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         

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

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

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

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

  .العالقة معهما

ـات الجزئیــة شــبه اال :الكلمــات المفتاحـیـة ـةالمقاسـ ـة مــن الـــنمط  -ولیــة التامـ ـات الجزئیـ المقاســات شــبه االولیــة التامــة المقاسـ

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

On Fully Semiprime Submodules and 
Fully Semiprime Modules 

 

I.M.A.Hadi and  B.N. Shihab 
Department of Mathematics, Ibn-Al-Haitham, College of Education , University 
of Baghdad  
 

Received in: 23, August, 2010   
Accepted in: 9, November, 2010 

 
Abstract 
        Let R be a commutative ring with unity  and let M be a unitary R-module. In this paper 
we study fully semiprime submodules and fully semiprime modules, where a proper fully 
invariant R-submodule W of M is called fully semiprime in M if whenever XXW for all 
fully invariant R-submodule X of M , implies XW. 
        M is called fully semiprime if (0) is a fully semiprime submodule of M. We give basic 
properties of these concepts. Also we study the relationships between fully semiprime 
submodules (modules) and other related submodules (modules) respectively. 
Key words: Fully semiprime submodule, fully semiprime modules, fully invariant 
submodule, fully prime modules. 

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

semiprime module as projects, where a proper fully invarianr R-module W

M is fully 

semiprime 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 semiprime if whenever XX=0 for all fully invariant R-

submodule X of M , it follows that X=0; that is M  is a fully semiprime module if 0

M is fully 

semiprime. 
        Also for R-submodules X, Y  M, the internal product XY is defined by 
{f(X):fHom(M,Y)}. 
        Notice 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 product of submodules of a given module over an associative not 
necessarily commutative ring was first introduced by Bican et.al, [2] to present the notion of 
prime modules. The definition is modified in [3], where arbitrary submodules are replaced by 
fully invariant ones. To avoid any possible confusion, such modules are referred to as fully 

prime modules, where a proper fully invariant submodule W

M is fully prime, if whenever 

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

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

XY=(0) for all fully invariant R-submodules XM, YM, it follows that X=(0) or Y=(0). 
        In this paper we give a comprehensive study of the concepts fully semiprime submodules 
and fully semiprime modules, where this paper consists of two sections. In section one, we 
give the basic properties of fully semiprime submodules and fully semiprime modules. 
Section two is devoted to study the relationships between fully semiprime modules and other 
modules such as uniform module, chained module, Z-regular module, quasi-Dedekind 
module, multiplication module and retractable module. 



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

Next throughout this paper, R is commutative ring with unity  and M be a unitary R-module. 
1- Fully Semiprime Submodules and Fully Semiprime Modules-Basic Results 
        In this section we study the concepts of fully semiprime submodules and fully semiprime 
modules which are introduced in [1] as projects. The concepts are generalizations of fully 
prime submodules and fully prime modules which are studied in [3]. 
        We give characterizations about theses concepts and establishe some basic properties 
about them. 
        We begin with the following definition. 
1.1 Definition, [2]: 
        Let K, L be two fully invariant submodules of R-module M. Then  

KL={f(K): f:ML} 
 

        A proper 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 Definition, [3]: 
        A fully invariant submodule N of an R-module M is called fully prime 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 Definition, [1]: 
        A fully invariant submodule N of an R-module M is called fully semiprime if for all fully 
invariant submodules K of M such that KKN, implies KN. 
We call M fully prime (fully semiprime) module if (0) is fully prime (fully semiprime) 
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 semiprime if and only if annRN is a semiprime ideal of R for 
each non-zero R-submodule N of M , see [6]. 
        Next, we give some remarks and examples. 
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 semiprime ideal if and only if I is a 
semiprime ideal. 
1.5 Remarks and Examples: 
1. Let N be a submodule of an R-module M. If N is a fully prime submodule, then N is a fully 

semiprime submodule. 
2. If an R-module M is fully prime module, then M is prime module. 
3. A submodule N of an R-module M is semiprime, if N is fully semiprime submodule. 
proof: Suppose 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

2x>N. Thus KKN, implies KN, so rxN. 
4. If an R-module M is a fully semiprime module, then M is a semiprime module. 
5. Z6 as a Z-module is fully semiprime, since for all submodule N, N(0), then NN(0). 

Thus Z6 is a semiprime Z-module. But it is not a fully prime because it is not prime. 
6. Z4 as a Z-module is not semiprime module, since annZZ4=4Z is not a semiprime ideal of Z. 

Hence Z4 is not fully semiprime. 
7. 6Z as a Z-submodule of Z is semiprime, so it is fully semiprime. 
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 semiprime in K. That is (0) is the only fully 
semiprime submodule in K, because if  N<K, N(0), N is fully semiprime submodule, 
then N is semiprime, which is a contradiction. 



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

9. 
p

Z   as a Z-module has no fully semiprime submodule. 

10. The homomorphic image of a fully semiprime module need not fully semiprime 
module, for example: Z as a Z-module is a fully prime. Then Z is fully semiprime. Now,                            
let :ZZ/(4) Z4. Z4 as a Z-module is not a fully semiprime. 

        Now, we have the following proposition. 
1.6 Proposition: 

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

N

  K. 

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

R
: K] is semiprime ideal for all 

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

R
: M] is a semiprime 

ideal. 
        The following result is a characterization of fully semiprime 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 Proposition: 
Let N be a submodule of an R-module M. Then N is a fully semiprime submodule if and only 
if [N

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

proof: () suppose that N is a fully semiprime submodule of M, let K be a fully invariant 

submodule of M such that KK[N
M
: I], implies I(KK)N, then by lemma (1.8), IKIK N, 

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

M
: I] is a 

fully semiprime submodule. 

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

        Next, we have the following proposition. 
1.10 Proposition: 
        Let N be a fully invariant submodule of M. If N is a fully semiprime submodule, then 
M/N is fully semiprime 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 semiprime submodule). That is M/ N

K
O

N
 . 

        The converse of proposition (1.10) holds under the condition M is self projective. 
1.11 Proposition: 
        Let M be a self projective module and let N be a fully invariant submodule of M. If M/N 
is fully semiprime, then N is a fully semiprime submodule in M. 
 
 

 



IBN AL- HAITHAM J. FOR PURE & APPL. SCI.         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. Thus N is a fully invariant semiprime submodule. 
 
        As an application of proposition (1.10) we give the following corollary. 
1.12 Corollary: 
        Let :M M' be an epimorphisim. If ker  is a fully semiprime submodule, then M' 
is a fully semiprime module. 
proof: Since  is an epimorphisim, then M/ker M'. But ker is a fully semiprime 
submodule, so by  proposition (1.10), M/ker is a fully semiprime module. This completes the 
proof. 
        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 the following proposition. 
1.14 Proposition: 
        If M and M' are two isomorphic R-modules, then M' is fully semiprime module if and 
only if M is fully semiprime module. 
proof: Let :MM',  is an isomorphism, and let L be a fully invariant submodule of M' 
such that LL=0. To prove 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), implies 
(KK)=0. Since  is one to one, we have KK=0 and since M is fully semiprime, we get 
K=0. This implies (K)=(0)=0. Therefore L=(0) and hence M' is a fully semiprime module. 
1.15 Proposition: 
        Let N and K be two fully semiprime submodules of an R-module M. Then NK is a 
fully semiprime 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 semiprime). 
Therefore LNK. Hence NK is a fully semiprime submodule of M. 
        By using the mathematical induction, we obtain the following result. 
1.16 Corollary: 
        The intersection of a finite collection of fully semiprime submodules of an R-module is a 
fully semiprime submodule. 
1.17 Proposition: 
        Let M be an R-module. Then M is fully semiprime module if and only if for all mM, 
(m)(m)=(0), implies m=0. 
proof: () It is clear. 
 
 



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

To prove the other side, let NN=(0). Suppose N(0), so there 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 completes the proof. 
 
        For our next proposition, 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 respectively, then (N1N1)(N2N2)(N1N2)   (N1N2). 
proof: (N1N1)(N2N2) = {(f(N1),g(N2)): f:M1N1, g:M2N2}. For any 
f:M1N1, g:M2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 homomorphism. 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 Proposition: 
        Let M 1, M 2 be two R-modules and M=M 1M 2 such that annM 1+annM2=R. If N1 and N2 
are fully semiprime R-submodules of M 1 and M 2 respectively, then N1N2 is also fully 
semiprime. 
proof: to prove N1N2 is fully semiprime, 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 semiprime, then KN1, LN2. Thus KLN1N2 and 
hence N1N2 is fully semiprime. 
        The converse of proposition (1.16) holds if (N1N2)   (N1N2)= (N1N1)(N2N2). 
1.20 Proposition: 
        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 semiprime implies, N1 and N2 are fully semiprime. 
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 semiprime, KLN1N2. Hence 
KN1 and LN2. Thus K and L are fully semiprime. 
1.21 Proposition: 
        Let M 1, M 2 be two R-modules such that annM1+annM2=R. Then M 1M 2 is fully 
semiprime module if and only if M1 and M2 are fully semiprime modules. 
proof: Let N be a fully invariant submodule of M 1M 2.If NN=0, to prove 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), implies 
N1N1=(0) and N2N2=(0). Therefore N1=(0) and N2=(0) (because M 1, M 2 are fully 
semiprime). Thus N1N2=(0)(0) and hence M is fully semiprime. 
Conversely, suppose that M 1M 2 is a fully semiprime module, let N1 be a fully invariant 
submodule of M 1 such that N1N1=(0). We can show that there is one to one correspondence 
between N1N1 and (N1(0))(N1(0)) as follows. For any f:M1N1, f(N1) N1N1, f can 

be extended to f : M1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 :M1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. Thus N1=0. 
        Similarly if N2 is an invariant submodule of M 2 such that, N2N2=0, implies N2=(0) and 
hence M2 is fully semiprime. 
        Next, we prove the following. 



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

1.22 Proposition: 
        Let M=M 1M 2 be a direct sum of two R-modules M 1 and M 2 such that 
annM1+annM2=R. If L1 is a fully semiprime submodule of M 1. Then L1M 2 is a fully 
semiprime 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 semiprime 
submodule, then N1L1. Thus N1N2L1M 2. Therefore L1M 2 is fully semiprime 
submodule of M. 
2- The Relationships Between Fully Semiprime Modules and Certain Types of 
Modules 
        In this section, we establishe some relationships between fully semiprime submodules 
and some type 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 provided that 
NK0 for every non-zero submodule K of M, see [7]. 
        Hence, we have the following proposition. 
2.1 Proposition: 
        Let M be a uniform R-module. Then M is a fully semiprime module if and only if M is a 
fully prime module. 
proof: Suppose that M is a fully semiprime 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, 
implies KL is fully invariant. Then (KL)(KL)KL=(0). Thus (KL)(KL)(0). 
But M is fully semiprime so KL=(0) which is a contradiction. Thus either K=(0) or L=(0). 
Then M is fully prime. 
The converse is obvious. 
        The following results are consequences of proposition (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-empty 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 semiprime if and only if M is fully prime. 
proof: It is known that every chained R-module M is uniform, then the result follows from 
proposition (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 semiprime R-module, then M is a quasi-Dedekind R-module. 
proof: From proposition (2.1) and remark (1.4), we get M is a prime module. Thus by 
[6,theorem (3.11),ch.3], M is quasi-Dedekind. 
        As an application of corollary (2.3), we give the following example. 
2.4 Example: 
        Z as a Z-module is uniform and fully prime 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 projective 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 proposition. 
 



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

2.5 Proposition: 
        If M is a Z-regular R-module, then M is fully semiprime module. 
proof: Let K be a fully invariant submodule of M such that KK=(0). Suppose 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 proposition. 
2.6 Proposition: 
        Let I be an ideal of R. Then the following statements are equivalent: 
1. I is a fully semiprime submodule in R. 
2. I is a semiprime submodule of R. 
3. R/I is a semiprime 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 Notes, [11]: 
1. For any R-module M and for any ideals I, J of R. Then (IM)(JM )(IJ)M . And the reverse 

inclusion is also easily established provided M is self generator, that is Trac(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 Proposition: 
        Let M be a faithful multiplication R-module M. Then M is fully semiprime if and only if 
R is semiprime ring. 
proof: Let I be a proper ideal of R such that I

2=0, let N=IM, N is a fully invariant submodule 
of M. Then NN=I2M=0 by corollary (2.9), so NN=0. But M is fully semiprime, implies 
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 

semiprime), and hence N=IM=(0). 
        Now, we have the following proposition. 
2.11 Proposition: 
        Let M be a multiplication R-module with annR(M) is semiprime. Then M is fully 
semiprime 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 implies that I

2M=0. Thus 
I2annR(M) and hence IannR(M).Then IM=0. Therefore N=(0). This completes the proof. 
        The following is an immediate consequence of proposition (2.11). 
2.12 Corollary: 
        Let M be a multiplication R-module. Then M is a semiprime R-module if and only if 
annR(M) is semiprime ideal. 
proof: Assume that M is semiprime R-module. Then (0) is a semiprime submodule, so 
(O:M)=annR(M) is semiprime ideal. 
Conversely, let annR(M) be a semiprime ideal. Then by proposition (2.11), we get M  is fully 
semiprime and hence M is semiprime by remarks (1.4), see [4]. 



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

  By proposition (2.11), we have the following. 
2.13 Corollary: 
        Let M be a faithful multiplication R-module. The following are equivalent: 
1. M is fully semiprime. 
2. R is semiprime ring. 
3. M is semiprime module. 
proof: 12 it is obvious. 
23 R is semiprime ring  (0) is semiprime ideal  annR(M) is semiprime ideal  M is 
semiprime module (by proposition (2.11)). 
        An R-module M is called coprime if for every proper 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 concepts, we can prove the following. 
2.14 Proposition: 
        Let M be a coprime scalar and fully semiprime R-module. Then M is simple. 
proof: Assume that N be a proper 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, implies 

r[N
R
: M]. But M is a coprime 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 semiprime, 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 state and prove the following result. 
2.15 Theorem: 
        Let M be an R-module, if M is fully semiprime, then M is retractable and the converse is 
true if EndR(M) is semiprime. 
proof: Assume that M is fully semiprime 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 semiprime , let N be a fully invariant 
submodule of M such that NN=0. Suppose 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 semiprime 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 semiprime. 
2. M is fully semiprime. 
3. M is semiprime. 
4. annR(M) is semiprime. 
proof: It is obvious. 
 

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