2011) 2(  24مجلة ابن الهیثم للعلوم الصرفة والتطبیقیة               المجلد

  المقاسات الجزئیة شبه االولیة التامة

  والمقاسات شبه االولیة التامة

 

  

 بثینة نجاد شهاب هادي و انعام محمد علي

 جامعة بغداد،ابن الهیثم  -كلیة التربیة،قسم الریاضیات 
 

 2010 ،اب، 23: استلم البحث في 

   2010،تشرین الثاني، 9: قبل البحث في 
 
 

  الخالصة

في هذا البحـث درسـنا مفهـومي المقاسـات الجزئیـة . مقاساً  احادیا ً  Mحلقة ابدالیة ذا عنصر محاید ولیكن  Rلتكن         

انـه مقـاس جزئـي شـبه  Wالتـام شبه االولیة التامة والمقاسات شبه االولیة التامة إذ یقال عن المقاس الجزئـي الفعلـي المتغیـر 

مقاسـاً  شـبه اولـي تـام اذا  Mویسـمى  XWمقاس جزئي متغیر تام یـؤدي الـى ان  Xلكل  XXWاولي تام اذا كان 

اعطینــا الخــواص االساسـیة لهــذین المفهـومین وكــذلك درسـنا العالقــات بــین . مقــاس شـبه اولــي تـام (0)كـان المقــاس الجزئـي 

ذات ) المقاسـات(مع انواع اخـرى مـن المقاسـات الجزئیـة ) المقاسات شبه االولیة التامة(المقاسات الجزئیة شبه االولیة التامة 

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المقاســات شــبه االولیــة التامــة المقاســات الجزئیــة مــن الـــنمط  -ولیــة التامــةالمقاســات الجزئیــة شــبه اال :الكلمــات المفتاحـیـة

invarian المقاسات االولیة التامة  -التامة  

  

  

 

 

 

 

 

 

 

 

 

 

 

 

 



IBN AL- HAITHAM J. FOR PURE & APPL. S CI.         VOL.24 ( 2) 2011 

On Fully Semiprime Submodules and 
Fully Semip rime Modules 

 

I.M.A.Hadi and  B.N. Shihab 
Departme nt of Mathematics, Ibn-Al-Haitham, College of Education , Unive rsity 
of Baghdad  
 

Received in: 23, August, 2010   
Accepte d in: 9, November, 2010 

 
Abstract 
        Let R be a commutative ring with unity  and let M  be a unitary R-module. In this p aper 
we st udy  fully  semip rime submodules and fully  semip rime modules, where a p rop er fully  
invariant R-submodule W of M  is called fully  semip rime in M  if whenever XXW for all 
fully  invariant R-submodule X of M , imp lies XW. 
        M  is called fully  semip rime if (0) is a fully  semip rime submodule of M . We give basic 
p rop erties of these concepts. Also we st udy  the relationship s between fully  semip rime 
submodules (modules) and other related submodules (modules) resp ectively. 
Key words: Fully semip rime submodule, fully  semip rime modules, fully  invariant 
submodule, fully  p rime modules. 

 
Introduction 
        J.Abuihlail in [1], suggested the definition of fully  semip rime submodule and fully  

semip rime module as p rojects, where a p rop er fully  invarianr R-module W

M  is fully  

semip rime in M , if whenever XXW for all fully invariant R-submodules XM , it follows 
that XW. 
        An R-module M  is called fully  semip rime if whenever XX=0 for all fully  invariant R-

submodule X of M , it follows t hat X=0; that is M  is a fully semip rime module if 0

M  is fully 

semip rime. 
        Also for R-submodules X, Y  M , the internal p roduct XY is defined by  
{f(X):fHom(M ,Y)}. 
        Not ice that, if YM  is fully  invariant, then XYM  is also fully invariant, and if XM  
is fully  invariant, then XY XY. 
        The internal p roduct of submodules of a given module over an associative not 
necessarily  commutative ring was first  introduced by  Bican et.al, [2] to p resent the notion of 
p rime modules. The definition is modified in [3], where arbitrary submodules are replaced by  
fully  invariant ones. To avoid any p ossible confusion, such modules are referred to as fully  

p rime modules, where a p rop er fully invariant submodule W

M  is fully  p rime, if whenever 

XYW, for all fully  invariant R-submodule XM , YM , it follows t hat XW or YW. An 

R-module is called fully  p rime if (0) 

 M  is a fully  p rime submodule; that is whenever 

XY=(0) for all fully  invariant R-submodules XM , YM , it follows t hat X=(0) or Y=(0). 
        In this p aper we give a comprehensive study  of the concep ts fully semip rime submodules 
and fully  semip rime modules, where this p aper consists of two sections. In section one, we 
give the basic p rop erties of fully  semip rime submodules and fully  semip rime modules. 
Section two is devoted to st udy  the relationships between fully  semip rime modules and other 
modules such as uniform module, chained module, Z-regular module, quasi-Dedekind 
module, multip lication module and retractable module. 



 IBN AL- HAITHAM J. FOR PURE & APPL. S CI.         VOL.24 ( 2) 2011 

Next throughout this p aper, R is commutative ring with unity  and M  be a unitary  R-module. 
1- Fully S emiprime  S ubmodul es and Fully S emiprime  Modul es-Basic Results 
        In this section we st udy  the concep ts of fully semip rime submodules and fully semip rime 
modules which are introduced in [1] as p rojects. The concepts are generalizations of fully  
p rime submodules and fully p rime modules which are st udied in [3]. 
        We give characterizations about theses concepts and establishe some basic p rop erties 
about t hem. 
        We begin with the following definition. 
1.1 De fini tion, [2]: 
        Let K, L be two fully  invariant submodules of R-module M . Then  

KL={f(K): f:ML} 
 

        A p rop er submodule N of an R-module M  is called invariant if for each f
R

End(M) , 

f(N)N. M  is called fully invariant if every submodule of M  is invariant, see [4]. 
        Invariant submodule is called fully  invariant submodule by  some authors, see [3,p .14]. 
 
1.2 De fini tion, [3]: 
        A fully  invariant submodule N of an R-module M  is called fully  p rime if for all fully  
invariant submodules K and L of M  such that KLN, implies KN or LN. 
        Now, we give the following concept. 
1.3 De fini tion, [1]: 
        A fully invariant submodule N of an R-module M  is called fully semip rime if for all fully  
invariant submodules K of M  such that KKN, implies KN. 
We call M  fully  p rime (fully  semip rime) module if (0) is fully  p rime (fully  semip rime) 
submodule, see [1]. 
        Recall that:An R-module M  is said to be a prime module if annRM =annRN for every non-
zero submodule N of M , where annRM ={rR:rx=0 for each xM }, see [5]. 
        An R-module M  is called semip rime if and only if annRN is a semip rime ideal of R for 
each non-zero R-submodule N of M , see [6]. 
        Next, we give some remarks and examp les. 
1.4 Note : 
        Consider R as a left R-module, let I, J be two ideals of R. Then IJ=IJ, since every  ideal 
of R is a fully  invariant R-submodule. Thus I is a fully  semip rime ideal if and only if I is a 
semip rime ideal. 
1.5 Remarks and Examples: 
1. Let N be a submodule of an R-module M . If N is a fully  p rime submodule, then N is a fully  

semip rime submodule. 
2. If an R-module M  is fully  p rime module, then M  is p rime module. 
3. A submodule N of an R-module M  is semip rime, if N is fully semip rime submodule. 
proof: Sup p ose that rR, xM  such that r

2
xN. Let K=<rx>, K is a fully  invariant 

submodule, then KK={f(K): f:MK= <rx>}. 
Now, f(K)=f<rx>=r<f(x)><r

2
x>N. T hus KKN, implies KN, so rxN. 

4. If an R-module M  is a fully semip rime module, then M  is a semiprime module. 
5. Z6 as a Z-module is fully  semip rime, since for all submodule N, N(0), then NN(0). 

Thus Z 6 is a semiprime Z-module. But it is not  a fully p rime because it is not  p rime. 
6. Z4 as a Z-module is not semip rime module, since annZZ4=4Z is not a semiprime ideal of Z. 

Hence Z4 is not fully  semip rime. 
7. 6Z as a Z-submodule of Z is semiprime, so it is fully semip rime. 
8. Let R be an integral domain and K be the quotient field of R. Then K is an R-module and 

the zero R-submodule of K is the only semip rime in K. That is (0) is the only fully  
semip rime submodule in K, because if  N<K, N(0), N is fully  semip rime submodule, 
then N is semip rime, which is a contradiction. 



IBN AL- HAITHAM J. FOR PURE & APPL. S CI.         VOL.24 ( 2) 2011 

9. 
p

Z   as a Z-module has no fully  semip rime submodule. 

10. The homomorp hic image of a fully  semip rime module need not fully  semip rime 
module, for examp le: Z as a Z-module is a fully  p rime. Then Z is fully  semip rime. Now,                            
let :ZZ/(4) Z4. Z4 as a Z-module is not a fully semip rime. 

        Now, we have the following p rop osition. 
1.6 Proposi tion: 

       If N is fully  semip rime R-submodule of M , then [N
R
: K] is a  semip rime ideal of R for all 

N

 K. 

proof: We have N is fully  semip rime submodule, then N is semip rime submodule (by  
remarks and examp les (1.4),3). Then it is easy  to show that [N

R
: K] is semip rime ideal for all 

KN. 
        The following result is a consequence of p rop osition (1.6). 
1.7 Corollary: 
        If N is a fully  semip rime submodule of an R-module M , then [N

R
: M ] is a semip rime 

ideal. 
        The following result is a characterization of fully  semip rime submodule, but first the 

following lemma is needed. 
1.8 Lemma: 
        Let K be a fully invariant submodule of an R-module M . Then I(KK)IKIK for every  
ideal I of R. 
proof: IKIK={f(IK):f:MIK} 
                      = I{f(K):f:MIKK} 
Since KK=={g(K):g:MK}. It follows that IKIK  I(KK) 
1.9 Proposi tion: 
Let N be a submodule of an R-module M . Then N is a fully semip rime submodule if and only 
if [N

M
: I] is a fully semip rime submodule of M  for every  ideal I of R. 

proof: () sup p ose that N is a fully  semip rime submodule of M , let K be a fully  invariant 

submodule of M  such that KK[N
M
: I], imp lies I(KK)N, t hen by lemma (1.8), IKIK N, 

but N is a fully  semip rime submodule. Thus IKN. Therefore K[N
M
: I]. Hence [N

M
: I] is a 

fully  semip rime submodule. 

The converse follows by  taking I=R, because [N
M
: R]=N. 

        Next, we have the following p rop osition. 
1.10 Proposi tion: 
        Let N be a fully  invariant submodule of M . If N is a fully  semip rime submodule, then 
M /N is fully  semip rime module. 
proof: Let K/N be a fully  invariant submodule of M /N such that K/NK/N=N=OM/N. Then K 

is a fully  invariant submodules of M , with K
M
 KN [3, corollary  (1.1.21)]. Hence KN 

(since N is a fully  semip rime submodule). That is M / N
K

O
N
 . 

        The converse of prop osition (1.10) holds under the condition M  is self projective. 
1.11 Proposi tion: 
        Let M  be a self p rojective module and let N be a fully invariant submodule of M . If M /N 
is fully  semip rime, then N is a fully  semip rime submodule in M . 
 
 

 



IBN AL- HAITHAM J. FOR PURE & APPL. S CI.         VOL.24 ( 2) 2011 

proof: Let K be a fully  invariant submodule of N such that KKN. Then K'=
K N

N


 is a 

fully  invariant submodule of 
M

N
 [3, lemma (1.1.20)(ii)] and K'

M / N
 K'=0. Hence K'=0, that is 

K+NN and so KN. T hus N is a fully invariant semip rime submodule. 
 
        As an ap p lication of prop osition (1.10) we give the following corollary . 
1.12 Corollary: 
        Let :M M ' be an epimorp hisim. If ker  is a fully  semip rime submodule, then M ' 
is a fully semip rime module. 
proof: Since  is an epimorp hisim, then M /ker  M '. But ker is a fully  semip rime 
submodule, so by  p rop osition (1.10), M /ker is a fully semip rime module. This comp letes the 
p roof. 
        Before we give the next result, we introduce the following lemma. 
1.13 Lemma: 
        Let :MM ' be an isomorphism, where M , M ' be two R-modules, let K be a fully  

invariant submodule of M . Then (KK)(K)(K). 
proof: (KK) = ({f(K):f:MK}) 
                         = {f(K):f:MK} 
                         = {( f)(K):f:MK}. 
But 

- 1
:M 'M  is an isomorphism and KM , so there exists K'M ' such that 

- 1
(K')=K; 

that is (K)=K'. Hence (KK)={ f 
- 1

(K'):f:MK}. 

But 
1 fM ' M K
       , hence ( f 

- 1
):M 'K'=(K). Hence  

(KK)={ f 
- 1

(K'):  f 
 - 1

:M '(K)=K'}.  
Now (K)(K)=K'K'={h(K'):h:M 'K'=(K)}. It follows that (KK)(K)(K). 
        However, we get t he following p rop osition. 
1.14 Proposi tion: 
        If M  and M ' are two isomorphic R-modules, then M ' is fully  semip rime module if and 
only if M  is fully  semip rime module. 
proof: Let :MM ',  is an isomorp hism, and let L be a fully  invariant submodule of M ' 
such that LL=0. To p rove L=0. Let K=

- 1
(L), that is (K)=L. Then 

LL=(K)(K)(KK) (by  lemma (1.13). But (K)(K)=0 (since LL=0), imp lies 
(KK)=0. Since  is one to one, we have KK=0 and since M  is fully  semip rime, we get 
K=0. This imp lies (K)=(0)=0. T herefore L=(0) and hence M ' is a fully semip rime module. 
1.15 Proposi tion: 
        Let N and K be two fully  semip rime submodules of an R-module M . Then NK is a 
fully  semip rime submodule of M . 
proof: Let L be a fully  invariant submodule of M  such that LLNK. But NKK and 
NKN. LLK and LLN. Thus LK and LN (since K and L are fully  semip rime). 
Therefore LNK. Hence NK is a fully  semip rime submodule of M . 
        By  using the mathematical induction, we obtain the following result. 
1.16 Corollary: 
        The intersection of a finite collection of fully  semip rime submodules of an R-module is a 
fully  semip rime submodule. 
1.17 Proposi tion: 
        Let M  be an R-module. Then M  is fully  semip rime module if and only if for all mM , 
(m)(m)=(0), imp lies m=0. 
proof: () It is clear. 
 
 



IBN AL- HAITHAM J. FOR PURE & APPL. S CI.         VOL.24 ( 2) 2011 

To p rove the other side, let NN=(0). Sup p ose N(0), so t here exists mN, m(0) such that 
(m)N, implies (m)(m)NN=(0). Then (m)(m)=(0). Hence m=0, which is a contradiction. 
Thus N=0 and this comp letes the proof. 
 
        For our next p rop osition, the following lemma is needed. 
1.18 Lemma: 
        If M 1, M 2 are two R-modules and N1, N2 are two fully invariant submodules of M 1 and 
M 2 resp ectively, then (N1N1)(N2N2)(N1N2)  (N1N2). 
proof: (N1N1)(N2N2) = {(f(N1),g(N2)): f:M 1N1, g:M 2N2}. For any 
f:M 1N1, g:M 2N2, define h: M 1M 2  N1N2 by  h(x,y )=(f(x),g(y )) for all (x,y ) 
M 1M 2. It is clear that h is well defined homomorp hism. Now, h(N1N2)=(f(N1),g(N2)). But 
(N1N2)  (N1N2)={(N1N2):: M 1M 2 N1N2}.It  follows that h(N1N2)=               
 (f(N1),g(N2)) be in (N1N2)  (N1N2). Thus we have 
(N1N1)(N2N2)(N1N2)(N1N2). 
1.19 Proposi tion: 
        Let M 1, M 2 be two R-modules and M =M 1M 2 such that annM 1+annM 2=R. If N1 and N2 
are fully  semip rime R-submodules of M 1 and M 2 resp ectively, then N1N2 is also fully  
semip rime. 
proof: to p rove N1N2 is fully  semip rime, let N be a fully  invariant submodule of M  such 
that NNN1N2. But N=KL for some KM 1, LM 2, by  [4,theorem (4.2),ch.1]. Then 
(KL) KLN1N2. Thus by  lemma (1.15), we get (KK)  (LL)  N1N2 so KKN1, 
LLN2. But N1 and N2 are fully  semip rime, then KN1, LN2. Thus KLN1N2 and 
hence N1N2 is fully  semip rime. 
        The converse of prop osition (1.16) holds if (N1N2)  (N1N2)= (N1N1)(N2N2). 
1.20 Proposi tion: 
        Let M 1, M 2 be two R-modules, let (N1N2)  (N1N2)= (N1N1)(N2N2) for each 
N1M 1, N2M 2. Then N1N2 is fully  semip rime implies, N1 and N2 are fully  semip rime. 
proof: Let KM 1, LM 2 such that KKN1 and LLN2. Hence (KK)(LL)N1N2. It 
follows that (KL)(KL)N1N2. Since N1N2 is fully  semip rime, KLN1N2. Hence 
KN1 and LN2. Thus K and L are fully  semip rime. 
1.21 Proposi tion: 
        Let M 1, M 2 be two R-modules such that annM 1+annM 2=R. Then M 1M 2 is fully  
semip rime module if and only if M 1 and M 2 are fully  semip rime modules. 
proof: Let N be a fully  invariant submodule of M 1M 2.If NN=0, to p rove N=0. Since                
N= N1N2 [4,theorem (2.4)], then NN=(N1N2)  (N1N2). By  lemma (1.15).  
(N1N1)(N2N2)(N1N2)  (N1N2)=(0)(0). Then (N1N1)(N2N2)=(0)(0), imp lies 
N1N1=(0) and N2N2=(0). Therefore N1=(0) and N2=(0) (because M 1, M 2 are fully  
semip rime). Thus N1N2=(0)(0) and hence M  is fully  semip rime. 
Conversely , sup p ose that M 1M 2 is a fully  semip rime module, let N1 be a fully  invariant 
submodule of M 1 such that N1N1=(0). We can show that t here is one to one corresp ondence 
between N1N1 and (N1(0))(N1(0)) as follows. For any  f:M 1N1, f(N1) N1N1, f can 

be extended to f : M 1M 2 N1(0) by  f
 (m1,m2)=(f(m1),0) for all (m1,m2) M 1M 2 it is 

clear that f (N1(0))=f(N1)(0), hence if f(N1)=0, then f
 (N1(0))=(0). 

Similarly , if g: M 1M 2 N1(0), g(N1(0))(N1(0))(N1(0)), then we define 
g :M 1N1 by  g (m1)=g(m1,0),  mM 1, hence g (N1)=g(N1(0)) and so g (N1) N1N1. 

Thus N1N1=0  (N1(0))(N1(0))=0. It follows that (N1(0))(N1(0))=0, and hence 
N1(0)=0. T hus N1=0. 
        Similarly  if N2 is an invariant submodule of M 2 such that, N2N2=0, imp lies N2=(0) and 
hence M 2 is fully  semip rime. 
        Next, we p rove the following. 



IBN AL- HAITHAM J. FOR PURE & APPL. S CI.         VOL.24 ( 2) 2011 

1.22 Proposi tion: 
        Let M =M 1M 2 be a direct sum of two R-modules M 1 and M 2 such that 
annM 1+annM 2=R. If L1 is a fully  semip rime submodule of M 1. Then L1M 2 is a fully  
semip rime submodule of M . 
proof: Let N be a fully  invariant submodule of M = M 1M 2 such that NNLM 2. By                       
[4,theorem (4.2),ch.1], there exists N1M 1, N2M 2 such that N=N1N2. Hence  
(N1N2)  (N1N2)L1M 2. But (N1N1)(N2N2)(N1N2)  (N1N2) by  lemma (1.15). 
Therefore (N1N1)(N2N2)L1M 2. It is clear that N1N1L1, but L1 is fully  semip rime 
submodule, then N1L1. Thus N1N2L1M 2. Therefore L1M 2 is fully  semip rime 
submodule of M . 
2- The  Relationshi ps Between Fully S emiprime  Modul es and Ce rtain Types of 
Modul es 
        In this section, we establishe some relationship s between fully  semip rime submodules 
and some ty p e of modules. 
        Recall that an R-module M  is said to be uniform module if every non-zero submodule of 
M  is essential see [7], where a submodule N of an R-module M  is essential p rovided that 
NK0 for every  non-zero submodule K of M , see [7]. 
        Hence, we have the following p rop osition. 
2.1 Proposi tion: 
        Let M  be a uniform R-module. Then M  is a fully semip rime module if and only  if M  is a 
fully  p rime module. 
proof: Sup p ose that M  is a fully  semip rime R-module, let K, L be two fully  invariant 
submodules of M  such that KL=(0). Assume that K(0), L(0). Then KL(0) (since M  is 
uniform). On the other hand KLK, KLL. Also, note that K, L are fully  invariant, 
imp lies KL is fully  invariant. Then (KL)(KL)KL=(0). Thus (KL)(KL)(0). 
But M  is fully  semip rime so KL=(0) which is a contradiction. Thus either K=(0) or L=(0). 
Then M  is fully  p rime. 
The converse is obvious. 
        The following results are consequences of p rop osition (2.1), but first we need to recall 
the following definition. 
        An R-module M  is said to be chained module if and only if every  non-emp ty  set of 
submodule of M  is ordered by  inclusion, [8]. 
        Hence, we have the following consequence of (2.1). 
2.2 Corollary: 
        Let M  be a chained R-module. Then M  is fully  semip rime if and only  if M  is fully  p rime. 
proof: It is known that every  chained R-module M  is uniform, then the result follows from 
p rop osition (2.1). 
        An R-module M  is called quasi-Dedekind if every  submodule N of M  is quasi-invertible 
[9,definition (1.1), ch.2], where a submodule N of M  is called quasi-invertible if 
Hom(M /N,M )=0 [9,definition (1.1),ch.1]. 
2.3 Corollary: 
        If M  is a uniform fully semip rime R-module, then M  is a quasi-Dedekind R-module. 
proof: From p rop osition (2.1) and remark (1.4), we get M  is a p rime module. Thus by  
[6,theorem (3.11),ch.3], M  is quasi-Dedekind. 
        As an ap p lication of corollary (2.3), we give the following examp le. 
2.4 Example: 
        Z as a Z-module is uniform and fully p rime module, also it is quasi-Dedekind. 
        Recall that an R-module M  is called Z-regular if and only  if each cyclic submodule of M  
is p rojective direct summand of M . Equivalently  if for each aM , fM *=Hom(M ,R) such 
that a=f(a)a, [10]. 
        By  using this concept, we give the following p rop osition. 
 



IBN AL- HAITHAM J. FOR PURE & APPL. S CI.         VOL.24 ( 2) 2011 

2.5 Proposi tion: 
        If M  is a Z-regular R-module, then M  is fully  semip rime module. 
proof: Let K be a fully  invariant submodule of M  such that KK=(0). Sup p ose K0. Then 
there exists xK, x0. Since M  is Z-regular, then there exists f:MR such that x=f(x)x. 

Define g:RK by  g(r)=rx for each rR. Then f gM R K  , so g f:MK such 
that (g f)(x)=g(f(x))=f(x)x0. Thus 0g f and hence KK0 which is a contradiction. Thus 
K=0. 
        Now, we have the following p rop osition. 
2.6 Proposi tion: 
        Let I be an ideal of R. Then the following st atements are equivalent: 
1. I is a fully semip rime submodule in R. 
2. I is a semiprime submodule of R. 
3. R/I is a semip rime ring. 
proof: 12, let J be an ideal of R such that J

2
I. Then JJI (by  note (1.4)). Thus JI, by  

(1). 
21, let JJI. Then J

2
I (by  note (1.4)) implies JI. 

23, it is obvious. 
2.7 Note s, [11]: 
1. For any  R-module M  and for any  ideals I, J of R. Then (IM )(JM )(IJ)M . And t he reverse 

inclusion is also easily established p rovided M  is self generator, t hat is T rac(M ,JM )=JM . 
2. The multiplication modules over commutative ring are self generator whose submodules 

are fully  invariant. 
        Recall that an R-module M  is called multiplication if for every  submodule N of M , there 
exists an ideal I of R such that N=IM , equivalently  for every  submodule N of M , 
N=[N

R
: M ]M , [7]. 

2.9 Corollary: 
        Let M  be a multiplication R-module. Then (IM )(JM )=(IJ)M . 
2.10 Proposi tion: 
        Let M  be a faithful multip lication R-module M . Then M  is fully  semip rime if and only  if 
R is semip rime ring. 
proof: Let I be a prop er ideal of R such that I

2
=0, let N=IM , N is a fully invariant submodule 

of M . Then NN=I
2
M =0 by  corollary  (2.9), so NN=0. But M  is fully  semip rime, imp lies 

N=0 and so IM =0. Then Iann(M )=0, since M  is faithful. Thus I=0. 
Conversely , let N be a fully invariant submodule and NN=0, since, N=IM  for some ideal I of 
R. Implies that NN=I

2
M , so I

2
M =0. Thus I

2
ann(M )=0. Therefore I=0 (since R is 

semip rime), and hence N=IM =(0). 
        Now, we have the following p rop osition. 
2.11 Proposi tion: 
        Let M  be a multiplication R-module with annR(M ) is semip rime. Then M  is fully  
semip rime module. 
proof: Let N be a fully invariant submodule of M  such that NN=0. But N=IM  for some ideal 
I of M , since M   is multiplication module. Hence IMIM =0 which imp lies that I

2
M =0. Thus 

I
2
annR(M ) and hence IannR(M ).Then IM =0. Therefore N=(0). T his comp letes the proof. 

        The following is an immediate consequence of p rop osition (2.11). 
2.12 Corollary: 
        Let M  be a multiplication R-module. Then M  is a semip rime R-module if and only if 
annR(M ) is semip rime ideal. 
proof: Assume that M  is semip rime R-module. Then (0) is a semip rime submodule, so 
(O:M )=annR(M ) is semip rime ideal. 
Conversely , let annR(M ) be a semiprime ideal. Then by  p rop osition (2.11), we get M  is fully  
semip rime and hence M  is semip rime by remarks (1.4), see [4]. 



      IBN AL- HAITHAM J. FOR PURE & APPL. S CI.         VOL.24 ( 2) 2011 

  By  p rop osition (2.11), we have the following. 
2.13 Corollary: 
        Let M  be a faithful multip lication R-module. The following are equivalent: 
1. M  is fully  semip rime. 
2. R is semip rime ring. 
3. M  is semip rime module. 
proof: 12 it is obvious. 
23 R is semip rime ring  (0) is semip rime ideal  annR(M ) is semip rime ideal  M  is 
semip rime module (by  p rop osition (2.11)). 
        An R-module M  is called coprime if for every  p rop er submodule N of M , 
annR(M )=annR(M /N), [2]. 
        Recall that an R-module M  is called a scalar module if for all fEndR(M );f0, there 
exists rR, r0 such that f(m)=rm, see [12]. 
        By  using these concep ts, we can prove the following. 
2.14 Proposi tion: 
        Let M  be a coprime scalar and fully semip rime R-module. Then M  is simple. 
proof: Assume that N be a p rop er R-submodule of M , let f:MN. Then there exists rR 
such that f(m)=rm for all mM  (since M  is a scalar module). Therefore f(m)=rm  N, imp lies 

r[N
R
: M ]. But M  is a cop rime module, so annRM =[N

R
: M ]. Thus rannRM . Then rM =0. Thus 

f(N)=rN=0. Hence {f(N):f:MN}=0. Then NN=0 and since N is a fully  semip rime, we 
get N=0. 
        Recall that an R-module M  is called retractable if Hom(M ,N)0, for every  non-zero 
submodule N of M , see [13]. 
        Now, we st ate and prove the following result. 
2.15 The orem: 
        Let M  be an R-module, if M  is fully  semip rime, then M  is retractable and the converse is 
true if EndR(M ) is semip rime. 
proof: Assume that M  is fully  semip rime and let N be a submodule of M , N0. Assume M  is 
not retractable, that is Hom(M ,N)=0, then NN=0 and so N=0 which is a contradiction. Hence 
M  is retractable. 
Conversely , if M  is retractable and EndR(M ) is semip rime , let N be a fully  invariant 
submodule of M  such that NN=0. Sup p ose N0. Since M  is retractable, then there exists 
f:MN, f0. But 0= NN={f(N):f:MN}=0, hence f(N)=0. Then for any mM , f

 

2
(m)=f(f(m))=0. Then  f

 2
=0 and hence f=0 which is a contradiction. Thus N=0. Therefore M  

is a fully semip rime R-module. 
        We end this section by the following corollary . 
2.16 Corollary: 
        Let M  be a finitely  generated multiplication R-module. The following are equivalent: 
1. M  is retractable and EndR(M ) is semip rime. 
2. M  is fully  semip rime. 
3. M  is semip rime. 
4. annR(M ) is semip rime. 
proof: It is obvious. 
 

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