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.