On Fully-M-Cyclic Modules

Received: November 26, 2010 Accepted: December 7, 2010 doi:10.5539/jmr.v3n2p23 Abstract The aim of this work was to generalize generator, M-generated modules in order to apply them to a wider class of rings and modules. We started by establishing a new concept which is called a fully-M-cyclic module. We defined this notation by using HomR(M, ∗) operators which are helpful to contract the new construction and describe their properties. Finally, we could see the structure of fully-M-cyclic module and quasi-fully-cyclic module by the structure of M.


Introduction
Throughout this paper, R is an associative ring with identity and M R is the category of unitary right R-modules.Let M be a right R-module and S = End R (M), its endomorphism ring.A right R-module N is called M-generated if there exists an epimorphism M (I) −→ N for some index set I. If I is finite, then N is called finitely M-generated.In particular, N is called M-cyclic if it is isomorphic to M/L for some submodule L of M. Following Wisbauer [1991], σ[M] denotes the full subcategory of Mod-R, whose objects are the submodules of M-generated modules.A module M is called a self-generator if it generates all of its submodules.M is called a subgenerator if it is a generator of σ[M].

On Fully-M-cyclic module
In this part, a module M be given as a right R-module.
Remark 2.2.Dealing directly from definition, the following statements are routine: (1) Submodule of a fully-M-cyclic module is a fully-M-cyclic module.
(2) If M is simple module and N is fully-M-cyclic module, then any nonzero submodule of N is simple submodule.
Obviously, every semi-simple module is a quasi-fully-cyclic module.
Lemma 2.4.Let N be a fully-M-cyclic module.If M is a noetherian module then S oc(M) S oc(N).
Proof.Since N is a fully-M-cyclic module, a simple submodule B of N is of the form s(M) for some s ∈ Hom R (M, N).By the simply property of B, there is b ∈ B such that B = bR.Suppose that s(a) = b for some a ∈ M. In noetherian module aR, there exists a simple submodule A containing a.It is easily to see that A B. Conversely, if A is a simple submodule of M then s(A) = B is a simple submodule of N and then A B for all s ∈ Hom R (M, N).This shows that S oc(M) S oc (N). 2 Lemma 2.5.If N is a fully-M-cyclic module then N has no nonzero small submodule.
Proof.In a contrary, we suppose that there is a nonzero submodule A which is small in N. Let B be a submodule of N Leading directly from definition, the following properties in Lemma 2.9 are routine, Lemma 2.9.Let N be a fully-M-cyclic module and A be a submodule of N and s its a presented homomorphism.
(1) If M is an epimorphism image of M then N is also a fully-M -cyclic module.
(2) If M is a fully-M -cyclic module then N is also a fully-M -cyclic module.
(3) A is an essential in N if and only if for any nonzero element t of Hom R (M, N), Im(t) ∩ Im(s) 0.
(4) A is uniform if and only if every t ∈ Hom R (M, N) with 0 Im(t) ⊂ > Im(s) then Im(t) is an essential in Im(s).
(5) A is a direct summand of N if and only if there exists t ∈ Hom R (M, N) such that Im(s) ∩ Im(t) = 0 and s + t is an epimorphism.

Quasi-fully-cyclic module
In this part, we put S = End R (M).We have known that for any right R-module M, the direct summand A of M is image of a presented homomorphism which is an idempotent of S but not all.Which is case of the form submodules such that every its presented homomorphisms are idempotents?.The following lemma is a clear answer: Lemma 3.1.Let M be a quasi-fully-cyclic module.If A is a simple submodule of M with s its a presented homomorphism then s is an idempotent of S = End R (M).
Proof.Let s be a presented homomorphism of A. Because A is a simple submodule of M then s 2 (A) 0. Therefore, we have 0 s 2 (M) ⊂ > s(M) = A and s 2 (M) must be equal to A = s(M), showing that s is an idempotent of S . 2 Right now, we suppose that M be a quasi-fully-cyclic module.If e 2 = e, the one gets a direct sum decomposition M = e(M) ⊕ (1 − e)(M).Conversely, if M = A ⊕ B then we can write 1 = π A + π B with π A (resp.π B ) being a natural projection map from M to A ( resp.B). π A ( resp.π B ) is an idempotent element of S which is a presented homomorphism of A ( resp.B) so that we can get the following corollary.
Corollary 3.2.In a quasi-fully-cyclic module, every simple submodule is a direct summand.
Theorem 3.3.Let M be a quasi-fully-cyclic module.M is a Noetherian (resp.Artinian) if and only if S is a right self Noetherian (resp.Artinian) ring.
Proof.Suppose that M is Noetherian.We may easily analogize our self the proof of the case Artinian.Take any ascending chain of the right ideals s 1 S ⊂ > s 2 S ⊂ > s 3 S ⊂ > ...s n S ⊂ > ... of the ring S .Since s i S ⊂ > s i+1 S , s i = s i+1 t for some t ∈ S .We have s i (M) ⊂ > s i+1 (M) for all i ∈ N. The ascending chain of the submodules s .. must be stationary in the noetherian module M so that s n 0 (M) = s n 0 + j (M) for some n 0 and all j ≥ 0. This implies that for i ≥ n 0 there is a permutation function t ∈ S such that s i+1 = s i t for some t ∈ S .It follows that s i+1 S ⊂ > s i S , and hence s i+1 S ⊂ > s i S for all i ≥ n 0 .It says that the ascending chain of the right ideals s 1 S ⊂ > s 2 S ⊂ > s 3 S ⊂ > ...s n S ⊂ > ... must be exact stationary at n 0 .Conversely, if S is a right self noetherian ring.Take any ascending chain of the submodules A 1 ⊂ > A 2 ⊂ > A 3 ⊂ > ... ⊂ > A n ⊂ > ... of M. Since M is a quasi-fully-cyclic module, for every i index, there is s i ∈ S such that s i (M) = A i .Following that s 1 S ⊂ > s 2 S ⊂ > s 3 S ⊂ > ...s n S ⊂ > ... is a ascending chain of the right ideals of S .By assumption, this ascending chain must be stationary at some n 0 index.
is an epimorphism and hence s is an epimorphism.It follows that A = N, a contradiction, showing that N has no nonzero small submodule.LetA = s(M) ⊂ > Im(t) in N. Since A is a maximal submodule of N then Im(t) must be N, and hence t is an epimorphism.Conversely, let A = s(M) and A ⊂ > B. Since N is a fully-M-cyclic module, there is an element t ∈ Hom R (M, N) such that B = t(M).By assumption, the non equality s(M) ⊂ > t(M) follows that t is an epimorphism, and hence B = N.2 2Corollary 2.6.If N is a fully-M-cyclic module then Rad(N) = 0. Definition 2.7.Let N be a fully-M-cyclic module.For a submodule A of N there exists a homomorphism s ∈ Hom R (M, N) such that s(M) = A. s is called a presented homomorphism of A.