Throughout, all rings R are associative with identity and all modules are unitary right R-modules. For a module M, we denote by Rad(M), Soc(M) and E(M) the Jacobson radical, the socle and the injective hull of M, respectively. We write N⊆M if N is a subset of M, and N≤M if N is a submodule of M. The notation N≤dM means that N is a direct summand of M. For two modules X, Y over a ring R, the set of R-homomorphisms from X to Y is denoted by HomR(X,Y). For a ring R, we denote by J(R) the Jacobson radical of R. Let M be a module over a ring R. Recall that M is called a C2-module if every submodule of M which is isomorphic to a direct summand is itself a direct summand of M. In addition, the module M is said to be a C3-module if the sum of any two direct summands of M with zero intersection is again a direct summand of M. It was shown in [4, Proposition 2.1] that the "simple" versions of C2 and C3-modules coincide. Camillo et al. [4] called these modules simple-direct-injective modules. Then M is a simple-direct-injective module if, whenever A and B are simple submodules of M with A≅B and B a direct summand of M, then A is a direct summand of M. Equivalently, for any simple direct summands A and B of M with A∩B=0, A⊕B is a direct summand of M. A number of examples of simple-direct-injective modules appeared in some research papers. Among others, note that every indecomposable right R-module is simple-direct-injective by [4, Example 2.3(1)]. Moreover, according to [4, Example 2.3(2)], every cyclic R-module is simple-direct-injective if R is a commutative ring. It is clear that every R-module M which has no simple direct summands is simple-direct-injective. For example, we can take a module M with M=Rad(M). Also, if a module M=M1⊕M2 is a direct sum of two nonsimple indecomposable R-modules M₁ and M₂, then M has no simple direct summands (see [8, Example 2.6(1)]). Next, we present some known characterizations of various rings in terms of simple-direct-injective modules.

• 1. Given a ring R, the right R-module R_R is simple-direct-injective if and only if every projective right R-module is simple-direct-injective ([4, Corollary 2.15]).

• 2. A ring R is an artinian serial ring with J(R)2=0 if and only if every simple-direct-injective right R-module is a C3-module (a quasi-injective module) ([4, Theorem 3.4]).

• 3. A ring R is semisimple if and only if every simple-direct-injective right R-module is injective ([4, Corollary 3.5]).

• 4. A ring R is a right V-ring if and only if every 2-generated (finitely cogenerated) right R-module is simple-direct-injective ([4, Proposition 4.1]).

• 5. A von Neumann regular ring R is a right V-ring if and only if every cyclic right R-module is simple-direct-injective ([4, Theorem 4.4]).

Recently, Büyükaṣik et al. [3] gave a complete characterization of simple-direct-injective abelian groups. They proved in [3][Theorem 2] that an abelian group M is simple-direct-injective if and only if for each prime number p, the p-primary component Tp(M) is semisimple or Soc(Tp(M))⊆ Rad(Tp(M)).

The motivation of this paper comes from these three works: [3], [4] and [8]. Our goals are to extend the preceding characterization of simple-direct-injective abelian groups to modules over commutative Dedekind domains, investigate some properties of simple-direct-injective modules and rings and construct some useful examples.

In Section 2, we present some easy examples. In Section 3, we obtain several equivalent conditions for modules to being simple-direct-injective and give some properties of this type of modules. Among others, we prove that a module M is simple-direct-injective if and only if for every pair of idempotents e,f∈S=EndR(M) such that e(M) and f(M) are simple and e(M)∩f(M)=0, there exist orthogonal idempotents g,h∈S such that eS = gS and fS = hS. It is also shown that a module M is simple-direct-injective if and only if for every submodule K of M such that K=K1⊕K2, K₁ and K₂ are simple (perspective) direct summands of M, every homomorphism φ:K→M can be extended to an endomorphism θ:M→M. In addition, we prove that for an R-module M=⊕i∈IMi which is a direct sum of submodules Mi(i∈I) such that for every direct summand X of M, X=⊕i∈I(X∩Mi) (for example, if HomR(Mi,Mj)=0, for all distinct i,j∈I), then M is simple-direct-injective if and only if M_i is simple-direct-injective for all i∈ I. This result is useful to investigate simple-direct-injective modules over commutative Dedekind domains. In Section 4, we fully characterize simple-direct-injective modules over commutative Dedekind domains. In particular, we prove an extension of [3, Theorem 2] to modules over commutative Dedekind domains. In Section 5, we investigate simple-direct-injective rings. A ring R is called left (right) simple-direct-injective if the left (right) R-module RR (RR) is simple-direct-injective. We show that being a simple-direct-injective ring is not left-right symmetric. Moreover, we prove that the class of left (right) simple-direct-injective rings is closed under direct products. Some necessary conditions for the endomorphism ring of a module to be right simple-direct-injective are investigated.

In this section, we provide some other examples of simple-direct-injective modules. It is well known that a simple submodule of a module M is either small in M or a direct summand of M. This clearly implies that any module M with Soc(M)∩ Rad(M)=0 is simple-direct-injective since all simple submodules of M are direct summand. Next, we present an important class of modules satisfying this condition.

Example 2.1. Let R be any ring and let M be a regular right R-module (i.e. every cyclic (finitely generated) submodule of M is a direct summand of M (see [15, Remark 6.1])). Therefore M is a simple-direct-injective module since every simple submodule of M is a direct summand of M. In particular, any projective module over a von Neumann regular ring is simple-direct-injective by [15, Proposition 6.7(4)].

A module M is called dual Rickart if for every endomorphism f of M, f(M) is a direct summand of M (see [9] and [11]). It is easily seen that every dual Rickart module is simple-direct-injective, but the converse is not true, in general. For example, the Z-module Z is simple-direct-injective but it is not dual Rickart. Now using [9, Theorem 3.2 and Corollary 3.3], we obtain the following two examples of simple-direct-injective modules. For the undefined notions here we refer to [7].

Example 2.2. Let R be a prime right Goldie ring such that R is not right primitive and let an R-module M be a direct sum of a torsion-free divisible submodule X and a torsion semisimple submodule Y. Then M is simple-direct-injective.

Example 2.3. Let R be a prime PI-ring which is not artinian and let an R-module M be a direct sum of a torsion-free divisible submodule X and a torsion semisimple submodule Y. Then M is simple-direct-injective.

It is clear that any module M with Soc(M)⊆ Rad(M) (for example, Soc(M)=0 or Rad(M)=M) is simple-direct-injective since every simple submodule of M is small in M.

Example 2.4. Let R be a local right artinian ring which is not semisimple (for example, we can take R=ℤ/pnℤ for some prime number p and some integer n≥2). Clearly the right R-module R_R has no simple direct summands. Therefore R_R is a simple-direct-injective R-module. Note that Soc(RR)⊆Rad(RR), Soc(RR)≠0 and Rad(RR)≠RR.

It is well known that the notion of simple-direct-injective modules is closed under direct summands. To construct counterexamples showing that the simple-direct-injective property is not inherited by submodules and direct sums, we need the following lemma.

Lemma 2.5. Let M be a module having a simple submodule S such that S is not a direct summand of M (e.g., M is indecomposable with Soc(M)≠0 and Soc(M)≠M). Then M⊕Soc(M) is not simple-direct-injective.

Proof. Assume that M⊕Soc(M) is simple-direct-injective. Then M⊕S is simple-direct-injective by [3][Lemma 6]. Now consider the inclusion map i:S→M. By [4][Proposition 2.1], ℑ(i)=S is a direct summand of M. This contradicts our assumption. Therefore M⊕Soc(M) can not be simple-direct-injective.

Example 2.6. Let a module M=⊕i∈IMi be a direct sum of indecomposable nonsimple submodules Mi(i∈I) such that Soc(Mi0)≠0 for some i0∈I. Then the module N=M⊕M is simple-direct-injective since N has no simple direct summands. However, its submodule M⊕Soc(M) is not simple-direct-injective by Lemma 2.5. Moreover, note that both M and Soc(M) are simple-direct-injective. As an explicit example, we can take for M any direct sum of indecomposable nonsimple ℤ-modules (e.g., ℤ, ℚ) such that at least one of them must be isomorphic to ℤ/pnℤ or to ℤ(p∞) for some prime number p and some integer n ≥ 2.

In this section we provide some new equivalent formulations of being a simple-direct-injective module and establish some properties of this type of modules. In the proof of [4, Proposition 2.1], one can easily see that the implication (1) ⇒ (2) is true even if the condition "B is simple" is deleted from the statement (2). We thus get the following proposition. Its proof is given for completeness.

Proposition 3.1. The following conditions are equivalent for a module M:

• (i) M is simple-direct-injective;

• (ii) For any direct summands A and B of M such that A is simple and A∩B=0, A⊕B≤dM.

Proof. (i) ⇒ (ii) Let A and B be direct summands of M such that A is simple and A∩B=0. Then M=B⊕B′ for some submodule B′ of M. Let π:B⊕B′→B′ be the projection map with Kerπ=B. Since A∩B=0, π(A)≅A. Since M is simple-direct-injective, π(A)≤dM. Hence π(A)≤dB′. Therefore there exists a submodule X of B′ such that B′=X⊕π(A). It follows that M=B⊕B′=π(A)⊕B⊕X. Note that π(A)⊕B=A⊕B. Thus M=A⊕B⊕X and hence A⊕B≤dM.

(ii) ⇒ (i) This follows from [4, Proposition 2.1].

Recall that two direct summands A and B of a module M are called perspective (see [5]) if M=A⊕X=B⊕X for some submodule X of M. Following [1], two idempotents e and f of a ring R are called perspective if eR and fR are perspective direct summands of the right R-module R_R (i.e. there exists a right ideal C of R such that R=eR⊕C=fR⊕C). It was shown in [1, Proposition 2.13] that a module M is simple-direct-injective if and only if for any simple perspective direct summands A, B of M with A∩B=0, A⊕B is a direct summand of M. Using the notion of perspectivity of idempotents, we obtain the following characterization of simple-direct-injective modules. The proof of this result is similar to that of [12, Lemma 4.5] (see also [1, Lemma 3.1]).

Theorem 3.2. The following are equivalent for a module M and S=EndR(M):

• (i) M is simple-direct-injective;

• (ii) For every pair of idempotents e,f∈S such that e(M) and f(M) are simple and e(M)∩f(M)=0, there exist orthogonal idempotents g,h∈S such that eS=gS and fS=hS;

• (iii) For every pair of perspective idempotents e,f∈S such that e(M) and f(M) are simple and e(M)∩f(M)=0, there exist orthogonal idempotents g,h∈S such that e(M)=g(M) and f(M)=h(M);

• (iv) For every pair of perspective idempotents e,f∈S such that e(M) and f(M) are simple and e(M)∩f(M)=0, there exists an idempotent g of S such that e(M)=g(M) and f(M)⊆(1−g)(M).

Proof. (i) ⇒ (ii) Suppose that M is simple-direct-injective. Let e2=e,f2=f∈S with e(M)∩f(M)=0 such that e(M) and f(M) are simple modules. Then M=e(M)⊕f(M)⊕N for some submodule N of M (see Proposition 3.1). Let g:M→e(M) be the projection map with Kerg=f(M)⊕N and h:M→f(M) be the projection map with Kerh=e(M)⊕N. Clearly, g(M)=e(M) and h(M)=f(M). Then by [12, Lemma 1.1], gS=eS and hS=fS. It is not hard to see that h and g are orthogonal.

(ii) ⇒ (iii) Clear by [12, Lemma 1.1].

(iii) ⇒ (iv) Let e, f ∈ S be perspective idempotents such that e(M) and f(M) are simple and e(M)∩f(M)=0. By hypothesis, there exist orthogonal idempotents g, h∈ S such that e(M) = g(M) and f(M) = h(M). Note that f(M)=h(M)⊆ Kerg=(1−g)(M), as desired.

(iv) ⇒ (i) Let A and B be simple perspective direct summands of M with A∩B=0. Then A=e(M) and B=f(M) for some idempotents e and f of the ring S. By (iv), there exists an idempotent g of S such that A=g(M) and B⊆(1−g)(M). Hence B is a direct summand of (1−g)(M). Since M=g(M)⊕(1−g)M, it follows that A⊕B is a direct summand of M.

Now using [1, Proposition 2.13], we conclude that M is a simple-direct-injective module. This completes the proof.

The following lemma is needed to prove another characterization of simple-direct-injective modules.

Lemma 3.3. Let A and B be direct summands of a module M such that A∩B=0. Then the following are equivalent:

• (i)
  C=A⊕B is a direct summand of M;

• (ii) Every homomorphism φ:C→M can be extended to an endomorphism θ:M→M.

Proof (i) ⇒ (ii) This is clear.

(ii) ⇒ (i) Note that M=B⊕U for some submodule U of M. Let π:C→B be the projection of C on B along A and let μ:B→M denote the inclusion map. By (ii), the homomorphism μπ:C→M can be extended to an endomorphism θ:M→M. Since θ(A)=μπ(A)=0, we have A⊆Kerθ. Moreover, it is clear that θ(y)=y for all y∈B and hence Kerθ∩B=0. Now take an element m∈M. As M=B⊕U, θ(m)=b+u for some elements b∈B and u∈U. Therefore θ(m)=θ(b)+u. Thus u=θ(m−b) and so m−b∈θ−1(U). This yields m=(m−b)+b∈θ−1(U)+B. It follows that M=θ−1(U)+B.

Now to show that B∩θ−1(U)=0, take a∈B∩θ−1(U). Then θ(a)=a∈U∩B=0. Therefore M=θ−1(U)⊕B. Let V be a submodule of M such that M=A⊕V. Then θ−1(U)=θ−1(U)∩(A⊕V)=A⊕(θ−1(U)∩V) because A⊆Kerθ⊆θ−1(U). Hence M=θ−1(U)⊕B=A⊕(θ−1(U)∩V)⊕B=C⊕(θ−1(U)∩V). This completes the proof.

Proposition 3.4. The following are equivalent for a module M:

• (i) M is simple-direct-injective;

• (ii) For every submodule K of M such that K=K1⊕K2 and K1 and K2 are simple perspective direct summands of M, every homomorphism φ:K→M can be extended to an endomorphism θ:M→M.

Proof. This follows by combining the preceding lemma with [1, Proposition 2.13].

Proposition 3.5. The following are equivalent for a module M:

• (i) M is simple-direct-injective;

• (ii) For every endomorphisms α,β of M such that Kerβ=Kerα and Kerα is a direct summand of M which is a maximal submodule of M, there exists an endomorphism γ of M such that γα=β.

Proof. (i) ⇒ (ii) Let α and β be endomorphisms of M such that Kerβ=Kerα and Kerα is a direct summand of M which is a maximal submodule of M. Then M=Kerα⊕L for some simple submodule L of M. Clearly α∣L:L→M is a monomorphism. Since M is simple-direct-injective, it follows that α(L) is a direct summand of M. So there exists a homomorphism η:M→L such that ηα∣L=1L. Now since Kerα=Kerβ, the homomorphism β¯:M/Kerα→M given by m+Kerα↦β(m) is well defined. In addition, consider the isomorphism α¯:L→M/Kerα defined by l↦l+Kerα for all l∈ L. Set γ=β¯α¯η. Note that γ is an endomorphism of M. To show that γα=β, take m∈ M. Then m=x+l for some x∈Kerα and l∈ L. Therefore,

This implies that γ is the desired endomorphism of M.

(ii) ⇒ (i) Let K be a simple direct summand of M. Hence M=K⊕K′ for some submodule K′ of M. Assume that K is isomorphic to a simple submodule S of M. So there exists an isomorphism μ:K→S. Consider the endomorphisms φ and θ of M defined respectively by k+k′↦μ(k) and k+k′↦k for any k∈K and k′∈K′. Then φθ is an endomorphism of M such that Kerφθ=Kerθ=K′ is a maximal submodule of M. By hypothesis, there exists an endomorphism γ:M→M such that γφθ=θ. Hence (γφ−1M)θ=0. Since 𝜃 is an epimorphism, we have γφ=1M. Thus ℑφ=S is a direct summand of M (see [2, Lemma 5.1]). Therefore M is a simple-direct-injective module.

Recall that any module M is called pseudo-N-injective, if every monomorphism f: K → M, where K ≤ N, can be extended to a homomorphism from N to M. The next proposition characterizes modules whose submodules are simple-direct-injective in terms of the pseudo-injectivity.

Proposition 3.6. The following conditions are equivalent for a module M:

• (i) Every submodule of M is simple-direct-injective;

• (ii) For any submodules A and B of M with B simple and A∩B=0, B is pseudo-A-injective.

Proof. (i) ⇒ (ii) Let A and B be submodules of M such that B is simple and A∩B=0. Let us show that B is pseudo-A-injective. Let 0≠f:X→B be a monomorphism with X≤A. Since B is simple, f is an isomorphism. Now X≅B≤dA⊕B and B and X are submodules of A⊕B. By hypothesis, A⊕B is simple-direct-injective and B is simple. Therefore X≤dA⊕B, and hence X≤dA.

Thus f can be extended to A.

(ii) ⇒ (i) Let L be a submodule of M. We will show that L is simple-direct-injective. For, let L=A⊕B with B simple and let f:B→A be a nonzero homomorphism. By hypothesis, B is pseudo-A-injective. Since B≅f(B), f(B) is pseudo-A-injective. This implies that the identity homomorphism 1f(B):f(B)→f(B) can be extended to a homomorphism g:A→f(B). Now g∘i=1f(B), where i:f(B)→A is the inclusion map. Therefore i(f(B))=f(B)≤dA. From [4, Proposition 2.1 (1) ⇔ (3)], it follows that L is simple-direct-injective.

Next, we present some examples of modules satisfying the conditions in the hypothesis of Proposition 3.6.

Example 3.7. (i) Let M be a module such that Soc(M)=0. Then every submodule of M is simple-direct-injective.

(ii) Consider the ℤ-module M=ℤ(p∞), where p is a prime number. It is well known that any nonzero proper submodule of M is isomorphic to ℤ/pkℤ for some integer k≥1. It follows that every submodule of M is simple-direct-injective.

(iii) Let R be a commutative principal ideal ring. By [4, Example 2.3(2)], every cyclic R-module is simple-direct-injective. Thus, every submodule of a cyclic R-module is simple-direct-injective.

In the following proposition, we provide sufficient conditions under which a direct sum of simple-direct-injective modules is simple-direct-injective. This result should be contrasted with Lemma 2.5 and Example 2.6.

Proposition 3.8. Let R be any ring and let an R-module M=⊕i∈IMi be a direct sum of submodules Mi(i∈I). Suppose that one of the following conditions is fulfilled:

• (i) For every simple direct summand X of M, X⊆Mi for some i∈I;
  or

• (ii) HomR(Mi,Mj)=0 for all distinct i,j∈I; or

• (iii) For every direct summand X of M, X=⊕i∈I(X∩Mi).

Then M is simple-direct-injective if and only if M_i is simple-direct-injective for all i ∈ I.

Proof. If M is simple-direct-injective, then M_i is simple-direct-injective for all i ∈ I by [3, Lemma 6]. Conversely, assume that each M_i (i ∈ I) is simple-direct-injective.

(i) Let A and B be simple direct summands of M with A∩B=0. By hypothesis, there exist j and k in I such that A⊆Mj and B⊆Mk. Since A and B are direct summands of M, A≤dMj and B≤dMk. Assume that j≠k. Then clearly A⊕B≤dMj⊕Mk, and hence A⊕B≤dM. Now assume that j=k. Since Mj is simple-direct-injective, it follows that A⊕B is a direct summand of Mj and hence it is a direct summand of M. Therefore M is simple-direct-injective by [4, Proposition 2.1].

(ii) Let S be a simple direct summand of M. It is well known that S has the exchange property (see [16, Proposition 1]). Then M=S⊕(⊕i∈IMi′) for some submodules Mi′≤Mi(i∈I). It follows that Mi0′≠Mi0 for some i0∈I. Moreover, since ⊕i∈IMi′ is a maximal submodule of M, we have Mi′=Mi for every i≠ i₀. This implies that M=(S⊕Mi0′)⊕(⊕i≠i0Mi). Thus S⊕Mi0′≅Mi0. But Mi0 is fully invariant in M as HomR(Mi0,Mj)=0 for every j≠i0. Then S⊕Mi0′⊆Mi0 and hence S⊆Mi0. The result now follows from (i).

(iii) This follows from the fact that the condition (iii) implies (i).

The following example shows that the conditions (i), (ii) and (iii) in the hypothesis of Proposition 3.8 are not superfluous.

Example 3.9. Consider the ℤ-module M=ℤ/pℤ⊕ℤ(p∞), where p is a prime number. By Lemma 2.5, M is not a simple-direct-injective module. Let 0≠a∈ℤ(p∞) with pa=0 and set S=(1¯,a)ℤ. It is clear that S⊕ℤ(p∞)=M. On the other hand, S⊈ℤ/pℤ⊕0 and S⊈0⊕ℤ(p∞).

In the next example, we present another simple-direct-injective module.

Example 3.10. Let M=M1⊕M2 be an R-module such that Rad(M1)=M1 and Rad(M2)=0. It is clear that M₁ has no simple direct summands. In addition, all simple submodules of M₂ are direct summand. Therefore M₁ and M₂ are simple-direct-injective. Assume that either Soc(M1)=0 or Soc(M2)=0. Then for any simple submodule S of M, we have either S⊆M1 or S⊆M2. Using Proposition 3.8, we conclude that M is simple-direct-injective.

Next, we will be concerned with factor modules of simple-direct-injective modules. We begin with a result which characterizes the class of commutative rings over which the class of simple-direct-injective modules is closed under factor modules.

Proposition 3.11. Let R be a right simple-direct-injective ring (for instance, R is commutative). Then the following conditions are equivalent:

• (i) All factor modules of simple-direct-injective
  R-modules are simple-direct-injective;

• (ii) R is a right V-ring.

Proof. (i) ⇒ (ii) By [4][Theorem 2.14], every free right R-module is simple-direct-injective. Now the condition (i) implies that every right R-module is simple-direct-injective. Thus R is a right V-ring by [4, Proposition 4.1].

(ii) ⇒ (i) This follows from [4, Proposition 4.1].

From the preceding proposition, it follows that any commutative ring R which is not von Neumann regular has a simple-direct-injective R-module M such that M/N is not simple-direct-injective for some submodule N of M. Next, we provide an explicit example.

Example 3.12. Let R be a commutative ring having a maximal ideal 𝔪 such that 𝔪k≠𝔪k+1 for some integer k≥1 (for example, we can take a discrete valuation ring with maximal ideal 𝔪). Consider the R-module M=R/𝔪⊕R/𝔪k+1. It is easily seen that 0≠mk/mk+1⊆Soc(R/mk+1). So R/𝔪k+1 contains a simple submodule S which is isomorphic to R/𝔪⊕0 but 0⊕S is not a direct summand of M. Therefore M is not simple-direct-injective. On the other hand, it is clear that M≅R(2)/(𝔪⊕𝔪k+1). Moreover, since R is commutative, R is simple-direct-injective and so R(2) is asimple-direct-injective R-module by [4, Theorem 2.14].

This small section is devoted to the study of simple-direct-injective modules over commutative Dedekind domains.

Let M be a module over a commutative domain R. We denote by T(M) the set of all elements x of M for which AnnR(x)≠0. It is well known that T(M) is a submodule of M which is called the torsion submodule of M. The module M is said to be a torsion module if T(M)=M. If T(M) = 0, the module M is said to be torsion-free. We begin by some examples of simple-direct-injective modules over a commutative domain.

Remark 4.1. Let R be a commutative domain with quotient field Q such that R ≠ Q.

• (i) Since RR is an indecomposable R-module, it follows that RR is a simple-direct-injective R-module (see [4, Example 2.3(1)]). Hence, every projective R-module is simple-direct-injective by [4, Corollary 2.15].

• (ii) All torsion-free R-modules are simple-direct-injective since they have no simple submodules.

• (iii) If R is a Dedekind domain, then for any index set I, the R-module M=(Q/R)(I) is simple-direct-injective since M is injective.

Let R be a commutative Dedekind domain with quotient field Q and let M be an R-module. Let p be a nonzero prime ideal of R. The set TpM={x∈M∣xpk=0 for some non-negative integer k} is a submodule of M which is called the p-primary component of M. It is well known that every torsion R-module is a direct sum of its p-primary components. The set of all nonzero prime ideals of R is denoted by ℙ.

In [3, Theorem 2], the authors characterized simple-direct-injective abelian groups. The next theorem is an extension of this characterization.

Theorem 4.2. Let R be a commutative Dedekind domain. Let ℙ be the set of all nonzero prime ideals of R.

Then the following are equivalent for an R-module M:

• (i) M is simple-direct-injective;

• (ii) T(M) is simple-direct-injective;

• (iii) For every p∈ℙ, the p-primary component Tp(M) is simple-direct-injective;

• (iv) For every p∈ℙ, Tp(M) is semisimple or Soc(Tp(M))⊆Rad(Tp(M));

• (v) For every p∈ℙ, Tp(M) is semisimple or Tp(M) has no simple direct summands.

Proof. (i) ⇔ (ii) This follows from [3, Corollary 5].

(ii) ⇒ (iii) This follows by using [3, Lemma 6] since each Tp(M) (p∈ℙ) is a direct summand of T(M).

(iii) ⇒ (ii) By Proposition 3.8.

(iv) ⇔ (v) This follows from the fact that any simple submodule S of a module M is either small in M or a direct summand of M.

(iii) ⇒ (v) Let p∈ℙ. Assume that Tp(M) has a simple direct summand S but Tp(M) is not semisimple. Then S≅R/p and Tp(M)=S⊕E for some submodule E of Tp(M). It is clear that E is not semisimple. Therefore there exists x ∈ E such that xR≅R/pn for some integer n ≥ 2. Note that xR is an indecomposable R-module which is not simple. It is easily seen that xR contains a simple submodule S′ which is isomorphic to R/p. Since Tp(M) is simple-direct-injective and S≅S′, it follows that S′ is a direct summand of Tp(M) and hence S′ is a direct summand of xR, a contradiction.

(v) ⇒ (iii) This is immediate.

Let R be a commutative domain with field of fractions Q. Recall that an R-submodule F of Q is called a fractional ideal of R if rF⊆R for some nonzero element r of R. As an application of Theorem 4.2, the next corollary characterizes finitely generated simple-direct-injective modules over commutative Dedekind domains.

Corollary 4.3. Let R be a commutative Dedekind domain and let M be a finitely generated R-module. Then M is simple-direct-injective if and only if

where k, s and t are non-negative integers, qi(1≤i≤s) and pi(1≤i≤t) are nonzero prime ideals of R such that qi≠pj for all (i,j)∈{1,…,s}×{1,…,t}, mi(1≤i≤s) and ni(1≤i≤t) are positive integers, and Ij(1≤j≤k) are nonzero fractional ideals of R.

Proof. The necessity follows from [14, Theorem 6.16], Example 3.12 and [3, Lemma 6]. Conversely, suppose that M satisfies the stated conditions. Then clearly,

It is easily seen that for each nonzero prime ideal p of R, we have either Soc(Tp(M))=Tp(M) or Soc(Tp(M))⊆ Rad(Tp(M)). Thus M is simple-direct-injective by Theorem 4.2.

Example 4.4. Let R be a commutative Dedekind domain with quotient field Q. Let ℙ denote the set of all nonzero prime ideals of R.

(i) Let an R-module M=M1⊕M2 be a direct sum of submodules M₁ and M₂ such that M₁ is torsion-free and M₂ is semisimple. Then T(M)=M2. From Theorem 4.2, we infer that M is a simple-direct-injective module.

(ii) Consider the R-module M=∏ p∈ℙR/p. Clearly, T(M)=⊕p∈ℙR/p is simple-direct-injective. Thus M is also simple-direct-injective by Theorem 4.2.

A ring R is called left (right) simple-direct-injective if the left (right) R-module RR (R_R) is simple-direct-injective. We begin by providing examples of left (and right) simple-direct-injective rings.

Example 5.1. It is clear that semisimple rings, local rings and von Neumann regular rings are left and right simple-direct-injective.

(ii) It is clear that a left (right) V-ring is a left (right) simple-direct-injective ring.

(iii) Recall that a ring R is said to be left (right) Kasch if every simple left (right) R-module can be embedded in RR (R_R).

By [13, Proposition 1.46], every left (right) Kasch ring is right (left) simple-direct-injective.

Next, we exhibit some examples to illustrate that the property of being a simple-direct-injective ring is not left-right symmetric.

Example 5.2. (i) Here we are using the ring R given in [10, Examples 8.29(6)]; that is, R is the ring of matrices of the form

over a division ring D. It is shown in [10, Examples 8.29(6)] that Soc(RR)=Rad(R). Thus the left R-module RR has no simple direct summands. Therefore R is left simple-direct-injective. On the other hand, let us show that R is not a right simple-direct-injective ring. Consider the elements r1=0000000000000001 and r2=0001000000000000 of R. It is easily seen that r₁ is an idempotent of R and r₁R is a simple direct summand of R_R. In addition, r₂R is a simple submodule of R_R which is isomorphic to r₁R. Note that r₂R is not a direct summand of R_R since r₂R contains no nonzero idempotents of R. This proves that the ring R is not right simple-direct-injective.

(ii) Consider the ring R=ℤ2ℤ20ℤ where ℤ2=ℤ/2ℤ. Then the set of all idempotents of R is 0¯0¯00,0¯0¯01, 1¯0¯00,0¯ 1¯01, 1¯ 1¯00, 1¯0¯01. Thus the direct summands of the right R-module R_R are: 0¯0¯00R, 0¯0¯01R=000ℤ, 0¯ 1¯01R=0¯n¯ 0n∣n∈ℤ,  1¯0¯00R= 1¯ 1¯00R=ℤ2ℤ200 and 1¯0¯01R=R. It is obvious that none of them is simple. Therefore R is a right simple-direct-injective ring. On the other hand, note that R1¯0¯00=ℤ2000 is a simple direct summand of the left R-module RR. Moreover, the submodules ℤ2000 and 0ℤ200 of RR are isomorphic. But 0ℤ200 is not a direct summand of RR since it contains no nonzero idempotents of R. Therefore R is not a left simple-direct-injective ring.

Next, we present an example of a ring R which is neither left simple-direct-injective nor right simple-direct-injective. This example shows also that a direct sum of two simple-direct-injective modules need not be simple-direct-injective, in general (see also Example 2.6).

Example 5.3. Let R=FF0F, where F is a field. Then the set of all idempotents of R is

Note that 0001R=000F≅0F00 is a simple right R-module. In addition, 000F is a direct summand of the right R-module R_R but 0F00 is not. This implies that the ring R is not right simple-direct-injective. Similarly, 0F00≅R1000=F000 is a simple left R-module. Moreover, F000 is a direct summand of the left R-module RR but 0F00 is not. Therefore R is not a left simple-direct-injective ring.

On the other hand, note that RR=000F⊕FF00. Moreover, U=000F is a simple right R-module and V=FF00 is an injective right R-module as V is a direct summand of E(RR)=FFFF. In particular, U and V are simple-direct-injective right R-modules.

Let x be an element of a ring R. We denote by Ann_R(x) the right annihilator of x in R, that is, AnnR(x)={r∈R∣xr=0}. Next, we show that the class of left (right) simple-direct-injective rings is closed under direct products.

Proposition 5.4. Let {Ri∣i∈I} be a family of rings and R=∏ i∈IRi. Then R is a right simple-direct-injective ring if and only if each Ri(i∈I) is a right simple-direct-injective ring.

Proof. Suppose that R is a right simple-direct-injective ring and fix j∈ I. Then R_j regarded as a right R-module is simple-direct-injective since it is a direct summand of R_R (see [3, Lemma 6]). Moreover, it is easily seen that the submodules of R_j are the same whether it is regarded as a right R-module or as a right R_j-module. Thus R_j regarded as a right R_j-module is also a simple-direct-injective module. Conversely, assume that each R_i is a right simple-direct-injective ring. Let S and E be simple submodules of the right R-module R_R with S≅E≤dRR. Clearly, E=eR for some nonzero idempotent e of R. Let f:S→E be an R-isomorphism and let s=(si)i∈I∈R such that s∈S and f(s)=e. Then s≠0 and sR=S. Since R/AnnR(s)≅S, AnnR(s)=∏i∈IAnnRi(si) is a maximal right ideal of R. It is easily seen that the right R-modules R/AnnR(s) and ∏i∈IRi/AnnRi(si) are isomorphic. So there exists j∈I such that AnnRj(sj) is a maximal right ideal of Rj and AnnRi(si)=Ri for all i≠j. Hence si=0 for all i≠j and sjRj is a simple submodule of the right Rj-module Mj=RjRj. Now let (ei)i∈I∈R such that e=(ei)i∈I. It is clear that AnnR(e)=AnnR(s) and hence ei=0 for all i≠j and sjRj≅ejRj (as right Rj-modules). Since Rj is right simple-direct-injective and ej is an idempotent of Rj, it follows that sjRj is a direct summand of Mj. Therefore sR is a direct summand of RR. Consequently, R is right simple-direct-injective.

As an application of the preceding proposition, we obtain the following examples of left (and right) simple-direct-injective rings.

Example 5.5. (i) Every direct product of right indecomposable rings (e.g., local rings) is a right simple-direct-injective ring.

(ii) Every direct product of copies of the ring R given in Example 5.2(i) is a left simple-direct-injective ring which is not a right simple-direct-injective ring.

In the next proposition, we provide some necessary conditions for the endomorphism ring of a module to be right simple-direct-injective. Its proof is similar in spirit to that of [12, Proposition 4.6]. We first prove the following lemma.

Lemma 5.6. Let R be a ring and let M be an R-module with S=End_R(M). Let e and f be idempotents in S. Then:

• (i) If
  e(M) is a simple R-module, then eS is a minimal right ideal of S.

• (ii) If e(M)∩f(M)=0, then eS∩fS=0.

Proof. (i) Suppose that e(M) is a simple R-module. Let 0≠s=es∈eS. Then 0≠s(M)⊆e(M). As e(M) is simple, we have s(M)=e(M) is a direct summand of M. Hence sS=eS by [12, Lemma 1.1]. This implies that eS is a simple S-module.

(ii) Assume that e(M)∩f(M)=0. Suppose on the contrary that eS∩fS≠0.

Let 0≠s∈eS∩fS. Then es=s and fs=s and hence s(M)⊆e(M)∩f(M). But s(M)≠0. So e(M)∩f(M)≠0, a contradiction. It follows that eS∩fS=0.

Proposition 5.7. Let R be a ring and let M be an R-module with S=End_R(M). If the ring S is right simple-direct-injective, then M and S satisfy the following two conditions:

• (a)
  M is a simple-direct-injective R-module, and

• (b) for every pair of idempotents e,f∈S with eS∩fS=0, we have e(M)∩f(M)=0.

The converse holds when the following condition is satisfied:

• (c) For any idempotent e∈S such that eS is a minimal right ideal of S, e(M) is a simple R-module.

Proof. Assume that S is a right simple-direct-injective ring. Let e and f be idempotents in S such that e(M) and f(M) are simple and e(M)∩f(M)=0. Then eS and fS are minimal right ideals of S with eS∩fS=0 by Lemma 5.6. Applying Theorem 3.2, there exist orthogonal idempotents g,h∈S such that eS=gS and fS=hS. By using again Theorem 3.2, we deduce that M is a simple-direct-injective R-module. This proves (a). Now to prove (b), let e and f be idempotents in S such that eS∩fS=0. Since the right S-module S_S is simple-direct-injective, there exist orthogonal idempotents g, h∈ S such that eS = gS and fS = hS by Theorem 3.2. Thus e(M)=g(M) and f(M)=h(M) by [12, Lemma 1.1]. Hence e(M)∩f(M)=g(M)∩h(M)=0 as g and h are orthogonal.

The converse follows by Theorem 3.2.

The following example shows that the condition (c) in Proposition 5.7 is not necessary for the endomorphism ring of a module to be right simple-direct-injective.

Example 5.8. Consider the ring R=FF0F, where F is a field. Then the right R-module M=FF00 is an indecomposable module which is not simple. Moreover, we have S=EndR(M)≅F (as rings). Clearly, S is a right simple-direct-injective ring. On the other hand, the right ideal 1_MS is minimal, but 1M(M)=M is not a simple R-module.

Let R be a ring and M an R-R-bimodule. Then the trivial extension R∝ M is a ring whose underlying group is R×M with the multiplication defined by (r,m)(s,n)=(rs,rn+ms), where r, s ∈ R and m, n ∈ M.

Proposition 5.9. Let R be a ring and let M be an R-R-bimodule such that eM(1-e) = 0 for any idempotent e ∈ R. If R is a right simple-direct-injective ring, then so is the ring T=R∝M.

Proof. Suppose that R is right simple-direct-injective and let A be a simple direct summand of the right T-module T_T. From [6, Proof of Proposition 4.5], it follows that A=(eR, eM) for some nonzero idempotent e of R. But (0, eM) is a proper T-submodule of A. Then (0, eM)=0 and hence A=(eR, 0). Let B be another simple direct summand of T_T such that A∩B=0. As above B=(fR, 0) for some nonzero idempotent f of R with fM=0. Thus A⊕B=(eR,0)⊕(fR,0)=(eR+fR,0). Moreover, it is easily checked that eR and fR are minimal right ideals of R with eR∩fR=0. Since R is a right simple-direct-injective ring, eR +fR is a direct summand of the right R-module R_R. Therefore there exists an idempotent g of R such that (eR+fR)⊕gR=R. This implies that (A⊕B)⊕(gR,M)=T. It follows that T is a right simple-direct-injective ring.