arXiv:1712.00845v1 [] 3 Dec 2017

Second Representable Modules over Commutative
Rings *
Jawad Abuhlail†
Department of Mathematics and Statistics
King Fahd University of Petroleum & Minerals
31261 Dhahran, KSA

Hamza Hroub‡
Department of Mathematics
King Saud University
11451 Riyad, KSA

December 5, 2017

Abstract
Let R be a commutative ring. We investigate R-modules which can be written as finite
sums of second R-submodules (we call them second representable). We provide sufficient
conditions for an R-module M to be have a (minimal) second presentation, in particular
within the class of lifting modules. Moreover, we investigate the class of (main) second
attached prime ideals related to a module with such a presentation.

Introduction
Throughout, R is a commutative ring. We consider second representable modules, i.e. modn

ules which can be written as finite sums M = ∑ Mi of second submodules M1 , · · · , Mn of R M
i=1

(recall that N ≤ M is said to be second iff IN = N or IN = 0 for every ideal I ≤ R [14], [1]).
The paper is divided in three sections. In Section 1, and for the convenience of the reader, we
collect some preliminaries from Module Theory. In Section 2, is devoted to the study of second
representable modules. In particular, we provide sufficient conditions for the existence of second
representations for R M are provided, among others, in Proposition 2.18 and Theorem 2.28.
* MSC2010: Primary 13A15; Secondary 13C13.
Key Words: second submodules, second representations, secondary representations, second attached prime ideals,
lifting modules
† Corresponding Author; Email: abuhlail@kfupm.edu.sa.
The authors would like to acknowledge the support provided by the Deanship of Scientific Research (DSR) at King
Fahd University of Petroleum & Minerals (KFUPM) for funding this work through projects No. RG1213-1 &
RG1213-2
‡ The paper is extracted from the Ph.D. dissertation of Dr. Hamza Hroub under the supervision of Prof. Jawad
Abuhlail

1

Second and semisimple modules are trivially second representable, and Example 2.26 provides examples modules which are second representable but neither second not semisimple. Several other examples showing that some of the sufficient conditions in the results mentioned above
are not necessary (e.g. Examples 2.24 and 2.25). Since every second module is secondary, the
First and the Second Uniqueness Theorems (Theorems 2.9 and 2.10, respectively) for a second
representable R-module follow from the corresponding ones on secondary representations ([10],
[6]). As a byproduct, we introduced a new class of modules lying properly between the classes
of semisimple and lifting modules, namely the class of s-lifting modules (see Figure 1 at the end
of Section 2).
Section 3 is devote to the study of second attached prime ideals for a given second representable R-module. As a consequence of Theorem 3.6, it follows that a second representable
Noetherian R-module is a finite direct sum of second submodules. Theorem 3.10 investigates the
relation between semisimple, multiplication and second representable modules.

1 Preliminaries
In this section, and for the convenience of the reader, we collect some definitions and results
from the literature.
Throughout this paper, R is a commutative ring, M a non-zero R-module, LAT (R M) is the
canonical lattice of R-submodules of M and we write N ≤ M to indicate that N ∈ LAT (R M). For
N, K ≤ M and I ≤ R, we set
(K :R N) := {r ∈ R | rN ⊆ K} and (N :M I) := {x ∈ M | Ix ⊆ N}.
In particular, we set Ann(N) := (0 :R N).
1.1. ([14], [1]) Let M be an R-module. An R-submodule K ≤R M is said to be second [1] iff
K 6= 0 and for any ideal I ≤ R we have
IK = K or IK = 0.
By Specs (M), we denote the spectrum of second R-submodules of M.
1.2. ([17, Sec. 41]) We say that an R-submodule N ≤ M has a supplement K in M iff there is
an R-submodule K ≤ M minimal with respect to N + K = M. The R-module M is said to be
supplemented iff every R-submodule of M has a supplement in M. We say that N ≤ M has ample
supplements in M [17] iff for each submodule U ≤ M with N + U = M there is a supplement
K ⊆ U of N in M. The R-module M is called amply supplemented iff every R-submodule of M
has ample supplements in M. For example, every Artinian module is amply supplemented.
1.3. A submodule N ≤ M is called small (or superfluous) in M [17, 19.1] iff N +K 6= M for any Rsubmodule K M. An epimorphism of R-modules f : M −→ M ′ is called a small epimorphism iff
Ker( f ) is small M. An R-submodule N ≤ M is called large (or essential) [17, 17.1] iff N ∩ K 6= 0
for any R-submodule 0 6= K ≤ M. A monomorphism of R-modules g : M −→ M ′ is called a large
monomorphism iff f (M) is large in M ′ .
2

1.4. We say that M is a lifting R-module [7, 22.2] iff any R-submodule N ≤ M contains a direct
summand X ≤ M such that N/X is small in M/X . An R-module M is called extending [7, p. 265]
iff every nonzero submodule of M is essential in a direct summand of M.
1.5. An R-module M is called uniform [7] iff every nonzero R-submodule of M is large in M
(equivalently, 0 ∈ LAT (R M) is irreducible). An R-module M has uniform dimension n [7], and
we write u.dim(M) = n, iff there exists a large monomorphism from a direct sum of n uniform
R-modules to M. An R-module M is hollow iff every proper R-submodule of M is small in M
(equivalently, 1 ∈ LAT (R M) is hollow). We say that M has hollow dimension n [7] iff there
exists a small epimorphism from M to a direct sum of n hollow R-modules, in this case we write
h.dim(M) = n.
Lemma 1.6. ([7, Proposition 22.11]) Let M be a nonzero R-module with finite hollow (uniform)
dimension.
(1) If R M is lifting, then M =

n
L

Hi where each Hi is a hollow R-module and n = h.dim(M).

i=1

(2) If R M is extending, then M =

n
L

Ui where each Ui is a uniform R-module and n = u.dim(M).

i=1

Lemma 1.7. ([7, 22.2], [7, 20.34])
(1) Every lifting R-module is amply supplemented.
(2) The following are equivalent for an amply supplemented R-module M :
(a) M has finite hollow dimension.
(b) M has the DCC on supplements.
(c) M has the ACC on supplements.
1.8. Let L = (L, ∧, ∨, 0, 1) be a bounded lattice.
Let X ⊆ L\{0}, and denote by Min(X ) the set of minimal elements of X . We say that X is
atomic iff for every p ∈ X there is q ∈ Min(X ) such that q ≤ p;
Let X ⊆ L\{1}, and denote by Max(X ) the set of maximal elements X . We say that X is
coatomic iff for every element p ∈ X there is q ∈ Max(X ) such that p ≤ q.
Let M be an R-module. We say that M is atomic (resp. coatomic) iff the class of non-zero
(resp. proper R submodules) of M is atomic (resp. coatomic).
Lemma 1.9. ([17, 41.5, 41.6])
(1) If R M is coatomic and every maximal R-submodule of M has a supplement in M, then M is
a sum of hollow submodules of M.
(2) Let R M be finitely generated. Then M is supplemented if and only if M is a sum of hollow
submodules.
3

Primary and Secondary Representations
∈ N and x ∈
/ N we
1.10. A proper R-submodule N M is called primary [2] iff whenever axp
n
have a M ⊆ N for some n ∈ N. If N is a primary submodule of M, then p := (N :R M) is prime
ideal of R and we say that N is p-primary. A submodule K ≤ M has a primary decomposition
T
[2] iff there are primary submodules N1 , · · · , Nn of M with K = ni=1 Ni . Such a decomposition
of K, if it exists, is called minimal iff:
p
p
(1) (Ni :R M) 6= (N j :R M) for i 6= j;
(2)

T

Ni * N j ∀ j ∈ {1, 2, · · · , n}.

i6= j

Theorem 1.11. (Lasker-Noether Theorem [2, Theorem 18.20]) Every submodule of a finitely
generated module over a Noetherian ring has a primary decomposition.
Theorem 1.12. (First uniqueness Theorem of Primary Decompositions [2, Theorem 18.19]) Let
R be Noetherian and M an R-module. If

n
T

Ni = N =

i=1

m
T

K j are two minimal primary decom-

j=1

positions of N ≤ M, where Ni is pi - primary for all i ∈ {1, 2, · · · , n} and K j is q j -primary for all
j ∈ {1, · · · , m}, then n = m and {p1 · · · , pn } = {q1 , · · · , qm }.
Theorem 1.13. (Second Uniqueness Theorem of Primary Decomposition ([2, Theorem 18.24])
Let M be a finitely generated module over a Noetherian ring R and

n
T

i=1

Ni = N =

n
T

Ki be two

i=1

minimal primary decompositions of N ≤ M, where Ni and Ki are pi - primary submodules of M
for all i ∈ {1, · · · , n}. If p j is minimal among {p1 , · · · , pn } for some j ∈ {1, · · · , n}, then N j = K j .
Dual to primary submodules and primary decompositions are secondary submodules and
secondary representations:
1.14. An R-module M is called secondary ([10], [11]) iff for any ap
∈ R we have aM = M or
n
a M = 0 for some n ∈ N. If M is a secondary R-module, then p := Ann(M) is a prime ideal
of R and M is called p-secondary. An R-module M is called representable ([10], [11]) iff M =
n

n

∑ Ni , where N1 , · · · , Nn are secondary R-module. Moreover, M = ∑ Ni is said to be a minimal
i=1
i=1
p
p
secondary representation iff Ann(Ni ) 6= Ann(N j ) whenever i 6= j and N j * ∑ Ni for all
i6= j
p
j ∈ {1, · · · , n}. For each i ∈ {1, · · · , n}, the prime ideal pi := Ann(Ni ) is called an attached
prime [11] and we set Att(M) := {p1 , · · · , pn }. A subset A ⊆ Att(M) is called isolated iff q ∈ A
whenever q ∈ Att(M) and q ⊆ p for some p ∈ A. Examples of representable modules are artinian
modules ([10, Theorem 1]) and injective modules over Noetherian rings ([12, Theorem 2.3]).
Every quotient Q of a representable module R M is representable and Att(Q) ⊆ Att(M) ([15,
Theorem 1.10]).
1.15. A prime ideal p ≤ R is called a coassociated prime [9] to R M iff there is a hollow factor
M ′ of M such that p = {a ∈ R | aM ′ 6= M ′ }. The set of coassociated primes of an R-module M is
denoted by Coass(M). If R M is representable, then Att(M) = Coass(M) ([15, Theorem 1.14]).
4

Proposition 1.16. ([11]) Let M1 , · · · , Mn be secondary R-submodules of the R-module M. Then
M1 ⊕ M2 ⊕ · · · ⊕ Mn is a p-secondary R-module if and only if Mi is a p-secondary R-submodule
of M for all i.
Theorem 1.17. (First Uniqueness Theorem of Secondary Representations) ([10, Theorem 2]) If
m

n

∑ Ki and ∑ N j are two minimal secondary representations for R M, with Ki is pi -secondary for
j=1

i=1

i = 1, · · · , n and N j is q j -secondary for j = 1, · · · , m, then n = m and {p1 , · · · , pn } = {q1 , · · · , qn }.
Theorem 1.18. (Second Uniqueness Theorem of Secondary Representation [6, Theorem 3.2.7])
n

Let M be representable, A ⊆ Att(M) an isolated subset and M = ∑ Ki a minimal secondary
i=1

representation for M with Ki is pi -secondary for i ∈ {1, · · · , n}. Then ∑ Ki is independent of
pi ∈A

the choice of the minimal second representation.
Theorem 1.19. ([15, Theorem 1.10]) If N is a representable R-submodule of the representable
module M, then Att(M/N) ⊆ Att(M) ⊆ Att(N) ∪ Att(M/N).
Theorem 1.20. ([15, Theorem 1.11]) If M1 , · · · , Mn are representable R-modules, then
representable and Att(

n
L

i=1

Mi ) =

n
S

n
L

Mi is

i=1

Att(Mi).

i=1

2 Second Representations
Recall that R is a commutative ring. Yassemi [14] introduced the notion of second submodules
of a given non-zero module over a commutative ring. Annin [3] called these coprime modules
(see also [16]) and used them to dualize the notion of attached primes.
2.1. A nonzero submodule K ≤ M is called second [14] iff for any ideal I ≤ R, we have IK = 0 or
IK = K. The spectrum of second R-submodules of M is denoted by Specs (M). If K ∈ Specs (M),
then p := (0 :R K) is a prime ideal, called a second attached prime of M and K is called p-second.
By
Att s(M) := {(0 :R K) | K ∈ Specs (M)}
we denote the set of second attached primes of M.
Lemma 2.2. Let {Ki }i∈A be family of second R-submodule of M such that K j *

∑

Ki for all

i∈A\{ j}

j ∈ A. Let p be a prime ideal of R. Then Ki is p-second for all i ∈ A if and only if ∑i∈A Ki is
p-second.
Proof. (⇒) Assume that Ki is p-second for all i ∈ A. Clearly, p = (0 :R ∑ Ki ). Let I ≤ R. If
i∈A

IK j = 0 for some j ∈ A, then I ⊆ p, whence I ∑i∈A Ki = 0. Otherwise, IK j = K j for all j ∈ A and
so I ∑ Ki = ∑ IKi = ∑ Ki . Consequently, ∑i∈A Ki is second.
i∈A

i∈A

i∈A

5

(⇐) Assume that ∑ Ki is p-second and that Ki is pi -second for i ∈ A. Clearly p ⊆ pi for all
i∈A

i ∈ A. For any j ∈ A, we have p j ∑ Ki =
i∈A

p j Ki ⊆

∑
i∈A\{ j}

∑
i∈A\{ j}

Ki 6= ∑ Ki , whence p j ∑ Ki = 0,
i∈A

i∈A

i.e. p j ⊆ p. Hence p = p j for all j ∈ A.
n

Definition 2.3. We say that an R-module M is (directly) second representable iff M = ∑ Ki
(M =

n
L

i=1

Ki ) where Ki is a second R-submodule of M for all i = 1, 2, · · · , n; in this case we call

i=1
n
L

n

∑ Ki (
i=1

Ki ) a (direct) second representation of M. An R-module M is called semisecond iff M

i=1

is a (not necessarily finite) sum of second submodules of M.
Example 2.4.
Let p be a prime number. Any divisible p-group is a semisecond Z-module but not semisimple.
This follows from the fact that every divisible p-group is a direst sum of copies of the Prüfer
group Z(p∞ ) which is a 0-second Z-module but not simple (see [17, p. 124], [8, p. 96] for the
properties of the Prüfer group).
n

2.5. A (direct) second representation M = ∑ Ki (M =
i=1

n
L

Ki ) is called a minimal (direct) second

i=1

representation for M iff it satisfies the following conditions:
(1) (0 :R Ki ) 6= (0 :R K j ) for i 6= j.
(2) K j *

n

∑
i=1,i6= j

Ki for all j = 1, 2, · · · , n.

2.6. Let R M be second representable. It is clear that M has a minimal second representation
n

say ∑ Ki . Each Ki in such a minimal representation is called a main second submodule of M and
i=1

(0 :R Ki ) is called a main second attached prime of M. So, the set of main second attached primes
is
att s(M) = {Ann(Ki ) | i = 1, · · · , n}.
By Theorem 2.9 below, att s(M) is independent of the choice of the minimal second representation of M.
The result follows by Lemma 2.2:
Theorem 2.7. (Existence Theorem for Minimal Second Representations) Let M be second representable. Then M has a minimal second representation.
Example 2.8. The Abelian group Z18 has a minimal secondary representation as a Z-module,
namely Z18 = 2Z18 + 9Z18 . However, Z18 has no second representation (9Z18 is the unique
second Z-submodule of Z18 ).
6

In the light of 3.1, we obtain as a direct consequence of Theorem 1.17 and 1.18 the First
& Second Uniqueness Theorems for Second Representations:
Theorem 2.9. (First Uniqueness Theorem of Second Representations) Let M be an R-module
n

with two minimal second representations ∑ Ki = M = ∑mj=1 N j , where Ki is pi -second for all i ∈
i=1

{1, 2, · · · , n} and N j is q j -second for all j ∈ {1, 2, · · · , m}. Then {p1 , p2 , · · · , pn } = {q1, q2 , · · · , qm }.
Theorem 2.10. (Second Uniqueness Theorem of Second Representations) Let M be second repn

n

i=1

i=1

resentable. If ∑ Ki = M = ∑ Ni are minimal second representations for M with K j and N j
are p j -second submodules of M and p j is minimal in {p1 , · · · , pn } for some j ∈ {1, · · · , n}, then
K j = N j.
Remarks 2.11. Let M, M1 , ..., Mn be second representable submodules of an R-module L.
n

n

i=1

i=1

(1) ∑ Mi is second representable and att s( ∑ Mi ) ⊆

n
S

att s(Mi ).

i=1

(2) Any quotient Q of M is second representable and att s(Q) ⊆ att s(M). To see this, let K1 +
· · · + Kn be a minimal second representation for M and Q = M/N for some R-submodule
n

N ≤ M, then M/N = ∑ (Ki + N)/N. It is easy to see that (Ki + N)/N is second for i ∈
i=1

{1, · · · , n}. The result is obtained now by applying (1).
(3) If N is a second representable submodule of M, then it follows from (2) and Theorem 1.19
that
att s(M/N) ⊆ att s(M) ⊆ att s(N) ∪ att s(M/N).
(4) If M j ∩ ∑ Mi = 0 for all j, then it follows from Theorem 1.20 that
i6= j

Ln

i=1 Mi

is second

representable and
s

att (

n
M

Mi ) =

n
[

att s(Mi ).

i=1

i=1

(5) For any multiplicatively closed subset of S of R, the S−1 R-module S−1 M is second representable and
att s(S−1 M) = {pS | p ∈ att s(M) such that p ∩ S = 0}.
/
Proposition 2.12. Let M = ∑ Ki (resp. M =
i∈Λ

L

Ki ), where Ki is second for every i ∈ Λ. If

i∈Λ

Att s(M) is finite, then M is second representable (resp. directly second representable).
Proof. Let M = ∑ Ki (resp. M =
i∈Λ

L

Ki ) such that each Ki is second for every i ∈ Λ. Assume

i∈Λ

that Att s(M) = {p1 , p2 , · · · , pn }. For j ∈ {1, 2, · · · , n}, set A j = {Ki : i ∈ Λ such that (0 :R Ki ) =
p j }. Notice that N j = ∑ Ki is second by Proposition 2.2 for each j ∈ {1, 2, · · · , n}. Moreover,
n

Ki ∈A j
n
L

M = ∑ N j (resp. M =
j=1

N j ).

j=1

7

2.13. We say that a submodule K ≤ M satisfies the IS-condition iff for every I ≤ R for which
IK 6= 0, the submodule IK ≤ M has a proper supplement in M.
Remark 2.14. Let R M be supplemented, K ≤ M and 0 6= H ≤ K. The following conditions are
equivalent:
(1) K is not contained in any supplement of H in M.
(2) H has a proper supplement in M.
Proof. (1 ⇒ 2) Assume that K is not contained in any supplement of H in M. Since M is
supplemented, H has a supplement L in M, i.e. H + L = M. Indeed L 6= M as K * L.
(2 ⇒ 1) Assume that H has a proper supplement L in M. Then K * L; otherwise, H + L =
L 6= M.
Lemma 2.15. Every hollow R-submodule 0 6= K ≤ M satisfying the IS-condition is second.
Proof. Let 0 6= K ≤ M be a hollow R-submodule satisfying the IS-condition. Let I ≤ K and
suppose that IK 6= 0. By the IS-condition, IK has a proper supplement L M. It is easy to
show that IK + (L ∩ K) = K. Since K is hollow, IK = K (notice that L ∩ K 6= K; otherwise,
IK + L = L 6= M).
Example 2.16. The Abelian group Z18 , considered as a Z-module, is supplemented but not
semisimple. The submodule K1 = 9Z18 is hollow and satisfies the IS-condition. Notice that
K2 := 6Z18 is hollow and second but does not satisfy the IS-condition (i.e. the IS-condition is
not necessary for a hollow submodule module to be second).
2.17. We say that an R-module M is (directly) hollow representable iff M is a finite (direct) sum
of hollow R-submodules.
Proposition 2.18. Let R M be (directly) hollow representable. If every maximal hollow non-zero
submodule of M is second, then M is (directly) second representable.
n

Proof. Let M = ∑ Hi be a sum of hollow R-submodules. Assume, without loss of generality, that
i=1

this sum is irredundant. Claim: H1 , · · · , Hn are maximal hollow submodules of M. To see this,
suppose that H is a hollow submodule of M with Hi ≤ H for some i ∈ {1, · · · , n} and consider
N := ∑ H j . For any x ∈ H, there are y ∈ N and z ∈ Hi such that x = y + z. But z ∈ H implies that
j6=i

y ∈ H. So, H = (N ∩ H) + Hi . Since H is hollow, either H ∩ N = H whence H ⊆ N, or Hi = H.
n

But Hi ⊆ H and M = ∑ Hi is an irredundant sum, whence H = Hi . Hence Hi is maximal hollow.
i=1

n

By our assumption, H1 , · · · , Hn are second, whence M = ∑ Hi is a second representation of M.
If M =

n
L

i=1

Hi is a direct sum of hollow R-submodules, then one can show similarly that each

i=1

Hi is a maximal hollow R-submodule of M, whence M is a direct sum of hollow R-submodules.
8

Example 2.19. Every Artinian left R-module is hollow representable (see [13, Lemma 3.2]). Let
p ∈ Z be a prime number. The Prüfer group, considered as a Z-module, is Artinian and the
unique maximal hollow Z-submodule of Z(p∞ ) is second.
Example 2.20. A lifting R-module M is directly hollow representable if it satisfies any of the
following additional conditions:
(1) R M has a finite hollow dimension [7, Proposition 22.11] (e.g. R M is finitely generated [7,
Corollary 22.12]).
(2) R M has a finite uniform dimension [7, Proposition 22.11] (e.g. R M is finitely cogenerated
[7, Corollary 22.12]).
Inspired by Example 2.20 and Proposition 2.18, we introduce the notion of s-lifting modules.
Definition 2.21. We call R M s-lifting iff R M is lifting and every maximal hollow submodule of
M is second.
Examples 2.22. (1) Consider the Abelian group M = Z8 as a Z-module. Notice that N =
{0, 4} is the unique second submodule in M, hence M is not second representable. Notice
that Z M is Artinian and lifting but not s-lifting.
(2) Every semisimple module is s-lifting and trivially semisecond (every simple submodule is
second).
(3) Every second hollow module is s-lifting but not necessarily simple. Consider the Prüfer
group M = Z(p∞ ), considered as Z-module. Notice that Z M is not simple. Moreover, Z M
is hollow and second whence s-lifting hollow but not semisimple.
As a direct consequence of Proposition 2.18 and Example 2.20, we obtain the following class
of directly second representable modules.
Example 2.23. If R M is an s-lifting module and has a finite hollow dimension, then M is directly
second representable. Clearly, this class is nonempty; e.g. any finite direct sum of Prüfer groups
is s-lifting with finite hollow dimension.
The following example is an s-lifting second module with infinite hollow dimension which
is not semisimple.
Example 2.24.
Let P be the set of prime numbers, A ⊆ P infinite and consider M :=

L

Z(p∞ ) considered as a

p∈A

Z-module.
Claim: Z M is lifting. This can be obtained by applying [5, Theorem 2] (see the second
paragraph on page 60). However, we provide here our own proof.

9

Let N ≤ M. Assume, without loss of generality that N is not a direct summand of M (if N is a
L
direct summand of M, then N/N = 0 is small in M/N and we are done). Notice that N =
L p,
p∈A

where L p ≤ Z(p∞ ) for all p ∈ A.
Case 1: L p 6= Z(p∞ ) for all p ∈ A. In this case, N is small in M as the set of submodules of
Z(p∞ ) form a chain for all p ∈ A. Indeed, for every p ∈ A : if L p +Wp = M, then L p ⊆ Wp = M.
L
Case 2: L p = Z(p∞ ) for all p ∈ B ( A and L p 6= Z(p∞ ) ∀p ∈
/ B. Let K =
L p . In this case,
p∈B

N/K is small in M/K as the set of submodules of Z(p∞ ) form a chain for all p ∈ A.
Notice that the maximal hollow Z-submodules of M are {Z(p∞ ) | p ∈ A} and they are second,
whence Z M is s-lifting.
Notice that Z M is second, not semisimple and that h. dim(Z M) = ∞.
Example 2.25. Let n = p1 · · · pn be a product of distinct prime numbers and consider M = Zn [x]
as a Z-module. Then M is second representable semisimple. Indeed, let m j = pnj for all j =
1, 2, ..., n. Set K jk = Zm j xk for all j = 1, 2, ..., n and k ∈ {0, 1, 2, ...}. Then K jk is simple for
all j = 1, 2, ..., n and k ∈ {0, 1, 2, ...} and K j =
M=

n
L

∞
L

k=0

K jk is p j -second for all j = 1, 2, ..., n. Hence

K j is second representable while it is semisimple with infinite length.

j=1

The above two examples show also that the finiteness condition on the hollow dimension in
Example 2.23 is not necessary.
Example 2.26. Let n = p1 · · · pn be a product of distinct prime numbers, p any prime number
and consider the Abelian group M = Zn [x] ⊕ Z(p∞ ) as a Z-module. Since Zn [x] is second representable (see Example 2.25) and Z(p∞ ) is second, it follows that Z M is second representable.
Notice that Z M is neither second nor semisimple.
As a direct consequence of Lemma 1.7 and Example 2.23 we obtain:
Corollary 2.27. Let R M be s-lifting.
(1) If M has the ACC on supplements (e.g. Noetherian), then M is directly second representable.
(2) If M has the DCC on supplements (e.g. Artinian), then M is directly second representable.
Theorem 2.28. Let M be an R-module.
(1) If R M is finitely generated, supplemented and every maximal hollow R-submodule of M is
second, then M is second representable.
(2) If R M is coatomic, every maximal R-submodule of M has a supplement in M, every maximal
hollow R-submodule of M is second and Att s(M) is finite, then M is second representable.

10

Proof.

(1) Since R M is finitely generated and supplemented, R M = ∑ Mλ a sum of hollow
λ ∈Λ

R-submodules by Lemma 1.9 (2). Since R M is finitely generated, this sum can be taken to
be finite and it follows that M is second representable by Proposition 2.18.
(2) Since R M is coatomic and every maximal R-submodule of M is has a supplement in M,
it follows by Lemma 1.9 (1) that R M = ∑ Mλ a sum of hollow, whence second, Rλ ∈Λ

submodules of M. Since Att s(M) is finite, it follows by Proposition 2.12 that M is second
representable.
Example 2.29. Theorem 2.28 provides a non-empty class of examples of second representable
modules. For example, let p be a prime number and consider the Prüfer group M = Z(p∞ ), as a
Z-module. Notice that
p

p

p

Z(p∞ ) = g1 , g2 , g3 , ... | g1 = 1, g2 = g1 , g3 = g2 , · · · .
Clearly, MZ is second and supplemented but not finitely generated. This example shows that the
finiteness condition of Theorem 2.28 (1) is not necessary.
Moreover, consider N = hgk i ≤ M for some k ∈ N. Observe that NZ is finitely generated and
supplemented and so, by Theorem 2.28, NZ is second representable if and only if N is second as
it is hollow.
Definition 2.30. We define a semisecondary module as one which is a (possible infinite) sum of
secondary submodules.
Example 2.31. Assume that the prime spectrum of R be finite (e.g. R = Zn ). Assume that M
is coatomic and amply supplemented over R (e.g. an Artinian module over an Artinian ring) in
which the maximal hollow submodules are second. Then M is second representable by Theorem
2.28 (2). To show this, let K M be maximal submodule, whence there is element x ∈ M\K. So,
K + Rx = M as K is maximal. Since MR is amply supplemented, there is a supplement N ≤ Rx
of K.
Example 2.32. The Abelian group M = Z12 , considered as a Z-module, has a secondary representation M = (4) ⊕ (3) but no second representation, it has a finite hollow dimension (notice
that the epimorphism
φ : M −→ (4) ⊕ (6), x 7→ 2x
is small and so h.dim(M) = 2). Observe that M is not s-lifting as the submodule (3) is maximal
hollow but not second. This example shows that the assumption that M is s-lifting in Theorem
2.23 cannot be dropped.
Example 2.33. Let R M have an infinite number of distinct simple R-submodules {S1 , · · · } such
that A := {Ann(Si ) | i ∈ N} is also infinite. The semisimple module N =

∞
L

Si is not second rep-

i=1

resentable. This example shows that the finiteness condition on the hollow (uniform) dimension
of R M in Theorem 2.23 cannot be dropped.
11

Semisecondary

Lifting

s-Lifting

Supplemented

Artinian

Semisimple

Amply supplemented

Semisecond

Figure 1: s-lifting position chart
Example 2.34. A multiplication semisimple module M =

∞
L

Si , with infinite number of distinct

i=1

simple submodules {Si | i ∈ N} is not second representable. To prove this, we claim that A :=
{Ann(Si ) | i ∈ N} is infinite. Suppose that Ann(Si ) = P = Ann(S j ) for some i 6= j. Since R M is
multiplication, Si = IM for some ideal I ≤ R, whence I ⊆ Ann(S j ) = P = Ann(Si ). But this would
mean that Si * IM (a contradiction). Thus, A is infinite as {Si | i ∈ N} is infinite.

3 Second Attached Primes
Recall that R is a commutative ring. In this section, we investigate the class of (main) second
attached primes of a second representable R-module M (with second spectrum Specs (M)).
For every R-module M, set
Att s(M) := {Ann(K) | K ∈ Specs (M)}
n

If R M is second representable, with a minimal second representation M = ∑ Ki , then the class
i=1

of main second attached primes of M is given by
att s(M) = {Ann(K1), · · · , Ann(Kn )}

12

3.1. Every p-second R-module is p-secondary and every (minimal) second representation is a
(minimal) secondary representation. So, every second representable R-module is secondary representable and att s(M) = Att(M). A subset A ⊆ att s(M) is called isolated iff for any p ∈ att s(M)
with p ⊆ q for some q ∈ A, we have p ∈ A.
Proposition 3.2. Let R M be second representable, with a minimal second representation M =
n

∑ Ki , and consider
i=1

att s(M) := {Ann(K1 ), · · · , Ann(Kn )};
Att s(M) := {Ann(K) | K ∈ Specs (M)}
(1) Att s(M) is atomic and Min(att s(M)) = Min(Att s(M))
(2) If there is no small second submodule of M, then Att s(M) is coatomic and Max(att s(M)) =
Max(Att s(M)).
Proof.

(1) Claim:

T

p ⊆ q for every q ∈ Att s(M). Let a ∈

p∈att s (M)

T

p. Then a ∈ Ann(M).

p∈att s (M)

It is easy to show that Ann(M) ⊆

T

p, hence

\

p=

p∈Att s (M)

p

p∈Att s (M)

p∈att s (M)

as att s(M) ⊆ Att s(M). It follows that

\

p ⊆ q for all q ∈ Att s(M).

T

p∈att s (M)

Now, Suppose that q ∈ Min(Att s(M)). Then

T

p ⊆ q. Since att s(M) is finite and

p∈att s (M)

each element in Att s(M) is prime, it follows that p ≤ q for some p ∈ att s(M). By the minimality of q in Att s(M) and att s(M) ⊆ Att s(M), we have p = q. Therefore, Min(Att s(M)) ⊆
Min(att s(M)). So, Att s(M) is atomic.
For the inverse inclusion, let p ∈ Min(att s(M)). Suppose that p ∈
/ Min(Att s(M)). Then
T
s
s
p ⊆ q and att (M) is finite, p′ ⊆ q for
there is q ∈ Att (M) such that q ( p. Since
p∈att s (M)

some

p′

∈

att s(M),

i.e.

p′

⊆ q ( p, which contradicts the minimality of p in att s(M).

(2) Assume that there is no small second submodule in M.
Claim: For every p ∈ Att s(M), we have pM 6= M and p ⊆ q for some q ∈ att s(M): Let
p ∈ Att s(M). Then there is a p-second submodule K ≤ M. Since K is not small in M, there
is a proper submodule L M such that K + L = M and so pM = L 6= M.
Let p ∈ Max(Att s(M)) and assume, without loss of generality that pM = ∑m
i=1 Ki with m
n (as pM 6= M) and p ⊆ Ann(Ki ) for all i ∈ {m +1, m +2, · · · , n}. Since p ∈ Max(Att s(M)),
n = m + 1 and p = Ann(Kn ), i.e. p ∈ Max(att s(M)). Therefore, Att s(M) is coatomic and
Max(Att s(M)) ⊆ Max(att s(M)).
13

For the inverse inclusion, let q ∈ Max(att s(M)). Suppose that q ∈
/ Max(Att s(M)), so
s
′
′
s
that q ( p for some p ∈ Att (M). Then p ⊆ q for some q ∈ att (M), whence q ( p ⊆
q′ , which contradicts the maximality of p in att s(M). Consequently, Max(att s(M)) ⊆
Max(Att s(M)).
Example 3.3. Consider the Abelian group M = Zn as a Z-module. We describe the second
spectrum of M and find Att s(M) and att s(M).
Example 3.4. If n is prime, then
Att s(M) = att s(M) = {(n)}.
k

If n is not prime, then consider the prime factorization n = ∏ pni i and let mi := n/pi for i ∈
i=1

{1, · · · , n}. Notice that (mi ) is pi -second for all i ∈ {1, · · · , n} and Att s(M) = {(p1 ), · · · , (pk )}.
To find att s(M), we have the following cases:
k

Case 1: ni = 1 for all i ∈ {1, · · · , n}. In this case, M = ∑ (mi ) is a second representation and
i=1

att s(M) = {(p1 ), · · · , (pk )}.
Case 2: n j > 1 for some j ∈ {1, · · · , n}. In this case, M is not second representable since
k

∑ (mi ) ⊆ (p j ) 6= M.
i=1

Example 3.5. Let M be a second representable Z-module. Then either 0 ∈ att s(M) or 0 ∈
/
s
s
s
s
att (M), and so by Proposition 3.2 we have Min(Att (M)) = {0} or Att (M) = Min(Att (M)) =
att s(M). In particular, if M is a torsion module (e.g. M = Z p × Zq for some prime numbers p
and q), then
Att s(M) = Min(Att s(M)) = att s(M).
Theorem 3.6. Let R M be Noetherian.
(1) Let p be a prime ideal. Then M is p-secondary (p-second) if and only if every nonzero
submodule of M is p-secondary (p-second).
n

(2) If M = ∑ Ki is a minimal secondary representation with Ki is pi -secondary for some prime
i=1

ideals {p1 , · · · , pn } ⊆ Spec(R), then M =

n
L

Ki .

i=1

Proof. For R M, consider for every x ∈ R the endomorphism
aM : M −→ M, x 7→ ax.
(1) We prove the result for the case of p-secondary modules; the case of p-second modules
can be proved similarly.
14

(⇒) Let M be a p-secondary module for some prime ideal p ≤ R. Let 0 6= K ≤ M. For
any a ∈
/ p, we have aM = M. Since R M is Noetherian, every surjective endomorphism is
injective and so aM is injective. Hence anM is injective for any n, i.e. an K 6= 0 for all n ∈ N.
On the other hand, aK ⊆ K = aL for some submodule L (as aM is surjective), whence
K ⊆ L and
aK ◦ aL : L −→ aK
is an isomorphism of R-modules. Hence L = K and aK = K. Therefore K is p-secondary.
(⇐) trivial.
n

(2) M = ∑ Ki is a minimal secondary representation with Ki is pi -secondary for some prime
i=1

ideals {p1 , · · · , pn } ⊆ Spec(R). Let A = {1, · · · , n}.
Claim: For any j ∈ A, we have K j ∩

Ki = 0. Suppose that K j ∩

∑
i∈A\{ j}

j ∈ A. Notice that by (1), N = K j ∩

Ki 6= 0 for some

∑
i∈A\{ j}

Ki is p j -secondary. Set J := {m ∈ A : pm * p j }.

∑
i∈A\{ j}

/ p j (as
For any m ∈ J, there is am ∈ pm \p j . Consider a = ∏ am and notice that a ∈
m∈J

J is finite am ∈ pm \p j for all m ∈ J) and so aN = N. Suppose that

∑

xi ∈ N such

i∈A\{ j}

that xi ∈ Ki for all i ∈ A. Then al

xi =

∑
i∈A\{ j}

so N ⊆

∑

xi for some l. But al N = N and

i∈A\({ j}∪J)

Ki = ∑ Ki , whence N = K j ∩ ∑ Ki . Since N 6= 0, it follows that

∑
i∈A\({ j}∪J)

pi (p j

pi (p j

{i ∈ A : pi ( p j } =
6 0.
/ We have the following cases:
Case 1 : {i ∈ A : pi ( p j } = {h}. In this case, N = K j ∩ Kh , N is pi -secondary and ph secondary at the same time (a contradiction).
Case 2: {i ∈ A : pi ( p j } has more than one element. In this case, N = K j ∩ ∑ Ki . Let
pi (p j

ph be minimal among all pi ⊆ p j , ah ∈ p j \ph and ai ∈ pi \ph for all ph ( pi ( p j if it
exists. Consider b = ah ∏ ai . Then b ∈ pi for all ph ( pi ⊆ p j and so bKh = Kh .
Since R M is

ph (pi (p j
Noetherian, btKh is injective

bt

bk xh

∑ xi =

pi (p j

for every t. Hence, for any ∑ xi ∈ N, we have
pi (p j

for some k large enough. But bN is nilpotent as N is p j -secondary and

b ∈ p j . So, xh = 0 and N = K j ∩

∑

pi (p j ,pi 6= ph

Ki . Also, the set

{i ∈ A | pi ( p j and pi 6= ph }
has a minimal element as it is finite. We continue removing the minimal elements until we
arrive at a set containing exactly one element (i.e. Case 1) which yields a contradiction.
Therefore, N = 0.

15

Theorem 3.7. Let R M be Noetherian and Artinian and assume that for any p ∈ Max(R) the
canonical map φ p : M −→ M p is injective. Then M is (second) secondary representable if and
only if M p is (second) secondary representable R p -module for any p ∈ Max(R).
Proof. We prove the result for the case of secondary representation; the case of second representation can be proved similarly.
Assume that M p is a secondary representable R p -module for any maximal ideal p ≤ R, say
n

M p = ∑ Ki′ is a minimal secondary representation for M p where each Ki′ is a secondary submodi=1

ule of M p and set for all i ∈ A = {1, 2, · · · , n} :
K i := {x ∈ M | x/s ∈ Ki′ for some s ∈
/ p};
i
′
N := {x ∈ M | x/1 ∈ Ki }
Then N i = K i and K pi = Ki′ . We may write
M p = K p1 + K p2 + · · · + K pn .
Consider the canonical map:

φ : M −→ M p; x 7→ x/1.
n

Then φ −1 (Ki′ ) = K i for all i ∈ A and so M = ∑ K i .
i=1

Claim: for any a ∈ R and all i ∈ A, the map

φ a,i : K i −→ K i ; x 7→ ax
is nilpotent or surjective. To show this, suppose that φ a,i is not surjective. Then φ a,i is not
injective since K i is Artinian (every injective endomorphism of an Artinian module is surjective).
Since φ a,i is not injective, [4, Proposition 3.9] implies that φ a,i
p is not injective for some maximal
a,i
ideal p. So, φ p is not surjective (any surjective endomorphism of a Noetherian module is injecn
i
tive). It follows that φ a,i
p is nilpotent, i.e. for some n we have a x/1 = 0 for all x ∈ K whence
an x = 0 for all x ∈ K i by our assumption.
The converse is clear (see Remark 2.11(5) for the second representation case). 
∞

Example 3.8. Consider the Z-module M = ∏ Z pi p′i , where pi and p′i are primes and pi 6= p j ,
i=1

p′i 6= p′j for all i 6= j ∈ N and p′i 6= p j for any i and j. Let the simple Z-modules K pi and K p′i be
such that (0 : K pi ) = (pi ) and (0 : K p′i ) = (p′i ), so
M=

∞
M

K pi ⊕

∞
M

K p′i .

i=1

i=1

Every second Z-submodule of M is simple, while Z M is not multiplication.
16

Example 3.9. (1) There exists an R-module M which is p-secondary but not Noetherian, while
every submodule of M is p-secondary. Appropriate semisimple modules with infinite
lengths provide a source for such modules, see Example 3.8.
(2) If R M is a Noetherian or an Artinian R-module with no zero divisors, then M satisfies the
conditions of Theorem 3.7.
Theorem 3.10. (1) A multiplication R-module M is semisecond (resp. second representable)
if and only if each nonzero submodule of M is semisecond (resp. second representable).
(2) A multiplication R-module M which is not hollow is semisecond (resp. second representable) if and only if each non-small proper submodule of M is semisecond (resp. second
representable).
(3) The following conditions are equivalent for a second representable R-module M with a
n

minimal second representation ∑ Ki .
i=1

(a) M is multiplication.
(b) Att s(M) = att s(M) = Min(att s(M)) and every nonzero submodule of M has a second
representable ∑ K j for some A ⊆ {1, · · · , n}.
j∈A

(4) If M is semisimple second representable, then the following conditions are equivalent:
(a) M is multiplication.
(b) The elements of att s(M) are incomparable and every second submodule of M is simple.
(5) The following conditions are equivalent for an atomic module R M :
(a) M is semisimple.
(b) M = ∑ Ki where every submodule of Ki is pi -second for some prime ideal pi , i ∈ A.
i∈A

Proof.

(1) Let R M be multiplication. Suppose that M = ∑ Ki is a semisecond representation
i∈A

of M. Let 0 6= K ≤ M, whence K = IM for some ideal I ≤ R. Suppose that I  p j for all
j ∈ B ⊆ A and I ≤ pi for all i ∈ A\B. Then IM = ∑ K j and so K is semisecond.
i∈B

(2) Assume that R M is multiplication and not hollow. Then M = K1 + K2 , where K1 and K2
are proper non-small submodules of M. Notice that K1 and K2 are semisecond, whence M
is semisecond.

17

n

(3) Let M = ∑ Ki be a minimal second representation.
i=1

(a ⇒ b) Assume that M is a multiplication module. Let p ∈ Att s(M), whence there is
K ≤ M which is p-second. Since R M is multiplication, K = IM for some ideal I ≤ R.
m

Assume, without loss of generality, that IM = ∑ Ki where I ⊆ pi for all i ∈ {m + 1, · · · , n}.
i=1

Since K is p-second, it follows by Proposition 2.2 that m = 1 and p = pi for some i ∈
{1, · · · , n} and so Att s(M) = att s(M). If K 6= 0, then there is an ideal J ≤ R such that
n

K = JM = J ∑ Ki = ∑ Ki , where
i=1

i∈A

A = {i ∈ {1, 2, · · · , n} | J * pi }.
n

n

i=1

i=1

Assume that pi ( p j and K j = IM, but IM = I ∑ Ki . By the minimality of ∑ Ki and using
Proposition 2.2, I ⊆ pi and so I ⊆ p j (a contradiction).

(b ⇒ a) Assume that K ≤ M is a nonzero submodule. By our assumption, K = ∑ K j with
j∈A
m

some A ⊆ {1, · · · , n}. Assume, without loss of generality, that K = ∑ Ki . Let I =
i=1

n
T

pi .

i=m+1

Since I ⊆ pi for all i ∈ {m + 1, · · · , n} and since every element in Att s(M) is minimal
and strongly irreducible and Att s(M) is finite, it follows that I * pi for all i = 1, · · · , m.
Therefore, K = IM. Consequently, R M is multiplication.
n

(4) Let R M be semisimple with a second presentation M = ∑ Ki . It follows that Att s(M) =
i=1

att s(M). Apply now (3) and observe that every nonzero submodule of M is the second
representable module ∑ K j for some A ⊆ {1, · · · , n} if and only if every second submodule
j∈A

of M is simple.
(5) Assume that M is atomic and semisecond, say M = ∑ Ki where every submodule of Ki is
i∈A

pi -second for some prime ideal pi , i ∈ A. For each i ∈ {1, · · · , n} we set
Hi = {S : S is simple in Ki }.
Claim: Ki =

L

S. If not, then there exits x ∈ Ki \

S∈Hi

L

S∈Hi S

and so Rx must contain some

S ∈ Hi and so there is a ∈ R such that ax ∈ S, whence S = Rax. But Rx is pi -second and
Rax = S 6= 0, whence S = Rax = Rx and so x ∈ S (a contradiction). It follows that M is a
sum of simple submodules. The converse is trivial.
Example 3.11. Every semisecond atomic Noetherian R-module is semisimple. This follows
directly from Theorem 3.6 (1) and Theorem 3.10 (5).
Theorem 3.12. Let E be an injective R-module and

n
T

i=1

prime ideals.
18

pi = 0 where p1 , · · · , pn are incomparable

(1) E is second representable and att s(E) ⊆ {p1 , p2 , · · · , pn }.
(2) Ann(E) = 0 if and only if att s(E) = {p1 , · · · , pn }.
Proof. (1) Set E[I] := (0 :E I). By [12, Lemma 2.1], E[pi ] = 0 or E[pi ] is pi -secondary for all
i ∈ {1, 2, · · · , n}. Assume that E[pi ] 6= 0.
Claim: E[pi ] is second. Let I ≤ R. If I ⊆ pi , then IE[pi ] = 0. If I * pi , then there is a ∈
/ pi .
Since E[pi ] is pi -secondary, aE[pi ] = E[pi ] and so IE[pi ] = E[pi ].
n

Notice that E = ∑ E[pi ] follows from [12, Lemma 2.2] as E = E[0] = E[
i=1

Hence, E is second representable and att s(E) ⊆ {p1 , · · · , pn }.

n
T

i=1

n

pi ] = ∑ E[pi ].
i=1

n

(2) (⇒) Assume that Ann(E) = 0. We want to show that E = ∑ E[pi ] is a minimal second
i=1

representation. By [12, Lemma 2.2], E[

T

pi ] = ∑ E[pi ] for any j ∈ {1, · · · , n} and so E 6=

i6= j

i6= j

∑ E[pi ] for any j ∈ {1, · · · , n}. Otherwise, E = E[

i6= j

and so

T

i6= j

T

pi ] = ∑ E[pi ] for some j ∈ {1, · · · , n}
i6= j

pi annihilates every element in E for some j ∈ {1, · · · , n}. But Ann(E) = 0,

i6= j

whence i6= j pi = 0, which contradicts the fact that pi * p j for all i ∈ {1, · · · , n}\{ j}.
Therefore, att s(E) = {p1 , · · · , pn }.
T

(⇐) Assume that att s(E) = {p1 , · · · , pn }. Suppose that 0 6= a ∈ Ann(E). Then a ∈
/ p j for
some j ∈ {1, · · · , n} and so E[p j ] = 0 or aE[p j ] = E[p j ]. Since p j is a second attached
prime, E[p j ] 6= 0. Therefore, aE = ∑ni=1 aE[pi ] 6= 0.
Example 3.13. Injective modules over Artinian rings are second representable. So, any module
over an Artinian ring is embedded in a second representable one, namely, the injective hull of it.

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