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eISSN 0454-8124
pISSN 1225-6951

### Article

KYUNGPOOK Math. J. 2019; 59(3): 391-401

Published online September 23, 2019

### The 𝜋-extending Property via Singular Quotient Submodules

Department of Mathematics, Bursa Uludağ University, Bursa, 16059, Turkey
e-mail : yelizkara@uludag.edu.tr
Department of Mathematics, Hacettepe University, Ankara, 06532, Turkey
e-mail : tercan@hacettepe.edu.tr

Received: March 27, 2018; Revised: August 7, 2018; Accepted: August 13, 2018

### Abstract

A module is said to be π-extending provided that every projection invariant submodule is essential in a direct summand of the module. In this article, we focus on the class of modules having the π-extending property by looking at the singularity of quotient submodules. By doing so, we provide counterexamples, using hypersurfaces in projective spaces over complex numbers, to show that being generalized π-extending is not inherited by direct summands. Moreover, it is shown that the direct sums of generalized π-extending modules are generalized π-extending.

Keywords: extending modules, π-extending module, projection invariant.

### Introduction

All rings are associative with unity unless indicated otherwise. R and M will denote a ring and a right R-module, respectively. Recall that a module M is called extending or CS if every submodule of M is essential in a direct summand of M or equivalently; every complement submodule of M is a direct summand of M. There have been many generalizations of extending modules including the class of π-extending modules [15]. A module M is called π-extending [2] if every projection invariant submodule (i.e., every submodule which is invariant under all idempotent endomorphisms of M) is essential in a direct summand of M.

In this paper, we investigate the π-extending property of modules via singular quotient submodules. To this end, we call a module M generalized π-extending if for every projection invariant submodule N of M, there exists a direct summand K of M such that NK and K/N is singular. Furthermore, a ring R is called right generalized π-extending if RR is a right generalized π-extending module. The class of extending modules and the class of π-extending modules are proper subclasses of generalized π-extending modules. We obtain fundamental results related to the notion of generalized π-extending and make connections with the notions of extending and π-extending. Moreover, we determine conditions which ensure the equivalence of the extending, π-extending and generalized π-extending properties. We then examine direct summands and direct sums properties for the aforementioned class. We provide several counterexamples including hypersurfaces in projective spaces over complex numbers to show that being generalized π-extending is not inherited by direct summands. Having done this, we deal with the question of when a direct summand of a generalized π-extending module is generalized π-extending. We prove that the class of generalized π-extending modules is closed under direct sums. Moreover, we exhibit several applications on right generalized π-extending rings including polynomial rings.

Recall from [11], a module M has C2 (C3) condition if for each direct summand K of M and each monomorphism α : KM, the submodule α(K) is a direct summand of M (if for all direct summand K and L of M with KL = 0, the submodule KL is also a direct summand of M). It is clear that C2 implies C3 but not conversely [11]. Throughout this paper, if XM, then XM, Xe M, Xp M, Z(M), Z2(M) and E(M) denote X is a submodule of M, X is an essential submodule of M, X is a projection invariant submodule of M, the singular submodule of M, the second singular submodule of M and the injective hull of M, respectively. A ring R is called Abelian if every idempotent of R is central. Let e2 = eR. Recall from [3], e is called a left (right) semicentral idempotent if xe = exe (ex = exe) for all xR. Sl(R) and Sr(R) denote the set of left semicentral idempotents and right semicentral idempotents, respectively. Other terminology can be found in [5, 9, 11, 15].

### Basic Results

In this section, we deal with the class of generalized π-extending modules. We obtain fundamental results and make connections with the notions of extending and π-extending. Moreover, we determine conditions which ensure the equivalence of the π-extending and generalized π-extending properties. Let us begin with a basic fact about projection invariant submodules.

### Lemma 2.1

([6, Exercise 4]) Let M be a right R-module. Then

• Any sum or intersection of projection invariant submodule of M is projection invariant submodule of M.

• Let X and Y be submodules of M such that XYM. If X is projection invariant in Y and Y is projection invariant in M then X is projection invariant in M.

• Let M = ⊕iIMi and N be a projection invariant submodule of M. Then N = ⊕iI(πi(N)) = ⊕iI(NMi) where πi is the i-th projection map.

It can be easily seen that any singular module satisfies generalized π-extending condition. But the converse of this fact is not true. For example, M = is generalized π-extending, but it is nonsingular.

### Lemma 2.2

Let M be a module. Consider the following statements:

• M is extending.

• M is π-extending.

• M is generalized π-extending.

Then (i) ⇒ (ii) ⇒ (iii), but the reverse implications do not hold, in general.

Proof.
• (i) ⇒ (ii). It is clear from [2, Proposition 3.7].

• (ii) ⇒ (iii). Let N be a projection invariant submodule of M. Then there exists a direct summand K of M such that Ne K. Let kK. Then there exists an essential right ideal I = {rR | krN} of R. Hence (k + N)I = 0 implies that K/N is singular. Thus M is generalized π-extending.

• (ii) ⇏ (i). Let M be ℤ-module such that M = (ℤ/ℤp) ⊕ (ℤ/ℤp3) for any prime p. It is well known that M is not extending by [13, page 1814]. However it is π-extending.

• (iii) ⇏ (ii). Let R be a subring of $[FV0F]$ such that $R={[fv0f]:f∈F,v∈V}$where F is a field and VF is a vector space over F with dimension 2. It is clear that RR is a commutative and indecomposable module. Since the dimension of VF is 2, RR is not uniform. Hence RR is not π-extending by [2, Proposition 3.8]. On the other hand, let $N1=[0v1F⊕000]$, $N2=[00⊕v2F00]≤RR$for v1, v2V. It is clear that N1 and N2 are projection invariant in RR and N1, $N2⊆Z(RR)=[0V00]$. Thus N1 and N2 singular. It follows that R/N1 and R/N2 are singular which yields that RR is a generalized π-extending module.

### Proposition 2.3

The following statements are equivalent for a nonsingular module M.

• M is π-extending.

• M is generalized π-extending.

• Every projection invariant essentially closed submodule of M is a direct summand.

Proof
• (i) ⇒ (ii). It is obvious from Lemma 2.2.

• (ii) ⇒ (iii). Let X be a projection invariant essentially closed submodule of M. Then there exists a direct summand K of M such that XK and K/X is singular. Hence Xe K by [7, Proposition 1.21]. Since X has no proper essential extension, X = K which gives that X is a direct summand of M.

• (iii) ⇒ (i). Let N be a projection invariant submodule of M. Then there exists a submodule K of M such that K is an essential closure of N in M. Since M is nonsingular, K is projection invariant in M by [2, Proposition 2.4]. Now K is a direct summand of M by assumption. Then M is π-extending.

Recall that a module M is called C11-module [13] if each submodule of M has a complement that is a direct summand of M.

### Proposition 2.4

Let M be an indecomposable nonsingular R-module. Then the following statements are equivalent.

• M is uniform.

• M is extending.

• M is a C11-module.

• M is π-extending.

• M is generalized π-extending.

Proof
• (i) ⇒ (ii) ⇒ (iii) ⇒ (iv). These implications are obvious from [2, Proposition 3.7].

• (iv) ⇒ (v) It is clear from Lemma 2.2.

• (v) ⇒ (i). Let 0 ≠ XM. Since M is indecomposable, every submodule of M is projection invariant so Xp M. Then there exists a direct summand K of M such that XK and K/X is singular. It follows that Xe K by [7, Proposition 1.21]. Hence K = M, so M is uniform.

Observe that if R is an indecomposable right generalized π-extending ring, then for all nonzero right ideal I of R, R/I is singular. It follows that there exists an essential right ideal J of R such that x̄JI for all R/I. Hence I is an essential right ideal of R which gives that RR is uniform. Therefore Proposition 2.4 is true without nonsingularity condition for an indecomposable ring R.

The following result provides a number of characterizations for generalized π-extending modules.

### Proposition 2.5

The following statements are equivalent for a module M.

• M is a generalized π-extending module.

• For any projection invariant submodule N of M, there exists a direct summand K of M such that N ≤ K and M/(K′N) is singular where M = KK′ for some submodule K′ of M.

• For any projection invariant submodule N, M/N has a decomposition M/N = (K/N) ⊕ (K′/N) such that K is a singular direct summand of M where M = KK′ for some submodule K′ of M.

• For any projection invariant submodule N of M, there exists a direct summand K of M such that NK and for any xK, there is an essential right ideal I of R such that xIN.

Proof
• (i) ⇒ (ii). Let Np M. Then there exists a direct summand K of M such that NK and K/N is singular. Hence M = KK′ for some submodule K′ of M. It is clear that NK′ = 0. Thus K/NM/(K′N) is singular.

• (ii) ⇒ (iii). Take N = 0 in (ii) which yields the result.

• (iii) ⇒ (iv). Let Np M. Then M/K′K is singular by the condition (iii). Let xK. Then there exists an essential right ideal I of R such that xI = 0, as K is singular. Hence we get the result.

• (iv) ⇒ (i). Let Np M. Then there exists a direct summand K of M such that NK and for any xK, there is an essential right ideal I of R such that xIN. We need to show that K/N is singular. Since xIN, we have (x + N)IN. Thus x + NZ(K/N), which yields that K/N is singular.

Any submodule of generalized π-extending modules need not to be generalized π-extending as shown in the following example.

### Example 2.6

Let M be the Specker group, with Ai = ℤ for any positive integer i. Then M is not π-extending by [6], but M is nonsingular by [7, Proposition 1.22]. Hence M is not generalized π-extending module by Proposition 2.3. However M is a submodule of its injective hull E(M) while E(M) is a generalized π-extending module.

We focus whether the generalized π-extending property is inherited by submodules.

### Lemma 2.7

Let M be a generalized π-extending module and N any projection invariant submodule of M. Then N is a generalized π-extending module.

Proof

Let Xp N and Np M. Hence Xp M by Lemma 2.1. Then there exists a direct summand K of M such that XK and K/X is singular. Hence M = KK′ for some submodule K′ of M. Since Np M, N = (NK)⊕(NK′) by Lemma 2.1. It is clear that XNK where NK is a direct summand of N and (NK)/XK/X. Thus (NK)/X is singular, hence N is generalized π-extending module.

### Proposition 2.8

Let R be a nonsingular right R-module. Then M is generalized π-extending if and only if for any projection invariant submodule N of M, there existse2 = e ∈ End(E(M)) such that N ≤ e(E(M)), e(E(M))/N is singular and e(M) ≤ M.

Proof

Let M be generalized π-extending and Np M. Then there exists a direct summand K of M such that NK and K/N is singular. Hence M = KK′ for some submodule K′ of M. Let τ : E(M) → E(K) be a projection map. Then τ(M) ≤ M and K/NE(K)/N = τ(E(M))/N. Since K/N is singular and Z(RR) = 0, τ(E(M))/N is singular by [7, Proposition 1.23]. Conversely, let NpM. Then there exists e2 = e ∈ End(E(M)) such that Ne(E(M)), e(E(M))/N is singular and e(M) ≤ M by hypothesis. Since e(M) ≤ M, e(M) is a direct summand of M. It is clear that NMe(E(M)) ≤ e(M) and e(M) ≤ e(E(M)). Thus e(M)/Ne(E(M))/N gives that e(M)/N singular. Therefore M is generalized π-extending.

### Direct Sums and Direct Summands

In this section, we examine direct summands and direct sums properties of generalized π-extending modules. It is shown that a direct summand of generalized π-extending modules need not to be generalized π-extending. Further, we deal with when a direct summand of a generalized π-extending module is generalized π-extending. Moreover, we are able to show that the class of generalized π-extending modules is closed under direct sums.

It is well known that a direct summand of any extending module is extending. In contrast to extending modules, generalized π-extending property is not inherited by direct summands. The following results illustrate this fact.

### Example 3.1

([2, Example 5.5] or [14, Example 4]) Let ℝ be the real field and n any odd integer with n ≥ 3. Let S be the polynomial ring ℝ [x1, ..., xn] with indeterminates x1, ..., xn over ℝ. Let R be the ring S/Ss where $s=x12+…+xn2−1$. Then the free R-module $M=⊕i=1nR$ is generalized π-extending, but contains a direct summand KR which is not generalized π-extending.

Proof

MR is a π-extending module which contains a direct summand KR is not π-extending by [2, Example 5.5]. Hence MR is generalized π-extending module by Lemma 2.2. It is clear that R is a commutative Noetherian domain. Then MR is a nonsingular module, so is KR. Therefore KR is not generalized π-extending by Proposition 2.3.

We can construct more examples which based on hypersurfaces in projective spaces, over complex numbers.

### Theorem 3.2

([10, Theorem 1.5]) Let X be the hypersurface in, n ≥ 2, defined by the equation$x0m+x1m+⋯+xn+1m=0$. Let$R=ℂ[x1,…,xn+1]/(∑i=1n+1xim+1)$be the coordinate ring of X. Then there exist generalized π-extending R-modules but contain direct summands which are not generalized π-extending for m ≥ n + 2.

Proof

There are indecomposable projective R-modules of rank n over R by [12]. Then there exists a free R-module FR such that FR = KK′ where K is indecomposable and projective R-module of rank n. From Theorem 3.9, FR is generalized π-extending. However KR is not uniform. Thus KR is not generalized π-extending by Proposition 2.4.

The next proposition gives a condition which ensures that a direct summand of a module is generalized π-extending module.

### Proposition 3.3

Let M = M1M2. Then M1is generalized π-extending if and only if for every projection invariant submodule N of M1there exists a direct summand K of M such that M2K, KN = 0 and M/(KN) is singular.

Proof

Let N be a projection invariant submodule of M1. Then there exists a direct summand L of M1 such that M1 = LL′ with NL and M1/(L′N) is singular by Proposition 2.5. It is clear that L′M2 is a direct summand of M, M2L′M2 and (L′M2)∩N = 0. Moreover M1/(L′N) ≅ M/(L′NM2) is singular. Conversely, let M1 holds the assumptions. Let T be a projection invariant submodule of M1. By hypothesis there exists a direct summand K of M such that M2K, KT = 0 and M/(KT) is singular. Now K = K ∩ (M1M2) = M2 ⊕ (KM1) yields that KM1 is a direct summand of M1. Hence there exists a submodule X of M1 such that M1 = (KM1) ⊕ X. Since T is a projection invariant submodule of M1, T = (TKM1) ⊕ (TX) by Lemma 2.1. Note that KT = 0, hence we get TX. Furthermore it can be easily seen that M/(KT) ≅ M1/[(KM1) ⊕ T] is singular. Thus Proposition 2.5 yields the result.

### Theorem 3.4

Let M = M1M2be a generalized π-extending module. If M2is a projection invariant direct summand and for every direct summand K of M with KM2 = 0 and KM2is a direct summand of M, then M1and M2are generalized π-extending.

Proof

It is clear that M2 is generalized π-extending by Lemma 2.7. Now, let N be a projection invariant submodule of M1. Then NM2 is a projection invariant submodule of M by [2, Lemma 4.13]. By Proposition 2.5, there exists a direct summand K of M such that NM2K and M/(K′NM2) is singular where M = KK′ for some submodule K′ of M. Since K′M2K′ ∩ (NM2) = 0, K′M2 is a direct summand of M.

Now the result follows from Proposition 3.3.

### Corollary 3.5

Let M = M1M2be a generalized π-extending module with C3condition. If M2is a projection invariant direct summand, then M1and M2are generalized π-extending.

Proof

It is clear from Theorem 3.4.

The next result characterizes the direct summand of a generalized π-extending module in terms of relative injectivity.

### Proposition 3.6

Let R be a nonsingular right R-module and M a generalized π-extending module. Then M = Z(M) ⊕ T for some submodule T of M and T is Z(M)-injective.

Proof

If Z(M) = 0 or Z(M) = M, then the result holds trivially. Assume Z(M) 0 and Z(M) ≠ M. Since Z(M) is a fully invariant submodule of M, it is also projection invariant submodule of M. Hence there exists a direct summand K of M such that Z(M) ≤ K and K/Z(M) is singular where M = KT for some TM. Thus K is singular by [7, Proposition 1.23]. It follows that K = Z(M). Thus M = Z(M) ⊕ T for some TM. Now, let N be a submodule of Z(M). Note that Z(T) = 0, so HomR(N, T) = 0 by [7, Proposition 1.20]. Therefore T is Z(M)-injective.

### Proposition 3.7

Let M be a π-extending module and K a projection invariant submodule of M such that K is essentially closed in M. If M/K is nonsingular, then M = Z2(M) ⊕ XY where K = Z2(M) ⊕ X and Y are generalized π-extending.

Proof

Let M be π-extending, Kp M and K essentially closed in M. Then M = KN for some NM by [2, Corollary 3.2]. Since Kp M, K and N are π-extending by [2, Proposition 4.14] and hence K and N are generalized π-extending by Lemma 2.2. Note that Z2(M) is projection invariant closed submodule of M which follows that Z2(M) = eM for some eSl(EndR(M)) by [2, Proposition 4.12] and [2, Corollary 3.2]. Since M/K is nonsingular, Z2(M) ⊆ K. Thus K = Z2(M) ⊕ X where X = (1 − e)MK. Now, M = KN = Z2(M) ⊕ XN. So let N = Y, Y is the desired direct summand.

### Proposition 3.8

Let M be a generalized π-extending module with Abelian endomorphism ring. Then every direct summand of M is generalized π-extending.

Proof

Let M be generalized π-extending module and K be a direct summand of M. Let S = End(MR) and π : MK′ be the canonical projection where K′M such that M = KK′. It is clear that kerπ = K. Since S is Abelian, f(kerπ) ⊆ kerπ for all f2 = fS. Hence K is a projection invariant submodule of M. Therefore apply Lemma 2.7 to get the result.

It is well known that a direct sum of extending modules (even, for uniform modules) need not to be an extending module, in general. For example, let M be the ℤ-module (ℤ/ℤp) ⊕ (ℤ/ℤp3) for any prime p, and let R = ℤ[x] be the polynomial ring. Now, let us think of the free R-module T = RR. Then both M and TR is not extending (see, [15]). However, generalized π-extending property yields the following result.

### Theorem 3.9

Any direct sum of generalized π-extending modules is generalized π-extending.

Proof

Let M = ⊕iIMi where Mi is generalized π-extending for all iI. Let N be projection invariant submodule of M. Then N = ⊕iI (MiN) by Lemma 2.1. Note that NMip Mi for all iI. Hence there exists a direct summand Hi of Mi such that NMiHi and Hi/(NMi) is singular. Then H = ⊕iIHi is a direct summand of M such that NH. It is clear that H/N is singular. Thus M is generalized π-extending.

### Corollary 3.10

Let M = ⊕iIMi where Mi is projection invariant in M for all iI. ThenM is generalized π-extending if and only if Mi is generalized π-extending for all iI.

Proof

The result follows from Lemma 2.7 and Theorem 3.9.

### Corollary 3.11

Let M has an Abelian endomorphism ring. Then M = ⊕iIMiis generalized π-extending if and only if Mi is generalized π-extending for all iI.

Proof

It is a consequence of Proposition 3.8 and Theorem 3.9.

### Corollary 3.12

Let M be a nonsingular module. Then M is π-extending module if and only if M = Z2(M) ⊕ X where X and Z2(M) are generalized π-extending.

Proof

Let M be a π-extending module. Then take K = Z2(M) and apply Proposition 3.7 to get the result. Conversely, assume that M has stated property. It is clear from Theorem 3.9, M is generalized π-extending. Hence M is π-extending by Proposition 2.3.

Recall that extending property is not closed under essential extensions (see, [11, page 19]). To this end, the following example shows that π-extending and generalized π-extending modules behave same as extending modules with respect to the essential extensions.

### Example 3.13

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 [8, Theorem 19]. Hence there exist uniform submodules U1, U2 of M such that U1U2 is essential in M. Then U1U2 is generalized π-extending by Theorem 3.9. However MR is not generalized π-extending by Proposition 2.4.

Recall that a ring R is right generalized π-extending in case for every projection invariant right ideal I of R, there exists e2 = eR such that IeR and eR/I is singular. Finally, we obtain the following applications on generalized π-extending rings.

### Proposition 3.14

Let R be a right generalized π-extending ring. Then every cyclic R-module is generalized π-extending.

Proof

Let R be a right generalized π-extending ring and M a cyclic R-module. Then there exists a right ideal I of R such that MR/I. Let J/Ip R/I where IJR. Then it is clear that Jp R. Hence there exists e2 = eR such that JeR and eR/J is singular. Since J/IeR/I and (eR/I)/(J/I) ≅ eR/J is singular, M is generalized π-extending.

### Theorem 3.15

R is a right generalized π-extending ring if and only if R[x] is a right generalized π-extending ring.

Proof

Let R be a right generalized π-extending ring and I[x] be a projection invariant right ideal of R[x]. Then I is a projection invariant right ideal of R by [4, Lemma 4.1]. Thus there exists e2 = eR such that IeR and eR/I is singular. Notice that I = eI, so I[x] = eI[x]. It is clear that eR[x] is a direct summand of R[x] and I[x] = eI[x] ≤ eR[x]. It is easy to see that eR[x]/I[x] ≅ (eR/eI)[x]. Observe that ZR[x](eR[x]/I[x]) ≅ (ZR(eR/eI))[x] = (eR/eI)[x] ≅ eR[x]/I[x] which shows that eR[x]/I[x] is singular. Hence R[x] is right generalized π-extending.

Conversely, let R[x] be right generalized π-extending and J a projection invariant right ideal of R. Then J[x] is a projection invariant right ideal of R[x] by [4, Lemma 4.1]. It follows that J[x] ≤ fR[x] and fR[x]/J[x] is singular for some f2 = fR[x]. Note that fJ[x] = J[x], and let g2 = gR[x]. Then g(J[x]) ⊆ J[x], as J[x] is projection invariant right ideal of R[x]. Hence we obtain that fgf = gf, so fSl(R[x]). Observe from [1, Proposition 2.4] that fR[x] = f0R[x] for some f0Sl(R). Since J[x] ≤ fR[x] = f0R[x], so Jf0R. Further, ZR(f0R/J) ≤ ZR[x]((f0R/J)[x]) ≅ ZR[x](f0R[x]/J[x]) = ZR[x](fR[x]/J[x]). Hence, ZR(f0R/J) is singular, as ZR[x](fR[x]/J[x]) is singular. Therefore R is right generalized π-extending.

### Corollary 3.16

Let R be a right generalized π-extending (or uniform) ring and R[x] the polynomial ring. Then every free right R[x]-module is generalized π-extending.

Proof

It is clear from Theorem 3.15 and Theorem 3.9.

### Proposition 3.17

Let R be a right nonsingular generalized π-extending ring and FR a free right R-module. Then the endomorphism ring End(FR) of FR is generalized π-extending.

Proof

Let R be a right nonsingular generalized π-extending ring and FR a free right R-module. Then FR is generalized π-extending by Theorem 3.9. Since R is right nonsingular, we obtain that End(FR) is right π-extending by [15, Theorem 4.157], so is generalized π-extending by Lemma 2.2.

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