Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 21 (2016). Number 2, pp. 214–238
© Journal “Algebra and Discrete Mathematics”
Generalizations of semicoprime preradicals
Ahmad Yousefian Darani and Hojjat Mostafanasab
Communicated by R. Wisbauer
Abstract. This article introduces the notions quasi-co-
n-absorbing preradicals and semi-co-n-absorbing preradicals, gen-
eralizing the concept of semicoprime preradicals. We study the
concepts quasi-co-n-absorbing submodules and semi-co-n-absorbing
submodules and their relations with quasi-co-n-absorbing preradi-
cals and semi-co-n-absorbing preradicals using the lattice structure
of preradicals.
1. Introduction
The notion of 2-absorbing ideals of commutative rings was introduced
by Badawi in [2], where a proper ideal I of a commutative ring R is called
a 2-absorbing ideal of R if whenever a, b, c ∈ R and abc ∈ I, then ab ∈ I
or ac ∈ I or bc ∈ I. Anderson and Badawi [1] generalized the concept of
2-absorbing ideals to n-absorbing ideals. According to their definition, a
proper ideal I of R is called an n-absorbing (resp. strongly n-absorbing)
ideal if whenever x1 · · ·xn+1 ∈ I for x1, . . . , xn+1 ∈ R (resp. I1 · · · In+1 ⊆ I
for ideals I1, . . . , In+1 of R), then there are n of the xi’s (resp. n of the
Ii’s) whose product is in I. In [24], the concept of 2-absorbing ideals
was generalized to submodules of a module over a commutative ring.
A proper submodule N of an R-module M is said to be a 2-absorbing
submodule of M if whenever a, b ∈ R and m ∈ M with abm ∈ N , then
ab ∈ (N :R M) or am ∈ N or bm ∈ N . For more studies concerning
2010 MSC: 16N99, 16S99, 06C05, 16N20.
Key words and phrases: lattice, preradical, quasi-co-n-absorbing, semi-co-n-
absorbing.
A. Yousefian Darani, H. Mostafanasab 215
2-absorbing (submodules) ideals we refer to [3],[9],[24],[25]. In [13], Raggi
et al. introduced the concepts of prime preradicals and prime submodules
over noncommutative rings, and Raggi, Ríos and Wisbauer [18], studied
the dual notions of these, coprime preradicals and coprime submodules. A
generalization of prime preradicals and submodules, “2-absorbing preradi-
cals and submodules” was investigated by Yousefian and Mostafanasab in
[23]. In [14], Raggi et al. defined and investigated semiprime preradicals,
and Mostafanasab and Yousefian [10], studied the concepts of quasi-n-
absorbing and semi-n-absorbing preradicals. Raggi et al. [11] defined the
notions of semicoprime preradicals and submodules. In this paper, we
introduce the concepts of “quasi-co-n-absorbing preradicals” and “semi-
co-n-absorbing preradicals”. As well we investigate“quasi-co-n-absorbing
submodules” and “semi-co-n-absorbing submodules” in this paper.
2. Preliminaries
Throughout this paper R is an associative ring with nonzero identity,
and R-Mod denotes the category of all the unitary left R-modules. We
denote by R-simp a complete set of representatives of isomorphism classes
of simple left R-modules. For M ∈ R-Mod, we denote by E(M) the
injective hull of M . Let U,N ∈ R-Mod, we say that N is generated by
U (or N is U -generated) if there exists an epimorphism U (Λ) → N for
some index set Λ. Dually, we say that N is cogenerated by U (or N is
U -cogenerated) if there exists a monomorphism N → UΛ for some index
set Λ. Also, we say that an R-module X is subgenerated by M (or X is
M -subgenerated) if X is a submodule of an M -generated module. The
category of M -subgenerated modules (the Wisbauer category) is denoted
σ[M ] (see [21]). A preradical over the ring R is a subfunctor of the identity
functor on R-Mod. Denote by R-pr the class of all preradicals over R.
There is a natural partial ordering in R-pr given by σ  τ if σ(M) 6 τ(M)
for every M ∈ R-Mod. It is proved in [15] that with this partial ordering,
R-pr is an atomic and co-atomic big lattice. The smallest and the largest
elements of R-pr are denoted, respectively, 0 and 1.
Let M ∈ R-Mod. Recall ([5] or [15]) that a submodule N of M is
called fully invariant if f(N) 6 N for each R-homomorphism f : M → M .
In this paper, the notation N 6fi M means that “N is a fully invariant
submodule of M”. Obviously the submodule K of M is fully invariant if
and only if there exists a preradical τ of R-Mod such that K = τ(M).
If N 6 M , then the preradicals αMN and ωMN are defined as follows: For
K ∈ R-Mod,
216 Generalizations of semicoprime preradicals
1) αMN (K) =
∑{f(N)|f ∈ HomR(M,K)}.
2) ωMN (K) =
⋂{f−1(N)|f ∈ HomR(K,M)}.
Notice that for σ ∈ R-pr and M, N ∈ R-Mod we have that σ(M) = N
if and only if N 6fi M and αMN  σ  ωMN . We have also that if
K 6 N 6 M with K, N 6fi M , then αMK  αMN and ωMK  ωMN .
The atoms and coatoms ofR-pr are, respectively, {αE(S)S | S ∈ R-simp}
and {ωRI | I is a maximal ideal of R} (See [15, Theorem 7]).
There are four classical operations in R-pr, namely, ∧,∨, · and : which
are defined as follows. For σ, τ ∈ R-pr and M ∈ R-Mod:
1) (σ ∧ τ)(M) = σM ∩ τM ,
2) (σ ∨ τ)(M) = σM + τM ,
3) (στ)(M) = σ(τM) and
4) (σ : τ)(M) is determined by (σ : τ)(M)/σM = τ(M/σM).
The meet ∧ and join ∨ can be defined for arbitrary families of preradicals
as in [15]. The operation defined in (3) is called product, and the operation
defined in (4) is called coproduct. It is easy to show that for σ, τ ∈ R-pr,
στ  σ ∧ τ  σ ∨ τ  (σ : τ). It is clear that in R-pr the operations
(1)-(3) are associative, and in [22] it was shown that the coproduct “ :′′ is
associative. Notice the fact that coproduct of preradicals preserves order
on both sides, see [8, Remark 2.1]. We denote σσ · · ·σ (n times) by σn and
(σ : σ : · · · : σ) (n times) by σ[n]. Recall that σ ∈ R-pr is an idempotent
if σ2 = σ, while σ is a radical if σ[2] = σ. Note that σ is a radical if and
only if, σ(M/σ(M)) = 0 for each M ∈ R-Mod. We say that σ is nilpotent
if σn = 0 for some n > 1, while σ is unipotent if σ[n] = 1 for some n > 1.
Using the preradical ωMN , in the papers [6], [7] and [18], the following
operation was introduced and studied:
ω-coproduct of submodules K, N 6 M : (K :M N) = (ωMK : ωMN )(M).
Henceforward, for brevity, (K : N) is written instead of (K :M N).
For any σ ∈ R-pr, we will use the following class of R-modules:
Tσ = {M ∈ R−Mod | σ(M) = M}.
Let σ ∈ R-pr. By [18, Theorem 8.2], the following classes of modules
are closed under taking arbitrary meets and arbitrary joins:
Ae = {τ ∈ R-pr | τσ = σ} and At = {τ ∈ R-pr | (σ : τ) = 1}.
As in [16], we define, for σ ∈ R-pr, the following preradicals:
• e(σ) = ∧{τ ∈ Ae} the equalizer of σ;
A. Yousefian Darani, H. Mostafanasab 217
• t(σ) = ∧{τ ∈ At} the totalizer of σ.
Clearly e(σ)σ = σ and (σ : t(σ)) = 1. For undefined notions we refer
the reader to [13,15–17].
In [18], Raggi et al. defined the notions of coprime preradicals and
coprime submodules as follows:
Let σ ∈ R-pr. σ is called coprime in R-pr if σ 6= 0 and for any
τ, η ∈ R-pr, σ  (τ : η) implies that σ  τ or σ  η. Let M ∈ R-Mod
and let N 6 M be a nonzero fully invariant submodule of M . The
submodule N is said to be coprime in M if whenever K, L are fully
invariant submodules of M with N 6 (K : L), then N 6 K or N 6 L.
Also, Raggi et al. [11] defined a preradical σ semicoprime in R-pr if σ 6= 0
and for any τ ∈ R-pr, σ  (τ : τ) implies that σ  τ . They said that
a nonzero fully invariant submodule N of M is semicoprime in M if
whenever K is a fully invariant submodule of M with N 6 (K : K),
then N 6 K. In special case, M is called a coprime (resp. semicoprime)
module if M is a coprime (resp. semicoprime) submodule of itself.
Yousefian and Mostafanasab in [22] defined the notions of co-2-absorb-
ing preradicals and co-2-absorbing submodules. The preradical σ ∈ R-pr
is called co-2-absorbing if σ 6= 0 and, for each η, µ, ν ∈ R-pr, σ  (η : µ : ν)
implies that σ  (η : µ) or σ  (η : ν) or σ  (µ : ν). More generally,
a preradical 0 6= σ in R-pr is said to be a co-n-absorbing preradical if
whenever σ  (η1 : η2 : · · · : ηn+1) for η1, η2, . . . , ηn+1 ∈ R-pr, there
are i1, i2, . . . , in ∈ {1, 2, . . . , n + 1} such that i1 < i2 < · · · < in and
σ  (ηi1 : ηi2 : · · · : ηin). They denoted by R-co-ass the class of all
R-modules M that the operation ω-coproduct is associative over fully
invariant submodules of M , i.e., for any fully invariant submodules K,N,L
of M , ((K : N) : L) = (K : (N : L)). Let M ∈ R-co-ass and K be a
fully invariant submodule of M . Then (K : K : · · · : K) (n times)
is simply denoted by K[n]. By Proposition 5.4 of [7], we can see that
if an R-module M is injective and artinian, then M ∈ R-co-ass. Let
M ∈ R-co-ass and N a nonzero fully invariant submodule of M . The
submodule N is said to be co-2-absorbing in M if whenever J,K,L are
fully invariant submodules of M with N 6 (J : K : L), then N 6 (J : K)
or N 6 (J : L) or N 6 (K : L). The generalization of co-2-absorbing
submodules is that, the submodule N is said co-n-absorbing in M if
whenever N 6 (K1 : K2 : · · · : Kn+1) for fully invariant submodules
K1,K2, . . . ,Kn+1 of M , there are i1, i2, . . . , in ∈ {1, 2, . . . , n + 1} such
that i1 < i2 < · · · < in and N 6 (Ki1 : Ki2 : · · · : Kin). An R-module M
is called a co-n-absorbing module if M is a co-n-absorbing submodule of
itself.
218 Generalizations of semicoprime preradicals
We say that a preradical 0 6= σ ∈ R-pr is called a quasi-co-n-absorbing
preradical if whenever σ  (µ[n] : ν) for µ, ν ∈ R-pr, then σ  µ[n] or
σ  (µ[n−1] : ν). A preradical 0 6= σ ∈ R-pr is called a semi-co-n-absorbing
preradical if whenever σ  µ[n+1] for µ ∈ R-pr, then σ  µ[n]. Let M ∈ R-
co-ass. We say that a nonzero fully invariant submodule N of M is quasi-
co-n-absorbing in M if for every fully invariant submodules K, L of M ,
N 6 (K[n] : L) implies thatN 6 K[n] orN 6 (K[n−1] : L). A nonzero fully
invariant submodule N ofM is called semi-co-n-absorbing in M if for every
fully invariant submodule K of M , N 6 K[n+1] implies that N 6 K[n].
An R-module M satisfies the ω-property if (τ(M) :M η(M)) = (τ : η)(M)
for every τ, η ∈ R-pr, see [22].
We recall the definition of relative epi-projectivity (see [20]). Let M
and N be modules. N is said to be epi-M -projective if, for any submodule
K of M , any epimorphism f : N → MK can be lifted to a homomorphism
g : N → M
Proposition 1 ([22, Proposition 2.9 (1)]). Let M ∈ R-Mod. If for any
fully invariant submodule K of M , MK is epi-M-projective, then M has
the ω-property.
In the next sections we frequently use the following proposition.
Proposition 2 ([12, Proposition 1.2]). Let {Mγ}γ∈I and {Nγ}γ∈I be
families of modules in R-Mod such that for each γ ∈ I, Nγ 6 Mγ. Let
N = ⊕γ∈I Nγ, M =
⊕
γ∈I Mγ, N ′ =
∏
γ∈I Nγ and M ′ =
∏
γ∈I Mγ.
(1) If N 6fi M , then for each γ ∈ I, Nγ 6fi Mγ and αMN =
∨
γ∈I α
Mγ
Nγ .
(2) If N ′ 6fi M ′, then for each γ ∈ I, Nγ 6fi Mγ and ωM
′
N ′ =
∧
γ∈I ω
Mγ
Nγ .
3. Quasi-co-n-absorbing preradicals
Suppose that m, n are positive integers with n > m. A preradical
σ 6= 0 is called a quasi-co-(n,m)-absorbing preradical if whenever σ 
(µ[n−1] : ν) for µ, ν ∈ R-pr, then σ  µ[m] or σ  (µ[m−1] : ν).
Proposition 3. Let σ ∈ R-pr and let m > 0. The following conditions
are equivalent:
(1) σ is quasi-co-(n,m)-absorbing for every n > m;
(2) σ is quasi-co-(n,m)-absorbing for some n > m;
(3) σ is quasi-co-m-absorbing.
Proof. (1)⇒(2) Is trivial.
A. Yousefian Darani, H. Mostafanasab 219
(2)⇒(3) Assume that σ is quasi-co-(n,m)-absorbing for some n > m.
Let σ  (µ[m] : ν) for some µ, ν ∈ R-pr. Since m 6 n−1, then (µ[m] : ν) 
(µ[n−1] : ν) and so σ  (µ[n−1] : ν). Therefore σ  µ[m] or σ  (µ[m−1] : ν).
Consequently σ is quasi-co-m-absorbing.
(3)⇒(1) Suppose that σ is quasi-co-m-absorbing and get n > m. Let
σ  (µ[n−1] : ν) for some µ, ν ∈ R-pr. Therefore σ  (µ[m] : (µ[n−1−m] :
ν)). Hence σ  µ[m] or σ  (µ[m−1] : (µ[n−1−m] : ν)) = (µ[n−2] : ν).
Repeating this method implies that σ  µ[m] or σ  (µ[m−1] : ν). Thus σ
is quasi-co-(n,m)-absorbing.
Remark 1. Let σ ∈ R-pr.
(1) σ is coprime if and only if σ is quasi-co-1-absorbing if and only if σ
is co-1-absorbing.
(2) If σ is quasi-co-n-absorbing, then it is quasi-co-i-absorbing for all
i > n.
(3) If σ is coprime, then it is quasi-co-n-absorbing for all n > 1.
(4) If σ is quasi-co-n-absorbing for some n > 1, then there exists the
least n0 > 1 such that σ is quasi-co-n0-absorbing. In this case, σ is
quasi-co-n-absorbing for all n > n0 and it is not quasi-co-i-absorbing
for n0 > i > 0.
Proposition 4. Let C be a family of coprime preradicals. Then ∨σ∈C σ
is a quasi-co-i-absorbing preradical for every i > 2.
Proof. Let τ = ∨σ∈C σ. By Remark 1(2), it is sufficient to show that
τ is a quasi-co-2-absorbing preradical. Suppose that τ  (µ[2] : ν) for
some µ, ν ∈ R-pr. Since every σ ∈ C is coprime and σ  (µ[2] : ν), then
σ  µ or σ  ν. Hence τ  (µ : ν), and so we conclude that τ is a
quasi-co-2-absorbing preradical.
Let ζ = ∨{αSS | S ∈ R-simp}. Note that for every R-module M ,
ζ(M) = Soc(M). As in [14], ζ is called the socle preradical. Also, let
κ = {αR/IR/I | I a maximal ideal of R}. We call κ the ultrasocle preradical,
see [11].
As a direct consequence of Proposition 4 we have the following result.
Proposition 5. ζ, κ are quasi-co–i-absorbing preradicals for every i > 2.
Proof. By [18, p. 57], for each simple R-module S, αSS is coprime. Also, for
every maximal ideal I of R, αR/IR/I is a coprime preradical, [11, Remark 6].
Then by Proposition 4, the claim holds.
220 Generalizations of semicoprime preradicals
Proposition 6. If R is a semisimple Artinian ring, then every nonzero
preradical σ ∈ R-pr is a quasi-co-i-absorbing preradical for every i > 2.
Proof. Suppose that R is a semisimple Artinian ring. According to [18,
Proposition 3.2], every atom αE(S)S is a coprime preradical. On the other
hand [15, Theorem 11] implies that σ = ∨{αE(S)S | S∈R-simp, α
E(S)
S  σ}.
Therefore every nonzero preradical σ in R-pr is quasi-co-i-absorbing for
every i > 2, by Proposition 4.
Remark 2. Let S1, S2, . . . , Sn+1 ∈ R-simp be distinct. Then by Proposi-
tion 4,αS1S1∨α
S2
S2∨· · ·∨α
Sn+1
Sn+1 is a quasi-co-i-absorbing preradical in R-pr for
every i > 2. But, [22, Proposition 3.6] implies that αS1S1∨α
S2
S2∨· · ·∨α
Sn+1
Sn+1 is
not a co-n-absorbing preradical. This remark shows that the two concepts
of quasi-co-n-absorbing preradicals and of co-n-absorbing preradicals are
different in general.
Corollary 1. If R is a ring such that every quasi-co-n-absorbing prerad-
ical in R-pr is co-n-absorbing, then |R-simp| 6 n.
Notice the fact that coproduct of preradicals preserves order on both
sides.
Proposition 7. Let R be a ring. The following statements are equivalent:
(1) for every µ, ν ∈ R-pr, (µ[n] : ν) = µ[n] or (µ[n] : ν) = (µ[n−1] : ν);
(2) for every σ1, σ2, . . . , σn+1 ∈ R-pr,
(σ1 : σ2 : · · · : σn+1)  (σ1 ∨ σ2 ∨ · · · ∨ σn)[n]
or
(σ1 : σ2 : · · · : σn+1)  ((σ1 ∨ σ2 ∨ · · · ∨ σn)[n−1] : σn+1);
(3) every preredical 0 6= σ ∈ R-pr is quasi-co-n-absorbing.
Proof. (1)⇒(2) If σ1, σ2, . . . , σn+1 ∈ R-pr, then by part (1) we have that,
(σ1 : σ2 : · · · : σn+1)  ((σ1 ∨ σ2 ∨ · · · ∨ σn)[n] : σn+1)
= (σ1 ∨ σ2 ∨ · · · ∨ σn)[n],
or
(σ1 : σ2 : · · · : σn+1)  ((σ1 ∨ σ2 ∨ · · · ∨ σn)[n] : σn+1)
= ((σ1 ∨ σ2 ∨ · · · ∨ σn)[n−1] : σn+1).
A. Yousefian Darani, H. Mostafanasab 221
(2)⇒(1) For preradicals µ, ν ∈ R-pr, we have from (2),
(µ[n] : ν)  (
n times︷ ︸︸ ︷
µ ∨ · · · ∨ µ)[n] = µ[n]
or
(µ[n] : ν)  ((
n times︷ ︸︸ ︷
µ ∨ · · · ∨ µ)[n−1] : ν) = (µ[n−1] : ν).
Thus we have that (µ[n] : ν) = µ[n] or (µ[n] : ν) = (µ[n−1] : ν).
(1)⇔(3) Is evident.
In the next proposition we use (µ1 : · · · : µ̂i : · · · : µn+1) when the i-th
term is excluded from (µ1 : · · · : µn+1).
Proposition 8. Let 0 6= σ ∈ R-pr be an idempotent radical.
(1) If σ is such that for any µ, ν ∈ R-pr, we have
µ ∨ ν  σ  (µ[n] : ν) ⇒ [σ  µ[n] or σ  (µ[n−1] : ν)],
then σ is quasi-co-n-absorbing.
(2) If σ is such that for any µ1, µ2, . . . , µn+1 ∈ R-pr, we have
µ1 ∨ µ2 ∨ · · · ∨ µn+1  σ  (µ1 : µ2 : · · · : µn+1) ⇒
[σ  (µ1 : · · · : µ̂i : · · · : µn+1), for some 1 6 i 6 n+ 1],
then σ is a co-n-absorbing preradical.
Proof. (1) Let σ 6= 0 be an idempotent radical that satisfies the hypothesis
in part (1). Let σ  (τ[n] : λ) for some τ, λ ∈ R-pr. Then, by [15,
Theorem 8(3)] we have
τσ ∨ λσ  σ = σ2  (τ[n] : λ)σ = (τ[n]σ : λσ) = ((τσ)[n] : λσ).
So, by hypothesis we have σ  (τσ)[n] = τ[n]σ  τ[n] or σ  ((τσ)[n−1] :
λσ) = (τ[n−1] : λ)σ  (τ[n−1] : λ). Therefore σ is quasi-co-n-absorbing.
(2) The proof is similar to that of (1).
Proposition 9. Let C be a chain of quasi-co-n-absorbing preradicals, that
is, a subclass of quasi-co-n-absorbing preradicals which is linearly ordered.
Then
∨
σ∈C σ is a quasi-co-n-absorbing preradical.
222 Generalizations of semicoprime preradicals
Proof. Let τ = ∨σ∈C σ and assume that τ  (µ[n] : ν) for some µ, ν ∈ R-
pr. If σ  µ[n] for each σ ∈ C, then τ  µ[n]. If there exists σ0 ∈ C
such that σ0  µ[n], then σ  µ[n] for each σ0  σ. Since all preradicals
in C are quasi-co-n-absorbing, it follows that σ  (µ[n−1] : ν) for each
σ0  σ. Thus σ  (µ[n−1] : ν) for each σ ∈ C, so that τ  (µ[n−1] : ν).
Consequently, we deduce that τ is a quasi-co-n-absorbing preradical.
Proposition 10. If σi is a quasi-co-ni-absorbing preradical in R-pr for
every 1 6 i 6 k, then σ1∨σ2∨· · ·∨σk is a quasi-co-n-absorbing preradical
for n = n1 + · · ·+ nk.
Proof. For k = 1 there is nothing to prove. Then, suppose that k > 1.
Assume that σ1 ∨ σ2 ∨ · · · ∨ σk  (µ[n] : ν) for some µ, ν ∈ R-pr. Notice
that for every 1 6 i 6 k, σi  (µ[n] : ν) = (µ[ni] : µ[n−ni] : ν). Then,
for every 1 6 i 6 k, either σi  µ[ni] or σi  (µ[ni−1] : µ[n−ni] : ν) =
(µ[n−1] : ν), because σi is quasi-co-ni-absorbing. On the other hand,
for every 1 6 i 6 k, µ[ni]  µ[n−1] and so µ[ni]  (µ[n−1] : ν). Hence
σ1 ∨ σ2 ∨ · · · ∨ σk  (µ[n−1] : ν) which shows that σ1 ∨ σ2 ∨ · · · ∨ σk is a
quasi-co-n-absorbing preradical.
Proposition 11. Let σ1, σ2, . . . , σt ∈ R-pr.
(1) If σ1 is a quasi-co-n-absorbing preradical and σ2 is a quasi-co-m-
absorbing preradical for m 6 n, then σ1 ∨ σ2 is a quasi-co-(n+ 1)-
absorbing preradical.
(2) If σ1, σ2, . . . , σt are quasi-co-n-absorbing preradicals, then σ1 ∨ σ2 ∨
· · · ∨ σt is a quasi-co-(n+ t− 1)-absorbing preradical.
(3) If σi is a quasi-co-ni-absorbing preradical for every 1 6 i 6 t with
n1 < n2 < · · · < nt and t > 2, then σ1 ∨ σ2 ∨ · · · ∨ σt is a quasi-co-
(nt + 1)-absorbing preradical.
Proof. (1) Let µ, ν ∈ R-pr be such that σ1 ∨ σ2  (µ[n+1] : ν). Since σ1
is quasi-co-n-absorbing and σ1  (µ[n] : µ : ν), then either σ1  µ[n] or
σ1  (µ[n−1] : µ : ν) = (µ[n] : ν). Also, σ2 is quasi-co-m-absorbing and
σ2  (µ[m] : µ[n+1−m] : ν), so either σ2  µ[m] or σ2  (µ[m−1] : µ[n+1−m] :
ν) = (µ[n] : ν). There are four cases.
Case 1. Assume that σ1  µ[n] and σ2  µ[m]. Then σ1 ∨ σ2  µ[n].
Case 2. Assume that σ1µ[n] and σ2(µ[n] : ν). Then σ1 ∨ σ2  (µ[n] : ν).
Case 3. Assume that σ1(µ[n] : ν) and σ2µ[m]. Then σ1 ∨ σ2  (µ[n] : ν).
Case 4. Assume that σ1  (µ[n] : ν) and σ2  (µ[n] : ν). Then σ1 ∨ σ2 
(µ[n] : ν). Hence σ1 ∨ σ2 is quasi-co-(n+ 1)-absorbing.
A. Yousefian Darani, H. Mostafanasab 223
(2) We use induction on t. For t = 1 there is nothing to prove. Let
t > 1 and assume that for t−1 the claim holds. Then σ1∨σ2∨· · ·∨σt−1 is
quasi-co-(n+ t− 2)-absorbing. Since σt is quasi-co-n-absorbing, then it is
quasi-co-(n+ t−2)-absorbing, by Remark 1(2). Therefore σ1∨σ2∨· · ·∨σt
is quasi-co-(n+ t− 1)-absorbing, by part (1).
(3) Induction on t: For t = 3 apply parts (1) and (2). Let t > 3
and suppose that for t− 1 the claim holds. Hence σ1 ∨ σ2 ∨ · · · ∨ σt−1 is
quasi-co-(nt−1 + 1)-absorbing. We consider the following cases:
Case 1. Let nt−1+1 < nt. In this case σ1∨σ2∨· · ·∨σt is quasi-co-(nt+1)-
absorbing, by part (1).
Case 2. Let nt−1 + 1 = nt. Thus σ1 ∨ σ2 ∨ · · · ∨ σt is quasi-co-(nt + 1)-
absorbing, by part (2).
Case 3. Let nt−1 + 1 > nt. Then σ1 ∨ σ2 ∨ · · · ∨ σt is quasi-co-(nt−1 + 2)-
absorbing, by part (1). Since nt−1 + 2 6 nt + 1, then σ1 ∨ σ2 ∨ · · · ∨ σt is
quasi-co-(nt + 1)-absorbing.
Proposition 12. Let σ ∈ R-pr be a radical. If σ is quasi-co-n-absorbing,
then e(σ) is quasi-co-n-absorbing.
Proof. Assume that σ is quasi-co-n-absorbing, and let e(σ)  (µ[n] : ν) for
some µ, ν ∈ R-pr. Then σ = e(σ)σ  (µ[n] : ν)σ  ((µσ)[n] : νσ). Since σ
is quasi-co-n-absorbing and radical, [15, Theorem 8(3)] implies that either
σ  (µσ)[n] = µ[n]σ  µ[n] or σ  ((µσ)[n−1] : νσ) = (µ[n−1] : ν)σ 
(µ[n−1] : ν). Consequently e(σ) is quasi-co-n-absorbing.
Definition 1. For τ, ρ ∈ R-pr define the totalizer of ρ relative to τ as
tτ (ρ) =
∧{η ∈ R-pr| (ρ : η)  τ}. Note that t1(ρ) = t(ρ).
Proposition 13. Let τ ∈ R-pr. If τ is quasi-co-2-absorbing, then for
each λ ∈ R-pr, either τ  λ[n] or tτ (λ[n]) = tτ (λ[n−1]). In particular,
if 1 is a quasi-co-2-absorbing preradical, then for each λ ∈ R-pr, either
λ[n] = 1 or t(λ[n]) = t(λ[n−1]).
Proof. Suppose that τ is quasi-co-2-absorbing and let λ ∈ R-pr such that
τ  λ[n]. If ν ∈ R-pr is such that τ  (λ[n] : ν), then τ  (λ[n−1] : ν), since
σ is quasi-co-2-absorbing. Therefore tτ (λ[n−1])  tτ (λ[n]). On the other
hand λ[n−1]  λ[n] and so tτ (λ[n])  tτ (λ[n−1]). Consequently tτ (λ[n]) =
tτ (λ[n−1]).
224 Generalizations of semicoprime preradicals
4. Semi-co-n-absorbing preradicals
Suppose that m, n are positive integers with n > m. A more general
concept than semi-co-n-absorbing preradicals is the concept of semi-co-
(n,m)-absorbing preradicals. A preradical σ 6= 0 is called a semi-co-(n,m)-
absorbing preradical if whenever σ  µ[n] for µ ∈ R-pr, then σ  µ[m].
Note that a semicoprime preradical is just a semi-co-1-absorbing
preradical.
Theorem 1. Let σ ∈ R-pr and m, n be positive integers with n > m.
(1) If σ is quasi-co-m-absorbing, then it is semi-co-(k,m)-absorbing for
every k > m.
(2) If σ is semi-co-(n,m)-absorbing, then it is semi-co-(i,m)-absorbing
for every m < i < n, in particular it is semi-co-m-absorbing.
(3) σ is semi-co-(n,m)-absorbing if and only if σ is semi-co-(n, k)-
absorbing for each n > k > m if and only if σ is semi-co-(i, j)-
absorbing for each n > i > j > m.
(4) If σ is semi-co-(n,m)-absorbing, then it is semi-co-(nk,mk)-absorb-
ing for every positive integer k.
(5) If σ is semi-co-(n,m)-absorbing and semi-co-(r, s)-absorbing for
some positive integers r > s, then it is semi-co-(nr,ms)-absorbing.
Proof. (1) Is trivial.
(2) Is easy.
(3) Straightforward.
(4) Suppose that σ is semi-co-(n,m)-absorbing. Let µ ∈ R-pr and let
k be a positive integer such that σ  µ[nk]. Then σ 
(
µ[k]
)
[n]
. Since σ
is semi-co-(n,m)-absorbing, σ 
(
µ[k]
)
[m]
= µ[mk], and so σ is semi-co-
(nk,mk)-absorbing.
(5) Assume that σ is semi-co-(n,m)-absorbing and semi-co-(r, s)-
absorbing for some positive integers r > s. Let σ  µ[nr]. Since σ is
semi-co-(n,m)-absorbing, then σ  µ[mr]; and since σ is semi-co-(r, s)-
absorbing, σ  µ[ms]. Hence σ is semi-co-(nr,ms)-absorbing.
Corollary 2. Let σ ∈ R-pr and n be a positive integer.
(1) If σ is quasi-co-n-absorbing, then it is semi-co-n-absorbing.
(2) Let t 6 n be an integer. If σ is semi-co-(n+ 1, t)-absorbing, then it
is semi-co-(nk + i, tk)-absorbing for all k > i > 1.
(3) If σ is semi-co-n-absorbing, then it is semi-co-(nk+ i, nk)-absorbing
for all k > i > 1.
A. Yousefian Darani, H. Mostafanasab 225
(4) If σ is semi-co-n-absorbing, then it is semi-co-(nk + j)-absorbing
for all k > j > 0.
(5) If σ is semi-co-n-absorbing, then it is semi-co-(nk)-absorbing for
every positive integer k.
(6) If σ is semicoprime, then it is semi-co-k-absorbing for every positive
integer k.
(7) If σ is semicoprime, then for every k > 1 and every µ ∈ R-pr,
σ  µ[k] implies that σ  µ.
(8) If σ is semi-co-n-absorbing, then it is semi-co-((n+ 1)t, nt)-absorb
-ing for all t > 1.
(9) If σ is semicoprime, then it is quasi-co-k-absorbing for every k > 1.
Proof. (1) By parts (1), (2) of Theorem 1.
(2) Let σ be semi-co-(n+ 1, t)-absorbing. Then by Theorem 1(4), σ
is semi-co-(nk + k, tk)-absorbing, for every positive integer k. Hence by
Theorem 1(2), σ is semi-co-(nk + i, tk)-absorbing for every k > i > 1.
(3) In part (2) get t = n.
(4) By part (3).
(5) Is a special case of (4).
(6) Is a direct consequence of (5).
(7) By part (6).
(8) By Theorem 1(5).
(9) Assume that σ is semicoprime. Let σ  (µ[k] : ν) for some µ, ν ∈
R-pr and some k > 1. Then σ  (µ[k] : ν)  (µ : ν)[k]. Therefore
σ  (µ : ν), by part (7). So σ is quasi-co-k-absorbing.
In the following remark we prove Proposition 4 in another way.
Remark 3. Clearly, an arbitrary join of a family of semicoprime (coprime)
preradicals is semicoprime, and so it is quasi-co-k-absorbing for every
k > 1, by Corollary 2(9).
Proposition 14. Let σ1, σ2, . . . , σn ∈ R-pr. If for every 1 6 i 6 n, σi is
a semicoprime preradical, then (σ1 : σ2 : · · · : σn) is a semi-co-n-absorbing
preradical. In particular, if σ is a semicoprime preradical, then σ[n] is a
semi-co-n-absorbing preradical.
Proof. Apply Corollary 2(7).
Lemma 1. Let σ ∈ R-pr. If σ[n+1] is a semi-co-n-absorbing preradical,
then σ[n+1] = σ[n]. In particular, if σ[2] is a semicoprime preradical, then
σ is radical.
226 Generalizations of semicoprime preradicals
Proposition 15. Let σ ∈ R-pr, σ 6= 0 be an idempotent radical. If σ is
such that for any µ ∈ R-pr, we have µ  σ  µ[n+1] ⇒ σ  µ[n], then σ
is semi-co-n-absorbing.
Proof. The proof is similar to that of Proposition 8(1).
Proposition 16. Let σ1, σ2, . . . , σn ∈ R-pr be semi-co-2-absorbing pre-
radicals. Then (σ1 : σ2 : · · · : σn) is a semi-co-(3n−1)-absorbing preradical.
Proof. Suppose that (σ1 : σ2 : · · · : σn)  µ[3n] for some µ ∈ R-pr. For
every 1 6 i 6 n, σi  µ[3n] =
(
µ[3n−1]
)
[3] and σi is semi-co-2-absorbing,
then σi 
(
µ[3n−1]
)
[2] = µ[2·3n−1] =
(
µ[2·3n−2]
)
[3]. Again, since σi is semi-
co-2-absorbing, we conclude that σi  µ[22·3n−2]. Repeating this method
implies that σi  µ[2n]. So (σ1 : σ2 : · · · : σn)  µ[n2n]. On the other
hand n2n 6 3n − 1. So (σ1 : σ2 : · · · : σn)  µ[3n−1] which shows that
(σ1 : σ2 : · · · : σn) is semi-co-(3n − 1)-absorbing.
Proposition 17. If σi is a semi-co-ni-absorbing preradical in R-pr for
every 1 6 i 6 k, then σ1 ∨ σ2 ∨ · · · ∨ σk is a semi-co-(n − 1)-absorbing
preradical for n =
k∏
i=1
(ni + 1).
Proof. Let µ ∈ R-pr be such that σ1 ∨σ2 ∨ · · · ∨σk  µ[n]. Thus for every
1 6 i 6 k, σi 
(
µ[m]
)
[ni+1]
, where m =
k∏
j=1, j 6=i
(nj + 1). Since σi’s are
semi-co-ni-absorbing, then, for each 1 6 i 6 k, σi  µ[nim]. Note that for
every 1 6 i 6 k,
nim 6
k∏
i=1
(ni + 1)− 1 = n− 1.
So we have σi  µ[n−1] for every 1 6 i 6 k. Hence σ1∨σ2∨· · ·∨σk  µ[n−1]
which implies that σ1 ∨ σ2 ∨ · · · ∨ σk is a semi-co-(n − 1)-absorbing
preradical.
Proposition 18. Let σ1, σ2 ∈ R-pr and m, n be positive integers.
(1) If σ1 is quasi-co-m-absorbing and σ2 is semi-co-n-absorbing, then
(σ1 : σ2) is semi-co-(n(m+ 1) +m)-absorbing.
(2) If σ1 is quasi-co-(2m)-absorbing and σ2 is semi-co-m-absorbing,
then (σ1 : σ2) is semi-co-(m2 + 2m)-absorbing.
Proof. (1) Suppose that (σ1 : σ2)  µ[(n+1)(m+1)] for some µ ∈ R-pr.
Since σ1 is quasi-co-m-absorbing and σ1  µ[(n+1)(m+1)], then σ1  µ[m].
A. Yousefian Darani, H. Mostafanasab 227
On the other hand σ2 is semi-co-n-absorbing and σ2  µ[(n+1)(m+1)], then
σ2  µ[n(m+1)]. Consequently (σ1 : σ2)  µ[n(m+1)+m], and so (σ1 : σ2) is
semi-co-(n(m+ 1) +m)-absorbing.
(2) Suppose that (σ1 : σ2)  µ[(m+1)2] for some µ ∈ R-pr. Since σ1
is quasi-co-(2m)-absorbing and σ1  µ[(m+1)2], then σ1  µ[2m]. Since
σ2 is semi-co-m-absorbing and σ2  µ[(m+1)2], then σ2  µ[m2]. Hence
(σ1 : σ2)  µ[m2+2m] which shows that (σ1 : σ2) is semi-co-(m2 + 2m)-
absorbing.
Proposition 19. Let R be a ring. The following statements are equiva-
lent:
(1) for every preradical σ ∈ R-pr, σ[n+1] = σ[n];
(2) for all preradicals σ1σ2, . . . , σn+1 ∈ R-pr we have
(σ1 : σ2 : · · · : σn+1)  (σ1 ∨ σ2 ∨ · · · ∨ σn+1)[n];
(3) every preredical 0 6= σ ∈ R-pr is semi-co-n-absorbing.
Proof. (1)⇒(2) If σ1, σ2, . . . , σn+1 ∈ R-pr, then we get from (1),
(σ1 : σ2 : · · · : σn+1)  (σ1∨σ2∨· · ·∨σn+1)[n+1] = (σ1∨σ2∨· · ·∨σn+1)[n].
(2)⇒(1) For a preradical σ ∈ R-pr, we have from (2),
σ[n+1]  (
n+1 times︷ ︸︸ ︷
σ ∨ · · · ∨ σ)[n] = σ[n].
So we have that σ[n+1] = σ[n].
(1)⇔(3) Is clear.
Remark 4. Let {σα}α∈I ⊆ R-pr. If σα is semi-co-n-absorbing for every
α ∈ I, then ∨α∈I σα is semi-co-n-absorbing.
Proposition 20. Let σ ∈ R-pr be radical. If σ is semi-co-n-absorbing,
then e(σ) is semi-co-n-absorbing.
Proof. Is similar to the proof of Proposition 12.
In Proposition 23 of [11], it was shown that σ0 := ∨{σ ∈ R-pr | σ is
semicoprime} is the unique greatest semicoprime preradical.
Proposition 21. There exists in R-pr a unique greatest semi-co-n-
absorbing preradical.
228 Generalizations of semicoprime preradicals
Proof. Set σ0(n) =
∨{σ ∈ R-pr | σ is semi-co-n-absorbing}. By Remark 4,
σ0(n) is the greatest semi-co-n-absorbing preradical.
By notation in the the proof of the previous proposition we have that
σ0(1) = σ0.
Remark 5. As ζ  κ  σ0 are semicoprime preradicals, then ζ[n], κ[n],
σ0[n] are semi-co-n-absorbing preradicals, by Proposition 14. Therefore
ζ[n]  κ[n]  σ0[n]  σ0(n).
Proposition 22. The following statements hold:
(1) σ0 = ∧
n>1
σ0(n).
(2) σ0(n)  σ0[nk] for every positive integer k.
(3) σ[n]  σ0(n) for every semicoprime preradical σ.
Proof. (1) By Corollary 2(6) every semicoprime preradical is semi-co-n-
absorbing for every n > 1. Then σ0  σ0(n) for every n > 1.
(2) By Corollary 2(5).
(3) By Proposition 14.
In Proposition 26 of [11] it was shown that σ0  ν0, where ν0 =∧{τ | τ ∈ R-pr, τ is unipotent}.
The following proposition is straightforward.
Proposition 23. Suppose that ν(n)0 :=
∧{τ[n] | τ ∈ R-pr, τ[n+1] = 1}.
Then:
(1) σ0(n)  ν
(n)
0 ;
(2) ν0  ν(1)0 .
Corollary 3. The following statements hold:
(1) If ζ[n+1] = 1, then ζ[n] = κ[n] = σ0[n] = σ0(n) = ν
(n)
0 ;
(2) If ζ[2] = 1, then ζ = κ = σ0 = ν0 = ν(1)0 .
Proof. (1) By Remark 5 and Proposition 23 we have that ζ[n]  κ[n] 
σ0[n]  σ0(n)  ν
(n)
0 . If ζ[n+1] = 1, then ν
(n)
0  ζ[n], and so ζ[n] = κ[n] =
σ0[n] = σ0(n) = ν
(n)
0 .
(2) By part (1) and [11, Corollary 27].
Proposition 24. For a ring R the following statements are equivalent:
(1) For every µ ∈ R-pr, µ[n+1] = 1 implies that µ[n] = 1;
A. Yousefian Darani, H. Mostafanasab 229
(2) 1 is a semi-co-n-absorbing preradical;
(3) σ0(n) = 1;
(4) ν(n)0 = 1.
Proof. Is easy.
For τ ∈ R-pr define
C(n)(τ) =
∨
{σ ∈ R-pr | σ  τ, σ semi-co-n-absorbing},
which is the unique greatest semi-co-n-absorbing preradical less than or
equal to τ . Notice that in [11], C(1) is denoted by C.
Proposition 25. Let R be a ring.
(1) σ0(n) = C(n)(1) =
∨
τ∈R-pr
C(n)(τ).
(2) For each τ ∈ R-pr, C(n)(τ)  τ .
(3) For each τ, σ ∈ R-pr we have τ  σ ⇒ C(n)(τ)  C(n)(σ).
(4) For each τ ∈ R-pr, C(n)(τ[n+1]) = C(n)(τ[n]).
(5) For each τ ∈ R-pr, τ is semi-co-n-absorbing if and only if τ =
C(n)(τ).
(6) {τ ∈ R-pr | τ is semi-co-n-absorbing} = Im C(n) = {C(n)(σ) | σ ∈
R-pr}.
(7)
[
C(n)
]2
= C(n). Thus, C(n) is a closure operator on R-pr.
(8) For each family {τα}α∈I ⊆ R-pr, we have
C(n)(
∧
α∈I
τα) = C(n)(
∧
α∈I
C(n)(τα)).
(9) C(n) = ∧
k>1
C(nk), in particular C = ∧
k>1
C(k).
(10) C(n)(σ[n+1]) = C(n)(σ[n]) = σ[n] for any semicoprime preradical σ.
Proof. The proofs of (1), (2), (3), (5) and (6) is easy.
(4) For any τ ∈ R-pr, part (3) implies that C(n)(τ[n])  C(n)(τ[n+1]).
Since C(n)(τ[n+1]) is semi-co-n-absorbing (by Remark 4) and C(n)(τ[n+1])
 τ[n+1], then C(n)(τ[n+1])  τ[n]. Hence C(n)(τ[n+1])  C(n)(τ[n]). So the
equality holds.
(7) Is a direct consequence of part (5).
(8) The proof is similar to that of [11, Proposition 31](5).
(9) By Corollary 2(5).
(10) Apply Proposition 14 and parts (4), (5).
230 Generalizations of semicoprime preradicals
Now consider the operator (−) in R-pr that assigns to each preradical
σ the least radical over σ (see [19, p. 137]).
Lemma 2. Let σ, τ ∈ R-pr be such that σ is radical and τ is semi-co-n-
absorbing. Then:
(1) C(n)(σ)  C(n)(σ)  σ.
(2) C(n)(σ) = C(n)(C(n)(σ)).
(3) τ  C(n)(τ)  τ .
(4) τ = C(n)(τ).
Proof. Similar to the proof of [11, Lemma 32].
Proposition 26. Let R be a ring.
(1) The operator C(n)(−) defines an interior operator on the ordered
class of radicals.
(2) The operator C(n)((−)) defines a closure operator on the ordered
class of semi-co-n-absorbing preradicals.
Notice that the “open” radicals associated with the interior operator
C(n)(−) are
O(n)rad = {σ radical | σ = τ for some semi-co-n-absorbing τ}.
The “closed” semi-co-n-absorbing preradicals associated with the closure
operator C(n)((−)) are
C(n)sca = {τ semi-co-n-absorbing | τ = C(n)(σ) for some radical σ}.
The following result is immediate.
Corollary 4. For a ring R the operators C(n)(−) and (−) restrict to
mutually inverse maps between O(n)rad and C
(n)
sca.
Definition 2. Let τ ∈ R-pr. Define
C(n)1 (τ) =
∧
{σ[n] | σ ∈ R-pr, τ  σ[n+1]}.
Proposition 27. For a ring R the following conditions hold:
(1) For each τ ∈ R-pr, C(n)1 (τ)  τ[n].
(2) For each τ ∈ R-pr, τ is semi-co-n-absorbing if and only if τ 
C(n)1 (τ).
(3) 1 is a semi-co-n-absorbing preradical if and only if C(n)1 (1) = 1.
A. Yousefian Darani, H. Mostafanasab 231
(4) Let τ, σ ∈ R-pr. If τ  σ, then C(n)1 (τ)  C
(n)
1 (σ).
(5) For each family {τα}α∈I ⊆ R-pr, C(n)1 (
∧
α∈I
τα) 
∧
α∈I
C(n)1 (τα) and
∨
α∈I
C(n)1 (τα)  C
(n)
1 (
∨
α∈I
τα).
Proof. The assertions have straightforward verifications.
We apply an “Amitsur construction” to C(n)1 as follows:
Definition 3. Let τ ∈ R-pr. We define recursively the preradical C(n)λ (τ)
for each ordinal λ as follows:
(1) C(n)0 (τ) = τ .
(2) C(n)λ+1(τ) = C
(n)
1 (C
(n)
λ (τ)).
(3) If λ is a limit ordinal, then C(n)λ (τ) =
∧
β<λ
C(n)β (τ).
(4) C(n)Ω (τ) =
∧
λ ordinal
C(n)λ (τ).
Proposition 28. Let τ ∈ R-pr. Then the following statements are equiv-
alent:
(1) τ is semi-co-n-absorbing;
(2) For each ordinal λ, τ  C(n)λ (τ);
(3) C(n)Ω (τ) = τ .
Proof. By Proposition 27 and using transfinite induction we have the
claim.
As is the case with C(n)1 , all of the operators C
(n)
λ preserve order
between preradicals.
Proposition 29. Let τ, σ ∈ R-pr be such that τ  σ. Then:
(1) For each ordinal λ, C(n)λ (τ)  C
(n)
λ (σ).
(2) C(n)Ω (τ)  C
(n)
Ω (σ).
Proposition 30. For each τ ∈ R-pr, C(n)(τ)  C(n)Ω (τ).
Proof. Let τ ∈ R-pr. We use transfinite induction. First, note that
C(n)(τ)  τ = C(n)0 (τ). Assume that λ is an ordinal such that C(n)(τ) 
C(n)λ (τ). Since C(n)(τ) is semi-co-n-absorbing, C(n)(τ)  C
(n)
1 (C(n)(τ))
 C(n)1 (C
(n)
λ (τ)) = C
(n)
λ+1(τ), by parts (2) and (4) of Proposition 27.
If λ is a limit ordinal and C(n)(τ)  C(n)β (τ) for each β < λ, then
C(n)(τ)  ∧
β<λ
C(n)β (τ) = C
(n)
λ (τ).
232 Generalizations of semicoprime preradicals
In the following result we give equivalent conditions for the equality
C(n)Ω (τ) = C(n)(τ).
Proposition 31. For each τ ∈ R-pr the following statements are equiv-
alent:
(1) C(n)Ω (τ) is semi-co-n-absorbing;
(2) C(n)Ω (τ)  C
(n)
1 (C
(n)
Ω (τ));
(3) For each ordinal λ we have C(n)Ω (τ)  C
(n)
λ (C
(n)
Ω (τ));
(4) C(n)Ω (C
(n)
Ω (τ)) = C
(n)
Ω (τ);
(5) C(n)Ω (τ) = C(n)(τ).
Proof. (1)⇒(2) By Proposition 27(2).
(2)⇒(3) It follows by using transfinite induction on λ.
(3)⇒(4) Is easy.
(4)⇒(1) By Proposition 28.
(1)⇒(5) Assume thatC(n)Ω (τ) is semi-co-n-absorbing. Since C
(n)
Ω (τ)τ ,
the definition of C(n)(τ) implies that C(n)Ω (τ)  C(n)(τ). On the other
hand C(n)(τ)  C(n)Ω (τ), by Proposition 30. So the equality holds.
(5)⇒(1) Is straightforward.
5. Quasi-co-n-absorbing and semi-co-n-absorbing
submodules
Remark 6. Let M ∈ R-co-ass and N be a nonzero fully invariant
submodule of M . Then we have:
(1) N is co-n-absorbing in M ⇒ N is quasi-co-n-absorbing in M ⇒ N
is semi-co-n-absorbing in M .
(2) N is a quasi-co-1-absorbing submodule of M if and only if N is a
coprime submodule of M .
(3) N is a semi-co-1-absorbing submodule of M if and only if N is a
semicoprime submodule of M .
Proposition 32. Let σ ∈ R-pr. If for every M ∈ R-Mod, σ(M) is a
semicoprime submodule of M , then σ is a semicoprime preradical.
Proof. By hypothesis, [11, Proposition 19] implies that αMσ(M) is a semi-
coprime preradical. So σ = ∨{αMσ(M) | M ∈ R-Mod} (see [17, Remark 1])
is a semicoprime preradical.
Corollary 5. Let R be a ring. If every nonzero R-module is semicoprime,
then 1 is a semicoprime preradical in R-pr.
A. Yousefian Darani, H. Mostafanasab 233
Lemma 3 ([7, Lemma 2.5]). Let M ∈ R-Mod. Then for any submodules
N, K of M , αMN+K = αMN ∨ αMK .
Proposition 33. Let M ∈ R-Mod. Suppose that {Ni}i∈I is a family
of semicoprime submodules of M . Then N = ∑
i∈I
Ni is a semicoprime
submodule of M .
Proof. Let {Ni}i∈I be a family of semicoprime submodules of M . Then,
by [11, Proposition 19], αMNj ’s are semicoprime preradicals. Thus α
M
N =∨
i∈I αMNi is a semicoprime preradical. Again by [11, Proposition 19],
N = ∑
i∈I
Ni is a semicoprime submodule of M .
Proposition 34. Let R be a ring and {Mi}i∈I be a family of semicoprime
R-modules. Then M = ⊕
i∈I
Mi is a semicoprime R-module.
Proof. Since for every i ∈ I, Mi is a semicoprime R-module, then for
every i ∈ I, αMiMi is a semicoprime preradical, by [11, Proposition 19].
Therefore ∨
i∈I
αMiMi = α
M
M is a semicoprime preradical, and so again by
[11, Proposition 19], M = ⊕
i∈I
Mi is a semicoprime R-module.
Proposition 35. For a ring R the following statements are equivalent:
(1) R is a finite product of simple rings;
(2) κ = 1;
(3) 1 is a semicoprime preradical;
(4) RR is a semicoprime R-module;
(5) There exists a semicoprime R-module that is a generator in R-Mod.
Proof. (1)⇔(2) By [11, Theorem 10].
(1)⇔(3) By [11, Theorem 29].
(3)⇔(4) Notice the fact that an R-module G is a generator in R-Mod
if and only if αGG = 1. Since R is a generator in R-Mod, then αRR = 1.
Now, use [11, Proposition 19].
(4)⇒(5) Is trivial.
(5)⇒(3) See the proof of (3)⇔(4).
Theorem 2. Let M ∈ R-co-ass and N a fully invariant submodule of M .
Consider the following statements.
(a) N is co-n-absorbing in M .
(b) αMN is a co-n-absorbing preradical.
Then (b) ⇒ (a), and if M satisfies the ω-property, then (a) ⇒ (b).
234 Generalizations of semicoprime preradicals
Proof. The proof is similar to that of [22, Theorem 4.2].
Theorem 3. Let M ∈ R-co-ass and N a fully invariant submodule of M .
Consider the following statements:
(1) N is quasi-co-n-absorbing (resp. semi-co-n-absorbing) in M .
(2) αMN is a quasi-co-n-absorbing (resp. semi-co-n-absorbing) preradical.
Then (2) ⇒ (1), and if M satisfies the ω-property, then (1) ⇒ (2).
Proof. (1) ⇒ (2) Assume that N is quasi-co-n-absorbing in M and that
(η(M) : µ(M)) = (η : µ)(M) for every η, µ ∈ R-pr. Since N 6= 0 we have
αMN 6= 0. Now let η, µ ∈ R-pr be such that αMN  (η[n] : µ). In this case
we have
N = αMN (M) 6 (η[n] : µ)(M) = (η(M)[n] : µ(M)).
Since N is quasi-co-n-absorbing in M , by hypothesis we have that N 6
η(M)[n] = η[n](M) or N 6 (η(M)[n−1] : µ(M)) = (η[n−1] : µ)(M). It
follows from [15, Proposition 5] that αMN  αMη[n](M)  η[n] or α
M
N 
αM(η[n−1]:µ)(M)  (η[n−1] : µ), and so α
M
N is quasi-co-n-absorbing.
(2) ⇒ (1) Assume that αMN is a quasi-co-n-absorbing preradical. Since
αMN 6= 0, we have N 6= 0. Suppose that J, K are fully invariant submodules
of M such that N 6 (J[n] : K). Then we have N 6
(
(ωMJ )[n] : ωMK
)
(M).
By [15, Proposition 5], we get
αMN  αM((ωMJ )[n]:ωMK )(M) 
(
(ωMJ )[n] : ωMK
)
.
Since αMN is quasi-co-n-absorbing, we have αMN  (ωMJ )[n] or αMN (
(ωMJ )[n] : ωMK
)
. Therefore N = αMN (M)  (ωMJ )[n](M) = J[n] or N =
αMN (M) 
(
(ωMJ )[n] : ωMK
)
(M) = (J[n−1] : K). Hence N is a quasi-co-
n-absorbing submodule. A similar proof can be stated for semi-co-n-
absorbing preradicals.
Remark 7. Note that in Theorem 3, for the case n = 2 we can omit
the condition M ∈ R-co-ass, by the definition of quasi-co-2-absorbing
(semi-co-2-absorbing) submodules.
Definition 4. LetM ∈ R-co-ass. We say thatM is a quasi-co-n-absorbing
(resp. semi-co-n-absorbing) module if M is a quasi-co-n-absorbing (resp.
semi-co-n-absorbing) submodule of itself.
A. Yousefian Darani, H. Mostafanasab 235
Corollary 6. Let M1,M2, . . . ,Mt be injective Artinian R-modules. Sup-
pose that Mi’s are quasi-co-n-absorbing modules that satisfy the ω-property.
Then M = ⊕ti=1 Mi is a quasi-co-(n+ t− 1)-absorbing R-module.
Proof. Let M1,M2, . . . ,Mt be quasi-co-n-absorbing R-modules. Then,
by Theorem 3, αM1M1 , α
M2
M2 , . . . , α
Mt
Mt are quasi-co-n-absorbing preradicals,
and so αMM = αM1M1 ∨ α
M2
M2 ∨ · · · ∨ α
Mt
Mt is a quasi-co-(n+ t− 1)-absorbing
preradical, by Proposition 11(2). Again by Theorem 3, M = ⊕ti=1 Mi is
a quasi-co-(n+ t− 1)-absorbing R-module.
Corollary 7. Let R be a ring. The following statements hold:
(1) If the preradical 1 is quasi-co-2-absorbing (resp.semi-co-2-absorbing),
then every generator R-module is a quasi-co-2-absorbing (resp. semi-
co-2-absorbing) R-module.
(2) If R is a semisimple Artinian ring, then every R-module is quasi-
co-i-absorbing for every i > 2.
Proof. (1) Suppose that 1 is a quasi-co-2-absorbing (resp. semi-co-2-
absorbing) preradical and G is a generator R-module. Since αGG = 1, the
result follows from Theorem 3.
(2) By Proposition 6 and Theorem 3.
Example 1. Let R be a semisimple Artinian ring and S1, S2, . . . , Sn+1 ∈
R-simp be distinct. Then the injective Artinian R-module ⊕n+1i=1 Si is
quasi-co-n-absorbing, by Corollary 7(2). But note that, by [22, Proposi-
tion 3.6] and Theorem 2, ⊕n+1i=1 Si is not co-n-absorbing. This example
shows that the two concepts of quasi-co-n-absorbing modules and of
co-n-absorbing modules are different in general.
The following two propositions can be proved similar to [22, Proposi-
tion 4.10] and [22, Theorem 4.11], respectively.
Proposition 36. Let N, H ∈ R-co-ass such that H be a fully invariant
submodule of N and N be self-injective. For a fully invariant submodule
K of H,
(1) If K is quasi-co-n-absorbing in N , then K is quasi-co-n-absorbing
in H.
(2) If K is quasi-co-n-absorbing in N and K ∈ R-co-ass, then K is a
quasi-co-n-absorbing module.
(3) If αNK is a quasi-co-n-absorbing preradical and N satisfies the ω-
property, then αHK is a quasi-co-n-absorbing preradical.
236 Generalizations of semicoprime preradicals
Proposition 37. Let N, Q ∈ R-co-ass such that N be a fully invariant
submodule of Q and Q be self-injective. Then N is a quasi-co-n-absorbing
module if and only if N is quasi-co-n-absorbing in Q.
Theorem 4. Let M ∈ R-co-ass that satisfies the ω-property. The follow-
ing statements are equivalent:
(1) M is quasi-co-n-absorbing;
(2) αMM is quasi-co-n-absorbing;
(3) For each τ, η ∈ R-pr, M ∈ T(τ[n]:η) ⇒ M ∈ Tτ[n] or M ∈ T(τ[n−1]:η).
Proof. (1) ⇔ (2) Is clear by Theorem 3.
(2) ⇒ (3) Suppose that αMM is quasi-co-n-absorbing. Let τ, η ∈ R-pr
such that M ∈ T(τ[n]:η). Then (τ[n] : η)(M) = M , and so αMM  (τ[n] : η).
Therefore αMM  τ[n] or αMM  (τ[n−1] : η). Hence τ[n](M) = M or
(τ[n−1] : η)(M) = M . Consequently M ∈ Tτ[n] or M ∈ T(τ[n−1]:η).
(3) ⇒ (2) has a routine verification.
Similarly we can prove the following theorem.
Theorem 5. Let M ∈ R-co-ass that satisfies the ω-property. The follow-
ing statements are equivalent:
(1) M is semi-co-n-absorbing;
(2) αMM is semi-co-n-absorbing;
(3) For each τ ∈ R-pr, M ∈ Tτ[n+1] ⇒ M ∈ Tτ[n].
Theorem 6. Let M ∈ R-Mod be such that, for each pair K, L of fully
invariant submodules of M , we have
(
ωMK : ωML
)
= ωM(K:L). Then, for each
quasi-co-n-absorbing (resp. semi-co-n-absorbing) preradical σ such that
σ(M) 6= 0, we have that σ(M) is quasi-co-n-absorbing (resp. semi-co-n-
absorbing) in M .
Proof. By hypothesis M ∈ R-co-ass, [22, Lemma 4.12]. Let σ be a quasi-
co-n-absorbing preradical such that σ(M) 6= 0. If K,L are fully invariant
submodules of M such that σ(M) 6 (K[n] : L), then
σ  ωMσ(M)  ωM(K[n]:L) =
(
(ωMK )[n] : ωML
)
.
Since σ is quasi-co-n-absorbing, then
σ  (ωMK )[n] or σ 
(
(ωMK )[n−1] : ωML
)
.
In the first case we have σ(M) 6 (ωMK )[n](M) = K[n]; in the second case
we have σ(M) 6
(
(ωMK )[n−1] : ωML
)
(M) = (K[n−1] : L). Consequently
σ(M) is quasi-co-n-absorbing.
A. Yousefian Darani, H. Mostafanasab 237
Acknowledgement
The authors would like to thank the referee for the careful reading of
the manuscript and all the suggestions that improved the paper.
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Contact information
A. Yousefian
Darani,
H. Mostafanasab
Department of Mathematics and Applications,
University of Mohaghegh Ardabili, P. O. Box
179, Ardabil, Iran
E-Mail(s): yousefian@uma.ac.ir,
h.mostafanasab@gmail.com
Web-page(s): www.yousefiandarani.com
Received by the editors: 21.09.2015
and in final form 27.11.2015.