EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 10, No. 2, 2017, 211-230
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

A module whose second spectrum has the surjective or
injective natural map

H. Ansari-Toroghy1, S. S. Pourmortazavi 1,∗

1 Department of Pure Mathematics, Faculty of Mathematical Science, University of Guilan,
Rasht, Iran

Abstract. Let R be a commutative ring and M be an R-module. Let Specs(M) be the set of
all second submodules of M . In this article, we topologize Specs(M) with Zariski and classical
Zariski topologies and study the classes of all modules whose second spectrum have the surjective
or injective natural map. Moreover, we investigate the interplay between the algebraic properties
of M and the topological properties of Specs(M).

2010 Mathematics Subject Classifications: 13C13, 13C99

Key Words and Phrases: Cotop module, second submodule, Xs-injective module, Zariski
topology

1. Introduction

Throughout this article, R denotes a commutative ring with identity and all modules
are unitary. Also the notation Z (resp. Q) will denote the ring of integers (resp. the
field of fractions of Z). If N is a subset of an R-module M , then N ≤ M denotes N
is an R-submodule of M . For any ideal I of R containing AnnR(M), R̄ and Ī denote
R/AnnR(M) and I/AnnR(M), respectively. The colon ideal of M into N is defined to be
(N : M) = {r ∈ R : rM ⊆ N} = AnnR(M/N).

Let M be an R-module. A proper submodule N of M is said to be prime if for any
r ∈ R and m ∈ M with rm ∈ N , we have m ∈ N or r ∈ (N :R M). This implies that
(N :R M) = p is prime ideal of R and we say that N is a p-prime submodule of M .

The dual notion of prime submodules (i.e., second submodules) was introduced and
studied in [30]. A non-zero submodule S of M is said to be second if for each a ∈ R
the homomorphism S

a→ S is either surjective or zero. This implies that AnnR(S) = p
is a prime ideal of R and S is said to be p-second. More information about this class of
modules can be seen in [4, 5, 7, 8, 14, 16, 15].

∗Corresponding author.

Email addresses: ansari@guilan.ac.irr (H. Ansari-Toroghy), mortazavi@phd.guilan.ac.ir (S. S.
Pourmortazavi)

http://www.ejpam.com 211 c© 2017 EJPAM All rights reserved.



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 212

The concept of prime submodule has led to the development of topologies on the
spectrum of modules. A brief history of this development can be seen in [23, p. 808].
More information concerning the spectrum of rings, posets, and modules can be found in
[1, 2, 7, 8, 12, 14, 20, 22, 26].

The second spectrum of M is defined as the set of all second submodules of M and
denoted by Specs(M) or Xs. We call the map ψ : Xs → Spec(R) given by S 7→ AnnR(S)
as the natural map of Xs.

Let N be a submodule of M . Define V s(N) := {S ∈ Specs(M) : AnnR(N) ⊆
AnnR(S)} and set ζs(M) := {V s(N) : N ≤M}. Then there exists a topology, τ s say, on
Specs(M) having ζs as the family of all its closed sets. This topology is called the Zariski
topology on Specs(M) (see [7]).

For any submodule N of M , define V s∗(N) = {S ∈ Specs(M) : S ⊆ N}. Set ζs∗(M) =
{V s∗(N) : N ⊆M}. Then ζs∗(M) contains the empty set and Specs(M), and it is closed
under arbitrary intersections. In general ζs∗(M) is not closed under finite unions. A
module M is called a cotop module if ζs∗(M) is closed under finite unions. In this case,
ζs∗(M) is called the quasi Zariski topology (see [7]).

Now for a submodule N of M , define W s(N) = Specs(M)−V s∗(N) and set Ωs(M) =
{W s(N) : N ≤M}. Let ηs(M) be the topology on Specs(M) by the sub-basis Ωs(M). In
fact ηs(M) is the collection U of all unions of finite intersections of elements of Ωs(M) . We
call this topology the classical Zariski topology on Specs(M) (or second classical Zariski
topology) (see [8]). It is clear that if M is a cotop module, then its related topology, as it
was mentioned in the above paragraph, coincide with the second classical Zariski topology.

The dual notion of the prime radical of a submodule (i.e, second socle or second radical)
have been introduced by H. Ansari-Toroghy and F. Farshadifar in [4]. For a submodule
N of M , the second radical of N is defined as the sum of all second submodules of M
contained in N and denoted by sec(N) (or soc(N)). In case N does not contain any
second submodule, the second radical of N is defined to be (0) . Also, N 6= (0) is said
to be a socle submodule of M if sec(N) = N . One can see that M is cotop if and only if
for every two socle submodules N and K of M and every second submodule S of M with
S ⊆ N +K, we have S ⊆ N or S ⊆ K.

In this article, we introduce the concept of Xs-injective modules and investigate some
of their basic properties. We say that M is Xs-injective if the natural map of Xs is
injective (see Definition 3.9).

The primeful R-modules was introduced and studied by C.P. Lu in several papers.
The dual of this notion (i.e., secondful modules) was studied by H. Ansari-toroghy and
F. Farshadifar in [3] and [17]. An R-module M is secondful if the natural map of Xs is
surjective. In section two, we explore more properties of this class of modules. In Theorem
2.3, we provide a useful characterization for Artinian secondful modules and by using this,
we prove that if M is an Artinian secondful module with Noetherian spectrum, then
HomR(Rp,M) has also a Noetherian second spectrum with Zariski topology. Moreover,
in Theorem 2.9 we provide a useful characterization for secondful modules.

In section three, among other results, we show that if every closed subset of (Specs(M),
τ s∗) has a finite number of irreducible components, then every submodule of M has a finite



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 213

number of maximal second submodules. We provide an example which shows the converse
is not true in general. In [8, Proposition 3.13], it is proved that if M has dcc condition
on its socle submodules, then every irreducible closed subset of Specs(M) (with second
classical Zariski topology) has a generic point. In Theorem 3.2, we remove this restriction
and prove that every irreducible closed subset of Specs(M) has a generic point. Further it
is shown that if Specs(M) is a Noetherian space, then it is a spectral space (see Corollary
3.3). Proposition 3.13 states that if (Mi)i∈I is a family of R-modules and M =

⊕
i∈IMi is

an Xs-injective module, then Specs(M) can be specified in terms of second submodules of
Mi. Theorem 3.15 says that M =

⊕
i∈IMi is an Xs-injective (resp. a cotop) R-module if

and only if (Mi)i∈I is a family of second-compatible Xs-injective (resp. cotop) R-modules.
Moreover, in Theorem 3.18, we prove that if R is a perfect ring, then the classes of cotop,
weak comultiplication, and Xs-injective modules are all equal.

In the rest of this article, Xs := Specs(M) will denote the set of all second submodules
of M . Also the map ψ : Xs → Spec(R) given by S 7→ AnnR(S) is called the natural map
of Xs.

2. Secondful modules

We recall that an R-module M is secondful if the natural map Xs is surjective. For
example, for each positive integer n (n > 1), Zn as Z-module is secondful.

Throughout this section, we assume that Specs(M) is topologized with Zariski topol-
ogy.

Remark 2.1. Let M be an R-module. Then we have the following.

(a) Let M be an Artinian module. Then M is secondful ⇔ HomR(Rp, (0 :M p)) 6= (0)
for every p ∈ V (AnnR(M)) [7, Theorem 3.8].

(b) Let M be a secondful R-module. Then M has a Noetherian second spectrum if and
only if the ring R has Noetherian spectrum [17, Theorem 4.3].

Lemma 2.2. Let M be an R-module and p, q be two prime ideals of R such that q ⊆ p.
Then

HomR(Rq, (0 :M q)) ∼= HomRp((Rp)qRp , (0 :HomR(Rp,M) qRp)).

Proof.

HomR(Rq, (0 :M q)) ∼= HomR(Rp ⊗Rp Rq, (0 :M q))
∼= HomRp(Rq, HomR(Rp, HomR(R/q,M)))
∼= HomRp(Rq, HomR(Rp ⊗R R/q,M))
∼= HomRp(Rq, (0 :HomR(Rp,M) qRp))

∼= HomRp((Rp)qRp , (0 :HomR(Rp,M) qRp)).



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 214

Theorem 2.3. Let M be an Artinian R-module. Then the following statements are equiv-
alent:

(a) M is a non-zero secondful R-module;

(b) HomR(Rp,M) is non-zero secondful Rp-module for every p ∈ V (AnnR(M)).

Proof. (a) ⇒ (b) Let p be a prime ideal of R such that p ⊇ AnnR(M). Since M
is a non-zero secondful module, HomR(Rp, (0 :M p)) 6= (0) by Remark 2.1 (a). Hence
we have HomR(Rp,M) 6= (0). To prove that HomR(Rp,M) is secondful Rp-module, let
qRp ∈ V (AnnRp(HomR(Rp,M))). Since Rp is quasi-local ring,

pRp ⊇ qRp ⊇ AnnRp(HomR(Rp,M)) ⊇ AnnR(M)Rp.

Taking the contraction of each term of this sequence, we have that

p ⊇ q ⊇ Sp(AnnR(M)) ⊇ AnnR(M).

Hence HomR(Rq, (0 :M q)) 6= (0) by Remark 2.1 (a). It follows that HomR(Rp,M) is a
secondful Rp-module by Remark 2.1 (a) and Lemma 2.2.

(b)⇒ (a) Let p ∈ V (AnnR(M)). By part (b), HomR(Rp,M) 6= (0). ThusAnnRp(HomR(Rp,M)) 6=
Rp. Therefore, AnnRp(HomR(Rp,M)) ⊆ pRp and so pRp ∈ V (AnnRp(HomR(Rp,M))).
Now by Remark 2.1 (a),

HomRp((Rp)p, (0 :HomR(Rp,M) pRp)) 6= (0),

and so HomR(Rp, (0 :M p)) 6= (0). Thus M is a secondful R-module.

Corollary 2.4. Let M be a non-zero Artinian secondful R-module and suppose p ∈
V (AnnR(M)). Then if M has Noetherian second spectrum, so does the Rp-module
HomR(Rp,M).

Proof. Let p ∈ Spec(R) and AnnR(M) ⊆ p. By Theorem 2.3, HomR(Rp,M) is a
secondful Rp-module. Hence, in order to show that HomR(Rp,M) has Noetherian second
spectrum, we need to prove that Rp/AnnRp(HomR(Rp,M)) has Noetherian spectrum by
Remark 2.1 (b). Let S = R\p. Then S = {s + Ann(M) : s ∈ S} is a multiplicative
closed subset of R = R/AnnR(M). Let φ : R→ (R)S be the natural homomorphism and
φ∗ : Spec((R)S)→ Spec(R) the associated mapping. Set Σ = {p ∈ Spec(R) : p ∩ S = ∅}.
Since M is secondful and has Noetherian spectrum, Spec(R) is Noetherian by Remark 2.1
(b). Further Σ as a subspace of Spec(R) is a Noetherian space and it is homeomorphic to
Spec((R)S) by [21, p. 81, Proposition 4.12 (b)]. Hence Spec((R)S) is a Noetherian space.
Also, we have (R)S

∼= Rp/(AnnR(M))p. Thus Spec(Rp/(AnnR(M))p) is a Noetherian
space. On other hand, we have (AnnR(M))p ⊆ AnnRp(HomR(Rp,M)). Now

Rp
AnnRp(HomR(Rp,M))

∼=
(

Rp
(AnnR(M))p

)
/

(
AnnRp(HomR(Rp,M))

(AnnR(M))p

)
.



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 215

From this, we deduce that Rp/AnnRp(HomR(Rp,M)) has Noetherian spectrum.

Let T be a topological space and T ′ ⊆ T . Then T is irreducible if T 6= ∅ and for every
decomposition T = A1 ∪ A2 with closed subsets Ai ⊆ T , i = 1, 2, we have A1 = T or
A2 = T . The subset T ′ of T is irreducible if it is irreducible as a space with the relative
topology. Equivalently, T ′ is irreducible if and only if for every pair of sets F,G which are
closed in T , it holds that T ′ ( F ∪G, T ′ ( F or T ′ ( G [13, p. 94].

An irreducible component of T is a maximal irreducible subset of T . Every irreducible
subset of T is contained in an irreducible component of T , and T is the union of its
irreducible components.

Let Z be a subset of a topological space W . Then the notion cl(Z) will denote the
closure of Z in W .

Let Y be a closed subset of a topological space. An element y ∈ Y is called a generic
point of Y if Y = cl({y}). If the topological space is a T0-space, then a generic point of
every closed subset is unique.

Proposition 2.5. Let M be a secondful module over R and N ≤M .

(a) Let Y be a nonempty subset of V s(N). Then Y is an irreducible closed subset of
V s(N) if and only if Y has a generic point in V s(N).

(b) The mapping ρ : S 7→ V s(S) is a surjection of V s(N) onto the set of irreducible
closed subsets of V s(N).

(c) The mapping φ : V s(S) 7→ AnnR(S) ∈ Spec(R) is a bijection of the set of irreducible
components of V s(N) onto the set of minimal prime divisors of AnnR(N) in R =
R/AnnR(M).

Proof.

(a) This is clear.

(b) This follows directly from part (a).

(c) Let φ : V s(S) 7→ AnnR(S). Then φ is a well defined injective mapping by part (a).
We show that φ is surjective. Let p be a minimal prime divisor of AnnR(N) in R
and let p be the prime ideal of R such that p/AnnR(M) = p. Then AnnR(M) ⊆
AnnR(N) ⊆ p. Since M is secondful, there exists a p-second submodule S ∈ Xs.
Now AnnR(N) ⊆ p = AnnR(S) implies that S ∈ V s(N), and so V s(S) ⊆ V s(N).
Thus V s(S) is an irreducible closed subset of V s(N). Note that the minimality
of p ∈ V (AnnR(N)) implies the maximality of V s(S) among all irreducible closed
subsets V s(S′), S′ ∈ V s(N), as AnnR(N) ⊆ AnnR(S′). Therefore, V s(S) is an
irreducible component of V s(N). This proves that φ is surjective.



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 216

Proposition 2.6. Let M be a secondful R-module. Then Specs(M) has a chain of
irreducible closed subsets of length r if and only if R has a chain of prime ideals of length
r.

Proof.
Let Z0 ( Z1 ( ... ( Zr be a strictly increasing chain of irreducible closed subset

Zi of Specs(M) of length r. By [7, Theorem 5.8], each Zi has a generic point, so that
Zi = V s(Si) for some Si ∈ Specs(M). Hence, we have V s(S0) ( V s(S1) ( ...V s(Sr).
Thus AnnR(S0) ) AnnR(S1) ) ... ) AnnR(Sr), a strictly decreasing chain of prime
ideals R of length r. Conversely, let q0 ) q1 ) ... ) qr be a strictly decreasing chain
of prime ideals in R of length r and, for each i (1 ≤ i ≤ r), let qi be a prime ideal of
R containing AnnR(M) such that qi = qi/AnnR(M). Since M is a secondful R-module,
there exists a qi-second submodule Qi of M for each i (1 ≤ i ≤ r). Hence we have
AnnR(Q0) ) AnnR(Q1) ) ... ) AnnR(Qr) so that V s(Q0) ( V s(Q1) ( ... ( V s(Qr).
Thus, we obtain a strictly increasing chain of length r of irreducible closed subsets of
Specs(M).

We consider strictly decreasing (or strictly increasing) chain Z0, Z1, ..., Zr of length r
of irreducible closed subsets Zi of T . The supremum of the lengths, taken over all such
chains, is called the combinatorial dimension of T and denoted by dimT (for T = ∅, define
dimT = −1).

A submodule N of M is said to be cocyclic if N ⊆ E(R/m) for some maximal ideal
m of R (here E(R/m) denote the injective envelope of R/m) (see [29]). The cosupport
of M , denoted by Cosupp(M), is defined as the set of all prime ideals p of R such that
p ⊇ AnnR(N) for some cocyclic homomorphic image N of M (see [28]).

For every finitely cogenerated module M , Cosupp(M) = V (AnnR(M)) [28, Lemma
2.3]. The equality holds also if M is a secondful module by [17, Theorem 2.5]. Using this
fact and Proposition 2.6, we can establish the next theorem.

Theorem 2.7. Let M be a secondful R-module and equip Specs(M) and Spec(R) with
their Zariski topologies. Then the combinatorial dimension of Specs(M), the Krull di-
mension of R = R/AnnR(M), and the combinatorial dimension of the closed subspace
Cosupp(M) = V (AnnR(M)) of Spec(R) are all equal.

As a result of Theorem 2.7, the definition of the classical Krull dimension of modules,
which is independent of Specs(M), becomes more significant for secondful modules M .

Let p be a prime ideal of R. For an R-module M , Specsp(M) denotes the set of all
p-second submodules of M .

Corollary 2.8. Let M be a secondful R-module such that Specs(M) has zero combina-
torial dimension. Then we have the following.

(a) Every irreducible closed subset of Specs(M) is an irreducible component of Specs(M).

(b) For every p ∈ V (AnnR(M)) and for every p-second submodule S of M , Specsp(M) =
V s(S).



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 217

(c) If M has Noetherian second spectrum, then the set of irreducible components of
Specs(M) is {V s((0 :M m1)), V s((0 :M m2)), ..., V s((0 :M mk))} for some positive
integer k, where mi, for i = 1, 2, ..., k, are all the minimal prime divisors of AnnR(M).

Proof.

(a) Let V s(N) be an irreducible closed subset of Specs(M). By Proposition 2.5 (a), there
exists S ∈ Specs(M) such that V s(N) = V s(S). Now if V s(S) is not component,
then V s(S) is contained properly in some component V s(S′) for some S′ ∈ Specs(M)
by [13, p. 95, Proposition 5]. It follows that AnnR(S) ) AnnR(S′) and hence
dim(Specs(M)) ≥ 1, a contradiction.

(b) Let p ∈ V (AnnR(M)) and let S be a p-second submodule ofM . Since dim(Specs(M))
= 0, we have dim(R) = 0. Hence Spec(R) = Max(R). Now if S′ ∈ Specs(M), then
AnnR(S′) is also a maximal ideal. Therefore, we have S′ ∈ V s(S) ⇔ AnnR(S′) =
AnnR(S) = p⇔ S′ ∈ Specsp(M). Thus V s(S) = Specsp(M).

(c) Since Specs(M) is a Noetherian space with dim(Specs(M)) = 0, we have dim(R) = 0
and R has a Noetherian spectrum. Now Spec(R) = Max(R) and Spec(R) has only
finitely many elements m1,m2, ...,mk, each of which is both maximal and minimal
prime ideal of R by [21, p. 41, Examples 1.4 (c) and (d)]. Then mi is a minimal
prime divisor of AnnR(M), which is also a maximal ideal of R for every i (1 ≤ i ≤ k).
Thus {m1,m2, ...,mk} is the set of all minimal prime divisor of AnnR(M). Since M
is a secondful R-module, (0 :M mi) 6= (0) and AnnR((0 :M mi)) = mi is a maximal
ideal of R. Thus (0 :M mi) is an mi-second submodule and V s((0 :M mi)) is an
irreducible component of Specs(M) for every i by part (a). Applying Proposition
2.5 (c), we can conclude that {V s((0 :M m1)), V s((0 :M m2)), ..., V s((0 :M mk))} is
the set of all irreducible components of Specs(M).

Theorem 2.9. Let M be a non-zero secondful R-module. Then the following are equiva-
lent:

(a) Specs(M) = Min(M) (here Min(M) denotes the set of all minimal submodules of
M);

(b) (Specs(M), τ s) is a T1-space;

(c) (Specs(M), τ s) is a T4-space;

(d) (Specs(M), τ s) is a T0-space and Cosupp(M) ≈ Spec(R) = Max(R);

(e) Specs(M) = {(0 :M p) | p ∈ V (AnnR(M)) ∩Max(R)}.

Proof.
(a) ⇒ (b). Let {S} be a singleton subset of Specs(M). Let S′ ∈ V s(S) with

AnnR(S′) = q. Then by [30, Proposition 1.6], q ∈ Max(R) and (0 :M q) 6= (0), so



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 218

AnnR(0 :M q) = q. By [30, Proposition 1.4], it follows that (0 :M q) ∈ Specs(M) and
hence (0 :M q) ∈Min(M). But AnnR(S) = AnnR(S′) = q implies that S ⊆ (0 :M q) and
S′ ⊆ (0 :M q). Hence S = S′ = (0 :M q). It turns out that {S} = V s(S) is a closed subset,
as required.

(b)⇒ (c). Let Specs(M) be a T1-space. Then Specs(M) is homeomorphic to Spec(R)
by [7, Theorem 6.3]. Thus Spec(R) is a T1-space. Now by [10, p.44, Exer. 11], Spec(R)
is a T1-space if and only if it is a Hausdorff topological space (i.e. T2-space). Hence by
the above arguments, Specs(M) is a Hausdorff topological space. On the other hand, by
[27, chap. 3, Exer. 3.5], every Hausdorff quasi-compact topological space is a T4-space.
Therefore, the result follows from the fact that (Specs(M), τ s) is a compact topological
space [7, Theorem 4.4].

(c) ⇒ (d). Since (Specs(M), τ s) is a T0-space, it is homeomorphic to Spec(R) by [7,
Theorem 6.3]. Thus Spec(R) is T4-space. This implies that Spec(R) = Max(R) by [10,
p. 44, Exer. 11], but Spec(R) is homeomorphic to V (AnnR(M)) by [10, p. 13, Exer. 21].
On the other hand, V (AnnR(M)) = Cosupp(M) by [17, Theorem 2.5]. It follows that
Cosupp(M) ≈ Spec(R) = Max(R), as desired.

(d)⇒ (e). Set T = {(0 :M p) | p ∈ V (AnnR(M))∩Max(R)}. Let p ∈ V (AnnR(M))∩
Max(R). Since M is secondful, (0 :M p) 6= (0). Thus (0 :M p) ∈ Specs(M) by [30,
Proposition 1.4]. Therefore, T ⊆ Specs(M). Conversely, let S ∈ Specs(M). Then
AnnR(S) ∈ Spec(R). This implies that AnnR(S) ∈ V (AnnR(M)) ∩ Max(R). Set
p = AnnR(S). Then S ⊆ (0 :M p), so AnnR(0 :M p) = AnnR(S) = p ∈ Max(R).
Thus (0 :M p) ∈ Specs(M). Now since (Specs(M), τ s) is a T0-space, the natural map of
Specs(M) is injective by [7, Theorem 6.3]. It follows that S = (0 :M p) as required.

(e) ⇒ (a). Let S ∈ Specs(M). Then there is p ∈ V (AnnR(M)) ∩ Max(R) such
that (0 :M p) = S. Now let N be a non-zero submodule of M with N ⊆ S. Then
AnnR(S) = AnnR(N) = p ∈Max(R). Hence we have N ∈ Specsp(M) by [30, Proposition
1.4], so N = (0 :M p) = S. It follows that S is a minimal submodule of M and the proof
is completed.

Corollary 2.10. Let M be a non-zero secondful R-module. Then Specs(M) = Min(M)
if and only if Specs(M) is a singleton set provided that R contains no idempotent other
than 0 and 1; R is such a ring if R is a quasi-local ring or AnnR(M) ∈ Spec(R). (We
recall that R is a quasi-local ring if |Max(R)| = 1.)

Proof. Let Specs(M) be a singleton set. Then Specs(M) is a T1-space. Therefore,
Specs(M) = Min(M) by Theorem 2.9. Conversely if Specs(M) = Min(M), then we have
Spec(R) = Max(R) by Theorem 2.9. But by [10, p. 44, Exer. 11], Spec(R) is totally
disconnected topological space. Thus Specs(M) ≈ Spec(R) is a totally disconnected space.
On the other hand, Specs(M) is a connected space by [7, Corollary 3.13]. Thus Specs(M)
is a single set and the proof is completed.



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 219

3. On the second classical Zariski topology

Throughout this section, Specs(M) is equipped with classical Zariski topology. We
recall that if M is a cotop module, then its related topology (i.e. quasi-Zariski topology)
coincide with classical Zariski topology.

A spectral space is a topological space homeomorphic to the prime spectrum of a
commutative ring equipped with the Zariski topology.

Spectral spaces have been characterized by M. Hochster as quasi-compact T0-space
having a quasi-compact open base closed under finite intersections and each irreducible
closed subset has a generic point (see [19]).

Lemma 3.1. Let M be an R-module and Y be a nonempty subset of SpecsR(M). Then
cl(Y ) = cl(

⋃
S∈Y V

s∗(S)). In particular, when Y is closed we have Y =
⋃
S∈Y V

s∗(S).

Proof. This is straightforward.

Theorem 3.2. For any R-module M , every irreducible closed subset of SpecsR(M) has a
generic point. In particular, this is true when M is a cotop module.

Proof. Let Y be an irreducible closed subset of Specs(M) and
∑

S∈Y S = S1. Then
by [8, Theorem 3.5 (a)], S1 is a second submodule of M . We claim that Y = V s∗(S1). By
Lemma 3.1, it is enough to show that

⋃
S∈Y V

s∗(S) = V s∗(S1). Clearly,
⋃
S∈Y V

s∗(S) ⊆
V s∗(S1). To see the reverse inclusion, let F be a closed subset of Specs(M) containing⋃
S∈Y V

s∗(S). Since F is closed, F =
⋂
i∈Λ

⋃ni
j=1 V

s∗(Ni,j) for some submodules Ni,j of M .
Without loss of generality, We can assume F = V s∗(N1)∪ V s∗(N2), where N1 and N2 are
submodules of M . Now by Lemma 3.1, we have

⋃
S∈Y V

s∗(S) = Y is irreducible. Since⋃
S∈Y V

s∗(S) ⊆ V s∗(N1) ∪ V s∗(N2), we have
⋃
S∈Y V

s∗(S) ⊆ V s∗(N1) or
⋃
S∈Y V

s∗(S) ⊆
V s∗(N2). It follows that S1 =

∑
S∈Y S ⊆ N1 or S1 =

∑
S∈Y S ⊆ N2. Thus V s∗(S1) ⊆

N1 or V s∗(S1) ⊆ N2. Therefore V s∗(S1) ⊆ F . This in turn implies that V s∗(S1) =⋃
S∈Y V

s∗(S). Therefore Y = V s∗(S1) =
⋃
S∈Y V

s∗(S).

Corollary 3.3. Let M be an R-module and suppose Specs(M) is a Noetherian space.
Then Specs(M) is a spectral space. In particular, this is true when M is a cotop module.

Proof. Since Specs(M) is Noetherian, it is quasi-compact and the quasi-compact open
subsets of Specs(M) are closed under finite intersection and form an open base by [10, p.
79, Exer. 6]. Also Specs(M) is a T0 space by [8, Lemma 3.7 (a)]. Now the result follows
from Theorem 3.2 and Hochster’s characterizations.

Let N be a non-zero submodule of an R-module M and let S be a second submodule
of M such that S ≤ N . S is said to be a maximal second submodule of N if there doesn’t
exist S′ ∈ Specs(M) with S � S′ � N (see [4]).

Definition 3.4. Let N be a non-zero submodule of an R-module M and let S be a second
submodule of M . We say that S is a second summand (resp. maximal second summand)
of N if S ∈ V s∗(N) (resp. S ∈ Max(V s∗(N))). If V s∗(N) 6= ∅, then by using Zorn’s
lemma, one can see that N contains a maximal second summand.



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 220

The second submodule dimension of an R-module M , denoted by S.dimM , is defined
to be the supremum of the length of chains of second submodules of M if Specs(M) 6= ∅
and −1 otherwise (see [6]).

Theorem 3.5. Let M be an R-module.

(a) Let Y be a closed subset of Specs(M). Then Z is an irreducible component of Y if
and only if Z = V s∗(S) for some maximal element S of Y .

(b) If every closed subset of Specs(M) has a finite number of irreducible components,
then every submodule of M has a finite number of maximal second summands. How-
ever, the converse is not true in general.

(c) dim(Specs(M)) = S.dimM .

Proof.

(a) Let Y be a closed subset of Specs(M) and let Z be an irreducible component of Y .
Since every irreducible component is closed, Z is closed in Y . But every irreducible
closed subset of Y is an irreducible closed subset of Specs(M). Therefore Z =
V s∗(S1) for some second submodule S1 of M by Theorem 3.2. Since Z = V s∗(S1) ⊆
Y is an irreducible component of Y , it follows that S1 is a maximal element in
Y (note that if S1, S2 ∈ Specs(M), then S1 ≤ S2 ⇔ V s∗(S1) ⊆ V s∗(S2) by [8,
Corollary 3.2 (b)]). Conversely, suppose S is a maximal element of Y . Then V s∗(S)
is an irreducible closed subset of Specs(M). Since S ∈ Y , V s∗(S) ⊆ cl(Y ) = Y
by Lemma 3.1. Now let V s∗(S) ⊆ T , where T is an irreducible subset of Y . This
implies that cl(T ) be an irreducible closed subset of Specs(M) and so (T ) = V s∗(S1)
for some second submodule of M by Theorem 3.2. It follows that S ≤ S1 so that
S = S1. Hence T = V s∗(S) is a maximal irreducible subset of Y .

(b) Let N be a non-zero submodule of M and let S be a maximal second summand of
N . Then V s∗(S) is an irreducible component of V s∗(N) by part (a). Hence every
submodule of M has a finite number of maximal second submodules by hypothesis.
To see the second assertion, set M = Q⊕Q⊕Q⊕ · · · and regard M as Q-module.
The second submodules of a vector space are just the non-zero submodules, so every
submodule of M has a finite number of maximal second summands. Now let S1 =
(0) ⊕ Q ⊕ Q ⊕ · · · , S2 = Q ⊕ (0) ⊕ Q ⊕ · · · , S3 = Q ⊕ Q ⊕ (0) ⊕ Q ⊕ · · · , · · · ,
S′1 = Q⊕(0)⊕(0)⊕· · · , S′2 = (0)⊕Q⊕(0)⊕· · · , S′3 = (0)⊕(0)⊕Q⊕(0)⊕· · · , · · · , and
Y =

⋂
i∈N(V s∗(Si) ∪ V s∗(S′i)). One can see that S′1, S

′
2, · · · are maximal elements

of Y . Hence Y is a closed subset of Specs(M) with infinitely many irreducible
components by part (a).

(c) Let Z0 ( Z1 ( ... ( Zt be a strictly increasing chain of irreducible closed subsets Zi
of Specs(M) of length t. By Theorem 3.2, for each i, 0 ≤ i ≤ t, we have Zi = V s∗(Si)
for some Si ∈ Specs(M). On the other hand V s∗(Si) ( V s∗(Sj) if and only if Si ( Sj .



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 221

Hence S0 ) S1 ) ... ) St, is an strictly decreasing chain of second submodules of
M of length t. Conversely, for every strictly decreasing chain S0 ) S1 ) ... ) St of
second submodules of M of length t, V s∗(S0) ( V s∗(S1) ( ... ( V s∗(St) is a strictly
increasing chain of irreducible closed subsets of Specs(M) of length t. This in turn
implies that dim(Specs(M)) = S.dimM and the proof is completed.

A proper submodule N of an R-module M is said to be completely irreducible if
N =

⋂
i∈I Ni, where {Ni}i∈I is a family of submodules of M , implies that N = Ni for

some i ∈ I (see [18]).
Let p be a prime ideal of R and let N be a submodule of an R-module M . Then

N ec = {m ∈ M : cm ∈ N for some c ∈ R\p} and it is called the p-closure of N and
denoted by clp(N) (see [24, p. 92]). The dual of this notion, i.e., p-interior of N relative
to M is defined as the set IMp (N) :=

⋂
{L|L is a completely irreducible submodule of M

and rN ⊆ L for some r ∈ R\p} (see [4]). It is easy to see that IMp (N) =
⋂
r∈R\p rN .

Let R be an integral domain. A submodule N of an R-module M is said to be
cotorsion-free (resp. cotorsion) if IM0 (N) = N (resp. IM0 (N) = (0)) (see [5]).

We need the following lemma.

Lemma 3.6. Let p and q be prime ideals of R with q ⊆ p. Let M be an R-module with
clp(0) = (0) and let φ : HomR(Rp,M) → M be the natural homomorphism given by
f 7→ f(1/1). Then we have the following.

(i) If S is an Rp-submodule of HomR(Rp,M), then we have Sec = S = HomR(Rp, L),
where L = Se. (Here T e, where T ⊆ HomR(Rp,M), and N c, where N ⊆M , denote
φ(T ) and φ−1(N), respectively.)

(ii) IfM is an ArtinianR-module andK is a q-second submodule ofM , thenHomR(Rp,K)
is a qRp-second submodule of HomR(Rp,M) and Kc = HomR(Rp,K). Further we
have (HomR(Rp,K))e = φ(HomR(Rp,K)) = Ip(K) = K and Kce = K.

Proof.

(i) Clearly, S ⊆ Sec. So let g ∈ Sec. Then there exists f ∈ S such that g(1/1) =
f(1/1) ∈ Se and so f(r/1) = g(r/1) for every r ∈ R. Now let λ ∈ Rp, then λ = r/s
for some r ∈ R and s ∈ R − p. Then we have s(f(λ)− g(λ)) = 0. Since clp(0) = 0,
f(λ) = g(λ), so g ∈ S. Now HomR(Rp, L) = Sec, where L = Se, follows directly
from the above arguments.

(ii) LetK be a q-second submodule ofM . Then by [25, Theorem 3.1 (2)], φ : HomR(Rp,K)
→ K given by f 7→ f(1/1) is a surjective homomorphism, so HomR(Rp,K) 6= (0).
To see HomR(Rp,K) is a qRp-second, let r

s ∈ RP \qRp and so r ∈ R\q. Thus
rK = K because K is q-second. Since clp(0) = (0) and K is an Artinian R-module,
we see that HomR(Rp, rK) = rHomR(Rp,K) by [25, Proposition 2.4]. This implies
that

r

s
(HomR(Rp,K)) =

1

s
(HomR(Rp, rK)) = HomR(Rp,K).



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 222

If r
s ∈ qRp, then r

s(HomR(Rp,K)) = 1
s (HomR(Rp, rK)) = (0). Further clp(0) = (0)

implies that Kc = HomR(Rp,K). The last assertion follows from [7, Lemma 2.15].

Proposition 3.7. Let p be a prime ideal of R and let M be a cotop R- module with
clp(0) = 0. Then HomR(Rp,M) is a cotop Rp-module.

Proof. Let S be a second submodule of the Rp-module HomR(Rp,M). Then we have
Se = φ(S) 6= (0). Clearly, Se is a second submodule of M . Now let S1 and S2 be socle
submodules of Rp-module HomR(Rp,M) with S ⊆ S1 + S2. Then Se1 and Se2 are socle
submodules of M with Se ⊆ (S1 +S2)e = Se1 +Se2. Since M is a cotop R-module, we have
Se ⊆ Se1 or Se ⊆ Se2. Thus we have S = Sec ⊆ Sec1 = S1 or S = Sec ⊆ Sec2 = S2 by Lemma
3.6 (i). This implies that HomR(Rp,M) is a cotop Rp-module.

Example 3.8. Let R be an integral domain and let Q be the field of fractions of R.
Then clearly, Q is a torsion-free cotop R-module and hence for every prime ideal p of R,
HomR(Rp, Q) is a cotop Rp-module by Proposition 3.7. Note that since Q is a torsion-free
R-module, clp(0) = (0) for every prime ideal p of R.

Let M be an R-module. The class of X-injective modules, where X is the set of all
prime submodules of M , was studied by H. Ansari-Toroghy and R. Ovlyaee-Sarmazdeh
in [9]. M is X-injective if the map φ : X → R̄ given by P 7→ (P :R M) = (P :R
M)/AnnR(M) is injective.

Definition 3.9. Let M be an R-module. We say that M is an Xs-injective if the nat-
ural map of Xs is injective. Equivalently, M is Xs-injective if and only if AnnR(S1) =
AnnR(S2), S1, S2 ∈ Xs, implies that S1 = S2 if and only if for every p ∈ Spec(R),
|Specsp(M)| ≤ 1.

Let M be an R-module. M is said to be a comultiplication module if for every sub-
module N of M there exists an ideal I of R such that N = (0 :M I) (see [3]).

Example 3.10. Let p be a prime integer.

(a) Every comultiplication module is Xs-injective. In particular Zp∞ = E(Z/pZ) is an
Xs-injective Z-module. Note that Specs(Zp∞) = {〈1/p+ Z〉,Zp∞}.

(b) Let M = Q ⊕ Zp. Then M is Xs-injective Z-module. Note that Specs(M) =
{Q⊕ (0), (0)⊕ Zp}.

(c) Every submodule of an Xs-injective R-module is an Xs-injective module.

The following example shows that not every homomorphic image of an Xs-injective
R-module is Xs-injective.



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 223

Example 3.11. Let p be a positive prime integer and let M = Q ⊕ Zp. Then M is an
Xs-injective Z-module but its homomorphic image Z∞p ⊕ Zp is not. note that

Specs(Z∞p ⊕ Zp) = {Z∞p ⊕ (0), (0)⊕ Zp, < 1/p+ Z >}.

Remark 3.12. Let S be a commutative ring with identity. S is said to be a perfect ring
if it satisfies DCC on principal ideals. Clearly, every Artinian ring is perfect. Note that if
S is a perfect ring and p ∈ Spec(S), then by [11, Lemma 2.2], R/p is a perfect domain so
that it is a field. Hence dim(S) = 0. Furthermore, every perfect ring is a semilocal ring
by [11, Theorem P or p. 475, Examples (6)].

Proposition 3.13. (i) If M is a cotop module over a semilocal (for example a perfect)
ring R. Then (0 :M Jac(R)) is cyclic.

(ii) Let (Mi)i∈I be a family of R-modules and let M =
⊕

i∈IMi. If M is an Xs-injective
module, then

Specs(M) =

{
S ⊕

(⊕
j 6=i∈I

(0)

)
| j ∈ I, S ∈ Specs(Mj)

}
.

Proof.

(i) Suppose that M is a cotop module. Let m1, m2, ..., mn denote the distinct maximal
ideals of R, where n is a positive integer. By [7, Corollary 2.6 (e)], (0 :M mi) is
cyclic for each 1 ≤ i ≤ n. We show that (0 :M Jac(R)) =

⊕n
i=1(0 :M mi). Set

I = m2 ∩m3 ∩ ... ∩mn, so that Jac(R) = m1 ∩ I. Note that m1 + I = R implies
that 0 = (0 :M R) = (0 :M m1 + I) = (0 :M m1) ∩ (0 :M I). Moreover,

(0 :M m1) + (0 :M I) = ((0 :M m1) + (0 :M I) :M m1 + I)

⊇ ((0 :M m1) :M I) ∩ ((0 :M I) :M m1)

= (0 :M m1I) = (0 :M Jac(R))

⊇ (0 :M m1) + (0 :M I).

Therefore (0 :M m1) + (0 :M I) = (0 :M Jac(R)). By induction, this implies that
(0 :M Jac(R)) =

⊕n
i=1(0 :M mi). Without loss of generality, we may assume that

(0 :M mi) 6= 0. Thus we have

(0 :M Jac(R)) =
n⊕
i=1

(0 :M mi) =
n⊕
i=1

R/mi
∼= R/Jac(R).

Hence (0 :M Jac(R)) is cyclic.

(ii) The inclusion ⊇ is trivial. Conversely, let S be a p-second submodule of M . Then
there exists j ∈ I such that S * (0) ⊕ (

⊕
j 6=i∈I(Mi)). By [7, Lemma 2.2 (b)],

S+((0)⊕(
⊕

j 6=i∈I(Mi)))

(0)⊕(
⊕

j 6=i∈I(Mi))
∈ Specsp(Mj). This implies that S = Sj ⊕ (

⊕
j 6=i∈I(0)), where

Sj ∈ Specs(Mj).



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 224

Definition 3.14. A family (Mi)i∈I of R-modules is said to be second-compatible if for all
i 6= j in I, there doesn’t exist a prime ideal p in R with Specsp(Mi) and Specsp(Mj) both
nonempty.

Theorem 3.15. Let (Mi)i∈I be a family of R-modules and let M =
⊕

i∈IMi. Then M
is an Xs-injective (resp. a cotop) R-module if and only if (Mi)i∈I is a family of second-
compatible Xs-injective (resp. cotop) R-modules.

Proof. We consider the proof for two cases Xs-injective and cotop modules.

Case(I). Xs-injective modules.
(⇒). Since every submodule of an Xs-injective module is Xs-injective, Mi’s are Xs-
injective. Let i 6= j and let S ∈ Specsp(Mi) andK ∈ Specsp(Mj). Then S⊕(

⊕
i 6=k∈I(0)),K⊕

(
⊕

j 6=k∈I(0)) ∈ Specsp(M). Since M is Xs-injective, S ⊕ (
⊕

i 6=j∈I(0)) = K ⊕ (
⊕

j 6=i∈I(0)),
a contradiction. Hence Mi’s are second-compatible.
(⇐). Let S be a second submodule of M and let p = AnnR(S). Since S 6= 0, it follows that

there exists j ∈ I with S * (
⊕

j 6=i∈IMi) ⊕ (0) and so
S+((

⊕
j 6=i∈I Mi)⊕(0))

(
⊕

j 6=i∈I Mi)⊕(0) ∈ Specsp(Mj).

By hypothesis, Specsp(Mi) is empty for all j 6= i ∈ I and hence S ⊆ (
⊕

i 6=k∈IMk) ⊕ (0).
This implies that

S ⊆
⋂

j 6=i∈I

(( ⊕
i 6=k∈I

Mk

)
⊕ (0)

)
= Mj ⊕

(⊕
j 6=i∈I

(0)

)
.

Thus there exist Sj ∈ Specsp(Mj) such that S = Sj ⊕ (
⊕

j 6=i∈I(0)). Now let S,K ∈
Specsp(M). By the above arguments, there exist Si ∈ Specsp(Mi) and Sj ∈ Specsp(Mj) such
that S = Si ⊕ (

⊕
i 6=j∈I(0)) and K = Sj ⊕ (

⊕
j 6=i∈I(0)). This implies that i = j because

Mi’s are second compatible. Hence K = S, i.e., M is Xs-injective.

Case(II). Cotop modules.
(⇒). Since every submodule of a cotop module is cotop, Mi’s are cotop. Let i 6= j
and let S ∈ Specsp(Mi) and K ∈ Specsp(Mj). Set S1 = S ⊕ (

⊕
i 6=k∈I(0)) and K1 =

K ⊕ (
⊕

j 6=k∈I(0)). Thus S1 + K1 ∈ Specsp(M). Since M is cotop, K1 ⊆ S1 or S1 ⊆ K1,
which is a contradiction. Therefore Mi’s are second-compatible.
(⇐). Let S be p-second and let S1 and S2 be socle submodules of M such that S ⊆ S1+S2.
Then by similar arguments in case(I), for each i ∈ I, there exist submodules S1i and
S2i of Mi, such that Sk =

⊕
i∈I Ski (k = 1, 2) and each submodule Ski is either so-

cle submodule or equals (0). Also there exists a second submodule Sj of Mj such that
S = Sj⊕ (

⊕
j 6=i∈I(0)). Therefore Sj ⊆ S1j +S2j . Since Mj is cotop, Sj ⊆ S1j or Sj ⊆ S2j .

It follows that S ⊆ S1 or S ⊆ S2 and hence M is a cotop module.

An R-module M is said to have the double annihilator conditions if for each ideal I of
R, we have I = AnnR(0 :M I).

Corollary 3.16. (a) Let R be a domain with field of fractions Q and suppose M is an
R-module such that M/I0(M) is finitely cogenerated. Then the R-module Q⊕M is



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 225

a cotop (resp. an Xs-injective) module if and only if M is a cotorsion cotop (resp.
Xs-injective) module.

(b) Let (R,m) be a local ring, Iλ (λ ∈ Λ) a family of ideals of R, and (0 :M m) 6= 0. If
M =

⊕
λ∈Λ(0 :M Iλ) is a cotop (resp. an Xs-injective) R-module, then the ideals Iλ

(λ ∈ Λ) are comaximal.

(c) Let M be a weak comultiplication module and let (Iλ)λ∈Λ be a family of ideals of
R. If M =

⊕
λ∈Λ(0 :M Iλ) and the ideals Iλ (λ ∈ Λ) are comaximal, then M is a

cotop (resp. an Xs-injective) R-module.

Proof. We just consider the proof for cotop modules. We have similar arguments when
M is an Xs-injective module.

(a) (⇐). We can see that Specs(Q) = Specs0(Q) = {Q}. We show that R-module
M and R-module Q are second-compatible. Let S ∈ Specs0(M). By [5, Theorem
2.10], IM0 (S) = S. Since IM0 (S) ⊆ IM0 (M) and M is cotorsion, we have S = (0), a
contradiction. Therefore Q⊕M is cotop by Theorem 3.15.
(⇒). By Theorem 3.15, Q and M are cotop modules and second-compatible. If M
is not cotorsion, then IM0 (M) belongs to Specs0(M) by [4, Corollary 2.10] which is a
contradiction. Thus M is cotorsion.

(b) Let M be a cotop R-module, and λ, λ′ ∈ Λ. If Iλ + Iλ′ 6= R, then Iλ + Iλ′ ⊆ m.
Therefore (0 :M m) ⊆ (0 :M Iλ + Iλ′) = (0 :M Iλ) ∩ (0 :M Iλ′). But (0 :M m) is an
m-second submodule of M . It follows that Specsm((0 :M Iλ)) and Specsm((0 :M Iλ′))
are both nonempty sets which is a contradiction by Theorem 3.15.

(c) We show that for every λ ∈ Λ, (0 :M Iλ) is a cotop R-module. Let S ∈ Specsp((0 :M
Iλ)) and S1, S2 are socle submodules of (0 :M Iλ) such that S ⊆ S1 + S2. Thus
AnnR(S1+S2) = AnnR(S1)∩AnnR(S2) ⊆ AnnR(S) ∈ Spec(R). ThereforeAnnR(S1)
⊆ p or AnnR(S2) ⊆ p. Since M is weak comultiplication, S = (0 :M p) ⊆ (0 :M
AnnR(S1)) = S1 or S = (0 :M p) ⊆ (0 :M AnnR(S1)) = S2. We show that
(0 :M Iλ)’s are second-compatible. Let λ1, λ2 ∈ Λ and q ∈ Spec(R) and let
S1 ∈ Specsq((0 :M Iλ1)) and S2 ∈ Specsq((0 :M Iλ2)). Since Si ⊆ (0 :M Iλi), we
have Iλi ⊆ AnnR((0 :M Iλi)) ⊆ AnnR(Si) = q (i = 1, 2). This implies that
R = Iλ1 + Iλ2 ⊆ q, a contradiction by hypothesis.

Theorem 3.17. Let M be a non-zero Xs-injective Artinian R-module.

(a) Specs(M) = {IMp ((0 :M p)) | p ∈ V (AnnR(M)), IMp (0 :M p) 6= (0)},

Min(M) = {(0 :M p) | p ∈Max(R), (0 :M p) 6= (0)}.

(b) If M is secondful, then

Specs(M) = {IMp ((0 :M p)) | p ∈ V (AnnR(M))},

Min(M) = {(0 :M p) | p ∈ V (AnnR(M)) ∩Max(R)}.



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 226

(c) If R is PID and M is faithful secondful, then

Specs(M) = Min(M) ∪ {IM0 (M)}, where

Min(M) = {(0 :M p) | p ∈Max(R)}.

Proof.

(a) Put T = {IMp ((0 :M p)) | p ∈ V (AnnR(M)), IMp ((0 :M p)) 6= (0)}. Then T ⊆
Specs(M) by [4, Lemma 2.9]. Let S ∈ Specs(M). Then Specs(M) 6= ∅ for p =
AnnR(S). Thus IMp ((0 :M p)) ∈ Specsp(M) by [4, Lemma 2.9] as IMp ((0 :M p)) 6= (0).

Since M is Xs-injective, S = IMp ((0 :M p)). This implies that S ∈ T . To prove
the second assertion, put Ω = {(0 :M p) | p ∈ Max(R), (0 :M p) 6= (0)} and let
(0 :M p) ∈ Ω. Then (0 :M p) ∈ Specs(M) by [30, Proposition 1.4]. Moreover,
(0 :M p) ∈Min(M) because if K ⊆ (0 :M p) for some non-zero submodule K of M ,
then AnnR((0 :M p)) = AnnR(K) = p and hence (0 :M p) = K because M is Xs-
injective. Thus Ω ⊆ Min(M). Conversely, let S ∈ Min(M). Then S ∈ Specs(M)
and we have p = AnnR(S) ∈ Max(R). Whence p = AnnR(S) = AnnR((0 :M p)).
It follows that S = (0 :M p) 6= (0). Thus S ∈ Ω, and we can conclude that
Min(M) = Ω.

(b) Let p ∈ V (AnnR(M)). Since M is secondful, there exists S ∈ Specs(M) such that
AnnR(S) = p. Thus IMp ((0 :M p)) 6= (0) by [4, Corollary 2.10]. Now the proof
follows from part (a).

(c) Use part (b), the fact that AnnR(M) = (0), and that if M is secondful, then for
every maximal ideal p of R, IMp ((0 :M p)) = (0 :M p).

Let M be an R-module. M said to be a weak comultiplication module if M does not
have any second submodule or for every second submodule S of M , S = (0 :M I) for some
ideal I of R (see [5]).

Assume that V s(sec(N)) = V s∗(sec(N)) for every N ≤ M . Then clearly, τ s∗M ⊆ τ sM
and hence M is a cotop module.

Theorem 3.18. (a) Let R be a perfect ring. Then the three classes of cotop, weak
comultiplication, and Xs-injective modules are all equal.

(b) Let R be a one-dimensional integral domain and let M be an Artinian R-module.
Then M is cotorsion or cotorsion-free Xs-injective if and only if M is weak comul-
tiplication.

(c) Let M be a secondful Xs-injective R-module. If V s(sec(N)) = V s∗(sec(N)) for every
submodule N of M , then (Specs(M), τ s∗) is homeomorphic to Spec(R). Therefore,
(Specs(M), τ s∗) is a spectral space.



H. Ansari-Toroghy, S. S. Pourmortazavi / Eur. J. Pure Appl. Math, 10 (2) (2017), 211-230 227

(d) Let M be an Xs-injective Artinian R-module. If for any p ∈ V (AnnR(M)) and
every family {pi}i∈I , where pi ∈ V (AnnR(M)),

⋂
i∈I pi ⊆ p implies that IMp ((0 :M

p)) ⊆
∑

i∈I I
M
pi ((0 :M pi)), then we have V s(sec(N)) = V s∗(sec(N)).

Proof.

(a) First we assume that M is a cotop R-module and show that it is weak comultipli-
cation. To see this, let S ∈ Specs(M). Clearly, S ⊆ (0 :M AnnR(S)). Since R
is perfect, (0 :M AnnR(S)) a simple R/AnnR(S)-module by [7, Corollary 2.6 (a)
and (d)] and hence a simple R-module. This implies that S = (0 :M AnnR(S)),
as desired. Also, it is clear every weak comultiplication module is an Xs-injective
module. To complete the proof, we assume that M is an Xs-injective module and
show that M is a cotop module. To see this, let S be a second submodule and
let N,L are socle submodules of M such that S ⊆ N + L. Put N =

∑
α∈I Nα

and L =
∑

β∈J Lβ, where Nα and Lβ are second submodules of M for each α ∈ I
and β ∈ J . We have AnnR(N + L) = AnnR(N) ∩ AnnR(L) ⊆ AnnR(S) and so
AnnR(N) ⊆ AnnR(S) or AnnR(L) ⊆ AnnR(S). Thus

⋂
α∈I AnnR(Nα) ⊆ AnnR(S)

or
⋂
β∈J AnnR(Lβ) ⊆ AnnR(S). Now since R is a perfect ring, both I and J

are finite index sets. Consequently, there exists α0 ∈ I or β0 ∈ J such that
AnnR(Nα0) = AnnR(S) or AnnR(Lβ0) = AnnR(S). As M is an Xs-injective mod-
ule, we have Nα0 = S or Lβ0 = S. This implies that S ⊆ N or S ⊆ L. Hence M is
a cotop module.

(b) Let M be a cotorsion-free Xs-injective, and let S be a second submodule of M .
Then we have S = IMp (S) = IMp (0 :M p) for p = AnnR(S) by [4, Corollary 2.10]. If

p = (0), then S = IM0 (S) = IM0 (M) = M = (0 :M 0). If p 6= (0), then p is a maximal
ideal so that S = IMp (0 :M p) = (0 :M p) by [30, Proposition 1.4]. Now let M be a
cotorsion Xs-injective R-module and S a p-second submodule of M . If p = (0), then
S = IM0 (S) = IM0 (M) = (0), a contradiction. This implies that p 6= (0) and hence
S = IMp (S) = IMp (0 :M p) = (0 :M p). Thus M is a weak comultiplication module.
The reverse implication is clear.

(c) Let ψ : Specs(M)→ Spec(R) be the natural map of Specs(M). As M is a secondful
Xs-injective R-module, ψ is a bijective map. Now let I be an ideal of R. Then by
[7, Lemma 3.3 (c) and Proposition 3.6], (ψ)−1(V (I)) = V s(0 :M I) = V s∗(0 :M I),
so ψ is continuous. Now let N be a submodule of M . Then we have

ψ(V s∗(N)) = ψ(V s∗(sec(N))) = ψ(V s(sec(N))) = V (AnnR(sec(N))).

It follows that ψ is a closed map and the proof is completed.

(d) If M = (0), there is nothing to prove. Hence we assume that M 6= (0). Let
N be a submodule of M . Clearly, V s∗(sec(N)) ⊆ V s(sec(N)). So we assume
that S ∈ V s(sec(N)). Then by Theorem 3.17 (a), S = IMp ((0 :M p)), where p =
AnnR(S) ∈ V (AnnR(M)). If set

I = {q ∈ V (AnnR(M)) | (0) 6= IMq ((0 :M q)) ⊆ N},



REFERENCES 228

then sec(N) =
∑

q∈I I
M
q ((0 :M q)). It follows that

AnnR(sec(N)) = AnnR(
∑
q∈I

IMq ((0 :M q))) =
⋂
q∈I

q ⊆ p.

Now by using the assumption, we have

S = IMp ((0 :M p)) ⊆
∑
q∈I

IMq ((0 :M q)) = sec(N).

So S ∈ V s∗(sec(N)), as desired.

Example 3.19. Set M =
⊕n

i=1 Zpi , where pi are distinct positive prime integers. Then
by using Proposition 3.13 (ii) and Theorem 3.15, we see that M is a secondful Xs-injective
Z- module and we have

Specs(M) = {Zpj ⊕ (
⊕

1≤i 6=j≤n
(0)) | 1 ≤ j ≤ n}.

Moreover, V s(sec(N)) = V s∗(sec(N)) for every submodule N of M . Hence (Specs(M),
τ s∗) is a spectral space by Theorem 3.18. This example shows that for each n > 1, when
n is square free, (Specs(Zn), τ s∗) is a spectral space.

Remark 3.20. (a) Let M = Zp∞ ⊕ Zq, where p and q are distinct positive prime inte-
gers. Then M is an Artinian Xs-injective over the one dimensional integral domain
Z; but it is not a weak comultiplication Z-module. This shows that the condition “M
is cotorsion or cotorsion-free” in part (b) of Theorem 3.18 is a necessary condition
and can not be omitted.

(b) Set M = (
⊕

p Zp)⊕Q, where p runs over all distinct positive prime integers. Then
by using Proposition 3.13 (ii) and Theorem 3.15, M is a secondful Xs-injective Z-
module. Moreover, we see that (Specs(M), τ s∗) is not a quasi-compact space and
hence not a spectral space by Hochster’s characterizations. This shows that the
condition “V s(sec(N)) = V s∗(sec(N)) for every submodule N of M” in Theorem
3.18 (d) is a necessary condition and can not be omitted.

We end this section with the following question.

Question 3.21. Is every cotop R-module an Xs-injective R-module?

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