on small injective rings and modules - CSDL Khoa học

c World Scientific Publishing Company

ON SMALL INJECTIVE RINGS AND MODULES^∗

LE VAN THUYET^†and TRUONG CONG QUYNH^‡ Department of Mathematics

Hue University, Vietnam

†lvthuyethue@gmail.com

‡tcquynh@dce.udn.vn

Received 26 April 2007 Accepted 19 November 2008 Communicated by D. V. Huynh

A rightR-module MR is calledsmall injective if every homomorphism from a small right ideal toMR can be extended to anR-homomorphism fromRR toMR. A ringR is called right small injective, if the rightR-moduleRRis small injective. We prove that Ris semiprimitive if and only if every simple right (or left)R-module is small injective.

Further we show that the Jacobson radicalJof a ringRis a noetherian rightR-module if and only if, for every small injective moduleER,E^(N)is small injective.

Keywords: Small injective rings (modules); PF rings; QF rings.

2000 Mathematics Subject Classification: 16D50, 16D70, 16D80

1. Introduction

Throughout the paper,Rrepresents an associative ring with identity 1= 0 and all modules are unitaryR-modules. We writeM_R (resp.,_RM) to indicate thatM is a right (resp., left)R-module. ByJ (resp.,Z_r,S_r) we denote the Jacobson radical (resp., the right singular ideal, the right socle) of R. For a module M_R, E(M_R) stands for its injective hull. IfX is a subset ofR, the right (resp., left) annihilator of X in R is denoted by r_R(X) (resp.,l_R(X)) or simplyr(X) (resp., l(X)) if no confusion appears. IfN is a submodule of M (resp., proper submodule) we denote it byN ≤M (resp., N < M). Moreover, we write N ≤^e M, N M, N ≤^⊕ M andN ≤^max M to indicate thatN is an essential submodule, a small submodule, a direct summand and a maximal submodule ofM, respectively. A module M is called uniform ifM = 0 and every nonzero submodule ofM is essential inM.

Recall that a ringRis called right mininjectiveif every homomorphism from a minimal right ideal toRis given by left multiplication by an element ofR.Ris called

∗The work was supported by the Natural Science Council of Vietnam.

379

right Kaschif every simple right R-module embeds in R; or equivalently,l(I)= 0 for every maximal right ideal I of R. A ring R is called quasi-Frobenius (briefly, QF-ring) if it is right (or left) artinian and right (or left) self-injective. R is said to be right pseudo-Frobenius (briefly,PF-ring) ifR_Ris an injective cogenerator in the category of rightR-modules.

Yousif and Zhou [13] proved that, for a semiperfect ring R with an essential right socle,R is right self-injective if and only ifR is right small injective. In [10], Shen and Chen proved that if R is semilocal, then R is right self-injective if and only ifR is right small injective. They also gave some new characterizations of QF rings and right PF rings in terms of small injectivity.

In this paper, we show that R is PF if and only if R is right small injective, right Kasch and left min-CS. We also prove that if every simple right (resp., left) R-module is small injective, thenRis semiprimitive. Finally, using the small injec- tivity, we give some characterizations for the Jacobson radical of a ring R to be a noetherian rightR-module.

General background materials can be found in [2, 4, 8, 12].

2. Results

A module M_R is called small injectiveif every homomorphism from a small right ideal toM_R can be extended to anR-homomorphism fromR_RtoM_R. A ringRis called right small injective ifR_R is small injective.

Examples.(i) LetR=Zbe the ring of integers, thenRis small injective but not self-injective.

(ii) LetR={(^{n x}_{0 n})|n∈Z, x∈Z2}(see [13, Example 1.6]). ThenR is a com- mutative ring andJ =S_r={(⁰_{0 0}^x)|x∈Z2}. ThereforeR is small injective.

We claim that R is not injective. Let I ={(²ⁿ_{0 2n}⁰ )|n ∈Z} be an ideal of R and g : I → R with f((²ⁿ_{0 2n}⁰)) = (^{0 ¯}_{0 0}ⁿ), then g is a homomorphism. Assume that ¯g : R → R be a homomorphism that extends g. There exists (ⁿ₀¹ _n^x1

1) ∈ R such that ¯g((^{n x}_{0 n})) = (ⁿ₀¹ _n^x1

1)(^{n x}_{0 n}) for all (^{n x}_{0 n}) ∈ R. Thus f((²ⁿ_{0 2n}⁰ )) = (ⁿ₀¹ _n^x1

1)(²ⁿ_{0 2n}⁰ ). It implies that (^{0 ¯}_{0 0}ⁿ) = (^2nx₀^{1 0}₀). This is a contradiction.

A ringR is defined to be right minsymmetric (cf. [8]) if, fork ∈R, whenever kRis a minimal right ideal ofR thenRk is a minimal left ideal ofR.

Next, we show the following lemmas.

Lemma 2.1 (McCoy’s Lemma). Let R be a ring anda, c∈R. Ifb=a-aca is a regular element of R,then so is a.

Proof. This follows easily from the definition.

Lemma 2.2. LetRbe a right minsymmetric ring withS_r≤^eR_R. If the ascending chain r(a₁) ≤ r(a₂a₁) ≤ · · ·r(a_na_n−1· · ·a₁) ≤ · · · terminates for every infinite sequence a₁, a₂, . . . inR,thenR is right perfect.

Proof. SinceRis right minsymmetric,S_r≤S_l. ThenJ ≤l(S_r) =Z_rsinceS_r≤^e R_R. Now we prove thatZ_ris rightT-nilpotent. In fact that, letx₁, x₂, . . . , x_n, . . .∈ Z_r. We have

r(x₁)≤r(x₂x₁)≤ · · ·.

By hypothesis, there exists k ∈ N such that r(x_k· · ·x₂x₁) = r(x_k+1x_k· · ·x₂x₁).

If x_k· · ·x₂x₁ = 0, then x_k· · ·x₂x₁R∩r(x_k+1) = 0 (since x_k+1 ∈ Z_r). There exists r ∈ R such that 0 = x_k· · ·x₂x₁r ∈ r(x_k+1) and so x_k+1x_k· · ·x₂x₁r = 0.

That means r ∈r(x_k+1x_k· · ·x₂x₁) = r(x_k· · ·x₂x₁) or x_k· · ·x₂x₁r= 0, this is a contradiction. Hencex_k· · ·x₂x₁= 0. Thus Z_r is rightT-nilpotent. It implies that Z_r≤J. ThusJ =l(S_r) =Z_r. And thenJ is also right T-nilpotent.

Now we prove that R/J is a von Neumann regular ring. Let a₁ ∈ J, then r(a₁) is not an essential right ideal of R. Hence there exists I ≤ R_R such that I = 0 and I ∩r(a₁) = 0. But S_r ≤^e R_R, there exists a simple right ideal bR of R such that bR ≤ I. Then bR∩r(a₁) = 0, that is a₁b = 0. Therefore R=l(bR∩r(a₁)) =l(b) +Ra₁ by [8, Proposition 2.26], write 1 =c₁a₁+t,where t∈l(b), c₁ ∈R. So b=c₁a₁b. It is easy to see that 0=b∈r(a₁−a₁c₁a₁)\r(a₁), r(a₁) < r(a₁−a₁c₁a₁). Put a₂ =a₁−a₁c₁a₁. We denote by ¯a =a+J ∈ R/J. If a₂ ∈ J, then we have ¯a₁ = ¯a₁¯c₁¯a₁, i.e., ¯a₁ is a regular element of R/J. If a₂ ∈ J, there exists a₃ ∈ R such that r(a₂) < r(a₃) with a₃ = a₂−a₂c₂a₂ for somec₂ ∈R by the preceding proof. Repeating the above-mentioned process, we get a strictly ascending chain r(a₁) < r(a₂) < · · ·, where a_i+1 = a_i−a_ic_ia_i for somec_i∈R,i= 1,2, . . . .Letb₁=a₁, b₂= 1−a₁c₁, . . . , b_i+1= 1−a_ic_i, . . . ,then a₁ = b₁, a₂ = b₂b₁, . . . , a_i+1 = b_i+1b_i· · ·b₂b₁, . . . , whence we have the following strictly ascending chainr(b₁)< r(b₂b₁)<· · ·,which contradicts the hypothesis. So there exists a positive integer m such that a_m+1 ∈ J, i.e., a_m−a_mc_ma_m ∈ J.

This shows that ¯a_m is a regular element of R/J, and hence ¯a_m−1,¯a_m−2, . . . ,

¯a₁ are regular elements of R/J by Lemma 2.1, i.e., R/J is von Neumann regular.

To show that R/J is semisimple, by [7, Corollary 2.16], we only need to prove thatR/J contains no infinite sets of nonzero orthogonal idempotents. This can be proved by arguing as [3, p. 2107].

From above lemma, we give some properties of PF and QF rings.

Corollary 2.3. LetRbe a right small injective ring with S_r≤^eR_R. If the ascend- ing chain r(a₁)≤r(a₂a₁)≤ · · ·r(a_na_n−1· · ·a₁)≤ · · ·terminates for every infinite sequence a₁, a₂, . . . inR,thenR is right PF.

Proof. By Lemma 2.2, R is right semiperfect. But sinceR is right small injec- tive, R is right self-injective by [10, Theorem 3.16]. Thus R is right self-injective, semiperfect andS_r≤^eR_R; i.e.,Ris right PF by [6, Theorem 2.1].

Corollary 2.4. A ring R is QF if and only if R is right small injective with S_r ≤^e R_R, l(J²) is countable generated as a left ideal and the ascending chain r(a₁) ≤ r(a₂a₁) ≤ · · · ≤ r(a_na_n−1· · ·a₁) ≤ · · · terminates for every infinite sequence a₁, a₂, . . .in R.

Proof. By Lemma 2.2, [9, Lemma 2.2] and [13, Theorem 2.18].

With a left and right small injective ring, we have:

Proposition 2.5. LetR be a ring. Then the following conditions are equivalent:

(1) Ris QF.

(2) R is a right and left small injective ring with S_r ≤^e R_R and the ascending chain r(a₁) ≤ r(a₂a₁) ≤ · · · ≤ r(a_na_n−1· · ·a₁) ≤ · · · terminates for every infinite sequencea₁, a₂, . . .inR.

Proof. (1)⇒(2) is clear.

(2)⇒(1). By Lemma 2.2,R is right perfect. But R is also left small injective which yieldsR is left self-injective. ThusR is QF by [6, Corollary 2.3].

Remark. The condition “S_r ≤^e R_R” in Proposition 2.5 can be not omitted. Let R=Zbe the ring of integer numbers, thenR is small injective, noetherian butR is not QF.

A ringR is calledleft CS(resp.,left min-CS) if every left ideal (resp., minimal left ideal) is essential in a direct summand of_RR. It is well-known that R is right PF if and only ifRis right self-injective, right Kasch. But it is unknow whether a right small injective, right Kasch ring is right PF.

Theorem 2.6. LetR be a ring. Then the following conditions are equivalent:

(1) Ris right PF.

(2) Ris right small injective,right Kasch and left min-CS.

(3) Ris right small injective,right Kasch andlr(a)is essential in a direct summand of_RR for every simple right(resp., left)idealaR (resp., Ra)ofR.

Proof. (1)⇒(2) is clear.

(2)⇒(3).Assume thatRais a simple left ideal. If (Ra)²= 0 thenRa≤^⊕_RR which implies that lr(a) =Ra. Otherwise,a² = 0 and so a∈ J. Since R is right small injective,Ra=lr(a) is a simple left ideal. Then by (2), lr(a) is essential in a direct summand of_RR. On the other hand, ifbRis a simple right ideal thenRbis

a simple left ideal (becauseRis right minsymmetric). It is easy to see thatlr(b) is essential in a direct summand of_RR.

(3)⇒(1).LetTbe a maximal right ideal ofR. SinceRis right Kasch,l(T)= 0.

There exists 0=a∈l(T) orT ≤r(a) which yieldsT =r(a). ButaR∼=R/r(a) and soaRis a right simple ideal. Thereforel(T) =lr(a)≤^eRefor somee²=e∈Rby hypothesis. ThusR is semiperfect by [8, Lemma 4.1]. This implies that Ris right self-injective and so right PF by [8, Corollary 7.32].

Corollary 2.7. If R is right small injective, right Kasch and left CS, then R is right PF.

Recall that a ringRis called semiprimitiveifJ = 0.

Theorem 2.8. Let R be a ring. Then the following conditions are equivalent:

(1) R is semiprimitive.

(2) Every right (or left)R-module is small injective.

(3) Every simple right (or left) R-module is small injective.

Proof. It is clear that (1)⇒(2)⇒(3).

(3)⇒(1). Suppose thatJ = 0, let 0=a∈J, that isaRR_R. IfJ+r(a)< R, then we take a maximal right ideal I of R such that J +r(a) ≤I. Then R/I is small injective by (3). We define ϕ : aR → R/I by ϕ(ar) = r+I. Then ϕ is a well-defined R-homomorphism. So there exists c ∈ R such that 1 +I = ca+ I and then 1−ca ∈ I. But ca ∈ J ≤ I which yields 1 ∈ I, a contradiction.

ThereforeJ+r(a) =R and sor(a) =R (because J R_R). Soa= 0, which is a contradiction.

Proposition 2.9. If every simple singular right R-module is small injective, then for everya∈J, r(a)≤^⊕ R_R andaR is projective.

Proof. For everya ∈ J, letL = RaR+r(a). There exists K_R ≤ R_R such that L⊕K ≤^e R_R. Assume thatL⊕K =R, then there exists a maximal right ideal I of R such that L⊕K ≤ I and so I ≤^e R_R. Therefore R/I is small injective by hypothesis. We define ϕ : aR → R/I by ϕ(ar) = r+I. Then ϕ is a well- definedR-homomorphism. So there existsc∈R such that 1 +I=ca+I and then 1−ca∈I. Butca∈RaR≤Iwhich yields 1∈I, a contradiction. ThusL⊕K=R or RaR+ (r(a)⊕K) = R which implies that r(a)⊕K =R (since RaR R_R).

Then r(a) = (1−e)R for some e² = e ∈ R and it follows that a_sx = a_sx. Let ψ :eR →aeR be defined by ψ(er) =aer for all r ∈R. Thenψ is a well-defined R-epimorphism. It is easy to see that Ker(ψ) = eR∩r(a) = 0. Hence ψ is an isomorphism and thenaR=aeRis projective.

Corollary 2.10. If every simple singular right R-module is small injective, then Z_r∩J = 0.

Recall that a ringRis calledzero insertive(briefly ZI) [11], if fora, b∈R, ab= 0 impliesaRb= 0. Note that ifR is a ZI ring, then every idempotent inR is central andr(a), l(a) are two-sided ideals witha∈R.

Lemma 2.11. If R is a ZI ring, thenRaR+r(a)is an essential right ideal of R, for every a∈R.

Proof. Given a∈Rand assume that (RaR+r(a))∩I= 0, whereI≤R_R. Then aI≤I∩RaR= 0. HenceI≤r(a); whence I= 0.

Proposition 2.12. Let R be a ZI ring. If every simple singular right (or left) R-module is small injective, thenR is semiprimitive.

Proof. Suppose that there exists 0=a∈J; whenceRaRR_R. IfRaR+r(a)<

R, then there exists a maximal right idealI of R such that RaR+r(a) ≤I. By Lemma 2.11,I is an essential right ideal ofR. Therefore R/Iis small injective by hypothesis. Let ϕ : aR → R/I be defined viaϕ(ar) = r+I for all r ∈ R. It is easy to see thatϕis a well-definedR-homomorphism. SinceR/Iis small injective, there existsc∈Rsuch that 1 +I=ϕ(a) =ca+I. Butca∈RaR≤I which yields 1 ∈ I, a contradiction. Therefore RaR+r(a) = R which implies that r(a) = R (since RaRR_R). Soa= 0, which is a contradiction.

Finally we consider the direct-sum representation of small injective module.

Let M be a right R-module. We denote that r_J(N) = {a ∈ J|Na = 0} and l_M(I) = {m ∈ M|mI = 0} where N ⊆ M and I ⊆ J. Then r_J(X) ≤ J_R and l_M(I)≤SM whereS= End(M_R).

Lemma 2.13. The following conditions are equivalent for a right R-moduleM: (1) Rsatisfies the ACC for right ideals of form the r_J(X), whereX ⊆M.

(2) For each right ideal I_R ≤J_R there corresponds a finitely generated right ideal I₁≤I_R such that l_M(I) =l_M(I₁).

Proof. (1) ⇒(2). The condition that R satisfies the ACC for right ideals of the form r_J(X), where X ⊆ M is equivalent to that R satisfies the DCC for l_M(A) where A ⊆J. LetI₁ be a finitely generated right ideal of I_R such that l_M(I₁) is minimal in the set

Ω ={l_M(K)|K_Ris finitely generated andK_R≤I_R}.

Ifx∈I, thenH =I₁+xR≤I_R is finitely generated andl_M(H)≤l_M(I₁). By the choice ofI₁, we havel_M(H) =l_M(I₁), sol_M(I₁)x= 0. It implies thatl_M(I₁)I= 0, that is l_M(I₁)≤l_M(I). ButI₁≤I impliesl_M(I₁)≥l_M(I), sol_M(I) =l_M(I₁).

(2) ⇒ (1). Let I₁ ≤ I₂ ≤ · · · ≤ I_n· · · be a chain of right ideals, where I_i =r_J(M_i) and M_i ⊆ M for each i, let X_i = l_M(I_i) for each i = 1,2, . . . , and

I = _∞

i=1I_i, then I ≤ J_R. By (2), there exists a finitely generated right ideal I₁ ofI such thatl_M(I) = l_M(I₁). Since I₁ is finitely generated, there is an integerk such that I₁ ≤I_m for all m ≥k, that is l_M(I₁) ≥l_M(I_m) = X_m for allm ≥k.

Butl_M(I₁) = _∞

i=1l_M(I_i) = _∞

i=1X_i, that is l_M(I₁) = X_m for all m ≥k. Then I_m=r_J(X_m) =I_k for allm≥k, proving (1).

Now an argument of the proof of Faith [5, Proposition 3], we have:

Proposition 2.14. The following conditions on a small injective module E_R are equivalent:

(1) E^(N) is small injective.

(2) R satisfies the ACC for right ideals of formr_J(X), whereX⊆E.

(3) E^(S)is small injective for any index set S.

Proof. (3)⇒(1) is clear.

(1)⇒(2). Assume thatr_J(X₁)< r_J(X₂)<· · ·< r_J(X_n)<· · ·, X_i ⊆E. Then l_Er_J(X₁)> l_Er_J(X₂)>· · ·> l_Er_J(X_n)>· · ·

because r_Jl_Er_J(X) = r_J(X) for any X ⊆ E. For each i ≥ 1, choose m_i ∈ l_Er_J(X_i)\l_Er_J(X_i+1). Hence there exists a_i+1 ∈ r_J(X_i+1) such thatm_ia_i+1 = 0.

Let T = _∞

i=1r_J(X_i) which yields T R_R. Then, for all t ∈ T there exists n_t ≥ 1 such that t ∈ r_J(X_i) for all i ≥ n_t. Then m_it = 0 for all i ≥ n_t, and if ¯m = (m_i)_i, then ¯mt ∈ E^(N) for every t ∈ T. Hence ϕ_m¯ : T → E^(N) is well- defined byϕ_m¯(t) = ¯mt. SinceE^(N) is small injective by hypothesis,ϕ_m¯ extends to ψ: R→ E^(N). So ϕ_m¯(t) = ¯mt=ψ(t) =ψ(1)t for all t ∈T. But ψ(1)∈E^(N), so there exists k ≥ 1 such that m_it = 0 for all i ≥ k and all t ∈ T. In particular, m_ia_i+1= 0 for alli > k, which is a contradiction.

(2) ⇒ (3). Let I be a right ideal and I_R ≤ J_R. By Lemma 2.13, there is I₁ =r₁R+r₂R+· · ·r_nR ≤I_R such that l_M(I) = l_M(I₁). Letϕ : I → E^(S) be an R-homomorphism. Since E^S is small injective by [10, Proposition 3.5], there exists an element a∈E^S such that ϕ(r) =arfor all r∈I. In particularϕ(r_i) = ar_i ∈ E^(S), i = 1,2, . . . , there exists an elementa ∈ E^(S) such that a_sr_i = a_sr_i for all s ∈ S, i = 1,2, . . . , where g_s is the s th-coordinate of anyg ∈ E^S. Since {r₁, r₂, . . . , r_n} generates I₁, this implies that ar = ar for all r ∈ I₁, whence (a_s−a_s)∈l_M(I₁) for alls∈S. Sincel_M(I) =l_M(I₁), it follows thata_sx=a_sxfor alls∈S, x∈I, that isax=axfor all x∈I. Thus ϕ(x) =axfor allx∈I with a∈E^(S), so E^(S)is small injective.

Lemma 2.15 ([1]). Let M be a right R-module. Then Rad(M) is noetherian if and only if M has ACC on small submodules.

Lemma 2.16. Every direct summand of a small injective module is small injective.

Proof. It is clear.

It is well-know thatRis right noetherian if and only ifE^(N)is injective for every injective moduleE_R. We also have:

Theorem 2.17. For a ring R,the following conditions are equivalent:

(1) J is noetherian as a rightR-module.

(2) Every direct sum of small injective rightR-modules is small injective.

(3) IfM₁, M₂, . . . are simple right modules then_∞

i=1E(M_i)is small injective.

(4) E^(N)is small injective for every small injective module E_R. Proof. (2)⇒(3) and (2)⇒(4) are clear.

(1)⇒ (2). Let E =

i∈IE_i, where eachE_i is small injective, T_R ≤J_R, and ϕ : T → E an R-homomorphism. Since T is finitely generated by (1), we can write ϕ : T → E₁ =

i∈I1E_i, where I₁ ⊆I is a finite subset. Since E₁ is small injective, let ¯ϕ:R →E₁ extends ϕ. Thenιϕ¯ extends ϕ, whereι :E₁ →E is the inclusion.

(3) ⇒ (1). Let I₀ < I₁ < · · · be a strictly ascending chain of small finitely generated right ideals of R. Let I = _∞

i=0I_i, then I is a small right ideal of R.

For each i ≥ 1 choose M_i ≤^max I_i such thatI_i−1 ≤ M_i. Thus K_i = I_i/M_i is a simple right R-module. We define η_i : I_i/I_i−1 → K_i by η_i(x+I_i−1) = x+M_i, and write ι_i : K_i → E(K_i) for the inclusion. Since E(K_i) is injective, let ϕ_i : I/I_i−1 → E(K_i) be the homomorphism such that ϕ_i = ι_iη_i (see the following diagram):

I_i/I_i−1 → I/I_i−1 η_i↓

K_i ϕ_i ι_i ↓

E(K_i)

SinceM_i< I_i, there existsc_i∈I_i\M_i such that ϕ_i(c_i+I_i−1)= 0. For eacht∈I, choosen_t≥1 such thatt∈I_i−1for alli≥n_tand soϕ_i(t+I_i−1) = 0 for alli≥n_t, so we can defineα:I→_∞

i=1E(K_i) byα(t) = (ϕ_i(t+I_i−1))_i. Since_∞

i=1E(K_i) is small injective by (3), αextends to ¯α: R → _∞

i=1E(K_i). Write ¯α(1) = (b_i)_i, so there exists n ≥ 1 such that b_i = 0 for all i ≥ n. Given any t ∈ I, we have (ϕ_i(t+I_i−1))_i =α(t) = ¯α(t) = ¯α(1)t= (b_it)_i. Thusϕ_i(t+I_i−1) = 0 for alli≥n and allt∈I. Butϕ_n(c_n+I_n−1)= 0 by the definition ofϕ_i, and this contradiction proves (1).

(4) ⇒ (1). Let I₁ ≤ I₂ ≤ · · · be a chain of small right ideals. For each i, let E_i = E(R/I_i), and E = _∞

i=1E_i. For every i ≥ 1,_∞

j=1E_j = E_i (

j=iE_j).

Let M_i =_∞

j=1E_j, then M_i is small injective by [10, Proposition 3.5]. By above

notation, we have ∞

i=1

M_i= _∞

i=1

E_i



^∞

i=1

j=i

E_j



. By assumption, _∞

i=1M_i is small injective. Thus E itself is small injective by Lemma 2.16. Now theR-homomorphismf :_∞

i=1I_i→Edefined byf(t) = (t+I_i)_i extends to ¯f : R →E. Let n≥1 such that ¯f(1) ∈_n

j=1E_j. Then f(_∞

i=1I_i)≤ _n

j=1E_j. So, ift ∈_∞

i=1I_i then t∈I_m for allm > n, and so _∞

i=1I_i =I_n+1 and the chain should terminate.

Corollary 2.18. The following conditions are equivalent for a ring R:

(1) J is noetherian as a right R-module.

(2) Every direct sum of injective right R-modules is small injective.

(3) E^(N) is small injective for every injective modulesE_R. Acknowledgment

The authors wish to thank the referee for many useful suggestions and comments.

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