Quasi-morphic modules Ngo Sy Tung (a) , Tran Giang Nam (b) Ngo Ha Chau Loan (c) Abstract. In this paper, we prove that the matrix ring M n (R) is left quasi-morphic if and only if the left R-module R n is quasi-morphic. Then, we consider a general problem: For a quasi-morphic module M , when is end(M ) left quasi-morphic, and conversely? Using this result, we show that a ring R is regular if and only if it is a left quasi{morphic, left P P ring. 1 Introduction Nicholson - Campos ( [5], p. 2630) call a left module M morphic if M/Im(α) ∼ = ker(α) for all endomorphism α in end(M), equivalently if there exists β ∈ end(M) such that Im(β) = ker(α) and Im(α) = ker(β). In this paper, we only need the existence of β and γ such that Im(β) = ker(α) and Im(α) = ker(γ), and we call M quasi-morphic if for every element α in end(M), α satisfies the above condition. We use the notion to characterize the classes of quasi{morphic rings{these rings were introduced and studied by Camillo - Nicholson in [2]. More precisely, we answer the question: For each a ring R, when is the ring M n (R) left quasi{morphic? And we obtain that the matrix ring M n (R) is left quasi-morphic if and only if R n is quasi- morphic as a left R-module (Theorem 2.2). More generally, we investigate when M being quasi-morphic implies that end(M) is left quasi-morphic, and conversely (Proposition 3.1). Furthermore, we also investigate when M being quasi-morphic implies that end(M) is regular (Theorem 3.1). Then, applying this result, we obtain that a ring R is regular if and only if it is a left quasi{morphic, left PP ring (Corollary 3.4). Throughout this paper, every ring R is associative with unity and all modules are unitary. We denote left annihilator of a set X ⊆ R by l R (X). Finally, all notions used here without any comments, can be found in [4]. ( 1) NhËn bµi ngµy 08/4/2009. Söa ch÷a xong ngµy 03/6/2009 2 Quasi morphic modules Recall ([5], p. 2630) that a left module M is called morphic if M/Im(α) ∼ = ker(α) for all endomorphism α in end(M); equivalently if for each α ∈ end(M) there exists β ∈ end(M) such that Im(β) = ker(α) and Im(α) = ker(β) (see [5], Lemma 1). More generally, we call an endomorphism α of a left module M quasi–morphic if there exist β and γ in end(M) such that Im(β) = ker(α) and Im(α) = ker(γ). The left module M is called quasi–morphic if every endomorphism of M is quasi{morphic. Clearly, every morphic module is quasi{morphic, but the converse is false. In order to prove this fact, we first need the following result{a characterization of a quasi-morphic module in terms of the lattice of submodules. Theorem 2.1 Let M be a left module. Then the following conditions are equivalent: (1) M is quasi{morphic; (2) If M/K ∼ = N where K and N are submodules of M, then there exist β, γ ∈ end(M) such that K = Im(β) and N = ker(γ). Proof. (1) =⇒ (2). If η : M/K → N is an isomorphism, define α : M −→ M by α(m) = η(m + K). It is easy to see that α is an endomorphism of M. Since M is quasi{morphic, so there exist β and γ in end(M) such that Im(β) = ker(α) and Im(α) = ker(γ). Then, Im(β) = K and Im(α) = N. (2) =⇒ (1). It is clear. Using Theorem 2.1, we immediately have that Z 2 ⊕ Z 4 is a quasi-morphic Z- module, but it is well known that Z 2 ⊕ Z 4 isn't a morphic Z-module (see [5], p. 2631). Corollary 2.1 Every semisimple left module is quasi{morphic. Proof. Let M be a semisimple left module. Then, for each submodule N of M, since N is a direct summand of M, so that there exist α, β ∈ end(M) such that K = Im(α) and N = ker(β). Using this and Theorem 2.1, we immediately have M is quasi-morphic. Recall ([5], Theorem 23) that every direct summand of a morphic left module is again morphic. Is this true for the class of quasi-morphic modules? We can't answer this question yet, however we partially answer by using Theorem 2.1 as follows: Corollary 2.2 Let M and N be left R-modules such that Hom R (M, N) = 0 = Hom R (N, M). Then, M ⊕ N is quasi-morphic if and only if M and N are quasi- morphic. It is well known that there exists a natural isomorphism ϕ : R −→ end( R R) is defined by ϕ(r)(x) = xr for all r, x ∈ R. Using this, we immediately obtain that if R is a ring, R R is quasi-morphic if and only if for each a ∈ R there exist b, c ∈ R such that Ra = l(b) and l(a) = Rc. These rings were introduced and studied by Camillo - Nicholson in [2], and called left quasi-morphic. Camillo - Nicholson showed that there exists the ring R such that the ring M n (R) is not left (right) quasi{morphic (see, [2], Example 4). In light of this fact, it is natural to bring up the following problem: When is M n (R) left quasi{morphic? In the following theorem, we give the answer to the question. Theorem 2.2 Let R be a ring. Then, the following conditions are equivalent: (1) R R n is quasi-morphic; (2) M n (R) is left quasi{morphic. Proof. We prove it for n = 2; the general case is analogous. (1) =⇒ (2). Let K and N be left ideals of S = M 2 (R) such that S/N ∼ = K. By ([5], p. 2635), so there exist submodules X and Y of R R 2 such that K =  X X  and N =  Y Y  . Since S/N ∼ = K and ([5], Lemma 16), X ∼ = R 2 /Y. By R R 2 is quasi- morphic and Theorem 2.1, so there exist β, γ ∈ end( R R 2 ) such that Y = Im(β) and X = ker(γ). Then, there are the homomorphisms α, δ : S −→ S defined by α(  x y z t  ) =  β(x, y) β(z, t)  and δ(  x y z t  ) =  γ(x, y) γ(z, t)  . It is easy to see that Im(α) = N and ker(δ) = K. Using again Theorem 2.1, we have S is left-morphic. (2) =⇒ (1). It is similarly proved as the direction (1) =⇒ (2). 3 Endomorphism rings Let M be a left R-module. The set of all endomorphisms of M will be denoted by end(M). Further, end(M) has the structure of a ring, when we define (f + g)(x) = f(x) + g(x) and fg(x) = g(f(x)) for all x ∈ M and f, g ∈ end(M). Recall (see, for example, [5], p. 2640) that a left module M is called image– projective if, whenever Im(γ) ⊆ Im(α) where γ, α ∈ E = end(M), then γ ∈ Eα. Of course, every projective left module is image{projective. Also, in ([5], p. 2641) a left module M is said to generate a submodule K ⊆ M if K =  {Im(λ)|λ ∈ end(M), Im(λ) ⊆ K}, and we say that M generates its kernels if M generates ker(β) for all β ∈ end(M). One has the following remark. Remark 3.1 Let R be a ring and F a free left R-module. Then, F generates all its submodules. In particular, F generates its kernels. Proof. Assume K is a submodule of F . Since F is free, so that F ∼ = R (I) for some set I. For each x ∈ K, there exists the homomorphism f x : F −→ F defined by f x ((r i ) i∈I ) =  i∈I r i x for all (r i ) i∈I ∈ F . Clearly, x ∈ Im(f x ). Hence, K ⊆  x∈K Im(f x ) ⊆  {Im(λ)|λ ∈ end(F ), Im(λ) ⊆ K}. It follows that M generates K. Proposition 3.1 Let M be a left R-module and E = end(M). Then, (1) If M is quasi-morphic and image{projective, then E is left quasi{morphic; (2) If M is quasi-morphic, then it generates its kernels; (3) If E is left quasi{morphic and M generates its kernels, then M is quasi- morphic. Proof. (1) If M is quasi-morphic and image–projective, and given α ∈ E, choose β, γ ∈ E such that Im(α) = ker(β) and Im(γ) = ker(α). Then, since αβ = 0, Eα ⊆ l E (β). Conversely, if ϕ ∈ l E (β) then Im(ϕ) ⊆ ker(β) = Im(α), so ϕ ∈ Eα because M is image–projective. Thus Eα = l E (β), and Eγ = l E (α) follows in the same way. Hence E is left quasi–morphic. (2) It is clear. (3) Give α ∈ E, choose β, γ ∈ E such that Eα = l E (β) and Eγ = l E (α). Then Im(α) ⊆ ker(β) because αβ = 0. By M generates its kernels, so ker(β) =  {Im(λ)|λ ∈ end(M), Im(λ) ⊆ ker(β)}. But Im(λ) ⊆ ker(β) implies λ ∈ l E (β) = Eα, say λ = δα, δ ∈ E. Hence, Im(λ) = Im(δα) ⊆ Im(α). It follows that ker(β) ⊆ Im(α). Thus ker(β) = Im(α), and ker(α) = Im(γ) is proven in the same way. Remark 3.2 Let R be a ring. It is well known that M n (R) ∼ = end( R R n ). Further, since R n is free left R-module, it is image{projective. And, applying Remark 3.1, we have R n generates its kernels. Therefore, taking M = R n in Proposition 3.1 provides another proof of Theorem 2.2. Let M be a left R-module and α ∈ end(M). In [1], Azumaya proved that α is regular if and only if both Im(α) and ker(α) are direct summands of M". Using this fact, we immediately obtain the following remark. Remark 3.3 Let M be a left R-module and E = end(M). Then, for any α ∈ end(M), if α is regular then it is quasi-morphic. In particular, if E is a regular ring then M is quasi-morphic. In the next theorem, we can give a partial converse for Remark 3.3. First, recall ([5], p. 2643) that a left module M is called kernel-direct if ker(α) is a summand of M for every α ∈ end(M), and call M image-direct if Im(α) is a summand of M for every α ∈ end(M). Theorem 3.1 The following conditions are equivalent for a left module M: (1) end(M) is regular; (2) M is quasi{mophic and kernel-direct; (3) M is quasi{mophic and image-direct. Proof. (1) =⇒ (2) This follows from Remark 3.3. (2) =⇒ (3) This follows from the definitions. (3) =⇒ (1) Given α ∈ end(M), Im(α) is a summand of M by (3). Since M is quasi-morphic, there exists β ∈ end(M) such that ker(α) = Im(β). Using again (3), Im(β) is a summand of M; hence ker(α) is also. Thus α is regular. If R is a ring then R R is kernel-direct if and only if l R (a) is a direct summand of R R for all a ∈ R, that is if and only if every principal left ideal Ra is projective. These are called left P P rings. Using this fact and Theorem 3.1, we immediately have the following corollary. Corollary 3.2 A ring R is regular if and only if it is a left quasi{morphic, left PP ring. References [1] G. Azumaya, On generalized semi-primary rings and Krull-Remak-Schmidt's theorem, Japan J. Math., 19, 1960, 525 - 547. [2] V. Camillo and W. K. Nicholson, Quasi-morphic rings, Journal of Algebra and its Appl., Vol. 6, No. 5, 2007, 789 - 799. [3] G. Erlich, Units and one-sided units in regular rings, Trans A.M.S. 216, 1976 81-90. [4] T. Y. Lam, A first course in noncommutative rings, 2nd Ed., Springer-Verlag, New York-Berlin, 2001. [5] W. K. Nicholson and E. S. Campos, Morphic modules, Comm. in Algebra, 33, 2005, 2629 - 2647. Tóm tắt Môđun tựa cấu xạ Trong bài báo này, chúng tôi chứng minh rằng vành ma trận M n (R) là tựa cấu xạ khi và chỉ khi R-môđun trái R n là tựa cấu xạ. Tiếp đó chúng tôi xét bài toán tổng quát: với môđun tựa cấu xạ M, khi nào vành end(M ) là tựa cấu xạ trái, và ngợc lại? áp dụng kết quả này, chúng tôi chỉ ra rằng một vành R là chính qui khi và chỉ khi nó là tựa cấu xạ trái, P P trái. (a) Departrment of Mathematics, University of Vinh (b) Department of Mathematics, University of Dong Thap (c) 46A, Departrment of Mathematics, University of Vinh. . tựa cấu xạ Trong bài báo này, chúng tôi chứng minh rằng vành ma trận M n (R) là tựa cấu xạ khi và chỉ khi R-môđun trái R n là tựa cấu xạ. Tiếp đó chúng tôi xét bài toán tổng quát: với môđun tựa. quát: với môđun tựa cấu xạ M, khi nào vành end(M ) là tựa cấu xạ trái, và ngợc lại? áp dụng kết quả này, chúng tôi chỉ ra rằng một vành R là chính qui khi và chỉ khi nó là tựa cấu xạ trái, P P trái. (a)