GENERALIZED LIFTING MODULES

We introduce the concepts of lifting modules and (quasi-)discrete modules relative to a given left module. We also introduce the notion of SSRS-modules. It is shown that (1) if M is an amply supplemented module and 0→N ′ →N →N ′′ → 0 an exact sequence, then M isN-lifting if and only if it isN ′-lifting andN ′′-lifting; (2) ifM is a Noetherian module, then M is lifting if and only if M is R-lifting if and only if M is an amply supplemented SSRS-module; and (3) let M be an amply supplemented SSRS-module such that Rad(M) is finitely generated, then M = K ⊕K ′, where K is a radical module and K ′ is a lifting module.


Introduction and preliminaries
Extending modules and their generalizations have been studied by many authors (see [2,3,8,7]).The motivation of the present discussion is from [2,8], where the concepts of extending modules and (quasi-)continuous modules with respect to a given module and CESS-modules were studied, respectively.In this paper, we introduce the concepts of lifting modules and (quasi-)discrete modules relative to a given module and SSRSmodules.It is shown that (1) if 0 → N → N → N → 0 is an exact sequence and M an amply supplemented module, then M is N-lifting if and only if it is both N -lifting and N -lifting; (2) if M is a Noetherian module, then M is lifting if and only if M is R-lifting if and only if M is an amply supplemented SSRS-module; and (3) let M be an amply supplemented SSRS-module such that Rad(M) is finitely generated, then M = K ⊕ K , where K is a radical module and K is a lifting module.
Throughout this paper, R is an associative ring with identity and all modules are unital left R-modules.We use N ≤ M to indicate that N is a submodule of M. As usual, Rad(M) and Soc(M) stand for the Jacobson radical and the socle of a module M, respectively.
Let M be a module and S ≤ M. S is called small in M (notation S M) if M = S + T for any proper submodule T of M. Let N and L be submodules of M, N is called a supplement of L in M if N + L = M, and N is minimal with respect to this property.Equivalently, supplement of some submodule of M. M is called an amply supplemented module if for any two submodules A and B of M with A + B = M, B contains a supplement of A. M is called a weakly supplemented module (see [5]) if for each submodule A of M there exists a submodule B of M such that Let M be a module.M is called a lifting module (or satisfies (D 1 )) (see [9]) if for every submodule A of M, there exists a direct summand K of M such that K ≤ A and A/K M/K, equivalently, M is amply supplemented and every supplement submodule of M is a direct summand.M is called discrete if M is lifting and has the following condition.
lifting and has the following condition: (D 3 ) For each pair of direct summands A and For more details on these concepts, see [9].
Lemma 1.1 (see [12, 19.3]).Let M be a module and If M is a module and φ : M → M a homomorphism, then φ(L) M whenever L M.
Lemma 1.2 (see Lemma 1.1 in [5]).Let M be a weakly supplemented module and N ≤ M.
Then the following statements are equivalent.
(1) N is a supplement submodule of M.
(3) For all X ≤ N, X M implies X N.
Lemma 1.3 (see Proposition 1.5 in [5]).Let M be an amply supplemented module.Then every submodule of M has an s-closure.

Relative lifting modules
To define the concepts of relative lifting and (quasi-)discrete modules, we dualize the concepts of relative extending and (quasi-)continuous modules introduced in [8] in this section.We start with the following.
Let N and M be modules.We define the family Proposition 2.1.$(N,M) is closed under small submodules, isomorphic images, and coessential extensions.
Proof.We only show that $(N,M) is closed under coessential extensions.Let A∈$(N,M), A ≤ A ≤ M, and A /A M/A.There exist X ≤ N and f ∈ Hom(X,M) such that f (X Proof.There exist Proof.It is clear by Lemma 1.1(1).
Proposition 2.4.Let M be an amply supplemented module.Then every . Note that M is amply supplemented, and so f (X) has an s-closure A in M by Lemma 1.3.Thus A is also an s-closure of A by Lemma 2.3.The verification for A ∈ $(N,M) is analogous to that for B ∈ $(N,M) in Lemma 2.2.
Let N be a module.Consider the following conditions for a module M. ($(N,M)-D 1 ) For every submodule A ∈ $(N,M), there exists a direct summand K of M such that and $(N,M)-D 3 , respectively.
One easily obtains the hierarchy: Clearly, the notion of relative discreteness generalizes the concept of discreteness.For any module N, lifting modules are N-lifting.But the converse is not true as shown in the following examples.
Example 2.6.Since, for any module M, $(0,M) = {A | A M} and 0 is a direct summand of M such that A/0 M/0 for any A ∈ $(0,M), all modules are 0-lifting.However, the Z-module Z/2Z × Z/8Z is not lifting since the supplement submodule (1,2) ( (1,2) is a supplement of (1,1) ) and is not a direct summand of it though it is amply supplemented.
Example 2.7.Let M be a module with zero socle and S a simple module.Then M is Slifting since $(S,M) is a family only containing all small submodules of M. So all torsionfree Z-modules are S-lifting for any simple Z-module S (see [12,Exercise 21.17]).In particular, Z Z and Z Q are S-lifting for any simple Z-module, but each one is not a lifting module.Proof.Let M satisfy $(N,M)-D 1 and H be a coclosed submodule of M. H is amply supplemented by Lemma 1.4.For any Corollary 2.11.Let M be an amply supplemented module.Then the condition $(N,M)-D 1 is inherited by direct summands of M. Proposition 2.12.Let M be an amply supplemented module.Then $(N,M)-D i (i = 2,3) is inherited by direct summands of M.
Proof.(1) Let M satisfy $(N,M)-D 2 and H be a direct summand of M. We will show that H satisfies $(N,H)-D 2 . Let and hence A is a direct summand of H.
(2) Let A ∈ $(N,H) ⊆ $(N,M) and A, L be direct summands of H with A + L = H.We will show that A ∩ L is a direct summand of H. Since H is a direct summand of M, there exists Theorem 2.13.Let M be an amply supplemented module and Proof.The proof follows from Lemma 2.2, Corollary 2.11, and Proposition 2.12.
Proof.Without loss of generality we can assume that N ≤ N and N = N/N .By definition, N ≤ N implies $(N ,M) ⊆ $(N,M).Next, let A 2 ∈ $(N ,M).Then there exist ).The rest is obvious.
Dual to [8, Proposition 2.7], we have the following.
Theorem 2.15.Let 0 → N → N → N → 0 be an exact sequence and M an amply supplemented module.Then M is N-lifting if and only if it is both N -lifting and N -lifting.
Proof.Let M be N-lifting.Then it is both N -lifting and N -lifting by Proposition 2.14.Conversely suppose that M is both N -lifting and N -lifting.For any submodule Corollary 2.17.Let M be an amply supplemented module.Then M is lifting if and only if M is N-lifting and M/N-lifting for every submodule N of M if and only if M is N-lifting and M/N-lifting for some submodule N of M.
Recall that a module M is said to be distributive if [4]) if the sum of any pair of direct summands of M is a direct summand of M.
Corollary 2.18.Let 0 → N → N → N → 0 be an exact sequence and let M be a distributive and amply supplemented module with SSP.If M is both N -quasidiscrete and Nquasidiscrete, then M is N-quasidiscrete.
Proof.We only need to show that M satisfies $(N,M)-D 3 when M satisfies $(N ,M)-D 3 and $(N ,M)-D 3 by Theorem 2.15.Let A ∈ $(N,M) and A, H be direct summands of M with A + H = M.We know that A = A 1 ⊕ A 2 , where A 1 ∈ $(N ,M), A 2 ∈ $(N ,M) from the proof of Theorem 2.15.Since M is a distributive module with SSP, A 1 ∩ H and A 2 ∩ H are direct summands of M. This implies that A ∩ H is a direct summand of M. Thus M satisfies $(N,M)-D 3 .

SSRS-modules
In [2], a module is called a CESS-module if every complement with essential socle is a direct summand.As a dual of CESS-modules, the concept of SSRS-modules is given in this section.It is proven that: (1) let M be an amply supplemented SSRS-module such that Rad(M) is finitely generated, then M = K ⊕ K , where K is a radical module and K is a lifting module; (2) let M be a finitely generated amply supplemented module, then M is an SSRS-module if and only if M/K is a lifting module for every coclosed submodule K of M. Definition 3.1.A module is called an SSRS-module if every supplement with small radical is a direct summand.
Lifting modules are SSRS-modules, but the converse is not true.For example, Z Z is an SSRS-module which is not a lifting module.Proposition 3.2.Let M be an SSRS-module.Then any direct summand of M is an SSRSmodule.
Proof.Let K be a direct summand of M and N a supplement submodule of K such that Rad(N) N. Let N be a supplement of L in K, that is, N + L = K and N ∩ L N. Since K is a direct summand of M, there exists Thus N is a direct summand of M since M is an SSRS-module.So N is a direct summand of K.The proof is complete.

Call for Papers
Thinking about nonlinearity in engineering areas, up to the 70s, was focused on intentionally built nonlinear parts in order to improve the operational characteristics of a device or system.Keying, saturation, hysteretic phenomena, and dead zones were added to existing devices increasing their behavior diversity and precision.In this context, an intrinsic nonlinearity was treated just as a linear approximation, around equilibrium points.Inspired on the rediscovering of the richness of nonlinear and chaotic phenomena, engineers started using analytical tools from "Qualitative Theory of Differential Equations," allowing more precise analysis and synthesis, in order to produce new vital products and services.Bifurcation theory, dynamical systems and chaos started to be part of the mandatory set of tools for design engineers.
This proposed special edition of the Mathematical Problems in Engineering aims to provide a picture of the importance of the bifurcation theory, relating it with nonlinear and chaotic dynamics for natural and engineered systems.Ideas of how this dynamics can be captured through precisely tailored real and numerical experiments and understanding by the combination of specific tools that associate dynamical system theory and geometric tools in a very clever, sophisticated, and at the same time simple and unique analytical environment are the subject of this issue, allowing new methods to design high-precision devices and equipment.
Authors should follow the Mathematical Problems in Engineering manuscript format described at http://www .hindawi.com/journals/mpe/.Prospective authors should submit an electronic copy of their complete manuscript through the journal Manuscript Tracking System at http:// mts.hindawi.com/according to the following timetable:

Lemma 2 . 8 .
Let M be a module.Then $(M,M) = {A | A ≤ M} = N∈R-Mod $(N,M), where R-Mod denotes the category of left R-module.Proof.It is straight forward.Proposition 2.9.Let M be a module.Then M is lifting or (quasi-)discrete if and only if M is M-lifting or M-(quasi-)discrete if and only if M is N-lifting or N-(quasi-)discrete for any module N. Proof.It is clear by Lemma 2.8.Proposition 2.10.Let M be an amply supplemented module.Then the condition $(N,M)-D 1 is inherited by coclosed submodules of M.

Proposition 3 . 3 .
Let M be a weakly supplemented SSRS-module and K a coclosed submodule of M. Then K is an SSRS-module.Proof.It follows from the assumption and [4, Lemma 2.6(3)].Proposition 3.4.Let M be an amply supplemented module.Then M is an SSRS-module if and only if for every submodule N with small radical, there exists a direct summand K of M such that K ≤ N and N/K M/K.Proof."⇐."Let N be a supplement submodule with small radical.By assumption, there exists a direct summand K of M such that K ≤ N and N/K M/K.Since N is coclosed in M, N = K.Thus N is a direct summand of M. "⇒."Let N ≤ M with Rad(N) N.There exists an s-closure N of N since M is amply supplemented.Since Rad(N) M (for Rad(N) N) and Rad(N) ≤ Rad(N),