(1)ON (m, n)-QUASI-INJECTIVE MODULES Z

ON (m, n)-QUASI-INJECTIVE MODULES

Z. M. ZHU, J. L. CHEN and X. X. ZHANG

Abstract. LetRbe a ring. For two fixed positive integersmandn, anR-moduleM is called (m, n)-quasi-injective if eachR-homomorphism from ann-generated submodule ofM^mtoM extends to one fromM^mtoM. It is showed that MR is (m, n)-quasi-injective if and only if the right R^n×n-moduleM^m×n is principally quasi-injective. Many properties of (m, n)-injective rings and principally quasi-injective modules are extended to these modules. Moreover, some properties of (m, n)-quasi-injective Kasch modules are investigated.

Throughout this paperRandSare associative rings with identities, and all modules are unitary. Unless specified otherwise,m and n will be two fixed positive integers. For an Abelian group G, we writeG^m×n for the set of all formalm×n-matrices with entries in G, and writeGⁿ ( resp. G_n) forG^1×n(resp. for G^n×1). Multiplication maps x 7→ ax and x 7→ xa will be denoted by a· and ·a, respectively. For A = (a_ij)_m×n ∈ G^m×n (resp.

a= (a1, . . . , an)^T ∈Gn), we write πij(A) (resp. πi(a)) foraij (resp. ai). For anyx∈G, we write lij(x) (resp.

li(x))for them×n-matrices (resp. the m×1-matrices) whose (i, j) entry (resp. i-th entry) isxand the others are 0’s. Let SMR be a bimodule. For x ∈ M^m×n, u ∈ S^l×m and v ∈ R^n×k, under the usual multiplication of matrices, ux ( resp. xv) is a well-defined element in M^l×n (resp. M^m×k). If X ⊆ M^l×n, U ⊆ S^l×m and

Received December 7, 2002.

2000Mathematics Subject Classification. Primary 16D50, 16D90.

Key words and phrases. (m, n)-quasi-injective modules, Kasch modules.

V ⊆R^n×k, define

r_R^n×k(X) =

v∈R^n×k |xv= 0,∀x∈X , l_Sm×l(X) =

u∈S^m×l| ux= 0,∀x∈X , r_Mm×n(U) =

y∈M^m×n |uy= 0,∀u∈U , l_Mm×n(V) =

z∈M^m×n |zv= 0,∀v∈V .

1. Characterizations of (m, n)-quasi-injective modules

Firstly, we recall some concepts. A rightR-moduleM_Ris calledprincipally quasi-injective(orPQ-injective in brief) [5] if eachR-homomorphism from a cyclic submodule ofM toM can be extended to an endomorphism ofM. A ringRis said to beright (m, n)-injective[3] in case each rightR-homomorphism from ann-generated submodule ofR^mtoRextends to one fromR^mtoR. A rightR-moduleMRis said to befinitely quasi-injective [8] if eachR-homomorphism from a finitely generated submodule ofM toM extends to an endomorphism ofM. Motivated by these concepts, we introduce the following definition.

Definition 1.1. An R-module M is called (m, n)-quasi-injective in case each R-homomorphism from an n-generated submodule ofM^mtoM extends to one fromM^mtoM. AnR-moduleM is calledn-quasi-injective if it is (1, n)-quasi-injective.

Examples.(1) Every quasi-injective module is (m, n)-quasi-injective for all positive integersmandn[2, Propo- sition 16.13(2)].

(2) R is right (m, n)-injective if and only if RR is (m, n)-quasi-injective.

(3) MR is PQ-injective if and only ifMR is (1,1)-quasi-injective.

(4) MR is finitely quasi-injective if and only ifMRisn-quasi-injective for all positive integers n.

It is easy to see thatMR is (m, n)-quasi-injective if and only if MR is (l, k)-quasi-injective for all 1≤l≤m and 1≤k≤n.

Definition 1.2. A bimodule _SM_R is called left balancedin case every right R-endomorphism of M is left multiplication by an element ofS.

Remark. (1)End(M_R)MR is left balanced for every rightR-moduleMR.

(2) Given a moduleSM, then the bimodule SMEnd(_SM)is left balanced if and only ifSMEnd(_SM) is balanced [2, p. 60].

Theorem 1.3. Let SMR be a left balanced bimodule, then the following statements are equivalent:

(1) MR is(m, n)-quasi-injective.

(2) lMⁿrR_n{α1, α2,· · · , αm}=Sα1+Sα2+· · ·+Sαm for anym-element subset{α1, α2,· · ·, αm} ofMⁿ. (2)⁰ lMⁿrR_n(A) =S^mA for allA∈M^m×n.

(3) If rR_n(A)⊆rR_n(B)where A, B∈M^m×n, then S^mB⊆S^mA.

(4) If z∈Mⁿ andA∈M^m×n satisfyrR_n(A)⊆rR_n(z), thenz∈S^mA.

(5) l_Ml[CRn∩rR_l(A)] =l_Ml(C) +S^mA for all positive integersl,A∈M^m×l andC∈R^l×n. (5)⁰ lMⁿ[CRn∩rRn(A)] =lMⁿ(C) +S^mA for allA∈M^m×n andC∈R^n×n.

(6) The right R-moduleM^m (orM_m) is n-quasi-injective.

Proof. (1)⇔(6), (2)⇔(2)⁰ and (5)⇒(5)⁰ ⇒(2)⁰ ⇒(3) are trivial.

(1)⇔(2). Argue as the proof of [3, Theorem 2.4].

(3) ⇒ (4). Let B = z

0

∈M^m×n. Then rR_n(A)⊆ rR_n(z) =rR_n(B) and S^mB =Sz. By (3), we have Sz=S^mB⊆S^mA. Thereforez∈S^mA.

(4)⇒(5). Letx∈l_Ml[CRn∩rR_l(A)]. For ally∈rR_n(AC),ACy= 0 implies thatCy∈CRn∩rR_l(A). Hence xCy= 0, i.e.,y∈rR_n(xC). Thus

rR_n(AC)⊆rR_n(xC).

By (4),xC =uAC for some u∈S^m. So

x= (x−uA) +uA∈l_Ml(C) +S^mA.

Therefore,

l_Ml[CRn∩rR_l(A)]⊆l_Ml(C) +S^mA.

The inverse inclusion is clear.

Corollary 1.4. LetSMR be a left balanced bimodule. Then

(1) MRis PQ-injective if and only iflMrR(a) =Safor anya∈M if and only ifrR(x)⊆rR(y)wherex, y∈M impliesy∈Sx;

(2) MR is n-quasi-injective if and only if lMⁿrRn(α) =Sα for anyα∈Mⁿ if and only ifrRn(A)⊆rRn(B) where A, B∈Mⁿ impliesB ∈SA;

(3) M_R is(m,1)-quasi-injective if and only if M_R^m(or(M_m)_R)is PQ-injective if and only if l_Mr_R(N) =N for any m-generated submodule N of _SM;

(4) M_R is finitely-quasi-injective if and only ifl_Mnr_Rn(α) =Sαfor all positive integersnand any α∈Mⁿ if and only ifr_Rn(A)⊆r_Rn(B)whereA, B∈Mⁿ impliesB∈SA for all positive integers n.

Theorem 1.5. Let SMR be a left balanced bimodule. Then the following conditions are equivalent.

(1) M_R is(m, n)-quasi-injective.

(2) M_R is (m,1)-quasi-injective and l_Sm(I∩K) = l_Sm(I) +l_Sm(K), where I, K are submodules of (M_m)_R such that I+K is n-generated.

(3) M_R is (m,1)-quasi-injective and l_Sm(I∩K) = l_Sm(I) +l_Sm(K), where I, K are submodules of (M_m)_R such that I is cyclic and K is(n−1)-generated (K= 0 if n= 1).

Proof. (1)⇒(2). It is obvious thatMRis (m,1)-quasi-injective andlS^m(I∩K)⊇lS^m(I)+lS^m(K). Conversely, let x∈ lS^m(I∩K) and define f : I+K → M byf(c+b) = xc for all c ∈ I and b ∈ K. Then f is a right

R-homomorphism. Since MR is (m, n)-quasi-injective and SMR is left balanced, f = y· for some y ∈ S^m. Therefore, for anyc∈Iandb∈K, we haveyc=f(c) =xcandyb=f(b) = 0. This means that

x= (x−y) +y∈l_Sm(I) +l_Sm(K).

(2)⇒(3) is obvious.

(3)⇒(1). We proceed by induction onn. LetK=α1R+α2R+· · ·+αnR be ann-generated submodule of (M_m)_R and f :K→M be a rightR-homomorphism. Write K₁=α₁R, K₂=α₂R+· · ·+α_nR. By induction hypothesis,f|K1=y₁·andf|K2=y₂·for some y₁, y₂∈S^m. Clearly,

y1−y2∈lS^m(K1∩K2) =lS^m(K1) +lS^m(K2).

Supposey1−y2=z1+z2withzi∈lS^m(Ki) (i= 1,2) and lety=y1−z1=y2+z2. Then for anyx=x1+x2∈K withxi ∈Ki (i= 1,2),

f(x) =f(x₁) +f(x₂) =y₁x₁+y₂x₂= (y₁−z₁)x₁+ (y₂+z₂)x₂=y(x₁+x₂) =yx.

Sof =y·and (1) follows.

Corollary 1.6. Given a left balanced bimoduleSMR. (1) The following statements are equivalent:

(i) MR is n-quasi-injective.

(ii) MR is PQ-injective and lS(I∩K) = lS(I) +lS(K), where I, K are submodule of MR andI+K is n-generated.

(iii) MR is PQ-injective and lS(I∩K) =lS(I) +lS(K), where I is a cyclic submodules ofMR andK is an(n−1)-generated submodule ofMR.

(2) MR is finitely quasi-injective if and only if lMrR(x) =Sx for all x∈M and lS(I∩K) =lS(I) +lS(K) for any finitely generated submodules I and K ofMR.

(3) MR is(m,2)-quasi-injective if and only if(Mm)R is PQ-injective and lS^m(αR∩βR) =lS^m(α) +lS^m(β)

for all α, β∈M_m. In particular,M_R is2-quasi-injective if and only if M_R is PQ-injective and lS(xR∩yR) =lS(x) +lS(y)

for all x, y∈M.

Lemma 1.7. Let M be a rightR-module. Iff ∈End(M_R^m×nn×n), then (1) πijf(X) =πijf(Pm

k=1lkj(xkj))for eachX = (xij)∈M^m×n and all 1≤i≤m,1≤j≤n.

(2) πijf lkj=πi1f lk1 for all1≤i≤m,1≤j≤nand1≤k≤m.

Proof. (1) Since

f

m

X

k=1

lkt(xkt)

!

=f(XEtt) =f(X)Ett=

m

X

k=1

lkt(πktf(X)),

we haveπ_ijf _m

P

k=1

l_kt(x_kt)

= 0 in case t6=j. Thus

πijf(X) =πij

" _n X

t=1

f(

m

X

k=1

lkt(xkt))

#

=πijf

m

X

k=1

lkj(xkj)

! .

(2) For anyx∈M,

πijf lkj(x) =πijf(lk1(x)P(1, j)) =πij[f(lk1(x))P(1, j)] =πi1f lk1(x).

So

π_ijf l_kj=π_i1f l_k1.

Corollary 1.8. Given a module MR with S = End(MR). Then a map f : M^m×n → M^m×n is a right R^n×n-homomorphism if and only if f =C· for someC∈S^m×m.

Proof. (⇒) Supposef ∈End(M_R^m×nn×n) and take C= (π_i1f l_k1)_m×m∈S^m×m. Then for eachX = (x_ij)_m×n ∈ M^m×n and all 1≤i≤m, 1≤j≤n, by Lemma1.7, we have

πijf(X) =πijf

m

X

k=1

lkj(xkj)

!

=

m

X

k=1

πijf lkj(xkj) =

m

X

k=1

πi1f lk1(xkj) =πij(CX).

Therefore

f(X) =CX.

(⇐) It is clear.

Theorem 1.9. Given a module MR with S= End(MR). MR is(m, n)-quasi-injective if and only if the right R^n×n-moduleM^m×n is PQ-injective.

Proof. (⇒). Let A, B∈M^m×n withr_R^n×n(A)⊆r_R^n×n(B) and write

B=





 B1

... B_m





.

Then for eachi = 1,2,· · · , m, r_R^n×n(A) ⊆ r_R^n×n(Bi). Consequently rR_n(A)⊆ rR_n(Bi). SinceMR is (m, n)- quasi-injective, by Theorem1.3(4),Bi ∈S^mA(i= 1,2,· · ·, m). SoB=CAfor someC∈S^m×m. Now we define f :M^m×n →M^m×n byf(X) =CX. Thenf ∈End(M_R^m×nn×n) and B=f(A), whenceM_R^m×nn×n is PQ-injective by Corollary1.4(1).

(⇐) Suppose z ∈ Mⁿ, A ∈ M^m×n and rR_n(A) ⊆ rR_n(z). Let B = z

0

∈ M^m×n. Then r_Rn×n(A) ⊆ r_R^n×n(B). SinceM_R^m×nn×n is PQ-injective,B =CAfor someC∈S^m×mby Corollary1.4(1) and Corollary 1.8. It follows thatz∈S^mA. By Theorem1.3(4), we see thatMR is (m, n)-quasi-injective.

Corollary 1.10. A ring R is right(m, n)-injective if and only if the rightR^n×n-moduleR^m×n is PQ-injective.

In particular, R is right(n, n)-injective if and only if M_n(R) is P-injective.

By Theorem1.9, Corollary 1.4and Corollary1.8, we have

Corollary 1.11. MR is finitely quasi-injective if and only if the right R^n×n-module Mⁿ is PQ-injective for all positive integers n if and only if lMⁿr_Rn×n(x) = Sx for all positive integers n and all x ∈ Mⁿ, where S= End(MR).

2. Properties of(m, n)-quasi-injective modules

In this section, some known results on PQ-injective modules and principally injective rings are extended to (m, n)-quasi-injective modules.

We begin with the following theorem, which extends [5, Proposition 1.2].

Theorem 2.1. Given a left balanced bimoduleSMR with MR (m, n)-quasi-injective andA, B∈M^m×n. (1) If (BR_n)_R embeds in (AR_n)_R, then_S(S^mB)is an image of _S(S^mA).

(2) If (AR_n)_R is an image of (BR_n)_R, then_S(S^mA)embeds in _S(S^mB).

(3) If (BRn)R∼= (ARn)R, thenS(S^mA)∼=S(S^mB).

Proof. Ifσ : BR_n → AR_n is a rightR-homomorphism, then the (m, n)-quasi- injectivity of M_R forces σ = g|BRn for someg ∈End((M_m)_R). Let D= (π_igl_j)_m×m. Theng =D·. But SM_R is let balanced, sog =C·for some C ∈ S^m×m. Choose u1, u2,· · ·, un ∈ Rn such that σ(Bei) =Aui, where ei= (0,· · ·,0,1,0,· · · ,0)^T∈Rn

(with 1 in theith position and 0’s in all the other positions), i= 1,2,· · · , n. LetU = (u1, u2,· · · , un). Then AU = (Au₁, Au₂,· · · , Au_n) = (σ(Be₁), σ(Be₂),· · ·, σ(Be_n))

= (CBe1, CBe2,· · ·, CBen) =CB.

Now we defineϕ:S^mA→S^mB byyA7→yAU. Thenϕis a leftS-homomorphism.

(1) Ifσis a monomorphism, then for anyx= (x1, x2,· · · , xn)^T ∈rR_n(AU),

σ(Bx) =σ

n

X

i=1

Beixi

!

=

n

X

i=1

σ(Bei)xi=

n

X

i=1

(Aui)xi=0

follows that

Bx= 0.

Thusr_Rn(AU)⊆r_Rn(B). By Theorem 1.3(3),S^mB⊆S^mAU. ButS^mAU =S^mCB⊆S^mB, soS^mB=S^mAU. Henceϕis an epimorphism.

(2) Supposeσ is an epimorphism. LetAe_i =σ(Bv_i),v_i∈R_n,i= 1,2,· · ·, n, and write V= (v₁, v₂,· · ·, v_n).

ThenV ∈R^n×n and A=CBV. Thus, if ϕ(yA) = 0, then yAU = 0, i.e.,yCB = 0, whenceyA =yCBV = 0.

Thereforeϕis a monomorphism.

(3) By (1) and (2).

The next theorem extends [5, Lemma 1.2].

Theorem 2.2. Suppose thatSMR is left balanced andMR is(m, n)-quasi-injective. Then l_Sk[r_Mk(A)∩BR_n] =S^mA+l_Sk(B)

for all positive integers k,A∈S^m×k andB ∈M^k×n.

Proof. Let x ∈ l_Sk[rM_k(A)∩BRn]. For all y ∈ rR_n(AB), we have ABy = 0. This implies that By ∈ rM_k(A)∩BRn. SoxBy= 0, i.e.,y ∈rR_n(xB). ThusrR_n(AB)⊆rR_n(xB). Since MR is (m, n)-quasi-injective, by Theorem1.3(4),xB=u(AB) for someu∈S^m. Thenx−uA∈l_Sk(B). Hence

x=uA+ (x−uA)∈S^mA+l_Sk(B).

Therefore

l_Sk[rM_k(A)∩BRn]⊆S^mA+l_Sk(B).

The inverse inclusion is obvious.

Corollary 2.3. Let MR be (m, n)-quasi-injective. If α1, α2, . . . , αm ∈ S = End(MR), x1, x2,· · · , xn ∈ M, then

l_S



(

m

\

i=1

Kerα_i)∩

n

X

j=1

x_jR)



=

m

X

i=1

Sα_i+

n

\

j=1

l_S(x_j).

Proof. Takek= 1,A= (α₁, . . . , α_m)^T andB= (x₁, x₂,· · ·, x_n) in Theorem2.2and then the result follows.

Corollary 2.4. LetMR be an n-generated(m, n)-quasi-injective module withS= End(MR). Then (1) lS

m

\

i=1

Kerαi

!

=

m

X

i=1

Sαi for any α1, α2, . . . , αm∈S.

(2) If α_i, β_i ∈ S (i = 1,2,· · ·, m) satisfy

m

\

i=1

Kerα_i ⊆

m

\

i=1

Kerβ_i, then

βi∈

m

X

i=1

Sαi (i= 1,2,· · · , m).

Take MR = xR, k = n, A =





 α1

... αm





 and B =





 x

. .. x







n×n

in Theorem 2.2. Then we have the

following corollary.

Corollary 2.5. LetMR be a cyclic(m, n)-quasi-injective module with S= End(MR). Then

lSⁿrMn{α1, α2,· · ·, αm}=

m

X

i=1

Sαi

for any α1,α2,· · ·, αm∈Sⁿ.

LetMR be a module withS = End(MR), write W(S) ={w∈S|Ker(w)M}. Then W(S) =J(S) in case MR is a cyclic PQ-injective module [5, Proposition 2.4]. For the case ofn-quasi-injective modules, we have

Lemma 2.6. If M_R isn-quasi-injective andn-generated, then W(S) =J(S), whereS= End(M_R).

Proof. Ifa ∈W(S), then rM(a) = KeraM, and this forces rM(1−a) = 0, i.e.,lSrM(1−a) = S. Since MR isn-quasi-injective andn-generated, we haveS(1−a) =S by Corollary2.4. This means thatW(S)⊆J(S).

Conversely, leta∈J(S). For anyx∈M, ifrM(a)∩xR= 0, thenlS[rM(a)∩xR] =S. So we haveSa+lS(x) =S by Corollary2.3. It follows thatlS(x) =S, i.e., x= 0. ThereforerM(a)M, that is,a∈W(S).

Given a module M_R. We call U(6=0)∈M^m×n aright uniform element if U R_n is a uniform submodule of (M_m)_R, and writeM_U={x∈S^m|rMm(x)∩U R_n6=0}.

Lemma 2.7. Let MR be(m, n)-quasi-injective withS= End(MR). IfU ∈M^m×n is a right uniform element, thenMU is the unique maximal submodule of SS^m which containslS^m(U).

Proof. Since U R_n is a uniform submodule of (M_m)_R, M_U is a submodule of _SS^m. It is easy to see that lS^m(U) ⊆MU 6= S^m. If A ∈ S^m\MU, then rM_m(A)∩U Rn = 0. SolS^m(rM_m(A)∩U Rn) = S^m. Let A =

A 0

∈S^m×m. ThenrM_m(A) =rM_m(A) and S^mA=SA. But MR is (m, n)-quasi-injective, by Theorem 2.2, SA+lS^m(U) = S^m. Hence SA+MU =S^m. ThereforeMU is a maximal submodule of SS^m which contains lS^m(U). Now, iflS^m(U)⊆SL$S^m, thenL⊆MU (otherwise, ifA∈L\MU, thenlS^m(U) +SA=S^mas before.

So we haveL=S^m, a contradiction). This completes the proof.

Lemma 2.8. Let MR be (m, n)-quasi-injective with S = End(MR) and W = U1Rn ⊕ · · · ⊕UtRn, where Ui ∈M^m×n are right uniform elements, i= 1,2,· · · , t. If SL is a maximal submodule ofSS^m not of the form MU for any right uniform elementU ∈M^m×n, thenrM_m(Em−A)∩WW for someA∈Lm.

Proof. Since L 6= MU₁, so rM_m(x)∩U1Rn = 0 for some x ∈ L, thus rR_n(xU1) ⊆ rR_n(U1). Let B = (xU1,0)^T ∈M^m×n. ThenrR_n(B) =rR_n(xU1)⊆rR_n(U1). Since MR is (m, n)-quasi-injective, S^mU1⊆S^mB by Theorem 1.3(3). Letε1 = (1,0,· · ·,0), ε2 = (0,1,0,· · ·,0), · · ·, εm = (0,· · ·,0,1) ∈S^m and supposeεiU1 = sixU1 for some si ∈ S (i = 1,2,· · · , m). WriteA1 = (s1x, . . . , smx)^T. Then A1 ∈ Lm and (Em−A1)U1 = 0.

So rM_m(Em−A1)∩U1Rn 6= 0. If rM_m(Em−A1)∩U2Rn = 0, then (Em−A1)U2Rn ∼= U2Rn is a unform right R-module. Hence (Em−A2)(Em−A1)U2 = 0 for some A2 ∈ Lm. Let A3 = A1+A2−A2A1. Then (E_m−A₃)U₁ = (E_m−A₃)U₂ = 0. Thusr_Mm(E_m−A₃)∩U_iR_n 6= 0, i = 1,2. Continue in this way to obtain

A∈L_m such thatr_Mm(E_m−A)∩WW.

The following theorem extends [6, Theorem 3.3]. We complete this section with it and two corollaries.

Theorem 2.9. Let MR be an n-generated n-quasi-injective and finite dimensional module withS= End(MR).

(1) If L⊆S is a maximal left ideal, thenL=MU for some right uniform element U ∈Mⁿ. (2) S/J(S)is semisimple artinian.

Proof. SinceMRis finite dimensional, we may assumeW =U1Rn⊕ · · · ⊕UtRnMR, whereU1,· · · , Ut∈Mⁿ and eachUiRn is uniform [4, Proposition 3.19]. IfSLis a maximal left ideal ofSS not of the formMU for any right uniform elementU ∈Mⁿ, thenrM(1−a)∩WW for somea∈Lby Lemma2.8. So 1−a∈J(S)⊆Lby

Lemma2.6, a contradiction. Thus (1) follows. As to (2), ifa∈MU₁∩MU₂∩ · · · ∩MU_t, thenrM(a)∩UiRn6= 0, i= 1,2,· · · , t. Hence

t

M

i=1

[rM(a)∩UiRn]MR

because eachUiRn is uniform. This meansrM(a)MR. By Lemma2.6,a∈J(S). But eachMU_i is maximal in

SS by Lemma 2.7, so

J(S) =M_U1∩M_U2∩ · · · ∩M_Ut.

ThereforeS/J(S) is semisimple artinian.

Corollary 2.10. If M_R is finitely quasi-injective finite dimensional and finitely generated, then S/J(S) is semisimple artinian, whereS= End(M_R).

Corollary 2.11. If M_R is an n-quasi-injective and n-generated uniform module, thenS= End(M_R)is local.

3. (m, n)-quasi-injective Kasch modules

Following Albu and Wisbauer [1], a rightR-moduleMRis called aKaschmodule if any simple module inσ[MR] embeds inMR, whereσ[M] is the category consisting of allM-subgenerated rightR-modules [9, p. 118]. In this section, we study some properties of (m, n)-quasi-injective (in particular, n-quasi-injective) Kasch modules.

Recall that a bimoduleSMRis said to befaithfully balanced[2] in case the canonical ring homomorphisms λ:S→End(MR) andρ:R→End(SM) are isomorphisms.

Proposition 3.1. If SMR is faithfully balanced and MR is an(n, m+ 1)-quasi-injective Kasch module, then

SM is(m, n)-quasi-injective.

Proof. Letα1, α2,· · · , αm∈Mn. Then

N =α1R+· · ·+αmR⊆rM_nlSⁿ{α1, . . . , αm}.

Assume β ∈ rM_nlSⁿ{α1, . . . , αm} but β∈N. Then NR ⊆ LR for some maximal submodule LR of βR+NR. Since (βR+N)/L is a simple module inσ[MR], there exists a monomorphism δ : (βR+N)/L→MR. Define f : βR+N → MR by f(x) = δ(x+L). Then f(αi) = 0 for all i = 1,2,· · · , m, but f(β) 6= 0. Note that MR is (n, m+ 1)-quasi-injective and βR+N is an (m+ 1)-generated submodule of (Mn)R, so f(x) = ux for some u ∈ (End(MR))ⁿ. And hence there exists v ∈ Sⁿ such that f(x) = vx for SMR is balanced. Thus vαi = 0, i = 1,2,· · · , m, i.e., v ∈lSⁿ{α1, α2,· · ·, αm}. This implies that f(β) = vβ = 0, a contradiction. So N =rM_nlSⁿ{α1,· · · , αm}, whenceSM is (m, n)-quasi-injective.

Corollary 3.2. [3, Theorem 2.7] If R is right Kasch and right (n, m+ 1)-injective, then R is left (m, n)- injective.

Our next theorem extends [6, Lemma 2.3].

Theorem 3.3. Given a left balanced bimoduleSMR. If MR is l-generated and ln-quasi-injective and Kasch, thenlSⁿ(Jn)SSⁿ, whereJ = Rad(MR).

Proof. If 06=a∈Sⁿ, then choose a maximal submoduleAof the rightR-moduleaMn. Letσ:aMn/A→MR

be a monomorphism and defineα:aMn→MR byα(x) =σ(x+A). SinceaMn is anln-generated submodule of theln-quasi-injective module MR,αextends to an endomorphism ofM. Thenα=s0· for somes0∈S because

SMR is left balanced. Choose y ∈Mn such thatay∈A. Then s0ay =α(ay) =σ(ay+A)6= 0. So s0a6= 0. If aJn *A, thenaTn+A=aMn. Now, leta= (s1,· · ·, sn). Then si(Rad(MR))siM (i= 1,2,· · ·, n) forMR

is finitely generated. This follows that

n

X

i=1

si(RadMR)

n

X

i=1

si(MR), i.e., aJn aMn.

HenceA=aMn, a contradiction. ThusaJn ⊆A and it implies that (s0a)Jn=α(aJn) =σ(0) = 0.

So 06=s0a∈Sa∩lSⁿ(Jn). ThereforelSⁿ(Jn)SSⁿ. Corollary 3.4. Given a cyclic module MR with S = End(MR), if MR is PQ-injective and Kasch, then lS(J)SS, whereJ = Rad(MR).

Corollary 3.5. Given a finitely generated moduleMR with S = End(MR). If MR is finitely quasi-injective and Kasch, thenlSⁿ(Jn)SSⁿ for all positive integers n, whereJ = Rad(MR).

Lemma 3.6. Given a module M_R with S = End(M_R). If Rad(M_R) 6= M_R and consider the following conditions:

(1) M_R is a Kasch module.

(2) l_Sn(T)6= 0 for all positive integers nand for any maximal submodule T of(M_n)_R. (3) l_Sn(T)6= 0 for some positive integern and for any maximal submoduleT of (M_n)_R. (4) lS(T)6= 0 for any maximal submoduleT of MR.

Then we always have the following implications:

(1) ⇒ (2) ⇒ (3) ⇒ (4).

IfMRgenerates all simple modules inσ[M](in particular, ifMRis a generator inσ[M]), then we have(4)⇒(1).

Proof. Since Rad(M)6=M, soM (and henceM_n) has maximal submodules.

(1) ⇒ (2). Let ϕ : M_n/T → M_R be a monomorphism, define f : M_n → M by x 7→ ϕ(x+T), and write a= (f l1, f l2,· · ·, f ln). Then 06=a∈Sⁿ andaT =f(T) = 0. SolSⁿ(T)6= 0.

(2)⇒(3) is clear.

(3)⇒ (4). Ifn= 1, the implication holds. Now we assume n >1. Let T be any maximal submodule ofM, writeK=

T Mn−1

, and defineϕ:Mn/K →M/T via x

y

+K7→x+T, where x∈M,y∈M_n−1. Then ϕis a right R-isomorphism. This means thatK is a maximal submodule ofMn. Hence lSⁿ(K)6= 0. Suppose 06= (u, v)∈lSⁿ(K), whereu∈S andv∈Sⁿ⁻¹. Then 06=u∈lS(T).

Lastly, assumeM generates all simple R-modules inσ[M] and (4) holds. Then for every simple moduleAR

in σ[M], there exists a maximal submodule T of M such that A ∼= M/T. Suppose 0 6= s0 ∈ lS(T). Then T ⊆rM(s0)6=M. HenceT =rM(s0). Now we defineϕ:M/T →M byx+T 7→s0x. Then it is easy to see that

ϕis anR-monomorphism.

The following theorem is an extension of [7, Theorem 1.2].

Theorem 3.7. Let MR be an n-quasi-injective cyclic Kasch module with S = End(MR). Then the map K7→rM_n(K)andT 7→lSⁿ(T)are mutually inverse bijections between the set of all minimal submodules ofSSⁿ and the set of all maximal submodules of(Mn)R. In particular,

(1) lSⁿrM_n(K) =K for all minimal submodulesK of SSⁿ. (2) rM_nlSⁿ(T) =T for all maximal submodules T of (Mn)R.

Proof. (1) follows from Corollary2.5. As to (2), observe that T ⊆r_Mnl_Sn(T) and thatr_Mnl_Sn(T)6=M_n by Lemma3.6. The proof is completed by establishing the following claims.

Claim 1. rM_n(K)is a maximal submodule of (Mn)R for each minimal submoduleK of SSⁿ.

Proof. Letr_Mn(K)⊆T, whereT is a maximal submodule ofM_n. Then 06=l_Sn(T)⊆l_Snr_Mn(K) =K by (1).

Sol_Sn(T) =K becauseK is minimal in_SSⁿ. Hencer_Mn(K) =r_Mnl_Sn(T) =T by (2).

Claim 2. lSⁿ(T)is a minimal submodule ofSSⁿ for all maximal submodulesT of (Mn)R.

Proof. SinceM_Ris Kasch, by Lemma3.6(2), we may choose 06=x∈l_Sn(T). ThenT ⊆r_Mn(x)6=M_n, whence T =r_Mn(x). AsM_R isn-quasi-injective and cyclic, this givesl_Sn(T) =l_Snr_Mn(x) =Sxby Corollary 2.5and it

follows thatl_Sn(T) is a minimal submodule of_SSⁿ.

Acknowledgment. This work was supported in part by the Teaching and Research Award Program for Outstanding Young Teachers in Higher Education Institutions of MOE, China, and the Foundation of Graduate Creative Program of JiangSu (No. xm04-10),

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Z. M. Zhu, Department of Mathematics, Jiaxing University, Jiaxing, Zhejiang 314001, P. R. China, e-mail:zhanmin [email protected]

J. L. Chen, Department of Mathematics, Southeast University Nanjing 210096, P. R. China, e-mail:[email protected]

X. X. Zhang, Department of Mathematics, Southeast University Nanjing 210096, P. R. China, e-mail:[email protected]