Supplemented and π-Projective Semimodules

In modules there is a relation between supplemented and π-projective semimodules. This relation was introduced, explained and investigated by many authors. This research will firstly introduce a concept of "supplement subsemimodule" analogues to the case in modules: a subsemimodule Y of a semimodule W is said to be supplement of a subsemimodule X if it is minimal with the property X+Y=W. A subsemimodule Y is called a supplement subsemimodule if it is a supplement of some subsemimodule of W. Then, the concept of supplemented semimodule will be defined as follows: an S-semimodule W is said to be supplemented if every subsemimodule of W is a supplement. We also review other types of supplemented semimodules. Previously, the concept of π-projective semimodule was introduced. The main goal of the present study is to explain the relation between the two concepts, supplemented semimodule and π-projective semimodules, and prove these relations by many results.

(ii) q(W) ={q(w)|wϵ W}. (iii) Im (q) = {nϵ N | n +q(w) =q(w′) for some w, w′ W As previously described [9], ker( ) is a subtractive subsemimodule of W Im( ) is a subtractive subsemimodule of W and ( ) is a subsemimodule of W. In the module theory, ( ) = Im ( ), whereas in the semimodule theory this is not true in general. It is clear that (W) Im(q), where the equality is satisfied if q(W is subtractive subsemimodule of K.
According to the same study [9], End In this work, S-semimodule will be cancellative, subtractive and semisubtractive.

Supplemented semimodules:
The following concept was introduce [1] and then studied [12].We will introduce this concept for semimodules and give some results related to this concept.  In the coming results, we need to define the concept "lies above a direct summand" for semimodules, where this concept was given previously for modules [1, p.357]. Definition 3.11. A subsemimodule A of a semimodule W lies above a direct summand if there exists a decomposition W=K Kʹ, with K A and Kʹ∩A Kʹ. Example 3.12. As an -semimodule, 3 12 4 12 = 12 , 4 12 2 12 and 3 12 ∩2 12 3 12 , hence 2 12 lies above a direct summand. The following lemma is needed for the next results. Lemma 3.13. Let C, D and K be subsemimodules of a U-semimodule W then: 1) If C D and D K, then C K.
2) If C W and q:W→Y is a homomorphism, then q(C) q(W).

3) If C W,
, and D is a direct summand of W, then C D. ▓ Lemma 3.14. If W is a supplemented semimodule and W Wʹ, then Wʹ is supplemented.
Proof: Let W and Wʹ be semimodules such that W Wʹ, then W →Wʹ, β(X)=Xʹ, X W and Xʹ Wʹ. Since W is supplemented, then there exists a subsemimodule Y such that X+Y=W and X∩Y Y then  and f+ g =1 W , then we have y ϵf(W)∩ g (w) implies that y=f(w)= g (wʹ), but w= g (w)+ g (wʹ)→ g (w) = g 2 (wʹ)+ g 2 (w) → g (w)= g (wʹ)+ g (w)→ g (wʹ)=0→y=0, that is, f(W)∩ g (W)=0. f+ g =1 W , W =f (W) .  (3.18) it is amply supplemented. Now, suppose that C is a supplement of a subsemimodule D of W, since D lies above a direct summand of W by Lemma (3.17), D∩C is a direct summand of D and again by Lemma (3.17)(2→1 W=K C for some K W. Hence C is a direct summand of W. ▓

Supplemented and π-projective semimodules
The class of supplemented π-projective modules was studied by many references. In this section, this class will be converted to semimodules. The next proposition is a generalization for the proposition (41.15) which was previously introduced [1]. Proposition 4.1. Let A and B be mutual supplements in a π-projective semimodule W, then W=A B. Proof: Assume that A and B are mutual supplements in W, then W=A+B such that A∩B A and A∩B B. It is enough to prove that A∩B=0. Consider the map h: A B→W defined by h(a, b)=a +b, it is a split epimorphism [12], so ker h={(a, b)ϵ A B: h(a, b)=0} ={(a, b)ϵ A B: a +b = 0} is a direct summand of A B, hence there is a submodule K such that A B= ker h K. On the other hand, (a, b) ϵ ker h implies a, b ϵ A∩B, then ker h (A∩B) (A∩B) A B, hence ker h A B and so ker h={(0, 0)}, that is, h is an injective map. Now, a ϵ A∩B implies h(a, 0)=h(0, a)= a, so (a, 0)= (0, a) which implies that a=0. Therefore, A∩B=0. ▓ The next proposition is a generalization for the proposition (2.3.1) in which was described earlier [12].  (3.17) to prove that every supplement subsemimodule of W is a direct summand (since W is amply supplemented). Let D be a supplement subsemimodule of W, then D is a supplement of some subsemimodule C of W, implies that W=C+D and C∩D D. But W is amply supplemented, thus C contains a supplement Cʹ of D in W, then W= Cʹ+D and Cʹ∩D Cʹ. Since Cʹ∩D C∩D D, hence Cʹ∩D D, therefore Cʹ and D are mutual supplements, then by 2(b), Cʹ∩D=0, thus W=Cʹ D. (b) Suppose that C and D are direct summands of W with W=C+D. W is amply supplemented, then there exists a supplement Dʹ of C in W such that Dʹ D and Cʹ a supplement of Dʹ in W such that Cʹ C. Therefore, C∩Dʹ Dʹ and Cʹ∩Dʹ Cʹ. Now, Cʹ∩Dʹ C∩Dʹ and since C∩Dʹ Dʹ, then Cʹ∩Dʹ Dʹ, that is, Cʹ and Dʹ are mutual supplements, and by 2(b) Cʹ∩Dʹ=0, hence W=Cʹ Dʹ. W∩C=C=Cʹ C∩Dʹ) and C is a direct summand in W, then W=C H=Cʹ C∩Dʹ H, where H is a suitable subsemimodule of W, implies that C∩Dʹ is a direct summand of W, since C∩Dʹ Dʹ, then C∩Dʹ W, then C∩Dʹ=0. Therefore, W=C , where Dʹ D. Similarly, W=Cʹ D, where Cʹ C. Since W=C+D, then C=W∩C=(Cʹ D(∩C=Cʹ (D∩C), also D=W∩D=(C Dʹ)∩D=(C∩D) Dʹ, hence W=Cʹ Dʹ C∩D, that is, (C∩D) is a direct summand of W. (3 4) Let every subsemimodule of W lies above a direct summand, by Proposition(3.20) W is amply supplemented and every supplement is a direct summand in W. Let W=C+D, since W is amply supplemented, then there is a supplement Dʹ C in W such that Dʹ D. This means that W=C+Dʹ and C∩Dʹ is small in Dʹ. Also, there is a supplement Cʹ C of Dʹ in W, then W=Cʹ+Dʹ and Cʹ∩Dʹ is small in Cʹ, which means that Cʹ and Dʹ are mutual supplements. Thus Cʹ∩Dʹ=0, then W=Cʹ Dʹ. Now, let h:W→Cʹ and q:W→Dʹ are the natural projections, so h+ q=1 W , then h(W) Cʹ, then h(W) C and q(W) Dʹ D, also q(C) q(W) Dʹ and q(C) C, then q(C) C∩Dʹ. Now, to prove that C∩Dʹ q(C), let aϵ C∩Dʹ, then aϵ Dʹ implies a= q(a) and aϵ C implies q(a) ϵ q(C), hence a ϵ q(C), so C∩Dʹ q(C), then C∩Dʹ q(C). But C∩Dʹ Dʹ, then q(C) Dʹ. By Proposition (3.19), q(C) q(W). (4 1) Let C and D be subsemimodules of W such that C+D=W. By (4) there exist idempotent h and q ϵEnd(W) with h(W) C and q(W) D such that h+ q=1 W . Thus, W is π-projective. Now to show that W is supplemented. Let A be subsemimodule of W, then we can write W=A+W and by (4) there exist idempotent f and g ϵEnd(W) such that g + f=1 W f(W) A, g(W) W and g(A) g(W). Claim that g(W) is a supplement of A. To verify this claim :1) to satisfy that W=g(W)+A, let w ϵW, since f+ g=1 W , then w=f(w)+g(w), but g(w) ϵg(W) and f(w)ϵA implies that W g(W)+A and it is clear that g(W)+A W, hence W=g(W)+A. Now we must show that g(W)∩A g(W), here we must prove that g(W)∩A g(A) . Let cϵ g(W)∩A, then c=g(w)ϵA, for some wϵ W, since f(W) A, then f(w)ϵ A and since f(w)+g(w)=w, then wϵ A, so that cϵ g(A), thus g(W)∩A g(A), but g(A) g(W), therefore g(W)∩A g(W), hence g(W) is a supplement of A, thus W is supplemented semimodule. ▓ The next corollary is a generalization for the corollary (2.3.2) in [12].