SummandsIn 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 [x_{1}, ..., x_{n}] with indeterminates x_{1}, ..., x_{n} over ℝ. Let R be the ring S/Ss where $s={x}_{1}^{2}+\dots +{x}_{n}^{2}-1$. Then the free R-module $M={\oplus}_{i=1}^{n}R$ is generalized π-extending, but contains a direct summand K_{R} which is not generalized π-extending.
ProofM_{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.
Theorem 3.2
([10, Theorem 1.5]) Let X be the hypersurface in, n ≥ 2, defined by the equation${x}_{0}^{m}+{x}_{1}^{m}+\cdots +{x}_{n+1}^{m}=0$. Let$$R=\u2102\left[{x}_{1},\dots ,{x}_{n+1}\right]/\left(\sum _{i=1}^{n+1}{x}_{i}^{m}+1\right)$$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.
ProofThere 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.
Proposition 3.3
Let M = M_{1} ⊕ M_{2}. Then M_{1}is generalized π-extending if and only if for every projection invariant submodule N of M_{1}there exists a direct summand K of M such that M_{2} ⊆ K, K ∩ N = 0 and M/(K ⊕ N) is singular.
ProofLet N be a projection invariant submodule of M_{1}. Then there exists a direct summand L of M_{1} such that M_{1} = L ⊕ L′ with N ≤ L and M_{1}/(L′ ⊕ N) is singular by Proposition 2.5. It is clear that L′ ⊕ M_{2} is a direct summand of M, M_{2} ⊆ L′ ⊕M_{2} and (L′ ⊕M_{2})∩N = 0. Moreover M_{1}/(L′ ⊕ N) ≅ M/(L′ ⊕ N ⊕M_{2}) is singular. Conversely, let M_{1} holds the assumptions. Let T be a projection invariant submodule of M_{1}. By hypothesis there exists a direct summand K of M such that M_{2} ⊆ K, K ∩ T = 0 and M/(K ⊕ T) is singular. Now K = K ∩ (M_{1} ⊕ M_{2}) = M_{2} ⊕ (K ∩ M_{1}) yields that K ∩ M_{1} is a direct summand of M_{1}. Hence there exists a submodule X of M_{1} such that M_{1} = (K ∩ M_{1}) ⊕ X. Since T is a projection invariant submodule of M_{1}, T = (T ∩ K ∩ M_{1}) ⊕ (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) ≅ M_{1}/[(K ∩ M_{1}) ⊕ T] is singular. Thus Proposition 2.5 yields the result.
Theorem 3.4
Let M = M_{1} ⊕ M_{2}be a generalized π-extending module. If M_{2}is a projection invariant direct summand and for every direct summand K of M with K ∩ M_{2} = 0 and K ⊕ M_{2}is a direct summand of M, then M_{1}and M_{2}are generalized π-extending.
ProofIt is clear that M_{2} is generalized π-extending by Lemma 2.7. Now, let N be a projection invariant submodule of M_{1}. Then N ⊕ M_{2} 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 ⊕ M_{2} ≤ K and M/(K′ ⊕ N ⊕ M_{2}) is singular where M = K ⊕ K′ for some submodule K′ of M. Since K′ ∩ M_{2} ⊆ K′ ∩ (N ⊕ M_{2}) = 0, K′ ⊕ M_{2} is a direct summand of M.
Now the result follows from Proposition 3.3.
Corollary 3.5
Let M = M_{1} ⊕ M_{2}be a generalized π-extending module with C_{3}condition. If M_{2}is a projection invariant direct summand, then M_{1}and M_{2}are generalized π-extending.
ProofIt 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.
ProofIf 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.
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 = Z_{2}(M) ⊕ X ⊕ Y where K = Z_{2}(M) ⊕ X and Y are generalized π-extending.
ProofLet 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 Z_{2}(M) is projection invariant closed submodule of M which follows that Z_{2}(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, Z_{2}(M) ⊆ K. Thus K = Z_{2}(M) ⊕ X where X = (1 − e)M ∩ K. Now, M = K ⊕ N = Z_{2}(M) ⊕ X ⊕ N. 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.
ProofLet 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 f^{2} = 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/ℤp^{3}) 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.
Theorem 3.9
Any direct sum of generalized π-extending modules is generalized π-extending.
ProofLet M = ⊕_{i∈I}M_{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∈I}H_{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.
Corollary 3.10
Let M = ⊕_{i∈I}M_{i} where M_{i} is projection invariant in M for all i ∈ I. ThenM is generalized π-extending if and only if M_{i} is generalized π-extending for all i ∈ I.
ProofThe result follows from Lemma 2.7 and Theorem 3.9.
Corollary 3.11
Let M has an Abelian endomorphism ring. Then M = ⊕_{i∈I}M_{i}is generalized π-extending if and only if M_{i} is generalized π-extending for all i ∈ I.
ProofIt 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 = Z_{2}(M) ⊕ X where X and Z_{2}(M) are generalized π-extending.
ProofLet M be a π-extending module. Then take K = Z_{2}(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 U_{1}, U_{2} of M such that U_{1} ⊕ U_{2} is essential in M. Then U_{1} ⊕ U_{2} is generalized π-extending by Theorem 3.9. However M_{R} 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 e^{2} = e ∈ R such that I ≤ eR 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.
ProofLet 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 e^{2} = 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.
Theorem 3.15
R is a right generalized π-extending ring if and only if R[x] is a right generalized π-extending ring.
ProofLet 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 e^{2} = 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 f^{2} = f ∈ R[x]. Note that fJ[x] = J[x], and let g^{2} = 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] = f_{0}R[x] for some f_{0} ∈ S_{l}(R). Since J[x] ≤ fR[x] = f_{0}R[x], so J ≤ f_{0}R. Further, Z_{R}(f_{0}R/J) ≤ Z_{R[x]}((f_{0}R/J)[x]) ≅ Z_{R[x]}(f_{0}R[x]/J[x]) = Z_{R[x]}(fR[x]/J[x]). Hence, Z_{R}(f_{0}R/J) is singular, as Z_{R[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.
ProofIt is clear from Theorem 3.15 and Theorem 3.9.
Proposition 3.17
Let R be a right nonsingular generalized π-extending ring and F_{R} a free right R-module. Then the endomorphism ring End(F_{R}) of F_{R} is generalized π-extending.
ProofLet 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.