Gamma Modules R. Ameri, R. Sadeghi Department of Mathematics, Faculty of Basic Science University of Mazandaran, Babolsar, Iran e-mail: ameri@umz.ac.ir Abstract Let R be a Γ-ring. We introduce the notion of gamma modules over R and study important properties of such modules. In this regards we study submodules and homomorphism of gamma modules and give related basic results of gamma modules. Keywords: Γ-ring, RΓ-module, Submodule, Homomorphism. 1 Introduction The notion of a Γ-ring was introduced by N. Nobusawa in [6]. Recently, W.E. Barnes [2], J. Luh [5], W.E. Coppage studied the structure of Γ-rings and obtained various gen- eralization analogous of corresponding parts in ring theory. In this paper we extend the concepts of module from the category of rings to the category of RΓ-modules over Γ-rings. Indeed we show that the notion of a gamma module is a generalization of a Γ-ring as well as a module over a ring, in fact we show that many, but not all, of the results in the Ratio Mathematica 20, 2010 127 theory of modules are also valid for RΓ-modules. In Section 2, some definitions and re- sults of Γ − ring which will be used in the sequel are given. In Section 3, the notion of a Γ-module M over a Γ − ring R is given and by many example it is shown that the class of Γ-modules is very wide, in fact it is shown that the notion of a Γ-module is a general- ization of an ordinary module and a Γ − ring. In Section 3, we study the submodules of a given Γ-module. In particular, we that L(M ), the set of all submodules of a Γ-module M constitute a complete lattice. In Section 3, homomorphisms of Γ-modules are studied and the well known homomorphisms (isomorphisms) theorems of modules extended for Γ-modules. Also, the behavior of Γ-submodules under homomorphisms are investigated. 2 Preliminaries Recall that for additive abelian groups R and Γ we say that R is a Γ − ring if there exists a mapping · : R × Γ × R −→ R (r, γ, r′) 7−→ rγr′ such that for every a, b, c ∈ R and α, β ∈ Γ, the following hold: (i) (a + b)αc = aαc + bαc; a(α + β)c = aαc + aβc; aα(b + c) = aαb + aαc; (ii) (aαb)βc = aα(bβc). A subset A of a Γ-ring R is said to be a right ideal of R if A is an additive subgroup of R and AΓR ⊆ A, where AΓR = {aαc| a ∈ A, α ∈ Γ, r ∈ R}. A left ideal of R is defined in a similar way. If A is both right and left ideal, we say that A is an ideal of R. Ratio Mathematica 20, 2010 128 If R and S are Γ-rings. A pair (θ, ϕ) of maps from R into S such that i) θ(x + y) = θ(x) + θ(y); ii) ϕ is an isomorphism on Γ; iii) θ(xγy) = θ(x)ϕ(γ)θ(y). is called a homomorphism from R into S. 3 RΓ -Modules In this section we introduce and study the notion of modules over a fixed Γ-ring. Definition 3.1. Let R be a Γ-ring. A (left) RΓ-module is an additive abelian group M together with a mapping . : R × Γ × M −→ M ( the image of (r, γ, m) being denoted by rγm), such that for all m, m1, m2 ∈ M and γ, γ1, γ2 ∈ Γ, r, r1, r2 ∈ R the following hold: (M1) rγ(m1 + m2) = rγm1 + rγm2; (M2) (r1 + r2)γm = r1γm + r2γm; (M3) r(γ1 + γ2)m = rγ1m + rγ2m; (M4) r1γ1(r2γ2m) = (r1γ1r2)γ2m. A right RΓ − module is defined in analogous manner. Definition 3.2. A (left) RΓ-module M is unitary if there exist elements, say 1 in R and γ0 ∈ Γ, such that, 1γ0m = m for every m ∈ M . We denote 1γ0 by 1γ0 , so 1γ0 m = m for all m ∈ M . Remark 3.3. If M is a left RΓ-module then it is easy to verify that 0γm = r0m = rγ0 = 0M . If R and S are Γ-rings then an (R, S)Γ-bimodule M is both a left RΓ-module and right SΓ-module and simultaneously such that (rαm)βs = rα(mβs) ∀m ∈ M, ∀r ∈ R, ∀s ∈ S and α, β ∈ Γ. Ratio Mathematica 20, 2010 129 In the following by many examples we illustrate the notion of gamma modules and show that the class of gamma module is very wide. Example 3.4. If R is a Γ-ring, then every abelian group M can be made into an RΓ- module with trivial module structure by defining rγm = 0 ∀r ∈ R, ∀γ ∈ Γ, ∀m ∈ M . Example 3.5. Every Γ-ring R, is an RΓ-module with rγ(r, s ∈ R, γ ∈ Γ) being the Γ-ring structure in R, i.e. the mapping . : R × Γ × R −→ R. (r, γ, s) 7−→ r.γ.s Example 3.6. Let M be a module over a ring A. Define . : A × R × M −→ M , by (a, s, m) = (as)m, being the R-module structure of M . Then M is an AA-module. Example 3.7. Let M be an arbitrary abelian group and S be an arbitrary subring of Z, the ring of integers. Then M is a ZS-module under the mapping . : Z × S × M −→ M (n, n′, x) 7−→ nn′x Example 3.8. If R is a Γ-ring and I is a left ideal of R .Then I is an RΓ-module under the mapping . : R × Γ × I −→ I such that (r, γ, a) 7−→ rγa . Example 3.9. Let R be an arbitrary commutative Γ-ring with identity. A polynomial in one indeterminate with coefficients in R is to be an expression P (X) = anX n + + a2X 2 + a1X + a0 in which X is a symbol, not a variable and the set R[x] of all polynomials is then an abelian group. Now R[x] becomes to an RΓ-module, under the mapping . : R × Γ × R[x] −→ R[x] (r, γ, f (x)) 7−→ r.γ.f (x) = ∑n i=1(rγai)x i. Ratio Mathematica 20, 2010 130 Example 3.10. If R is a Γ-ring and M is an RΓ-module. Set M [x] = { ∑n i=0 aix i | ai ∈ M}. For f (x) = ∑n j=0 bjx j and g(x) = ∑m i=0 aix i, define the mapping . : R[x] × Γ × M [x] −→ M [x] (g(x), γ, f (x)) 7−→ g(x)γf (x) = ∑m+n k=1 (ak.γ.bk)x k. It is easy to verify that M [x] is an R[x]Γ-module. Example 3.11. Let I be an ideal of a Γ-ring R. Then R/I is an RΓ-module, where the mapping . : R × Γ × R/I −→ R/I is defined by (r, γ, r′ + I) 7−→ (rγr′) + I. Example 3.12. Let M be an RΓ-module, m ∈ M . Letting T (m) = {t ∈ R | tγm = 0 ∀γ ∈ Γ}. Then T (m) is an RΓ-module. Proposition 3.12. Let R be a Γ-ring and (M, +, .) be an RΓ-module. Set Sub(M ) = {X| X ⊆ M}, Then sub(M ) is an RΓ-module. proof. Define ⊕ : (A, B) 7−→ A ⊕ B by A ⊕ B = (A\B) ∪ (B\A) for A, B ∈ sub(M ). Then (Sub(M ), ⊕) is an additive group with identity element ∅ and the inverse of each element A is itself. Consider the mapping: ◦ : R × Γ × Sub(M ) −→ sub(M ) (r, γ, X) 7−→ r ◦ γ ◦ X = rγX, where rγX = {rγx | x ∈ X}. Then we have (i) r ◦ γ ◦ (X1 ⊕ X2) = r · γ · (X1 ⊕ X2) = r · γ · ((X1\X2) ∪ (X2\X1)) = r · γ · ({a | a ∈ (X1\X2) ∪ (X2\X1)} = {r · γ · a | a ∈ (X1\X2) ∪ (X2\X1)}. And r ◦ γ ◦ X1 ⊕ r ◦ γ ◦ X2 = r · γ · X1 ⊕ r · γ · X2 = (r · γ · X1\r · γ · X2) ∪ (r · γ · X2\r · γ · X1) Ratio Mathematica 20, 2010 131 = {r · γ · x | x ∈ (X1\X2)} ∪ {r · γ · x | x ∈ (X2\X1)}. = {r · γ · x | x ∈ (X1\X2) ∪ (X2\X1)}. (ii) (r1 + r2) ◦ γ ◦ X = (r1 + r2) · γ · X = {(r1 + r2) · γ · x | x ∈ X} = {r1 · γ · x + r2 · γ · x | x ∈ X} = r1 · γ · X + r2 · γ · X = r1 ◦ γ ◦ X + r2 ◦ γ ◦ X. (iii) r ◦ (γ1 + γ2) ◦ X = r · (γ1 + γ2) · X = {r · (γ1 + γ2) · x | x ∈ X} = {r · γ1 · x + r · γ2 · x | x ∈ X} = r · γ1 · X + r · γ2 · X = r ◦ γ1 ◦ X + r ◦ γ2 ◦ X. (iv) r1 ◦ γ1 ◦ (r2 ◦ γ2 ◦ X) = r1 · γ1 · (r2 ◦ γ2 ◦ X) = {r1.γ1.(r2 ◦ γ2 ◦ x)|x ∈ X} = {r1.γ1.(r2.γ2.x) | x ∈ X} = {(r1.γ1.r2).γ2.x | x ∈ X} = (r1.γ1.r2).γ2.X. Corollary 3.13. If in Proposition 3.12, we define ⊕ by A ⊕ B = {a + b|a ∈ A, b ∈ B}. Then (Sub(M ), ⊕, ◦) is an RΓ-module. Proposition 3.14. Let (R, ◦) and (S, •) be Γ-rings. Let (M, .) be a left RΓ-module and right SΓ-module. Then A = {   r m 0 s   | r ∈ R, s ∈ S, m ∈ M} is a Γ-ring and AΓ-module under the mappings ? : A × Γ × A −→ A (   r m 0 s   , γ,   r1 m1 0 s1  ) 7−→   r ◦ γ ◦ r1 r.γ.m1 + m.γ.s1 0 s • γ • s1   . � Proof. Straightforward. Example 3.15. Let (R, ◦) be a Γ-ring . Then R ⊕ Z = {(r, s) | r ∈ R, s ∈ Z} is an left RΓ-module, where ⊕ addition operation is defined (r, n) ⊕ (r′, n′) = (r +R r′, n +Z n′) and the product · : R × Γ × (R ⊕ Z) −→ R ⊕ Z is defined r′ · γ · (r, n) −→ (r′ ◦ γ ◦ r, n). Ratio Mathematica 20, 2010 132 Example 3.16. Let R be the set of all digraphs (A digraph is a pair (V, E) consisting of a finite set V of vertices and a subset E of V × V of edges) and define addition on R by setting (V1, E1) + (V2, E2) = (V1 ∪ V2, E1 ∪ E2). Obviously R is a commutative group since (∅, ∅) is the identity element and the inverse of every element is itself. For Γ ⊆ R consider the mapping · : R × Γ × R −→ R (V1, E1) · (V2, E2) · (V3, E3) = (V1 ∪ V2 ∪ V3, E1 ∪ E2 ∪ E3 ∪ {V1 × V2 × V3}), under condition (∅, ∅) = (∅, ∅) · (V1, E1) · (V2, E2)(V1, E1) · (∅, ∅) · (V2, E2) = (V1, E1) · (∅, ∅) · (V2, E2) = (V1, E1) · (V2, E2) · (∅, ∅). It is easy to verify that R is an RΓ-module . Example 3.17. Suppose that M is an abelian group. Set R = Mmn and Γ = Mnm, so by definition of multiplication matrix subset R (t) mn = {(xij) | xtj = 0 ∀ j = 1, ...m} is a right RΓ-module. Also, C (k) mn = {xij) | xik = 0 ∀i = 1, ..., n} is a left RΓ-module. Example 3.18. Let (M, •) be an RΓ-module over Γ-ring (R, .) and S = {(a, 0)|a ∈ R}. Then R × M = {(a, m)|a ∈ R, m ∈ M} is an SΓ-module, where addition operation is defined by (a, m) ⊕ (b, m1) = (a +R b, m +M m1). Obviously, (R × M, ⊕) is an additive group. Now consider the mapping ◦ : S × Γ × (R × M ) −→ R × M ((a, 0), γ, (b, m)) 7−→ (a, 0) ◦ γ ◦ (b, m) = (a.γ.b, a • γ • m). Then it is easy to verify that R × M is an SΓ-module. Example 3.19 Let R be a Γ-ring and (M, .) be an RΓ-module. Consider the mapping α : M −→ R. Then M is an MΓ-module, under the mapping Ratio Mathematica 20, 2010 133 ◦ : M × Γ × M −→ M (m, γ, n) 7−→ m ◦ γ ◦ n = (α(m)).γ.n. Example 3.20. Let (R, ·) and (S, ◦) be Γ- rings. Then (i) The product R × S is a Γ- ring, under the mapping ((r1, s1), γ, (r2, s2)) 7−→ (r1 · γ · r2, s1 ◦ γ ◦ s2). (ii) For A = {   r 0 0 s   | r ∈ R, s ∈ S} there exists a mapping R × S −→ A, such that (r, s) −→   r 0 0 s   and A is a Γ- ring. Moreover, A is an (R × S)Γ- module under the mapping (R × S) × Γ × A −→ A ((r1, s1), γ,   r2 0 0 s2  ) −→   r1 · γ · r2 0 0 s1 ◦ γ ◦ s2   . Example 3.21. Let (R, ·) be a Γ-ring. Then R×R is an RΓ-module and (R×R)Γ- module. Consider addition operation (a, b)+(c, d) = (a+R c, b+R d). Then (R×R, +) is an additive group. Now define the mapping R×Γ×(R×R) 7−→ R×R by (r, γ, (a, b)) 7−→ (r·γ·a, r·γ·b) and (R×R)×Γ×(R×R) −→ R×R by ((a, b), γ, (c, d)) 7−→ (a·γ·c+b·γ·d, a·γ·d+b·γ·c). Then R × R is an (R × R)Γ- module. 4 Submodules of Gamma Modules In this section we study submodules of gamma modules and investigate their properties. In the sequel R denotes a Γ-ring and all gamma modules are RΓ-modules Definition 4.1. Let (M, +) be an RΓ-module. A nonempty subset N of (M, +) is said to be a (left) RΓ-submodule of M if N is a subgroup of M and RΓN ⊆ N , where Ratio Mathematica 20, 2010 134 RΓN = {rγn|γ ∈ Γ, r ∈ R, n ∈ N}, that is for all n, n′ ∈ N and for all γ ∈ Γ, r ∈ R; n − n′ ∈ N and rγn ∈ N . In this case we write N ≤ M . Remark 4.2. (i) Clearly {0} and M are two trivial RΓ-submodules of RΓ-module M , which is called trivial RΓ-submodules. (ii) Consider R as RΓ-module. Clearly, every ideal of Γ-ring R is submodule, of R as RΓ-module. Theorem 4.3. Let M be an RΓ-module. If N is a subgroup of M , then the factor group M/N is an RΓ-module under the mapping . : R × Γ × M/N −→ M/N is defined (r, γ, m + N ) 7−→ (r.γ.m) + N. Proof. Straight forward. Theorem 4.4. Let N be an RΓ-submodules of M . Then every RΓ-submodule of M/N is of the form K/N , where K is an RΓ-submodule of M containing N . Proof. For all x, y ∈ K, x + N, y + N ∈ K/N ; (x + N ) − (y + N ) = (x − y) + N ∈ K/N , we have x − y ∈ K, and ∀r ∈ R ∀γ ∈ Γ, ∀x ∈ K, we have rγ(x + N ) = rγx + N ∈ K/N ⇒ rγx ∈ K. Then K is a RΓ-submodule M . Conversely,it is easy to verify that N ⊆ K ≤ M then K/N is RΓ-submodule of M/N . This complete the proof. � Proposition 4.5. Let M be an RΓ-module and I be an ideal of R. Let X be a nonempty subset of M . Then IΓX = { ∑n i=1aiγixi | ai ∈ Irγ i ∈ Γ, xi ∈ X, n ∈ N} is an RΓ-submodule of M . Proof. (i) For elements x = ∑n i=1aiαixi and y = ∑m j=1xa′j βj yj of IΓX, we have x − y = ∑m+n k=1 bkγkzk ∈ IΓX. Now we consider the following cases: Case (1): If 1 ≤ k ≤ n, then bk = ak, γk = αk, zk = xk. Ratio Mathematica 20, 2010 135 Case(2): If n + 1 ≤ k ≤ m + n, then bk = −a′k−n, γk = βk−n, zk = yk−n. Also (ii) ∀r ∈ R, ∀γ ∈ Γ, ∀a = ∑n i=1 aiγixi ∈ IΓX, we have rγx = ∑n i=1 rγ(aiγixi) =∑n i=1(rγai)γixi. Thus IΓX is an RΓ-submodule of M. � Corollary 4.6. If M is an RΓ-module and S is a submodule of M . Then RΓS is an RΓ-submodule of M . Let N ≤ M . Define N : M = {r ∈ R|rγm ∀γ ∈ Γ ∀m ∈ M}. It is easy to see that N : M is an ideal of Γ ring R. Theorem 4.7. Let M be an RΓ-module and I be an ideal of R. If I ⊆ (0 : M ), then M is an (R/I)Γ-module. proof. Since R/I is Γ-ring, def inethemapping• : (R/I) × Γ × M −→ M by (r + I, γ, m) 7−→ rγm.. The mapping • is well-defined since I ⊆ (0 : M ). Now it is straight forward to see that M is an (R/I)Γ-module. � Proposition 4.8. Let R be a Γ-ring, I be an ideal of R, and (M, .) be a RΓ-module. Then M/(IΓM ) is an (R/I)Γ- module. Proof. First note that M/(IΓM ) is an additive subgroup of M . Consider the mapping γ • (m + IΓM ) = r.γ.m + IΓM )N owitisstraightf orwardtoseethatMisan(R/I)Γ-module. � Proposition 4.9. Let M be an RΓ-module and N ≤ M , m ∈ M . Then (N : m) = {a ∈ R | aγm ∈ N ∀γ ∈ Γ} is a left ideal of R. Proof. Obvious. Proposition 4.10. If N and K are RΓ-submodules of a RΓ-module M and if A, B are nonempty subsets of M then: (i) A ⊆ B implies that (N : B) ⊆ (N : A); (ii) (N ∩ K : A) = (N : A) ∩ (K : A); (iii) (N : A) ∩ (N : B) ⊆ (N : A + B), moreover the equality hold if 0M ∈ A ∩ B. Ratio Mathematica 20, 2010 136 proof. (i) Easy. (ii) By definition, if r ∈ R, then r ∈ (N ∩ K : A) ⇐⇒ ∀a ∈ Ar ∈ (N ∩ K : a) ⇐⇒ ∀γ ∈ Γ; rγa ∈ N ∩ K ⇐⇒ r ∈ (N : A) ∩ K : A). (iii) If r ∈ (N : A) ∩ (N : B). Then ∀γ ∈ Γ, ∀a ∈ A, ∀b ∈ B, rγ(a + b) ∈ N and r ∈ (N : A + B). Conversely, 0M ∈ A + B =⇒ A ∪ B ⊆ A + B =⇒ (N : A + B) ⊆ (N : A ∪ B) by(i). Again by using A, B ⊆ A ∪ B we have (N : A ∪ B) ⊆ (N : A) ∩ (N : B). � Definition 4.11. Let M be an RΓ-module and ∅ 6= X ⊆ M . Then the generated RΓ-submodule of M , denoted by < X > is the smallest RΓ-submodule of M containing X, i.e. < X >= ∩{N|N ≤ M}, X is called the generator of < X >; and < X > is finitely generated if |X| < ∞. If X = {x1, ...xn} we write < x1, ..., xn > instead < {x1, ..., xn} >. In particular, if X = {x} then < x > is called the cyclic submodule of M , generated by x. Lemma 4.12. Suppose that M is an RΓ-module. Then (i) Let {Mi}i∈I be a family of RΓ-submodules M . Then ∩Mi is the largest RΓ-submodule of M , such that contained in Mi, for all i ∈ I. (ii) If X is a subset of M and |X| < ∞. Then < X >= { ∑m i=1 nixi + ∑k j=1 rjγjxj|k, m ∈ N, ni ∈ Z, γj ∈ Γ, rj ∈ R, xi, xj ∈ X} . Proof. (i) It is easy to verify that ∩i∈I Mi ⊆ Mi is a RΓ-submodule of M . Now suppose that N ≤ M and ∀ i ∈ I, N ⊆ Mi, then N ⊆ ∩Mi. (ii) Suppose that the right hand in (b) is equal to D. First, we show that D is an RΓ-submodule containing X. X ⊆ D and difference of two elements of D is belong to D and ∀r ∈ R ∀γ ∈ Γ, ∀a ∈ D we have rγa = rγ( ∑m i=1 nixi + ∑k j=1 rjγjxj) = ∑m i=1 ni(rγxi) + ∑k j=1(rγrj)γjxj ∈ D. Also, every submodule of M containing X, clearly contains D. Thus D is the smallest Ratio Mathematica 20, 2010 137 RΓ-submodules of M , containing X. Therefore < X >= D. � For N, K ≤ M , set N + K = {n + k|n ∈ N, K ∈ K}. Then it is easy to see that M + N is an RΓ-submodules of M , containing both N and K. Then the next result immediately follows. Lemma 4.13. Suppose that M is an RΓ-module and N, K ≤ M . Then N + K is the smallest submodule of M containing N and K. Set L(M ) = {N|N ≤ M}. Define the binary operations ∨ and ∧ on L(M ) by N ∨ K = N + K andN ∧ K = N ∩ K. In fact (L(M ), ∨, ∧) is a lattice. Then the next result immediately follows from lemmas 4.12. 4.13. Theorem 4.13. L(M ) is a complete lattice. 5 Homomorphisms Gamma Modules In this section we study the homomorphisms of gamma modules. In particular we investigate the behavior of submodules od gamma modules under homomorphisms. Definition 5.1. Let M and N be arbitrary RΓ-modules. A mapping f : M −→ N is a homomorphism of RΓ-modules ( or an RΓ-homomorphisms) if for all x, y ∈ M and ∀r ∈ R, ∀γ ∈ Γ we have (i) f (x + y) = f (x) + f (y); (ii) f (rγx) = rγf (x). A homomorphism f is monomorphism if f is one-to-one and f is epimorphism if f is onto. f is called isomorphism if f is both monomorphism and epimorphism. We denote the set of all RΓ-homomorphisms from M into N by HomRΓ (M, N ) or shortly by Ratio Mathematica 20, 2010 138 HomRΓ (M, N ). In particular if M = N we denote Hom(M, M ) by End(M ). Remark 5.2. If f : M −→ N is an RΓ-homomorphism, then Kerf = {x ∈ M|f (x) = 0}, Imf = {y ∈ N|∃x ∈ M ; y = f (x)} are RΓ-submodules of M . Example 5.3. For all RΓ-modules A, B, the zero map 0 : A −→ B is an RΓ-homomorphism. Example 5.4. Let R be a Γ-ring. Fix r0 ∈ Γ and consider the mapping φ : R[x] −→ R[x] by f 7−→ f γ0x. Then φ is an RΓ-module homomorphism, because ∀r ∈ R, ∀γ ∈ Γ and ∀f, g ∈ R[x] : φ(f + g) = (f + g)γ0x = f γ0x + gγ0x = φ(f ) + φ(g) and φ(rγf ) = rγf γ0x = rγφ(f ). Example 5.5. If N ≤ M , then the natural map π : M −→ M/N with π(x) = x + N is an RΓ-module epimorphism with kerπ = N . Proposition 5.6. If M is unitary RΓ-module and End(M ) = {f : M −→ M|f is RΓ − homomorphism}. Then M is an End(M )Γ-module. Proof. It is well known that End(M ) is an abelian group with usual addition of functions. Define the mapping . : End(M ) × Γ × M −→ M (f, γ, m) 7−→ f (1.γ.m) = 1γf (m), where 1 is the identity map. Now it is routine to verify that M is an End(M )Γ-module.� Lemma 5.7. Let f : M −→ N be an RΓ-homomorphism. If M1 ≤ M and N1 ≤. Then (i) Kerf ≤ M , Imf ≤ N ; (ii) f (M1) ≤ Imf ; (iii) Kerf−1(N1) ≤ M . Ratio Mathematica 20, 2010 139 Example 5.8. Consider L(M ) the lattice of RΓ-submodules of M . We know that (L(M ), +) is a monoid with the sum of submodules. Then L(M ) is RΓ-semimodule under the mapping . : R × Γ × T −→ T , such that (r, γ, N ) 7−→ r.γ.N = rγN = {rγn|n ∈ N}. Example 5.9. Let θ : R −→ S be a homomorphism of Γ-rings and M be an SΓ-module. Then M is an RΓ-module under the mapping • : R × Γ × M −→ M by (r, γ, m) 7−→ r • γ • m = θ(r). Moreover if M is an SΓ-module then M is a RΓ-module for R ⊆ S. Example 5.10. Let (M, .) be an RΓ-module and A ⊆ M . Letting M A = {f|f : A −→ M is a map}. Then M A is an RΓ-module under the mapping ◦ : R × Γ × M A −→ M A defined by (r, γ, f ) 7−→ r ◦ γ ◦ f = rγf (a), since M A is an additive group with usual addition of maps. Example 5.11. Let(M, .) and (N, •) be RΓ-modules. Then Hom(M, N ) is a RΓ-module, under the mapping ◦ : R × Γ × Hom(M, N ) −→ Hom(M, N ) (r, γ, α) 7−→ r ◦ γ ◦ α, where (r • γ • α)(m) = rγα)(m). Example 5.12. Let M be a left RΓ-module and right SΓ-module. If N be an RΓ-module, then (i) Hom(M, N ) is a left SΓ-module. Indeed ◦ : S × Γ × Hom(M, N ) −→ Hom(M, N ) (s, γ, α) −→ s ◦ γ ◦ α : M −→ N m 7−→ α(mγs) (ii) Hom(N, M ) is right SΓ-module under the mapping Ratio Mathematica 20, 2010 140 ◦ : Hom(N, M ) × Γ × S −→ Hom(N, M ) (α, γ, s) 7−→ α ◦ γ ◦ s : N −→ M n 7−→ α(n).γ.s Example 5.13. Let M be a left RΓ-module and right SΓ-module and α ∈ End(M ) then α induces a right S[t]Γ-module structure on M with the mapping ◦ : M × Γ × S[t] −→ M (m, γ, ∑n i=0 sit i) 7−→ m ◦ γ ◦ ( ∑n i=0 sit i) = ∑n i=0(mγsi)α i Proposition 5.14. Let M be a RΓ-module and S ⊆ M . Then SΓM = { ∑ siγiai | si ∈ S, ai ∈ M, γi ∈ Γ} is an RΓ-submodule of M . Proof. Consider the mapping ◦ : R × Γ × (SΓM ) −→ SΓM (r, γ, ∑n i=1 siγiai) 7−→ ∑n i=1 siγi(rγai). Now it is easy to check that SΓM is a RΓ-submodule of M . Example 5.16. Let (R, .) be a Γ-ring. Let Z2, the cyclic group of order 2. For a nonempty subset A, set Hom(R, BA) = {f : R −→ BA}. Clearly (Hom(R, BA), +) is an abelian group. Consider the mapping ◦ : R × Γ × Hom(R, BA) −→ Hom(R, BA) that is defined (r, γ, f ) 7−→ r ◦ γ ◦ f, where (r ◦ γ ◦ f )(s) : A −→ B is defied by (r ◦ γ ◦ f (s))(a) = f (sγr)(a). Now it is easy to check that Hom(R, BA) is an Γ-ring. Example 5.17. Let R and S be Γ-rings and ϕ : R −→ S be a Γ-rings homomorphism. Then every SΓ-module M can be made into an RΓ-module by defining Ratio Mathematica 20, 2010 141 rγx (r ∈ R, γ ∈ Γ, x ∈ M ) to be ϕ(r)γx. We says that the RΓ-module structure M is given by pullback along ϕ. Example 5.18. Let ϕ : R −→ S be a homomorphism of Γ-rings then (S, .) is an RΓ-module. Indeed ◦ : R × Γ × S −→ S (r, γ, s) 7−→ r ◦ γ ◦ s = ϕ(r).γ.s Example 5.19. Let (M, +) be an RΓ-module. Define the operation ◦ on M by a ⊕ b = b.a. Then (M, ⊕) is an RΓ-module. Proposition 5.20. Let R be a Γ-ring. If f : M −→ N is an RΓ-homomorphism and C ≤ kerf , then there exists an unique RΓ-homomorphism f̄ : M/C −→ N , such that for every x ∈ M ; Kerf̄ = Kerf /C and Imf̄ = Imf and f̄ (x + C) = f (x), also f̄ is an RΓ-isomorphism if and only if f is an RΓ-epimorphism and C = Kerf . In particular M/Kerf ∼= Imf . Proof. Let b ∈ x + C then b = x + c for some c ∈ C, also f (b) = f (x + c). We know f is RΓ-homomorphism therefore f (b) = f (x + c) = f (x) + f (c) = f (x) + 0 = f (x) (since C ≤ kerf ) then f̄ : M/C −→ N is well defined function. Also ∀ x + C, y + C ∈ M/C and ∀ r ∈ R, γ ∈ Γ we have (i) f̄ ((x + C) + (y + C)) = f̄ ((x + y) + C) = f (x + y) = f (x) + f (y) = f̄ (x + C) + f̄ (y + C). (ii) f̄ (rγ(x + C)) = f̄ (rγx + C) = f (rγx) = rγf (x) = rγf̄ (x + C). then f̄ is a homomorphism of RΓ-modules, also it is clear Imf̄ = Imf and ∀(x + C) ∈ kerf̄ ; x + C ∈ kerf̄ ⇔ f̄ (x + C) = 0 ⇔ f (x) = 0 ⇔ x ∈ kerf then kerf̄ = kerf /C. Then definition f̄ depends only f , then f̄ is unique. f̄ is epimorphism if and only if f is epimorphism. f̄ is monomorphism if and only if kerf̄ be trivial RΓ-submodule of M/C. In actually if and only if Kerf = C then M/Kerf ∼= Imf .� Ratio Mathematica 20, 2010 142 Corollary 5.21. If R is a Γ-ring and M1 is an RΓ-submodule of M and N1 is RΓ-submodule of N , f : M −→ N is a RΓ-homomorphism such that f (M1) ⊆ N1 then f make a RΓ-homomorphism f̄ : M/M1 −→ N/N1 with operation m + M1 7−→ f (m) + N1. f̄ is RΓ-isomorphism if and only if Imf + N1 = N, f −1(N1) ⊆ M1. In particular, if f is epimorphism such that f (M1) = N1, kerf ⊆ M1 then f is a RΓ-isomorphism. proof. We consider the mapping M −→f N −→π N/N1. In this case; M1 ⊆ f−1(N1) = kerπf (∀m1 ∈ M1, f (m1) ∈ N1 ⇒ πf (m1) = 0 ⇒ m1 ∈ kerπf ). Now we use Proposition 5.20 for map πf : M −→ N/N1 with function m 7−→ f (m) + N1 and submodule M1 of M . Therefore, map f̄ : M/M1 −→ N/N1 that is defined m + M 7−→ f (m) + N1 is a RΓ-homomorphism. It is isomorphism if and only if πf is epimorphism, M1 = kerπf . But condition will satisfy if and only if Imf + N1 = N , f −1(N1) ⊆ M1. If f is epimorphism then N = Imf = Imf + N1 and if f (M1) = N1 and kerf ⊆ M1 then f−1(N1) ⊆ M1 so f̄ is isomorphism.� Proposition 5.22. Let B, C be RΓ-submodules of M . (i) There exists a RΓ-isomorphism B/(B ∩ C) ∼= (B + C)/C. (ii) If C ⊆ B, then B/C is an RΓ-submodule of M/C and there is an RΓ-isomorphism (M/C)/(B/C) ∼= M/B . Proof. (i) Combination B −→j B + C −→π (B + C)/C is an RΓ-homomorphism with kernel= B ∩ C, because kerπj = {b ∈ B|πj(b) = 0(B+C)/C} = {b ∈ B|π(b) = C} = {b ∈ B|b + C = C} = {b ∈ B|b ∈ C} = B ∩ C therefore, in order to Proposition 5.20., B/(B ∩ C) ∼= Im(πj)(?), every element of (B + C)/C is to form (b + c) + C, thus (b + c) + C = b + C = πj(b) then πj is epimorphism and Imπj = (B + C)/C in attention (?), B/(B ∩ C) ∼= (B + C)/C. (ii) We consider the identity map i : M −→ M , we have i(C) ⊆ B, then in order to Ratio Mathematica 20, 2010 143 apply Proposition 5.21. we have RΓ-epimorphism ī : M/C −→ M/B with ī(m + C) = m + B by using (i). But we know B = ī(m + C) if and only if m ∈ B thus ker ī = {m + C ∈ M/C|m ∈ B} = B/C then kerī = B/C ≤ M/C and we have M/B = Imī ∼= (M/C)/(B/C).� Let M be a RΓ-module and {Ni|i ∈ Ω} be a family of RΓ-submodule of M . Then ∩i∈ΩNi is a RΓ-submodule of M which, indeed, is the largest RΓ-submodule M contained in each of the Ni. In particular, if A is a subset of a left RΓ-moduleM then intersection of all submodules of M containing A is a RΓ-submodule of M , called the submodule generated by A. If A generates all of the RΓ-module, then A is a set ofgenerators for M . A left RΓ-module having a finite set of generators is finitely generated. An element m of the RΓ-submodule generated by a subset A of a RΓ-module M is a linear combination of the elements of A. If M is a left RΓ-module then the set ∑ i∈Ω Ni of all finite sums of elements of Ni is an RΓ-submodule of M generated by ∪i∈ΩNi. RΓ-submodule generated by X = ∪i∈ΩNi is D = { ∑s i=1 riγiai + ∑t j=1 njbj|ai, bj ∈ X, ri ∈ R, nj ∈ Z, γi ∈ Γ} if M is a unitary RΓ-module then D = RΓX = { ∑s i=1 riγiai|ri ∈ R, γi ∈ Γ, ai ∈ X}. Example 5.23. Let M, N be RΓ-modules and f, g : M −→ N be RΓ-module homomorphisms. Then K = {m ∈ M | f (m) = g(m)} is RΓ-submodule of M . Example 5.24. Let M be a RΓ-module and let N, N ′ be RΓ- submodules of M . Set A = {m ∈ M | m + n ∈ N ′ f or some n ∈ N} is an RΓ-module of M containing N ′. Proposition 5.25. Let (M, ·) be an RΓ- module and M generated by A. Then there exists an RΓ-homomorphism R (A) −→ M , such that f 7−→ ∑ a∈A,a∈supp(f ) f (a) · γ · a. Remark 5.26. Let R be a Γ- ring and let {(Mi, oi)|i ∈ Ω} be a family of left RΓ- modules. Then ×i∈ΩMi, the Cartesian product of Mi’s also has the structure of a left RΓ-module under componentwise addition and mapping Ratio Mathematica 20, 2010 144 · : R × Γ × (×Mi) −→ ×Mi (r, γ, {mi}) −→ r · γ · {mi} = {roiγoimi}Ω. We denote this left RΓ-module by ∏ i∈Ω Mi. Similarly,∑ i∈Ω Mi = {{mi} ∈ ∏ Mi|mi = 0 for all but finitely many indices i} is a RΓ-submodule of ∏ i∈Ω Mi. For each h in Ω we have canonical RΓ- homomorphisms πh : ∏ Mi −→ Mh and λh : Mh −→ ∑ Mi is defined respectively by πh :< mi >7−→ mh and λ(mh) =< ui >, where ui =   0 i 6= h mh i = h The RΓ-module ∏ Mi is called the ( external)direct product of the RΓ- modules Mi and the RΓ- module ∑ Mi is called the (external) direct sum of Mi. It is easy to verify that if M is a left RΓ-module and if {Mi|i ∈ Ω} is a family of left RΓ-modules such that, for each i ∈ Ω, we are given an RΓ-homomorphism αi : M −→ Mi then there exists a unique RΓ- homomorphism α : M −→ ∏ i∈Ω Mi such that αi = απi for each i ∈ Ω. Similarly, if we are given an RΓ-homomorphism βi : Mi −→ M for each i ∈ Ω then there exists an unique RΓ- homomorphism β : ∑ i∈Ω Mi −→ M such that βi = λiβ for each i ∈ Ω . Remark 5.27. Let M be a left RΓ-module. Then M is a right R op Γ -module under the mapping ∗ : M × Γ × Rop −→ M (m, γ, r) 7−→ m ∗ γ ∗ r = rγm. Definition 5.28. A nonempty subset N of a left RΓ-module M is subtractive if and only if m + m′ ∈ N and m ∈ N imply that m′ ∈ N for all m, m′ ∈ M . Similarly, N is strong subtractive if and only if m + m′ ∈ N implies that m, m′ ∈ N for all m, m′ ∈ M . Ratio Mathematica 20, 2010 145 Remark 5.29. (i) Clearly, every submodule of a left RΓ-module is subtractive. Indeed, if N is a RΓ-submodule of a RΓ-module M and m ∈ M, n ∈ N are elements satisfying m + n ∈ N then m = (m + n) + (−n) ∈ N . (ii) If N, N ′ ⊆ N are RΓ-submodules of an RΓ-module M , such that N ′ is a subtractive RΓ-submodule of N and N is a subtractive RΓ-submodule of M then N ′ is a subtractive RΓ-module of M . Note. If {Mi|i ∈ Ω} is a family of (resp. strong) subtractive RΓ-submodule of a left RΓ-module M then ∩i∈ΩMi is again (resp. strong) subtractive. Thus every RΓ -submodule of a left RΓ-module M is contained in a smallest (resp. strong) subtractive RΓ-submodule of M , called its (resp. strong) subtractive closure in M . Proposition 5.30 Let R be a Γ-ring and let M be a left RΓ -module. If N, N ′ and N ′ ′ ≤ M are submodules of M satisfying the conditions that N is subtractive and N ′ ⊆ N , then N ∩ (N ′ + N ′′) = N ′ + (N ∩ N ′′). Proof. Let x ∈ N ∩ (N ′ + N ′′). Then we can write x = y + z, where y ∈ N ′ and z ∈ N ′′. by N ′ ⊆ N , we have y ∈ N and so, z ∈ N , since N is subtractive. Thus x ∈ N ′ + (N ∩ N ′′), proving that N ∩ (N ′ + N ′′) ⊆ N ′ + (N ∩ N ′′). The reverse containment is immediate.� Proposition 5.31. If N is a subtractive RΓ-submodule of a left RΓ-module M and if A is a nonempty subset of M then (N : A) is a subtractive left ideal of R. Proof. Since the intersection of an arbitrary family of subtractive left ideals of R is again subtractive, it suffices to show that (N : m) is subtractive for each element m. Let a ∈ R and b ∈ (N : M ) (for γ ∈ Γ ) satisfy the condition that a + b ∈ (N : M ). Then aγm + bγm ∈ N and bγm ∈ N so aγm ∈ N , since N is subtractive. Thus a ∈ (N : M ).�. proposition 5.32. If I is an ideal of a Γ-ring R and M is a left RΓ-module. Then Ratio Mathematica 20, 2010 146 N = {m ∈ M | IΓm = {0}} is a subtractive RΓ-submodule of M. Proof. Clearly, N is an RΓ-submodule of M . If m, m ′ ∈ M satisfy the condition that m and m + m′ belong to N then for each r ∈ I and for each γ ∈ Γ we have 0 = rγ(m + m′) = rγm + rγm′m′ = rγm′, and hence m′ ∈ N . Thus N is subtractive. � proposition 5.33. Let (R, +, ·) be a Γ-ring and let M be an RΓ-module and there exists bijection function δ : M −→ R. Then M is a Γ-ring and MΓ-module. Proof. Define ◦ : M × Γ × M −→ M by (x, γ, y) 7−→ x ◦ γ ◦ y = δ−1(δ(x) · γδ(y)). It is easy to verify that R is a Γ- ring. If M is a set together with a bijection function δ : X −→ R then the Γ-ring structure on R induces a Γ-ring structure (M, ⊕, �) on X with the operations defined by x ⊕ y = δ−1(δ(x) + δ(y)) and x � γ � y = δ−1(δ(x) · γ · δ(y)).� Acknowledgements The first author is partially supported by the ”Research Center on Algebraic Hyperstructures and Fuzzy Mathematics, University of Mazandaran, Babolsar, Iran”. References [1] F.W.Anderson ,K.R.Fuller , Rings and Categories of Modules, Springer Verlag ,New York ,1992. [2] W.E.Barens,On the Γ-ring of Nobusawa, Pa- cific J.Math.,18(1966),411-422. [3] J.S.Golan,Semirinngs and their Applications.[4] T.W.Hungerford ,Algebra.[5] J.Luh,On the theory of simple gamma rings, Michigan Math .J.,16(1969),65-75.[6] N.Nobusawa ,On a generalization of the ring theory ,Os- aka J.Math. 1(1964),81-89. Ratio Mathematica 20, 2010 147