S-second submodules of a module
Let R be a commutative ring with identity and let M be an R-module. The main purpose of this paper is to introduce and study the notion of S-second submodules of an R-module M as a generalization of second submodules of M.
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Інститут прикладної математики і механіки НАН України
2021
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| Цитувати: | S-second submodules of a module / F. Farshadifar // Algebra and Discrete Mathematics. — 2021. — Vol. 32, № 2. — С. 197-210. — Бібліогр.: 16 назв. — англ. |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1860190526926487552 |
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| author | Farshadifar, F. |
| author_facet | Farshadifar, F. |
| citation_txt | S-second submodules of a module / F. Farshadifar // Algebra and Discrete Mathematics. — 2021. — Vol. 32, № 2. — С. 197-210. — Бібліогр.: 16 назв. — англ. |
| collection | DSpace DC |
| container_title | Algebra and Discrete Mathematics |
| description | Let R be a commutative ring with identity and let M be an R-module. The main purpose of this paper is to introduce and study the notion of S-second submodules of an R-module M as a generalization of second submodules of M.
|
| first_indexed | 2025-12-07T18:06:17Z |
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© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 32 (2021). Number 2, pp. 197ś210
DOI:10.12958/adm1437
S-second submodules of a module
F. Farshadifar
Communicated by V. A. Artamonov
Abstract. Let R be a commutative ring with identity and let
M be an R-module. The main purpose of this paper is to introduce
and study the notion of S-second submodules of an R-module M
as a generalization of second submodules of M .
1. Introduction
Throughout this paper, R will denote a commutative ring with identity
and Z will denote the ring of integers.
Consider a nonempty subset S of R. We call S a multiplicatively closed
subset (brieŕy, m.c.s.) of R if (i) 0 /∈ S, (ii) 1 ∈ S, and (iii) sś ∈ S for all
s, ś ∈ S [15]. Note that S = R− P is a m.c.s. of R for every prime ideal
P of R. Let M be an R-module. A proper submodule P of M is said to
be prime if for any r ∈ R and m ∈ M with rm ∈ P , we have m ∈ P or
r ∈ (P :R M) [9]. A non-zero submodule N of M is said to be second if for
each a ∈ R, the homomorphism N
a
→ N is either surjective or zero [16].
Let S be a m.c.s. of R and P a submodule of an R-module M with
(P :R M) ∩ S = ∅. Then the submodule P is said to be an S-prime
submodule of M if there exists an s ∈ S, and whenever am ∈ P , then
sa ∈ (P :R M) or sm ∈ P for each a ∈ R, m ∈ M [14]. Particularly, an
ideal I of R is said to be an S-prime ideal if I is an S-prime submodule
of the R-module R.
2020 MSC: 13C13, 13C05, 13A15, 16D60, 13H15.
Key words and phrases: second submodule, S-second submodule, S-cotorsion-
free module, simple module.
https://doi.org/10.12958/adm1437
198 S-second submodules of a module
Let S be a m.c.s. of R and M be an R-module. The main purpose
of this paper is to introduce the notion of S-second submodules of an
R-module M as a generalization of second (dual notion of S-prime) sub-
modules of M and provide some useful information concerning this class of
modules. Moreover, we obtain some results analogous to those for S-prime
submodules considered in [14].
2. Main results
Let M be an R-module. A proper submodule N of M is said to be com-
pletely irreducible if N =
⋂
i∈I Ni, where {Ni}i∈I is a family of submodules
of M , implies that N = Ni for some i ∈ I. It is easy to see that every
submodule of M is an intersection of completely irreducible submodules
of M [11].
Remark 2.1. Let N and K be two submodules of an R-module M . To
prove N ⊆ K, it is enough to show that if L is a completely irreducible
submodule of M such that K ⊆ L, then N ⊆ L [4].
Theorem 2.2. Let S be a m.c.s. of R. For a submodule N of an R-module
M with AnnR(N) ∩ S = ∅ the following statements are equivalent:
(a) There exists an s ∈ S such that srN = sN or srN = 0 for each
r ∈ R;
(b) There exists an s ∈ S and whenever rN ⊆ K, where r ∈ R and K
is a submodule of M , implies either that rsN = 0 or sN ⊆ K;
(c) There exists an s ∈ S and whenever rN ⊆ L, where r ∈ R and
L is a completely irreducible submodule of M , implies either that
rsN = 0 or sN ⊆ L.
(d) There exists an s ∈ S, and JN ⊆ K implies sJ ⊆ AnnR(N) or
sN ⊆ K for each ideal J of R and submodule K of M .
Proof. (a) ⇒ (b) and (b) ⇒ (c) are clear.
(c) ⇒ (a) By part (c), there exists an s ∈ S. Assume that srN ̸= 0 for
some r ∈ R. Then s2rN ≠ 0. If rsN ⊆ L for some completely irreducible
submodule L of M , then by assumption, sN ⊆ L. Hence, by Remark 2.1,
sN ⊆ rsN , as required.
(b) ⇒ (d) Suppose that JN ⊆ K for some ideal J of R and submodule
K of M . By part (b), there is an s ∈ S so that rN ⊆ K implies sr ∈
AnnR(N) or sN ⊆ K for each r ∈ R. Assume that sN ̸⊆ K. Then by
Remark 2.1, there exists a completely irreducible submodule L of M such
that K ⊆ L but sN ̸⊆ L. Then note that for each a ∈ J , we have aN ⊆ L.
By part (b), we can conclude that sa ∈ AnnR(N) and so sJ ⊆ AnnR(N).
F. Farshadifar 199
(d) ⇒ (b) Take a ∈ R and K a submodule of M with aN ⊆ K. Now,
put J = Ra. Then we have JN ⊆ K. By assumption, there is an s ∈ S
such that sJ = Ras ⊆ AnnR(N) or sN ⊆ K and so either sa ∈ AnnR(N)
or sN ⊆ K as needed.
Deőnition 2.3. Let S be a m.c.s. of R and N be a submodule of an
R-module M such that AnnR(N)∩S = ∅. We say that N is an S-second
submodule of M if satisőes the equivalent conditions of Theorem 2.2. By
an S-second module, we mean a module which is an S-second submodule
of itself.
The following lemma is known, but we write it here for the sake of
reference.
Lemma 2.4. Let M be an R-module, S a m.c.s. of R, and N be a őnitely
generated submodule of M . If S−1N ⊆ S−1K for a submodule K of M ,
then there exists an s ∈ S such that sN ⊆ K.
Proof. This is straightforward.
Let S be a m.c.s. of R. Recall that the saturation S∗ of S is deőned
as S∗ = {x ∈ R : x/1 is a unit of S−1R}. It is obvious that S∗ is a m.c.s.
of R containing S [12].
Proposition 2.5. Let S be a m.c.s. of R and M be an R-module. Then
we have the following.
(a) If N is a second submodule of M such that S ∩AnnR(N) = ∅, then
N is an S-second submodule of M . In fact if S ⊆ u(R) and N is an
S-second submodule of M , then N is a second submodule of M .
(b) If S1 ⊆ S2 are m.c.s.s of R and N is an S1-second submodule of M ,
then N is an S2-second submodule of M in case AnnR(N)∩S2 = ∅.
(c) N is an S-second submodule of M if and only if N is an S∗-second
submodule of M
(d) If N is a őnitely generated S-second submodule of M , then S−1N
is a second submodule of S−1M
Proof. (a) and (b) These are clear.
(c) Assume that N is an S-second submodule of M . We claim that
AnnR(N)∩S∗ = ∅. To see this assume that there exists an x ∈ AnnR(N)∩
S∗ As x ∈ S∗, x/1 is a unit of S−1R and so (x/1)(a/s) = 1 for some a ∈ R
and s ∈ S. This yields that us = uxa for some u ∈ S. Now we have that
us = uxa ∈ AnnR(N)∩S, a contradiction. Thus, AnnR(N)∩S∗ = ∅. Now
200 S-second submodules of a module
as S ⊆ S∗, by part (b), N is an S∗-second submodule of M . Conversely,
assume that N is an S∗-second submodule of M . Let rN ⊆ K. As N is an
S∗-second submodule of M , there is an x ∈ S∗ such that xr ∈ AnnR(N)
or xN ⊆ K. As x/1 is a unit of S−1R, there exist u, s ∈ S and a ∈ R such
that us = uxa. Then note that (us)r = uaxr ∈ AnnR(N) or us(xN) ⊆ K.
Therefore, N is an S-second submodule of M .
(d) If S−1N = 0, then as N is őnitely generated, there is an s ∈ S such
that s ∈ AnnR(N) by Lemma 2.4. This implies that AnnR(N) ∩ S ̸= ∅,
a contradiction. Thus S−1N ≠ 0. Now let r/t ∈ S−1R. As N is an
S-second submodule of M , there is an s ∈ S such that rsN = sN or
rsN = 0. If rsN = sN , then (r/s)S−1N = S−1N . If rsN = 0, then
(r/s)S−1N = 0, as needed.
Corollary 2.6. Let M be an R-module and set S = {1}. Then every
second submodule of M is an S-second submodule of M .
Proof. Let N be a second submodule of M . Then as N ≠ 0, we have
1 ̸∈ AnnR(N). Hence S ∩ AnnR(N) = ∅ and the result follows from
Proposition 2.5 (a).
The following examples show that the converses of Proposition 2.5 (a)
and (d) are not true in general.
Example 2.7. Take the Z-module M = Zp∞ ⊕ Z2 for a prime number p.
Then 2(Zp∞⊕Z2) = Zp∞⊕0 implies that M is not a second Z-module. Now,
take the m.c.s. S = Z \ {0} and put s = 2. Then 2rM = Zp∞ ⊕ 0 = 2M
for all r ∈ Z and so M is an S-second Z-module.
Example 2.8. Consider the Z-module M = Q ⊕ Q, where Q is the
őeld of rational numbers. Take the submodule N = Z⊕ 0 and the m.c.s.
S = Z \ {0}. Then one can see that N is not an S-second submodule
of M . Since S−1Z = Q is a őeld, S−1(Q ⊕ Q) is a vector space so that
a non-zero submodule S−1N is a second submodule of S−1(Q⊕Q).
An R-module M is said to be a comultiplication module if for every sub-
module N of M there exists an ideal I of R such that N = (0 :M I), equiv-
alently, for each submodule N of M , we have N = (0 :M AnnR(N)) [2].
Proposition 2.9. Let M be an R-module and S be a m.c.s. of R. Then
the following statements hold.
(a) If N is an S-second submodule of M , then AnnR(N) is an S-prime
ideal of R.
F. Farshadifar 201
(b) If M is a comultiplication R-module and AnnR(N) is an S-prime
ideal of R, then N is an S-second submodule of M .
Proof. (a) Let ab ∈ AnnR(N) for some a, b ∈ R. As N is an S-second
submodule of M , there exists an s ∈ S such that asN = sN or asN = 0
and bsN = sN or bsN = 0. If asN = 0 or bsN = 0 we are done. If
asN = sN , then 0 = basN = bsN , a contradiction. If bsN = sN , then
0 = absN = asN , a contradiction. Thus in any case, asN = 0 or bsN = 0,
as needed.
(b) Assume that M is a comultiplication R-module and AnnR(N)
is an S-prime ideal of R. Let r ∈ R and K be a submodule of M with
rN ⊆ K. Then AnnR(K)rN = 0. As AnnR(N) is an S-prime ideal ofR, by
[14, Corollary 2.6], there is an s ∈ S such that sAnnR(K) ⊆ AnnR(N) or
sr ∈ AnnR(N). If sr ∈ AnnR(N), we are done. If sAnnR(K) ⊆ AnnR(N),
then AnnR(K) ⊆ AnnR(sN). Now as M is a comultiplication R-module,
we have sN ⊆ K, as desired.
An R-module M satisőes the double annihilator conditions (DAC for
short) if for each ideal I of R we have I = AnnR(0 :M I) [10]. An R-module
M is said to be a strong comultiplication module if M is a comultiplication
R-module and satisőes the DAC conditions [6].
Theorem 2.10. Let M be a strong comultiplication R-module and N be
a submodule of M such that AnnR(N) ∩ S = ∅, where S is a m.c.s. of R.
Then the following are equivalent:
(a) N is an S-second submodule of M ;
(b) AnnR(N) is an S-prime ideal of R;
(c) N = (0 :M I) for some S-prime ideal I of R with AnnR(N) ⊆ I.
Proof. (a) ⇒ (b) This follows from Proposition 2.9.
(b) ⇒ (c) As M is a comultiplication R-module, N = (0 :M AnnR(N)).
Now the result is clear.
(c) ⇒ (a) As M satisőes the DAC conditions, AnnR((0 :M I)) = I.
Now the result follows from Proposition 2.9.
Let Ri be a commutative ring with identity, Mi be an Ri-module
for each i = 1, 2, . . . , n, and n ∈ N. Assume that M = M1 × M2 ×
· · · ×Mn and R = R1 ×R2 × · · · ×Rn. Then M is clearly an R-module
with componentwise addition and scalar multiplication. Also, if Si is
a multiplicatively closed subset of Ri for each i = 1, 2, . . . , n, then S =
S1 × S2 × · · · × Sn is a multiplicatively closed subset of R. Furthermore,
202 S-second submodules of a module
each submodule N of M is of the form N = N1 ×N2 × · · · ×Nn, where
Ni is a submodule of Mi.
Theorem 2.11. Let M = M1 × M2 be an R = R1 × R2-module and
S = S1×S2 be a m.c.s. of R, where Mi is an Ri-module and Si is a m.c.s.
of Ri for each i = 1, 2. Let N = N1 ×N2 be a submodule of M . Then the
following are equivalent:
(a) N is an S-second submodule of M ;
(b) N1 is an S1-second submodule of M1 and AnnR2
(N2) ∩ S2 ̸= ∅ or
N2 is an S2-second submodule of M2 and AnnR1
(N1) ∩ S1 ̸= ∅.
Proof. (a) ⇒ (b) Let N = N1 × N2 be an S-second submodule of M .
Then AnnR(N) = AnnR1
(N1) × AnnR2
(N2) is an S-prime ideal of R
by Proposition 2.9. By [14, Lemma 2.13], either AnnR(N1) ∩ S1 ̸= ∅ or
AnnR(N2)∩ S2 ≠ ∅. We may assume that AnnR(N1)∩ S1 ̸= ∅. We show
that N2 is an S2-second submodule of M2. To see this, let r2N2 ⊆ K2
for some r2 ∈ R2 and a submodule K2 of M2. Then (1, r2)(N1 ×N2) ⊆
M1×K2. As N is an S-second submodule of M , there exists an(s1, s2) ∈ S
such that (s1, s2)(N1 × N2) ⊆ M1 ×K2 or (s1, s2)(1, r2)(N1 × N2) = 0.
It follows that s2N2 ⊆ K2 or s2r2N2 = 0 and so N2 is an S2-second
submodule of M2. Similarly, if AnnR2
(N2) ∩ S2 ̸= ∅, one can see that N1
is an S1-second submodule of M1.
(b) ⇒ (a) Assume that N1 is an S1-second submodule of M1 and
AnnR2
(N2) ∩ S2 ≠ ∅. Then there exists an s2 ∈ AnnR2
(N2) ∩ S2. Let
(r1, r2)(N1 ×N2) ⊆ K1 ×K2 for some ri ∈ Ri and submodule Ki of Mi,
where i = 1, 2. Then r1N1 ⊆ K1. As N1 is an S1-second submodule of
M1, there exists an s1 ∈ S1 such that s1N1 ⊆ K1 or s1r1N1 = 0. Now we
set s = (s1, s2). Then s(N1 ×N2) ⊆ K1 ×K2 or s(r1, r2)(N1 ×N2) = 0.
Therefore, N is an S-second submodule of M . Similarly one can show
that if N2 is an S2-second submodule of M2 and AnnR1
(N1) ∩ S1 ≠ ∅,
then N is an S-second submodule of M .
Theorem 2.12. Let M = M1×M2×· · ·×Mn be an R = R1×R2×· · ·×Rn-
module and S = S1 × S2 × · · · × Sn be a m.c.s. of R, where Mi is an
Ri-module and Si is a m.c.s. of Ri for each i = 1, 2, . . . , n. Let N =
N1×N2×· · ·×Nn be a submodule of M . Then the following are equivalent:
(a) N is an S-second submodule of M ;
(b) Ni is an Si-second submodule of Mi for some i ∈ {1, 2, . . . , n} and
AnnRj
(Nj) ∩ Sj ̸= ∅ for all j ∈ {1, 2, . . . , n} − {i}.
Proof. We apply induction on n. For n = 1, the result is true. If n = 2,
then the result follows from Theorem 2.11. Now assume that parts (a)
F. Farshadifar 203
and (b) are equal when k < n. We shall prove (b) ⇔ (a) when k = n.
Let N = N1 × N2 × · · · × Nn. Put Ń = N1 × N2 × · · · × Nn−1 and
Ś = S1 × S2 × · · · × Sn−1. Then by Theorem 2.11, the necessary and
sufficient condition for N is an S-second submodule of M is that Ń is an
Ś-second submodule of Ḿ and AnnRn
(Nn)∩Sn ̸= ∅ or Nn is an Sn-second
submodule of Mn and AnnŔ(Ń)∩ Ś ̸= ∅, where Ŕ = R1×R2×· · ·×Rn−1.
Now the result follows from the induction hypothesis.
Lemma 2.13. Let S be a m.c.s. of R and N be an S-second submodule
of an R-module M . Then the following statements hold for some s ∈ S.
(a) sN ⊆ śN for all ś ∈ S.
(b) (AnnR(N) :R ś) ⊆ (AnnR(N) :R s) for all ś ∈ S.
Proof. (a) Let N be an S-second submodule of M . Then there is an s ∈ S
such that rN ⊆ K for each r ∈ R and a submodule K of M implies that
sN ⊆ K or srN = 0. Let L be a completely irreducible submodule of M
such that śN ⊆ L. Then sN ⊆ L or śsN = 0. As AnnR(N) ∩ S = ∅, we
get that sN ⊆ L. Thus sN ⊆ śN by Remark 2.1.
(b) This follows from Proposition 2.9 (a) and [14, Lemma 2.16 (ii)].
Proposition 2.14. Let S be a m.c.s. of R and N be a őnitely generated
submodule of M such that AnnR(N) ∩ S = ∅. Then the following are
equivalent:
(a) N is an S-second submodule of M ;
(b) S−1N is a second submodule of S−1M and there is an s ∈ S
satisfying sN ⊆ śN for all ś ∈ S.
Proof. (a) ⇒ (b) This follows from Proposition 2.5 (d) and Lemma 2.13.
(b) ⇒ (a) Let aN ⊆ K for some a ∈ R and a submodule K of
M . Then (a/1)(S−1N) ⊆ S−1K. Thus by part (b), S−1N ⊆ S−1K or
(a/1)(S−1N) = 0. Hence by Lemma 2.4, s1N ⊆ K or s2aN = 0 for some
s1, s2 ∈ S. By part (b), there is an s ∈ S such that sN ⊆ s1N and
sN ⊆ s2N ⊆ (0 :M a). Therefore, sN ⊆ K or asN = 0, as desired.
Theorem 2.15. Let S be a m.c.s. of R and N be a submodule of an R-
module M such that AnnR(N)∩S = ∅. Then N is an S-second submodule
of M if and only if sN is a second submodule of M for some s ∈ S.
Proof. Let sN be a second submodule of M for some s ∈ S. Let aN ⊆ K
for some a ∈ R and a submodule K of M . As asN ⊆ K and sN is a second
submodule of M , we get that sN ⊆ K or asN = 0, as needed. Conversely,
assume that N is an S-second submodule of M . Then there is an s ∈ S
204 S-second submodules of a module
such that if aN ⊆ K for some a ∈ R and a submodule K of M , then
sN ⊆ K or saN = 0. Now we show that sN is a second submodule of M .
Let a ∈ R. As asN ⊆ asN , by assumption, sN ⊆ asN or as2N = 0. If
sN ⊆ asN , then there is nothing to show. Assume that sN ̸⊆ asN . Then
as2N = 0 and so a ∈ (AnnR(N) :R s2) ⊆ (AnnR(N) :R s) by Lemma 2.13.
Thus, we can conclude that asN = 0, as desired.
The set of all maximal ideals of R is denoted by Max(R).
Theorem 2.16. Let S be a m.c.s. of R and N be a submodule of an
R-module M such that AnnR(N) ⊆ Jac(R), where Jac(R) is the Jacobson
radical of R. Then the following statements are equivalent:
(a) N is a second submodule of M ;
(b) AnnR(N) is a prime ideal of R and N is an (R \M)-second sub-
module of M for each M ∈ Max(R).
Proof. (a) ⇒ (b) Let N be a second submodule of M . Clearly, AnnR(N)
is a prime ideal of R. Since AnnR(N) ⊆ Jac(R), AnnR(N) ⊆ M for each
M ∈ Max(R) and so AnnR(N) ∩ (R \M) = ∅. Now the result follows
from Proposition 2.5 (a).
(b) ⇒ (a) Let AnnR(N) be a prime ideal of R and N be an (R \M)-
second submodule of M for each M ∈ Max(R). Let a ∈ R and a /∈
AnnR(N). We show that aN = N . Let M ∈ Max(R). Then as aN ⊆ aN ,
there exists an sM ∈ R \ M such that sMN ⊆ aN or sMaN = 0. As
AnnR(N) is a prime ideal of R and sM /∈ AnnR(N), we have asM /∈
AnnR(N) and so sMN ⊆ aN . Now consider the set
Ω = {sM : ∃M ∈ Max(R), sM /∈ M and sMN ⊆ aN}.
Then we claim that Ω = R. To see this, take any maximal ideal Ḿ
containing Ω. Then the deőnition of Ω requires that there exists an
s
Ḿ
∈ Ω and s
Ḿ
/∈ Ḿ. As Ω ⊆ Ḿ, we have s
Ḿ
∈ Ω ⊆ Ḿ, a contradiction.
Thus, Ω = R and this yields
1 = r1sM1
+ r2sM2
+ · · ·+ rnsMn
for some ri ∈ R and sMi
∈ R\Mi with sMi
N ⊆ aN , where Mi ∈ Max(R)
for each i = 1, 2, . . . , n. This yields that
N = (r1sM1
+ r2sM2
+ · · ·+ rnsMn
)N ⊆ aN.
Therefore, N ⊆ aN as needed.
F. Farshadifar 205
Now we determine all second submodules of a module over a quasilocal
ring in terms of S-second submodules.
Corollary 2.17. Let S be a m.c.s. of a quasilocal ring (R,M) and N
be a submodule of an R-module M . Then the following statements are
equivalent:
(a) N is a second submodule of M ;
(b) AnnR(N) is a prime ideal of R and N is an (R\M)-second submodule
of M .
Proof. This follows from Theorem 2.16.
Proposition 2.18. Let S be a m.c.s. of R and f : M → Ḿ be a monomor-
phism of R-modules. Then we have the following.
(a) If N is an S-second submodule of M , then f(N) is an S-second
submodule of Ḿ .
(b) If Ń is an S-second submodule of Ḿ and Ń ⊆ f(M), then f−1(Ń)
is an S-second submodule of M .
Proof. (a) As AnnR(N) ∩ S = ∅ and f is a monomorphism, we have
AnnR(f(N)) ∩ S = ∅. Let r ∈ R. Since N is an S-second submodule
of M , there exists an s ∈ S such that srN = sN or srN = 0. Thus
srf(N) = sf(N) or srf(N) = 0, as needed.
(b) AnnR(Ń) ∩ S = ∅ implies that AnnR(f
−1(Ń)) ∩ S = ∅. Now
let r ∈ R. As Ń is an S-second submodule of M , there exists an s ∈ S
such that srŃ = sŃ or srŃ = 0. Therefore srf−1(Ń) = sf−1(Ń) or
srf−1(Ń) = 0, as requested.
Proposition 2.19. Let S be a m.c.s. of R,M a comultiplication R-module,
and let N be an S-second submodule of M . Suppose that N ⊆ K +H for
some submodules K,H of M . Then sN ⊆ K or sN ⊆ H for some s ∈ S.
Proof. As N ⊆ K +H, we have AnnR(K)AnnR(H) ⊆ AnnR(N). This
implies that there exists an s ∈ S such that sAnnR(K) ⊆ AnnR(N) or
sAnnR(H) ⊆ AnnR(N) since by Proposition 2.9, AnnR(N) is an S-prime
ideal of R. Therefore, AnnR(K) ⊆ AnnR(sN) or AnnR(H) ⊆ AnnR(sN).
Now as M is a comultiplication R-module, we have sN ⊆ K or sN ⊆ H
as needed.
Let M be an R-module. The idealization R(+)M = {(a,m) : a ∈
R,m ∈ M} of M is a commutative ring whose addition is componentwise
and whose multiplication is deőned as (a,m)(b, ḿ) = (ab, aḿ+ bm) for
206 S-second submodules of a module
each a, b ∈ R, m, ḿ ∈ M [13]. If S is a m.c.s. of R and N is a submodule
of M , then S(+)N = {(s, n) : s ∈ S, n ∈ N} is a m.c.s. of R(+)M [1].
Proposition 2.20. Let M be an R-module and let I be an ideal of R
such that I ⊆ AnnR(M). Then the following are equivalent:
(a) I is a second ideal of R;
(b) I(+)0 is a second ideal of R(+)M .
Proof. This is straightforward.
Theorem 2.21. Let S be a m.c.s. of R, M be an R-module, and I be an
ideal of R such that I ⊆ AnnR(M) and I ∩S = ∅. Then the following are
equivalent:
(a) I is an S-second ideal of R;
(b) I(+)0 is an S(+)0-second ideal of R(+)M ;
(c) I(+)0 is an S(+)M -second ideal of R(+)M .
Proof. (a) ⇒ (b) Let (r,m) ∈ R(+)M . As I is an S-second ideal
of R, there exists an s ∈ S such that rsI = sI or rsI = 0. If
rsI = 0, then (r,m)(s, 0)(I(+)0) = 0. If rsI = sI, then we claim that
(r,m)(s, 0)(I(+)0) = (s, 0)(I(+)0). To see this let (sa, 0) = (s, 0)(a, 0) ∈
(s, 0)(I(+)0). As rsI = sI, we have sa = rsb for some b ∈ I. Thus as
b ∈ I ⊆ AnnR(M),
(sa, 0) = (srb, 0) = (srb, smb) = (sr, sm)(b, 0) = (s, 0)(r,m)(b, 0).
Hence (s, 0)(a, 0) ∈ (r,m)(s, 0)(I(+)0), and so
(s, 0)(I(+)0) ⊆ (r,m)(s, 0)(I(+)0).
Since the inverse inclusion is clear we reach the claim.
(b) ⇒ (c) Since S(+)0 ⊆ S(+)M , the result follows from Proposi-
tion 2.5(b).
(c) ⇒ (a) Let r ∈ R. As I(+)0 is an S(+)M -second ideal of
R(+)M , there exists an (s,m) ∈ S(+)M such that (r, 0)(s,m)(I(+)0) =
(s,m)(I(+)0) or (r, 0)(s,m)(I(+)0) = 0. If (r, 0)(s,m)(I(+)0) = 0, then
0 = (r, 0)(s,m)(a, 0) = (rs, rm)(a, 0) = (rsa, rma) = (rsa, 0)
far each a ∈ I. Thus rsI = 0. If (r, 0)(s,m)(I(+)0) = (s,m)(I(+)0), then
we claim that rsI = sI. To see this let sa ∈ sI. Then for some b ∈ I, we
have
(sa, 0) = (sa, am) = (s,m)(a, 0) = (s,m)(r, 0)(b, 0)
= (srb, rmb) = (srb, 0).
Hence, sa ∈ rsI and so sI ⊆ srI, as needed.
F. Farshadifar 207
Let P be a prime ideal of R and N be a submodule of an R-module M .
The P -interior of N relative to M is deőned (see [3, 2.7]) as the set
IMP (N) =
⋂
{L | L is a completely irreducible submodule of M
and rN ⊆ L for some r ∈ R− P}.
Let R be an integral domain. A submodule N of an R-module M is
said to be a cotorsion-free submodule of M (the dual of torsion-free) if
IM
0
(N) = N and is a cotorsion submodule of M (the dual of torsion) if
IM
0
(N) = 0. Also, M said to be cotorsion (resp. cotorsion-free) if M is
a cotorsion (resp. cotorsion-free) submodule of M [5].
One can see that if M is a cotorsion-free R-module, then R is an
integral domain and M is a faithful R-module. In [5, Proposition 2.9 (e)],
it is shown that if M is a comultiplication R-module the reverse is true.
The following example shows that sometimes the reverse of this statement
may not be true.
Example 2.22. Consider the Z-module M =
∏
∞
i=1
Zpi , where p is a prime
number. Then it is easy to see that M is a faithful Z-module. But the
Z-module M is not second since (1̄, 0̄, 0̄, . . . ) /∈ pM and so M ≠ pM .
Therefore, by [5, Theorem 2.10], IM
0
(M) ̸= M and so the Z-module M is
not a cotorsion-free module.
Deőnition 2.23. Let M be an R-module and S be a m.c.s. of R with
AnnR(M) ∩ S = ∅. We say that M is an S-cotorsion-free module in the
case that we can őnd s ∈ S such that if rM ⊆ L, where r ∈ R and L is
a completely irreducible submodule of M , then sM ⊆ L or rs = 0.
Proposition 2.24. Let M be an R-module and S be a m.c.s. of R. Then
the following statements are equivalent.
(a) M is an S-second R-module.
(c) P = AnnR(M) is an S-prime ideal of R and the R/P -module M is
an S-cotorsion-free module.
Proof. (a) ⇒ (b). We can assume that P = 0. By Proposition 2.9 (a),
AnnR(M) is an S-prime ideal of R. Now let L be a completely irreducible
submodule of M and r ∈ R such that rM ⊆ L. Then there exists an
s ∈ S such that sM ⊆ L or srM = 0 because M is S-second. Therefore,
sM ⊆ L or rs ∈ AnnR(M) = 0 , as needed.
(b) ⇒ (a). As AnnR(M) is an S-prime ideal of R, AnnR(M) ∩ S = ∅.
Suppose that there exist r ∈ R and completely irreducible submodule L of
M such that rM ⊆ L. By assumption, there is an s ∈ S such that sM ⊆ L
or rs = 0R/P . Thus sM ⊆ L or rs ∈ P = AnnR(M), as desired.
208 S-second submodules of a module
Theorem 2.25. Let M be a module over an integral domain R. Then the
following are equivalent:
(a) M is a cotorsion-free R-module;
(b) M is an (R \ P )-cotorsion-free for each prime ideal P of R;
(c) M is an (R \M)-cotorsion-free for each maximal ideal M of R.
Proof. (a) ⇒ (b) This is clear.
(b) ⇒ (c) This is obvious.
(c) ⇒ (a) Let M be (R \M)-cotorsion-free for each maximal ideal M
of R. Let aM ⊆ L for some a ∈ R and a completely irreducible submodule
L of M . Assume that a ≠ 0. Take a maximal ideal M of R. As M is
(R \M)-cotorsion-free, there exists an sM ∈ R \M such that sMM ⊆ L
or asM = 0. As R is an integral domain, asM ≠ 0 and so sMM ⊆ L. Now
set
Ω = {sM : ∃M ∈ Max(R), sM /∈ M and sMM ⊆ L}.
A similar argument as in the proof of Theorem 2.16 shows that Ω = R.
Thus we have ⟨sM1
⟩+ ⟨sM2
⟩+ · · ·+ ⟨sMn
⟩ = R for some sMi
∈ Ω. This
implies that M = (⟨sM1
⟩+ ⟨sM2
⟩+ · · ·+ ⟨sMn
⟩)M ⊆ L and hence M = L.
This means that M is a cotorsion-free R-module.
Let M be an R-module. The dual notion of ZR(M), the set of zero
divisors of M , is denoted by WR(M) and deőned by
W (M) = {a ∈ R : aM ̸= M}.
Theorem 2.26. Let S be a m.c.s. of R and M be a őnitely generated
comultiplication R-module with AnnR(M) ∩ S = ∅. Then the following
statements are equivalent:
(a) Each non-zero submodule of M is S-second;
(b) M is a simple R-module.
Proof. (a) ⇒ (b) Assume that every non-zero submodule of M is an
S-second submodule of M . First we show that WR(M) = AnnR(M).
Let a ∈ WR(M). Then aM ≠ M . Since M is S-second, there exists
an s ∈ S such that saM = sM or saM = 0. If saM = sM , then
s ∈ (saM :R M). Now put N = (0 :M (saM :R M)) and note that
s ∈ S ∩ AnnR(N) ̸= ∅. Thus, N is not S-second and so by part (a),
we have N = (0 :M (saM :R M)) = 0. Now as M is a comultiplication
R-module, one can see that M(saM :R M) = M . By [7, Corollary 2.5],
1 − x ∈ AnnR(M) ⊆ (saM :R M) for some x ∈ (saM :R M) since M
is a őnitely generated R-module. This implies that (saM :R M) = R
F. Farshadifar 209
and so saM = M . It follows that aM = M , which is a contradiction.
Therefore, saM = 0. Then s ∈ AnnR(aM) and so S ∩ AnnR(aM) ̸= ∅.
Hence by assumption, aM = 0. Thus, we get WR(M) = AnnR(M). Let
a /∈ WR(M). Now we will show that (0 :M a) = 0. If (0 :M a2) = 0, then
(0 :M a) = 0. Suppose (0 :M a2) ̸= 0. Since (0 :M a2) is an S-second
submodule of M and a(0 :M a2) ⊆ (0 :M a), there is an s ∈ S such
that sa(0 :M a2) = 0. This implies that (0 :M a2) ⊆ (0 :M as). Let
m1 ∈ (0 :M a). Then am1 = 0. We have m1 = am2 since aM = M . Thus
am1 = a2m2 = 0. Hence m2 ∈ (0 :M a2) ⊆ (0 :M as). This implies that
0 = sam2 = sm1 and so m1 ∈ (0 :M s). Therefore, (0 :M a) ⊆ (0 :M s)
and so s ∈ AnnR((0 :M a)). Hence S ∩ AnnR((0 :M a)) ̸= ∅. So by
assumption, we have (0 :M a) = 0. Now take a submodule H of M .
If AnnR(H) = AnnR(M), then H = M since M is a comultiplication
R-module. Take an element a ∈ AnnR(H) \ AnnR(M). As WR(M) =
AnnR(M), a /∈ WR(M) and so (0 :M a) = 0. Then we get
H = (0 :M AnnR(H)) ⊆ (0 :M a) = 0.
Therefore, M is a simple R-module.
(b) ⇒ (a) Note that every simple R-module M is a second submodule
of M . Since AnnR(M)∩S = ∅, by Proposition 2.5 (a), M is an S-second
submodule of M .
An R-module M is said to be a multiplication module if for every
submodule N of M there exists an ideal I of R such that N = IM [8].
Corollary 2.27. Let S be a m.c.s. of R. If M is a őnitely generated
multiplication and comultiplication R-module with AnnR(M) ∩ S = ∅,
then the following statements are equivalent:
(a) Each non-zero submodule of M is S-second;
(b) M is a simple R-module;
(c) Each proper submodule of M is an S-prime submodule of M .
Proof. This follows from Theorem 2.26 and [14, Theorem 2.26].
Example 2.28. Consider the Z-module Zn. Take S = Z−0. We know that
Zn is a őnitely generated multiplication and comultiplication Z-module.
Then by Corollary 2.27, if n is not a prime number, the Z-module Zn
has a non-zero submodule which is not S-second and a proper submodule
which is not S-prime.
210 S-second submodules of a module
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Contact information
Faranak
Farshadifar
Assistant Professor, Department of Mathematics,
Farhangian University, Tehran, Iran.
E-Mail(s): f.farshadifar@cfu.ac.ir
Received by the editors: 15.08.2019
and in őnal form 25.01.2020.
mailto:f.farshadifar@cfu.ac.ir
F. Farshadifar
|
| id | nasplib_isofts_kiev_ua-123456789-188747 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1726-3255 |
| language | English |
| last_indexed | 2025-12-07T18:06:17Z |
| publishDate | 2021 |
| publisher | Інститут прикладної математики і механіки НАН України |
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| spelling | Farshadifar, F. 2023-03-14T16:50:08Z 2023-03-14T16:50:08Z 2021 S-second submodules of a module / F. Farshadifar // Algebra and Discrete Mathematics. — 2021. — Vol. 32, № 2. — С. 197-210. — Бібліогр.: 16 назв. — англ. 1726-3255 DOI:10.12958/adm1437 2020 MSC: 13C13, 13C05, 13A15, 16D60, 13H15 https://nasplib.isofts.kiev.ua/handle/123456789/188747 Let R be a commutative ring with identity and let M be an R-module. The main purpose of this paper is to introduce and study the notion of S-second submodules of an R-module M as a generalization of second submodules of M. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics S-second submodules of a module Article published earlier |
| spellingShingle | S-second submodules of a module Farshadifar, F. |
| title | S-second submodules of a module |
| title_full | S-second submodules of a module |
| title_fullStr | S-second submodules of a module |
| title_full_unstemmed | S-second submodules of a module |
| title_short | S-second submodules of a module |
| title_sort | s-second submodules of a module |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/188747 |
| work_keys_str_mv | AT farshadifarf ssecondsubmodulesofamodule |