Direct sum of π-projective semimodules

Let A, and N are a semiring ,and a left Asemimodule, respectively. In this work we will discuss two cases: 1. The direct summand of π-projective semi module is π-projective, while the direct sum of two π-projective semimodules in general is not π-projective . The details of the proof will be given. 2. We will give a condition under which the direct sum of two π-projective semi modules is π-projective, as well as we also set conditions under which π-projective semi modules are projective.


1.Introdiction
We assume that A, and N are a semiring, and a left A-semimodule respectively. The direct sum of semimodules, and π-projective semimodules were studied by several authors [1,3], that means if for every two subsemimodules M and L with M+L=N , then there exist f, g ϵ End (N) such that f+ g=1 N , f(N) M and g(N) . The main goal of this work is to study the direct sum of π-projective semimodules.
The paper is divided into three sections. In section 2 , we give preliminaries that are used throughout this work. In section 3, we will give proofs for some facts and results in details.

2-
We give a condition under which the direct sum of two π-projective semimodules is π-projective, as well as we also set conditions under which projective semimodules are projective.

Preliminaries
In this section, we will give the basic definitions, and concepts that will be used throughout this paper. We will start with Some concepts that will use to complete this research:

Definition 2.12.[7]
A semimodule N is said to be quasi-projective if it is projective relative to N. Lemma 2.13.[6, p.14] Assume that N and P are semimodules , and H is a subsemimodule of N. . If P is N-projective, then P is H-projective and N/H-projective. Definition 2.14.
[1] An A-semimodule N is said to be π-projective if for every two subsemimodules B and C of N, with B+C=N, there exist f and g ϵEnd(N) such that f+ g=1 N , f(N) B, and g(N) C. Every projective semimodule is π-projective([1, 3.10]).   In this work we assume that N will be subtractive, semisubtractive and cancellative semimodule 3. Direct sum of π-projective semimodule.
In [11, 4.9], the authors proved that the direct summand of π-projective semimodule is πprojective semimodule. But the direct sum of two π-projective semimodules in general is not π-projective semimodule , for example ℤ 2 and ℤ 4 are π-projective semimodules because both of them are hollow [1, 3.5], while in the other hand ℤ 2 ℤ 4 is not π-projective semimodule, In order to prove this case suppose that ℤ 2 ℤ 4 is π-projective semimodule. In [1,4,3], authors proved that ℤ 2 is ℤ 4 -projective which is impossible.Now let us consider the next diagram: is an isomorphism and p is the natural epimorphism. If ℤ 2 is ℤ 4 -projective, then there exists a non-zero homomorphism q:ℤ 2 →ℤ 4 such that pq= h. But the homomorphism which defined by f( )= is the only non-zero homomorphism from ℤ 2 →ℤ 4 , then h=0. Therefore we get contradiction.

proposition[1,4.3]
Let N=K D be a π-projective semimodule, then D is K-projective(and K is D-projective). Now a condition under which the direct sum of two π-projective semimodules is π-projective , as well as conditions under which π-projective semimodules are projective will be given. The following result for semimodules is converted from a module version in [12, 2.2.5].
Proposition 3.1. If 0→N 1 →N 2 →N 3 →0 be a short exact sequence, such that N 2 N 3 is πprojective,then the sequence splits and N 3 is quasi-projective. Proof: See the next diagram: Since N 2 N 3 is π-projective, then by [1, 4.3 q where X and Y are two semimodules, h is the epimorphism from: M onto N and g is a right inverse of h(since M is projective, hence h splits) λ:X→Y any epimorphism and f :N→Y any homomorphism. since M is projective, then there exists a homomorphism q:M→X with λq=fh. Let ϕ =qg, then λϕ=(λq)g=f hg=f, so λϕ=f. Therefore N is projective. ▓ Proposition 3.3. For a semiring A, the following conditions are equivalent: (i) The direct sum of any two π-projective A-semimodules is π-projective. (ii) Every π-projective A-semimodule is projective. Proof:To prove i . Assume that N is a π-projective semimodule, then there exists a short exact sequence M →N→0, where M is a free semimodule, since N is a π-projective, by assumption then M and N are π-projective and so by (i), M N is π-projective, then by Proposition (3.2) N is projective. To prove . Suppose that N and L are π-projective semimodules, then N and L are projective by assumption. Hence N L is a projective semimodule [1,4.2] , then M L is a πprojective [1, 3.11]. ▓ The following result for semimodules is converted from a module version in [14], but its proof for modules was given in [13, p.49].
be a direct summand of a π-projective semimodule, and a nonzero factor semimodule of the semimodule L, then: 1-N is a π-projective semimodule and N is a -projective semimodule. 2-If there exists an epimorphism q: →N, then it splits and N is a quasi-projective semimodule isomorphic to a direct summand of . 3-If is indecomposable and q: →N an epimorphism, then q is an isomorphism. Proof: 1-N is a direct summand of a π-projective semimodule, then by [11, 4.9] N L is πprojective semimodule and by [1, 4.3] N is L-projective. Let = , where D is a subsemimodule of L. Now consider the next diagram: N h λq q L nn L 0 π 1 λ 0 q: N→ be any homomorphism with is subsemimodule of , π, π 1 , π 2 are the natural epimorphisms. Let λ: → an isomorphism ( the natural way). In fact, as λ(l+ D)+ =l+L 1 , that is λ(π(π 1 (l)))= π 2 (l). N is L-projective and there exists a homomorphism h: N→L such that π 2 h=λq. Claim that π 1 h: N→ is the required homomorphism. We note that λππ 1 =π 2 , thus λππ 1 h=π 2 h=λq, and since λ is an isomorphism, then ππ 1 h= q. 2-Suppose that q: →N is an epimorphism , consider the following diagram: N λ I q N 0 Where I is the identity homomorphism. By(1) N is -prpjective, hence there exists a homomorphism λ: N→ such that qλ=I and that means q has a right inverse, that is, q splits. So, =kerq , where is a suitable subsemimodule of . It is clear that N /kerq and /kerq, hence N . To show that N is quasi-projective; let C be a subsemimodule of N and let χ:N→ be any homomorphism. We considerthe next diagram: where π is natural epimorphism. Surly that πβ is an epimorphism (where β is an isomorphism). By (1) N is -projective, which implies that there exists a homomorphism h: N→ such that πβh=χ. Let g=βh, then πg=πβh=χ. Thus N is a quasi-projective semimodule. 3-Assume that there exists an epimorphism q: →N, where is indecomposable, by (2) kerq is a direct summand of , but is indecomposable, which implies that either kerq=0 or kerq= . If kerq= , then q=0, this means N=0, and this a contradiction, hence kerq=0 , it implies that q is monomorphism, then it is an isomorphism. ▓ The condition under which a π-projective module is quasi-projective was given in [14] , and the proof was given in [13, p.74-75-76].. Here, this condition will be given for semimodule. Proposition 3.5. If a semimodule N is a direct sum of a finitely generated semimodules {N j } jϵJ , then the following cases are equivalent: 1-N is a quasi-projective semimodule. 2-N is a π-projective semimodule and every N j is quasi-projective. 3-N j is N i -projective for every subscripts i and j in J. Proof: To prove 1 2 . We have N is quasi-projective, then by [1, 3.10] N is π-projective. See the following diagram Now for proving 2 3. Suppose that j ,k ϵ J and k≠j (since N j is quasi-projective, then N j is N j -projective for all jϵ J ), but N is a π-projective semimodule so by Proposition(3.4)(1)N j is -projective, where N k . Hence N j is N k -projective for all j . Therefore N j is N iprojective for all j and iϵ J. Finally to prove 3 1, we have N j is N i -projective for all j and i in J, and since N is finitely generated, then by [1] N j is N j -projective for every jϵ J , then also by [1] N j is N jprojective. Hence N is quasi-projective semimodule. ▓ The next proposition also gives conditions under which π-projective semimodules are projective. We will compare it with an analogues one in [15, 2.9] for modules. Before this proposition we need to define two concepts hereditary semimodule, and finite dimensional . They are analogous to the concepts in modules for details see [15,16].  Definition 3.7. An A-semimodule N is called finite dimensional if N contains no infinite independent families of non-zero subsemimodules. Recall that, a family of subsemimodules is independent if its sum is a direct sum. Example: Z 6 is a finite dimensional semimodule. Proposition 3. 8. Let N be a π-projective semimodule such that every cyclic subsemimodule of N is hereditary, N is a finite dimensional and a non-zero direct summand of N has a maximal subsemimodule, then N is projective. Proof: Let N be a π-projective semimodule such that every cyclic subsemimodule of N is hereditary. By assumption N has a maximal subsemimodule B. let d ϵ N and b d, then the cyclic subsemimodule 〈 〉≠ B. Since B is maximal, then 〈 〉+B=N. Since N is π-projective, hence there exist a homomorphisms and λϵ End(N) such that (N) 〈 〉, λ(N) B and +λ=1 N . (N) is a subsemimodule of the cyclic subsemimodule 〈 〉, thus (N) is a projective semimodule and the short exact sequence N→ (N)→0 splits, then N=C 1 H 1 , where C 1 (N), hence C 1 is a projective semimodule. Now letC 1 ≠0, since N= (N)+λ(N)= (N)+B≠ B, therefore (N)≠0, consequently C 1 ≠0. If H 1 =0, then N=C 1 ,so that N is projective.Assume that H 1 ≠0,thus hypothesis H 1 has a maximal subsemimodule and H 1 is πprojective by [1]. Similarly ,if we take H 1 = C 2 H 2 , where H 2 is a subsemimodule of H 1 , then N=C 1 C 2 H 2 , and if H 2 =0, then C 2 is projective, and so on. But N is finite dimensional, then there exits bϵ B such that H n =0 and N=C 1 C 2 C 3 … C n-1 , and for i=1, 2, …,n-1, C i is projective. Therefore N is projective.