On Properties of Graded Rings and Graded Modules
aa r X i v : . [ m a t h . A C ] J a n ON PROPERTIES OF GRADED RINGS AND GRADEDMODULES
MASHHOOR REFAI
AND
RASHID ABU-DAWWAS
Abstract.
Let R be a G -graded ring. In this article, we introduce two new con-cepts on graded rings, namely, weakly graded rings and invertible graded rings, andwe discuss the relations between these concepts and several properties of gradedrings. Also, we study the concept of weakly crossed products and study someproperties defined on weakly crossed product and give the relationship betweenthis new concept and several properties of graded rings. Moreover, in this article,we give a generalization for the concept of graded essential submodules, and in-troduce the concept of graded semi-uniform modules which is a generalization forthe concept of graded uniform modules. Introduction
Let G be a group with identity e and R be a ring with unity 1. Then R is saidto be G -graded ring if there exist additive subgroups R g of R such that R = M g ∈ G R g and R g R h ⊆ R gh for all g, h ∈ G . A G -graded ring R is denoted by ( R, G ). Theelements of R g are called homogeneous of degree g and R e (the identity componentof R ) is a subring of R and 1 ∈ R e . For x ∈ R , x can be written uniquely as X g ∈ G x g where x g is the component of x in R g . The support of ( R, G ) is defined by supp ( R, G ) = { g ∈ G : R g = 0 } and h ( R ) = [ g ∈ G R g .Let R be a G -graded ring and I an ideal of R . Then I is said to be G -gradedideal if I = M g ∈ G ( I ∩ R g ), i.e., if x ∈ I and x = X g ∈ G x g , then x g ∈ I for all g ∈ G . Anideal of a G -graded ring need not be G -graded, (see [1]).The concepts of faithful and non-degenerate graded rings have been introduced in[14]; ( R, G ) is said to be faithful if for all g, h ∈ G , a g ∈ R g −{ } we have a g R h = { } and R h a g = { } . Clearly, if ( R, G ) is faithful, then supp ( R, G ) = G . ( R, G ) is saidto be non-degenerate if for all g ∈ G , a g ∈ R g − { } we have a g R g − = { } and R g − a g = { } . Otherwise, ( R, G ) is said to be degenerate. Clearly, if (
R, G ) isfaithful, then (
R, G ) is non-degenerate. However, the converse is not true in generaland one can look in [14].Regular graded rings have been introduced in [14]; (
R, G ) is said to be regular iffor all g ∈ G , a g ∈ R g − { } we have a g ∈ a g R g − a g . It is easily to prove that if( R, G ) is regular, then (
R, G ) is non-degenerate.
Mathematics Subject Classification.
Key words and phrases. weakly graded rings, weakly crossed products, invertible graded rings,graded semi-essential submodules, graded essential submodules, graded uniform modules, gradedsemi-uniform modules.
Strongly graded rings have been introduced in [14]; (
R, G ) is said to be strong if R g R h = R gh for all g, h ∈ G . Clearly, if ( R, G ) is strong, then (
R, G ) is faithful.However, the converse is not true in general and one can look in [14].First strongly graded rings have been introduced and studied in [15]; (
R, G ) issaid to be first strong if 1 ∈ R g R g − for all g ∈ supp ( R, G ). Clearly, if (
R, G ) isstrong, then (
R, G ) is first strong. However, the converse is not true in general andone can look in [15].Clearly, if (
R, G ) is strong, then supp ( R, G ) = G . On the other hand, if ( R, G )is first strong, then supp ( R, G ) is a subgroup of G . In fact, it has been proved that( R, G ) is first strong if and only if supp ( R, G ) is a subgroup of G and R g R h = R gh for all g, h ∈ supp ( R, G ), (see [15]).Also, in [15], the concept of second strongly graded rings was introduced; (
R, G )is second strong if supp ( R, G ) is a monoid in G and R g R h = R gh for all g, h ∈ supp ( R, G ). Clearly, if (
R, G ) is strong, then (
R, G ) is second strong. However, theconverse is not true in general and one can look in [15]. If (
R, G ) is first strong,then (
R, G ) is second strong. However, the converse is not true in general and onecan look in [15].In fact, if (
R, G ) is second strong and supp ( R, G ) is a subgroup of G , then ( R, G )is first strong. Moreover, (
R, G ) is first strong if and only if (
R, G ) is second strongand non-degenerate. Also, (
R, G ) is strong if and only if (
R, G ) is second strong andfaithful, (see [15]).In this article, we introduce the concept of weakly graded rings; (
R, G ) is said tobe weak if whenever g ∈ G with R g = { } , then R g − = { } . We discuss the relationsbetween weakly graded rings and the concepts of strong, first strong, second strong,faithful, nondegenerate and regular graded rings, and then we give an analogousstudy to [8].Also, we introduce the concept of invertible graded rings; ( R, G ) is said to beinvertible if the identity component R e is a field. We introduce an example oninvertible graded rings which is not strong and an example on strongly graded ringswhich is not invertible. On the other hand, we prove that every invertible weaklygraded domain is first strong. Several results are investigated, and then we studyinvertible graded rings as a vector space over R e .The concept of crossed product was introduced in [14]; ( R, G ) is said to be crossedproduct if R g contains a unit for all g ∈ G . Clearly, if ( R, G ) is a crossed product,then supp ( R, G ) = G . In this article, we study the concept of weakly crossed prod-ucts (crossed products over the support) that was mentioned and used in [3]; ( R, G )is said to be weakly crossed product if R g contains a unit for all g ∈ supp ( R, G ). Westudy some properties defined on weakly crossed product and give the relationshipbetween this new concept and several properties of graded rings.Let M be a left R - module. Then M is a G -graded R -module if there existadditive subgroups M g of M indexed by the elements g ∈ G such that M = M g ∈ G M g and R g M h ⊆ M gh for all g, h ∈ G . The elements of M g are called homogeneousof degree g . If x ∈ M , then x can be written uniquely as X g ∈ G x g , where x g is thecomponent of x in M g . Clearly, M g is R e -submodule of M for all g ∈ G . Also, wewrite h ( M ) = [ g ∈ G M g and supp ( M, G ) = { g ∈ G : M g = 0 } . Let M be a G -graded R -module and N be an R -submodule of M . Then N is called G -graded R -submodule N PROPERTIES OF GRADED RINGS AND GRADED MODULES 3 if N = M g ∈ G (cid:16) N \ M g (cid:17) , i.e., if x ∈ N and x = X g ∈ G x g , then x g ∈ N for all g ∈ G .Not all R -submodules of a G -graded R -module are G -graded, (see [1] and [14]). Lemma 1.1. ( [12] ) Let R be a G -graded ring and M be a G -graded R -module. (1) If I and J are graded ideals of R , then I + J and I T J are graded ideals of R . (2) If N and K are graded R -submodules of M , then N + K and N T K aregraded R -submodules of M . (3) If N is a graded R -submodule of M , r ∈ h ( R ) , x ∈ h ( M ) and I is a gradedideal of R , then Rx , IN and rN are graded R -submodules of M . Moreover, ( N : R M ) = { r ∈ R : rM ⊆ N } is a graded ideal of R . In [11], it has been proved that if N is a graded R -submodule of M , then Ann ( N ) = { r ∈ R : rN = 0 } is a graded ideal of R .A G -graded R -module M is said to be strongly graded if R g M h = M gh for all g, h ∈ G . Clearly, ( R, G ) is strong if and only if every graded R -module is stronglygraded (see [14]). Also, ( M, G ) is called first strong if supp ( R, G ) is a subgroup of G and R g M h = M gh for all g ∈ supp ( R, G ), h ∈ G . Moreover, ( R, G ) is first strongif and only if every graded R -module is first strongly graded (see [18]).Let M be a G -graded R -module and N be an R -submodule of M . Then M/N may be made into a graded module by putting (
M/N ) g = ( M g + N ) /N for all g ∈ G (see [14]). In fact, the proof of the following will be a routine. Lemma 1.2.
Let M be a graded R -module, K and N be R -submodules of M suchthat K ⊆ N . Then N is a graded R -submodule of M if and only if N/K is a graded R -submodule of M/K . Graded essential submodules have been introduced and studied in [14], a nonzerograded R -submodule K of M is said to be graded essential if K T N = { } for everynonzero graded R -submodule N of M . Also, graded essential submodules have beenstudied in [7].Graded prime ideals have been introduced and studied in [17]; a proper gradedideal B of a graded ring R is said to graded prime if whenever a, b ∈ h ( R ) suchthat ab ∈ B , then either a ∈ B or b ∈ B . Graded prime submodules have beenintroduced and studied in [9], a proper graded R -submodule P of M is said to begraded prime if whenever r ∈ h ( R ) and m ∈ h ( M ) such that rm ∈ P , then either m ∈ P or r ∈ ( P : R M ). Graded prime submodules have been studied by severalauthors, for example [4] and [6]. It has been proved in [9] that if N is a gradedprime R -submodule of M , then ( N : R M ) is a graded prime ideal of R .In this article, we introduce and study the concept of graded semi-essential sub-modules, a nonzero graded R -submodule K of M is said to be graded semi-essentialif K T P = { } for every nonzero graded prime R -submodule P of M . Clearly, ev-ery graded essential submodule is graded semi-essential, we prove that the converseis not true in general. We prove that the intersection of two graded semi-essentialsubmodules is not graded semi-essential in general, however, it will be true if one isgraded essential submodule. Also, we introduce another case where the intersectionof two graded semi-essential submodules will be graded semi-essential.Let N be a graded R -submodule of M . Then the graded radical of N is denotedby Grad M ( N ) and it is defined to be the intersection of all graded prime submodules M. REFAI AND R. ABU-DAWWAS of M containing N . If there is no graded prime submodule containing N , then wetake Grad M ( N ) = M .Let M and M ′ be two G -graded R -modules. An R -homomorphism f : M → M ′ is said to be graded R -homomorphism if f ( M g ) ⊆ M ′ g for all g ∈ G . One can provethat if K is a graded R -submodule of M and L is a graded R -submodule of M ′ ,then f ( K ) is a graded R -submodule of M ′ and f − ( L ) is a graded R -submodule of M (see [14]). Let f : M → M ′ be a graded R -epimorphism and K be a graded R -submodule of M such that Ker ( f ) ⊆ K . Then there exists a one to one orderpreserving correspondence between the proper graded R -submodules of M contain-ing K and the proper graded R -submodules of M ′ containing f ( K ). Moreover, forany graded R -submodule L of M ′ , there exists a graded R -submodule N of M suchthat Ker ( f ) ⊆ N and f ( N ) = L .Based on analogous results for graded rings, the proof of the following will beroutine. Lemma 1.3.
Let f : M → M ′ be a graded R -epimorphism and K be a graded R -submodule of M such that Ker ( f ) ⊆ K . (1) If N is a graded prime R -submodule of M containing K , then f ( P ) is agraded prime R -submodule of M ′ containing f ( K ) . (2) If L is a graded prime R -submodule of M ′ containing f ( K ) , then f − ( L ) isa graded prime R -submodule of M containing K . We study graded semi-essential submodules under homomorphisms.In [10], a graded R -module M is said to be graded multiplication if for everygraded R -submodule N of M , N = IM for some graded deal I of R . In this case,we can take I = ( N : R M ). Graded multiplication modules have been studied byseveral authors, for example, see [2], [5] and [13]. We prove that if M is a gradedmultiplication faithful R -module and K a graded R -submodule of M such that( K : R M ) is a graded semi-essential ideal of R , then K is a graded semi-essential R -submodule of M .The concept of graded uniform modules have been introduced and studied in[14], a graded R module M is said to be graded uniform if every nonzero graded R -submodule of M is graded essential. In this work, we introduce and study theconcept of graded semi-uniform modules, a graded R module M is said to be gradedsemi-uniform if every nonzero graded R -submodule of M is graded semi-essential.Clearly, every graded uniform module is graded semi-uniform, we prove that theconverse is not true in general. It is known that the graded uniform property ishereditary, we prove that the graded semi-uniform property is not hereditary. Weprove that if R is a graded semi-uniform ring, then every graded multiplicationfaithful R -module is graded semi-uniform.2. Weakly Graded Rings
In this section, we introduce the concept of weakly graded rings and discuss therelations between weakly graded rings and several properties of graded rings.
Definition 2.1.
Let R be a G -graded ring. Then ( R, G ) is said to be weak if when-ever g ∈ G with R g = { } , then R g − = { } . Example 2.2.
The trivial graduation of a ring R by a group G is weak, that is R e = R and R g = { } otherwise. N PROPERTIES OF GRADED RINGS AND GRADED MODULES 5
Example 2.3.
Let R = M ( K ) and G = Z . Then R is G -graded by R = (cid:18) K K (cid:19) , R = (cid:18) KK (cid:19) and R = R = { } . ( R, G ) is first strongsince I ∈ R R and I ∈ R R but ( R, G ) is not strong since R R = { } 6 = R .Clearly, ( R, G ) is weak. Example 2.4.
Let R = K [ x ] (the ring of polynomials with coefficients from afield K ) and G = Z . Then R is G -graded by R j = Kx j , j ≥ and R j = { } otherwise. ( R, G ) is second strong but it is not first strong since ∈ supp ( R, G ) with R R − = { } 6 = R . ( R, G ) is not weak since R − = { } but R = { } . Proposition 2.5. If ( R, G ) is non-degenerate, then ( R, G ) is weak.Proof. Let g ∈ G with R g = { } . If R g − = { } , then there exists a g − ∈ R g − − { } and since ( R, G ) is non-degenerate, a g − R g = { } and then R g = { } a contradiction.So, R g − = { } and hence ( R, G ) is weak. (cid:3)
The next example shows that the converse of Proposition 2.5 is not true in general.
Example 2.6.
Let R = M ( K ) and G = D = h a, b : a = b = e, ba = a − b i .Then R is G -graded by R e = K K K
00 0 0 K , R a = K K
00 0 0 00 0 0 0 , R a = K
00 0 0 00 0 0 00 0 0 0 , R a = K , R b = K K , R ab = K K , R a b = K K , R a = K K and R a b = R a b = { } . ( R, G ) isweak but it is degenerate since x ab = ∈ R ab − { } with x ab R ( ab ) − = x ab R a b = { } . Since every faithful graded ring is non-degenerate and every regular graded ringis non-degenerate, we have the next two results.
Corollary 2.7. If ( R, G ) is faithful, then ( R, G ) is weak. Corollary 2.8. If ( R, G ) is regular, then ( R, G ) is weak. Proposition 2.9.
Let R be a G -graded ring. If R has no zero divisors, then ( R, G ) is weak if and only if ( R, G ) is non-degenerate.Proof. Suppose that (
R, G ) is weak. Let g ∈ G and a g ∈ R g − { } . Then R g = { } and then R g − = { } since ( R, G ) is weak. Since R has no zero divisors, a g R g − = { } and R g − a g = { } and hence ( R, G ) is non-degenerate. The converse holds byProposition 2.5. (cid:3)
M. REFAI AND R. ABU-DAWWAS
Proposition 2.10.
Let R be a weakly G -graded ring. If I is an ideal of R with I T R e = { } and R has no zero divisors, then I T R g = { } for all g ∈ G .Proof. Let g ∈ G . If g / ∈ supp ( R, G ), then R g = { } and then I T R g = { } .Suppose that g ∈ supp ( R, G ). Then { } = I T R e ⊇ I T ( R g R g − ) ⊇ ( I T R g ) R g − .Since R has no zero divisors, either I T R g = { } or R g − = { } . If R g − = { } ,then R g = { } since ( R, G ) is weak and that is a contradiction since g ∈ supp ( R, G ).So, I T R g = { } . (cid:3) Proposition 2.11.
Let R be a G -graded ring. If supp ( R, G ) is a subgroup of G ,then R is weakly graded.Proof. Let g ∈ G with R g = { } . If R g − = { } , then g − ∈ supp ( R, G ) and since supp ( R, G ) is a subgroup, g ∈ supp ( R, G ) and then R g = { } a contradiction. So, R g − = { } and hence R is weakly graded. (cid:3) The next example shows that the converse of Proposition 2.11 is not true ingeneral.
Example 2.12.
Let R = M ( K ) and G = Z . Then R is G -graded by R = K K
00 0 K , R = K
00 0 K , R = K , R = K , R = K K and R = R = { } . Clearly, ( R, G ) is weak but supp ( R, G ) = { , , , , } is not a subgroup of G . The next result states that the converse of Proposition 2.11 is true if R has nozero divisors. Proposition 2.13.
Let R be a G -graded ring. If R has no zero divisors, then ( R, G ) is weak if and only if supp ( R, G ) is a subgroup of G .Proof. Suppose that (
R, G ) is weak. Let g, h ∈ supp ( R, G ). Then { } 6 = R g R h ⊆ R gh and then gh ∈ supp ( R, G ). Since 0 = 1 ∈ R e , e ∈ supp ( R, G ). Let g ∈ supp ( R, G ). Then g − ∈ supp ( R, G ) since (
R, G ) is weak. So, supp ( R, G ) is asubgroup of G . The converse holds by Proposition 2.11. (cid:3) Corollary 2.14.
Let R be an integral domain. If R is weakly G -graded, then supp ( R, G ) is an abelian subgroup of G .Proof. By Proposition 2.13, supp ( R, G ) is a subgroup of G . Let g, h ∈ supp ( R, G ).Then { } 6 = R g R h ⊆ R gh and { } 6 = R h R g ⊆ R hg . Since R is commutative, R g R h = R h R g and then { } 6 = R g R h ⊆ R gh T R hg and hence gh = hg . So, supp ( R, G ) is anabelian subgroup of G . (cid:3) Corollary 2.15.
Let R be an integral domain. If R is G -graded such that ( R, G ) isfaithful, then G is abelian.Proof. By Corollary 2.7, (
R, G ) is weak and then by Corollary 2.14, supp ( R, G ) is anabelian subgroup of G . On the other hand, since ( R, G ) is faithful, supp ( R, G ) = G and hence G is abelian. (cid:3) In [15], it is proved that if (
R, G ) is first strong, then supp ( R, G ) is a subgroup of G . Combining this with Proposition 2.11 to have the next result. N PROPERTIES OF GRADED RINGS AND GRADED MODULES 7
Corollary 2.16. If R, G ) is first strong, then ( R, G ) is weak. The converse of Corollary 2.16 is not true in general; in Example 2.12, (
R, G ) isweak but it is not first strong since supp ( R, G ) is not a subgroup of G . Also, it isclear that if ( R, G ) is strong, then supp ( R, G ) = G . Combining this with Proposition2.11 to have the next result. Corollary 2.17. If R, G ) is strong, then ( R, G ) is weak. The converse of Corollary 2.17 is not true in general; in Example 2.12, (
R, G ) isweak but it is not strong since Since R R = { } 6 = R .On the other hand, if ( R, G ) is second strong, then (
R, G ) need not be weak; inExample 2.4, (
R, G ) is second strong but it is not weak. Also, the next exampleshows that if (
R, G ) is weak, then (
R, G ) need not be second strong.
Example 2.18.
Let R = K [ x ] and G = Z . Then R is G -graded by R = h , x , x , .... i , R = h x, x , x , .... i and R = h x , x , x , .... i . ( R, G ) is weakbut it is not second strong since R R = R since ∈ R such that / ∈ R R . In fact, if (
R, G ) is second strong and weak, then (
R, G ) is first strong and hence supp ( R, G ) is a subgroup of G . This is what we are going to prove. Proposition 2.19.
Let R be a G -graded ring. If ( R, G ) is second strong, then ( R, G ) is weak if and only if ( R, G ) is first strong.Proof. Suppose that (
R, G ) is weak. Let g ∈ supp ( R, G ). Then since (
R, G ) isweak, g − ∈ supp ( R, G ) and then since (
R, G ) is second strong, R g R g − = R e andthen 1 ∈ R g R g − and hence ( R, G ) is first strong. The converse holds by Corollary2.16. (cid:3)
Corollary 2.20.
Let R be a G -graded ring. If ( R, G ) is second strong, then ( R, G ) is weak if and only if supp ( R, G ) is a subgroup of G . In the rest of this section, we introduce an analogous study to that given in[8]. The question we deal with is: For a group G and a field K , can we write R = K [ x , x , ...., x m ] as strongly G -graded ring?.The next example shows that it is possible for G = Z and K = C (the field ofcomplex numbers). Example 2.21.
Let R = C [ x , x , ....x m ] and G = Z . Then R is strongly G -graded by R = R [ x , x , ....x m ] and R = i R [ x , x , ....x m ] where R is he field of realnumbers. Lemma 2.22.
Let R be a weakly G -graded ring ( R is any ring). If R has no zerodivisors and R e is a division ring, then R is simple.Proof. Let I be a nonzero ideal of R . If I T R e = { } , then by Proposition 2.10, I T R g = { } for all g ∈ G and then I = { } a contradiction. So, I T R e is anonzero ideal of R e and then since R e is division ring, I T R e = R e . Let g ∈ G .Then R g ⊆ R g R e = R g ( I T R e ) ⊆ R g I T R g R e ⊆ R g I T R g ⊆ I T R g and hence I T R g = R g , i.e., R g ⊆ I for all g ∈ G . Thus, I = R and hence R is simple. (cid:3) By Lemma 2.22, we can introduce the following:
Lemma 2.23.
Let R be a weakly G -graded ring ( R is any ring). If R has no zerodivisors and R e is a division ring, then R is Artinian (Noetherian). M. REFAI AND R. ABU-DAWWAS
Proposition 2.24.
Let R = K [ x , x , ....x m ] be a G -graded. If R e ⊆ K , then ( R, G ) is not weak.Proof. Firstly, we prove that R e is a subfield of K . Clearly, R e is a commutative ringwith unity. Let k ∈ R e − { } and k − = r g + r g + .... + r g n where r g i ∈ R g i − { } .Then 1 = kk − = kr g + kr g + .... + kr g n with kr g i ∈ R g i and 1 ∈ R e . So, n = 1 and g = e . Hence, k − ∈ R e . Let f be a nonconstant homogeneous element of R . Then I a = h rf n + a : r ∈ R, n ∈ N i ( a ∈ N ) is an ideal of R with I ⊃ I ⊃ .... but f n / ∈ I n for all n ∈ N ; let a ∈ N and g ∈ I a +1 . Then g = rf n +( a +1) for some r ∈ R and n ∈ N and then g = rf ( n +1)+ a ∈ I a . Suppose f n ∈ I n for some n ∈ N . Then f n = rf t + n for some t ∈ N and then since R is integral domain, 1 = rf t a contradiction since t ∈ N and f is nonconstant. Therefore, there is no s ∈ N with I n = I s for all n ≥ s ,i.e., R is not Artinian. Hence, by Lemma 2.22, ( R, G ) is not weak. (cid:3)
Corollary 2.25.
Let R = K [ x , x , ....x m ] be a G -graded. If R e ⊆ K , then ( R, G ) is not strong.Proof. If (
R, G ) is strong, then by Corollary 2.17, (
R, G ) is weak and this contradictsProposition 2.24. Hence, (
R, G ) is not strong. (cid:3) Invertible Graded Rings
In this section, we introduce the concept of invertible graded rings and we studythis concept assuming that R is a weakly graded domain. Definition 3.1.
Suppose that R is a G -graded ring. Then ( R, G ) is said to beinvertible if the identity component R e is a field. Example 3.2.
Consider R = C (the field of complex numbers) and G = Z . Then R is G -graded by R = R and R = i R . Since R is a field, ( R, G ) is invertible. The next example shows that if (
R, G ) is invertible, then R need not to be a fieldin general. Example 3.3.
Consider R = K [ x ] (where K is a field) and G = Z . Then R is G -graded by R j = Kx j for j ≥ and R j = { } otherwise. Since R = K is a field, ( R, G ) is invertible. However, R it self is not a field. Lemma 3.4.
Suppose that R is a G -graded ring. If g ∈ G and r ∈ R g is an unit,then r − ∈ R g − .Proof. By ([14], Proposition 1.1.1), r − ∈ R h for some h ∈ G and then rr − ∈ R g R h ⊆ R gh . On the other hand, rr − = 1 ∈ R e . So, 0 = rr − ∈ R e T R gh and then gh = e , i.e., h = g − . Hence, r − ∈ R g − . (cid:3) Lemma 3.5.
Let R be a weakly G -graded domain. If ( R, G ) is invertible, then everynonzero homogeneous element is unit.Proof. Let g ∈ G and r ∈ R g − { } . Then since ( R, G ) is weak, R g − = { } andthen since R is a domain, I = R g − r is a nonzero ideal of R e and since R e is a field, I = R e and then 1 ∈ I . It follows that 1 = xr for some x ∈ R g − and hence r isunit. (cid:3) Proposition 3.6.
Let R be a weakly G -graded domain. If ( R, G ) is invertible, then R g is cyclic R e -module for all g ∈ G . N PROPERTIES OF GRADED RINGS AND GRADED MODULES 9
Proof.
Let g ∈ G . If R g = { } , then it is done. Suppose that R g = { } . Thenthere exists a nonzero r ∈ R g and then by Lemma 3.4 and Lemma 3.5, r is unitwith r − ∈ R g − . Let x ∈ R g . Then x = x. xr − r ∈ R g R g − r ⊆ R e r and then R g ⊆ R e r ⊆ R e R g ⊆ R g . Hence, R g = R e r , i.e., R g is cyclic R e -module. (cid:3) Corollary 3.7.
Let R be a weakly G -graded domain. If ( R, G ) is invertible and supp ( R, G ) is finite, then R is Noetherian.Proof. Since supp(
R, G ) is finite, R = L ni =1 R g i . By Proposition 3.6, R g i is cyclic R e -module for all i ≤ i ≤ n . On the other hand, since R e is a field, R e is Noetherian.So, each R g i is Noetherian and hence R is Noetherian. (cid:3) Proposition 3.8.
Let R be a weakly G -graded domain. If ( R, G ) is invertible, then R g is simple R e module for all g ∈ G .Proof. Let g ∈ G . If R g = { } , then it is done. Suppose that R g = { } . Then byProposition 3.6, R g = R e r for some r ∈ R g . Let N be a nonzero R e -submodule of R g and let I = { x ∈ R e : xr ∈ N } . Then clearly, I is an ideal of R e . We show that N = Ir . Let n ∈ N ⊆ R g . Then n = xr for some x ∈ R e and then x ∈ I and hence n ∈ Ir . Thus, N ⊆ Ir . Let n ∈ Ir . Then n = xr for some x ∈ I . Since x ∈ I , xr ∈ N and then n ∈ N . Thus, Ir ⊆ N . Therefore, N = Ir . Now, if I = { } ,then N = { } a contradiction. So, I is a nonzero ideal of R e and since R e is a field, I = R e and then N = R e r = R g . Hence, R g is simple R e -module. (cid:3) If (
R, G ) is invertible, then R need not to be strongly G -graded. To see this, inExample 3.3, ( R, G ) is invertible but R is not strongly G -graded since 1 / ∈ R R − = { } . Also, if R is strongly G -graded, then ( R, G ) need not to be invertible, see thefollowing example.
Example 3.9.
Consider R = Z [ i ] and G = Z . Then R is G -graded by R = Z and R = i Z . Since ∈ R R and ∈ R R , R is strongly G -graded. However, ( R, G ) is not invertible since R = Z is not a field. In fact, we prove that every invertible weakly graded domain is first stronglygraded.
Proposition 3.10.
Let R be a weakly G -graded domain. If ( R, G ) is invertible,then R is first strongly G -graded.Proof. Let g ∈ supp ( R, G ). Then there exists a nonzero r ∈ R g and then by Lemma3.4 and Lemma 3.5, r is unit with r − ∈ R g − . So, 1 = rr − ∈ R g R g − and hence R is first strongly G -graded. (cid:3) Definition 3.11. ( [14] ) A graded ring R is said to be graded simple if the onlygraded ideals of R are { } and R itself. Proposition 3.12.
Let R be a weakly G -graded domain. If ( R, G ) is invertible,then R is graded simple.Proof. Let I be a nonzero graded ideal of R . Then there exists g ∈ G such that I g = { } , i.e., I g is a nonzero R e -submodule of R g . By Proposition 3.8, R g is simple R e -module and then I g = R g . On the other hand, I g = I T R g and then R g ⊆ I .Now, by Proposition 3.10, R is first strongly G -graded and then 1 ∈ R e = R g − R g ⊆ R g − I ⊆ I and hence I = R . Therefore, R is graded simple. (cid:3) In the rest of this section, we suppose that R is a G -graded ring such that ( R, G )is invertible. Then we can consider R as a vector space over R e , i.e., R e is the fieldof scalars. Proposition 3.13.
Suppose that R is a non-trivially G -graded ring such that ( R, G ) is invertible. Define T : R → R e by T ( x ) = x e . Then (1) T is Linear transformation. (2) KerT T R e = { } . (3) R g ⊆ Ker ( T ) for all g ∈ G − { e } . (4) T is not injective. (5) T is surjective.Proof. (1) Let x, y ∈ R and r ∈ R e . Then T ( x + y ) = ( x + y ) e = x e + y e = T ( x ) + T ( y ) and T ( rx ) = ( rx ) e = rx e = rT ( x ) (since r ∈ R e ). Hence, T islinear.(2) Let x ∈ Ker ( T ) T R e . Then x ∈ R e with T ( x ) = 0 and then 0 = T ( x ) = x e = x (since x ∈ R e ). Hence, Ker ( T ) T R e = { } .(3) Let g ∈ G − { e } and x ∈ R g . Then T ( x ) = x e = 0 (since x ∈ R g and g = e ),i.e., x ∈ Ker ( T ). Hence, R g ⊆ Ker ( T ).(4) Since R is non-trivially graded, there exists g ∈ G − { e } such that R g = { } and then by (3), { } 6 = R g ⊆ Ker ( T ), i.e., Ker ( T ) = { } and hence T isnot injective.(5) Let x ∈ R e . Then x ∈ R such that T ( x ) = x e = x (since x ∈ R e ). Hence, T is surjective. (cid:3) Lemma 3.14. If ( R, G ) is invertible, then R g is a subspace of R for all g ∈ G .Proof. Let g ∈ G . Since R g is additive subgroup, x + y ∈ R g for all x, y ∈ R g . Let r ∈ R e and x ∈ R g . Then rx ∈ R e R g ⊆ R g . Hence, R g is a subspace of R . (cid:3) Proposition 3.15. If ( R, G ) is invertible, then ( R g + R h ) /R g ≈ R h for all g, h ∈ G , g = h .Proof. Let g, h ∈ G such that g = h . Clearly, R g is a subspace of R g + R h . Define T : R h → ( R g + R h ) /R g by T ( x ) = x + R g . Then it is easy to prove that T is well-defined linear transformation. To prove that T is surjective, let y ∈ ( R g + R h ) /R g .Then y = r + R g for some r ∈ R g + R h and then y = ( a + b ) + R g for some a ∈ R g and b ∈ R h and so y = b + R g and hence T ( b ) = y . Therefore, T issurjective. So, R h /Ker ( T ) ≈ ( R g + R h ) /R g . On the other hand, Ker ( T ) = { x ∈ R h : x + R g = { }} = { x ∈ R h : x ∈ R g } = R h T R g = { } (since g = h ). Thus,( R g + R h ) /R g ≈ R h / { } ≈ R h . (cid:3) Proposition 3.16. If ( R, G ) is invertible, then for every g ∈ G there exists a lineartransformation T g such that Ker ( T g ) = R g .Proof. Let g ∈ G . Define T g : R → R/R g by T g ( x ) = x + R g . Then it is easy toprove that T g is well-defined linear transformation with Ker ( T g ) = R g . (cid:3) Weakly Crossed Products
In this section, we study the relations between weakly crossed products and sev-eral properties of graded rings, and we introduce several results concerning weaklycrossed products.
N PROPERTIES OF GRADED RINGS AND GRADED MODULES 11
Definition 4.1.
Let R be a G -graded ring. Then ( R, G ) is said to be weakly crossedproduct if R g contains a unit for all g ∈ supp ( R, G ) . Clearly, if (
R, G ) is a crossed product, then (
R, G ) is a weakly crossed product.However, the converse is not true in general; if | G | ≥
2, then the trivial graduationof R ( R e = R and R g = 0 otherwise) is a weakly crossed product but it is not acrossed product since for g = e , R g does not contain units. Example 4.2.
In Example 2.3, ( R, G ) is a weakly crossed product since (cid:18) (cid:19) ∈ R and (cid:18) (cid:19) ∈ R are units. However, ( R, G ) is not a crossed product since G = supp ( R, G ) . Example 4.3.
In Example 2.4, ( R, G ) is not a weakly crossed product since R doesnot contain units. Proposition 4.4. If ( R, G ) is a weakly crossed product, then ( R, G ) is first strong.Proof. Let g ∈ supp( R, G ). Since (
R, G ) is weakly crossed product, R g contains aunit, say r and then 1 = rr − ∈ R g R g − by Lemma 3.4. Hence, ( R, G ) is firststrong. (cid:3)
Corollary 4.5. If ( R, G ) is a weakly crossed product, then supp ( R, G ) is a subgroupof G . Also, since every first strongly graded ring is second strongly graded, we can statethe following.
Corollary 4.6. If ( R, G ) is a weakly crossed product, then ( R, G ) is second strong. As every first strongly graded ring is non-degenerate, we have the following.
Corollary 4.7. If ( R, G ) is a weakly crossed product, then ( R, G ) is non-degenerate. Example 4.8.
Let R = M ( K ) (the ring of all × matrices with entries from afield K ) and G = Z (the group of integers). Then R is G -graded by R = (cid:18) K K (cid:19) , R = (cid:18) K (cid:19) , R − = (cid:18) K (cid:19) and R j = { } otherwise. ( R, G ) is non-degenerate but it is not weakly crossed product since R does notcontain units. So, the converse of Corollary 4.7 is not true in general. Proposition 4.9. If ( R, G ) is a weakly crossed product, then ( R, G ) is weak.Proof. Let g ∈ G such that R g = { } . If R g − = { } , then g − ∈ supp ( R, G ),and then R g − contains a unit, say r . By Lemma 3.4, 0 = r − ∈ R g which is acontradiction. So, R g − = { } , and hence ( R, G ) is weak. (cid:3)
Example 4.10.
Let R = M ( K ) and G = Z (the group of integers modulo ).Then R is G -graded by R = K K K K K and R = K K K K . ( R, G ) is weak but not weaklycrossed product since R does not contain units. So, the converse of Proposition 4.9is not true in general. Proposition 4.11.
Let R be a weakly G -graded domain. If ( R, G ) is invertible,then ( R, G ) is a weakly crossed product. Proof.
Apply Lemma 3.5. (cid:3)
Remark 4.12.
In Example 3.2, ( R, G ) is weakly crossed product by Proposition4.11. Remark 4.13.
The converse of Proposition 4.11 is not true in general; if R is not afield, then the trivial graduation of R is a weakly crossed product but not invertible. Proposition 4.14. If ( R, G ) is a weakly crossed product, then for all g ∈ supp ( R, G ) ,there exists a unit r ∈ R g such that R g = R e r (i.e., R g is a cyclic R e -module).Proof. Let g ∈ supp ( R, G ). Since (
R, G ) is a weakly crossed product, R g contains aunit, say r . Let x ∈ R g . Then x = x. x.r − r ∈ R g R g − r ⊆ R e r . On the otherhand, R e r ⊆ R e R g ⊆ R g . Hence R g = R e r . (cid:3) Remark 4.15.
Similarly, one can prove that if ( R, G ) is a weakly crossed product,then for all g ∈ supp ( R, G ) , there exists a unit r ∈ R g such that R g = rR e . Corollary 4.16. If ( R, G ) is a weakly crossed product, then R g is isomorphic to R e as an R e - module for all g ∈ supp ( R, G ) .Proof. Let g ∈ suup ( R, G ). Then by Proposition 4.14, R g = R e r for some unit r ∈ R g , and then f : R e → R g such that f ( x ) = xr is an R e -isomorphism. (cid:3) Proposition 4.17.
Let R be a commutative G - graded ring and M be a G -graded R -module. If ( R, G ) is a weakly crossed product, then M g is isomorphic to M e asan R e - module for all g ∈ supp ( R, G ) .Proof. Let g ∈ supp ( R, G ). Since (
R, G ) is a weakly crossed product, we have R g = R e r for some unit r ∈ R g . Since ( R, G ) is first strong, (
M, G ) is first strong.So, M g = R g M e = R e rM e = rR e M e = rM e , and then f : M e → M g such that f ( x ) = rx is an R e -isomorphism. (cid:3) Proposition 4.18.
Let ( R, G ) be a weakly crossed product, M a G -graded R -moduleand f : M → R an R -homomorphism. Then f ( M e ) = R e if and only if f ( M g ) = R g for all g ∈ supp ( R, G ) .Proof. Suppose that f ( M e ) = R e . Let g ∈ supp ( R, G ). Since (
R, G ) is a weaklycrossed product, R g = rR e for some unit r ∈ R g and ( R, G ) is first strong, and hence(
M, G ) is first strong. So, f ( M g ) = f ( R g M e ) = f ( rR e M e ) = f ( rM e ) = rf ( M e ) = rR e = R g . The converse is obvious. (cid:3) Graded fixing maps have been introduced and studied in [16]; an R -homomorphism f : M → R is said to be a grade fixing map if f ( M g ) ⊆ R g for all g ∈ G . It has beenproved that if f is a surjective grade fixing map, then f ( M g ) = R g for all g ∈ G .Also, if f is an injective grade fixing map, then supp ( M, G ) ⊆ supp ( R, G ). Proposition 4.19.
Let ( R, G ) be a weakly crossed product, M a G -graded R -moduleand f : M → R an R -isomorphism. Then f is a grade fixing map if and only if f ( M e ) = R e .Proof. Suppose that f is a grade fixing map. Since f is surjective, f ( M e ) = R e .Conversely, by Proposition 4.18, f ( M g ) = R g for all g ∈ supp ( R, G ). Since f isinjective, supp ( M, G ) ⊆ supp ( R, G ). If g / ∈ supp ( R, G ), then g / ∈ supp ( M, G ) andthen f ( M g ) = f ( { } ) = { } = R g . Hence, f is a grade fixing map. (cid:3) N PROPERTIES OF GRADED RINGS AND GRADED MODULES 13
Graded Noetherian modules have been introduced and studied in [14]; a graded R -module M is said to be graded Noetherian if every descending chain of graded R -submodules of M terminates. Clearly, every Noetherian Module is graded Noe-therian. However, the converse is not true in general (see [14]) Proposition 4.20.
Let R be a G -graded domain such that ( R, G ) is a weakly crossedproduct, ( R, G ) is invertible and supp ( R, G ) is finite. Let M be a G -graded R -module. Suppose that there exists a bijective grade fixing map from M to R . If M e is R e -simple, then M is graded Noetherian.Proof. Let f be a bijective grade fixing map from M to R . Firstly, we prove that R e is R e -simple. Let I be a nonzero ideal of R e . Then f − ( I ) is a nonzero R e -submoduleof M e and M e is R e -simple, f − ( I ) = M e and then I = f ( M e ) = R e . By Proposition4.9 and Corollary 3.7, R is Noetherian and hence R is graded Noetherian. Let N ⊆ N ⊆ ..... be graded R -submodules of M . Then f ( N ) ⊆ f ( N ) ⊆ ..... aregraded ideals of R and since R is graded Noetherian, there exists m ∈ N such that f ( N n ) = f ( N m ) for all n ≥ m and then N n = N m for all n ≥ m . Hence, M is gradedNoetherian. (cid:3) Graded Semi-essential Submodules
In this section, we introduce and study the concepts of graded semi-essentialsubmodules and graded semi-uniform modules.
Definition 5.1.
Let M be a graded R -module and K a nonzero graded R -submoduleof M . Then K is said to be a graded semi-essential R -submodule of M if K T P = { } for every nonzero graded prime R -submodule P of M . A nonzero graded ideal I of a graded ring R is said to be graded semi-essential if I T B = { } for everynonzero graded prime ideal B of R . Clearly, every graded essential R -submodule is graded semi-essential. The con-verse is not true in general as we see in the following example. Example 5.2.
Consider R = Z (the ring of integers), G = Z (the group of integersmodulo ) and M = Z [ i ] = { a + ib : a, b ∈ Z , i = − } where Z is the ring ofintegers modulo . Then R is G -graded by R = Z and R = { } , M is G -gradedby M = Z and M = i Z . The graded R -submodule N = h i of M is gradedsemi-essential but not graded essential. The following proposition gives a necessary and sufficient condition for a graded R -submodule to be a graded semi-essential. The proof is clear and hence it isomitted. Proposition 5.3.
Let M be a graded R -module and K a nonzero graded R -submoduleof M . Then K is a graded semi-essential R -submodule of M if and only if for everygraded prime R -submodule P of M there exist r ∈ h ( R ) and m ∈ h ( M ) such that m ∈ P and = rm ∈ K . Also, the proof of the following proposition is straightforward and hence it isomitted.
Proposition 5.4.
Let M be a graded R -module and K , K two graded R -submodulesof M such that K is a graded R -submodule of K . If K is a graded semi-essential R -submodule of M , then K is a graded semi-essential R -submodule of M . The next example shows that the converse of Proposition 5.4 is not true in general.
Example 5.5.
Consider Example 5.2. Clearly, K = h i and K = h i are graded R -submodules of M such that K is a graded R -submodule of K . Since P = h i is a graded prime R -submodule of M satisfies K T P = { } , K is not a gradedsemi-essential R -submodule of M , but it is obvious that K is a graded semi-essential R -submodule of M . As a result of Proposition 5.4, we can state the following corollary.
Corollary 5.6.
Let M be a graded R -module and K , K two graded R -submodulesof M . If K T K is a graded semi-essential R -submodule of M , then K and K are graded semi-essential R -submodules of M . The next example shows that the converse of Corollary 5.6 is not true in general.
Example 5.7.
Consider R = Z , G = Z and M = Z [ i ] . Then R is G -gradedby R = Z and R = { } , M is G -graded by M = Z and M = i Z . Clearly, K = h i and K = h i are graded R -submodules of M . The only graded prime R -submodules of M are P = h i and P = h i . Since K T P = { } and K T P = { } , K is a graded semi-essential R -submodule of M . Similarly, K is a gradedsemi-essential R -submodule of M . But K T K = { } which is not a graded semi-essential R -submodule of M . The next Proposition gives a condition under which the converse of Corollary 5.6is true.
Proposition 5.8.
Let M be a graded R -module and K , K two graded R -submodulesof M . If K is a graded essential R -submodule of M and K is a graded semi-essential R -submodule of M , then K T K is a graded semi-essential R -submoduleof M .Proof. It is straightforward. (cid:3)
Before we give another condition under which the converse of Corollary 5.6 istrue, we need the following lemma.
Lemma 5.9.
Let M be a graded R -module, K a graded R -submodule of M and P a graded prime R -submodule of M . If ( K T P : R m ) = Ann ( M ) for all m ∈ h ( M ) − ( K T P ) , then K T P is a graded prime R -submodule of M .Proof. Let r ∈ h ( R ) and m ∈ h ( M ) such that rm ∈ K T P . Suppose that m / ∈ K T P . Since rm ∈ K T P , r ∈ ( K T P : R m ) and then r ∈ Ann ( M ) which impliesthat r ∈ ( K : R M ) T ( P : R M ) and hence r ∈ ( K T P : R M ). Therefore, K T P is agraded prime R -submodule of M . (cid:3) Proposition 5.10.
Let M be a graded R -module and K , K two graded semi-essential R -submodules of M . If ( K T P : R m ) = Ann ( M ) for every graded prime R -submodule P of M and for every m ∈ h ( M ) − ( K T P ) , then K T K is a gradedsemi-essential R -submodule of M .Proof. Let P be a nonzero graded prime R -submodule of M . Then by Lemma5.9, K T P is a graded prime R -submodule of M and then ( K T K ) T P = K T ( K T P ) = { } . Hence, K T K is a graded semi-essential R -submoduleof M . (cid:3) N PROPERTIES OF GRADED RINGS AND GRADED MODULES 15
Proposition 5.11.
Let M be a graded R -module, K a nonzero graded R -submoduleof M and T a nonzero graded prime R -submodule of M such that ( K + T ) /T is agraded semi-essential R -submodule of M/T . If P is a graded prime R -submodule of M such that T ⊆ P and K T P = { } , then T = P .Proof. Let P be a graded prime R -submodule of M such that T ⊆ P and K T P = { } . Assume that m ∈ ( K + T ) T P . Then m = k + t = p for some k ∈ K , t ∈ T and p ∈ P and then k = p − t ∈ K T P = { } which implies that k = 0 and hence m = t ∈ T . So, ( K + T ) T P = T which implies that (( K + T ) /T ) T ( P/T )) = { } .But ( K + T ) /T is a graded semi-essential R -submodule of M/T by assumption and
P/T is a graded prime R -submodule of M/T , so
P/T = { } which implies that P = T . (cid:3) Proposition 5.12.
Let f : M → M ′ be a graded R -epimorphism such that Ker ( f ) ⊆ Grad M ( M ) . If K is a graded semi-essential R -submodule of M ′ , then f − ( K ) is agraded semi-essential R -submodule of M .Proof. Let P be a graded prime R -submodule of M such that f − ( K ) T P = { } .Since Ker ( f ) ⊆ Grad M ( M ) ⊆ P , f ( P ) is a graded prime R -submodule of M ′ with K T f ( P ) = { } . Since K is a graded semi-essential R -submodule of M ′ , f ( P ) = { } and then P ⊆ f − ( { } ) = Ker ( f ) ⊆ f − ( K ) and hence f − ( K ) T P = P whichimplies that P = { } . Therefore, f − ( K ) is a graded semi-essential R -submoduleof M . (cid:3) Proposition 5.13.
Let f : M → M ′ be a graded R -isomorphism. If K is a gradedsemi-essential R -submodule of M , then f ( K ) is a graded semi-essential R -submoduleof M ′ .Proof. Let P be a nonzero graded prime R -submodule of M ′ . Since f is epimor-phism, f − ( P ) is a graded prime R -submodule of M and then K T f − ( P ) = { } and hence f ( K ) T P = { } since f is monomorphism. Hence, f ( K ) is a gradedsemi-essential R -submodule of M ′ . (cid:3) Proposition 5.14.
Let M be a graded multiplication faithful R -module and K agraded R -submodule of M . If ( K : R M ) is a graded semi-essential ideal of R , then K is a graded semi-essential R -submodule of M .Proof. Let P be a graded prime R -submodule of M such that K T P = { } . Since M is graded multiplication, K = ( K : R M ) M and P = ( P : R M ) M . Now, { } = K T P = (( K : R M ) M T ( P : R M ) M ) = (( K : R M ) T ( P : R M )) M and since M is faithful, ( K : R M ) T ( P : R M ) = { } . But ( K : R M ) is a graded semi-essentialideal of R and ( P : R M ) is a graded prime ideal of R , so ( P : R M ) = { } and hence P = { } . Therefore, K is a graded semi-essential R -submodule of M . (cid:3) Definition 5.15.
A nonzero graded R -module M is said to be graded semi-uniformif every nonzero graded R -submodule of M is graded semi-essential. A nonzerograded ring R is said to be graded semi-uniform if every nonzero graded ideal of R is graded semi-essential. Clearly, every graded uniform R -module is graded semi-uniform. The converse isnot true in general as we see in the following example. Example 5.16.
Consider Example 5.7, M = Z [ i ] is a graded semi-uniform Z -module. But M = Z [ i ] is not a graded uniform Z -module since h i T h i = { } . It is known that the graded uniform property is hereditary. The graded semi-uniform property is not hereditary as we see in the following example.
Example 5.17.
Consider Example 5.7, M = Z [ i ] is a graded semi-uniform Z -module and K = h i is a graded Z -submodule of M . The only graded prime Z -submodules of K are P = h i and P = h i . But N = h i is a graded Z -submoduleof K with N T P = { } . So, K is not a graded semi-uniform Z -module. Proposition 5.18.
Let R be a graded semi-uniform ring. Then every graded mul-tiplication faithful R -module is graded semi-uniform.Proof. Let M be a graded multiplication faithful R -module and K a nonzero graded R -submodule of M . Since R is graded semi-uniform, ( K : R M ) is a graded semi-essential ideal of R and then by Proposition 5.14, K is a graded semi-essential R -submodule of M and hence M is a graded semi-uniform R -module. (cid:3) References [1] R. Abu-Dawwas, On graded semi-prime rings, Proceedings of the Jangjeon Mathematical So-ciety, 20 (1) (2017), 19-22.[2] R. Abu-Dawwas, Multiplication Components of graded modules, Italian Journal of Pure andApplied Mathematics, 35 (2015), 389-392.[3] R. Abu-Dawwas, More on crrossed over the support of graded rings, International MathematicalForum, 5 (36) (2010), 3121-3126.[4] R. Abu-Dawwas, K. Al-Zoubi and M. Bataineh, Prime submodules of graded modules, Proyec-ciones Journal of Mathematics, 31 (4) (2012), 355-361.[5] R. Abu-Dawwas and M. Refai, Some remarks on R e -multiplication modules over first stronglygraded rings, East-West Journal of Mathematics, 13 (1) (2011), 57-61.[6] R. Abu-Dawwas and M. Refai, Further results on graded prime submodules, InternationalJournal of Algebra, 4 (28) (2010), 1413-1419.[7] S. Ceken and M. Alkan, On graded second and coprimary modules and graded secondaryrepresentations, Bulletin of the Malaysian Mathematical Sciences Society, 38 (4) (2015), 1317-1330.[8] E. C. Dade, Compounding clifford’s theory, Annals of Mathematics, Second Series, 91 (1)(1970), 236-290.[9] S. Ebrahimi Atani, On graded prime submodules, Chiang Mai Journal of Science, 33 (1) (2006),3-7.[10] J. Escoriza and B. Torrecillas, Multiplication objects in commutative Grothendieck category,Communications in Algebra, 26 (1998), 1867-1883.[11] F. Farzalipour and P. Ghiasvand, On graded weak multiplication modules, Tamkang Journalof Mathematics, 43 (2) (2012), 171-177.[12] F. Farzalipour and P. Ghiasvand, On the union of graded prime submodules, Thai Journal ofMathematics, 9 (1) (2011), 49-55.[13] K. Khaksari and F. R. Jahromi, Multiplication graded modules, International Journal ofAlgebra, 7 (1) (2013), 17-24.[14] C. Nastasescu and F. Van Oystaeyen, Methods of graded rings, Lecture Notes in Mathematics,1836, Springer-Verlag, Berlin, 2004.[15] M. Refai, Various types of strongly graded rings. Abhath Al-Yarmouk Journal (Pure Sciencesand Engineering Series) 4 (2) (1995), 9-19.[16] M. Refai and R. Abu-Dawwas, Grade fixing maps, International Journal of Algebra, 4 (19),939-944, (2010).[17] M. Refai, M. Hailat and S. Obiedat, Graded radicals and graded prime spectra, Far EastJournal of Mathematical Sciences, part 1 (2000), 59-73.[18] M. Refai and F. Moh’d, First strongly graded modules, International Journal of Mathematics(Game Theory and Algebra), 15 (4) (2006), 451-457. N PROPERTIES OF GRADED RINGS AND GRADED MODULES 17
Mashhoor Refai, President of Princess Sumaya University for Technology, Jordan.Email address : [email protected] Rashid Abu-Dawwas, Department of Mathematics, Yarmouk University, Jordan.Email address ::