Article
KYUNGPOOK Math. J. 2019; 59(3): 391401
Published online September 23, 2019
Copyright © Kyungpook Mathematical Journal.
The 𝜋extending Property via Singular Quotient Submodules
Yeliz Kara∗, Adnan Tercan
Department of Mathematics, Bursa Uludağ University, Bursa, 16059, Turkey
email : yelizkara@uludag.edu.tr
Department of Mathematics, Hacettepe University, Ankara, 06532, Turkey
email : tercan@hacettepe.edu.tr
Received: March 27, 2018; Revised: August 7, 2018; Accepted: August 13, 2018
A module is said to be
Keywords: extending modules, πextending module, projection invariant.
All rings are associative with unity unless indicated otherwise.
In this paper, we investigate the
Recall from [11], a module
In this section, we deal with the class of generalized
Lemma 2.1
([6, Exercise 4])

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 X ≤Y ≤M. If X is projection invariant in Y and Y is projection invariant in M then X is projection invariant in M. 
Let M = ⊕_{i∈I}M_{i} and N be a projection invariant submodule of M. Then N = ⊕_{i∈I}(π_{i} (N )) = ⊕_{i∈I}(N ∩M_{i} )where π_{i} is the ith projection map.
It can be easily seen that any singular module satisfies generalized
Lemma 2.2

M is extending. 
M is πextending. 
M is generalized πextending.

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

(ii) ⇒ (iii). Let
N be a projection invariant submodule ofM . Then there exists a direct summandK ofM such thatN ≤_{e} K . Letk ∈K . Then there exists an essential right idealI = {r ∈R kr ∈N } ofR . Hence (k +N )I = 0 implies thatK /N is singular. ThusM is generalizedπ extending. 
(ii) ⇏ (i). Let
M be ℤmodule such thatM = (ℤ/ℤp ) ⊕ (ℤ/ℤp ^{3}) for any primep . It is well known thatM _{ℤ} is not extending by [13, page 1814]. However it isπ extending. 
(iii) ⇏ (ii). Let
R be a subring of$\left[\begin{array}{cc}F& V\\ 0& F\end{array}\right]$ such that$$R=\left\{\left[\begin{array}{cc}f& v\\ 0& f\end{array}\right]:f\in F,v\in V\right\}$$ whereF is a field andV_{F} is a vector space overF with dimension 2. It is clear thatR_{R} is a commutative and indecomposable module. Since the dimension ofV_{F} is 2,R_{R} is not uniform. HenceR_{R} is notπ extending by [2, Proposition 3.8]. On the other hand, let${N}_{1}=\left[\begin{array}{cc}0& {v}_{1}F\oplus 0\\ 0& 0\end{array}\right]$ ,${N}_{2}=\left[\begin{array}{cc}0& 0\oplus {v}_{2}F\\ 0& 0\end{array}\right]\le {R}_{R}$ forv _{1},v _{2} ∈V . It is clear thatN _{1} andN _{2} are projection invariant inR_{R} andN _{1},${N}_{2}\subseteq Z({R}_{R})=\left[\begin{array}{cc}0& V\\ 0& 0\end{array}\right]$ . ThusN _{1} andN _{2} singular. It follows thatR/N _{1} andR/N _{2} are singular which yields thatR_{R} is a generalizedπ extending module.
Proposition 2.3

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

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

(ii) ⇒ (iii). Let
X be a projection invariant essentially closed submodule ofM . Then there exists a direct summandK ofM such thatX ≤K andK/X is singular. HenceX ≤_{e} K by [7, Proposition 1.21]. SinceX has no proper essential extension,X =K which gives thatX is a direct summand ofM . 
(iii) ⇒ (i). Let
N be a projection invariant submodule ofM . Then there exists a submoduleK ofM such thatK is an essential closure ofN inM . SinceM is nonsingular,K is projection invariant inM by [2, Proposition 2.4]. NowK is a direct summand ofM by assumption. ThenM isπ extending.
Recall that a module
Proposition 2.4

M is uniform. 
M is extending. 
M is a C _{11}module. 
M is πextending. 
M is generalized πextending.

(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 ≠
X ≤M . SinceM is indecomposable, every submodule ofM is projection invariant soX ⊴_{p} M . Then there exists a direct summandK ofM such thatX ≤K andK/X is singular. It follows thatX ≤_{e} K by [7, Proposition 1.21]. HenceK =M , soM is uniform.
Observe that if
The following result provides a number of characterizations for generalized
Proposition 2.5

M is a generalized πextending 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 =K ⊕K′ 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 =K ⊕K′ for some submodule K′ of M. 
For any projection invariant submodule N of M, there exists a direct summand K of M such that N ≤K and for any x ∈K, there is an essential right ideal I of R such that xI ≤N.

(i) ⇒ (ii). Let
N ⊴_{p} M . Then there exists a direct summandK ofM such thatN ≤K andK/N is singular. HenceM =K ⊕K′ for some submoduleK′ ofM . It is clear thatN ∩K′ = 0. ThusK/N ≅M/ (K′ ⊕N ) is singular. 
(ii) ⇒ (iii). Take
N = 0 in (ii) which yields the result. 
(iii) ⇒ (iv). Let
N ⊴_{p} M . ThenM/K′ ≅K is singular by the condition (iii). Letx ∈K . Then there exists an essential right idealI ofR such thatxI = 0, asK is singular. Hence we get the result. 
(iv) ⇒ (i). Let
N ⊴_{p} M . Then there exists a direct summandK ofM such thatN ≤K and for anyx ∈K , there is an essential right idealI ofR such thatxI ≤N . We need to show thatK/N is singular. SincexI ≤N , we have (x +N )I ≤N . Thusx +N ∈Z (K/N ), which yields thatK/N is singular.
Any submodule of generalized
Example 2.6
Let
We focus whether the generalized
Lemma 2.7
Let
Proposition 2.8
Let
Direct Sums and Direct Summands
In this section, we examine direct summands and direct sums properties of generalized
It is well known that a direct summand of any extending module is extending. In contrast to extending modules, generalized
Example 3.1
([2, Example 5.5] or [14, Example 4]) Let ℝ be the real field and
We can construct more examples which based on hypersurfaces in projective spaces, over complex numbers.
Theorem 3.2
([10, Theorem 1.5])
There are indecomposable projective
The next proposition gives a condition which ensures that a direct summand of a module is generalized
Proposition 3.3
Let
Theorem 3.4
It is clear that
Now the result follows from Proposition 3.3.
Corollary 3.5
It is clear from Theorem 3.4.
The next result characterizes the direct summand of a generalized
Proposition 3.6
If
Proposition 3.7
Let
Proposition 3.8
Let
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
Theorem 3.9
Let
Corollary 3.10
The result follows from Lemma 2.7 and Theorem 3.9.
Corollary 3.11
It is a consequence of Proposition 3.8 and Theorem 3.9.
Corollary 3.12
Let
Recall that extending property is not closed under essential extensions (see, [11, page 19]). To this end, the following example shows that
Example 3.13
Let
Recall that a ring
Proposition 3.14
Let
Theorem 3.15
Let
Conversely, let
Corollary 3.16
It is clear from Theorem 3.15 and Theorem 3.9.
Proposition 3.17
Let
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