys∈S.

Then for any 0/a∈S, by the AGQP-injectivity ofMR, there exist a positive integernand a

left idealXan ofSsuch thatan/0 andl_SKeran San⊕X_an. SinceanMisM-projective,

Keran eMfor somee2_e_∈_S_{. Then, we have}_S₁₋_e _l_S_eM _l_S_Ker_an _San_⊕_X_a_n_,

and so 1−ebanxfor someb∈Sandx∈Xan. Thus,anan1−e anbananxanban.

This proves thatSisπ-regular and hence generalizedπ-regular. Clearly,N1S {0/a ∈ S | a2 ₀_}_{is regular}_{in this case,}_n_{must be equal to 1}_{. Therefore or,}_S_{is regular by}₁₃_,

Theorem 2.2.

Recall that a moduleMRis called an IN-module15iflSA∩B lSA lSBfor any

submodulesAandBofM, whereSEndMR.

Proposition 2.10. LetMRbe an AGQP-injective IN-module withSEndMR. Then,Sis regular

if and only ifWS 0.

Proof. By Theorem 2.3, we need only to prove the suﬃciency. Let 0/a ∈ S. Since MR is

AGQP-injective, there exist a positive integernand a left idealXan ofSsuch thatan/0 and

lSKeran San⊕Xan. SinceWS 0, Keranis not essential inMand then there exists a

nonzero submoduleKsuch that Keran⊕Kis essential inM. Moveover, we also have

lSKeran lSK lSKeran∩K S,

lSKeran∩lSK⊆lSKeran K 0, 2.7

becauseMRis an IN-module andWS 0. Thus,

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Let 1banxwithb∈S,x∈Xan⊕l_SK, thenananban. It follows thatSis regular by the

last part of the proof ofProposition 2.9.

Lemma 2.11. LetMRbe an AGQP-injective module in which every nonzero submodule contains a

nonzeroM-cyclic submodule andSEndMR. Ifs /∈WS, then the inclusion Kers⊆Kers−

stsis strict for somet∈S.

Proof. Ifs /∈WS, then Kers∩K0 for some nonzero submoduleKofM, and so Kers∩

s_M _{0 for some 0}_/_s _∈ _S_{by hypothesis. Clearly,} _ss_/_{0. Since} _M_R _{is AGQP-injective,}

there exist a positive integernand a left idealXssn _{such that}ssn/_{0 and}l_S_Kerssn Sssn_⊕_X

ssn_{. Thus,}

s_ssn−1_∈_l

SKersssn−1 lSKerssn Sssn⊕Xssn. _2.9

Writesssn−1 tssnx,wheret∈Sandx∈Xssn_{, then}₁−tssssn−1x_{and hence}

1−stssn s−stssssn−1sx∈Sssn∩Xssn. _2.10

This means thats−stssssn−1 0. It is obvious that Kers ⊆ Kers−sts. Note that

s_ssn−1_M_{is contained in Ker}_s₋_sts_{but not contained in Ker}_s_{, the inclusion Ker}_s_⊆

Kers−stsis strict.

Theorem 2.12. LetMR be AGQP-injective withSEndMR. If every nonzero submodule ofM

contains a nonzeroM-cyclic submodule, then the following conditions are equivalent:

1Sis right perfect;

2for any sequence{s1, s2, . . .} ⊆S, the chain Kers1⊆Kers2s1⊆ · · · terminates.

Proof. ByTheorem 2.3,Lemma 2.11, and 16, Lemma 2.8, one can complete the proof in a similar way to that of16, Theorem 2.9.

Acknowledgment

The authors are very grateful to the referees for their useful comments and suggestions.

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