summary:Let $R$ be a ring with an identity (not necessarily commutative) and let $M$ be a left $R$-module. This paper deals with multiplication and comultiplication left $R$-modules $M$ having right $\operatorname{End}_R(M)$-module structures

Ansari-Toroghy, H.

Farshadifar, F.

English

Institute of Mathematics AS CR, v. v. i.

Archivum MathematicumH. Ansari-Toroghy; F. FarshadifarOn endomorphisms of multiplication and comultiplication modulesArchivum Mathematicum, Vol. 44 (2008), No. 1, 9--15Persistent URL: http://dml.cz/dmlcz/108091Terms of use:© Masaryk University, 2008Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document must containthese Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech Digital MathematicsLibrary http://project.dml.czARCHIVUM MATHEMATICUM (BRNO)Tomus 44 (2008), 9–15ON ENDOMORPHISMS OF MULTIPLICATIONAND COMULTIPLICATION MODULESH. Ansari-Toroghy and F. FarshadifarAbstract. Let R be a ring with an identity (not necessarily commutative)and let M be a left R-module. This paper deals with multiplication andcomultiplication left R-modules M having right EndR(M)-module structures.1. IntroductionThroughout this paper R will denote a ring with an identity (not necessarilycommutative) and all modules are assumed to be left modules. Further “⊂” willdenote the strict inclusion and Z will denote the ring of integers.Let M be a left R-module and let S := EndR(M) be the endomorphism ring ofM . Then M has a structure as a right S-module so that M is an R− S bimodule.If f : M → M and g : M → M , then fg : M → M defined by m(fg) = (mf)g.Also for a submodule N of M ,IN :={f ∈ S : Im(f) =Mf ⊆ N}andIN :={f ∈ S : N ⊆ Ker(f)}are respectively a left and a right ideal of S. Further a submodule N of M iscalled ([3]) an open (resp. a closed) submodule of M if N = N◦, where N◦ =∑f∈IN Im(f) (resp. N = N̄ , where N̄ = ∩f∈IN Ker(f)). A left R-module M issaid to self-generated (resp. self-cogenerated) if each submodule of M is open (resp.is closed).Let M be an R-module and let S = EndR(M). Recently a large body ofresearches has been done about multiplication left R-module having right S-modulestructures. An R-module M is said to be a multiplication R-module if for everysubmodule N of M there exists a two-sided ideal I of R such that N = IM .In [2], H. Ansari-Toroghy and F. Farshadifar introduced the concept of a comulti-plication R-module and proved some results which are dual to those of multiplicationR-modules. An R-module M is said to be a comultiplication R-module if for everysubmodule N of M there exists a two-sided ideal I of R such that N = (0 :M I).2000 Mathematics Subject Classification: Primary: 13C99.Key words and phrases: endomorphisms, multiplication modules, comultiplication modules.Received September 28, 2006, revised October 2007. Editor J. Trlifaj.10 H. ANSARI-TOROGHY AND F. FARSHADIFARThis paper deals with multiplication and comultiplication left R-modulesM havingright EndR(M)-modules structures. In section three of this paper, among the otherresults, we have shown that every comultiplication R-module is co-Hopfian andgeneralized Hopfian. Further if M is a comultiplication module satisfying ascendingchain condition on submodules N such that M/N is a comultiplication R-module,then M satisfies Fitting’s Lemma. Also it is shown that if R is a commutative ringand M is a multiplication R-module and S is a domain, then for every maximalsubmodule P of M , IP is a maximal ideal of S.2. Previous resultsIn this section we will provide the definitions and results which are necessary inthe next section.Definition 2.1.(a) M is said to be (see [9]) a multiplication R-module if for any submodule Nof M there exists a two-sided ideal I of R such that N = IM .(b) M is said to be a comultiplication R-module if for any submodule N of Mthere exists a two-sided ideal I of R such that N = (0 :M I). For exampleif p is a prime number, then Z(p∞) is a comultiplication Z-module but Z(as a Z-module) is not a comultiplication module (see [2]).(c) Let N be a non-zero submodule ofM . Then N is said to be (see [1]) large oressential (resp. small) if for every non-zero submodule L of M , N ∩ L 6= 0(resp. L+N =M implies that L =M).(d) M is said to be (see [7]) Hopfian (resp. generalized Hopfian (gH for short))if every surjective endomorphism f of M is an isomorphism (resp. hasa small kernel).(e) M is said to be (see [8]) co-Hopfian (resp. weakly co-Hopfian) if everyinjective endomorphism f of M is an isomorphism (resp. an essentialhomomorphism).(f) An R-moduleM is said to satisfy Fitting’s Lemma if for each f ∈ EndR(M)there exists an integer n ≥ 1 such that M = Ker(fn)⊕Im(fn) (see [5]).(g) Let M be an R-module and let I be an ideal of R. Then IM is called tobe idempotent if I2M = IM .3. Main resultsLemma 3.1. Let R be any ring. Every comultiplication R-module is co-Hopfian.Proof. Let M be a comultiplication R-module and let f : M →M be a monomor-phism. There exists a two-sided ideal I of R such that Im(f) = (0 :M I). Now letm ∈M so that mf ∈ Im(f). Then for each a ∈ I, we have (am)f = a(mf) = 0. Itfollows that am ∈ Ker(f) = 0. This implies that am = 0 so that m ∈ (0 :M I) =Mf . Hence we have M ⊆Mf so that f is epic. It follows that M is a co-HopfianR-module. ON ENDOMORPHISMS OF MULTIPLICATION AND COMULTIPLICATION MODULES 11The following examples shows that not every comultiplication (resp. Artinian)R-module is an Artinian (resp. a comultiplication) R-module.Example 3.2. Let p be a prime number. Then let R be the ring with underlyinggroupR = EndZ(Z(p∞))⊕ Z(p∞) ,and with multiplication(n1, q1) · (n2, q2) = (n1n2, n1q2 + n2q1) .Osofsky has shown that R is a non-Artinian injective cogenerator (see [6, Exa.24.34.1]). In fact R is a commutative ring. Hence R is a comultiplication R-moduleby [6, Prop. 23.13].Example 3.3. Let F be a field, and let M = ⊕ni=1Fi, where Fi = F for i =1, 2, . . . , n. Clearly M is an Artinian non-comultiplication F -module.Theorem 3.4. Let M be a comultiplication module satisfying ascending chaincondition on submodules N such that M/N is a comultiplication R-module. ThenM satisfies Fitting’s Lemma.Proof. Let f ∈ EndR(M) and consider the sequenceKer f ⊆ Ker f2 ⊆ · · · .Since every submodule of a comultiplication R-module is a comultiplicationR-module by [2], for each n we have M/Ker fn ∼= Imfn implies that M/Ker fnis a comultiplication R-module. Hence by hypothesis there exists a positive in-teger n such that Ker(fn) = Ker(fn+h) for all h ≥ 1. Set fn1 = fn |M(fn).Then fn1 ∈ EndR(M(fn)). Further we will show that fn1 is monic. To see this letx ∈ Ker(fn1 ). Then x = y(fn) for some y ∈M and we have x(fn) = 0. It followsthat y(f2n) = 0 so thaty ∈ Ker(f2n) = Ker(fn) .Hence we have x = 0. But (M)fn is a comultiplication R-module and everycomultiplication R-module is co-Hopfian by Lemma 3.1. So we conclude that fn1is an automorphism. In particular, M(fn) ∩Ker(fn) = 0. Now let x ∈M . Sincefn1 is epimorphism, then there exists y ∈ M such that x(fn) = y(f2n). Hence(x− y(fn))(fn) = 0. It follows that x− y(fn) ∈ Ker(fn). Now the result followsfrom this because x = y(fn) + (x− y(fn)). Corollary 3.5. Let M be an indecomposable comultiplication module satisfyingascending chain condition on submodules N such that M/N is a comultiplicationR-module. Let f ∈ EndR(M). Then the following are equivalent.(i) f is a monomorphism.(ii) f is an epimorphism.(iii) f is an automorphism.(iv) f is not nilpotent.12 H. ANSARI-TOROGHY AND F. FARSHADIFARProof. (i)⇒(ii). This is clear by Lemma 3.1.(iii)⇒(ii). This is clear.(iii)⇒(iv). Assume that f is an automorphism. Then M =Mf . Hence,M =Mf =M(f2) = · · · .If f were nilpotent, then M would be zero.(ii)⇒(i). Assume that f is an epimorphism. Then M =Mf . HenceM =Mf =M(f2) = · · · .By Theorem 3.4, there is a positive integer n such thatM = Ker(fn)⊕ Im(fn) .Hence M = Ker(fn)⊕M , so Ker(fn) = 0. Thus, Ker(f) = 0.(ii)⇒(iii). This follows from (ii)⇒(i).(iv)⇒(iii). Suppose that f is not nilpotent. By Theorem 3.4, there exists apositive integer n such that M = Mfn⊕Ker fn. Since M is indecomposableR-module, it follows that Ker fn = 0 or Mfn = 0. Since f is not nilpotent, wemust have Ker fn = 0. This implies that f is monic. This in turn implies that f isepic by Lemma 3.1. Hence the proof is completed. Example 3.6. Let A = K[x, y] be the polynomial ring over a field K in twoindeterminates x, y. Then A = A/(x2, y2) is a comultiplication A-module. ButA/Axy is not a comultiplication A-module (see [6, Exa. 24.4]). Therefore, not everyhomomorphic image of a comultiplication module is a comultiplication module.Remark 3.7. In the Corollary 3.5 the condition M satisfying ascending chaincondition on submodules N such that M/N is a comultiplication R-module cannot be omitted. For example M = Z(p∞) is an indecomposable comultiplicationZ-module but not satisfying ascending chain condition on submodules N such thatM/N is a comultiplication Z-module. Define f : Z(p∞)→ Z(p∞) by x→ px. Clearlyf is an epimorphism with Ker f = Z(1/p+ Z). Hence f is not a monomorphism.Lemma 3.8. Let M be a comultiplication R-module and let N be an essentialsubmodule of M . If the right ideal IN of EndR(M) is non-zero, then it is small inEndR(M).Proof. Let J be any right ideal of S = EndR(M) such that IN + J = S. Then1M = f + j for some f ∈ IN and j ∈ J . Since Ker(1M − f) ∩N = 0 and N is anessential submodule of M , it follows that j is a monomorphism. Hence by Lemma3.1, j is an automorphism so that J = S. Hence IN is a small right ideal of S. Proposition 3.9. Let M be a comultiplication R-module and let N be a submoduleof M such that M/N is a faithful R-module. Then M/N is a co-Hopfian R-module.Proof. Let f : M/N →M/N be an R-monomorphism and (M/N)f = K/N , withN ⊆ K ⊆ M . Since M is a comultiplication R-module there exists a two-sidedideal I of R such that K = (0 :M I). Now(I(M/N))f = I(M/N)f = I(K/N) = 0 .ON ENDOMORPHISMS OF MULTIPLICATION AND COMULTIPLICATION MODULES 13Since f is monic, it follows that I(M/N) = 0. This in turn implies that I ⊆AnnR(M/N) = 0. Hence we have K =M so that f is an epimorphism. Lemma 3.10. Every comultiplication R-module is gH.Proof. Let M be comultiplication R-module and let f : M → M be an epimor-phism and assume that Ker(f) + K = M , where K is a submodule of M . SoKf = Mf = M . Since M is a comultiplication module, there exists a two-sidedideal J of R such that K = (0 :M J). Now0 = 0f =(J(0 :M J))f = J(Kf) = JM .It follows that J ⊆ AnnR(M). Hence we have K = (0 :M J) =M . This shows thatKer(f) is a small submodule of M . So the proof is completed. Proposition 3.11.(a) Assume that whenever f, g ∈ EndR(M) with fg = 0 then we have gf = 0. IfM is a self-generated (resp. self-cogenerated) R-module, then M is Hopfian(resp. co-Hopfian).(b) Let M be a self-generated (resp. self-cogenerated) R-module and let S bea left Noetherian (resp. right Artinian) ring. Then M is a NoetherianS-module.Proof. (a) Let S = EndR(M) and let g : M → M be an epimorphism. Let fbe any element of IKer(g). Then Mf ⊆ Ker(g), so M(fg) = (Mf)g = 0. Hence,fg = 0. By our assumption, gf = 0. Since g is an epimorphism, we haveMf = (Mg)f =M(gf) = 0 .Thus, if M is self-generated,Ker(g) =∑f∈IKer(g)Im(f) = 0 .Hence M is a Hopfian R-module. The proof is similar when M is a self-cogeneratedR-module.(b) LetN1 ⊆ N2 ⊆ N3 ⊆ · · ·be an ascending chain of S-submodules of M . This induces the sequenceIN1 ⊆ IN2 ⊆ · · · ⊆ INk ⊆ · · · .Now there exists a positive integer s such that for each 0 ≤ i, INs = INi+s . SinceM is a self-generated R-module, we have Ns =MINs =MINi+s = Ni+s for every0 ≤ i. Thus M is a Noetherian S-module. For right Artinian case when M is aself-cogenerator R-module, the proof is similar. So the proof is completed. Theorem 3.12. Let M be a multiplication R-module and let N be a submoduleof M .(a) If R is a commutative ring, and I is an ideal of R such that IM is anidempotent submodule of M , then IM is gH.14 H. ANSARI-TOROGHY AND F. FARSHADIFAR(b) If R is a commutative ring and N is faithful, then N is weakly co-Hopfian.(c) If M is a quasi-injective, N is gH.Proof. (a) Let I be an ideal of R such that IM be an idempotent submodule ofM .Let f : IM → IM be an epimorphism and assume that Ker(f) + L = IM , whereL is a submodule of IM . Then we have I(Ker(f))+ IL = IM . Let Ker(f) = JMfor some ideal J of R. Since R is a commutative ring, we have0 = I(Ker(f))f = (IJM)f = J(IM)f = JIM = IJM = I(Ker(f)).Thus by the above arguments, IL = IM so that IM ⊆ L. It follows that IM = Lso that IM is a generalized Hopfian R-module.(b) Let I be an ideal of R such that N = IM . Let f : N → N be an injectivehomomorphism and assume that Nf ∩K = 0, where K is a submodule of N . Thenthere exist ideals J1 and J2 of R such that Nf = J1M and K = J2M . Then wehave0 = K ∩Nf = K ∩ (IM)f = (J2M) ∩ (IM)f = J2M ∩ J1M ⊇ J2J1M .Hence J2J1M = 0. Now we have(IJ2M)f = J2(IM)f = J2J1M = 0 .Since f is monic, J2N = IJ2M = 0. Since N is a faithful R-module, we haveJ2 = 0 so that K = 0. Hence Nf is essential in N . It implies that N is a weaklyco-Hopfian R-module as desired.(c) Let f : N → N be an epimorphism and let Ker(f) +K = N , where K isa submodule of N . Since M is quasi-injective, we can extend f to g : M →M . Butas M is a multiplication module, Kg ⊆ K, therefore Kf ⊆ K. On the other hand,Kf = N since f is epimorphism. Therefore K = N . Hence N is a generalizedHopfian R-module as desired. Proposition 3.13. Let R be a commutative ring and let M be a multiplicationR-module. Let S = EndR(M) be a domain. Then the following assertions hold.(a) Each non-zero element of S is a monomorphism.(b) If I and J are ideals of S such that I 6= J , then MI 6=MJ .Proof. (a) Assume that 0 6= g ∈ S. Then there exist ideals I and J of R such thatIm(g) = JM and Ker(g) = IM . Now we have0 =(Ker(g))g = (IM)g = I(Mg) = IJM .It implies that IJ ⊆ AnnR(M). Since S is a domain, AnnR(M) is a prime ideal ofR by [2, 2.3]. Hence I ⊆ AnnR(M) or J ⊆ AnnR(M) so that IM = 0 or JM = 0.It turns out that Ker(g) = 0 as desired.(b) Since R is a commutative ring, M is a multiplication S-module. Hence for0 6= m ∈ M there exists an ideal K of S such that mS = MK. Now we assumethat MI = MJ . Since R is a commutative ring, S is a commutative ring by [4].HencemI = mSI = (MK)I = (MI)K = (MJ)K = (MK)J = mSJ = mJ .ON ENDOMORPHISMS OF MULTIPLICATION AND COMULTIPLICATION MODULES 15Choose f ∈ I\J . Then since mf ∈ mI = mJ , there exists h ∈ J such thatmh = mf . Thus we have m(h − f) = 0. Further h − f 6= 0. So by using part(a), we have m ∈ Ker(h − f) = 0. But this is a contradiction and the proof iscompleted. Corollary 3.14. Let R be a commutative ring andM be a multiplication R-module.Set S = EndR(M) and Im(J) =∑f∈J Im(f), where J is an ideal of S. If J isa proper ideal of a domain S, then Im(J) is a proper submodule of M .Proof. This is an immediate consequence of Proposition 3.13 (b). Theorem 3.15. Let R be a commutative ring and let M be a multiplicationR-module such that S = EndR(M) is a domain. Then for every maximal submoduleP of M , IP is a maximal ideal of S.Proof. Since IdM ∈ S and IdM 6∈ IP , we have IP 6= S. Now assume that U isan ideal of S such that IP ⊆ U ⊆ S. Then if MU = M , then by Corollary 3.14,U = S. If MU = P , then U ⊆ IP , so U = IP . Hence IP is a maximal ideal of Sand the proof is completed. Example 3.16. Let R be a commutative ring and let P be a prime ideal of R. SetM = R/P . Then M is a multiplication R-module and S = EndR(M) is a domain.Hence by Theorem 3.15, for every maximal submodule N of M , IN is a maximalideal of S.Acknowledgement. The authors would like to thank the referee for his invaluablecomments.References[1] Anderson, W., Fuller, K. R., Rings and categories of modules, Springer-Verlag, NewYork-Heidelberg-Berlin, 1974.[2] Ansari-Toroghy, H., Farshadifar, F., The dual notion of multiplication modules, Taiwanese J.Math. (to appear).[3] Bae, Soon-Sook, On submodule inducing prime ideals of endomorphism ring, East AsianMath. 16 (1) (2000), 33–48.[4] Choi, C. W., Multiplication modules and endomorphisms, Math. J. Toyama Univ. 18 (1995),1–8.[5] Choi, C. W., Smith, P. F., On endomorphisms of multiplication modules, J. Korean Math.Soc. 31 (1) (1994), 89–95.[6] Faith, C., Algebra II: Ring theory, Springer-Verlag, New York-Heidelberg-Berlin, 1976.[7] Ghorbani, A., Haghang, A., Generalized Hopfian modules, J. Algebra 255 (2002), 324–341.[8] Haghang, A., Vedali, M. R., Modules whose injective endomorphism are essential, J. Algebra243 (2001), 765–779.[9] Lomp, Ch. E., Prime elements in partially ordered groupoid applied to modules and Hopfalgebra actions, J. Algebra Appl. 4 (1) (2005), 77–98.Department of MathematicsFaculty of Science, Guilan UniversityP. O. Box 1914, Rasht, IranE-mail: ansari@guilan.ac.ir, Farshadifar@guilan.ac.ir

On endomorphisms of multiplication and comultiplication modules

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On endomorphisms of multiplication and comultiplication modules

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Institute of Mathematics AS CR, v. v. i.