@
Appl. Gen. Topol. 23, no. 2 (2022), 345-361

doi:10.4995/agt.2022.16332

© AGT, UPV, 2022

The Zariski topology on the graded primary

spectrum of a graded module over a graded

commutative ring

Saif Salam and Khaldoun Al-Zoubi

Department of Mathematics and Statistics, Jordan University of Science and Technology, P.O.Box

3030, Irbid 22110, Jordan (smsalam19@sci.just.edu.jo, kfzoubi@just.edu.jo)

Communicated by P. Das

Abstract

Let R be a G-graded ring and M be a G-graded R-module. We define
the graded primary spectrum of M, denoted by PSG(M), to be the
set of all graded primary submodules Q of M such that (GrM (Q) :R
M) = Gr((Q :R M)). In this paper, we define a topology on PSG(M)
having the Zariski topology on the graded prime spectrum SpecG(M)
as a subspace topology, and investigate several topological properties
of this topological space.

2020 MSC: 13A02; 16W50.

Keywords: graded primary submodules; graded primary spectrum; Zariski
topology.

1. Introduction and Preliminaries

Let G be a multiplicative group with identity e and R be a commutative
ring with identity. Then R is called a G-graded ring if there exist additive
subgroups Rg of R indexed by the elements g ∈ G such that R = ⊕Rg

g∈G
and

RgRh ⊆ Rgh for all g,h ∈ G. The elements of Rg are called homogeneous of
degree g. If r ∈ R, then r can be written uniquely as

∑
rg

g∈G
, where rg is the

component of r in Rg. The set of all homogeneous elements of R is denoted

Received 20 September 2021 – Accepted 29 May 2022

http://dx.doi.org/10.4995/agt.2022.16332
https://orcid.org/0000-0002-1330-2556
https://orcid.org/0000-0001-6082-4480


S. Salam and K. Al-Zoubi

by h(R), i.e. h(R) =
⋃
Rg

g∈G
. Let R be a G-graded ring and I be an ideal of R.

Then I is called G-graded ideal of R if I = ⊕
g∈G

(I
⋂
Rg). By I �G R, we mean

that I is a G-graded ideal of R, (see [13]). The graded radical of I is the set of
all a =

∑
g∈G

ag ∈ R such that for each g ∈ G there exists ng > 0 with a
ng
g ∈ I.

By Gr(I) (resp.
√
I) we mean the graded radical (resp. the radical) of I, (see

[18]). The graded prime spectrum SpecG(R) of a graded ring R consists of all
graded prime ideals of R. It is known that SpecG(R) is a topological space
whose closed sets are V RG (I) = {p ∈ SpecG(R) | I ⊆ p} for each graded ideal I
of R (see, for example, [14, 16, 18]).

Let R be a G-graded ring and M a left R-module. Then M is said to be a
G-graded R-module if M = ⊕Mg

g∈G
with RgMh ⊆ Mgh for all g,h ∈ G, where

Mg is an additive subgroup of M for all g ∈ G. The elements of Mg are called
homogeneous of degree g. If x ∈ M, then x can be written uniquely as

∑
xg

g∈G
,

where xg is the component of x in Mg. The set of all homogeneous elements
of M is denoted by h(M), i.e. h(M) =

⋃
Mg

g∈G
. Let M = ⊕Mg

g∈G
be a G-graded

R-module. A submodule N of M is called a G-graded R-submodule of M
if N = ⊕

g∈G
(N

⋂
Mg). By N ≤G M (resp. N <G M) we mean that N is

a graded submodule (resp. a proper graded submodule) of M, (see [13]). If
N ≤G M, then (N :R M) = {r ∈ R | rM ⊆ N} is a graded ideal of R,
(see [3, Lemma 2.1]). A proper graded submodule P of M is called a graded
prime submodule of M if whenever r ∈ h(R) and m ∈ h(M) with rm ∈ P ,
then either m ∈ P or r ∈ (P :R M). It is easily seen that, if P is a graded
prime submodule of M, then (P :R M) is a graded prime ideal of R (see [3,
Proposition2.7]). The graded prime spectrum of M, denoted by SpecG(M), is
the set of all graded prime submodules of M. A proper graded submodule Q
of M is called a graded primary submodule of M, if whenever r ∈ h(R) and
m ∈ h(M) with rm ∈ Q, then either m ∈ Q or r ∈ Gr((Q :R M)). Graded
prime submodules and Graded primary submodules of graded modules have
been studied by various authors (see, for example [1, 2, 3, 4, 5, 15]). The graded
radical of a proper graded submodule N of M, denoted by GrM (N), is defined
to be the intersection of all graded prime submodules of M containing N. If
N is not contained in any graded prime submodule of M, then GrM (N) = M,
(see[5]).

A graded R-module M is called a multiplication graded R-module if any
N ≤G M has the form IM for some I �G R. If N is a graded submodule of a
multiplication graded module M, then N = (N :R M)M, (see [15]). A graded
submodule N of a graded module M is called graded maximal submodule of
M if N 6= M and there is no graded submodule L of M such that N ⊂ L ⊂ M.
A graded ring R is called graded integral domain, if whenever ab = 0 for
a,b ∈ h(R), then a = 0 or b = 0. A graded principal ideal domain R is a

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 346



The Zariski topology on the graded primary spectrum

graded integral domain in which every graded ideal of R is generated by a
homogeneous element. One can easily see that, if R is a graded principal ideal
domain, then every non-zero graded prime ideal of R is graded maximal.

Let M be a G-graded R-module and let ζ∗(M) = {V ∗G(N) | N ≤G M}
where V ∗G(N) = {P ∈ SpecG(M) | N ⊆ P} for any N ≤G M. Then M is
called a G-top module if the set ζ∗(M) is closed under finite union. In this case,
ζ∗(M) generates a topology on SpecG(M) and this topology is called the quasi
Zariski topology on SpecG(M). In contrast with ζ

∗(M), ζ(M) = {VG(N) |
N ≤G M} where VG(N) = {P ∈ SpecG(M) | (N :R M) ⊆ (P :R M)} for
any N ≤G M always generates a topology on SpecG(M). Let M be a G-
graded R-module. Then the map ϕ : SpecG(M) → SpecG(R/Ann(M)) by
ϕ(P) = (P :R M)/Ann(M) is called the natural map on SpecG(M). For
more details concerning the topologies on SpecG(M) and the natural map on
SpecG(M), one can look in [7, 14].

In this paper, we call the set of all graded primary submodules Q of a
graded module M satisfying the condition (GrM (Q) :R M) = Gr((Q :R M))
the graded primary spectrum of M and denote it by PSG(M). It is easy to
see that SpecG(M) ⊆ PSG(M). The converse inclusion is not always true.
For example, if F is a G-graded field and M is a G-graded F-module, then
SpecG(M) = PSG(M) = {N | N <G M}. But if we take the ring of integers
R = Z, G = Z2 and M = Z × Z, then R is a G-graded ring by R0 = R and
R1 = {0}. Also, M is a G-graded R-module by M0 = Z×{0} and M1 = {0}×Z.
By some computations, we can see that N = Z×4Z ∈PSG(M). However, N /∈
SpecG(M), since 2 ∈ h(R) and (2, 0) ∈ h(M) such that 2(0, 2) ∈ N but (2, 0) /∈
N and 2 /∈ (N :R M). For a G-graded R-module M, it is clear that GrM (Q) 6=
M for any Q ∈ PSG(M) as Gr((Q :R M)) ∈ SpecG(R). We introduce the
primary G-top module which is a generalization of the G-top module. For this,
we define the variety of any N ≤G M by ν∗G(N) = {Q ∈ PSG(M) | N ⊆
GrM (Q)} and we set Ω∗(M) = {ν∗G(N) | N ≤G M}. Then M is called a
primary G-top module if Ω∗(M) is closed under finite union. When this the
case, the topology generated by Ω∗(M) is called the quasi-Zariski topology on
PSG(M). In particular, every primary G-top module is a G-top module. Next,
we define another variety of any N ≤G M by νG(N) = {Q ∈PSG(M) | (N :R
M) ⊆ (GrM (Q) :R M)}. Then the collection Ω(M) = {νG(N) | N ≤G M}
satisfies the axioms for closed sets of a topology on PSG(M), which is called
the Zariski topology on PSG(M), or simply PZG-topology. We give some
properties of these topologies. We also relate some properties of the graded
primary spectrum PSG(M) and SpecG(R/Ann(M)) by introducing the map ρ :
PSG(M) → SpecG(R/Ann(M)) given by ρ(Q) = (GrM (Q) :R M)/Ann(M).
It should be noted that (GrM (Q) :R M) ∈ SpecG(R), since Q is a graded
primary submodule of M and (GrM (Q) :R M) = Gr((Q :R M)). In the last
two sections, we find a base for the Zariski topology on PSG(M) and we make
certain observations and obtain a few results involving some conditions under
which PSG(M) is compact, irreducible, T0-space or spectral space.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 347



S. Salam and K. Al-Zoubi

Throughout this paper, G is a multiplicative group, R is a commutative
G-graded ring with identity and M is a G-graded R-module. We assume that
SpecG(M) and PSG(M) are non-empty.

2. The Zariski topology on PSG(M)

In this section, we introduce different varieties for graded submodules of
graded modules. Using the properties of these varieties, we define the quasi
Zariski topology and the PZG-topology on PSG(M). We also give some rela-
tionships between PSG(M), SpecG(R/Ann(M)) and SpecG(M).

Theorem 2.1. Let M be a G-graded R-module. For any G-graded submodule
N of M, we define the variety of N by ν∗G(N) = {Q ∈ PSG(M) | N ⊆
GrM (Q)}. Then the following hold:

(1) ν∗G(0) = PSG(M) and ν
∗
G(M) = ∅.

(2) If N,N′ ≤G M and N ⊆ N′, then ν∗G(N
′) ⊆ ν∗G(N).

(3)
⋂
i∈I
ν∗G(Ni) = ν

∗
G(

∑
i∈I
Ni) for any indexing set I and any family of graded

submodules {Ni}i∈I .
(4) ν∗G(N) ∪ν

∗
G(N

′) ⊆ ν∗(N ∩N′) for any N,N′ ≤G M.
(5) ν∗G(N) = ν

∗
G(GrM (N)) for any N ≤G M.

Proof. (1) and (2) are obvious.
(3) Since Ni ⊆

∑
i∈I
Ni for all i ∈ I, then by (2) we have ν∗G(

∑
i∈I
Ni) ⊆ ν∗G(Ni)

for all i ∈ I. Therefore ν∗G(
∑
i∈I
Ni) ⊆

⋂
i∈I
ν∗G(Ni). Conversely, let Q ∈

⋂
i∈I
ν∗G(Ni).

Then Ni ⊆ GrM (Q) for all i ∈ I, which implies that
∑
i∈I
Ni ⊆ GrM (Q). Hence

Q ∈ ν∗G(
∑
i∈I
Ni).

(4) Since N ∩ N′ ⊆ N and N ∩ N′ ⊆ N′, then by (2) we have ν∗G(N) ⊆
ν∗G(N∩N

′) and ν∗G(N
′) ⊆ ν∗G(N∩N

′). Therefore ν∗G(N)∪ν
∗
G(N

′) ⊆ ν∗G(N∩N
′).

(5) As N ⊆ GrM (N), we obtain ν∗G(GrM (N)) ⊆ ν
∗
G(N). Conversely, let Q ∈

ν∗G(N). Then N ⊆ GrM (Q). So GrM (N) ⊆ GrM (GrM (Q)) = GrM (Q). Thus
Q ∈ ν∗G(GrM (N)). �

Note that the reverse inclusion in Theorem 2.1 (4) is not always true. Take
R = Z,G = Z2,M = Z × Z,N = 4Z ×{0} and N′ = {0}× 4Z. Then R is a
G-graded ring by R0 = Z and R1 = {0}. Also M is a G-graded R-module by
M0 = Z×{0} and M1 = {0}×Z. Moreover, N,N′ ≤G M. Now, ν∗G(N∩N

′) =
ν∗G({(0, 0)}) = PSG(M). Let P = {(0, 0)}. Then P ∈ SpecG(M) ⊆PSG(M).
It follows that P ∈ ν∗G(N ∩N

′) and GrM (P) = P . But N * P and N′ * P .
Thus P /∈ ν∗G(N) ∪ν

∗
G(N

′).
By Theorem 2.1 (1), (3), and (4), the collection Ω∗(M) = {ν∗G(N) | N ≤G

M} satisfies the axioms for closed sets of a topology on PSG(M) if and only if
Ω∗(M) is closed under finite union. When this is the case, we call M a primary
G-top module and we call the generated topology the quasi Zariski topology on
PSG(M), or PZ

q
G-topology for short. It is clear that, every G-graded simple

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 348



The Zariski topology on the graded primary spectrum

R-module is a primary G-top module. In the following theorem, we show that
every multiplication graded R-module is a primary G-top module.

Theorem 2.2. If M is a multiplication graded R-module, then M is a primary
G-top module.

Proof. Let N,N′ ≤G M. It is sufficient to show that ν∗G(N ∩N
′) ⊆ ν∗G(N) ∪

ν∗G(N
′). So let Q ∈ ν∗G(N ∩ N

′). Then N ∩ N′ ⊆ GrM (Q). It follows that
(N :R M) ∩ (N′ :R M) = (N ∩ N′ :R M) ⊆ (GrM (Q) :R M) = Gr((Q :R
M)) ∈ SpecG(R) as Q is a graded primary submodule. Therefore (N :R M) ⊆
(GrM (Q) :R M) or (N

′ :R M) ⊆ (GrM (Q) :R M). Since M is a multiplication
graded R-module, then N = (N :R M)M ⊆ (GrM (Q) :R M)M = GrM (Q)
or N′ = (N′ :R M)M ⊆ (GrM (Q) :R M)M = GrM (Q). Thus Q ∈ ν∗G(N) ∪
ν∗G(N

′). �

Now we define another variety for a graded submodule N of a G-graded R-
module M. We set νG(N) = {Q ∈ PSG(M) | (N :R M) ⊆ (GrM (Q) :R M)}.
We state some properties of this variety in the following theorem to construct
the Zariski topology on PSG(M).

Theorem 2.3. Let M be a G-graded R-module and let N,N′,Ni ≤G M for
any i ∈ I, where I is an indexing set. Then the following hold:

(1) νG(0) = PSG(M) and νG(M) = ∅.
(2)

⋂
i∈I
νG(Ni) = νG(

∑
i∈I

(Ni :R M)M).

(3) νG(N) ∪νG(N′) = νG(N ∩N′).
(4) If N ⊆ N′, then νG(N′) ⊆ νG(N).

Proof. (1) and (4) are trivial.
(2) Let Q ∈

⋂
i∈I
νG(Ni). Then (Ni :R M) ⊆ (GrM (Q) :R M) for all i ∈ I. So

(Ni :R M)M ⊆ (GrM (Q) :R M)M for all i ∈ I. This implies that
∑
i∈I

(Ni :R

M)M ⊆ (GrM (Q) :R M)M. Therefore (
∑
i∈I

(Ni :R M)M : M) ⊆ ((GrM (Q) :R

M)M :R M) = (GrM (Q) :R M). Hence Q ∈ νG(
∑
i∈I

(Ni :R M)M). For the

reverse inclusion, let Q ∈ νG(
∑
i∈I

(Ni :R M)M). Then (
∑
i∈I

(Ni :R M)M :R M) ⊆

(GrM (Q) :R M). But for any i ∈ I, we have (Ni :R M) = ((Ni :R M)M :R
M) ⊆ (

∑
i∈I

(Ni :R M)M :R M) ⊆ (GrM (Q) :R M). Thus Q ∈
⋂
i∈I
νG(Ni).

(3) For any Q ∈ PSG(M), we have Q ∈ νG(N ∩ N′) ⇔ (N ∩ N′ :R M) ⊆
(GrM (Q) :R M) = Gr((Q :R M)) ⇔ (N :R M) ∩ (N′ :R M) ⊆ Gr((Q :R
M)) ∈ SpecG(R) ⇔ (N :R M) ⊆ Gr((Q :R M)) or (N′ :R M) ⊆ Gr((Q :R
M)) ⇔ Q ∈ νG(N) ∪νG(N′). Hence νG(N ∩N′) = νG(N) ∪νG(N′). �

In view of Theorem 2.3 (1), (2) and (3), the collection Ω(M) = {νG(N) |
N ≤G M} satisfies the axioms for closed sets of a topology on PSG(M), which
is called the Zariski topology on PSG(M), or PZG-topology for short.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 349



S. Salam and K. Al-Zoubi

Now we state some relations between the varieties V ∗G(N), VG(N), ν
∗
G(N)

and νG(N) for any graded submodule N of a G-graded R-module M. These
relations will be used continuously throughout the rest of this paper.

Lemma 2.4. Suppose that N and N′ are graded submodules of a G-graded
R-module M and that I is a G-graded ideal of R. Then the following hold:

(1) VG(N) = νG(N) ∩SpecG(M).
(2) V ∗G(N) = ν

∗
G(N) ∩SpecG(M).

(3) If Gr((N :R M)) = Gr((N
′ :R M)), then νG(N) = νG(N

′). The
converse is also true if N,N′ ∈PSG(M).

(4) νG(N) = νG((N :R M)M) = ν
∗
G((N :R M)M) = ν

∗
G(Gr((N :R

M))M). In particular, ν∗G(IM) = νG(IM).

Proof. The proof is straightforward. �

Corollary 2.5. Every primary G-top module is a G-top module.

Proof. Let M be a primary G-top module and N,N′ ≤G M. By Lemma 2.4
(2), we have V ∗G(N)∪V

∗
G(N

′) = (SpecG(M)∩ν∗G(N))∪(SpecG(M)∩ν
∗
G(N

′)) =
SpecG(M)∩(ν∗G(N)∪ν

∗
G(N

′)) = SpecG(M)∩ν∗G(J) = V
∗
G(J) for some graded

submodule J of M and hence M is a G-top module. �

By Corollary 2.5, if M is a primary G-top module, then ζ∗(M) = {V ∗G(N) |
N ≤G M} induces the quasi Zariski topology on SpecG(M) which will be,
by Lemma 2.4 (2), a topological subspace of PSG(M) equipped with PZ

q
G-

topology. Also by Lemma 2.4 (1), SpecG(M) with the Zariski topology is
a topological subspace of PSG(M) equipped with the PZG-topology for any
G-graded R-module M.

Consider ϕ and ρ as described in the introduction. Let M be a G-graded R-
module. For p ∈ SpecG(R), we set PS

p
G(M) = {Q ∈ PSG(M) | (GrM (Q) :R

M) = p}.

Proposition 2.6. The following statements are equivalent for any G-graded
R-module M:

(1) If whenever Q,Q′ ∈ PSG(M) with νG(Q) = νG(Q′), then Q = Q′.
(2) |PSpG(M)| ≤ 1 for every p ∈ SpecG(R).
(3) ρ is injective.

Proof. (1)⇒(2): Let p ∈ SpecG(R) and Q,Q′ ∈ PS
p
G(M). Then Q,Q

′ ∈
PSG(M) and (GrM (Q) :R M) = (GrM (Q′) :R M) = p. By Lemma 2.4 (3), we
have νG(Q) = νG(Q

′). So, by the assumption (1), Q = Q′.
(2)⇒(3): Assume that ρ(Q) = ρ(Q′), where Q,Q′ ∈PSG(M). Then (GrM (Q) :R
M) = (GrM (Q

′) :R M). Let p = (GrM (Q) :R M) ∈ SpecG(R). Then we get
Q,Q′ ∈PSpG(M) and by the hypothesis we obtain Q = Q

′.
(3)⇒(1): Let Q,Q′ ∈ PSG(M) with νG(Q) = νG(Q′). Then, by Lemma 2.4
(3), we have (GrM (Q) :R M) = (GrM (Q

′) :R M). So ρ(Q) = ρ(Q
′). Since ρ is

injective, then Q = Q′. �

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 350



The Zariski topology on the graded primary spectrum

Corollary 2.7. If |PSpG(M)| = 1 for every p ∈ SpecG(R), then ρ is bijective.

Proof. It is clear by Proposition 2.8. �

Let M be a G-graded R-module. From now on, we will denote R/Ann(M)
by R and any graded ideal I/Ann(M) of R by I. In the following lemma, we
recall some properties of the natural map ϕ of SpecG(M). These properties
are important in the rest of this section.

Lemma 2.8 ([14, Proposition 3.13 and Proposition 3.15]). Let M be a G-graded
R-module. Then the following hold:

(1) ϕ is continuous and ϕ−1(V RG (I)) = VG(IM) for every graded ideal I of
R containing Ann(M).

(2) If ϕ is surjective, then ϕ is both open and closed with ϕ(VG(N)) =

VG
R((N :R M)) and ϕ(SpecG(M)−VG(N)) = SpecG(R)−VGR((N :R M))

for any N ≤G M.

In the next two propositions, we give similar results for ρ.

Proposition 2.9. Let M be a G-graded R-module. Then ρ−1(V RG (I)) =
νG(IM), for every graded ideal I of R containing Ann(M). Therefore ρ is
continuous mapping.

Proof. For any Q ∈ PSG(M), we have Q ∈ ρ−1(V RG (I)) ⇔ ρ(Q) ∈ V
R
G (I) ⇔

I ⊆ (GrM (Q) :R M) ⇔ I ⊆ (GrM (Q) :R M) ⇔ IM ⊆ (GrM (Q) :R M)M ⇔
(IM :R M) ⊆ ((GrM (Q) :R M)M :R M) = (GrM (Q) :R M) ⇔ Q ∈ νG(IM).
Hence ρ−1(V RG (I)) = νG(IM). �

Proposition 2.10. Let M be a G-graded R-module. If ρ is surjective, then
ρ is both open and closed; more precisely, for any N ≤G M, ρ(νG(N)) =
V RG ((N :R M)) and ρ(PSG(M) −νG(N)) = SpecG(R) −V

R
G ((N :R M)).

Proof. By Proposition 2.9, we have ρ−1(V RG (I)) = νG(IM) for every I �G R

containing Ann(M). So ρ−1(V RG ((N :R M))) = νG((N :R M)M) = νG(N)

for any N ≤G M. It follows that V RG ((N :R M)) = ρ(ρ
−1(V RG ((N :R M)))) =

ρ(νG(N)) as ρ is surjective. For the second part, note that PSG(M)−νG(N) =
PSG(M) − ρ−1(V RG ((N :R M))) = ρ

−1(SpecG(R)) − ρ−1(V RG ((N :R M))) =
ρ−1(SpecG(R) − V RG ((N :R M))). This implies that ρ(PSG(M) − νG(N)) =
ρ(ρ−1(SpecG(R) −V RG ((N :R M)))) = SpecG(R) −V

R
G ((N :R M)). �

Corollary 2.11. Let M be a G-graded R-module. Then ρ is bijective if and
only if ρ is a homeomorphism.

The following theorem is a result for Lemma 2.8, Proposition 2.9 and Propo-
sition 2.10.

Theorem 2.12. Let M be a G-graded R-module. Consider the following state-
ments:

(1) SpecG(M) is connected.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 351



S. Salam and K. Al-Zoubi

(2) PSG(M) is connected.
(3) SpecG(R) is connected.

(i) If ρ is surjective, then (1)⇒(2)⇔(3).
(ii) If ϕ is surjective, then all the three statements are equivalent.

Proof. (i) (1)⇒(2): Assume that SpecG(M) is connected. If PSG(M) is
disconnected, then there is a U clopen in PSG(M) such that U 6= ∅ and
U 6= PSG(M). Since U is clopen in PSG(M), then U = PSG(M)−νG(N1) =
νG(N2) for some N1,N2 ≤G M. By Proposition 2.10, we have ρ(U) is clopen
in SpecG(R). But ϕ is continuous. So ϕ

−1(ρ(U)) is clopen in PSG(M),
and so ϕ−1(ρ(U)) = ∅ or ϕ−1(ρ(U)) = SpecG(M). If ϕ−1(ρ(U)) = ∅,
then ϕ−1(ρ(νG(N2))) = ∅. Thus ϕ−1(V RG ((N2 :R M))) = ∅. It follows that
VG(N2) = ∅, which means that (N2 :R M) * (P :R M) for any P ∈ SpecG(M).
As U = νG(N2) 6= ∅, then ∃Q ∈PSG(M) such that (N2 :R M) ⊆ (GrM (Q) :R
M). Since GrM (Q) 6= M, then ∃P ′ ∈ SpecG(M) such that Q ⊆ P ′. There-
fore (GrM (Q) :R M) ⊆ (GrM (P ′) :R M) = (P ′ :R M). Hence (N2 :R
M) ⊆ (P ′ :R M) which is a contradiction. Now, if ϕ−1(ρ(U)) = SpecG(M),
then SpecG(M) = ϕ

−1(ρ(νG(N2))) = VG(N2) = SpecG(M) ∩ νG(N2). It
follows that SpecG(M) ⊆ νG(N2) = U = PSG(M) − νG(N1). As U =
PSG(M) − νG(N1) 6= PSG(M), then ∃Q ∈ PSG(M) ∩ νG(N1). Therefore
(N1 :R M) ⊆ (GrM (Q) :R M) and ∃P ∈ SpecG(M) such that Q ⊆ P. It fol-
lows that (N1 :R M) ⊆ (GrM (Q) :R M) ⊆ (GrM (P) :R M). Then P ∈ νG(N1).
But P ∈ SpecG(M) ⊆ PSG(M) −νG(N1). Therefore P /∈ νG(N1) which is a
contradiction. Consequently, PSG(M) is a connected space.
(2)⇒(3): Since ρ is continuous surjective map and PSG(M) is connected, then
SpecG(R) is connected.
(3)⇒(2): Assume by way of contradiction that PSG(M) is disconnected. Then
there is a U clopen in PSG(M) such that U 6= ∅ and U 6= PSG(M). Since
ρ is surjective, then ρ(U) is clopen in SpecG(R), and so ρ(U) = ∅ or ρ(U) =
SpecG(R) as SpecG(R) is connected. Also, since U is open in PSG(M), then
U = PSG(M) − νG(N) for some N ≤G M. Now, if ρ(U) = SpecG(R), then
SpecG(R) = ρ(PSG(M) − νG(N)) = SpecG(R) − V RG ((N :R M)) by Propo-
sition 2.10. Thus V RG ((N :R M)) = ∅ which implies that ∅ = ρ

−1(∅) =
ρ−1(V RG ((N :R M))) = νG(N). Therefore νG(N) = ∅ and hence U = PSG(M),
a contradiction. Also if ρ(U) = ∅, then U ⊆ ρ−1(ρ(U)) = ∅. It follows that
U = ∅ which is also a contradiction. Therefore PSG(M) is connected.

(ii) If ϕ is surjective, then it is clear that ρ is surjective and hence (1)⇒(2)⇔(3)
by (i). So it is enough to show that (3)⇒(1). We prove it in a similar way to
proof of (3)⇒(2) using the properties of ϕ. So again, we assume by way of con-
tradiction that SpecG(M) is disconnected, which means that there is a U clopen
in SpecG(M) such that U 6= ∅ and U 6= SpecG(M). As SpecG(R) is connected
and ϕ is surjective, then ϕ(U) = ∅ or ϕ(U) = SpecG(R). Also the open set U
can be written as U = SpecG(M) −VG(N) for some N ≤G M. It is clear that

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 352



The Zariski topology on the graded primary spectrum

if ϕ(U) = ∅, then U = ∅ and we have a contradiction. If ϕ(U) = SpecG(R),
then SpecG(R) = ϕ(SpecG(M) − VG(N)) = SpecG(R) − V RG ((N :R M)), and
so V RG ((N :R M)) = ∅. Thus VG(N) = ϕ

−1(V RG ((N :R M))) = ϕ
−1(∅) = ∅.

It follows that U = SpecG(M) which is a contradiction. Therefore SpecG(M)
is a connected space and this completes the proof. �

Let M and S be two G-graded R-modules. Recall that an R-module ho-
momorphism f : M → S is called a G-graded R-module homomorphism if
f(Mg) ⊆ Sg for all g ∈ G, see [13]. Let f : M → M′ be a G-graded module
epimorphism between the two graded modules M and M′. If N′ ≤G M′, then
(N′ :R M

′) = (f−1(N′) :R M). Also it is easy to check that, (N :R M) =
(f(N) :R M

′) and f(GrM (N)) = GrM′ (f(N)) for any N ≤G M containing the
kernel of f. We will denote the kernel of f by kerf.

Lemma 2.13. Let M and M′ be G-graded R-modules. Let f : M → M′ be a
G-graded module epimorphism. Then the following hold:

(1) If Q′ ∈PSG(M′), then f−1(Q′) ∈PSG(M).
(2) If Q ∈PSG(M) and kerf ⊆ Q, then f(Q) ∈PSG(M′).

Proof. (1) It is easy to verify that f−1(Q′) is graded primary submodule of
M and it remains to show that (GrM (f

−1(Q′)) :R M) = Gr((f
−1(Q′) :R

M)). Since kerf ⊆ f−1(Q′) ⊆ GrM (f−1(Q′)) and (GrM′ (Q′) :R M′) =
Gr((Q′ :R M

′)), we obtain (GrM (f
−1(Q′)) :R M) = (f(GrM (f

−1(Q′))) :R
M′) = (GrM′ (f(f

−1(Q′))) :R M
′) = (GrM′ (Q

′) :R M
′) = Gr((Q′ :R M

′)) =
Gr((f−1(Q′) :R M)) as required.
(2) First note that f(Q) is a graded proper submodule of M′, since Q is
a graded proper submodule of M containing kerf. Let rm′ ∈ f(Q) for
r ∈ h(R) and m′ ∈ h(M′). As f is a graded module epimorphism and
m′ ∈ h(M′), we get ∃m ∈ h(M) such that f(m) = m′, which implies that
f(rm) ∈ f(Q). Thus ∃t ∈ Q such that rm − t ∈ kerf ⊆ Q. So rm ∈ Q,
and so m ∈ Q or r ∈

√
(Q :R M) =

√
(f(Q) :R M′) . Hence f(Q) is a graded

primary submodule of M′. Moreover, (GrM′ (f(Q)) :R M
′) = (f(GrM (Q)) :R

M′) = (GrM (Q) :R M) = Gr((Q :R M)) = Gr((f(Q) :R M
′)). Therefore

f(Q) ∈PSG(M′). �

Theorem 2.14. Let M and M′ be G-graded R-modules and f : M → M′ be a
graded module epimorphism. Then the mapping π : PSG(M′) → PSG(M) by
π(Q′) = f−1(Q′) is an injective continuous map. Moreover, if π is surjective
map, then PSG(M) is homeomorphic to PSG(M′).

Proof. By Lemma 2.13, π is well-defined. Also, the injectivity of π is ob-
vious. Now for any O ∈ PSG(M′) and any closed set νG(N) in PSG(M),
where N ≤G M, we have O ∈ π−1(νG(N)) = π−1(ν∗G(Gr((N :R M))M)) ⇔
Gr((N :R M))M ⊆ GrM (f−1(O)) ⇔ Gr((N :R M)) ⊆ (GrM (f−1(O)) :R
M) = Gr((f−1(O) :R M)) = Gr((O :R M

′)) = (GrM′ (O) :R M
′) ⇔ Gr((N :R

M))M′ ⊆ GrM′ (O) ⇔ O ∈ ν∗G(Gr((N :R M))M
′) = νG(Gr((N :R M))M

′).
Therefore π−1(νG(N)) = νG(Gr((N :R M))M

′) and hence π is continuous.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 353



S. Salam and K. Al-Zoubi

For the last statement, we assume that π is surjective and it is enough to
show that π is closed. So let νG(N

′) be a closed set in PSG(M′), where
N′ ≤G M′. As we have seen, π−1(νG(N)) = νG(Gr((N :R M))M′) for any
N ≤G M. It follows that π−1(νG(f−1(N′))) = νG(Gr((f−1(N′) :R M))M′) =
νG(Gr((N

′ :R M
′))M′) = νG(N

′) and hence π−1(νG(f
−1(N′))) = νG(N

′).
Thus π(νG(N

′)) = νG(f
−1(N′)) as π is surjective. Therefore π is closed, and

so PSG(M) is homeomorphic to PSG(M′). �

Corollary 2.15. Let M and M′ be G-graded R-modules. Let f : M → M′ be a
G-graded module isomorphism. Then PSG(M) is homeomorphic to PSG(M′).

Proof. By Lemma 2.13 (2) and Theorem 2.14. �

3. A base for the Zariski topology on PSG(M)

Let M be a G-graded R-module. In [14, Theorem 2.3], it has been proved
that for each r ∈ h(R), the set Dr = SpecG(R)−V RG (rR) is open in SpecG(R)
and the family {Dr | r ∈ h(R)} is a base for the Zariski topology on SpecG(R).
In addition, each Dr is compact and thus D1 = SpecG(R) is compact. In this
section, we set Sr = PSG(M) − νG(rM) for each r ∈ h(R) and prove that
S = {Sr | r ∈ h(R)} forms a base for the Zariski-topology on PSG(M). Also,
we show that each Sr is compact and hence PSG(M) is compact.

Proposition 3.1. For any G-graded R-module M, the set S = {Sr | r ∈
h(R)} forms a base for the Zariski topology on PSG(M).

Proof. Let U = PSG(M)−νG(N) be an open set in PSG(M), where N ≤G M.
Let Q ∈ U and it is enough to find an element r ∈ h(R) such that Q ∈ Sr ⊆ U.
Since Q ∈ U, then (N :R M) * (GrM (Q) :R M), and so there exists x ∈ R
and g ∈ G such that xg ∈ (N :R M) − (GrM (Q) :R M). Take r = xg ∈ h(R).
Therefore (rM :R M) * (GrM (Q) :R M) and thus Q ∈ Sr. Now for any
Q′ ∈ Sr, we have (rM :R M) * (GrM (Q′) :R M), which implies that (N :R
M) * (GrM (Q′) :R M). Thus Q′ ∈ U and hence Q ∈ Sr ⊆ U which completes
the proof. �

Lemma 3.2 ([14, Theorem 2.3 (2)]). Let R be a G-graded ring. Then Dr ∩
Dt = Drt for any r,t ∈ h(R).

Let R be a G-graded ring. As usual, the nilradical of R and the set of all
units of R will be denoted by N(R) and U(R), respectively.

Proposition 3.3. Let M be a G-graded R-module and r ∈ h(R). Then,
(1) ρ−1(Dr) = Sr
(2) ρ(Sr) ⊆ Dr. If ρ is surjective, then the equality holds.
(3) Sr ∩St = Srt, for any r,t ∈ h(R).
(4) If r ∈ N(R), then Sr = ∅.
(5) If r ∈ U(R), then Sr = PSG(M).

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 354



The Zariski topology on the graded primary spectrum

Proof. (1) ρ−1(Dr) = ρ
−1(SpecG(R)−V RG (rR)) = PSG(M)−ρ

−1(V RG (rR)) =
PSG(M) −νG(rM) = Sr by Proposition 2.9.
(2) Trivial.
(3) For any r,t ∈ h(R), we have r,t ∈ h(R) and hence Dr ∩ Dt = Drt by
Lemma 3.2. It follows that Sr ∩St = ρ−1(Dr) ∩ρ−1(Dt) = ρ

−1(Drt) = Srt.
(4) Assume that r ∈ N(R). It follows that Dr = ∅ by [18, Proposition 3.6 (2)],
and thus Dr = ∅. Therefore Sr = ρ−1(Dr) = ∅ by (1).
(5) Assume that r ∈ U(R). By [18, Proposition 3.6 (3)], we have Dr =
SpecG(R) and hence Dr = SpecG(R). By (1), we obtain Sr = ρ

−1(Dr) =
ρ−1(SpecG(R)) = PSG(M). �

In part (a) of the next example, we see that if F is a G-graded field and M
is a G-graded F-module, then the Zariski topology on PSG(M) is the trivial
topology. However, a G-graded ring R for which for a G-graded R-module M,
the Zariski topology on PSG(M) is the indiscrete topology is not necessarily a
G-graded field and this will be discussed in part (b).

Example 3.4. (a) Let F be a G-graded field and M be a G-graded F-module.
Then any non-zero homogeneous element of F is unit. By Proposition 3.3 (5),
we have Sr = PSG(M) for any non-zero homogeneous element r of F. Also
S0 = PSG(M) −νG(0) = ∅ and hence S = {Sr | r ∈ h(F)} = {PSG(M),∅}.
Therefore, the Zariski topology on PSG(M) is the trivial topology on PSG(M).
(b) Let R = Z8 as a Z2-graded Z8 module by R0 = Z8 and R1 = {0}. Note
that 1, 3, 5, 7 ∈ h(Z8) ∩ U(Z8). So S1 = S3 = S5 = S7 = PSZ2 (Z8) by
Proposition 3.3 (5). Also S0 = S2 = S4 = S6 = ∅ by Proposition 3.3 (4), since
0, 2, 4, 6 ∈ N(Z8) ∩ h(Z8). Now S = {Sr | r ∈ h(R)} = {∅, PSZ2 (Z8)} and
hence the Zariski topology on PSZ2 (Z8) is the trivial topology. But Z8 is not
Z2-graded field.
Theorem 3.5. Let M be a G-graded R-module. If ρ is surjective, then the
open set Sr in PSG(M) for each r ∈ h(R) is compact; in particular, the space
PSG(M) is compact.
Proof. Let r ∈ h(R) and ζ = {St | t ∈ ∆} be a basic open cover for Sr, where ∆
is a subset of h(R). Then Sr ⊆

⋃
t∈∆

St and thus Dr = ρ(Sr) ⊆
⋃
t∈∆

ρ(St) =
⋃
t∈∆

Dt

by Proposition 3.3 (2). Then ζ = {Dt | t ∈ ∆} is a basic open cover for the
compact set Dr and hence it has a finite subcover ζ = {Dti | i = 1, ...,n},

where ti ∈ ∆ for any i = 1, ...,n. This means that Dr ⊆
n⋃

i=1

Dti and it follows

that Sr = ρ
−1(Dr) ⊆

n⋃
i=1

ρ−1(Dti ) =
n⋃

i=1

Sti by Proposition 3.3 (1). Therefore

ζ = {Sti | i = 1, ...,n} ⊆ ζ is a finite subcover for Sr. For the other part of
the theorem, since PSG(M) = S1, then PSG(M) is compact. �

Theorem 3.6. Let M be a G-graded R-module. If ρ is surjective, then the
compact open sets of PSG(M) are closed under finite intersection and form an
open base.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 355



S. Salam and K. Al-Zoubi

Proof. Let C1,C2 be quasi compact open sets of PSG(M) and ζ = {Sr | r ∈
∆} be a basic open cover for C1 ∩ C2, where ∆ is a subset of h(R). Since
S is a base for the Zariski topology on PSG(M), then the compact open sets
C1,C2 can be written as a finite union of elements of S. So let C1 =

n⋃
i=1

Sti

and C2 =
m⋃
j=1

Szj . By Proposition 3.3 (3), we have C1 ∩C2 =
⋃
i,j

(Sti ∩Szj ) =⋃
i,j

Stizj ⊆
⋃

r∈∆
Sr. Note that for any i,j we have tizj ∈ h(R) as ti,zj ∈ h(R).

So without loss of generality we can assume that C1 ∩ C2 =
L⋃

k=1

Shk where

hk ∈ h(R), for k = 1, ...,L. Then Shk ⊆
⋃

r∈∆
Sr for each k. Now each Shk

is compact by Theorem 3.5 and it follows that Shk ⊆
dk⋃
i=1

Srk,i , where dk ≥ 1

depends on k and rk,i ∈ ∆, for any k = 1, ...,L and i = 1, ...,dk. Therefore

C1 ∩ C2 =
L⋃

k=1

Shk ⊆
L⋃

k=1

dk⋃
i=1

Srk,i and thus ζ = {Srk,i | k = 1, ...,L,i =

1, ...,dk} is a finite subcover for C1 ∩ C2. The other part of the theorem is
trivially true. �

4. Irreducibility in PSG(M)

Let M be a G-graded R-module and Y be a subset of PSG(M). We will
denote the closure of Y in PSG(M) by Cl(Y ) and the intersection

⋂
Q∈Y

GrM (Q)

by η(Y ). If Z is a subset of SpecG(R) or SpecG(M), then the intersection of
all members of Z will be expressed by γ(Z).

Proposition 4.1. Let M be a G-graded R-module and Y ⊆ PSG(M). Then
Cl(Y ) = νG(η(Y )). Thus, Y is closed in PSG(M) if and only if νG(η(Y )) = Y .

Proof. Let νG(N) be any closed set containing Y , where N ≤G M. Note that
Y ⊆ νG(η(Y )), and so it is enough to show that νG(η(Y )) ⊆ νG(N). So let
Q ∈ νG(η(Y )). Then (η(Y ) :R M) ⊆ (GrM (Q) :R M). Note that for any
Q′ ∈ Y , we have (N :R M) ⊆ (GrM (Q′) :R M) and hence (N :R M) ⊆⋂
Q′∈Y

(GrM (Q
′) :R M) = (

⋂
Q′∈Y

GrM (Q
′) :R M) = (η(Y ) :R M) ⊆ (GrM (Q) :R

M). Thus Q ∈ νG(N). Therefore νG(η(Y )) is the smallest closed set containing
Y and hence Cl(Y ) = νG(η(Y )). �

Recall that a topological space X is irreducible if any two non-empty open
subsets of X intersect. Equivalently, X is irreducible if for any decomposition
X = F1 ∪ F2 with closed subsets Fi of X with i = 1, 2, we have F1 = X or
F2 = X. A subset X

′ of X is irreducible if it is an irreducible topological space
with the induced topology. Let X be a topological space. Then a subset Y of
X is irreducible if and only if its closure is irreducible. Also every singleton
subset of X is irreducible, (see [6]).

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 356



The Zariski topology on the graded primary spectrum

Theorem 4.2. Let M be a G-graded R-module. Then for each Q ∈PSG(M),
the closed set νG(Q) is irreducible closed subset of PSG(M). In particular, if
{0}∈PSG(M), then PSG(M) is irreducible.

Proof. For any Q ∈PSG(M), we have Cl({Q}) = νG(η({Q})) = νG(GrM (Q)) =
νG(Q). Now {Q} is irreducible in PSG(M), then its closure νG(Q) is irre-
ducible. The other part of the theorem follows from the equality νG({0}) =
PSG(M). �

In Theorem 4.2, if we drop the condition that Q ∈ PSG(M), then νG(Q)
might not be irreducible. Actually, PSG(M) itself is not always irreducible.
For this, if we take R = Z6 as Z2-graded Z6 module by R0 = R and R1 = {0}.
Then it can easily be checked that PSZ2 (Z6) = {{0, 3},{0, 2, 4}}, νZ2 (3Z6) =
{{0, 3}} and νZ2 (2Z6) = {{0, 2, 4}}. Now PSZ2 (Z6) = νZ2 (3Z6) ∪ νZ2 (2Z6).
But PSZ2 (Z6) 6= νZ2 (3Z6) and PSZ2 (Z6) 6= νZ2 (2Z6). Therefore PSZ2 (Z6) =
νZ2 ({0}) is not irreducible.

Now, we need the following lemma to prove the next theorem.

Lemma 4.3 ([10, Lemma 4.9]). A subset Y of SpecG(R) for any graded ring
R is irreducible if and only if γ(Y ) is a graded prime ideal of R.

Proof. ⇒: Let Y be irreducible subset of SpecG(R) and r1,r2 ∈ h(R) with
r1r2 ∈ γ(Y ). Then r1r2 ∈ p for any p ∈ Y . Let U1 = Y ∩(SpecG(R)−V RG (r1R))
and U2 = Y ∩ (SpecG(R) − V RG (r2R)). If U1,U2 are non-empty sets, then
U1 ∩ U2 6= ∅ as Y is irreducible and U1,U2 are open sets in Y . So ∃p ∈ Y
such that r1R * p and r2R * p. It follows that r1 /∈ p and r2 /∈ p and hence
r1r2 /∈ p as p ∈ SpecG(R), a contradiction. Therefore U1 = ∅ or U2 = ∅. If
U1 = ∅, then Y ⊆ V RG (r1R). This implies that r1R ⊆ Q for any Q ∈ Y and
thus r1R ⊆

⋂
Q∈Y

Q = γ(Y ). Therefore r1 ∈ γ(Y ). Similarly, if U2 6= ∅, then

r2 ∈ γ(Y ). Hence r1 ∈ γ(Y ) or r2 ∈ γ(Y ).
⇐: Assume that γ(Y ) is a graded prime ideal of R, where Y ⊆ SpecG(R). Let
Y = F1 ∪ F2, where F1,F2 are closed sets in Y . Now F1 = V RG (I1) ∩ Y and
F2 = V

R
G (I2) ∩ Y for some I1,I2 �G R. It follows that Y = Y ∩ V

R
G (I1 ∩ I2)

and hence Y ⊆ V RG (I1 ∩ I2), which implies that I1 ∩ I2 ⊆ p for any p ∈ Y .
Thus I1 ∩I2 ⊆ γ(Y ). Since γ(Y ) ∈ SpecG(R), then I1 ⊆ γ(Y ) or I2 ⊆ γ(Y ). If
I1 ⊆ γ(Y ), then V RG (γ(Y )) ⊆ V

R
G (I1) and thus Y ⊆ V

R
G (I1) as Y ⊆ V

R
G (γ(Y )).

It follows that F1 = Y . Similarly, if I2 ⊆ γ(Y ), we obtain F2 = Y . This proves
that F1 = Y or F2 = Y , hence, Y is irreducible. �

Theorem 4.4. Let M be a G-graded R-module and Y ⊆PSG(M). Then:
(1) If η(Y ) is graded primary submodule of M, then Y is irreducible.
(2) If Y is irreducible, then Υ = {(GrM (Q) :R M) | Q ∈ Y} is an irre-

ducible subset of SpecG(R), i.e., γ(Υ) = (η(Y ) :R M) ∈ SpecG(R).

Proof. (1) Assume that η(Y ) is a graded primary submodule of M, then it is
easy to see that η(Y ) ∈ PSG(M). By Theorem 4.2 and Proposition 4.1, we
have νG(η(Y )) = Cl(Y ) is irreducible and hence Y is irreducible.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 357



S. Salam and K. Al-Zoubi

(2) Assume that Y is irreducible. Then ρ(Y ) = Y ′ is irreducible subset of
SpecG(R) as ρ is continuous by Proposition 2.9. Note that γ(Y

′) = γ(ρ(Y )) =⋂
Q∈Y

(GrM (Q) :R M) = (
⋂

Q∈Y
GrM (Q) :R M) = (η(Y ) :R M) and hence γ(Y

′) =

(η(Y ) :R M) ∈ SpecG(R) by Lemma 4.3. It follows that (η(Y ) :R M) ∈
SpecG(R). Now γ(Υ) =

⋂
Q∈Y

(GrM (Q) :R M) = (
⋂

Q∈Y
GrM (Q) :R M) =

(η(Y ) :R M) ∈ SpecG(R) and thus Υ is irreducible subset of SpecG(R) by
Lemma 4.3 again. �

Let X be a topological space and Y be a closed subset of X. An element
y ∈ Y is called a generic point if Y = Cl({y}). An irreducible component of
X is a maximal irreducible subset of X. The irreducible components of X are
closed and they cover X, (see [9]).

Theorem 4.5. Let M be a G-graded R-module. Let Y ⊆ PSG(M) and ρ be
surjective. Then Y is an irreducible closed subset of PSG(M) if and only if
Y = νG(Q) for some Q ∈ PSG(M). Hence every irreducible closed subset of
PSG(M) has a generic point.

Proof. Assume that Y is an irreducible closed subset of PSG(M). Then Y =
νG(N) for some N ≤G M. Also (η(Y ) :R M) = (η(νG(N)) :R M) ∈ SpecG(R)
by Theorem 4.4. It follows that (η(Y ) :R M) ∈ SpecG(R) and hence ∃Q ∈
PSG(M) such that (GrM (Q) :R M) = (η(νG(N)) :R M) as ρ is surjective.
So Gr((Q :R M)) = Gr((η(νG(N)) :R M)) and so νG(Q) = νG(η(νG(N))) =
Cl(νG(N)) = νG(N) = Y by Proposition 4.1 and Lemma 2.4 (3). Conversely, if
Y = νG(Q) for some Q ∈PSG(M), then Y is irreducible by Theorem 4.2. �

Theorem 4.6. Let M be a G-graded R-module and Q ∈ PSG(M). If
(GrM (Q) :R M) is a minimal graded prime ideal of R, then νG(Q) is irre-
ducible component of PSG(M). The converse is true if ρ is surjective.

Proof. Note that νG(Q) is irreducible by Theorem 4.2 and it remains to show
that it is a maximal irreducible. Let Y be irreducible subset of PSG(M) with
νG(Q) ⊆ Y and if we show that Y = νG(Q), then we are done. Since Q ∈
νG(Q) ⊆ Y , then Q ∈ Y and thus (η(Y ) :R M) ⊆ (GrM (Q) :R M). It follows
that (η(Y ) :R M) = (GrM (Q) :R M) as (GrM (Q) :R M) is a minimal graded

prime ideal of R and (η(Y ) :R M) ∈ SpecG(R) by Theorem 4.4 (2). Hence
V RG ((η(Y ) :R M)) = V

R
G ((GrM (Q) :R M)), which implies that νG(η(Y )) =

ρ−1(V RG ((η(Y ) :R M))) = ρ
−1(V RG ((GrM (Q) :R M))) = νG(GrM (Q)) = νG(Q)

by Proposition 2.9 and Lemma 2.4 (4). Since Y ⊆ νG(η(Y )), then Y ⊆ νG(Q)
and thus Y = νG(Q). For the converse, we assume that ρ is surjective. Since

Q ∈ PSG(M), then (GrM (Q) :R M) ∈ SpecG(R). Let J ∈ SpecG(R) with
J ⊆ (GrM (Q) :R M) and it is enough to show that J = (GrM (Q) :R M). Note
that ∃Q′ ∈ PSG(M) such that ρ(Q′) = J as ρ is surjective. So we have
J = (GrM (Q

′) :R M). Now (GrM (Q
′) :R M) ⊆ (GrM (Q) :R M) and thus

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 358



The Zariski topology on the graded primary spectrum

νG(Q) ⊆ νG(Q′). Since νG(Q) is irreducible component and νG(Q′) is irre-
ducible by Theorem 4.2, then νG(Q) = νG(Q

′). By Lemma 2.4 (3), we get

(GrM (Q) :R M) = (GrM (Q
′) :R M) = J and thus J = (GrM (Q) :R M). �

Corollary 4.7. Let M be a G-graded R-module and K = {Q ∈ PSG(M) |
(GrM (Q) :R M) is a minimal graded prime ideal of R}. If ρ is surjective, then
the following hold:

(1) T = {νG(Q) | Q ∈ K} is the set of all irreducible components of
PSG(M).

(2) PSG(M) =
⋃

Q∈K
νG(Q).

(3) SpecG(R) =
⋃

Q∈K
V RG ((Q :R M)).

(4) SpecG(M) =
⋃

Q∈K
VG(Q).

(5) If {0} ∈ SpecG(M), then the only irreducible component subset of
PSG(M) is PSG(M) itself.

Proof. (1) follows from Theorem 4.5 and Theorem 4.6.
(2) Since any topological space is the union of its irreducible components, then
PSG(M) =

⋃
Y ∈T

Y =
⋃

Q∈K
νG(Q).

(3) Since ρ is surjective, then SpecG(R) = ρ(PSG(M)) = ρ(
⋃

Q∈K
νG(Q)) =⋃

Q∈K
ρ(νG(Q)) =

⋃
Q∈K

V RG ((Q :R M)) by Proposition 2.10.

(4) SpecG(M) = PSG(M)∩SpecG(M) = (
⋃

Q∈K
νG(Q))∩SpecG(M) =

⋃
Q∈K

VG(Q)

by Lemma 2.4 (1).

(5) Assume that {0} ∈ SpecG(M). Then ({0} :R M) ∈ SpecG(R). For any
Q ∈ K, we have ({0} :R M) ⊆ (GrM (Q) :R M) and hence ({0} :R M) =
(GrM (Q) :R M) as (GrM (Q) :R M) is a minimal graded prime ideal of R.
Therefore Gr((Q :R M)) = Gr(({0} :R M)) and thus νG(Q) = νG({0}) =
PSG(M) by Lemma 2.4 (3). By (1), the set of all irreducible components
of PSG(M) is T = {νG(Q) | Q ∈ K} = {PSG(M)} which completes the
proof. �

Proposition 4.8. Let R be a G-graded principal ideal domain and M be a
multiplication graded R-module. Let Y ⊆ PSG(M). If η(Y ) is a non-zero
graded primary submodule of M, then Y ⊆PSpG(M) for some graded maximal
ideal p of R.

Proof. Clearly, η(Y ) = GrM (η(Y )). Since η(Y ) is a graded primary submod-
ule of the graded multiplication module M, then η(Y ) ∈ SpecG(M) by [15,
Theorem 13] and hence (η(Y ) :R M) ∈ SpecG(R). If (η(Y ) :R M) = {0},
then η(Y ) = (η(Y ) :R M)M = {0}, a contradiction. So (η(Y ) :R M) is a
non-zero graded prime ideal in the graded principle ideal domain R and thus
(η(Y ) :R M) is a graded maximal ideal of R. It follows that η(Y ) is a graded

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 359



S. Salam and K. Al-Zoubi

maximal submodule of M as M is a graded multiplication module. Now for any
Q ∈ Y ⊆PSG(M), we have η(Y ) ⊆ GrM (Q) 6= M and thus η(Y ) = GrM (Q).
This implies that (GrM (Q) :R M) = (η(Y ) :R M) for any Q ∈ Y . Take
p = (η(Y ) :R M). Therefore Y ⊆PS

p
G(M). �

A topological space X is called a T1-space if every singleton subset of X is
closed. A G-graded R-module M is called graded finitely generated R-module

if there are m1,m2, ...,mk ∈ h(M) such that M =
k∑

i=1

Rmi.

Proposition 4.9. Let M be a G-graded finitely generated R-module. If PSG(M)
is a T1-space, then PSG(M) = MaxG(M) = SpecG(M), where MaxG(M) is
the set of all graded maximal submodule of M.

Proof. It is clear that MaxG(M) ⊆ PSG(M). Now, let Q ∈ PSG(M). Since
PSG(M) is a T1-space, then Cl({Q}) = {Q} and so νG(Q) = {Q} by Propo-
sition 4.1. As M 6= Q is a graded finitely generated module, we obtain
M/Q is a non-zero graded finitely generated module and hence ∃N ≤G M
with Q ⊆ N such that N/Q ∈ MaxG(M/Q) by [4, Lemma 2.7 (ii)]. Now,
it is easy to see that N ∈ MaxG(M) ⊆ PSG(M). Since (Q :R M) ⊆
(N :R M) = (GrM (N) :R M) and N ∈ PSG(M), then N ∈ νG(Q) = {Q}
and thus N = Q ∈ MaxG(M). Therefore MaxG(M) = PSG(M). Now,
SpecG(M) ⊆ PSG(M) = MaxG(M) ⊆ SpecG(M). Hence SpecG(M) =
PSG(M) = MaxG(M). �

A topological space is called a T0-space if the closure of any two distinct
points are distinct. A topological space is called spectral space if it is homeo-
morphic to the prime spectrum of a ring equipped with the Zariski topology.
Spectral spaces have been characterized by Hochster[9, Proposition 4] as the
topological spaces X which satisfy the following conditions:
(a) X is a T0-space.
(b) X is compact.
(c) The compact open subsets of X are closed under finite intersection and
form an open base.
(d) each irreducible closed subset of X has a generic point.

Theorem 4.10. Let M be a G-graded R-module and ρ be surjective. Then the
following statements are equivalent:

(1) PSG(M) is a T0-space.
(2) If whenever νG(Q) = νG(Q

′) with Q,Q′ ∈PSG(M), then Q = Q′.
(3) ρ is injective.
(4) |PSpG(M)| ≤ 1 for every p ∈ SpecG(R).
(5) PSG(M) is a spectral space.

Proof. The equivalence of (2), (3) and (4) is proved in Proposition 2.6. Also
(1), (5) are equivalent by Theorem 3.5, Theorem 3.6 and Theorem 4.5. For
(1)⇒(2), assume that νG(Q) = νG(Q′) for Q,Q′ ∈ PSG(M), then Cl({Q}) =
νG(Q) = νG(Q

′) = Cl({Q′}) and hence Q = Q′ as PSG(M) is T0 space. For

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 360



The Zariski topology on the graded primary spectrum

(2)⇒(1), let Q,Q′ ∈ PSG(M) with Q 6= Q′, then by the assumption (2) we
have νG(Q) 6= νG(Q′). Hence Cl({Q}) 6= Cl({Q′}). Therefore PSG(M) is a
T0-space. �

Acknowledgements. The authors wish to thank sincerely the referees for
their valuable comments and suggestions.

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https://arxiv.org/abs/2106.09519