RATIO MATHEMATICA 26 (2014), 21-38 ISSN:1592-7415 On multiplication Γ-modules A. A. Estaji1, A. As. Estaji2, A. S. Khorasani3, S. Baghdari4 1 2 3 4 Department of Mathematics and Computer Sciences, Hakim Sabzevari University, PO Box 397, Sabzevar, Iran. 1 aaestaji@hsu.ac.ir, 2 a$ -$aestaji@yahoo.com 3 saghafiali21@yahoo.com, 4 m.baghdari@yahoo.com Abstract In this article, we study some properties of multiplication MΓ- modules and their prime MΓ-submodules. We verify the conditions of ACC and DCC on prime MΓ-submodules of multiplication MΓ- module. Key words: Γ-ring, multiplication MΓ-module, prime MΓ-submodule, prime ideal. MSC 2010: 13A15, 16D25, 16N60. 1 Introduction The notion of a Γ-ring was first introduced by Nobusawa [17]. Barnes [5] weakened slightly the conditions in the definition of Γ-ring in the sense of Nobusawa. After the Γ-ring was defined by Barnes and Nobusawa, a lot of researchers studied on the Γ-ring. Barnes [5], Kyuno [15] and Luh [16] studied the structure of Γ-rings and obtained various generalizations analogous of corresponding parts in ring theory. Recently, Dumitru, Ersoy, Hoque, Öztürk, Paul, Selvaraj, have studied on several aspects in gamma- rings (see [10, 8, 12, 14, 18, 19, 20]). McCasland and Smith [14] showed that any Noetherian module M con- tains only finitely many minimal prime submodules. D. D. Anderson [2] generalized the well-known counterpart of this result for commutative rings, i.e., he abandoned the Noetherianness and showed that if every prime ideal minimal over an ideal I is finitely generated, then R contains only finitely many prime ideals minimal over I. Behboodi and Koohy [7] showed that this 21 A. A. Estaji, A. As. Estaji, A. S. Khorasani, S. Baghdari result of Anderson was true for any associative ring (not necessarily commu- tative) and also, they extended it to multiplication modules, i.e., if M is a multiplication module such that every prime submodule minimal over a sub- module K is finitely generated, then M contains only finitely many prime submodules minimal over K. In this paper, we study some properties of multiplication left MΓ-modules and their prime MΓ-submodules. This paper is organized as follows: In Section 2, we review some basic notions and properties of Γ-rings. In Section 3, the concept of a moltiplication MΓ-module is introduced and its basic properties are discussed. Also, we show that If L is a left operator ring of the Γ-ring M and A is a multiplication unitary left MΓ-module, then A is a multiplication left L-module. In Section 4, we proved that in fact this result was true for Γ-rings and MΓ-modules. 2 Preliminaries In this section we recall certain definitions needed for our purpose. Recall that for additive abelian groups M and Γ we say that M is a Γ-ring if there exists a mapping · : M × Γ ×M −→ M (m,γ,m′) −→ mγm′ such that for every a,b,c ∈ M and α,β ∈ Γ, the following hold: 1. (a+b)αc = aαc+bαc, a(α+β)c = aαc+aβc and aα(b+c) = aαb+aαc; 2. (aαb)βc = aα(bβc). Note that any ring R, can be regarded as an R-ring. A Γ-ring M is called commutative, if for any x,y ∈ M and γ ∈ Γ, we have xγy = yγx. M is called a Γ-ring with unit, if there exists elements 1 ∈ M and γ0 ∈ Γ such that for any m ∈ M, 1γ0m = m = mγ01. If A and B are subsets of a Γ-ring M and Θ ⊆ Γ, we denote AΘB, the subset of M consisting of all finite sums of the form ∑ aiγibi, where (ai,γi,bi) ∈ A × Θ × B. For singleton subsets we abbreviate this notation for example, {a}ΘB = aΘB. A subset I of a Γ-ring M is said to be a right ideal of R if I is an additive subgroup of M and IΓM ⊆ I. A left ideal of M is defined in a similar way. If I is both a right and left ideal, we say that A is an ideal of M. For each subset S of a Γ-ring M, the smallest right ideal containing S is called the right ideal generated by S and is denoted by |S〉. Similarly 22 On multiplication Γ-modules we define 〈S| and 〈S〉, the left and two-sided (respectively) ideals generated by S. For each a of a Γ-ring M, the smallest right ideal containing a is called the principal right ideal generated by a and is denoted by |a〉. We similarly define 〈a| and 〈a〉, the principal left and two-sided (respectively) ideals generated by a. We have |a〉 = Za + aΓM, 〈a| = Za + MΓa, and 〈a〉 = Za + aΓM + MΓa + MΓaΓM, where Za = {na : n is an integer}. Let I be an ideal of Γ-ring M. If for each a + I, b + I in the factor group M/I, and each γ ∈ Γ, we define (a + I)γ(b + I) = aγb + I, then M/I is a Γ-ring which we shall call the difference Γ-ring of M with respect to I. Let M be a Γ-ring and F the free abelian group generated by Γ × M. Then A = { ∑ i ni(γi,xi) ∈ F : a ∈ M ⇒ ∑ l niaγiXi = 0} is a subgroup of F. Let R = F/A, the factor group, and denote the coset (γ,x) + A by [γ,x]. It can be verified easily that [α,x] + [β,x] = [α + β,x] and [α,x] + [α,y] = [α,x + y] for all α, β ∈ Γ and x, y ∈ M. We define a multiplication in R by ∑ i[αi,xi] ∑ J[βj,yj] = ∑ iJ [αi,xiβjyj]. Then R forms a ring. If we define a composition on M×R into M by a ∑ l[αi,xi] = ∑ i aαixi for a ∈ M,∑ i[αi,xi] ∈ R, then M is a right R-module, and we call R the right operator ring of the Γ -ring M. Similarly, we may construct a left operator ring L of M so that M is a left L-module. Clearly I is a right (left) ideal of M if and only if I is a right R-module (left L- module) of M. Also if A is a right (left) ideal of R(L), then MA(AM) is an ideal of M. For subsets N ⊆ M, Φ ⊆ Γ, we denote by [Φ,N] the set of all finite sums ∑ i[γi,xi] in R, where γi ∈ Φ, xi ∈ N, and we denote by [(Φ,N)] the set of all elements [ϕ,x] in R, where ϕ ∈ Φ, x ∈ N. Thus, in particular, R = [Γ,M]. An ideal P of M is prime if, for any ideals U and V of M, UΓU ⊆ P implies U ⊆ P or V ⊆ P . A subset S of M is an m-system in M if S = ∅ or if a,b ∈ S implies < a > Γ < b > ∩S 6= ∅. The prime radical P(A) is the set of x in M such that every m-system containing x meets A. The prime radical of the zero ideal in a Γ-ring M is called the prime radical of the Γ-ring M which we denote by P(M). An ideal Q of M is semi-prime if, for any ideals U of M, UΓU ⊆ Q implies U ⊆ Q. Proposition 2.1. [15] If Q is an ideal in a commutative Γ-ring with unit M, then P(Q) is the smallest semi-prime ideal in M which contains Q, i.e. P(Q) = ⋂ P where P runs over all the semi-prime ideals of M such that Q ⊆ P . Let P be a proper ideal in a commutative Γ-ring with unit M. It is clear that the following conditions are equivallent. 23 A. A. Estaji, A. As. Estaji, A. S. Khorasani, S. Baghdari 1. P is semi-prime. 2. For any a ∈ M, if aγ0a ∈ P , then a ∈ P . 3. For any a ∈ M and n ∈ N, if (aγ0)na ∈ P , then a ∈ P . Proposition 2.2. [13] Let Q be an ideal in a commutative Γ-ring with unit M and A be the set of all x ∈ M such that (xγ0)nx ∈ Q for some n ∈ N∪{0}, where (xγ0) 0x = x. Then A = P(Q). 3 MΓ-module Let M be a Γ-ring. A left MΓ-module is an additive abelian group A together with a mapping · : M × Γ × A −→ A ( the image of (m,γ,a) being denoted by mγa), such that for all a,a1,a2 ∈ A, γ,γ1,γ2 ∈ Γ, and m,m1,m2 ∈ M the following hold: 1. mγ(a1 + a2) = mγa1 + mγa2; 2. (m1 + m2)γa = m1γm + m2γa; 3. m1γ1(m2γ2a) = (m1γ1m2)γ2a. A right MΓ-module is defined in analogous manner. If I is a left ideal of a Γ-ring M, then I is a left MΓ-module with rγa (r ∈ M,γ ∈ Γ,a ∈ I) being the ordinary product in M. In particular, {0} and M are MΓ-modules. Let A be a left MΓ-module and B a nonempty subset of A. B is a MΓ- submodule of A, which we denote by B ≤ A, provided that B is an additive subgroup of A and mγb ∈ B, for all (m,γ,b) ∈ M × Γ ×B. Definition 3.1. Let A be a left MΓ-module and X a subset of A. Let {Aλ}λ∈Λ be the family of all MΓ-submodule of A which contain X. Then ⋂ λ∈Λ Aλ is called the MΓ-submodule of A generated by the set X and denoted 〈X|. If B ⊆ A, N ⊆ M and Θ ⊆ Γ, we denote NΘB, the subset of A consisting of all finite sums of the form ∑ niγibi where (ni,γi,bi) ∈ N × Θ × B. For singleton subsets we abbreviate this notation for example, {n}ΘB = nΘB. If X = {a1, . . . ,an}, we write 〈a1, . . . ,an| in place of 〈X|. If A = 〈a1, . . . ,an|, (ai ∈ A), A is said to be finitely generated. If a ∈ A, the MΓ-submodule 〈a| of A is called the cyclic MΓ-submodule generated by a. We have 〈X| = ZX + MΓX, where ZS = { ∑k i=1 nixi : ni ∈ Z,xi ∈ S and k is an integer}. Finally, if {Bλ}λ∈Λ is a family of MΓ-submodules of A, then the MΓ- submodule generated by X = ⋃ λ∈Λ Bλ is called the sum of the MΓ-modules 24 On multiplication Γ-modules Bλ and usually denoted 〈X| = ∑ λ∈Λ Bλ. If the index set Λ is finite, the sum of B1, . . . , Bk is denoted B1 + B2 + ... + Bk. It is clear that if {Bλ}λ∈Λ is a family of MΓ-submodules of A, then ∑ λ∈Λ Bλ consists of all finite sums bλ1 + . . . + bλk with bλj ∈ Bλl. Proposition 3.1. Let M be a Γ-ring and {Iλ}λ∈Λ be a family of left ideals of M. If A is a left MΓ-module, then ( ∑ λ∈Λ Iλ)ΓA = ∑ λ∈Λ (IλΓA). Proof. Let x ∈ ( ∑ λ∈Λ Iλ)ΓA. Then there exists a1, . . . ,ak ∈ A and γ1, . . . ,γk ∈ Γ and x1, . . . ,xk ∈ ∑ λ∈Λ Iλ such that x = ∑k t=1 xtγtat, it follows that for 1 ≤ t ≤ k, xt = ∑kt j=1 iλjt with iλjt ∈ Iλjt. Hence x = ∑k t=1 ∑kt j=1 iλjtγtat ∈∑ λ∈Λ(IλΓA). Therefore ( ∑ λ∈Λ Iλ)ΓA ⊆ ∑ λ∈Λ(IλΓA). Also, Since for every λ ∈ Λ, IλΓA ⊆ ( ∑ λ∈Λ Iλ)ΓA, we conclude that ∑ λ∈Λ(IλΓA) ⊆ ( ∑ λ∈Λ Iλ)ΓA. Hence ( ∑ λ∈Λ Iλ)ΓA = ∑ λ∈Λ(IλΓA). Definition 3.2. If A is a left MΓ-module and S is the set of all MΓ-submodules B of A such that B 6= A, then S is partially ordered by set-theoretic inclu- sion. B is a maximal MΓ-submodule if and only if B is a maximal element in the partially ordered set S. Proposition 3.2. If A is a non-zero finitely generated left MΓ-module, then the following statements are hold. 1. If K is a proper MΓ-submodule of A, then there exists a maximal MΓ- submodule of A such that contain K. 2. A has a maximal MΓ-submodule. Proof. (1) Let A = 〈a1, . . . ,an| and S = {L : K ⊆ L and L is a proper MΓ-submodule of A}. S is partially ordered by inclusion and note that S 6= ∅, since K ∈ S. If {Lλ}λ∈Λ is a chain in S, then L = ⋃ λ∈Λ Lλ is a MΓ-submodule of A. We show that L 6= A. If L = A, then for every 1 ≤ i ≤ n, there exists λi ∈ Λ such that ai ∈ Lλi. Since {Lλ}λ∈Λ is a chain in S, we conclude that there exists 1 ≤ j ≤ n such that a1, . . . ,an ∈ Lλj . Therefore A = Lλj ∈ S which contradicts the fact that A 6∈ S. It follows easily that L is an upper bound {Lλ}λ∈Λ in S. By Zorn’s Lemma there exists a proper MΓ-submodule B of A that is maximal in S. It is a clear that B a maximal MΓ-submodule of A such that contain K. (2) By part (1), it suffices we put K = 〈0|. 25 A. A. Estaji, A. As. Estaji, A. S. Khorasani, S. Baghdari Definition 3.3. A left MΓ-module A is unitary if there exists an element, say 1 in M and an element γ0 ∈ Γ, such that, 1γ0a = a and 1γ0m = m = mγ01 for every (a,m) ∈ A×M. Corolary 3.1. If M is a unitary left (right) MΓ-module, then M has a left (right) maximal ideal. Proof. It is evident by Proposition 3.2. Let A be a left MΓ-module. let X ⊆ A and let B ≤ A. Then the set (B : X) := {m ∈ M : mΓX ⊆ B} is a left ideal of M. In particular, if a ∈ A, then (0 : a) := ((0) : {a}) is called the left annihilator of a and (0 : A) := ((0) : A) is an ideal of M called the annihilating ideal of A. Furthermore A is said to be faithful if and only if (0 : A) = (0). Definition 3.4. A left MΓ-module A is called a multiplication left MΓ-module if each MΓ-submodule of A is of the form IΓA, where I is an ideal of M. Proposition 3.3. Let B be a MΓ-submodule of multiplication left MΓ-module A. Then B = (B : A)ΓA. Proof. It is a clear that (B : A)ΓA ⊆ B. Since A is a multiplication left MΓ- module, we conclude that there exists ideal I of Γ-ring M such that B = IΓA, it follows that B = IΓA ⊆ (B : A)ΓA ⊆ B. Therefore B = (B : A)ΓA. Proposition 3.4. Let A be a left MΓ-module. A is multiplication if and only if for every a ∈ A, there exists ideal I in M such that 〈a| = IΓA. Proof. In view of Definition 3.4, it is enough to show that if for every a ∈ A, there exists ideal I in M such that 〈a| = IΓA, then A is multiplication. Let B be an MΓ-submodule of A. Then for every b ∈ B, there exists ideal Ib in M such that 〈b| = IbΓA. By Proposition 3.1, ( ∑ b∈B Ib)ΓA = ∑ b∈B(IbΓA) =∑ b∈B〈b| = B, it follows that A is multiplication. Proposition 3.5. Let M be a Γ-ring which has a unique maximal ideal Q and A be a unitary multiplication left MΓ-module. If every ideal I in M is contained in Q, then for every a ∈ A\QΓA, 〈a| = A. Proof. Suppose that a ∈ A\QΓA. Since A is multiplication left MΓ-module, we conclude that there exists ideal I in M such that 〈a| = IΓA. Clearly I 6⊆ Q and hence I = M, which implies 〈a| = MΓA = A. Corolary 3.2. Let Γ-ring M be a unitary left MΓ-module which has a unique maximal ideal Q and A be a unitary multiplication left MΓ-module. Then for every a ∈ A\QΓA, 〈a| = A. 26 On multiplication Γ-modules Proof. By Propositions 3.2 and 3.5, it is evident. Proposition 3.6. Let L be a left operator ring of the Γ-ring M and let A be a unitary left MΓ-module. If we define a composition on L × A into A by ( ∑ l[xi,αi])a = ∑ i xiαia for a ∈ A, ∑ i[xi,αi] ∈ L, then A is a left L- module. Also, for every B ⊆ A, B is a MΓ-submodule of A if and only if B is a L-submodule of A. Proof. Suppose that 1 ∈ M and γ0 ∈ Γ such that for every (a,m) ∈ A×M, 1γ0a = a and 1γ0m = m = mγ01. Let ∑t i=1[xi,αi] = ∑s j=1[yj,βj] ∈ L and a = b ∈ A. By definition of left operator ring of the Γ-ring M, we conclude that ∑t i=1 xiαi1 = ∑s j=1 yjβj1, it follows that ( ∑t i=1[xi,αi])a = ∑t i=1 xiαia = ∑t i=1(xiαi(1γ0a)) = ∑t i=1(xiαi1)γ0a = ( ∑t i=1 xiαi1)γ0a = ( ∑s j=1 yjβj1)γ0b = ∑s j=1 yjβjb = ( ∑s j=1[yj,βj])b Hence composition on L × A into A is a well-defined. Let r = ∑t i=1[xi,αi] and s = ∑s j=1[yj,βj]. Then for every a ∈ A, (rs)a = ( ∑ i,j[xiαiyj,βj])a = ∑ i,j(xiαiyj)βja = ∑ i,j xiαi(yjβja) = ∑t i=1 xiαi( ∑s j=1 yjβja) = ( ∑t i=1[xi,αi])( ∑s j=1 yjβja) = r(( ∑s j=1[yj,βj])a) = r(sa) The remainder of the proof is now easy. Proposition 3.7. Let L be a left operator ring of the Γ-ring M. If A is a multiplication unitary left MΓ-module, then A is a multiplication left L- module. Proof. Let B be a L-submodule of A. By Proposition 3.6, B is a MΓ- submodule of A and there exists ideal I of Γ-ring M such that B = IΓA. It well known that [Γ,I] is an ideal of L. We show that B = [I, Γ]A. Suppose that a1, . . . ,at ∈ A, and for every 1 ≤ i ≤ t, ∑ki j=1[xij,αij ] ∈ [I, Γ]. Then we 27 A. A. Estaji, A. As. Estaji, A. S. Khorasani, S. Baghdari have ∑t i=1( ∑ki j=1[xij,αij ])ai = ∑t i=1 ∑ki j=1 xijαijai) ∈ B and it follows that [I, Γ]A ⊆ B. Also, if b ∈ B, then there exists x1, . . . ,xt ∈ I,γ1, . . . ,γt ∈ Γ, and a1, . . . ,at ∈ A such that b = ∑t i=1 xiγiai = ∑t i=1[xi,γi]ai ∈ [I, Γ]A and we conclude that B = [I, Γ]A. Proposition 3.8. Let A be a unitary cyclic left MΓ-module. If L is a left operator ring of the Γ-ring M and for every l, l′ ∈ L, there exists l′′ ∈ L such that ll′ = l′′l, then A is a multiplication left L-module. Proof. Let B be a L-submodule of A and I = {l ∈ L : lA ⊆ B}, then IA ⊆ B. Since A is a unitary cyclic left MΓ-module, we conclude that there exists a ∈ A such that A = MΓa. Let b ∈ B. Hence there exists m1, . . . ,mt ∈ M and γ1, . . . ,γt ∈ Γ such that b = ∑t i=1 miγia. In view of operations of addition and multiplication in left L-module A, we have b = ∑t i=1[mi,γi]a = ( ∑t i=1[mi,γi])a. We put l = ∑t i=1[mi,γi] and it follows that b = la. If a′ ∈ A, then a similar argument shows that there exists l′ ∈ L such that a′ = l′a. By hypothesis, there exists l′′ ∈ L such that ll′ = l′′l. Therefore la′ = ll′a = l′′la = l′′b ∈ B and it follows that l ∈ I, this is b = la ∈ IA. Hence B = IA and the proof is now complete. Definition 3.5. Let A be a unitary left MΓ-module and B be a MΓ-submodule in A and P ∈ Max(M). A is called P -cyclic if there exist p ∈ P and b ∈ B such that (1−p)γ0B ⊆ MΓb and also, it is clear that (1−p)γ0B = (1−p)ΓB. Define TPB as the set of all b ∈ B such that (1−p)γ0b = 0, for some p ∈ P . Lemma 3.1. Let A be a unitary left MΓ-module and B be a MΓ-submodule in A and P ∈ Max(M). If M is a commutative Γ-ring, then TPB is a MΓ-submodule in A. Proof. Suppose b1,b2 ∈ TPB. So there exist p1,p2 ∈ P such that b1 = p1γ0b1 and b2 = p2γ0b2. Let p0 = p1 +p2−p1γ0p2. It is clear that (1−p0)γ0(b1−b2) = 0. Hence b1 − b2 ∈ TPB. Let x ∈ MΓ(TPB). So x = ∑n i=1 miγibi, where n ∈ N, bi ∈ TPB, γi ∈ Γ and mi ∈ M (1 ≤ i ≤ n). Suppose i ∈ {1, · · · ,n}. Since bi ∈ TPN, there exists pi ∈ P such that (1 − pi)γ0miγibi = 0. Hence miγibi ∈ TPN. Thus x ∈ TPB. Hence MΓTPB = TPB. Proposition 3.9. Let M be a commutative Γ-ring and let A be a unitary left MΓ-module. A is multiplication MΓ-module if and only if for any ideal P ∈ Max(M), either A = TPA or A is P -cyclic. Proof. Let A be a multiplication ideal and P ∈ Max(M). First suppose that A = PΓA. Since A is multiplication ideal, we conclude that for every a ∈ A, there exists an ideal I in M such that < a >= IΓA. Hence < a >= PΓ < 28 On multiplication Γ-modules a >. So there exists p ∈ P such that (1−p)γ0a = 0, it follows that a ∈ TPB and then A = TPA. Now suppose that A 6= PΓA and x ∈ A\PΓA. Then there exists an ideal I in M such that < x >= IΓA and P + I = M. Obviously, if we assume that p ∈ P , then (1 −p)γ0A ⊆ MΓx. Therefore A is P-cyclic. Conversely, suppose that B is a MΓ-submodule in A. Define I as the set of all m ∈ M, where mγ0a ∈ B for any a ∈ A. Clearly I is an ideal in M and IΓA ⊆ B. Let b ∈ B. Define K as the set of all m ∈ M, where mγ0b ∈ IΓA. We claim K = M. Assume that K ⊂ M. Hence by Zorns Lemma there exists Q ∈ Max(M) such that K ⊆ Q ⊂ M. By hypothesis A = TQA or A is Q-cyclic. If A = TQA, then there exists s ∈ Q such that (1−s)γ0b = 0. Hence (1−s) ∈ K ⊆ Q, it follows that 1 ∈ Q, a contradiction. If A is Q-cyclic, then there exist t ∈ Q and c ∈ A such that (1−t)γ0A ⊆ MΓc =< c >. Define L as the set of all m ∈ M such that mγ0c ∈ (1−t)γ0B. Clearly L is an ideal in M and Lγ0c ⊆ (1 − t)γ0B ⊆< c >. Hence (1 − t)γ0B ⊆ Lγ0c. So (1 − t)γ0B = Lγ0c, it follows that (1 − t)γ0Lγ0A ⊆ (1 − t)γ0B ⊆ B and (1 − t)γ0L ⊆ I. Therefore (1 − t)γ0(1 − t)γ0B ⊆ IΓA. Hence (1 − t)γ0(1 − t) ∈ K ⊆ Q. Thus 1 − t ∈ Q, it follows that 1 ∈ Q, a contradiction. Hence K = M and b ∈ IΓA. Thus A is a multiplication ideal. Let A be a left MΓ-module. A is said to have the intersection property provided that for every non-empty collection of ideals {Iλ}λ∈Λ of M,⋂ λ∈Λ IλΓA = ( ⋂ λ∈Λ Iλ)ΓA. If left MΓ-module of A has intersection property, then for every non-empty collection of ideals {Iλ}λ∈Λ of M,⋂ λ∈Λ IλΓA = ( ⋂ λ∈Λ (Iλ + Ann(A)))ΓA. Proposition 3.10. Let M be a commutative Γ-ring and let A be a unitary left MΓ-module. 1. If A has intersection property and for any MΓ-submodule N in A any ideal I in M which N ⊂ IΓA, there exists ideal J in M such that J ⊂ I and N ⊆ JΓA, then A is multiplication left MΓ-module. 2. If A is faithful left multiplication MΓ-module, then A has intersection property and for any MΓ-submodule N in A any ideal I in M which N ⊂ IΓA, there exists ideal J in M such that J ⊂ I and N ⊆ JΓA. 29 A. A. Estaji, A. As. Estaji, A. S. Khorasani, S. Baghdari Proof. (1) Let N be a MΓ-submodule in A and S = {I : I is an ideal of M and N ⊆ IΓA}. Clearly M ∈ S. Since A has intersection property, we conclude from Zorns Lemma that S has a minimal member I (say). Since N ⊆ IΓA and I is minimal element of S, we can conclude that N = IΓA. It follows that A is a multiplication ideal. (2) Let {Iλ}λ∈Λ be a nonempty collection of ideal in M and I = ⋂ λ∈Λ Iλ. Clearly IΓA ⊆ ⋂ λ∈Λ(IλΓA). Let x ∈ ⋂ λ∈Λ(IλΓA) and we put L = {m ∈ M : mγ0x ∈ IΓA}. We claim L = M. Assume that L ⊂ M. By Proposition 3.2, there exists P ∈ Max(M) such that L ⊆ P . It is clear that x 6∈ TPA. Hence TPA 6= A and by Proposition 3.9, A is P-cyclic. Hence there exist a ∈ A and p ∈ P such that (1−p)γ0A ⊆ MΓa =< a >. Thus (1−p)γ0x ∈ ⋂ λ∈Λ(Iλγ0a) and so for any λ ∈ Λ, (1 −p)γ0x ∈ Iλγ0a. It is clear that (1 −p)γ0(1 −p) ∈ L ⊆ P , in view of the fact that A is faithful. Hence 1 ∈ P , a contradiction. Therefore L = M, it follows that x = 1γ0x ∈ IΓA and A has intersection property. Now suppose N be a MΓ-submodule in A and I be an ideal in M which N ⊂ IΓA. Since A is multiplication MΓ-module, there exists an ideal J in M such that N = JΓA. Let K = I ∩ J. Clearly, K ⊂ I and since A has intersection property, we conclude that N ⊆ KΓA. The proof is now complete. Proposition 3.11. Let A be a faithful multiplication MΓ-module and I,J be two ideals in M. IΓA ⊆ JΓA if and only if either I ⊆ J or A = [J : I]ΓA. Proof. Let I 6⊆ J. Note that [J : I] = ⋂ i∈X[J :< i >] where X is the set of all elements i ∈ I with i 6∈ J. By Proposition 3.10, [J : I]ΓA = ⋂ i∈X ([J :< i >]ΓA) If for every i ∈ X, A = [J :< i >]ΓA, then A = [J : I]ΓA, which finishes the proof. Let i ∈ X and Q = [J :< i >]. It is clear that Q 6= M. Let Ω denote the collection of all semi-prime ideals P in M containing Q. Suppose that there exists P ∈ Ω such that A 6= PΓA and x ∈ A\PΓA. Since A is a multiplication MΓ-module, we conclude that there exists ideal D in M such that < x >= DΓA and D 6⊆ P . Thus cΓA ⊆< x > for some c ∈ D \ P . Now we have cΓaΓA ⊆ JΓ < x >. It is easily to show that for any γ ∈ Γ, there exists γ1 ∈ Γ and b ∈ J such that (cγa − 1γ1b)γ0x = 0, it follows that (cγa − 1γ1b)ΓcΓA = 0. Hence cγc ∈ [J :< i >] = Q. Since P is a semi-prime ideal containing Q, we conclude that c ∈ P , a contradiction. Therefore for every P ∈ Ω, A = PΓA and by Propositions 2.1 and 3.10, 30 On multiplication Γ-modules A = P(Q)ΓA. Let j ∈ A. It is easily to show that < j >= P(Q)Γ < j >. Then there exists s ∈ P(Q) such that for every n ∈ N, j = (sγ0)nj. By Proposition 2.2, there exists t ∈ N ∪{0} such that (sγ0)ts ∈ Q, it follows that j = (sγ0) tsγ0j ∈ QΓA, i.e., A ⊆ QΓA. Hence QΓA = A. The converse is evident. 4 Prime MΓ-submodule Through this section M and A will denote a commutative Γ-ring with unit and an unitary left MΓ-module, respectively. Definition 4.1. A prime ideal P in M is called a minimal prime ideal of the ideal I if I ⊆ P and there is no prime ideal P ′ such that I ⊆ P ′ ⊂ P . Let Min(I) denote the set of minimal prime ideals of I in Γ-ring M, and every element of Min((0)) is called minimal prime ideal. Proposition 4.1. If an ideal I of Γ-ring M is contained in a prime ideal P of M, then P contains a minimal prime ideal of I. Proof. Let A = {Q : Q is prime ideal of M and I ⊆ Q ⊆ P}. By Zorn’s Lemma, there is a prime ideal Q of R which is minimal member with respect to inclusion in A. Therefore Q ∈ Min(I) and I ⊆ Q ⊆ P . Lemma 4.1. Let Γ be a finitely generated group. If I and J are finitely generated ideals of M, then IΓJ is finitely generated ideal of M. Proof. Let I = 〈a1, . . . ,an〉, J = 〈b1, . . . ,bm〉, and Γ = 〈γ1, . . . ,γk〉. It is clear that IΓJ = 〈aiγtbj : 1 ≤ i ≤ n, 1 ≤ t ≤ k, 1 ≤ j ≤ m〉. Proposition 4.2. Let Γ be a finitely generated group. If I is a proper ideal of M and each minimal prime ideal of I is finitely generated, then Min(I) is finite set. Proof. Consider the set S = {P1ΓP2...Pn; n ∈ N and Pi ∈ Min(I), for each 1 ≤ i ≤ n} and set ∆ = {K; K is an ideal of M and Q 6⊆ K, for each Q ∈S} 31 A. A. Estaji, A. As. Estaji, A. S. Khorasani, S. Baghdari which is the non-empty set, since I ∈ ∆. (∆,⊆) is the partial ordered set. Suppose {Kλ}λ∈Λ is the chain of ∆ in which Λ 6= ∅ and set K = ⋃ λ∈Λ Kλ. It is clear that K is an ideal of M. Also, if there exits Q ∈S such that Q ⊆ K, then by Lemma 4.1, Q = P1ΓP2...Pn is finitely generated ideal of M, i.e., Q = 〈x1, . . . ,xn〉. But Q ⊆ K implies that x1,x2, ...,xn ∈ K. Thus there exists λ ∈ Λ such that x1,x2, ...,xn ∈ Kλ and so Q ⊆ Kλ, contradiction. Hence, for each Q ∈S, Q 6⊆ K and K ∈ ∆ is the upper band of this chain. By Zorhn’s lemma ∆ has maximal element such as Q. Now if a 6∈ Q and b 6∈ Q for a,b ∈ M, then Q ⊆〈Q∪{a}〉 and Q ⊆〈Q∪{b}〉. Maximality of Q implies that 〈Q∪{a}〉, 〈Q∪{b}〉 6∈ ∆. So there exists Q1 and Q2 in S such that Q1 ⊆〈Q∪{a}〉 and Q2 ⊆〈Q∪{b}〉. It is clear that Q1ΓQ2 ⊆ Q which is contradiction, since Q1ΓQ2 ∈S. Therefore 〈a〉Γ〈b〉 6⊆ Q and Q is a prime ideal of M contained I. By Proposition 4.1, there exists a minimal prime ideal P ⊆ Q. But P ∈ S, contradictory with Q ∈ ∆. Above contradicts show that there exists Q′ = P1ΓP2...Pm ∈S such that Q′ ⊆ I. Now for each P ∈ Min(I) we have Q′ ⊆ I ⊆ P and P1ΓP2...Pm ⊆ P . It is clear that Pj ⊆ P for some 1 ≤ j ≤ m. Thus Pj = P , since P is minimal. Hence Min(I) = {P1,P2, ...,Pm} is finite. Proposition 4.3. For proper MΓ-submodule B of A, the following state- ments equivalent: 1. For every MΓ-submodule C of A, if B ⊂ C, then (B : A) = (B : C). 2. For every (m,a) ∈ M ×A, if mΓa ⊆ B, then a ∈ B or m ∈ (B : A). Proof. (1) ⇒ (2) Let (m,a) ∈ M × A such that mΓa ⊆ B and a 6∈ B. It is clear that B ⊂ B + MΓa. Since mΓ(B + MΓa) ⊆ mΓB + mΓ(MΓa) = mΓB + MΓ(mΓa) ⊆ B, we conclude from statement (1) that m ∈ (B : B + MΓa) = (B : A) and the proof is now complete. (2) ⇒ (1) Let C be a MΓ-submodule of A such that B ⊂ C. It is clear that (B : A) ⊆ (B : C). Now, suppose that m ∈ (B : C). Since B ⊂ C, we infer that there exists a ∈ C \B such that mΓa ⊆ B. By statement (2), m ∈ (B : A) and the proof is now complete. Definition 4.2. A proper MΓ-submodule B of A is said to be prime if mΓa ⊆ B for (m,a) ∈ M ×A implies that either a ∈ B or m ∈ (B : A). Proposition 4.4. If B is a prime MΓ-submodule of A, then (B : A) is a prime ideal of Γ-ring M. 32 On multiplication Γ-modules Proof. Let x,y ∈ M such that 〈x〉Γ〈y〉 ⊆ (B : A) and x 6∈ (B : A). Then there exists γ ∈ Γ and a ∈ A such that xγa 6∈ B, and also, yΓ(xγa) = (yΓx)γa = (xΓy)γa ⊆ B. Since B is a prime MΓ-submodule of A and xγa 6∈ B, we conclude that yΓA ⊆ B, i. e., y ∈ (B : A). The proof is now complete. Proposition 4.5. Let A be a multiplication left MΓ-module, and B, B1, . . . , Bk be MΓ-submodules of A. If B is a prime MΓ-submodule of A, then the following statements are equivalent. 1. Bj ⊆ B for some 1 ≤ j ≤ k. 2. ⋂k i=1 Bi ⊆ B. Proof. (1) ⇒ (2) It is clear. (2) ⇒ (1) We have Bi = IiΓA for some ideals Ii, (1 ≤ i ≤ k) of Γ-ring M. Then ( ⋂k i=1 Ii)ΓA ⊆ ⋂k i=1(IiΓA) = ⋂k i=1 Bi ⊆ B and so ⋂k i=1 Ii ⊆ (B : A). Since M is a commutative Γ-ring, we infer that for every permutations θ of {1, 2, . . . ,k}, I1ΓI2 · · ·Ik = Iθ(1)ΓIθ(2) · · ·Iθ(k), it follows that I1ΓI2 · · ·Ik ⊆⋂k i=1 Ii ⊆ (B : A). Since by Proposition 4.4, (B : A) is prime ideal of Γ-ring M, we conclude that Ij ⊆ (B : A) for some 1 ≤ j ≤ k. Therefore, by Proposition 3.3, Bj = IjΓA ⊆ B for some 1 ≤ j ≤ k. Proposition 4.6. If A is a multiplication left MΓ-module, then for MΓ- submodule B of A, the following statements are equivalent. 1. B is prime MΓ-submodule of A. 2. (B : A) is prime ideal of Γ-ring M. 3. There exists prime ideal P of Γ-ring M such that B = PΓA and for every ideal I of M, IΓA ⊆ B implies that I ⊆ P . Proof. (1) ⇒ (2) By Proposition 4.4, It is evident. (2) ⇒ (3) We put M = {P : B = PΓA and P is an ideal of Γ-ring M } Since A is multiplication left MΓ-module, we conclude that (M,⊆) is a non- empty partial order set. Let {Pλ}λ∈Λ ⊆M be a chain. By Proposition 3.10,⋂ λ∈Λ Pλ ∈ M is an upper bound of {Pλ}λ∈Λ. By Zorn’s Lemma M has a maximal element. Thus, we can pick a P to be maximal element of M. Let x,y ∈ M and 〈x〉Γ〈y〉 ⊆ P . Hence (〈x〉Γ〈y〉)ΓA ⊆ PΓA = B and we infer that 〈x〉Γ〈y〉⊆ (B : A). Now, by statement (2), x ∈ (B : A) or y ∈ (B : A). Since A is multiplication left MΓ-module, we conclude from the Proposition 33 A. A. Estaji, A. As. Estaji, A. S. Khorasani, S. Baghdari 3.3 that B = (B : A)ΓA, it follows that (B : A) ∈ M. On the other hand, clearly P ⊆ (B : A) and so P = (B : A), i.e., x ∈ P or y ∈ P , Thus P is prime ideal of Γ-ring M. (3) ⇒ (1) Let prime ideal P of Γ-ring M such that B = PΓA and for every ideal I of Γ-ring M, IΓA ⊆ B implies that I ⊆ P . It is clear that P = (B : A). Let m ∈ M and a ∈ A such that mΓa ⊆ B. Since A is a multiplication S-act, we conclude that there exists an ideal I of Γ-ring M such that 〈a〉 = IΓA, it follows that (mΓI)ΓA = mΓ(IΓA) = mΓ(MΓa) = (mΓM)Γa = (MΓm)Γa = MΓ(mΓa) ⊆ B. Therefore mΓI ⊆ (B : A) = P and it is easy to see directly that 〈m〉ΓI ⊆ (B : A). Then mΓA ⊆ B or a ∈ IΓA ⊆ B and the proof is now complete. Lemma 4.2. Let A be a finitely generated left MΓ-module. If I is an ideal of M such that A = IΓA, then there exists i ∈ I such that (1 − i)γ0A = 0. Proof. If A =< a1, . . . ,an >, then for every 1 ≤ i ≤ n, there exists yi1, . . . ,yin ∈ I such that ai = ∑n j=1 yijγ0aj, it follows that −yi1γ0a1−···−yi(i−1)γ0ai−1 + (1−yii)γ0ai−yi(i+1)γ0ai+1−···−yinγ0an = 0. If B =  1 −y11 −y12 · · · −y1n... ... ... ... −yn1 −yn2 · · · 1 −ynn   , then there exists y ∈ I such that detΓ(B) = (1 + y), where detΓ(B) = ∑ sign(σ)b 1,σ(1) γ0b2,σ(2)γ0 · · ·γ0bn,σ(n) and σ runs over all the permutation on {1, 2, . . . ,n} (see [13]). Since for every 1 ≤ i ≤ n, detΓ(B)γ0ai = 0, we conclude that (1 + y)γ0A = 0 and by setting i = −y the proof will be completed. Proposition 4.7. Let A be a finitely generated faitfull multiplication left MΓ- module. For proper ideal of P in M, the following statements are equivalent. 1. P is a prime ideal of M. 2. PΓA is a prime MΓ-submodule of A. Proof. (1) ⇒ (2) Let I be an ideal of M such that IΓA ⊆ PΓA. Then by Proposition 3.11, either I ⊆ P or A = [P : I]ΓA. If A = [P : I]ΓA, then by Lemma 4.2, there exists i ∈ [P : I] such that (1 − i)γ0A = 0. Since A is a 34 On multiplication Γ-modules faitfull MΓ-module, we conclude that i = 1 and I ⊆ P . Hence by Proposition 4.6, PΓA is a prime MΓ-submodule of A. (2) ⇒ (1) Since A is a faitfull MΓ-module and [PΓA : A]ΓA ⊆ PΓA, we conclude from the Proposition 3.11 and Lemma 4.2 that [PΓA : A] ⊆ P . Hence [PΓA : A] = P and by Proposition 4.6, P is a prime ideal of M. Proposition 4.8. Let A be a multiplication left MΓ-module. Then 1. If M satisfies ACC (DCC) on prime ideals, then A satisfies ACC (DCC) on prime MΓ-submodules. 2. If A is faitfull MΓ-module and (B : A) is a minimal prime ideal in M, then B is a minimal prime MΓ-submodule of A. Proof. (1) Assume that B1 ⊆ B2 ⊆ . . . is a chain of prime MΓ-submodule of A. By Proposition 4.4, (B1 : A) ⊆ (B2 : A) ⊆ . . . is a chain of prime ideal of Γ-ring M. By hypothesis there exists k ∈ N such that for every i ≥ k, (Bi : A) = (Bk : A). It follows from Proposition 3.3 that Bi = (Bi : A)ΓA = (Bk : A)ΓA = Bk. Thus A satisfies ACC on prime MΓ-submodules. (2) assume that B′ is a prime MΓ-submodule of A such that B ′ ⊆ B. By Proposition 4.6, (B′ : A) ⊆ (B : A) is a chain of prime ideal of Γ-ring M. By hypothesis (B′ : A) = (B : A), it follows from Proposition 3.3 that B′ = (B′ : A)ΓA = (B : A)ΓA = B. Thus B is a minimal prime MΓ-submodule of A. Proposition 4.9. Let A be a finitely generated faitfull multiplication left MΓ-module. Then 1. If A satisfies ACC (DCC) on prime MΓ-submodules, then Γ-ring M satisfies ACC (DCC) on prime ideals. 2. If B is a minimal prime MΓ-submodule of A, then (B : A) is a minimal prime ideal of Γ-ring M. Proof. (1) Assume that P1 ⊆ P2 ⊆ . . . is a chain of prime ideals of Γ-ring M. By Proposition 4.7, P1ΓA ⊆ P2ΓA ⊆ . . . is a chain of prime MΓ-submodule of A. By hypothesis there exists k ∈ N such that for every i ≥ k, PkΓA = PiΓA. Since A is a finitely generated faitfull multiplication MΓ-module, we conclude from the Proposition 3.11 and Lemma 4.2 that Pk = Pi. (2) By Proposition 4.6, (B : A) is a prime ideal of Γ-ring M. Assume that P is a prime ideal of Γ-ring M such that P ⊆ (B : A). Hence by Proposition 3.3, PΓA ⊆ (B : A)ΓA = B. Since by Proposition 4.7, PΓA is a prime MΓ- submodule of A, we conclude from our hypothesis that PΓA = (B : A)ΓA. 35 A. A. Estaji, A. As. Estaji, A. S. Khorasani, S. Baghdari Since A is a finitely generated faitfull multiplication MΓ-module, we conclude from the Proposition 3.11 and Lemma 4.2 that P = (B : A). The proof is now complete. Proposition 4.10. Let Γ be a finitely generated group. Let A be a finitely generated faitfull multiplication left MΓ-module. 1. If every prime ideal of Γ-ring M is finitely generated, then A contains only a finitely many minimal prime MΓ-submodule. 2. If every minimal prime MΓ-submodule of A is finitely generated, then Γ-ring M contains only a finite number of minimal prime ideal. Proof. (1) Assume that {Bλ}λ∈Λ is the family of minimal prime MΓ-submodules of A. Set Iλ = (Bλ : A) for λ ∈ Λ. By Proposition 4.9, each Iλ is a min- imal prime ideal of Γ-ring M. On the other hand, by Proposition 4.2, M contains only a finite number of minimal prime ideal as {I1,I2, . . .In}. Now suppose that λ ∈ Λ. So Iλ = Ii, for some 1 ≤ i ≤ n and by Proposition 3.3, Bλ = IλΓA = IiΓA. Thus {I1ΓA,I2ΓA,. . . ,InΓA} is the finite family of minimal prime MΓ-submodule of A. (2) Suppose that I and J are two distinct minimal prime ideal of Γ-ring M. By Proposition 3.11 and Lemma 4.2, A 6= IΓA 6= JΓA and also, by Proposition 4.7, IΓA and JΓA are prime MΓ-submodules of A. Assume that B1 and B2 are two prime MΓ-submodules of A such that B1 ⊆ IΓA and B2 ⊆ JΓA. By Proposition 3.3, B1 = (B1 : A)ΓA and B2 = (B2 : A)ΓA. By Proposition 3.11 and Lemma 4.2, (B1 : A) ⊆ I and (B2 : A) ⊆ J. Since I and J are two distinct minimal prime ideal of Γ-ring M, we conclude from the Proposition 4.4 that (B1 : A) = I and (B2 : A) = J. This says that IΓA and JΓA are two distinct minimal prime MΓ-submodules of A. 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