• (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 N ≤_e K. Let k ∈ K. Then there exists an essential right ideal I = {r ∈ R | kr ∈ N} 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_ℤ 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 V_F is a vector space over F with dimension 2. It is clear that R_R is a commutative and indecomposable module. Since the dimension of V_F is 2, R_R is not uniform. Hence R_R is not π-extending by [2, Proposition 3.8]. On the other hand, let N1=[0v1F⊕000], N2=[00⊕v2F00]≤RRfor v₁, v₂ ∈ V. It is clear that N₁ and N₂ are projection invariant in R_R and N₁, N2⊆Z(RR)=[0V00]. Thus N₁ and N₂ singular. It follows that R/N₁ and R/N₂ are singular which yields that R_R is a generalized π-extending module.

• (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 X ≤ K and K/X is singular. Hence X ≤_e 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.

• (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. Since M is indecomposable, every submodule of M is projection invariant so X ⊴_p M. Then there exists a direct summand K of M such that X ≤ K and K/X is singular. It follows that X ≤_e 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 x̄J ⊆ I for all x̄ ∈ R/I. Hence I is an essential right ideal of R which gives that R_R 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.

• (i) ⇒ (ii). Let N ⊴_p M. Then there exists a direct summand K of M such that N ≤ K and K/N is singular. Hence M = K ⊕ K′ for some submodule K′ of M. It is clear that N ∩ K′ = 0. Thus K/N ≅ M/(K′ ⊕ N) is singular.

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

• (iii) ⇒ (iv). Let N ⊴_p M. Then M/K′ ≅ K is singular by the condition (iii). Let x ∈ K. 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 N ⊴_p M. Then 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. We need to show that K/N is singular. Since xI ≤ N, we have (x + N)I ≤ N. Thus x + N ∈ Z(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.

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

Let M be generalized π-extending and N ⊴_p M. Then there exists a direct summand K of M such that N ≤ K and K/N is singular. Hence M = K ⊕ K′ for some submodule K′ of M. Let τ : E(M) → E(K) be a projection map. Then τ(M) ≤ M and K/N ≤ E(K)/N = τ(E(M))/N. Since K/N is singular and Z(R_R) = 0, τ(E(M))/N is singular by [7, Proposition 1.23]. Conversely, let N⊴_pM. Then there exists e² = e ∈ End(E(M)) such that N ≤ e(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 N ≤ M ∩ e(E(M)) ≤ e(M) and e(M) ≤ e(E(M)). Thus e(M)/N ≤ e(E(M))/N gives that e(M)/N singular. Therefore M is generalized π-extending.

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

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

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

Let N be a projection invariant submodule of M₁. Then there exists a direct summand L of M₁ such that M₁ = L ⊕ L′ with N ≤ L and M₁/(L′ ⊕ N) is singular by Proposition 2.5. It is clear that L′ ⊕ M₂ is a direct summand of M, M₂ ⊆ L′ ⊕M₂ and (L′ ⊕M₂)∩N = 0. Moreover M₁/(L′ ⊕ N) ≅ M/(L′ ⊕ N ⊕M₂) 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 M₂ ⊆ K, K ∩ T = 0 and M/(K ⊕ T) is singular. Now K = K ∩ (M₁ ⊕ M₂) = M₂ ⊕ (K ∩ M₁) yields that K ∩ M₁ is a direct summand of M₁. Hence there exists a submodule X of M₁ such that M₁ = (K ∩ M₁) ⊕ X. Since T is a projection invariant submodule of M₁, T = (T ∩ K ∩ M₁) ⊕ (T ∩ X) by Lemma 2.1. Note that K ∩ T = 0, hence we get T ≤ X. Furthermore it can be easily seen that M/(K ⊕ T) ≅ M₁/[(K ∩ M₁) ⊕ T] is singular. Thus Proposition 2.5 yields the result.

It is clear that M₂ is generalized π-extending by Lemma 2.7. Now, let N be a projection invariant submodule of M₁. Then N ⊕ M₂ 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 N ⊕ M₂ ≤ K and M/(K′ ⊕ N ⊕ M₂) is singular where M = K ⊕ K′ for some submodule K′ of M. Since K′ ∩ M₂ ⊆ K′ ∩ (N ⊕ M₂) = 0, K′ ⊕ M₂ is a direct summand of M.

Now the result follows from Proposition 3.3.

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.

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 = K ⊕ T for some T ≤ M. Thus K is singular by [7, Proposition 1.23]. It follows that K = Z(M). Thus M = Z(M) ⊕ T for some T ≤ M. Now, let N be a submodule of Z(M). Note that Z(T) = 0, so Hom_R(N, T) = 0 by [7, Proposition 1.20]. Therefore T is Z(M)-injective.

Let M be π-extending, K ⊴_p M and K essentially closed in M. Then M = K ⊕ N for some N ≤ M by [2, Corollary 3.2]. Since K ⊴_p 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 e ∈ S_l(End_R(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)M ∩ K. Now, M = K ⊕ N = Z₂(M) ⊕ X ⊕ N. So let N = Y, Y is the desired direct summand.

Let M be generalized π-extending module and K be a direct summand of M. Let S = End(M_R) and π : M → K′ be the canonical projection where K′ ≤ M such that M = K ⊕ K′. It is clear that kerπ = K. Since S is Abelian, f(kerπ) ⊆ kerπ for all f² = f ∈ S. 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 = R ⊕ R. Then both M_ℤ and T_R is not extending (see, [15]). However, generalized π-extending property yields the following result.

Let M = ⊕_i∈IM_i where M_i is generalized π-extending for all i ∈ I. Let N be projection invariant submodule of M. Then N = ⊕_i∈I (M_i ∩ N) by Lemma 2.1. Note that N ∩ M_i ⊴_p M_i for all i ∈ I. Hence there exists a direct summand H_i of M_i such that N ∩ M_i ≤ H_i and H_i/(N ∩ M_i) is singular. Then H = ⊕_i∈IH_i is a direct summand of M such that N ≤ H. It is clear that H/N is singular. Thus M is generalized π-extending.

The result follows from Lemma 2.7 and Theorem 3.9.

It is a consequence of Proposition 3.8 and Theorem 3.9.

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.

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 M ≅ R/I. Let J/I ⊴_p R/I where I ≤ J ≤ R. Then it is clear that J ⊴_p R. Hence there exists e² = e ∈ R such that J ≤ eR and eR/J is singular. Since J/I ≤ eR/I and (eR/I)/(J/I) ≅ eR/J is singular, M is generalized π-extending.

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 = e ∈ R such that I ≤ eR 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 Z_R[x](eR[x]/I[x]) ≅ (Z_R(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 = f ∈ R[x]. Note that fJ[x] = J[x], and let g² = g ∈ R[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 f ∈ S_l(R[x]). Observe from [1, Proposition 2.4] that fR[x] = f₀R[x] for some f₀ ∈ S_l(R). Since J[x] ≤ fR[x] = f₀R[x], so J ≤ f₀R. Further, Z_R(f₀R/J) ≤ Z_R[x]((f₀R/J)[x]) ≅ Z_R[x](f₀R[x]/J[x]) = Z_R[x](fR[x]/J[x]). Hence, Z_R(f₀R/J) is singular, as Z_R[x](fR[x]/J[x]) is singular. Therefore R is right generalized π-extending.

It is clear from Theorem 3.15 and Theorem 3.9.

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