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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2021; 61(2): 239-248

Published online June 30, 2021

### Weak FI-extending Modules with ACC or DCC on Essential Submodules

Department of Mathematics, Hacettepe University, Beytepe Campus, Ankara 06532, Turkey
e-mail : tercan@hacettepe.edu.tr

Hacettepe-ASO 1.OSB Vocational School, Hacettepe University, 06938 Sincan Ankara, Turkey
e-mail : ryasar@hacettepe.edu.tr

Received: April 19, 2020; Accepted: December 14, 2020

In this paper we study modules with the WFI+-extending property. We prove that if M satisfies the WFI+-extending, pseudo duo properties and M/(Soc M) has finite uniform dimension then M decompose into a direct sum of a semisimple submodule and a submodule of finite uniform dimension. In particular, if M satisfies the WFI+-extending, pseudo duo properties and ascending chain (respectively, descending chain) condition on essential submodules then M=M1 ⊕ M2 for some semisimple submodule M1 and Noetherian (respectively, Artinian) submodule M2. Moreover, we show that if M is a WFI-extending module with pseudo duo, C2 and essential socle then the quotient ring of its endomorphism ring with Jacobson radical is a (von Neumann) regular ring. We provide several examples which illustrate our results.

Keywords: CS-module, uniform dimension, ascending chain condition on essential submodules, FI-extending, WFI-extending

### 1. Introduction

Assume that all rings are associative and have identity elements and all modules are unital right modules. Let R be any ring and M a right R-module. Recall that M is called CS-module ( or extending module, module with C1) if every submodule of M is essential in a direct summand of M. Equivalently, every complement in M is a direct summand of M (see [7, 10, 18]). The class of extending modules contains injective, semisimple and uniform modules (i.e., every non-zero submodule is essential in the module). We say that the module M has finite uniform (Goldie) dimension if M does not contain an infinite direct sum of non-zero submodules. It is well-known that a module M has finite uniform dimension if and only if there exists a positive integer n and uniform submodules Ui ($1≤i≤n$) of M such that $U1⊕U2⊕⋯⊕Un$ is an essential submodule of M. In this case n is an invariant of the module called the uniform dimension of M (see, [1, p.294 Example 2] or [18, p.81]).

Armendariz [2, Proposition 1.1] proved that a module M satisfies DCC (descending chain condition) on essential submodules if and only if is an Artinian module. On the other hand, Goodearl [8, Proposition 3.6] proved that the module M satisfies ACC (ascending chain condition) on essential submodules if and only if is a Noetherian module. It is proved in [13, Theorem 2.1] that the following statements are equivalent for a module M:

• (i) M/N has finite uniform dimension for every essential submodule N of M,

• (ii) every homomorphic image of has finite uniform dimension.

Camillo and Yousif [6, Corollary 3] proved that if M is a CS-module and has finite uniform dimension then $M=M1⊕M2$ for some semisimple submodule M1 of M and submodule M2 with finite uniform dimension, and in this case M is a direct sum of uniform modules. They deduced in [6, Proposition 5] that if M is a CS-module then M has ACC (respectively, DCC) on essential submodules if and only if $M=M1⊕M2$ for some semisimple submodule M1 and Noetherian (respectively, Artinian) submodule M2 of M.

A module M is called a weak CS-module if, for each semisimple submodule S of M, there exists a direct summand K of M such that S is essential in K. Clearly, CS-modules are weak CS-modules. Smith [12, Corollary 2.7, Theorem 2.8] showed that the result of [6] mentioned above can be extended to weak CS-modules. A module M is called C11-module if, every submodule of M has a complement which is a direct summand of M. Smith and Tercan [14, Theorem 5.2, Corollary 5.3] extended the result of [6] to modules with C11+ (i.e., every direct summand of the module satisfies C11 property).

A module M is called weak C11-module, denoted WC11, if each of its semisimple submodules has a complement which is a direct summand. Tercan [16, Theorem 11, Corollary 12] showed that aforementioned results of [14] can be extended to WC11+-modules.

A module M is called FI-extending if every fully invariant submodule (i.e., every submodule such that the image under all endomorphisms contained in itself) is essential in a direct summand of M (see [3, 4]). Recently, a weak version of FI-extending was introduced and investigated. To this end, following [19], a module is called Weak FI-extending (or, WFI-extending) if, each of its semisimple fully invariant submodules is essential in a direct summand of M. If M is any module and X is any simple submodule, the sum of all submodules of M that are isomorphic to X is a submodule called the homogeneous component of M generated by X. It is well known that the socle of M is the direct sum of the various homogeneous components and any homogeneous component of socle is a fully invariant submodule of the module M. Thus, if a module is WFI-extending then any homogeneous component of socle is essential in a direct summand of the module. Note that the following implications hold for a module M:

$CS⇒C11⇒FI−extending⇓⇓⇓WCS⇒WC11⇒WFI−extending$

No other implications can be added to this table in general. To see why this is the case, please consult [19]. Note that it is an open problem to determine whether the FI-extending (and also the WFI-extending, WC11, WCS) property is inherited by direct summands or not?

The purpose of this paper is to try to extend the result of [16, Theorem 11, Corollary 12] to WFI+-extending (and so also FI+-extending ) modules. To do this, we need to add the pseudo duo condition on the class of fully invariant submodules of the module. Moreover, we also extend a result on the endomorphism ring of continuous modules to WFI-extending modules with the pseudo duo condition which yields that the quotient ring of endomorphism ring with its Jacobson radical is a (von Neumann) regular ring. For any unexplained terminology and definitions, we refer to [1, 5, 10, 18].

### 2. Weak FI+-extending Modules

Let P be some module property of modules. Following [14], we shall say that a module M satisfies P+ if every direct summand of M satisfies P. For example, if a module has injective socle then it satisfies WC11+ and hence also it satisfies WFI+. Moreover, if R is a Dedekind domain then any R-module M with finite uniform dimension is a WFI+-extending module (see [19]). Recall that every direct summand of a non-zero C11+-module with finite uniform dimension is a (finite) direct sum of uniform modules [14, Proposition 4.4]. However, this is not true for WFI+-extending modules, in general. The following example clarifies the situation:

### Example 2.1.

Let R be a principal ideal domain. If R is not a complete discrete valuation ring then there exists an indecomposable torsion-free R-module M of rank 2 by [9, Theorem 19]. For M, . So that M satisfies WFI+ and MR has finite uniform dimension, namely 2. But M is not a direct sum of uniform modules.

For more examples similar to Example 2.1 (see [17, Corollary 16]). Surprisingly, Example 2.1 and [17, Corollary 16] also show that we can not replace WFI+ with FI+ in the former case. Since WFI-extending modules are based on the semisimple fully invariant submodules, the following companion condition works well with WFI-extending property. To this end, M is said to have pseudo duo property provided that any semisimple submodule of M has at least one fully invariant (in M) direct summand in its decomposition i.e., if N is a semisimple submodule of M whenever $N=N1⊕N2$ then at least one of the Ni (i=1,2) is a fully invariant submodule of M. Observe that any duo module clearly satisfies the pseudo duo property. However, there are several modules with the pseudo duo property which are not duo modules. In fact, any non duo module with zero socle would be an example. In particular, let $R=ℤℤ0ℤ.$ Then RR is not duo. But . Hence R satisfies the pseudo duo property. One might wonder whether WFI+-extending with the pseudo duo condition implies FI+-extending ( or C11+ ) or not. However, [19, Example 2.4] makes it clear that the aforementioned implication is not true, in general.

### Theorem 2.2.

Let M be a finitely generated WFI+-extending module with the pseudo duo property. Let N be a semisimple submodule of M such that M/N has finite uniform dimension. Then N is finitely generated.

Proof. Let n < ∞ be the uniform dimension of M/N. Suppose that N is not finitely generated. Then there exist non-finitely generated submodules N1 and N2 such that $N=N1⊕N2$. By hypothesis, at least one of the Ni ( i=1,2) is fully invariant in M, say N1. By WFI-extending, there exist submodules M1, $M′$ of M such that $M=M1⊕M′$ and N1 is essential in M1. Then . Hence by the modular law.

Now so the submodule is not finitely generated. Repeating this argument, there exist $Ni≤Mi≤M$ ($2≤i≤n+1$) such that for each $2≤i≤n+1$, Ni is not finitely generated, $M=M1⊕M2⊕⋯⊕Mn+1$. Let $L=N1⊕N2⊕⋯⊕Nn+1$. Then $M/L≅(M1/N1)⊕(M2/N2)⊕⋯⊕(Mn+1/Nn+1)$. Since M/L has finite uniform dimension then there exists $1≤i≤n+1$ such that Mi=Ni. But Mi is finitely generated and hence so is Ni, a contradiction. Thus N is finitely generated.

The next example shows that WFI+-extending property is not superfluous in Theorem 2.2.

### Example 2.3.

Let Kbe a field and Van infinite dimensional vector space over K. Let Rbe the trivial extension Kwith Vi.e. ,

$R=KV\0K=kv0k∣k∈K, v∈V.$

Then R is a commutative indecomposable ring with respect to the usual matrix operations. Moreover, RR is not WFI+-extending with the pseudo duo property and contains a semisimple submodule I such that R/I has finite uniform dimension but I is not finitely generated.

Proof. Let . It is straightforward to see that RR is not WFI-extending with pseudo duo property. Define $φ:R→K$ by $φ(kv0k)=k$ where k ∈ K, v ∈ V. Then ⊎ is an epimorphism with kernel I. Thus R/I has uniform dimension 1. Since V is infinite dimensional, I is not finitely generated.

### Corollary 2.4.

Let M be a finitely generated FI+-extending module with the pseudo duo property. If has finite uniform dimension then is finitely generated.

Proof. Immediate by Theorem 2.2.

Now, let us think of general modules over arbitrary rings. Since we require both the pseudo duo property and that has finite uniform dimension in our next results, it would be better to clarify these conditions are independent.

### Example 2.5.

• (i) Let M be the free $ℤ$-module of infinite rank i.e., $ℤM=⊕i=1∞ℤ$. Then . Hence M satisfies pseudo duo property. However, which has infinite uniform dimension.

• (ii) Let R be a prime ring and let $MR=(R⊕R)R$. Then, it is clear that which is essential in MR and hence has finite uniform dimension. Now, let . Define $f1:M→M$ by $f1(x,y)=(y,0)$ and $f2:M→M$ by $f2(x,y)=(0,x)$. Obviously f1, . Let , . So, we have $f1(N2)=N1⊆N2$ and $f2(N1)=N2⊆N1$. It follows that MR does not have pseudo duo property.

The following is a key lemma for our main theorem in this section.

### Lemma 2.6.

Let M be a module such that M satisfies WFI+ and has pseudo duo property, and such that has finite uniform dimension. Suppose that is contained in a finitely generated submodule of M. Then M has finite uniform dimension.

Proof. Suppose M does not have finite uniform dimension. Then is not finitely generated. Then there exist submodules S1, S2 of such that Si is not finitely generated for i=1,2, and . By the pseudo duo assumption, without loss of generality, we may assume that S1 is fully invariant in M. By hypothesis, there exist submodules K, $K′$ of M such that $M=K⊕K′$, and S1 is essential in K. By [1, Proposition 9.7, 9.119], . Thus and hence is not finitely generated. Also, , so that , and hence is not finitely generated. By hypothesis, there exists a finitely generated submodule N of M such that . Suppose that . Then is a direct summand of M and hence also a direct summand of N. It follows that is finitely generated which is a contradiction. Thus . Similarly, . Now, . It follows that the modules and each have smaller uniform dimension than . By induction on the uniform dimension of , we conclude that K and K \prime both have finite uniform dimension, and so does $M=K⊕K′$, a contradiction. Thus M has finite uniform dimension.

Now we have the following result which was pointed out in the introduction.

### Theorem 2.7.

Let M be a WFI+-extending module with the pseudo duo property such that has finite uniform dimension. Then M contains a semisimple submodule M1 and a submodule M2 with finite uniform dimension such that $M=M1⊕M2$.

Proof. If then there is nothing to prove. Suppose that . Let m∈ M, . Then for some module X of M. Now, by the pseudo duo property one of the or X is fully invariant in M. First assume that is fully invariant in M. By hypothesis, there exist submodules K, $K′$ of M such that $M=K⊕K′$ and is essential in K. Hence . By Lemma 2.6, K has finite uniform dimension. Now . Otherwise, K≤ mR and hence $mR=K⊕(mR∩K′)$, by the modular law. Since $mR∩K′≅K+mR/mR$, . It follows that and so which is a contradiction.

Now assume that X is a fully invariant submodule of M. By hypothesis, $M=K⊕K′$ and X is essential in $K′$ where K, $K′$ are submodules of M. So . Hence . It follows that i.e., there exists an isomorphism . Note that . So . By Lemma 2.6, K has finite uniform dimension. Observe that . If it were then $α−1(m)R=K⊕(α−1(m)R∩K′)$, by the modular law. Since $α−1(m)R∩K′≅K+α−1(m)R/α−1(m)R$, $α−1(m)R∩K′=0$. Therefore . Hence which yields that . So that , which is a contradiction. Now implies that the module has smaller uniform dimension than . By induction on the uniform dimension of , there exist submodules K1, K2 of K such that $K=K1⊕K2$, K1 is semisimple and K2 has finite uniform dimension. Then M is the direct sum of the semisimple submodule K1, and the submodule $K2⊕K′$, which has finite uniform dimension.

Next we apply the former result to WFI+-extending (and, also FI+-extending) modules which satisfies ACC (respectively, DCC) on essential submodules.

### Corollary 2.8.

Let M be a WFI+-extending module with the pseudo duo property which satisfies ACC (respectively, DCC) on essential submodules. Then $M=M1⊕M2$ for some semisimple submodule M1 and Noetherian (respetively, Artinian) submodule M2.

Proof. We prove the result in the ACC case, the DCC case is similar. Suppose M satisfies ACC on essential submodules. By [8, Proposition 3.6], is Noetherian. Hence by Theorem 2.7, $M=M1⊕M2$ for some semisimple submodule M1 and submodule M2 with finite uniform dimension. Now by [1, Proposition 1.19] and hence . Thus is Noetherian. But is Noetherian, because M2 has finite uniform dimension. Thus M2 is Noetherian.

Recall that a module M is said to have SIP if the intersection of every pair of direct summands is also a direct summand (see, for example [18]). So, we have the following corollaries:

### Corollary 2.9.

Let M be a WFI-extending module with the pseudo duo property which has SIP. Assume M satisfies ACC (respectively, DCC) on essential submodules. Then $M=M1⊕M2$ for some semisimple submodule M1 and Noetherian (respectively, Artinian) submodule M2.

Proof. By [19, Theorem 3.12], M is WFI+-extending module. Now, Corollary 2.8 yields the result

### Corollary 2.10.

Let M be an FI-extending module with the pseudo duo property which has SIP. Assume M satisfies ACC (respectively, DCC) on essential submodules. Then $M=M1⊕M2$ for some semisimple submodule M1 and Noetherian (respectively, Artinian) submodule M2.

Proof. Immediate by Corollary 2.9.

We close this section by giving an example which illustrates that the converse of Theorem 2.7 is not true, in general.

### Example 2.11.

Let F be a field and . Put $R=F+Fx¯2+Fx¯3$ which is a subring of T. Note that R is a commutative local ring and its ideals of are $0,R,Fx¯2,Fx¯3,Fx¯2⊕Fx¯3$ (see [5, Exercise 8.1.10]). Observe that which is essential in R. Clearly R is not WFI-extending.

Now, let $M=M1⊕M2$ be the right R-module where and M2 = R. So, $0⊕M2$ is not WFI-extending which gives that M is not WFI+-extending. Next, let us show that M does not satisfy the pseudo duo property. For, let $N=SocM=M1⊕M1$ and $N1=M1⊕0$, $N2=0⊕M1$. In a similar argument in Example 2.5, we have that neither N1 nor N2 is fully invariant in M. Therefore M does not satisfy the pseudo duo property.

### 3. Endomorphism Rings of Weak FI-extending Modules

In this section our concern is the endomorphism ring of weak FI-extending modules. We will use S and to denote the endomorphism ring of a module M and the Jacobson radical of S respectively. Further Δ will stand for the ideal . Recall that a CS-module M is called continuous if, for each direct summand N of M and each monomorphism $α:N→M$, the submodule ⊎ (N) is also a direct summand of M (see [10, 18]). It was proved in [10, Proposition 3.5] that if M is continuous, then S/Δ is a (von Neumann) regular ring and $Δ=J(S)$. This nice result was generalized to modules with C11 and C2 in [15, Theorem 3.3] as well as weak C11 modules with C2 and essential socle in [17, Theorem 12]. It is natural to expect that whether [10, Proposition 3.5] can be generalized to weak FI-extending modules with C2. However, [17, Example 11] eliminates this expectation. On the other side, let M be the $ℤ$-module $(ℤ/ℤp)⊕ℚ$ where p is any prime integer. Then M is a WFI-extending module with C2 (see [14]). Note that which is not essential in M. Observe that $Mℤ$ has the pseudo duo property. So, we have the following result.

### Theorem 3.1.

Let M be a WFI-extending module with the pseudo duo property, C2 and essential socle. Then $S/Δ$ is a (von Neumann) regular ring and .

Proof. Let α ∈ S. Let . Then K is a direct summand of . Hence for some submodule X of M. By the pseudo duo property, we think of submodules K and X seperately. First assume that K is fully invariant in M. By hypothesis, there exists a complement L of K such that L is a direct summand of M (see [19, Proposition 2.3]). Then $M=L⊕L′$ for some submodule $L′$ of M. Since is essential in M, . It follows that $α|L$ is a monomorphism. So, by C2, α (L) is a direct summand of M. Hence there exists β ∈ S such that$βα=1L$. Then $(α−αβα)(K⊕L)=(α−αβα)(L)=0$, and so . Since K ⊙ L is essential in M, $α−αβα∈Δ$. Therefore $S/Δ$ is a (von Neumann) regular ring.

Next assume that X is a fully invariant submodule of M. By WFI-extending property, there exist direct summands L, $L′$ of M such that $M=L⊕L′$ and X is essential in L. Since is essential in M, $X∩K$ is essential in which gives that . Therefore $α|L$ is a monomorphism. Thus α (L) is a direct summand of M, by C2. Then there exists γ ∈ S such that $γα=1|L$. It can be easily seen that is essential in M. Now, let . So, we have $(α−αγα)(W⊕L)=(α−αγα)(L)=0$, and so . Since $W⊕L$ is essential in M, $α−αγα∈Δ$. Thus $S/Δ$ is a von Neumann regular ring.

In any case, we have that $S/Δ$ is a regular ring. This also proves that . Now, let f∈ Δ . Since and is essential in M, . Hence (1-f)M is a direct summand of M, by C2. However, (1-f)M is essential in M since . Thus (1-f)M=M, and therefore 1-f is a unit in S. It follows that , and hence .

### Corollary 3.2.

Let M be an FI-extending module with the pseudo duo property, C2, and essential socle. Then S/Δ is a (von Neumann) regular ring and $Δ=J(S)$.

Proof. By Theorem 3.1.

### Corollary 3.3.

Let M be a right nonsingular module with the pseudo duo property and essential socle. If M is WFI-extending (or FI-extending ) with C2, then S is a regular ring.

Proof. Let f∈ Δ and . Then, for any x∈ M, $N={r∈R∣xr∈W}$ is an essential right ideal of R. Now f(x)N=0.

Since M is nonsingular, f(x)=0, and since x was arbitrary f=0 (see [20, Lemma 1.3]). It follows that Δ =0. Hence the result follows by the Theorem 3.1.

Note that there are commutative, local rings R such that is simple essential in R. These rings have all the stated properties in Theorem 3.1. For, such rings, see [11, Example 2.6].

In the sense of construction certain examples, Corollary 3.3 is a useful tool. For example, let R be any domain which satisfies C2 condition (i.e., division ring). Let M be the right R-module R. Then, it is easy to see that M has all of the assumptions of Corollary 3.3 except M has essential socle. However, is not (von Neumann) regular. Thus the condition essential socle in Theorem 3.1 is not superfluous.

Furthermore, the next example shows that the pseudo duo assumption in Theorem 3.1 is not unnecessary either.

### Example 3.4.

Let R be any local Kasch ring such that is simple and essential in R (see [11, Example 2.5]). Now, let M be the right R-module $R⊕R$. Observe that is essential in M. It is easy to check that M does not have the pseudo duo property. Moreover, M is a WFI-extending (actually FI-extending) module by [19, Theorem 2.8]. Since R is a Kasch ring, it has C2 property [11, Proposition 1.46]. It is well-known that being right Kasch is Morita invariant which yields that $M2(R)=RRRR$ has C2 condition. By [11, Theorem 7.16], MR satisfies C2 property. But and $S/J(S)$ is not a (von Neumann) regular ring.

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