2-Visible Submodules and Fully 2-Visible Modules

: Let X be a T -module, T is a commutative ring with identity and K be a proper submodule of X . In this paper we introduce the concepts of 2-visible submodules and fully 2-visible modules as a generalizations of visible submodules and fully visible modules resp., where K is said to be 2-visible whenever K=I 2 K for every nonzero ideal I of T and A T -module X is called fully 2-visible if for any proper submodule of it is 2-visible.Study some of the properties of these concepts also discuss the relationship 2-visible submodules and fully 2-visible modules with 2-pure submoules and other related submodules and modules resp. are given.


INTRODUCTION
In this study, T is a commutative ring with identity and X is unitary module.Anderson and Fuller [1] called the submodule N a pure submodule of X if IN = N ∩ IX for every ideal I of T. Ribenboim [2] defined N to be pure in M if rX ∩N = rN for each r ∈T.Mijbass in 1992 presented the concept of cancellation module "A T-module X is defined to be a cancellation module if IX=JX for ideals I and J of T implies I=J ", [3].
In [4] Mahmood and Buthyna presented the concept of visible submodule" A proper submodule W of a module X over a ring T, so that it achieves W = IW for every nonzero ideal I of T. Many of the properties which characterize this concept have been built add to a lot of important results and features have been submitted in [5].Also in [6] Mahmood presented the concept of fully visible module, which is defined as the module , in which all submodules are visible.
In this article we made a generalizations of visible submodules to a 2-visible submodules and fully visivle modules to fully 2-visible modules, many properties of these concepts have been proved.
Note that: we denote to submodule: subm.

2-VISIBLE SUBMS
Definition 2.1.Let X be a T-module and K be a proper subm. of X.Then K is said to be 2-visible whenever K = l 2 K for every nonzero ideal I of T.
An ideal l of a ring T is called 2-visible if it is 2-visible T-subm. of X.
Remarks and Examples 2.2. 1) A zero subm. of any T-module X is always 2-visible.

4)
Let K be a 2-visible subm. of X.Then for every proper subm.L of X is 2-visible whenever f : K → L is an epimorphisim.
Proof.Let f : K → L be an epimorphisim.Then f (K) = L, but K is 2-visible subm. of X, therefore for every nonzero ideal 5) Let X 1 and X 2 be two T-modules and ψ : Proof.Since K is 2-visible subm. of X 2 , then for every nonzero ideal l of T, we have Proposition 2.4.Let X be a T-module and D,L be two subms of X.If D,L are 2-visible subms, then D + L is 2-visible subm.
Proof.Let A be a nonzero ideal of T.
Remark 2.5.As a generalization of Proposition (2.4), we get: if {N k } n k=1 is a finite collection of subms of a T-module X and N k is 2-visible subm.for all k, then the sum of all these subms is 2-visible subm. of X.
Proposition 2.6.Let X be a T-module and N be a 2-visible subm. of X.Then every subm. of N is also 2-visible.
Proof.Let N be a 2-visible subm. of a T-module X and let K be a proper subm. of N, that is K ⊆ N. Since N = I 2 N for every nonzero ideal l of T by hypothesis.Then K ⊆ I 2 N which implies that Also, from the above inclusion, we get l 2 K ⊆ I 2 N.And hence Form (1) and (2), we get Remark 2.7.The converse of Proposition (2.6) need not to be correct for example: The module Z 12 as Z 36 -module.Since (0) is contains in any subm. of any T -module X and (0) is 2-visible subm.by Remarks and Examples(2.2).But a subm.(6) of module Z 12 is not 2-visible, since there is a nonzero ideal of Z 36 that is As a directly result of Corollary (2.8), we give the following generalization.
Corollary 2.9.Let {N i } n i=1 be a family of subms of a T -module X such that at least one of them is 2-visible, then ∩ n i=1 N i is 2-visible subm.
Proof.We know that n i=1 N i ⊆ N i ∀i.Then the result is directly from Proposition (2.6).
Proposition 2.10.For each nonzero ideal A of T and for each nonempty collection {W α } of 2-visible subm. of an Tmodule X.We have A odule for each ∞ and hence A 2 W x = W x for each ∞, also by Proposition (2.6), we get 4) The module Z p ⊕ Z over a ring Z is not fully 2-visible, P is prime number, since ( 0)⊕Z be a proper subm. of Z p ⊕ Z. Then (0) ⊕ Z ̸ = (P) 2 ((0) ⊕ Z) = (0) ⊕ P 2 Z.This means that we found an ideal in T does not achive definition.5) Z 12 as Z 36 -module is not fully 2-visible , every proper subm (  2) , ( 4) , (  8) there is no a nonzero ideal I of Z 36 such that I = IN.6) Let X be an T -module.Then every subm. of fully 2-visible module is also fully 2-visible.
Proof.Suppose that X is fully 2-visible module and let N be a proper subm. of X.To prove that N is also fully 2-visible.Now let K be a proper subm. of N. Hence K ⊂ X and by hypothesis on X, we get K is 2-visible subm., therefore N is fully 2-visible.Proposition 3.3.A T -module X is fully 2-visible iff X/K is fully 2-visible T -module for every proper subm.K of X.
Proof.Suppose that X is fully 2-visible and W /K be a proper subm. of X/K and let I be a nonzero ideal of T .To prove The opposite side is clear by taking K = 0. Corollary 3.4.Let X,X ′ be two T -modules and f : X → X ′ is an T -epimorphism.If X is a fully 2-visible module, then X ′ is a fully 2-visible module.
Proof.We have f : X → X ′ is an epimorphisim and X is a fully 2-visible module.Then by the first isomorphisim theorem, we get X/ker f ∼ = X.But X/ker f is a fully 2-visible module by Proposition 3.3.Therefore X ′ is a fully visible module.Proposition 3.5.Let X = L ⊕ N be an T -module , where L and N are two subms of X such that annL + annN = T .Then L and N are fully 2-visible modules iff X is a fully 2-visible module.
Proof.Let P be a proper subm. of X.Then P = L 1 ⊕ N 1 where L 1 and N 1 are subms. of L and N respectively , also let 0 ̸ = I be an ideal of T , then Proposition 3.6.Let X 1 and X 2 be two T -modules such that ann(X 1 ) + ann(X 2 ) = T , then X = X 1 ⊕ X 2 is fully 2-visible module iff X 1 and X 2 are fully 2-visible module.
Proof.Suppose that X = X 1 ⊕ X 2 is fully 2-visible module on T .Let ψ i : X → X i be the natural projection map , ∀i = 1, 2. Note ψ i is an epimorphism and X is fully 2-visible module then by Corollary 3.4, we get X i is fully 2-visible module.Proposition 3.7.Let X 1 , X 2 , be two T -modules and W 1 ,W 2 , are two subms. of X 1 , X 2 , resp.such that ann T X 1 +ann T X 2 = T.
Proof.Let K 1 ⊕ K 2 be a proper subm. of W 1 ⊕W 2 where K 1 and K 2 are subms of W , ,W 2 resp.and for each a nonzero ideal l of T .To prove The opposite side: let K 1 be a proper subm. of W 1 .Then K 1 ⊆ W 1 ⊆ W 1 ⊕ W 2 , but W 1 ⊕ W 2 is fully 2-visible module, then K 1 is 2-visible and hence W 1 is fully 2-visible module.Similarly, we have W 2 is fully visible module.
It is clear that every direct summand is a subm. of X, then by Remarks and Examples (3.1) part(6), we obtain the following known fact.Remark 3.8.A direct summand of fully 2-visible T -module X is also fully 2-visible.Remark 3.9.If the direct summand of X is fully 2-visible, then it is not necessary that the module X is fully 2-visible and the following example shows: The Z module X = Z ⊕ Z 3 is not fully 2-visible Z-module, since Z ⊕ (0) is a proper subm.ofX, but there is no a nonzero ideal l of Z such that (z ⊕ (0)) = I 2 (z ⊕ (0)) = I 2 Z, while Z 3 as a Z-module is fully 2-visible module by Remarks and Examples( 3.1).
1 and N 1 are subms of L and N each of L and N is fully 2-visible.Therefore X is fully visible module.The opposite side: since L ⊆ X = L ⊕ N and N ⊆ X = L ⊕ N but X is fully 2-visible module, then by Remarks and Examples (3.1) part(6), we obtain that L and N are fully 2-visible modules.