On graded I_{e}-prime submodules of graded modules over graded commutative rings
aa r X i v : . [ m a t h . A C ] J a n ON GRADED I e -PRIME SUBMODULES OF GRADEDMODULES OVER GRADED COMMUTATIVE RINGS SHATHA ALGHUEIRI AND KHALDOUN AL-ZOUBI*
Abstract.
Let G be a group with identity e . Let R be a G -gradedcommutative ring with identity and M a graded R -module. In thispaper, we introduce the concept of graded I e -prime submodule as ageneralization of a graded prime submodule for I = ⊕ g ∈ G I g a fixedgraded ideal of R . We give a number of results concerning of theseclasses of graded submodules and their homogeneous components.A proper graded submodule N of M is said to be a graded I e -prime submodule of M if whenever r g ∈ h ( R ) and m h ∈ h ( M )with r g m h ∈ N − I e N, then either r g ∈ ( N : R M ) or m h ∈ N. Introduction and Preliminaries
Throughout this paper all rings are commutative with identity andall modules are unitary.Graded prime submodules of graded modules over graded commuta-tive rings, have been introduced and studied by many authors, (see forexample [1, 3-4, 6-7, 10, 16]). Then, many generalizations of gradedprime submodules were studied such as graded primary, graded clas-sical prime, graded weakly prime and graded 2-absorbing submodules(see for example [2, 5, 9, 16]). Akray and Hussein in [8] introducedthe concept of I -prime submodule over a commutative ring as a newgeneralization of prime submodule.The scope of this paper is devoted to the theory of graded mod-ules over graded commutative rings. Here, we introduce the concept ofgraded I e -prime submodule as a new generalization of a graded primesubmodule. A number of results concerning of these classes of gradedsubmodules and their homogeneous components are given. For exam-ple, we give a characterization of graded I e -prime submodule (see The-orem 2.15). We also study the behaviour of graded I e -prime submoduleunder localization (see Theorem 2.16). Mathematics Subject Classification.
Key words and phrases. graded I e -prime submodules, graded prime submodules,graded I e -prime ideals. ∗ Corresponding author.
First, we recall some basic properties of graded rings and moduleswhich will be used in the sequel. We refer to [12-15] for these basicproperties and more information on graded rings and modules.Let G be a multiplicative group with identity element e. A ring R is called a graded ring (or G -graded ring) if there exist additivesubgroups R g of R indexed by the elements g ∈ G such that R = ⊕ g ∈ G R g and R g R h ⊆ R gh for all g, h ∈ G . The elements of R g arecalled homogeneous of degree g and all the homogeneous elements aredenoted by h ( R ), i.e. h ( R ) = ∪ g ∈ G R g . If r ∈ R , then r can be writtenuniquely as P g ∈ G r g , where r g is called a homogeneous component of r in R g . Moreover, R e is a subring of R and 1 ∈ R e . Let R = ⊕ g ∈ G R g be a G -graded ring. An ideal I of R is said to be a graded ideal if I = P g ∈ G ( I ∩ R g ) := P g ∈ G I g , (see [15]. Let R = ⊕ g ∈ G R g be a G -graded ring. A left R -module M is said to be a graded R -module(or G -graded R -module) if there exists a family of additive subgroups { M g } g ∈ G of M such that M = ⊕ g ∈ G M g and R g M h ⊆ M gh for all g, h ∈ G . Also if an element of M belongs to ∪ g ∈ G M g = h ( M ), then itis called a homogeneous. Note that M g is an R e -module for every g ∈ G .Let R = ⊕ g ∈ G R g be a G -graded ring. A submodule N of M is said tobe a graded submodule of M if N = ⊕ g ∈ G ( N ∩ M g ) := ⊕ g ∈ G N g . Inthis case, N g is called the g -component of N . Moreover, M/N becomesa G -graded R -module with g -component ( M/N ) g := ( M g + N ) /N for g ∈ G , (see [15]).Let R be a G -graded ring and S ⊆ h ( R ) be a multiplicatively closedsubset of R . Then the ring of fraction S − R is a graded ring whichis called the graded ring of fractions. Indeed, S − R = ⊕ g ∈ G ( S − R ) g where ( S − R ) g = { r/s : r ∈ R, s ∈ S and g = ( degs ) − ( degr ) } . Let M be a graded module over a G -graded ring R and S ⊆ h ( R ) be a mul-tiplicatively closed subset of R . The module of fractions S − M over agraded ring S − R is a graded module which is called the module of frac-tions, if S − M = ⊕ g ∈ G ( S − M ) g where ( S − M ) g = { m/s : m ∈ M, s ∈ S and g = ( degs ) − ( degm ) } . We write h ( S − R ) = ∪ g ∈ G ( S − R ) g and h ( S − M ) = ∪ g ∈ G ( S − M ) g . For any graded submodule N of M , thegraded submodule of S − N of S − M is defined by S − N = { α ∈ S − M : α = m/s for m ∈ N and s ∈ S } and S − N = S − M if andonly if S ∩ ( N : R M ) = ∅ , (see [15]).The graded radical of a graded ideal I , denoted by Gr ( I ), is the setof all r = P g ∈ G r g ∈ R such that for each g ∈ G there exists n g ∈ N with r n g g ∈ I . Note that, if x is a homogeneous element, then x ∈ Gr ( I )if and only if x n ∈ I for some n ∈ N , (see [17]). N GRADED I e -PRIME SUBMODULES 3 Let R be a G -graded ring and M be a graded R -module. It isshown in [10, Lemma 2.1] that if N is a graded submodule of M , then( N : R M ) = { r ∈ R : rN ⊆ M } is a graded ideal of R . Also, forany r g ∈ h ( R ), the graded submodule { m ∈ M : r g m ∈ N } will bedenoted by ( N : M r g ). The graded radical of a graded submodule N of M , denoted by Gr M ( N ), is defined to be the intersection of all gradedprime submodules of M containing N . If N is not contained in anygraded prime submodule of M , then Gr M ( N ) = M , (see [16]).2. RESULTS
Definition 2.1.
Let R be a G -graded ring, M a graded R -module, I = ⊕ g ∈ G I g a graded ideal of R , N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G. (i) We say that N g is a g - I e -prime submodule of the R e -module M g if N g = M g ; and whenever r e ∈ R e and m g ∈ M g with r e m g ∈ N g − I e N g , implies either m g ∈ N g or r e ∈ ( N g : R e M g ) . (ii) We say that N is a graded I e -prime submodule of M if N = M ;and whenever r h ∈ h ( R ) and m λ ∈ h ( M ) with r h m λ ∈ N − I e N ,implies either m λ ∈ N or r h ∈ ( N : R M ) . Definition 2.2.
Let R be a G -graded ring, I = ⊕ g ∈ G I g and J = ⊕ g ∈ G J g be graded ideals of R. Then J e is said to be an e - I e -prime idealof R e , if J e = R e ; and whenever r e s e ∈ J e − I e J e , where r e , s e ∈ R e ,implies either r e ∈ J e or s e ∈ J e . Let R be a G -graded ring, M a graded R -module and I = ⊕ g ∈ G I g a graded ideal of R. Recall from [10] that a proper graded submodule N = ⊕ g ∈ G N g of a graded R -module M is said to be a graded primesubmodule of M if whenever r g m h ∈ N where r g ∈ h ( R ) and m h ∈ h ( M ) , then either m h ∈ N or r g ∈ ( N : R M ) . It is easy to see thatevery graded prime submodule of M is a graded I e -prime submodule.The following example shows that the converse is not true in general. Example 2.3.
Let G = Z and R = Z be a G -graded ring with R = Z and R = { } . Let M = Z be a graded R -module with M = Z and M = { } . Now, consider a graded ideal I = 4 Z of R and a graded submodule N = h i of M, then N is not a gradedprime submodule of M since 2 · ∈ N but neither 2 ∈ N nor2 ∈ ( N : R M ) = 4 Z . However, N is a graded I e -prime submodule of M since N − I N = h i − Z · h i = ∅ . Recall from [10] that if N = ⊕ g ∈ G N g is a graded submodule of agraded R -module M and g ∈ G , then N g is called a g -prime submodule S. ALGHUEIRI AND K. AL-ZOUBI of an R e -module M g if N g = M g ; and whenever r e m g ∈ N g where r e ∈ R e and m g ∈ M g , then either m g ∈ N g or r e ∈ ( N g : R e M g ). Theorem 2.4.
Let R be a G -graded ring, M a graded R -module, I = ⊕ g ∈ G I g a graded ideal of R , N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G. If N g is a g - I e -prime submodule of M g , then either N g is a g -prime submodule of M g or ( N g : R e M g ) N g ⊆ I e N g . Proof.
Suppose that N g is a g - I e -prime submodule of M g such that( N g : R e M g ) N g I e N g . Now, let r e ∈ R e and m g ∈ M g with r e m g ∈ N g . If r e m g I e N g , then either m g ∈ N g or r e ∈ ( N g : R e M g ) as N g is a g - I e -prime submodule of M g . Assume that r e m g ∈ I e N g . If r e N g I e N g , then there exists x g ∈ N g such that r e x g I e N g , sowe get r e ( m g + x g ) ∈ N g − I e N g and then either m g + x g ∈ N g or r e ∈ ( N g : R e M g ) as N g is a g - I e -prime submodule of M g . Hence,either m g ∈ N g or r e ∈ ( N g : R e M g ) . If ( N g : R e M g ) m g I e N g , thereexists t e ∈ ( N g : R e M g ) such that t e m g I e N g , so we get ( r e + t e ) m g ∈ N g − I e N g and then either m g ∈ N g or r e + t e ∈ ( N g : R e M g ) as N g is a g - I e -prime submodule of M g . Hence, either m g ∈ N g or r e ∈ ( N g : R e M g ) . Now, we can assume that r e N g ⊆ I e N g and ( N g : R e M g ) m g ⊆ I e N g . But ( N g : R e M g ) N g I e N g , so there exist s e ∈ ( N g : R e M g ) and l g ∈ N g such that s e l g I e N g . Thus ( r e + s e )( m g + l g ) ∈ N g − I e N g gives either m g + l g ∈ N g or r e + s e ∈ ( N g : R e M g ) as N g is a g - I e -prime submoduleof M g . Hence, either m g ∈ N g or r e ∈ ( N g : R e M g ) . Therefore, N g is a g -prime submodule of M g . (cid:3) Corollary 2.5.
Let R be a G -graded ring, M a graded R -module, N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G. If N g is a g - -primesubmodule of M g such that ( N g : R e M g ) N g = 0 , then N g is a g -primesubmodule of M g . Proof.
Take I = 0 in the Theorem 2.4. (cid:3) Recall from [11] that a graded R -module M is called a graded mul-tiplication if for each graded submodule N of M , we have N = IM for some graded ideal I of R . If N is graded submodule of a gradedmultiplication module M , then N = ( N : R M ) M . Let N and K betwo graded submodules of a graded multiplication R -module M with N = I M and K = I M for some graded ideals I and I of R . Theproduct of N and K denoted by N K is defined by
N K = I I M. Corollary 2.6.
Let R be a G -graded ring, M a graded R -module, I = ⊕ g ∈ G I g a graded ideal of R , N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G. If M g is a multiplication R e -module and N g is a g - I e -primesubmodule which is not a g -prime submodule of M g , then N g ⊆ I e N g . N GRADED I e -PRIME SUBMODULES 5 Proof.
Suppose that N g is a g - I e -prime submodule of M g which is nota g -prime, so by Theorem 2.4, we get ( N g : R e M g ) N g ⊆ I e N g . Now,as M g is a multiplication R e -module we get N g = ( N g : R e M g ) M g ⊆ ( N g : R e M g ) N g ⊆ I e N g . Therefore, N g ⊆ I e N g . (cid:3) Theorem 2.7.
Let R be a G -graded ring, M a graded R -module, I = ⊕ g ∈ G I g a graded ideal of R , N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G. Then the following statements are equivalent: (i) N g is a g - I e -prime submodule of M g . (ii) For r e ∈ R e \ ( N g : R e M g ) , ( N g : M g r e ) = N g ∪ ( I e N g : M g r e ) . (iii) For r e ∈ R e \ ( N g : R e M g ) , either ( N g : M g r e ) = N g or ( N g : M g r e ) = ( I e N g : M g r e ) . Proof. ( i ) ⇒ ( ii ) Suppose that N g is a g - I e -prime submodule of M g and r e ∈ R e \ ( N g : R e M g ). Let m g ∈ ( N g : M g r e ) , so r e m g ∈ N g . If r e m g I e N g , then m g ∈ N g . Now, if r e m g ∈ I e N g , then m g ∈ ( I e N g : M g r e ) . Hence, ( N g : M g r e ) = N g ∪ ( I e N g : M g r e ) . ( ii ) ⇒ ( iii ) If a submodule is a union of two submodules, then it isequal one of them.( iii ) ⇒ ( i ) Let r e ∈ R e and m g ∈ M g such that r e m g ∈ N g − I e N g and r e ( N g : R e M g ), so m g ∈ ( N g : M g r e ) . Now, by ( iii ) we get either( N g : M g r e ) = N g or ( N g : M g r e ) = ( I e N g : M g r e ) . But r e m g I e N g ,so m g ( I e N g : M g r e ) . Hence, ( N g : M g r e ) = N g and then m g ∈ N g . Therefore, N g is a g - I e -prime submodule of M g . (cid:3) Recall from [9] that a proper graded submodule N = ⊕ g ∈ G N g of agraded R -module M is said to be a graded weakly prime submodule of M if whenever 0 = r g m h ∈ N where r g ∈ h ( R ) and m h ∈ h ( M ) , theneither m h ∈ N or r g ∈ ( N : R M ) . Theorem 2.8.
Let R be a G -graded ring, M a graded R -module, I = ⊕ g ∈ G I g a graded ideal of R and N a proper graded submodule of M. Then the following statements are equivalent: (i) N is a graded I e -prime submodule of M . (ii) N/I e N is a graded weakly prime submodule of M/I e N. Proof. ( i ) ⇒ ( ii ) Suppose that N is a graded I e -prime submodule of M . Let r g ∈ h ( R ) and ( m h + I e N ) ∈ h ( M/I e N ) with 0 M/I e N = ( r g m h + I e N ) ∈ N/I e N , this yields that r g m h ∈ N − I e N. Hence, either m h ∈ N or r g M ⊆ N as N is a graded I e -prime submodule of M . Then either( m h + I e N ) ∈ N/I e N or r g ( M/I e N ) ⊆ N/I e N. Therefore,
N/I e N is agraded weakly prime submodule of M/I e N. ( i ) ⇒ ( ii ) Suppose that N/I e N is a graded weakly prime submoduleof M/I e N. Let r g ∈ h ( R ) and m h ∈ h ( M ) such that r g m h ∈ N − I e N. S. ALGHUEIRI AND K. AL-ZOUBI
This follows that 0
M/I e N = ( r g m h + I e N ) = r g ( m h + I e N ) ∈ N/I e N. Thus, either r g ∈ ( N/I e N : R M/I e N ) or ( m h + I e N ) ∈ N/I e N and theneither r g ∈ ( N : R M ) or m h ∈ N. Therefore, N is a graded I e -primesubmodule of M . (cid:3) Lemma 2.9.
Let R be a G -graded ring, M a graded R -module and I = ⊕ g ∈ G I g be a graded ideal of R and g ∈ G. If M g is a multiplication R e -module and N g is a g - I e -prime submodule of M g with ( N g : R e M g ) ⊆ I e , then Gr (( I e N g : R e M g )) N g = I e N g . Proof.
Suppose that N g is a g - I e -prime submodule of M g with ( N g : R e M g ) ⊆ I e . Clearly, I e N g ⊆ ( I e N g : R e M g ) N g ⊆ Gr (( I e N g : R e M g )) N g . Now, let r e ∈ Gr (( I e N g : R e M g )) . If r e ∈ I e , then r e N g ⊆ I e N g . So wecan assume that r e I e , which follows that r e ( N g : R e M g ) , thenby Theorem 2.7, either ( N g : M g r e ) = N g or ( N g : M g r e ) = ( I e N g : M g r e ) . If ( N g : M g r e ) = ( I e N g : M g r e ) , then r e N g ⊆ r e ( N g : M g r e ) = r e ( I e N g : M g r e ) ⊆ I e N g . On the other hand, if ( N g : M g r e ) = N g , let n ≥ r ne ∈ ( I e N g : R e M g ) . Since I e N g ⊆ N g , r ne M g = r e ( r n − e M g ) ⊆ N g which yields that ( r n − e M g ) ⊆ N g . Thus, after a finite number of steps we get r e M g ⊆ N g whichis a contradiction. Hence, Gr (( I e N g : R e M g )) N g ⊆ I e N g . Therefore, Gr (( I e N g : R e M g )) N g = I e N g . (cid:3) Theorem 2.10.
Let R be a G -graded ring, M a graded R -module, I = ⊕ g ∈ G I g a graded ideal of R , N = ⊕ g ∈ G N g a graded submodule of M and g ∈ G with ( I e N g : R e M g ) = I e ( N g : R e M g ) . If N g is a g - I e -prime submodule of M g , then ( N g : R e M g ) is an e - I e -prime ideal of R e . Proof.
Suppose that N g is a g - I e -prime submodule of M g . Now, let r e , s e ∈ R e such that r e s e ∈ ( N g : R e M g ) − I e ( N g : R e M g ) and r e ( N g : R e M g ) . By Theorem 2.7 since r e ( N g : R e M g ) , we get either( N g : M g r e ) = N g or ( N g : M g r e ) = ( I e N g : M g r e ) . If r e s e M g ⊆ I e N g ,then r e s e ∈ ( I e N g : R e M g ) = I e ( N g : R e M g ) , a contradiction. Hence, r e s e M g I e N g . Since s e M g ⊆ ( N g : M g r e ) and s e M g ( I e N g : M g r e ) , we get s e M g ⊆ N g . So s e ∈ ( N g : R e M g ) . (cid:3) Theorem 2.11.
Let R be a G -graded ring, I = ⊕ g ∈ G I g a graded idealof R and J = ⊕ g ∈ G J g a proper graded ideal of R . Then the followingstatements are equivalent: (i) J e is an e - I e -prime ideal of R e . (ii) For r e ∈ R e − J e , ( J e : R e r e ) = J e ∪ ( I e J e : R e r e ) . (iii) For r e ∈ R e − J e , ( J e : R e r e ) = J e or ( J e : R e r e ) = ( I e J e : R e r e ) . N GRADED I e -PRIME SUBMODULES 7 (iv) For any two graded ideals K = ⊕ g ∈ G K g and L = ⊕ h ∈ G L h of R with K e L e ⊆ J e and K e L e I e J e , implies either K e ⊆ J e or L e ⊆ J e .Proof. ( i ) ⇒ ( ii ) Suppose that J e is an e - I e -prime ideal of R e and r e ∈ R e − J e . It is easy to see that J e ∪ ( I e J e : R e r e ) ⊆ ( J e : R e r e ) . Now, let s e ∈ ( J e : R e r e ), so r e s e ∈ J e . If r e s e ∈ J e − I e J e , then s e ∈ J e . If r e s e ∈ I e J e , then s e ∈ ( I e J e : R e r e ), which follows that( J e : R e r e ) ⊆ J e ∪ ( I e J e : R e r e ) and hence ( J e : R e r e ) = J e ∪ ( I e J e : R e r e ).( ii ) ⇒ ( iii ) Note that if a ideal is a union of two ideals, then it isequal to one of them.( iii ) ⇒ ( iv ) Let K = ⊕ g ∈ G K g and L = ⊕ h ∈ G L h be two graded idealsof R such that K e L e ⊆ J e , K e L e I e J e and neither K e ⊆ J e nor L e ⊆ J e . Let k e ∈ K e , if k e J e , then k e L e ⊆ J e gives L e ⊆ ( J e : R e k e ).Now, by ( iii ) since k e ∈ R e − J e , ( J e : R e k e ) = J e or ( J e : R e k e ) =( I e J e : R e k e ) . But L e J e , so L e ⊆ ( I e J e : R e k e ). Hence, k e L e ⊆ I e J e .Now, if k e ∈ J e , then since K e J e , there exists k ′ e ∈ K e − J e , so wehave ( k e + k ′ e ) L e ⊆ J e and by using the first case we get k ′ e L e ⊆ I e J e . Since k e + k ′ e ∈ R e − J e and L e J e , L e ⊆ ( I e J e : R e k e + k ′ e ) andthen ( k e + k ′ e ) L e ⊆ I e J e . Hence k e L e ⊆ I e J e since k ′ e L e ⊆ I e J e . Thus, K e L e ⊆ I e J e , a contradiction.( iv ) ⇒ ( i ) Let r e , s e ∈ R e such that r e s e ∈ J e − I e J e . Then K = ( r e )and L = ( s e ) are graded ideals of R generated by r e and s e , respectively.Now, K e L e ⊆ J e and K e L e I e J e . So either K e ⊆ J e or L e ⊆ J e andhence r e ∈ J e or s e ∈ J e . (cid:3) Theorem 2.12.
Let R be a G -graded ring, M a graded R -module, I = ⊕ g ∈ G I g a graded ideal of R , N = ⊕ g ∈ G N g a proper graded submoduleof M and g ∈ G such that ( I e N g : R e M g ) = I e ( N g : R e M g ) . If N g isa g - I e -prime submodule of M g and M g is a multiplication R e -module,then for any graded submodules K = ⊕ h ∈ G K h and L = ⊕ h ∈ G L h with K g L g ⊆ N g and K g L g I e N g , implies either K g ⊆ N g or L g ⊆ N g . Proof.
Suppose that N g is a g - I e -prime submodule of M g so by The-orem 2.10, we get ( N g : R e M g ) is an e - I e -prime ideal of R e . Now, let K = ⊕ h ∈ G K h and L = ⊕ h ∈ G L h be two graded submodules of M suchthat K g L g ⊆ N g and K g L g I e N g and neither K g ⊆ N g nor L g ⊆ N g . Hence, K g L g = ( K g : R e M g )( L g : R e M g ) M g ⊆ N g and then ( K g : R e M g )( L g : R e M g ) ⊆ ( N g : R e M g ) . Now, K g = ( K g : R e M g ) M g N g gives ( K g : R e M g ) ( N g : R e M g ) , and L g = ( L g : R e M g ) M g N g gives ( L g : R e M g ) ( N g : R e M g ) . So by Theorem 2.11, we get ( K g : R e M g )( L g : R e M g ) ⊆ I e ( N g : R e M g ) = ( I e N g : R e M g ) which yields that S. ALGHUEIRI AND K. AL-ZOUBI K g L g = ( K g : R e M g )( L g : R e M g ) M g ⊆ I e N g , a contradiction. There-fore, either K g ⊆ N g or L g ⊆ N g . (cid:3) Suppose M g is a multiplication R e -module and m g , m g ∈ M g . Thenwe can define the product of m g and m g as m g m g = R e m g R e m g =( R e m g : R e M g )( R e m g : M g ) M g . Thus we have the following corollary.
Corollary 2.13.
Let R be a G -graded ring, M a graded R -module, I = ⊕ g ∈ G I g a graded ideal of R , N = ⊕ g ∈ G N g a proper graded submoduleof M and g ∈ G such that ( I e N g : R e M g ) = I e ( N g : R e M g ) . If M g isa multiplication R e -module and N g is a g - I e -prime submodule of M g ,then for any m g , m g ∈ M g with m g m g ∈ N g − I e N g , implies either m g ∈ N g or m g ∈ N g . Proof.
Let m g , m g ∈ M g such that m g m g ∈ N g − I e N g . Then K = ( m g ) and L = ( m g ) are graded submodules of M generatedby m g and m g , respectively. Since K g L g ⊆ N g and K g L g I e N g , byTheorem 2.12, we get either m g ∈ N g or m g ∈ N g . (cid:3) Theorem 2.14.
Let R be a G -graded ring, M and M be two graded R -modules, I = ⊕ g ∈ G I g a graded ideal of R and N and N be twograded submodules of M and M , respectively. Then: (i) If N is a graded I e -prime submodule of M , then N × M is agraded I e -prime submodule of M × M . (ii) If N is a graded I e -prime submodule of M , then M × N is agraded I e -prime submodule of M × M . Proof. ( i ) Suppose that N is a graded I e -prime submodule of M . Now,let r g ∈ h ( R ) and ( m h , m h ) ∈ h ( M × M ) such that r g ( m h , m h ) =( r g m h , r g m h ) ∈ ( N × M ) − I e ( N × M ) = ( N − I e N ) × ( M − I e M ) , which follows that r g m h ∈ N − I e N . Hence, either m h ∈ N or r g M ⊆ N and then either ( m h , m h ) ∈ N × M or r g ( M × M ) ⊆ N × M . Therefore, N × M is a graded I e -prime submodule of M × M . ( ii ) The proof is similar to ( i ) . (cid:3) Theorem 2.15.
Let R be a G -graded ring, M a graded R -module, I = ⊕ g ∈ G I g a graded ideal of R and N a proper graded submodule of M. Let K = ⊕ h ∈ G K h be a graded submodule of M . Then the followingstatements are equivalent: (i) N is a graded I e -prime submodule of M. (ii) For any r g ∈ h ( R ) and h ∈ G with r g K h ⊆ N and r g K h I e N, implies either K h ⊆ N or r g ∈ ( N : R M ) . Proof. ( i ) ⇒ ( ii ) Suppose that N is a graded I e -prime submodule of M. Now, let r g ∈ h ( R ) and h ∈ G such that r g K h ⊆ N , r g K h I e N and N GRADED I e -PRIME SUBMODULES 9 r g ( N : R M ) . Let k h ∈ K h , if r g k h I e N, then r g k h ∈ N − I e N . So k h ∈ N as N is a graded I e -prime submodule of M .
Now, if r g k h ∈ I e N, since r g K h I e N, there exists k ′ h ∈ K h such that r g k ′ h I e N, but r g k ′ h ∈ N − I e N and r g ( N : R M ) , so k ′ h ∈ N. Hence, we get r g ( k h + k ′ h ) ∈ N − I e N , which yields that k h + k ′ h ∈ N and then k h ∈ N. Therefore, K h ⊆ N. ( ii ) ⇒ ( i ) Let r g ∈ h ( R ) and m h ∈ h ( M ) such that r g m h ∈ N − I e N. Then K = ( m h ) is a graded submodule generated by m h . Since r g K h ⊆ N and r g K h I e N, by ( ii ) , we get either K h ⊆ N or r g ∈ ( N : R M )and then either m h ∈ N or r g ∈ ( N : R M ) . Therefore, N is a graded I e -prime submodule of M. (cid:3) Recall that a graded zero-divisor on a graded R -module M is anelement r g ∈ h ( R ) for which there exists m h ∈ h ( M ) such that m h = 0but r g m h = 0. The set of all graded zero-divisors on M is denoted by G - Zdv R ( M ), see [4].The following result studies the behavior of graded I e -prime submod-ules under localization. Theorem 2.16.
Let R be a G -graded ring, M a graded R -module, S ⊆ h ( R ) be a multiplicatively closed subset of R and I = ⊕ h ∈ G I h agraded ideal of R . (i) If N is a graded I e -prime submodule of M with ( N : R M ) ∩ S = ∅ ,then S − N is a graded I e -prime submodule of S − M . (ii) If S − N is a graded I e -prime submodule of S − M with S ∩ G - Zdv R ( M/N ) = ∅ , then N is a graded I e -prime submodule of M .Proof. ( i ) Since ( N : R M ) ∩ S = ∅ , S − N is a proper graded submoduleof S − M. Let r g s ∈ h ( S − R ) and m h s ∈ h ( S − M ) such that r g s m h s ∈ S − N − I e S − N. Then there exists t ∈ S such that tr g m h ∈ N − I e N which yields that either tm h ∈ N or r g ∈ ( N : R M ) as N is a graded I e -prime submodule of M. Hence either m h s = tm h ts ∈ S − N or r g s ∈ S − ( N : R M ) = ( S − N : S − R S − M ) . Therefore, S − N is a graded I e -prime submodule of S − M .( ii ) Let r g ∈ h ( R ) and m h ∈ h ( M ) such that r g m h ∈ N − I e N . Then r g m h ∈ S − N − I e S − N . Since S − N is a graded I e -prime submodule of S − M , either m h ∈ S − N or r g ∈ ( S − N : S − R S − M ). If m h ∈ S − N ,then there exists t ∈ S such that sm h ∈ N . Which yields that m h ∈ N since S ∩ G - Zdv R ( M/N ) = ∅ . Now, if r g ∈ ( S − N : S − R S − M ) = S − ( N : R M ), then there exists s ∈ S such that sr g M ⊆ N and hence r g ∈ ( N : R M ) since S ∩ G - Zdv R ( M/N ) = ∅ . Therefore, N is a graded I e -prime submodule of M . (cid:3) References [1] K. Al-Zoubi, Some properties of graded 2-prime submodules, Asian-EuropeanJournal of Mathematics, 8 (2), (2015) 1550016-1– 1550016-5[2] K. Al-Zoubi and R. Abu-Dawwas, On graded 2-absorbing and weakly graded2-absorbing submodules, J. Math. Sci. Adv. Appl., 28 (2014), 45–60[3] K. Al-Zoubi and R. Abu-Dawwas, On graded quasi-prime submodules, Kyung-pook Math. J., 55 (2) (2015), 259-266.[4] K. Al-Zoubi and A. Al-Qderat, Some properties of graded comultiplicationmodules, Open Mathematics, 15( 2017),187-192.[5] K. Al-Zoubi, M. Jaradat and R. Abu-Dawwas, On graded classical prime andgraded prime submodules, Bull. Iranian Math. Soc., 41 (1) (2015), 217-225.[6] K. Al-Zoubi and J. Paseka, On Graded Coprimely Packed Modules, Adv. Stud.Contemp. Math. (Kyungshang), 29(2) (2019), 271 – 279.[7] K. Al-Zoubi and F. Qarqaz, An Intersection condition for graded prime sub-modules in Gr-multiplication modules, Math. Reports, 20 (3), 2018, 329-336.[8] I. Akray and H. S. Hussein, I-prime submodules, Acta. Math. Academic Paed-agogicae Nyiregyhaziensis, 33 ( 2017),165-173.[9] S.E Atani, On graded weakly prime submodules, Int. Math. Forum, 1 (2) 2006,61-66.[10] S.E. Atani, On graded prime submodules, Chiang Mai J. Sci., 33 (1) (2006),3-7.[11] J. Escoriza and B. Torrecillas, Multiplication Objects in CommutativeGrothendieck Categories, Comm. in Algebra, 26 (6) (1998), 1867-1883.[12] R. Hazrat, Graded Rings and Graded Grothendieck Groups, Cambridge Uni-versity Press, Cambridge, 2016.[13] C. Nastasescu and F. Van Oystaeyen, Graded and filtered rings and modules,Lecture notes in mathematics 758, Berlin-New York: Springer-Verlag, 1982.[14] C. Nastasescu, F. Van Oystaeyen, Graded Ring Theory, Mathematical Library28, North Holand, Amsterdam, 1982.[15] C. Nastasescu and F. Van Oystaeyen, Methods of Graded Rings, LNM 1836.Berlin-Heidelberg: Springer-Verlag, 2004.[16] K. H. Oral, U. Tekir and A. G. Agargun, On graded prime and primary sub-modules, Turk. J. Math., 35 (2011), 159–167.[17] M. Refai and K. Al-Zoubi, On graded primary ideals, Turk. J. Math. 28 (3)(2004), 217-229.
Shatha Alghueiri, Department of Mathematics and Statistics, Jordan Universityof Science and Technology, P.O.Box 3030, Irbid 22110, Jordan.Email address : [email protected] Khaldoun Al-Zoubi, Department of Mathematics and Statistics, Jordan Univer-sity of Science and Technology, P.O.Box 3030, Irbid 22110, Jordan.Email address ::