إبن الهيثم للعلوم الصرفة و التطبيقيةمجلة 

 2012 السنة 25 المجلد 2 العدد

Ibn Al-Haitham Journal for Pure and Applied S cience  

 No. 2 Vol. 25 Year 2012 

Purely co-Hopfian Modules 
Z. T.Salman     
Department Of Mathematics , College of science , University of Baghdad            
Received in: 22 Septembre 2011  Accepted in: 11January  2012 
 
Abstract 
       Let R be an associative ring with identity and M a non – zero unitary R-module.In this 
paper we introduce the definition of purely co-Hopfian module, where an R-module M is said 
to be purely co-Hopfian if for any monomorphism f ΠEnd (M), Imf is pure in M and we give  
some properties of this kind of modules. 
 
Keywords: co-Hopfian module, semi co-Hopfian module, purely co-Hopfian module 
 
 Introduction and Preliminaries  
       Let R be an associative ring with identity and M a non – zero unitary R – module, Recall 
that a module M is called co-Hopfian if any injective endomorphism of M is an isomorphism 
[1].A module M is called semi co-Hopfian if any injective endomorphism of M has a direct 
summand image that means any injective endomorphism of M splits [1].A ring R is semi co-
Hopfian if R is semi co-Hopfian R - module. Clearly, any co-Hopfian is semi co-Hopfian but 
the converse is not true in general as, for example  M = Q N = Q ≈  Q ≈  . . . ,as Z-module is 
semi co-Hopfian but it is not co-Hopfian [1]. A submodule N of M is called pure if IM∩N=IN 
for each ideal of R,[8].It is well–known every direct summand of a module M is pure 
submodule but the converse is not true in general [2].This leads us to introduce the following 
concept, namely purely co-Hopfian module. 
 
Definition 1.1 
        An R- module M is called purely co-Hopfian if for any monomorphism  f ΠEnd (M), 
Imf is pure in M. 
 
Remarks and examples 1.2 
1. Every semi co-Hopfian module is purely co-Hopfian. 
2. Every F- regular module M is purely co-Hopfian, where M is F- regular if every 

submodule of M is pure,[3]. 
3. Every semi simple R-module is purely co-Hopfian. 
4. If M is pure simple  ( that means M has only two pure submodules 0 ,M )   [ 2 ], then M is 

purely co-Hopfian. 
 

Icmma 1.3  
          The following are equivalent for an R-module M:  
1. M is purely co-Hopfian. 
2. Any submodule N of M such that N @  M, N is pure in M. 
Proof   (1)→(2) 
          Let N ≤ M, N @  M. Then there exists  α : M → N, α is an isomorphism. Hence 
M aææÆ  N iææÆ M where i : N → M  is the inclusion map , and this implies iоa Œ 
End ( M ) , i оa   is monomorphism. So (i о α ) ( M ) is pure in M . Thus i (α (M)) = i (N) = 
N is pure in M. 
(2)→(1):  let f  Œ End (M), f is monomorphism. Hence f(M) @  M and so by (2), f (M) is 
pure in M. 



 
 
 

 إبن الهيثم للعلوم الصرفة و التطبيقيةمجلة 

 2012 السنة 25 المجلد 2 العدد

Ibn Al-Haitham Journal for Pure and Applied S cience  

 No. 2 Vol. 25 Year 2012 

 
Proposition 1.4  
          The following are equivalent for a ring R  
1. R is purely co-Hopfian. 
2. R is semi co-Hopfian.  
Proof   (1)→(2) 
          Let f: R → R, f is R-monomorphism. Hence f (R) = < a > for some             a π  0 Œ R. 
Since R is purely co-Hopfian,  < a > is pure ideal at R , hence      < a > = < a2 > ( since < a > 

I  <a >  =  < a > < a > ) .Thus a = ra2 for some r ΠR, this implies ra is idempotent and< a > 
= < ra >. It follows that < a > is a direct summand . 
The proof of the part (2)→(1) is clear. 
 
By combining proposition 1.4 and proposition 2.3 from [1] we get the following result. 
 
Corollary 1.5 
       The following are equivalent for any a ring R :  
1. R is purely co-Hopfian. 
2. R is semi co-Hopfian. 
3. ann ( a ) = 0 , a ΠR then < a > is a direct summand . 
4. If ann ( a ) = 0 , a ΠR then < a > = R .  
5. Every R – isomorphism < a > → R, a Œ R, extends to R.  
Proof 
(1) ↔ (2): see proposition 1.4 
(2) ↔(3) ↔(4) ↔(5): (see proposition 2.3), [1].  
 
 
Corollary 1.6 
          If R is a ring with two idempotent 0,1 then the following statement are equivalent : -  
1. R is co-Hopfian. 
2. R is semi co-Hopfian. 
3. R is purely co-Hopfian. 
Proof  
(1) → (2): it is clear  
(2) ↔ (3) :by proposition 1.4  
(3)→(2): Let f : R → R , f is monomorphism then  f ( R ) = < a > for some a Œ R , a π  0, but 
I = < a >  is a direct summand of R (since R is Semi co-Hopfian) then < a > is generated by 
idempotent. Since a π  0, hence a = 1 and < a > = R.Thus f is onto and we get R is co-
Hopfian. 
 
          Recall that module M has C2 if for any submodule N of M which is isomorphic to a 
direct summand of M, is a direct summand of M [4]. 
 
Corollary 1.7 
          If R is a ring only idempotent 0 and 1 the following equivalent:  
1. R has C2. 
2. R is co-Hopfian. 
3. R is purely co-Hopfian.  
4. R is semi co-Hopfian. 
Proof     (1) → (2)  
       let f: R → R be monomorphism. To prove that R is co-Hopfian, we must prove f is an 
isomorphism.Since f is monomorphism,f ( R ) @  R . But R is C2 by (1) and R is direct 



 
 
 

 إبن الهيثم للعلوم الصرفة و التطبيقيةمجلة 

 2012 السنة 25 المجلد 2 العدد

Ibn Al-Haitham Journal for Pure and Applied S cience  

 No. 2 Vol. 25 Year 2012 

summand of R, hence f(R) is direct summand of R.It follows that f (R) is generated by 
idempotent.Since R has only 2 – idempotent namely 0 , 1 and f ( R ) π  0 ,then f ( R ) = < 1 > 
thus f ( R ) = R and so that f is an isomorphism .   
(2) → (3 :It is clear. 
(3) → (4):It follows by proposition (1.4) . 
(4) → (1): It follows by proposition 2.4 [1] . 
Corollary 1.8  
          Let R be an integral domain. Then the following are equivalent:  
1. R is co-Hopfian.  
2. R is semi co-Hopfian. 
3. R is purely co-Hopfian. 
4.  R is field. 
Proof  
          (1) ↔ (2) ↔ (3) :It follows by corollary 1.6   
(1) → (4) :Let a Œ R , a π  0 then ann ( a ) = 0  since R is an  integral domain .  By corollary 
1.5, < a > = R. Hence a is an invertible element.Then R is a field. 
(4) → (1) :Since R is a field, R has only two ideals namely R, (0). Hence for any f: R → R, f 
is R – monomorphism f (R) π  0. Hence f (R) = R.Thus f is onto then R is co-Hopfian.  
 
Proposition 1.9  
          Any direct summand of purely co-Hopfian module is purely co-Hopfian.  
Proof  
     Let N be a direct summand of M, so M = N ≈  A for some submodule A of M .Let f: N → 
N be monomorphism. Define g : M →M by g(n+a) = f ( n )+a where n Œ N , a ŒA it is easy 
to see that g is monomorphism Hence g (M) = f (A) ≈  N . Since M is purely co-Hopfian, g 
(M) is pure in M.To prove f (N) pure in N,let I be any ideal of R , 
IM ∩ g ( M ) = I g ( M ) , 
I ( N ≈  A) ∩  ( f ( N) ≈  A ) = I ( f ( N ) ≈  A ) , 
(IN ≈  I A)  ∩ ( f ( N ) ≈  A ) =(IN ∩ f (N) ) ≈  (IA ∩ A)= I f (N) ≈  IA , 
(IN ∩ f ( N )) ≈  IA = If ( N ) ≈  IA, IN ∩ f ( N ) = If ( N ) .  
Thus f(N)is pure in N and so N is purely copfian. 
 
      Recall that a submodule N of M is a non- summand if N is not direct summand of M [1]. 
 
Proposition 1.10 
         Let M be an R- module such that every non summand N of M is purely co-Hopfian , if  
for any non – summand  submodule N of M, N is purely co-Hopfian, then M is purely co-
Hopfian. 
Proof  
          Suppose M is not purely co-Hopfian then there exists N < M, N @   M, N is not pure in 
M by lemma (1.3). But N is not pure implies N is not summand. Hence by hypothesis N is 
purely co-Hopfian which implies M is purely co-Hopfian which is a contradiction.  
 
          Recall that M is fully stable if for any submodule N of M, f: N → M is  then f (N) £  N 
[5].  
Proposition 1.11 
Let M = M1 ≈  M2, M is fully stable.Then M is purely co-Hopfian if and only if M1, M2 are 
purely co-Hopfian 
 
Proof  



 
 
 

 إبن الهيثم للعلوم الصرفة و التطبيقيةمجلة 

 2012 السنة 25 المجلد 2 العدد

Ibn Al-Haitham Journal for Pure and Applied S cience  

 No. 2 Vol. 25 Year 2012 

     It follows by proposition 1.9.Conversely, Let f: M → M be monomorphism put f1 = f│M1 
,f2 = f│M2 .Since M is fully stable,f1(M1) £  M1 and f2 (M2) £  M2. Since f is monomorphism, 
f1, f2 are monomorphism. Hence f1 (M1), f2 (M2) are pure in M1, M2 respectively. Hencef1 
(M1) ≈  f2 (M2) is pure in M [2]. But it is easy to see that f (M) = f1 (M1) ≈ f2 (M2). Thus 
f(M)is pure in M . 
Corollary 1. 12  

          Let M= ŠiΠI Mi , M is fully stable M is purely co-Hopfian if and only if Mi is purely 
co-Hopfian for all i ΠI .  
 
          Recall that M is torsion free if rm = 0 then r = 0 or m = 0 for any r ΠR, m ΠM. Note 
that torsion free module needs not purely co-Hopfian, for example Z as Z-module. Now we 
have the following result which improves proposition 2.13 in [1].Which states that ,let R be a 
commutative domain and let M be  a torsion free semi co-Hopfian R-module .Then M is 
injective . 
Proposition 1.13  
          Let R be an integral domain and let M be a torsion free purely co-Hopfian R – module 
.Then M is injective R-module. 
Proof  
          Let a Œ R , a π  0. Define f : M → M by f ( m ) = am , for all a Œ M . Then f is 
monomorphism, hence f (M) = aM is pure submodule in M since M is purely co-
Hopfian.Thus IM I  f (M) = I f (M) for any ideal I of R. Take I = < a >. Hence (a) M I  aM 
= (a). aM thus aM = a2M .Now for any m Π M, am = a2m1, so a (m Рam1) = 0.Hence m-
am1=0 since M is torsion free and so m = am1. Thus we have M = aM, that is M divisible 
torsion free, hence M is injective. 
 
Proposition 1.14  
         If M has Dcc on non pure submodule (that means has Dcc on not pure submodule), then 
M is purely co-Hopfian. 
Proof  
          Suppose M is not purely co-Hopfian, then by lemma 1.3, there existsM1 (not pure 
submodule of M) such that M1 @  M. Hence M1 is not purely co-Hopfian and, so there exists 
M2 submodule of M1   which is not pure of M2 @ M1 . By repeating this argument we have 
strictly descending chain M1 …  M2 … …Moreover Mi is not pure in M, for all i = 1, 2,……  
.  To show this M1 is not pure in M (by proof). If M2  pure in M ,then M1 pure in M 
[2,Rem.7.2(1)] , which is a contradiction . Thus M2 is not pure in M.Similarly Mi is not pure 
in M, for all i = 3, 4,…  . Thus M1 …  M2 … …  . is strictly descending chain of non pure 
submodule of M, which is a contradiction.Thus M is purely co-Hopfian. 
Remark 1.15  
          The endomorphism ring of purely co-Hopfian module need not be purely co-Hopfian. 
Example 1.16 
          The Z – module Zp •  is co-Hopfian. S = End ( Zp • ) is the integral domain of P-adic 
integers is not co-Hopfian [6] , Then S is not purely co-Hopfian by Corollary (1.6). 
          Recall that an R-module M is called multiplication module if for each N≤M ,there 
exists ideal I of R such that N=IM. Equivalently, Mis multiplication if for each N≤M, 
N=(N:M)M,where (N:M)={r:rŒR, rM ÕN}[7]. 
Theorem 1.17 
          Let M be a faithful finitely generated multiplication R – module the following 
statements are equivalent:  
1. M is purely co-Hopfian. 
2. R is semi co-Hopfian. 



 
 
 

 إبن الهيثم للعلوم الصرفة و التطبيقيةمجلة 

 2012 السنة 25 المجلد 2 العدد

Ibn Al-Haitham Journal for Pure and Applied S cience  

 No. 2 Vol. 25 Year 2012 

3. R is purely co-Hopfian. 
4. M is co-Hopfian. 
5. M is Semi co-Hopfian. 
Proof    
          (1) →(2): Let a  ŒR , annRa = 0. Define f : M → M by f ( m ) = am for any mŒM .we 
can see that f is monomorphism as follows, let m ΠKerf then am = 0 and so m ΠannM ( a ) . 
But annM (a)  = (annR (a) ) M .Hence m Π(annRa) M = 0 . M = 0, then we get m = 0.Now f 
(M) = a M is pure in M. Hence < a > is pure in R , since M is faithful finitely generated 
multiplication .Thus < a > = < a 2 > so a = ra2, which implies a ( 1- ra ) = 0 ,  since ann ( a ) = 
0 , 1-ra =0 , 1 = ra,that is a is an inevitable element , so < a > = R . 
(2) ↔(3): It follows by proposition (1.4). 
(3) → (4): Let f : M → M be monomorphism , Since M is finitely generated multiplication, 
then M is a scalar module , there exists a Œ R , a π  0such that , f ( m ) = am for all m Œ M 
[8]. Since Kerf = {0}, annMa = 0.[ To prove this. Since annM(a) = { m : am = 0 } = { m : f ( m 
) = 0 }= { m : m = 0 }] .But annMa = (annRa) M, so annR(a) .M = 0 .Thus annR(a) Õ annM = 0 
.It follows that annR(a) = 0. But R is purely co-Hopfian so < a > = R by corollary (1.5) . 
(4) →(5): It is clear any co-Hopfian is semi co-Hopfian by [ 1]. 
(5) → (1): By [ Remark and Examples 1.2 ]  
corollary 1.18 
          Let M be a faithful finitely generated multiplication R-Module then the following are 
equivalent: 
1. M is purely co-Hopfian module. 
2. End R M is purely co-Hopfian ring ( semi co-Hopfian ,co-Hopfian )  
Proof   (1) ↔ (2)  
       Since M is a finitely generated multiplication R-module M is a scalar module by 
[8,prop.1.1.10].Hence End M @ R by [9,lemma 6.1,ch.3].Thus by previous theorem we 
obtained the result  .  
  
References 
1- Aydogdu, P.Ozcan, A. Cigdem; 10(2008), Semi co-hopfian and semi hopfian modules. 
East-West J. Math., no. 1: 55-70.  
2- Yaseen, S.M.,Msc.thesis,(1993),On F-regular modules , University of Baghdad,   College 
of science. 
3- Fieldhouse, D.J.(1969), pure theorie ,Math.Ann. 184:1-18 .  
4- Mohamed,  S.H.,and  Muller  B.J.,(1990), Continuous and Discrete Module , London Math. 
Soc. LNS 147 Cambridge Univ.Press, Cambridge . 
5- Abbas, M.S ,(1990), On fully stable module , ph.D. thesis .University of Baghdad . 
6- Lam,T.Y.,3(2004) Acrash course on stable range , cancellation, substitution and exchange . 
J.Algebra Appl., no. 3:301-343. 
7- EL.Bast ,Abd , Z and P.F. Smith (1988), Multiplication modules, comm. Algebra ,16(4): 
755-779. 
8- Shihab B.N. ,(2004), Scalar Reflexive module , ph.D. thesis .University of Baghdad  
college of science. 
9- Mohammed Ali, E.A.AL-Am .(2006), On Ikeda Nakayma module, ph.D. thesis .University 
of Baghdad . 

 

 



 
 
 

 إبن الهيثم للعلوم الصرفة و التطبيقيةمجلة 

 2012 السنة 25 المجلد 2 العدد

Ibn Al-Haitham Journal for Pure and Applied S cience  

 No. 2 Vol. 25 Year 2012 

 ةالنقي Éالمضادةالمقاسات الهوبفيني

     زينب طالب سلمان 
                قسم الرياضيات ، كلية العلوم ، جامعة بغداد                                                                                    

  2012 كانون الثاني  11 :قبل البحث في  2011ايلول  22 : استلم البحث في

 
                                                                                ةالخالص
. في هذا البحث نقدم مفهوم  مقاسا احاديا غير صفري معرفا عليها Mحلقه تجميعيه ذا عنصر محايد ،  Rلتكن         

 f Œ End (M)  مقاسا هوبفينيا مضادا اذا كان لكل Rعلى حلقه  Mيقال عن مقاس  ¡ÐÇ ةالنقي Éالمضادةالمقاسات الهوفيني
 ¡f فان  ةمتباين ةدالImf  نقي فيM  واعطينا بعض خواص هذا النوع من المقاسات . . 
 

 Éمضاد ة، مقاسات هوفيني  Éمضاد ةهوفيني ة، مقاسات شب ةنقي Éمضاد ةمقاسات هوفيني:ةالكلمات المفتاحي