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International Mathematical Forum, 1, 2006, no. 2, 61-66
ON GRADED WEAKLY PRIME
SUBMODULES
Shahabaddin Ebrahimi Atani
Department of Mathematics
University of Guilan
P.O. Box 1914, Rasht, Iran
ebrahimi@guilan.ac.ir
Abstract
Let Gbe a monoid with identity e, and let Rbe a G-graded com-
mutative ring. Graded weakly prime ideals in a G-graded commutative
ring have been introduced and studied in [3]. Here we study graded
weakly prime submodules of a G-graded R-module. A number of re-
sults concerning of these class of submodules are given. For example,
we give some characterizations of homogeneous components of graded
submodules.
Mathematics Subject Classification: 13A02, 16W50
Keywords: Graded rings, Graded weakly prime submodules
1 Introduction
Weakly prime ideals in a commutative ring with non-zero identity have been
introduced and studied by D. D. Anderson and E. Smith in [1]. Also, weakly
primary ideals in a commutative ring with non-zero identity have been intro-
duced and studied in [2]. Here we study the graded weakly prime submod-
ules of a G-graded R-module (see sec. 2). Before we state some results let
us introduce some notation and terminology. Let Gbe an arbitrary monoid
with identity e.ByaG-graded commutative ring we mean a commutative
ring Rwith non-zero identity together with a direct sum decomposition (as
an additive group) R=⊕g∈GRgwith the property that RgRh⊆Rgh for all
62 Shahabaddin Ebrahimi Atani
g, h ∈G. Also, we write h(R)=∪g∈gRg. The summands Rgare called homo-
geneous components and elements of these summands are called homogeneous
elements. If a∈R, then acan be written uniquely as g∈Gagwhere agis the
component of ain Rg. Moreover, Reis a subring of Rand 1R∈Re.
Let Rbe a G-graded ring and Man R-module. We say that Mis a G-
graded R-module if there exists a family of subgroups {Mg}g∈Gof Msuch
that M=⊕g∈GMg(as abelian groups), and RgMh⊆Mgh for all g, h ∈G,
here RgMhdenotes the additive subgroup of Mconsisting of all finite sums of
elements rgshwith rg∈Rgand sh∈Mh. Also, we write h(M)=∪g∈GMg.If
M=⊕g∈GMgis a graded R-module, then Mgis an Re-module for all g∈G.
Let M=⊕g∈GMgbe a graded R-module and Na submodule of M.Forg∈G,
let Ng=N∩Mg. Then Nis a graded submodule of Mif N=⊕g∈GNg. In this
case, Ngis called the g-component of Nfor g∈G. Moreover, M/N becomes
aG-graded module with g-component (M/N)g=(Mg+N)/N for g∈G.
Clearly, 0 is a graded submodule of M.IfNand Kare submodules of an
R-module M, the ideal {a∈R:aK ⊆N}will be denoted by (K:RN).
2 Graded weakly prime submodules
Let Rbe a G-graded ring, Ma graded R-module, Na graded submodule of
Mand let g∈G. We say that Ngis a g-prime submodule of the Re-module
Mgif Ng=Mg; and whenever a∈Reand m∈Mgwith am ∈Ng, then either
m∈Ngor b∈(Ng:ReMg). We say that Nis a graded prime submodule of M
if N=M; and whenever a∈h(R) and m∈h(M) with am ∈N, then either
m∈Nor b∈(N:RM) ([4]).
Our starting point is the following definitions:
Definition 2.1 Let Rbe a G-graded ring, Ma graded R-module, Na graded
submodule of Mand let g∈G.
(i) We say that Ngis a weakly g-prime submodule of the Re-module Mgif
Ng=Mg; and whenever a∈Reand m∈Mgwith 0=am ∈Ng, then either
m∈Ngor b∈(Ng:ReMg)
(ii) We say that Nis a graded weakly prime submodule of Mif N=M;
and whenever a∈h(R)and m∈h(M)with 0=am ∈N, then either m∈N
or b∈(N:RM).
Clearly, a graded prime submodule of M(resp. a g-prime submodule of Mg)
is a graded weakly prime submodule of M(resp. weakly g-prime submodule
ON GRADED WEAKLY PRIME SUBMODULES 63
of Mg). However, since 0 is always a graded weakly prime submodule of M
(resp. a weakly g-prime submodule of Mg) (by definition), a graded weakly
prime submodule (resp. a weakly g-prime submodule) need not be graded
prime (resp. g-prime).
Proposition 2.2 Let Rbe a G-graded ring and Ma graded R-module. As-
sume that Nand Kare graded submodules of Msuch that K⊆Nwith
N=M. Then the following hold:
(i) If Nis a graded weakly prime submodule of M, then N/K is graded
weakly prime.
(ii) If Kand N/K are graded weakly prime, then Nis graded weakly prime.
Proof. (i) Let 0 =a(m+K)=am +K∈N/K where a∈h(R) and
m∈h(M), so am ∈N.Ifam =0∈K, then a(m+K) = 0, which is a
contradiction. If am =0,Ngraded weakly prime gives either a∈(N:RM)
or m∈N; hence either m+K∈N/K or a∈(N/K :RM/K), as required.
(ii) Let 0 =am ∈Nwhere a∈h(R) and m∈h(M), so a(m+K)∈
N/K.Ifam ∈K, then Kgraded weakly prime gives either m∈K⊆N
or a∈(K:RM)⊆(N:RM). So we may assume that am /∈K. Then
0=a(m+K)∈N/K. Since N/K is a graded weakly prime, we get either
m∈Nor a∈(N/K :RM/K)⊆(N:RM), as needed.
Theorem 2.3 Let Rbe a G-graded ring and Ma graded R-module. Assume
that Nand Kare graded weakly prime submodules of Msuch that K+N=M.
Then N+Kis a graded weakly prime submodule of M.
Proof. Since (N+K)/K ∼
=K/(N∩K), we get (N+K)/K is a graded
weakly prime submodule by Propositin 2.2 (i). Now the assertion follows from
Proposition 2.2 (ii).
Lemma 2.4 Let Rbe a G-graded ring, Ma graded R-module and Na graded
submodule of M.IfNis a graded weakly prime submodule of M, then Ngis
a weakly g-prime submodule of Mgfor every g∈G.
Proof. Suppose that Nis a graded weakly prime submodule of M.For
g∈G, assume that 0 =am ∈Ng⊆Nwhere a∈Reand m∈Mg,soN
graded weakly prime gives either m∈Nor a∈(N:RM). If m∈N, then
m∈Ng.Ifa∈(N:RM), then aMg⊆aM ⊆N; hence a∈(Ng:ReMg). So
Ngis a weakly g-prime submodule of Mg.
64 Shahabaddin Ebrahimi Atani
Proposition 2.5 Let Rbe a G-graded ring, Ma graded R-module and Na
graded weakly prime submodule of M. Then for each g∈G, either Ngis a
g-prime submodule of Mgor (Ng:ReMg)Ng=0.
Proof. By Lemma 2.4, Ngis a weakly g-prime submodule of Mgfor every
g∈G. It is enough to show that if (Ng:ReMg)Ng= 0 for some g∈G,
then Ngis a g-prime submodule of Mg. Let am ∈Ngwhere a∈Reand
m∈Mg.Ifam = 0, then either m∈Ngor a∈(Ng:ReMg) since Ng
is weakly g-prime. So suppose that am =0. IfaNg= 0, then there is
an element nof Ngsuch that an =0,so0=an =a(m+n)∈Ng, and
hence Ngweakly g-prime gives either a∈(Ng:ReMg)or(m+n)∈Ng.
Then we have either a∈(Ng:ReMg)orm∈Ngsince n∈Ng. So we can
assume that aNg= 0. Suppose that (Ng:ReRg)m=0,saycm = 0 where
c∈(Ng:ReMg). Then 0 =cm =(a+c)m∈Ng,soNgweakly g-prime gives
either m∈Ngor a∈(Ng:ReMg) since c∈(Ng:ReMg). So we can assume
that (Ng:ReRg)m=0.
Since we assumed (Ng:ReMg)Ng= 0, there exist c∈(Ng:ReMg) and
t∈Ngsuch that ct = 0. Then (a+c)(m+t)=ct ∈Ng, so either a+c∈
(Ng:ReMg)orm+t∈Ng, and hence either a∈(Ng:ReMg)orm∈Ng.
Thus Ngis a g-prime submodule of Mg.
We next give three other characterizations of homogeneous components of
graded submodules.
Theorem 2.6 Let Rbe a G-graded ring, Ma graded R-module, Na graded
submodule of M, and g∈G. Then the following assertion are equivalent.
(i) If whenever 0=IK ⊆Ngwith Ian ideal of Reand Ka submodule of
Mgimplies that I⊆(Ng:ReMg)or K⊆Ng.
(ii) Ngis a weakly g-prime submodule of Mg.
(iii) For a∈Mg−Ng,(Ng:Rea)=(Ng:ReMg)∪(0 :Rea).
(iv) For a∈Mg−Ng,(Ng:Rea)=(Ng:ReMg)or (Ng:Rea) = (0 :Rea).
Proof. (i)=⇒(ii) Let 0 =am ∈Ngwhere m∈Mgand a∈Re. Take
I=Reaand K=Rem. Then 0 =IK ⊆Ng, so either I⊆(Ng:ReMg)
or K⊆Ng; hence either m∈Ngor a∈(Ng:ReMg). Thus Ngis a weakly
g-prime submodule of Mg.
(ii)⇒(i) Suppose that Ngis a weakly g-prime submodule of Mg. Let
0=IK ⊆Ngwith x∈K−Ng. We show that I⊆(Ng:ReMg). Let r∈I.
If rx = 0, then Ngweakly g-prime gives r∈(Ng:ReMg). So assume that
ON GRADED WEAKLY PRIME SUBMODULES 65
rx =0. IfrK = 0, then rd = 0 for some 0 =d∈K⊆Mg.Ifd∈Ng, then
r(d+x)∈Nggives either r∈(Ng:ReMg)ord+x∈Ng,sor∈(Ng:ReMg)
since x/∈Ng.Ifd/∈Ng, then rd ∈Nggives r∈(Ng:ReMg). So we can assume
that rK = 0. Suppose that Ix =0,sayax = 0 where a∈I. Then Ngweakly
g-prime gives a∈(Ng:ReMg). It follows from the equality (r+a)x=ax that
r∈(Ng:ReMg), so I⊆(Ng:ReMg). Therefore we can assume that Ix =0.
Since IK = 0, there exist s∈Iand y∈Ksuch that sy =0. As
0=s(y+x)=sy ∈Ngwe divided the proof into the following cases:
Case 1 s/∈(Ng:ReMg) and y+x/∈Ng.
Since s(y+x)=sy ∈Ng,Ngweakly g-prime gives either y+x∈Ngor
s∈(Ng:ReMg), which is a contradiction.
Case 2 s/∈(Ng:ReMg) and y+x∈Ng.
As 0 =sy ∈Ngwe have y∈Ng,sox∈Ng, which is a contradictin.
Case 3 s∈(Ng:ReMg) and y+x∈Ng.
Since y+x∈Ng, we obtain y/∈Ng(otherwise x∈Ng). As 0 =(r+s)y∈
Ng, we get r∈(Ng:ReMg). Thus I⊆(Ng:ReMg).
Case 4 s∈(Ng:ReMg) and y+x/∈Ng.
Since 0 =(r+s)(y+x)=sy ∈Ngit follows that r+s∈(Ng:ReMg), so
r∈(Ng:ReMg). Hence I⊆(Ng:ReMg).
(ii)⇒(iii) Clearly, if x∈Mg−Ng, then H=(Ng:ReMg)∪(0 :Re
x)⊆(Ng:Rex). For the reverse inclusion, assume that b∈(Ng:Rex) where
x∈Mg−Ng. Then bx ∈Ng.Ifbx = 0, then b∈(Ng:ReMg) since Ngis
weakly g-prime, so b∈H.Ifbx = 0, then b∈(0 :Rex), so b∈H, and hence
we have equality.
(iii)⇒(iv) Is obvious.
(iv)⇒(ii) Suppose that 0 =am ∈Ngwith a∈Reand m∈Mg−Ng.
Then a∈(Ng:Rem) and a/∈(0 :Rem). It follows from (iv) that a∈(Ng:Re
m)=(Ng:ReMg), as required.
References
[1] D. D. Anderson and R. Smith, Weakly prime ideals, Houston J. of
Mathematics, 29 (2003), 831-840.
[2] S. Ebrahimi Atani and F. Farzalipour, On weakly primary ideals,
Georgian Mathematical Journal, to appear.
[3] S. Ebrahimi Atani, On graded weakly primary ideals, Quasi-groups
and Related Systems, to appear.
66 Shahabaddin Ebrahimi Atani
[4] S. Ebrahimi Atani, On graded prime submodules, submitted.
[5] C. Nastasescu and F. Van Oystaeyen, Graded Ring Theory, Mathe-
matical Library 28, North Holand, Amsterdam, (1982).
[6] M. Refai and K. Al-Zoubi, On Graded Primary Ideals, Turkish Journal
of Mathematics, 28 (2004), 217-229.
Received: June 21, 2006