1. No category

Southeast Asian Bulletin of Mathematics (2007) 31: 1091–1096
Southeast Asian
Bulletin of
Mathematics
c SEAMS. 2007
Modules Whose Essentially Injective Endomorphisms
Are Automorphisms
Yang Gang∗
School of Mathematics, Physics and Software Engineering,
Lanzhou Jiaotong University, Lanzhou, 730070, China
E-mail: yg0280@tom.com.
Zhao Liang
College of Science, Jiangxi University of Technology and Science, Ganzhou, 341000,
China
AMS Mathematics Subject Classification (2000): 16D10, 16W20
Abstract. Let R be an associative ring not necessarily possessing an identity and (S, ≤)
a strictly totally ordered monoid which is also artinian. It is proved that if a left Rmodule M has property (F), then any essentially injective endomorphism of M is an
automorphism if and only if any essentially injective endomorphism of the left [[RS,≤ ]]module [M S,≤ ] is an automorphism.
Keywords: Co-Hopfian; Weakly Co-Hopfian; Essentially injective endomorphism.
1. Introduction
All rings in this paper are not necessarily have an identity. In [1-3], a left
R-module M is called Co-Hopfian if any injective endomorphism of M is an automorphism. As a generalization of Co-Hopfian modules, A.Haghany [3] shown
that a left R-module M is weakly Co-Hopfian if any injective endomorphism
f of M is essential, i.e., f (M ) ≤e M . As motivated by Z.Liu [3] and G.Yang
[5], it was proved that a left R-module M is Co-Hopfian, weakly Co-Hopfian,
respectively if and only if a left [[RS,≤ ]]-module [M S,≤ ] is Co-Hopfian, weakly
Co-Hopfian, respectively, where (S, ≤) is a strictly totally ordered monoid which
is also artinian. We call an endomorphism f of M be essentially injective, if f is
essential and injective. In this paper, we consider the modules whose essentially
injective endomorphisms are automorphisms. Such modules contain Co-Hopfian
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Yang Gang and Zhao Liang
and continuous modules [6]. It is shown that if a left R-module M has property
(F), then any essentially injective endomorphism of the left R-module M is an
automorphism if and only if any essentially injective endomorphism of the left
[[RS,≤ ]]-module [M S,≤ ] is an automorphism, where (S, ≤) is a strictly totally
ordered monoid which is also artinian.
It is said that R M has property (F) if for any submodule N of M , we have
{m ∈ M |Rm ⊆ N } = N . It is easy to see that M has property (F) if and
only if m ∈ Rm for each m ∈ M , i.e., M is an s-unital module in the sense of
Tominaga [7]. It follows from [7,Th.1] that M has property (F) if and only if for
any finitely many elements m1 , m2 , ..., mn ∈ M , there is an element e ∈ R such
that emi = mi , i = 1, 2, ..., n.
Let (S, ≤) be an ordered set. Recall that (S, ≤) is artinian if every strictly
decreasing sequence of elements of S is finite. Let S be a commutative monoid,
unless stated otherwise, the operation of S shall be denoted additively, and the
neutral element by 0. The following definition is from [8] or [9].
Let (S, ≤) be a strictly totally ordered monoid (that is , (S, ≤) is a totally
ordered monoid satisfying the condition that if s, s0 , t ∈ S and s < s0 , then
s + t < s0 + t), and R is a ring. Let [[RS,≤ ]] be the set of all maps f : S −→ R
such that Supp(f ) = {s ∈ S|f (s) 6= 0} is artinian, with pointwise addition,
[[RS,≤ ]] is an abelian additive group. For every s ∈ S and f, g ∈ [[RS,≤ ]], let
χs (f, g) = {(u, v) ∈ S × S|s = u + v, f (u) 6= 0, g(v) 6= 0}. Then χs (f, g) is
finite [10, 4.1]. By the above fact, we can define the operation of convolution as
following:
X
(f g)(s) =
f (u)g(v).
(u,v)∈χs (f,g)
With the above operation, and pointwise addition, [[RS,≤ ]] becomes a ring which
is called the ring of generalized power series with coefficients in R and exponents
in S.
For example, if S = N ∪ {0} and ≤ is the usual order, then [[RS,≤ ]] ' R[[x]],
the usual ring of power series. If S is a commutative monoid and ≤ is the trivial
order, then [[RS,≤ ]] = R[S], the monoid ring of S over R. Further examples are
given in [11].
We shall henceforth assume that (S, ≤) is a strictly totally ordered monoid
which is also artinian. If M is a left R-module, then we let [M S,≤ ] be the set
of all maps φ : S −→ M such that supp(φ) = {s ∈ S|φ(s) 6= 0} is finite. Now
[M S,≤ ] can be turned into a left [[RS,≤ ]]-module. The addition in [M S,≤ ] is
componentwise and the scalar multiplication is defined as follows:
X
f φ(s) =
f (t)φ(s + t),
t∈S
for every s ∈ S, where f ∈ [[RS,≤ ]], and φ ∈ [M S,≤ ]. Since the set supp(φ) is
finite, this multiplication is well defined.
For example, if S = N ∪{0} and ≤ is the usual order, then [M S,≤ ] ' M [x−1 ],
the usual left R[[x]]-module in [13]. If S is the multiplicative monoid (N, ·),
Modules Whose Essentially Injective Endomorphisms Are Automorphisms
1093
endowed with the usual order ≤, then [[RS,≤ ]] is the ring of arithmetical functions
with values in R, endowed with the Dirichlet convolution:
X
(f g)(n) =
f (d)g(n/d),
d|n
for each n ≥ 1. If M is a left R-module, then [M (N,·),≤ ] is the left [[R(N,·),≤ ]]module with scalar multiplication as below:
X
(f φ)(n) =
f (k)φ(nk),
k≥1
for every n ≥ 1.
If s ∈ S is such that s ≤ 0, then
· · · ≤ 3s ≤ 2s ≤ s ≤ 0
by [11]. This contradicts to assumption that (S, ≤) is artinian. Thus for any
s ∈ S, we have 0 ≤ s. This result will be frequently used throughout this paper.
We now introduce the following notations. Let r ∈ R, t ∈ S. Define a
mapping dtr ∈ [[RS,≤ ]] as following:
r, s = t;
t
dr (s) =
0, s 6= t,
where s ∈ S.
Let m ∈ M, t ∈ S. Define a mapping φt,m ∈ [M S,≤ ] as following:
m, s = t;
φt,m (s) =
0, s 6= t,
where s ∈ S.
For every 0 6= φ ∈ [M S,≤ ], we denote by σ(φ) the maximal element in
supp(φ).
2. Main results
Lemma 2.1. [14] A submodule K ≤ M is essential in M if and only if for each
0 6= m ∈ M , there exists an r ∈ R such that 0 6= rm ∈ K.
Lemma 2.2. [5, 12] Let α ∈ Hom[[RS,≤ ]] ([M S,≤ ], [N S,≤ ]), and N has property
(F). Then σ(α(φ)) ≤ σ(φ) for any φ ∈ [M S,≤ ] with α(φ) 6= 0. In particular, if
α is injective and M has property (F), then σ(α(φ)) = σ(φ).
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Yang Gang and Zhao Liang
The property that any essentially injective endomorphism of a module is to
be an automorphism is very important since a quasi-continuous module M is
continuous if and only if every monomorphism M −→ M with essential image
is an isomorphism [6, P.46].
Theorem 2.3. Let M be a left R-module having property (F). Then the following
conditions are equivalent:
(1) Any essentially injective endomorphism of R M is an automorphism.
(2) Any essentially injective endomorphism of [[RS,≤ ]] [M S,≤ ] is an automorphism.
Proof. (1)=⇒(2). Let α ∈ End[[RS,≤ ]] ([M S,≤ ]) be essentially injective. Define
the map f : M −→ M via f (m) = α(φ0,m )(0), where m ∈ M . It will now be
shown that f is an injective R-homomorphism. For any m ∈ M, x ∈ S we have
X
d0r φ0,m (x) =
d0r (y)φ0,m (y + x) = d0r (0)φ0,m (x) = rφ0,m (x) = φ0,rm (x).
y∈S
Thus d0r φ0,m = φ0,rm . Therefore
f (rm) = α(φ0,rm )(0) = α(d0r φ0,m )(0) = [d0r α(φ0,m )](0)
X
=
d0r (y)α(φ0,m )(y) = rα(φ0,m )(0) = rf (m).
y∈S
For any m ∈ Ker(f ), we have α(φ0,m )(0) = f (m) = 0, clearly α(φ0,m )(x) = 0
for 0 6= x ∈ S by Lemma 2.2. Thus, φ0,m = 0 since α is injective, and so m = 0.
Since α is essentially injective, and for any 0 6= m ∈ M , there are two
elements d ∈ [[RS,≤ ]] and φ ∈ [M S,≤ ] such that α(φ) = dφ0,m 6= 0. It is easy
to check that dφ0,m = d0r φ0,m = φ0,rm 6= 0 for some r ∈ R with r = d(0).
Hence, we conclude that σ(φ) = 0 and φ = φ0,m0 for some m0 ∈ M . It follows
that 0 6= rm = α(φ)(0) = α(φ0,m0 )(0) = f (m0 ) ∈ Im(f ), which implies that
Im(f ) ≤e R M by Lemma 2.1, and we have Im(f ) = M , by condition (1).
In the following, we will prove Im(α) = [M S,≤ ] by using induction. For any
element 0 6= φ ∈ [M S,≤ ], let σ(φ) = s.
Case 1. If s = 0, let m = f −1 (φ(0)) ∈ M . Then for any t ∈ S, it is clear that
0,
t 6= 0;
α(φ0,m )(t) =
f (m) = φ(0), t = 0.
= φ(t).
Hence, there is an element φ0,m ∈ [M S,≤ ] so that α(φ0,m ) = φ.
Case 2. If 0 < s. Suppose that there exists an element ψ 0 ∈ [M S,≤ ] such that
α(ψ 0 ) = ψ for any element 0 6= ψ ∈ [M S,≤ ] with σ(ψ) = u < s. Now, it suffices
to show that for φ ∈ [M S,≤ ] with σ(φ) = s, there exists an element φ0 ∈ [M S,≤ ]
and so α(φ0 ) = φ.
Modules Whose Essentially Injective Endomorphisms Are Automorphisms
1095
Since f is an automorphism, there exists an element m ∈ M such that
f (m) = φ(s). Let em = m and eα(φs,m )(s) = α(φP
s,m )(s) for some element
s
e ∈ R. Then, it is obvious that (dse φs,m )(t) =
y∈S de (y)φs,m (t + y) =
eφs,m (t + s) and dseP
φs,m = φ0,m . Thus, we have α(φ0,m )(t) = α(dse φs,m )(t) =
s
[dse α(φs,m )](t) =
y∈S de (y)α(φs,m )(y + t) = eα(φs,m )(s + t). Therefore,
α(φs,m )(s) = α(φ0,m )(0) = f (m) = φ(s), which implies that [φ − α(φs,m )](s) =
φ(s) − α(φs,m )(s) = φ(s) − f (m) = 0. Clearly, for any s ≤ t ∈ S,
[φ − α(φs,m )](t) = 0. If φ − α(φs,m ) 6= 0, then σ(φ − α(φs,m )) < s. By induction,
there is an element ψ 0 ∈ [M S,≤ ] such that φ − α(φs,m ) = α(ψ 0 ), which implies
that α(φ0 ) = φ, where φ0 = ψ 0 + φs,m ; If φ − α(φs,m ) = 0, then φ = α(φs,m ).
Hence, α is an automorphism.
(2)=⇒(1). Let f : M −→ M be an essentially injective R-homomorphism.
Define a map α : [M S,≤ ] −→ [M S,≤ ] via α(φ)(s) = f (φ(s)), where
P s ∈ S and φ ∈
[M S,≤ ].PFor any g ∈ [[RS,≤ ]], we have α(gφ)(s) = f (gφ(s)) = y∈S g(y)f (φ(s +
y)) = y∈S g(y)α(φ)(s + y) = (gα(φ))(s). If α(φ) = 0, then for every s ∈ S,
f (φ(s)) = α(φ)(s) = 0, which implies that φ(s) = 0 for any s ∈ S Since f
σ(ψ)
is essentially injective, for every 0 6= ψ ∈ [M S,≤ ], de ψ = φ0,m , where m =
ψ(σ(ψ)) and e ∈ R with em = m. Also, there exist elements m0 ∈ M and r ∈ R
such that 0 6= rm = f (m0 ) = f (φ0,m0 (0)) = α(φ0,m0 )(0). It is obvious that
σ(ψ)
α(φ0,m0 )(t) = 0, for any 0 < t ∈ S. Thus 0 6= d0r φ0,m = d0r de ψ = α(φ0,m0 ),
which implies that Im(α) ≤e [M S,≤ ]. Hence, α is an automorphism. For every
m ∈ M , there exists an element φ ∈ [M S,≤ ] such that m = φ0,m (0) = α(φ)(0) =
f (φ(0)). This implies that Im(f ) = M .
The following corollaries will give more examples of modules whose essentially
injective endomorphisms are automorphisms.
Corollary 2.4. Let (S1 , ≤1 ), · · · , (Sn , ≤n ) be strictly totally monoids which are
also artinian. Denote by (lex ≤) and (revlex ≤) the lexicographic order, the
reverse lexicographic order, respectively, on the monoid S1 × · · · × Sn . If R M
has property (F), then the following conditions are equivalent:
(1) Any essentially injective endomorphism of R M is an automorphism.
(2) Any essentially injective endomorphism of
[[RS1 ×···×Sn ,(lex≤) ]] [M
S1 ×···×Sn ,(lex≤)
]
is an automorphism.
(3) Any essentially injective endomorphism of
[[RS1 ×···×Sn ,(revlex≤) ]] [M
S1 ×···×Sn ,(revlex≤)
]
is an automorphism.
Proof. It can be easyly to observed that (S1 × · · · × Sn , (lex ≤)) and (S1 × · · · ×
Sn , (revlex ≤)) are strictly totally ordered monoids which are artinian. Hence,
the results follow from Theorem 2.3.
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Yang Gang and Zhao Liang
Corollary 2.5. If R M has property (F), then the following conditions are
equivalent:
(1) Any essentially injective endomorphism of R M is an automorphism.
(2) Any essentially injective endomorphism of [[R(N,·)≤ ]] [M (N,·)≤ ] is an
automorphism.
Proof. The proof follows from Theorem 2.3.
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