# AGQP-Injective Modules

Authors: Zhanmin Zhu; Xiaoxiang Zhang
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

LetR be a ring and let M be a right R-module with S = End(MR). M is called almost general quasi-principally injective (or AGQP-injective for short) if, for any 0≠s∈S, there exist a positive integer n and a left ideal Xsn
of S such that sn≠0 and lS(Ker(sn))=Ssn⊕Xsn. Some characterizations and properties of AGQP-injective modules are given, and some properties of
AGQP-injective modules with additional conditions are studied.


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## Body

## 1. Introduction

ThroughoutR is an associative ring with identity, and all
modules are unitary. Recall that a ring R is called right principally injective [1] (or rightP-injective
for short) if, every
homomorphism from a principal right ideal of R to R can be extended to an endomorphism of R,
or equivalently, lr(a)=Ra for all a∈R.
The concept of right P-injective rings has been generalized by many authors.
For example, in [2, 3], right P-injective rings are generalized in two
directions, respectively. Following [2], a ring R is called right GP-injective if, for
any 0≠a∈R,
there exists a positive integer n such that an≠0 and any right R-homomorphism from anR to R can be extended to an endomorphism of R.
Note that GP-injective rings are also called YJ-injective in
[4]. From [5], we know that GP-injective rings need not
to be P-injective. Following [3], a right R-module MR with  S=End(MR) is called quasiprincipally injective (or QP-injective for short) if, every homomorphism from an M-cyclic submodule of M to M can be extended to an endomorphism of M, or equivalently, lS(Ker(s))=Ss for all s∈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 lr(an)=Ran⊕Xan.
In [7], the first author introduced the notion of
GQP-injective modules which can be regarded as the generalization of
GP-injective rings and QP-injective modules. According to [7], a right R-module M with S=End(MR) is called GQP-injective if, for any 0≠s∈S,
there exists a positive integer n such that sn≠0 and any right R-homomorphism from sn(M) to M can be extended to an endomorphism of M,
or equivalently, for any 0≠s∈S,
there exists a positive integer n such that sn≠0 and lS(Ker(sn))=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.
LetMR be a right R-module with S=End(MR).
Then, M is said to be almost general quasiprincipally
injective (briefly, AGQP-injective) if, for any 0≠s∈S,
there exist a positive integer n and a left ideal Xsn of S such that sn≠0 and lS(Ker(sn))=Ssn⊕Xsn.Clearly, a ringR is right AGP-injective if and only if RR is AGQP-injective, GQP-injective modules are
AGQP-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.
LetMR be a right R-module with S=End(MR).
Then,
(1)
ifS is right AGP-injective, then MR is AGQP-injective;(2)
ifMR is AGQP-injective and M generates Ker(s) for each s∈S,
then S is right AGP-injective.Proof.
(1)
Suppose thatS is right AGP-injective then for any 0≠s∈S,
there exist a positive integer n and a left ideal Isn of S such that sn≠0 and lSrS(sn)=Ssn⊕Isn.
If a∈lS(Ker(sn)) and b∈rS(sn),
then snb=0,
that is, b(M)⊆Ker(sn).
Hence, (ab)M=0,
that is, ab=0.
This shows that lS(Ker(sn))⊆lSrS(sn).
Therefore, we have Ssn⊆lS(Ker(sn))⊆Ssn⊕Isn,
which guarantees that(2.1)lS(Ker(sn))=Ssn⊕(lS(Ker(sn))∩Isn).Thus, (1) is proved.
(2) Suppose thatMR is AGQP-injective then for any 0≠s∈S,
there exist a positive integer n and a left ideal Xsn of S such that sn≠0 and lS(Ker(sn))=Ssn⊕Xsn.
Assume that a∈lSrS(sn) and Ker(sn)=∑t∈Tt(M) for some subset T of S.
It is easy to see that at=0 for each t∈T,
so we have ax=0 for each x∈Ker(sn). This implies that lSrS(sn)⊆lS(Ker(sn)),
from which we have(2.2)Ssn⊆lSrS(sn)⊆lS(Ker(sn))=Ssn⊕Xsn,and hence(2.3)lSrS(sn)=Ssn⊕(lSrS(sn)∩Xsn).Therefore, S is right AGP-injective.Recall that a moduleN is called M-cyclic [3], if it is a homomorphic image of M.
Let S = End(MR), following [8], we write W(S)={s∈S∣Ker(s)⊆essM}.Theorem 2.3.
LetMR be an AGQP-injective module with  S=End(MR). Then,
(1)
W(S)⊆J(S),(2)
if every nonzero submodule ofM contains a nonzero  M-cyclic submodule, then  W(S)=J(S).Proof.
(1)
Lets∈W(S).
Then, for each t∈S, ts∈W(S) and so 1−ts≠0.
Since MR is AGQP-injective, there exist a positive
integer n and a left ideal X(1−ts)n such that (1−ts)n≠0 and lS(Ker(1−ts)n)=S(1−ts)n⊕X(1−ts)n.
Note that (1−ts)n=1−u for some u∈W(S).
Since Ker(u)∩Ker(1−u)=0,
we have Ker(1−u)=0, and then S=S(1−u)⊕X1−u.
So 1=e+x for some e∈S(1−u) and x∈X1−u,
it follows that e2=e and S(1−u)=Se⊕S(1−e)∩S(1−u)=Se.
Therefore, 1−u=ve for some v∈S,
since Ker(u) is essential in MR,
if e≠1,
then there exists a nonzero element (1−e)m∈(1−e)M∩Ker(u),
and hence (1−u)(1−e)m=(1−e)m.
But (1−u)(1−e)m=ve(1−e)m=0,
a contradiction. So e=1, and hence 1−u is left invertible, which implies s∈J(S).
(2) We need only to prove thatJ(S)⊆W(S).
Let s∈J(S).
If s∉W(S),
then there exists 0≠t∈S such that Ker(s)∩t(M)=0 by hypothesis. Clearly, st≠0 and Ker(st)=Ker(t). Since MR is AGQP-injective, there exist a positive
integer n and a left ideal X(st)n such that (st)n≠0 and(2.4)lS(Ker(st)n)=S(st)n⊕X(st)n.If m∈Ker(st)n,
then (st)n−1m∈Ker(st)=Ker(t), and so m∈Ker(t(st)n−1). This shows that Ker(st)n=Ker(t(st)n−1). Hence, t(st)n−1∈S(st)n⊕X(st)n.
Write t(st)n−1=u(st)n+v,
where u∈S,v∈X(st)n.
Then (1−us)t(st)n−1=v,
which gives that (st)n=s(1−us)−1v∈S(st)n∩X(st)n=0,
a contradiction.Corollary 2.4  (see [6, Corollary 2.3]).
IfR is a right AGP-injective ring, then  J(R)=Z(RR).Following [9], for a set X⊆ Hom (NR,MR),
the submodule(2.5)KerX=∩{Kerg∣g∈X} of N is called an M-annihilator submodule of N.
By  [7, Lemma 9] and
Theorem 2.3, we have the following
corollary.Corollary 2.5.
LetMR be an AGQP-injective module with  S=End(MR). If every nonzero submodule of  M contains a nonzero  M-cyclic submodule,   and  M/Soc(M) satisfies ACC on  M-annihilator submodules, then  J(S) is nilpotent.Recall that a moduleMR is said to be a GC2 module [10] if every submodule N≤M with N≅M is a direct summand of M.
For convenience, we write N∣M to denote that N is a direct summand of M.Theorem 2.6.
LetMR be an AGQP-injective module. Then,
(1)
ifM1 and  M2 are submodules of  M such that  M1⊆M2 and M1≅M2∣M, then  M1∣M. In particular  M is a GC2 module;(2)
ifM1 and  M2 are simple submodules of  M such that  M1≅M2∣M, then  M1∣M.Proof.
(1)
LetS=End(MR). It is trivial in case M1=0.
Now suppose that M1≠0 and M2≅fM1.
Then M1=aM and M2=eM,
where e2=e∈S and a=fe.
Since MR is AGQP-injective, there exist a positive
integer n and a left ideal Xan such that an≠0 and lS(Ker(an))=San⊕Xan.
Let a0=e,
then f−1(ai+1M)=aiM(i=0,1,…,n−1) since M1⊆M2=eM.
So we have(2.6)aiM∣ai−1M⟺f−1(ai+1M)∣f−1(aiM)⟺ai+1M∣aiM(i=1,…,n−1).Consequently, aM∣eM⇔a2M∣aM⇔⋯⇔anM∣an−1M.
Thus, to show aM∣M,
it suffices to show that anM∣M.
Note that a|eM:eM→eM is monic and an(m)=an(em) for every m∈M, eM≅ananM and hence Ker(an)  =  Ker(e). It follows that e∈lS(Ker(e))=lS(Ker(an))=San⊕Xan.
Now, let e=ban+x with b∈S and x∈Xan,
then an=ane=anban+anx=anban.
Finally, let g=anb,
then g2=g and anM=gM as required.
(2) LetM2=e1M,
where e12=e1∈S,
and let M2≅f1M1.
Then M1=a1M,
where a1=f1e1.
Since MR is AGQP-injective, there exist a positive
integer n1 and a left ideal Xa1n1 such that a1n1≠0 and lS(Ker(a1n1))=Sa1n1⊕Xa1n1.
Note that 0≠a1n1M⊆a1M, and a1M is simple. We have a1n1M=a1M.
Clearly, Ker(e1)=Ker(a1) because f1 is a monomorphism. Since a1M is simple, Ker(a1) is a maximal submodule of M.
But Ker(a1)⊆Ker(a1n1)≠M,
so Ker(a1)=Ker(a1n1) and then Ker(e1)=Ker(a1n1).
It follows that e1∈lS(Ker(e1))=lS(Ker(a1n1))=Sa1n1⊕Xa1n1.
Now, let e1=b1a1n1+y with b1∈S and y∈Xa1n1,
then a1n1=a1n1e1=a1n1b1a1n1+a1n1y=a1n1b1a1n1.
Finally, let g1=a1n1b1,
then g12=g1 and M1=a1M=a1n1M=g1M as required.Recall that a moduleM is said to be weakly injective [11] if, for any finitely generated submodule N≤E(M),
there exists X≤E(M) such that N⊆X≅M.Corollary 2.7.
LetM be a finitely generated module. Then,  M is injective if and only if  M is weakly injective and AGQP-injective. In particular, a ring  R is right self-injective if and only if  RR is weakly injective and AGP-injective.Proof.
We
need only to prove the sufficiency. Letx∈E(M).
Then, there exists X⊆E(M) such that M+xR⊆X≅M.
Hence, X is AGQP-injective and M∣X follows from
Theorem 2.6(1). But M is essential in E(M),
so M=X and hence x∈M.Corollary 2.8.
LetMR be an AGQP-injective module with  S=End(MR).
(1)
IfMR
is of finite Goldie dimension, then S is semilocal.(2)
IfMR is a noetherian self-generator, then  S is semiprimary.Proof.
(1)
SinceMR is AGQP-injective, it satisfies the
GC2-condition by Theorem2.6(1) and then (1) follows immediately by [12,
Lemma 1.1].
(2) By (1) and Corollary2.5.Recall that ifM and U are two right R-modules, then U is called M-projective in case for each
epimorphism g:MR→NR and each homomorphism γ:UR→NR,
there is an R-homomorphism γ¯:UR→MR such that γ=gγ¯.
A module MR is called quasiprojective if it is M-projective.LetR be a ring. Recall that an element a∈R is called π-regular if there exists a positive integer m such that am=ambam [13] for some b∈R.
An element x∈R is called generalizedπ-regular if there exists a positive integer n such that xn=xnyx for some y∈R.
A ring R is called  π-regular (resp., generalizedπ-regular) if every element in R is  π-regular (resp., generalized  π-regular). If A is a subset of R,
then we say that A is regular if every element in A is regular.Proposition 2.9.
LetMR be quasiprojective with  S=End(MR). Then,  S is regular if and only if  MR is AGQP-injective and  s(M) is  M-projective for every  s∈S.Proof.
Assume thatS 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 S also implies that s(M) is a direct summand of M by [14, Theorem 37.7]. But M is quasiprojective, so s(M) is M-projective for every s∈S.
Conversely, supposeMR is AGQP-injective and s(M) is M-projective for every s∈S.
Then for any 0≠a∈S,
by the AGQP-injectivity of MR,
there exist a positive integer n and a left ideal Xan of S such that an≠0 and lS(Ker(an))=San⊕Xan.
Since anM is M-projective, Ker(an) = eM for some e2=e∈S.
Then, we have S(1−e)=lS(eM)=lS(Ker(an))=San⊕Xan,
and so 1−e=ban+x for some b∈S and x∈Xan.
Thus, an=an(1−e)=anban+anx=anban.
This proves that S is π-regular and hence generalized π-regular. Clearly, N1(S)={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 moduleMR is called an IN-module [15] if lS(A∩B)=lS(A)+lS(B) for any submodules A and B of M,
where S=End(MR).Proposition 2.10.
LetMR be an AGQP-injective IN-module with  S=End(MR). Then,  S is regular if and only if  W(S)=0.Proof.
By
Theorem2.3, we need only to prove the sufficiency. Let 0≠a∈S.
Since MR is AGQP-injective, there exist a positive
integer n and a left ideal Xan of S such that an≠0 and lS(Ker(an))=San⊕Xan.
Since W(S)=0,
Ker(an) is not essential in M and then there exists a nonzero submodule K such that Ker (an)⊕K is essential in M.
Moveover, we also have(2.7)lS(Ker(an))+lS(K)=lS(Ker(an)∩K)=S,lS(Ker(an))∩lS(K)⊆lS(Ker(an)+K)=0,because MR is an IN-module and W(S)=0.
Thus,(2.8)S=lS(Ker(an))⊕lS(K)=San⊕Xan⊕lS(K).Let 1=ban+x with b∈S, x∈Xan⊕lS(K),
then an=anban.
It follows that S is regular by the last part of the proof of
Proposition 2.9.Lemma 2.11.
LetMR be an AGQP-injective module in which every
nonzero submodule contains a nonzero  M-cyclic submodule and  S=End(MR). If  s∉W(S), then the inclusion  Ker(s)⊆Ker(s−sts) is strict for some  t∈S.Proof.
Ifs∉W(S),
then Ker(s)∩K=0 for some nonzero submodule K of M,
and so Ker(s)∩s′(M)=0 for some 0≠s′∈S by hypothesis. Clearly, ss′≠0.
Since MR is AGQP-injective, there exist a positive
integer n and a left ideal X(ss′)n such that (ss′)n≠0 and lS(Ker(ss′)n)=S(ss′)n⊕X(ss′)n.
Thus,(2.9)s′(ss′)n−1∈lS(Ker(s′(ss′)n−1)=lS(Ker(ss′)n)=S(ss′)n⊕X(ss′)n.Write s′(ss′)n−1=t(ss′)n+x, where t∈S and x∈X(ss′)n,
then (1−ts)s′(ss′)n−1=x and hence(2.10)(1−st)(ss′)n=(s−sts)s′(ss′)n−1=sx∈S(ss′)n∩X(ss′)n.This means that (s−sts)s′(ss′)n−1=0.
It is obvious that Ker (s)⊆ Ker(s−sts). Note that s′(ss′)n−1M is contained in Ker (s−sts) but not contained in Ker(s), the inclusion Ker (s)⊆ Ker (s−sts) is strict.Theorem 2.12.
LetMR be AGQP-injective with  S=End(MR). If every nonzero submodule of  M contains a nonzero  M-cyclic submodule, then the
following conditions are equivalent:
(1)
S
is right perfect;(2)
for any sequence{s1,s2,…}⊆S, the chain Ker(s1)⊆Ker(s2s1)⊆⋯ terminates.Proof.
By
Theorem2.3, Lemma 2.11, and [16,
Lemma 2.8], one can complete the proof in a similar
way to that of [16, Theorem 2.9].



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