P-(S. P) Submodules and C1(Extending) Modules

In this work, we study C1-(extending) module M through the sub-module of M which have two properties namely prime and semi-prime (P. (S. P)). Indeed these properties with another concept like fully invariant of any sub-modules explain the main objective of this study. Also heart submodule of a module M and the socle of M are studied with fully invariant property to obtain the same objective.


INTRODUCTION
Throughout this paper all.rings will be multiplication with identity and all modules will be unitary. "Let M be an R-module. Any submodule N of M is prime when if rM N, r R, m M\N, then M ⊆ N and N M is said to be semi-prime if N≠M and whenever ∈ , ∈ , and r m ∈ N for n where n ∈ N , then rm ∈ N" [2]. "Any module M is called multiplication if N IM, such that N M and I is an ideal of R and Ann is a prime ideal, then N is a prime submodule of M" [3]. Moreover; to find more information about multiplication and cyclic modules one can see ([4]). In [5], "can find fully invariant property in details. Let (S,∘) be an algebraic structure which has an identity e S ". If x∈S has an inverse, then x is say be invertible for ∘. That is, x is invertible if and only if :∃y ∈ S: x ∘ y eS y ∘ x.
Recall that a module M is called D 1 -module if for each sub-module N of M; there exists a direct summand of M is is a coessential submodule of N; equivalently; if N, K ≤ M and H≤ N, then M is direct summand H and K, (N∩H)≪ M. So D 1 -module is C 1 (extending)module.
In this paper, we studied a famous module in algebra through other properties of any submodule of this module like prime (semi-prime) sub-module and so is duo submodule.

2.
Several concepts have been used in this section for the purpose of reaching the main objective of the paper, for example, heart submodule of M and of M. We also we will use the fully invariant property to achieve the same goal. Note that P ideal of a ring R is called maximal ideal if P R and J also is ideal of a ring R then P⊆J⊆R. So J=R. . . Let M be a D -module over an integral domain R. If M is simple(cyclic) and f(N)⊆ N and satisfy (*), then M is prime-duo-C -module.
Proof. Suppose that M is simple(cyclic) and generated by one element : ∈ , ∈ . Or ; ∈ . So M is a multiplicationR-module. So any N M is a prime sub-module with fully in variant property (N is duo sub-module) and N=SM S is prime ideal of a ring R and Ann M ⊂S imply that M is P-duo-C 1 -module.
. . Let R be a semi-local ring. If M is cyclic D 1 -module such that for any N ,N M are a fully invariant, N ⊈ N and (L: N )=N , then M is P-duo-C 1 -module.
. Recall that if any sub module N of M is a P-sub-module, this means M is prime module. Therefore we can introduce the following result: . . Let M be a D 1 -cyclic R-module If M is a prime module and has fully invariant property, then M is a P-duo--module.
. The proof is very easy, because prime module gives every sub module of M is prime with same way in with fully property, we obtain M is P-duo--module. Now we need to introduce two concepts namely heart sub- Proof. Suppose that I and I are maximal ideal of R ∋ I I . Hence Sincei ∈ I ; i is not invertible element and hence 1-i is invertible element. This implies is invertible element (i 1 i ) and this contradiction. Since i ∈ I implies that i is not invertible element. Then R is local ring (R semi-local ring). Hence M is cyclic R-module and then is multiplication R-module. Since M have h-closed sub-module, then H(M) is fully invariant. We have N ⊈ N and Ann(N )=N , therefore N is a P-submodule.Thus.M is P-duo-C -modul. Corollary 2.7. Every Semi Prime-sub-modules of a multiplication C (extending) module is ∩ p. h closed sub modules and so M is P-duo-C (extending) module.
Proof. We know h-closed sub-modules means intersection of all sub-module. But h-closed (H(M)) is.fully.invariant in M with prim property imply M is S-P-duo-C 1 -module.
Proof. First, we claim R is not field. So 0 N M is a prime. Hence R is adomain. So T(M)=0 (M is a Torsion free R-module) over R. Note that when R is not field this is means M=Rx ( M is not simple module ); such that 0 Rx M. If x 0 and is not invertible element of R. So Rxm is a prime and xM Rxm; contradiction (because M=Rx) or x Rxm, then Rx=R and this contradiction. Hence N is a prime sub-module of M. We have N is a fully invariant (duo sub-module).Thus M is a -duo-C (extending) module.
.  (1) (mn∩N≠ϕ) (see [1]). So N is a proper prime submodule of M. We have H(M) is fully invariant. submodule (Lemma12) with condition (3), Soc(M) is also fully invariant submodule (N is a duo submodule). Since M is a D 1 -module, then it is C 1 -module. Thus M is.P-duo-C 1 -module.
. .Let M be.a D 1 (extending) module and satisfy the.following conditions. 1-M semisimple module. 2-M is Cyclicmodule. 3-Let P M. For every m, n ∈ M, if mn subset of P, then m ∈ P or n ∈ P. 4-M closed-duomodule. Then M is.P-duo-C 1 (extending) module. Proof. Assume that N be any sub-module of a semisimple module M. By assumption N is a directsummand of M; hence, it is a closed sub-module. We have M is a closed-duo-module, then N is isfully invariant. Since M is cyclic module, then M is a multiplication module with condition (3), we obtain N is prime submodule, and we obtained the result.

Conclusion
In this study we proved many facts about C 1 (extending)module. Every S.P-sub-modules of a multiplication (Extending) module is ∩ p h closed sub-modules and so M is P-duo-C (extending) module. Also, we investigated a fact which say if Soc(M)=H(M) with prime sub-module this imply M is a P-duo-C (extending) module.