Microsoft Word - 282-291 Mathematics | 282 2016) عام 2العدد ( 29مجلة إبن الهيثم للعلوم الصرفة و التطبيقية المجلد Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 29 (2) 2016 -Prime Submodules Nuhad S. AL-Mothafar Dept. of Mathematics / College of Science/ University of Baghdad. Adwia J. Abdil .Al-Khalik Dept. of Mathematics / College of Science/ Al-Mustansiriya University. Received in:6/ March /2016,Accepted in:5/June /2016 Abstract Let R be a commutative ring with identity and M be an unitary R-module. Let (M) be the set of all submodules of M, and : (M)  (M)  {} be a function. We say that a proper submodule P of M is -prime if for each r  R and x  M, if rx  P, then either x  P + (P) or r M  P + (P) . Some of the properties of this concept will be investigated. Some characterizations of -prime submodules will be given, and we show that under some assumptions prime submodules and -prime submodules are coincide. Key Words: Prime submodule, weakly prime submodules , -prime submodules. Mathematics | 283 2016) عام 2العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 29 (2) 2016 1- Introduction Throughout this paper, R is a commutative ring with identity and M is an unitary R- module. A proper ideal P of a ring R is prime if for all elements a ,b  R ,ab  P implies either a  P or b  P [ 1, p.40].In the theory of rings, Prime ideals play important roles. One of the natural generalizations of prime ideals which have attracted the interest of several authors in the last two decades is the notion of prime submodule,[2],[3],[4]. These have led to more information on the structure of the R-module M. For an ideal I of R and a submodule N of M let I denote the radical of I, and [N R : M] = {r  R, rM  N} which is clearly an ideal of R. A proper submodule P of M is called a prime submodule if r  R and x  M with rx  P implies that r  [P:M] or x  P,[3]. There are several generalization of the notion of a prime submodules, such as Ebrahimi Atani, F. Farzalipour , introduced and studied weakly prime submodules ,where a proper submodule N of M is said to be weakly prime submodule of M if r  R and x  M 0  rx  N gives that r  [ N : M] or x  N ,[5]. Khaksari and Jafari extended the notion of prime submodule to -prime.Let M be an R-module and (M) be the set of all submodules of M and : (M)  (M)  {} be a function. A proper submodule P of M is said to be -prime if r  R and x  M, rx  P\(P) implies that r  [P:M] or x  P [6]. In this paper ,we define and study the notion of -prime submodules . Let (M) be the set of all submodules of M and : (M)  (M)  {} be a function. A proper submodule P of M is said to be -prime if for each r  R and x  M, if rx  P, then either x  P + (P) or r M  P + (P) . 2-Basic Properties of -Prime Submodules First we give the following definition. Definition (2.1): Let M be an R-module and (M) be the set of all submodules of M. Let :(M) (M)  {} be a function. A proper submodule N of M is said to be -prime if for each r  R and x  M, if rx  N, then x  N + (N) or r M  N + (N ) . Remarks and Examples (2.2) (1) It is clear that every prime submodule of an R-module M is -prime submodule of M ,but the convers is not true in general for example: Let M = Z8 as a Z-module, N =  4 . Then N is not prime submodule of M . But N is -prime submodule of M. Proof : Let : (Z8) (Z8)  {}, where (N) = N +  2  , N  M , then for each r Z, ̅ Z8, if r ̅ N ,then ̅ N + (N) = N + N +  2  =  2  or r Z8  N + (N) =N + N +  2  =  2  .Therefore N =  4  is a -prime submodule of Z8. (2) If (N)  N or (N) = 0, then every -prime submodule of M is a prime submodule. (3) Let N, W be two submodules of an R- module M and NW. If N is -prime submodule of M and (N)  '( N) ,where ': (W)  (W)  {} and  : (M)  (M)  {} ,then N is '-prime submodule of W . Proof : Let rR , mW such that rm N. Since N is -prime submodule of M , so either m N + (N) or r M  N + (N) . But (N)  '( N) , so either m N + '( N) or r M  N + '( N) , so either mN + '( N) or r W  N + '( N) .Therefore N is '-prime submodule of W . (4) Given two functions ,': (M)  (M)  {}such that   ' and (N)  '( N) for each N (M). If N is a -prime submodule of M implies N is '-prime submodule of M . Proof : Let rR , mM such that rm N. Since N is -prime submodule of M , so either m N + (N) or r M  N + (N) . But (N)  '( N) , so either m N + '( N) or r M  N + '( N) . Therefore N is '-prime submodule of M . Mathematics | 284 2016) عام 2العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 29 (2) 2016 (5) Let N and W be two submodules of an R –module M such that NW . If N is -prime submodule of M then it is not necessary that W is -prime submodule of M as the following example explains : Consider the Z – module Z ,the submodule 2Z is -prime submodule of Z(since it is prime ) but 2Z  30Z and 30Z is not -prime submodule of Z .Since if (N) = N , N M and 6.5 =30  30Z but 5 30Z +30Z =30Z and 6Z ⊈ 30 Z +30Z = 30Z. (6) Let M = Z4 as Z-module, N = {0, 2} .Since N is prime, then N is -prime submodule of M. (7) Let M = Z12 as a Z – module and N = 6 and let  : (Z12) (Z12)  {}. If (N) = N+  2 ,  N {6  ,  2  ,  4 , 0  } and (N) = N where N {M , 3}.Therefore 6 = 6 + 2 = 2 and  x Z12, rZ then either x 6 + 2 = 2 or r Z12  6 + 2 = 2 . Hence N is a -prime submodule of M. (8) {0} is the only -prime submoduleof a simple modules.Therefore (0 ) of a simple Z- module Zp (p is prime) is -prime submodule. (9 ) Let M = Z  Z as a Z-module, N = 2Z(0) and let :(M)  (M)  {}.If (N) = N,  N  M, then N is not -prime submodule of M. (10) I is a -prime ideal of R if and only if I is a -prime submodule of R. Now, if N is a prime submodule, then sometimes N is called P – prime submodule , where P = [N : M] ,[4 ]. For a -prime, we called P- -prime submodule ,where P = [N +(N) : M]. The following theorem gives some characterizations for -prime submodules. Theorem(2.3): Let N be a proper submodule of an R- module M and P= N  N :M .Then, the following statements are equivalent: 1. N is -prime submodule of M. 2. For every submodule K of M and for every an ideal I of R such that IK N, implies that either K N  N or I P N  N :M . Proof: (1) (2): Let IK N, where I be an ideal of R and K be a submodule of M. Suppose K⊈ N + (N), then there exists k K such that kN + (N). It is clear that for each y I, thus ykN . But N is -prime submodule of M and kN + (N), hence y P= [N + (N): M]. Therefore I P . (2) ) (1): Let r  R , m  M such that rm N. Then < r > < m>  N. So either < m>  N + (N) or < r>  P= N  N :M by(2)); i.e., either mN + (N) or r  P= N  N :M . Therefore N is -prime submodule of M . We can give the following result . Proposition (2.4): Let N be a proper submodule of an R- module M . If N  N :M = [N+  (N) : K] for each submodule K of M such that K⊋ N + (N) , then N is -prime submodule of M. Proof: Let r  R , m  M such that rm N and suppose m N + (N). Let K= N + (N)+ , then K⊋ N + (N), mK and so r [N:K]  [N + (N):K] = N  N :M .It follows that r N  N :M and hence N is -prime. However , we can give another corollary of proposition (2.4). But first we state and prove the following lemma which we needed . Mathematics | 285 2016) عام 2العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 29 (2) 2016 Lemma (2.5) Let N be a proper submodule of an R- module M . If N  N :M N  N ∶ c for each c M \ N + (N) , then N  N :M = [N+(N) : K] for each submodule K of M such that K⊋ N + (N) . Proof: Since K  M so N  N :M  [N+ (N) : K] .Let r  N  N ∶ K ,hence r K  N + (N) . But N + (N) ⊊ K, implies that there exists xK and x N + (N).Hence rx  N+(N) and then r [N+(N) : x] =[N+ (N) : M], which implies that r  N  N ∶ K  N  N :M .Therefore N  N :M = [N+ (N): K] for each submodule K of M such that K⊋ N + (N) . Corollary (2.6): Let N be a proper submodule of an R- module M . If N  :M N  N ∶ c for each c M \ N + (N), then N is -prime submodule of M. Now, the following proposition shows that under the condition (N)  N for all submodule N of M the convers of proposition ( 2.4) is true. Proposition(2.7): If N is a -prime submodule of an R– module M and (N)  N ,then N  :M = [N+ (N) : K] for each submodule K of M such that K ⊋ N + (N) . Proof:Since N is a -prime submodule of M and (N)  N ,so by (Remark 2.2,(2)) N is a prime submodule .Hence [N :M] = [N:K] , for each submodule K of M such that K ⊋ N,[2]. Since (N)  N ,then N  :M = [N+ (N) : K] for each submodule K of M such that K ⊋ N + (N) . It is well Know if N is a prime submodule of an R – module ,then [N : M] is a prime ideal of R ,see [4]. But for a -prime we have: Remark(2.8): (1) If N is -prime submodule of M , then it is not necessarily that [N:M] is a - prime ideal of R, for example: Let M = Z8 as a Z-module ,N =  4 . Then N is - prime submodule of M by (Remark 2.2,(1)). But [ 4 : Z8] = 4Z is not -prime submodule of Z ,where (I) = I for each I an ideal of Z and : (Z)  (Z)  {} be a function. Now, the following proposition shows that under the condition (N)  N for all submodule N of M the above statement is true. Proposition(2.9): If N is a -prime submodule of an R– module M and (N)  N ,then N:M is a - prime ideal of R. Proof:Since N is a -prime submodule of an R– module M and (N)  N, so N is a prime submodule by (2.2,(2)), then [N:M] is a prime ideal of R and hence is a -prime ideal of R.( .2 ,2), then N:M is a Remark (2.10): If [N :M] is - prime ideal of R , then it is not necessarily that N is -prime submodule of M , for example: Let M = Z  Z as a Z-module, N = 2Z(0), N is not -prime submodule of M ,by (2.2,9). But [N :M] = [2Z  (0) : Z  Z ] = 0 is a prime ideal of Z and hence is - prime ideal of Z, where (I) = I for each I an ideal of Z and : (Z)  (Z)  {} be a function. Now, we shall give characterization of -prime submoules, but first recall the following: Let R be any ring. A subset S of R is called multiplicatively closed if 1 S and ab  S for every a ,bS. We Know that every proper ideal P in R is prime if and only if R-P is Mathematics | 286 2016) عام 2العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 29 (2) 2016 multiplicatively closed subset of R,[1,p.42]. And if N is a submodule of an R-module M and S is multiplicatively closed sub set of R, then N(S) ={x M: t  S, such that tx N}be a submodule of M and N N(S). Proposition (2.11): Let N be a proper submodule of an R- module M .If [N + (N):M] is a prime ideal of R and N(S)  N + (N) for each multiplicatively closed subset of R such that S  [N + (N):M] = ,then N is -prime submodule of M . Proof: Let r  R , m  M such that rm N and suppose m N + (N), r  [N + (N):M] . Consider the set S ={1,r, r 2 ,……….},this is multiplicatively closed subset of R and it is clear that S [N + (N):M]  , since N  N :M is a prime ideal of R. But m N + (N) implies that m N(S) and so r m N which is a contradiction .Therefore either m N + (N) or r[N + (N):M] and hence N is -prime submodule of M . Conversely ,if N is -prime submodule of M, to prove N(S)  N + (N) . Let x  N(S) , so there exists t S such that tx  N . But N is -prime submodule of M, so either x  N + (N) or t N  N :M . But t  N  N :M implies that S  N  N :M =  which is a contradiction .Thus , xN +(N) and hence N(S) N+ (N). Proposition (2.12): If [N + (N):M] is maximal ideal of R , then N is -prime submodule of M . Proof: Let r  R , m  M such that rm N .If r  N  N :M ,then R= < r> + N  N :M .Therefore there exist sR and k N  N :M such that 1=s r +k and so m= srm +km  N + (N) .Therefore N is -prime submodule of M. Proposition (2.13): Let N be a proper submodule of an R- module M such that [K: M] ⊈ [N+ (N):M] for each submodule K of M and containing N+ (N) properly .If [N + (N):M] is a prime ideal of R, then N is -prime submodule of M . Proof: Suppose [N + (N):M] is a prime ideal of R, to prove N is -prime submodule of M. Let r  R , m  M such that rm N and suppose m N + (N). Let K = N + (N) + , it is clear that N + (N)⊊ K, and so [K: M] ⊈ [N+ (N): M].Then there exists s [K:M] and s  [N + (N):M].Thus, sM  K and sM ⊈ N+ (N).But, sM  K implies ,r s M  r K = r ( N + (N) + )  N + (N) and rs [N + (N):M]. Since [N + (N):M] is a prime ideal of R and s  [N + (N):M], so r [N + (N):M] . Therefore N is -prime submodule of M . Recall that an R- module M is called mulitplication module if for every submodule N of M ,there exists an ideal I of R such that IM=N ,equivalently ;for every submodule N of M N=[N:M]M, see[7]. Corollary (2.14): Let N be a proper submodule of a mulitplication R- module M. Then N is -prime submodule of M if [N + (N):M] is a prime ideal of R. Proof: Suppose [N + (N):M] is a prime ideal of R, to prove N is -prime submodule of M. Let r  R , m  M such that rm N and suppose m N + (N). Let K = N + (N) + < m>,it is clear that N + (N) ⊊K. Since M is mulitplication , so [K: M] ⊈ [N+ (N):M] by[9,remark (2-15),chapter one ]. Then there exists s [K:M] and s  [N + (N):M].Thus, sM  K and sM ⊈ N+ (N).But, sM  K implies r s M  r K = r ( N + (N) + )  N + (N) and rs [N + (N):M]. Since [N + (N):M] is a prime ideal of R and s  [N + (N):M], so r [N + (N):M] .Therefore N is -prime submodule of M . Mathematics | 287 2016) عام 2العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 29 (2) 2016 As anther consequence of ( 2.13) , we have the following result: Corollary (2.15): Let N be a proper submodule of a cyclic R- module M. Then N is -prime submodule of M if [N + (N):M] is a prime ideal of R. Proof:Since M is cyclic ,then M is a mulitplication . Hence the result follows immediately from corollary (2.14). Recall that an R – module M is said to be bounded module if there exists an element x  M such that annM = ann(x), where annM = {r  R: rm = 0,  m M}, [8].And an R- module M is said to be fully stable if each submodule is stable, where a submodule N of an R – module M is said to be stable if f(N)  N for each f  Hom(N,M), [9]. Corollary (2.16): Let N be a proper submodule of a bounded fully stable R- module M. Then N is -prime submodule of M if [N + (N):M] is a prime ideal of R. Proof:Since M is bounded fully stable R- module, then M is a cyclic R- module by [10,prop.1.1.4,ch.1]. Hence the result follows immediately from corollary (2.15). Proposition (2.17): Let M be an R-module and N,L be two submodules of M .If K be a P- -prime submodule of M such that N  L  K, then L  K + (K) or [N: M]  P K  N :M . Proof:Suppose [N:M]⊈ K  N :M =P, so there exists s  [N: M] and s P K  N :M . Let t L, then st  L N and so st  K. But K is P- -prime submodule of M and s K  N :M . Therefore t K + (K), thus L  K + (K) . Corollary (2.18): Let A an ideal of a ring R and N be a submodule of M . If K is a P- -prime submodule of M such that AM  N  K , then either AM K + (K) or N  K + (K) . proposition (2.19): Let M be an R-module and N be a submodules of M .If P = N  N :M is a prime ideal of R , then N  N :M N  N :rM , r  N  N :M . Proof: Since rM M ,so N  N :M  N  N :rM .Let a  N  N :rM ,so ar M  N  N .Which means that ar  N  N :M .But N  N :M is a prime ideal of R, so either a  N  N :M or r N  N :M , but r  N  N :M , so a  N  N :M .Thus N  N :M N  N :rM , r  N  N :M . Now, we can give the following proposition : Proposition (2.20): Let N be a submodule of an R – module M and P = N  N :M . If the ideal = N  N :e = P , for each e  M , e  N + (N) , then N is a - prime submodule of M . Proof: Let r R , x  M such that rx  N and suppose x N + (N). Thus r  N  N :x . But N  N :  x = P , so r  P . Therefore N is a  - prime submodule of M . However, we can give a corollary of proposition (2.20).But first we state and prove the following lemma which is needed . Mathematics | 288 2016) عام 2العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 29 (2) 2016 Lemma (2.21): Let N be a submodule of an R – module M . If the submodule N  N :  r = N +  (N), for each r  R , r P , then the ideal P = N  N : e , for each e  M ;e  N + (N) . Proof: Let e  M ; e N + (N) . It is clear that P  N  N : e . Let r  N  N : e , then re  N + (N) . Suppose r  P = N  N : :M .Since N  N :  r =N+(N) and e  [N+ (N) : r ], so e  N+ (N) which a contradicts our assumption. Thus r  P for each e  M such that e  N +(N).Therefore P = N  N : :e . Corollary (2.22): Let N be a submodule of an R – module M and P = N  N :M . If the submodule N  N :  r = N +  (N), for each r  R , then N is a - prime submodule of M . Note that, the intersection of two -prime submodules of an R – module M need not be a -prime submodule of M ,for example : The Z-module Z6 has two -prime submodules , N1 =  2 and N2 =  3  but N1 N2 = 0  is not a -prime submodule of Z6 ,where (N) = N,  N  M . However , we have the following proposition: Proposition (2.23): Let K is a -prime of an R-module M and let N < M such that (K)  K. Then either N  K or K  N is a '-prime in N, where ' :(N)  (N)  {} and :(M)  (M)  {}. Proof: Since K is a -prime of an R-module M and (K)  K, so K is a prime by (2.2,(1)). Hence either N  K or K  N is a prime in N,[11]. Therefore either N  K or K  N is a '- prime in N. proposition (2.24): Let P be an ideal of a ring R and M be an R – module . Then a proper submodule N of M is a P -  - Prime if and only if 1. P  N  N :M , and 2. cm  N ,for all c  R \ P ,m  M \ N + (N). Proof:Suppose N is a P -  - Prime. To prove that (1) and (2) are hold. It is clear that P = N  N :M .Therefore P  N  N :M . Now if c  R \ P and m  M \ N + (N), then c[N + (N):M] and m N + N), hence cm  N. Conversely, let c  R and m  M such that m N + (N) and c [N + (N):M]. Since P  [N + (N):M], then m  M \ N + (N) and c  P .Therefore , c  R \ P. Hence cm  N, which implies that N is a P -  - Prime. proposition (2.25): Let  : M  M' be an homomorphism . If N is ' - prime submodule of an R-module M', such that  (M) ⊈ N and ( – 1(N))=  – 1(' (N)) , then  – 1(N) is -prime submodule of M, where :(M)  (M)  {} and ' :(M')  (M')  {} . Proof: First, we must show that  – 1(N) is a proper submodule of M. Suppose that  – 1(N) = M, then (M)  N, which a contradiction to the assumption. Let r  R , m  M such that rm  – 1(N) .Then r (m)  N and N is ' ‐ prime submodule of an R‐module M', then either  m  N ' N or r M'  N ' N . If  m N ' N , then m – 1 N  – 1 ' N and hence m   – 1 N   – 1 N . If r M'  N ' N , then r M  N ' N Mathematics | 289 2016) عام 2العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 29 (2) 2016 since  M  M'. This implies r M   – 1 N  – 1 ' N  – 1 N   – 1 N .Therefore  – 1 N is ‐prime submodule of M. Theorem (2.26): Let f: M  M' be an epimorphism and let N < M such that ker f  N. If N is a -prime submodule of an R- module M and '( f (N)) = f ((N)), then f (N) is a '-prime submodule of a module of M', where :(M)  (M)  {} and ' :(M')  (M')  {}. Proof: First, we must show that f (N) is a proper submodule of a module M'. Suppose f (N) = M'. But f is an epimorphism, thus f (N) = f (M) and hence M = N + ker f. This implies that M = N. A contradiction. Now, let r m'  f (N), where r  R and m'  M' ,m'= f (m) for some m M since f is an epimorphism . Then r f (m)  f (N),so f (r m)= f (n) for some n N and hence f (r m) - f (n) = 0 .Thus we get that rm-n  ker f  N which implies that rm N .But N is a -prime ,so either m N + (N) or r M  N + (N). If m N + (N),then f (m)  f (N) + f ((N));that is m' f (N) + f ((N)) = f (N) + '( f (N)). If r M  N + (N),then r f (M )  f (N) + f ((N)) = f (N) + '( f (N) implies that r M'  f (N) + '( f (N) . Corollary (2.27): Let M be an R-module, let K < N < M and N be a -prime of M. Then N/K is a '- prime submodule of M/K, where ': ( M/ K)  ( M/ K) . Proof: Let :M  M/K be the natural mapping ,then the result follows by proposition(2.26). Proposition (2.28): Let M be an R-module and let K < N < M .If N is a -prime submodule of M, then N/K is a '-prime submodule of M/K, where ': ( M/ K)  ( M/ K) and '( N / K) = (N) /K. Proof: Let rR and m  N/K with r m  [N / K] ,where m= m+K ,for some m M . So we have rm  N ,which gives that either m N + (N) or r M  N + (N) .Therefor either m +K ( N + (N) ) /K =[ N /K] +[ (N) /K] = [N / K] + [ '( N / K)] or r [M /K] [( N + (N)) / K ]  [ N /K] +[ (N) /K] = [N / K] + [ '( N / K)] .Hence either m  [N / K] + [ '( N / K)] or r [M /K]  [N / K] + [ '( N / K)] .Therefore N / K is a '-prime submodule of M / K. Let R1 ,R2 be two commutative rings with identity and M1, M2 be R1 and R2 – module respectively, put R = R1 R2 . Then M = M1  M2 is an R – module and each submodule of M is of the form N = N1  N2 for some N1 of M1 and N2 of M2 .Furthermore N = N1  N2 is a prime submodule of M if and only if N = N1 M2 or N = M1 N2 for some prime submodules N1 of M1 and N2 of M2 . Theorem (2.29): Let R = R1 R2 that each Ri is a commutative rings with identity . Let Mi be Ri – module and M = M1  M2 with ( r1, r2) ( m1,m2) = ( r1m1,r2m2) , be an R – module ,where ri  Ri , mi  Mi, and let i : ( Mi )   (Mi)  {  } be functions ,  = 1 2. Then we have : 1) N1  N2 is a - prime submodule , where Ni is a i – prime submodule of Mi , with Ni  i ( Ni ). 2) N1 M2 is a - prime submodule of M , where N1 is a prime submodule of M1. 3) N1 M2 is a - prime submodule of M, whereN1 is a 1- prime submodule of M1 and 2 ( M2 ) = M2 . Mathematics | 290 2016) عام 2العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 29 (2) 2016 4) M1  N2 is a - prime submodule of M, whereN2 is a prime submodule of M2 . 5) M1 N2 is a - prime submodule of M, where N2 is a 2- prime submodule of M2 and 1 ( M1 ) = M1 . Proof: (1) Suppose Ni is a i – prime submodule of Mi , with Ni  i ( Ni ) and let ( r1, r2) ( m1,m2) = ( r1m1,r2m2 )  N1  N2 , then r1m1  N1 and r2m2  N2 and since Ni is a i – prime submodule of Mi , so either r1 [ N1 + 1 ( N1 ): M1 ] and . r2  [ N2 + 2 (N2 ) : M2 ] or m1  N1 + 1 ( N1 ) and m2  N2 + 2 (N2 ) . Hence either ( r1, r2)  [ N1 + 1 ( N1 ): M1 ]  [ N2 + 2 (N2 ) : M2 ] = [1 ( N1 ): M1 ]  [2 (N2 ) : M2 ] = [1 ( N1 ) 2 (N2 ) : M1  M2 ] = [N1  N2 + (N1  N2 ) : M1  M2 ] or ( m1,m2) = [N1 + 1 ( N1) ] [ N2 + 2 (N2 )] =[1 ( N1 ) 2 (N2 )  [N1  N2 ] + [  (N1  N2 )] . Therefore N1  N2 is a - prime submodule . (2) If N1 is a prime submodule of M1, then N1 M2 is a prime submodule of M by [6,Th.2.14,p.1446 ] and hence N1 M2 is a - prime submodule of M by (2.2,(1)). 3) Let N1 is a 1- prime submodule of M1 and 2 ( M2 ) = M2 .Let ( r1, r2) R and ( m1,m2)  M be such that ( r1, r2) ( m1,m2) = ( r1m1,r2m2) N1 M2. Then r1m1  N1 and r2m2  M2 and since N1 is a 1 – prime submodule of M1, so either r1 [ N1 + 1 ( N1 ): M1 ] or m1  N1 + 1 ( N1 ) .So either ( r1, r2)  [ N1 + 1 ( N1 ): M1 ]  [M 2 : M2 ] or ( m1,m2)  [N1 + 1 ( N1)]  M2 . Hence either ( r1, r2)  [( N1 + 1 ( N1 ))  M 2 : M1  M 2 ] = [N1  M2 +  (N1  M2 ) : M1  M 2 ] or ( m1,m2)  [N1  M2 + 1 (N1 )  2(N2 )] =N 1  M2 +  (N1  M2).Therefore N1  M2 is a - prime submodule of M . Parts (4) ,(5) are proved similar to (2) ,(3) respectively. References 1. Larsen, M.D. and McCarlthy, P.J., (1971), Multiplication Theory Of Ideals, Academic Press, New York. 2. McCasland, R .L. and Smith, P.F., (1993), Prime Submodules Of Noetherian Modules, Rocky Mtn.J.,23, 1041-1062. 3. L, u, C.P., (1981),Prime Submodule of Modules , Commutative Mathematics , University Spatula ,33,16-69. 4. Saymeh ,S. A.,(1979), On Prime R-Submodules, Univ. Nac. Tucuman Rev. Ser. ,A 29, 121-136. 5. Atani ,S.E. and Farzalipouur, F.,(2007), On Weakly Prime Submodules, Tamkang Journal of Mathematics,38(3),247-252. 6. Khaksari, A. and Jafari, A., (2011), -Prime Submodules, International Journal of Algebra, 5 (20), 1443-1440. 7. EI- Bast Z.A.and Smith P.F.,(1988),Multiplication Modules, Comm. In Algebra,16,755- 779. 8. Faith, C., (1976), Ring Theory ,Springer- Verlag, Berlin Heidelberg, New York. 9. Abbas, M.S., (1990),On Fully Stable Modules, Ph.D. Thesis, University of Baghdad. 10. Ameen, Sh. A., (2002), Bounded Modules, Msc. Thesis, University of Baghdad. )،الموديوالت الجزئية االولية والموديوالت الجزئية شبه االولية،رسالة ماجستير, 1996.عذاب، ايمان علي ,(11 0 0جامعةبغداد Mathematics | 291 2016) عام 2العدد ( 29لمجلد ا مجلة إبن الهيثم للعلوم الصرفة و التطبيقية Ibn Al-Haitham Jour. for Pure & Appl. Sci. Vol. 29 (2) 2016 --المقاسات الجزئية األولية من النمط نهاد سالم المظفر جامعة بغداد /كلية العلوم /قسم الرياضيات عدويه جاسم عبدالخالق الجامعة المستنصرية /كلية العلوم /قسم الرياضيات 2016حزيران//5،قبل في:2016اذار//9استلم في: خالصةال مجموعة كل المقاسات (M). لتكن Rمقاساً معرفا ً على الحلقة Mحلقة ابدالية ذات عنصر محايد، وليكن Rلتكن هو مقاس Mمن Pدالة. في هذا البحث، نقول ان المقاس الجزئي : (M)  (M)  {}ولتكن Mالجزئية من P + (P) m، فانه يؤدي الى اما rx  Pا ذ ان R r , M mا ذا كان لكل  --جزئي أولي من النمط لقد درسنا واعطينا بعض خواص و مميزات هذا النوع من المقاسات الجزئية وبرهنا تحت  r 0 [P + (P): M]أو 0شروط معينة ان المقاسات الجزئية االولية وهذا النوع من المقاسات الجزئية تكون متكافئة 0 -ية االولية من النمط المقاسات الجزئية االولية ، المقاسات الجزئية الضعيفة ، المقاسات الجزئ : الكلمات المفتاحية