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On Property (A) of the amalgamated duplication of a
ring along an ideal
Youssef Arssi and Samir Bouchiba∗†
Department of Mathematics, Faculty of Sciences, Moulay Ismail university of Meknes,
Morocco‡§
June 25, 2020
Abstract
The main purpose of this paper is to totally characterize when the amalgamated
duplication R ./ I of a ring Ralong an ideal Iis an A-ring as well as an SA-ring. In
this regard, we prove that R ./ I is an SA-ring if and only if Ris an SA-ring and I
is contained in the set of zero divisors Z(R)of R. As to the Property (A) of R ./ I, it
turns out that its characterization involves a new concept that we introduce in [1] and
that we term the Property (A) of a module Malong an ideal I. In fact, we prove that
R ./ I is an A-ring if and only if Ris an A-ring, Iis an A-module along itself and if p
is a prime ideal of Rsuch that p⊆ZR(I)∪ZI(R), then either p⊆ZR(I)or p⊆ZI(R),
where ZI(R) := {a∈R:a+I⊆Z(R)}.
1 Introduction
Throughout this paper, all rings are supposed to be commutative with unit element and all
R-modules are unital. Let Rbe a commutative ring and Man R-module. We denote by
ZR(M) = {r∈R:rm = 0 for some nonzero element m∈M}the set of zero divisors of R
on Mand by Z(R) := ZR(R)the set of zero divisors of the ring R. In [4], the notions of A-
module and SA-module are extensively studied. In fact, an R-module Msatisfies Property
(A), or Mis an A-module over R(or A-module if no confusion is likely), if for every finitely
generated ideal Iof Rwith I⊆ZR(M)), there exists a nonzero m∈Mwith Im = 0, or
equivalently, annM(I)6= 0.Mis said to satisfy strong Property (A), or is an SA-module
over R(or an SA-module if no confusion is likely), if for any r1,· · · , rn∈ZR(M), there exists
a nonzero m∈Msuch that r1m=· · · =rnm= 0. The ring Ris said to satisfy Property
(A), or an A-ring, (resp., SA-ring) if Ris an A-module (resp., an SA-module). One may
∗Corresponding author
†s.bouchiba@fs.umi.ac.ma
‡Mathematics Subject Classification 2010 : 13C13, 13A99
§Key words and phrases:A-ring; A-module along an ideal; amalgamated duplication; SA-ring; zero
divisor.
1
On Property (A) of the amalgamated duplication of a ring along an ideal 2
easily check that Mis an SA-module if and only if Mis an A-module and ZR(M)is an ideal
of R. It is worthwhile reminding the reader that the Property (A) for commutative rings
was introduced by Quentel in [22] who called it Property (C) and Huckaba used the term
Property (A) in [16, 17]. In [14], Faith called rings satisfying Property (A) McCoy rings. The
Property (A) for modules was introduced by Darani [12] who called such modules F-McCoy
modules (for Faith McCoy terminology). He also introduced the strong Property (A) under
the name super coprimal and called a module Mcoprimal if ZR(M)is an ideal. In [20],
the strong Property (A) for commutative rings was independently introduced by Mahdou
and Hassani in [20] and further studied by Dobbs and Shapiro in [13]. Note that a finitely
generated module over a Noetherian ring is an A-module (for example, see [18, Theorem 82])
and thus a Noetherian ring is an A-ring. Also, it is well known that a zero-dimensional ring
Ris an A-ring as well as any ring Rwhose total quotient ring Q(R)is zero-dimensional. In
fact, it is easy to see that Ris an A-ring if and only if so is Q(R)[9, Corollary 2.6]. Any
polynomial ring R[X]is an A-ring [16] as well as any reduced ring with a finite number of
minimal prime ideals [16]. In [6], we generalize a result of T.G. Lucas which states that if R
is a reduced commutative ring and Mis a flat R-module, then the idealization RnMis an
A-ring if and only if Ris an A-ring [19, Proposition 3.5]. In effect, we drop the reduceness
hypotheses and prove that, given an arbitrary commutative ring Rand any submodule M
of a flat R-module F,RnMis an A-ring (resp., SA-ring) if and only if Ris an A-ring
(resp., SA-ring). In [7], we present an answer to a problem raised by D.D. Anderson and S.
Chun in [4] on characterizing when is the idealization RnMof a ring Ron an R-module
Man A-ring (resp., an SA-ring) in terms of module-theoretic properties of Rand M. Also,
we were concerned with presenting a complete answer to an open question asked by these
two authors which reads the following: What modules over a given ring Rare homomorphic
images of modules satisfying the strong Property (A)? [4, Question 4.4 (1)]. The main
theorem of [8] extends a result of Hong, Kim, Lee and Ryu in [15] which proves that a direct
product QRiof rings is an A-ring if and only if so is any Ri. In this regard, we show that
if {Ri}i∈Iis a family of rings and {Mi}i∈Iis a family of modules such that each Miis an
Ri-module, then the direct product Q
i∈I
Miof the Miis an A-module over Q
i∈I
Riif and only if
each Miis an A-module over Ri,i∈I. Finally, our main concern in [9] is to introduce and
investigate a new class of rings lying properly between the class of A-rings and the class of
SA-rings. The new class of rings, termed the class of P SA-rings, turns out to share common
characteristics with both A-rings and SA-rings. Numerous properties and characterizations
of this class are given as well as the module-theoretic version of PSA-rings is introduced
and studied. For further works related to the Property (A) and (SA), we refer the reader to
[2, 3, 4, 5, 11, 15, 19].
The main purpose of this paper is to totally characterize when the amalgamated dupli-
cation R I of a ring Ralong an ideal Iis an A-ring as well as an SA-ring. In this regard,
it is worth reminding that Mahdou and Hassani, on the one hand, and Mahdou and Moutui,
on the other, have tackled this subject in [20] and [21]. They essentially proved that if R I
is an SA-ring (resp., an A-ring), then Ris an SA-ring (resp., an A-ring) [20, Theorem 3.1].
Also, they showed that if Iis a regular ideal, then R I is an A-ring if and only if Ris
an A-ring [20, Theorem 3.1]. Furthemore, it is proved in [21, Corollary 2.10], that if Ris a
total ring of quotients which is an A-ring, I2= (0) and Z(R)⊆I, then R I is an A-ring.
On Property (A) of the amalgamated duplication of a ring along an ideal 3
In Section 3, we prove, via Theorem 3.1, that R I is an SA-ring if and only if Ris an
SA-ring and I⊆Z(R). As to the characterization of the Property (A) of R I , it turns out
that it involves a new concept that we introduce in [1] and that we term the Property (A)
of a module Malong an ideal I. So, in Section 2, for the sake of self-containement, we recall
the main results of [1] concerning the new notions of an A-ring and A-module along an ideal
Iof R. In particular, any A-ring (resp., any A-module) is an A-ring (resp., an A-module)
along any ideal Iof R. Our investigation makes it possible to exhibit many contexts of rings
Rpossessing an ideal Iwhich is an A-module along itself. For instance, we prove that if R
is Noetherian or zero-dimensional, then any ideal Iis an A-module along itself. Also, if R
is a ring and Iis an ideal of Rsuch that either In= (0) for some positive integer n≥1
or ZR(I) = Z(R), then Iis an A-module along itself. Moreover, through Example 2.13, we
provide a case of a ring Radmitting an ideal Iwhich is an A-module along itself while Iis
not an A-module. In the light of this, the principal theorem of Section 4, Theorem 4.4, reads
the following: the amalgamated duplication R I is an A-ring if and only if the following
assertions hold:
1) Ris an A-ring.
2) Iis an A-module along itself.
3) If p∈Spec(R)such that p⊆ZR(I)∪ZI(R), then either p⊆ZR(I)or p⊆ZI(R),
where ZI(R) = {a∈R:a+I⊆Z(R)}.
As a consequence of this theorem, we prove that if I*Z(R), then the following assertions
are equivalent:
1) R I is an A-ring;
2) Ris an A-ring.
3) Iis an A-module;
4) Iis an A-module along itself.
Also, given a ring Rand an ideal Iof Rsuch that either In= (0) for some positive integer
n≥1or ZR(I) = Z(R), we show that R I is an A-ring if and only if Ris an A-ring.
2 Preliminaries
In this section, we recall from [1] the new concepts of an A-ring Rand A-module Malong
an ideal Iof R. The new classes turn out to encompass the classical ones of A-rings and
A-modules. Also, the Property (A) along an ideal proved to be a necessary condition for the
amalgamated duplication R I along Ito be an A-ring (see Theorem 4.4). More precisely,
we exhibit conditions under which an ideal Iof a ring Ris an A-module along itself.
Let Rbe a ring and Man R-module. Let Spec(Z(R)) (resp., Max(Z(R))) denote the
set of prime ideals (resp., maximal ideals) of Rcontained in Z(R)and Spec(ZR(M)) (resp.,
Max(ZR(M))) denote the set of prime ideals (resp., maximal ideals) of Rcontained in ZR(M).
According to [18], Max(ZR(M)) stands for the set of the maximal primes of the R-module M.
We begin by recalling the definition of the new concepts.
Definition 2.1. Let Rbe a ring, Ian ideal of Rand Man R-module. Let S:= R\Z(R)
and SR(M) = R\ZR(M). Then
On Property (A) of the amalgamated duplication of a ring along an ideal 4
1) Ris said to be an A-ring along Iif for each finitely generated ideal J⊆Z(R)such that
(J+I)∩S6=∅, we have ann(J)6= (0).
2) Mis said to be an A-module along Iif for each finitely generated ideal J⊆ZR(M)such
that (J+I)∩SR(M)6=∅, we have annM(J)6= (0).
The vacuous case of a ring Rand an ideal Iof Rsuch that (Z(R)+I)∩S=∅is considered
of course an A-ring along I. For instance, any ring Ris an A-ring along (0) as Z(R)∩S=∅.
Also, any R-module Msuch that (ZR(M) + I)∩SR(M) = ∅is (vacuously) an A-module
along I.
The following two propositions records the fact that the Property (A) along a fixed ideal
Iof a ring Ris a weaker notion than the Property (A) of Rand that in the Noetherian
setting any ideal Iof Ris an A-module along itself.
Proposition 2.2. Let Rbe a ring and Ian ideal of R. Then
1) Any A-module Mover Ris an A-module along I. In particular, if Ris an A-ring, then
Ris an A-ring along I.
2) If Ris Noetherian, then Iis an A-module, and thus Iis an A-module along itself.
3) If Ris zero-dimensional, then Iis an A-module along itself.
Proof. 1) It is clear from Definition 2.1.
2) Assume that Ris Noetherian. Then Iis a Noetherian module over R. Therefore, by [4,
Theorem 2.2(5)], Iis an A-module. Hence, by (1), Iis an A-module along itself.
3) Assume that dim(R)=0. By [4, Theorem 2.2], any R-module is an A-module. Then,
using (1), we get that Iis an A-module along itself.
Through the next bunch of results we seek conditions under which an ideal Iof a ring R
is an A-module along itself.
Let Rbe a ring and Man R-module. We denote by J(R) := T
m∈Max(R)
mthe Jacobson
radical of Rand by J(ZR(M)) := T
m∈Max(ZR(M))
mthe intersection of all maximal primes of
M.
Proposition 2.3. Let Rbe a ring and Ian ideal of R. Let Mbe an R-module such that
I⊆J(ZR(M)). Then Mis an A-module along I. In particular, if I⊆J(ZR(I)), then Iis
an A-module along itself.
Proof. Let Jbe a finitely generated ideal of Rsuch that J⊆ZR(M). Then there exists a
maximal prime m∈Max(ZR(M)) such that J⊆m. Hence, as I⊆m, we get J+I⊆m.
This means that J+I⊆ZR(M)and thus (J+I)∩SR(M) = ∅. It follows that Mis
(vacuously) an A-module along I, as desired.
Let Rbe a ring. We denote by Rad(R)the nilradical of R.
Proposition 2.4. Let Rbe a ring and Ian ideal of R. Assume that I⊆Rad(R). Then any
R-module Mis an A-module along I. In particular, Iis an A-module along itself.
On Property (A) of the amalgamated duplication of a ring along an ideal 5
We record the following lemma.
Lemma 2.5. Let Rbe a ring. Then
Rad(R)⊆J(ZR(M))
for any R-module M.
Proof. It is direct as Rad(R) = T
p∈Spec(R)
pand J(ZR(M)) := T
m∈Max(ZR(M))
mfor a given
R-module M.
Proof of Proposition 2.4. Let Mbe an R-module. By Lemma 2.5, Rad(R)⊆J(ZR(M)).
Then I⊆J(ZR(M)). It follows by Proposition 2.3 that Mis an A-module along Icompleting
the proof of the proposition.
Corollary 2.6. Let Rbe a ring and Ia nilpotent ideal of R, that is, there exists n≥1
such that In= (0). Then any R-module Mis an A-module along I. In particular, Iis an
A-module along itself.
Proof. I suffices to note that I⊆Rad(R)and then to apply Proposition 2.4.
Corollary 2.7. Let Rbe a ring and Man R-module. Consider the idealization RnMof
M. Then the ideal I:= (0) nMof RnMis an A-module along itself.
Proof. Note that I2= (0). Then, by Corollary 2.6, Iis an A-module along itself.
Proposition 2.8. Let Rbe an A-ring and Ian ideal of Rsuch that ZR(I) = Z(R). Then I
is an A-module along itself.
Proof. Let J⊆ZR(I)be a finitely generated ideal of Rsuch that (J+I)∩SR(I)6=∅and thus
J∩(SR(I) + I)6=∅. Then, as J⊆ZR(I)⊆Z(R)and Ris an A-ring, we get ann(J)6= (0)
and thus there exists a∈R\ {0}such that aJ = (0). Also, there exists i∈Iand s∈SR(I)
such that s+i∈J. Hence a(s+i) = 0, so that as =−ai. Note that, as ZR(I) = Z(R), we
have SR(I) = S. This ensures that s∈Sis a regular element of R. Assume that ai = 0.
Therefore as = 0 and thus, as sis a regular element, we get a= 0 which is absurd. Hence
j:= ai ∈I\ {0}. It follows that jJ = (0) and j∈I\ {0}which means that annI(J)6= (0).
Consequently, Iis an A-module along itself, as desired.
Corollary 2.9. Let Rbe an SA-ring. Put I:= Z(R). Then Iis an A-module along itself.
Proof. Note that, as Ris an SA-ring, I:= Z(R)is an ideal of R. Let us first prove that
ZR(I) = Z(R). In effect, the inequality ZR(I)⊆Z(R)always holds. Let 06=x∈Z(R). Then
there exists 06=a∈Rsuch that ax = 0. Hence a∈Z(R) = Iand thus x∈ZR(I). It
follows that Z(R)⊆ZR(I)yielding the desired equality ZR(I) = Z(R). Now, Proposition 2.8
completes the proof.
On Property (A) of the amalgamated duplication of a ring along an ideal 6
Proposition 2.10. Let Rbe an A-ring and Ian ideal of R. Assume that ann(I)⊆I. Then
Iis an A-module and thus Iis an A-module along itself.
Proof. Let J⊆ZR(I)be a finitely generated ideal of R. As Ris an A-ring, we get ann(J)6=
(0) and thus there exists a∈Rsuch that a6= 0 and aJ = (0). If a∈I, then annI(J)6= (0).
Assume that a6∈ I. Then, as ann(I)⊆I,a6∈ ann(I). Hence there exists i∈Isuch that
j:= ai 6= 0. It follows that jJ = (0) and j∈I\ {0}, so that annI(J)6= (0). Consequently,
Iis an A-module and thus Iis an A-module along itself.
Corollary 2.11. Let Rbe an A-ring and Ian ideal of Rsuch that ann(I) = (0). Then Iis
an A-module along itself.
Corollary 2.12. Let Rbe an A-ring and Ian ideal of R. Assume that ZR(I)⊆I. Then I
is an A-module along itself.
Proof. It is direct from Proposition 2.10 as ann(I)⊆ZR(I).
Next, we give an example of an A-ring Rwhich admits an ideal Isuch that Iis an
A-module along itself while Iis not an A-module.
Example 2.13. We resume the example [4, Example 2.14] of Anderson and Chun. Let kbe
a countable field and D=k[X1, X2,· · · , Xn]with n≥2. There exists D-modules A1and A2
such that A1⊕A2is an A-module as a D-module while neither A1nor A2has Property (A)
over D. Let R:= (DnA1)nA2∼
=Dn(A1⊕A2). By [4, Theorem 2.12], Ris an A-ring as D
is a domain and A1⊕A2is an A-module over D. Also, by Corollary 2.7, the ideal I= (0)nA2
of Ris an A-module along itself. Moreover, observe that the natural ring homomorphisms
DnA1−→ Dand R−→ DnA1are surjective. Then, by two applications of [4, Theorem
2.1(1)(b)], we get that A2is not an A-module as an R-module, where ((d, a1), a2)x=dx
for any d∈D,a1∈A1,a2∈A2and any x∈A2. On the other hand, it is easy to see
that the natural map ϕ:A2−→ I= (0) nA2such that ϕ(a) = (0, a)for each a∈A2is an
isomorphism of R-modules. It follows that I= (0) nA2is not an A-module, as desired.
Our next result characterizes the Property (A) along an ideal Iof a ring Rin the case
where I*Z(R). We prove that in this setting the two notions of Property (A) and Property
(A) along Icollapse.
Theorem 2.14. Let Rbe a ring and Ian ideal of R. Let Mbe an R-module.
1) Assume that I*Z(R). Then the following assertions are equivalent:
a) Ris an A-ring.
b) Ris an A-ring along I
c) Iis an A-module;
d) Iis an A-module along itself.
2) Assume that I*ZR(M). Then Mis an A-module along Iif and only if Mis an
A-module.
First, we prove the following lemma.
On Property (A) of the amalgamated duplication of a ring along an ideal 7
Lemma 2.15. Let Rbe a ring and Iis an ideal such that I*Z(R). Then
ZR(I) = Z(R).
Proof. First, note that ZR(I)⊆Z(R). Also, as I*Z(R), we get S−1I=Q(R). Let
a∈Z(R). Then there exists r∈Rsuch that r6= 0 and ra = 0. Hence there exists i
s∈S−1I
with i∈I\ {0}and s∈Ssuch that r
1=i
sand thus i
s
a
1= 0. Therefore there exists t∈S
such that tia = 0. Then ia = 0 and i∈I\ {0}. It follows that a∈ZR(I). Consequently,
ZR(I) = Z(R), as desired.
Proof of Theorem 2.14. 1) Assume that I*Z(R), that is I∩S6=∅. Then 0∈S+Iand
thus J∩(S+I)6=∅for any ideal Jof R. Hence, by Definition 2.1, the equivalence a) ⇔b)
holds. Also, since by Lemma 2.15, Z(R)=ZR(I), we get S=SR(I)and thus for any ideal
Jof R,J∩(SR(I) + I)6=∅. Therefore, by Definition 2.1, the equivalence c) ⇔d) holds as
well. It remains to prove the equivalence a) ⇔c).
a) ⇒c) Assume that Ris an A-ring. Let J⊆ZR(I)be a finitely generated ideal of R. As
ZR(I)⊆Z(R)and as Ris an A-ring, there exists r∈Rsuch that r6= 0 and rJ = (0).
Since I*Z(R), we get S−1I=Q(R). Hence there exists i∈I\ {0}and s∈Ssuch that
r
1=i
sand thus i
s
j
1= (0) for each j∈J. Therefore, for each j∈J, there exists tj∈S
such that tjij = (0), so that ij = 0 for each j∈J. Hence iJ = (0) and i∈I\ {0}yielding
annI(J)6= (0). It follows that Iis an A-module proving c).
c) ⇒a) Assume that Iis an A-module. Let J⊆Z(R)be a finitely generated ideal of R.
Then, as, by Lemma 2.15, Z(R) = ZR(I), we get J⊆ZR(I). Hence, since Iis an A-module,
there exists i∈I\ {0}such that iJ = (0), so that ann(J)6= (0). It follows that Ris an
A-ring proving a).
2) It is similar to the proof of the equivalence a) ⇔b) of (1) completing the proof.
We deduce the following cases of ideals which are A-modules along themselves.
Corollary 2.16. Let Rbe a ring and Ian ideal of R. Then any ideal Iof R[X]such that
X∈Iis an A-module along itself.
Proof. Let Ibe an ideal of R[X]such that X∈I. Then I*Z(R[X]). It follows, since R[X]
is an A-ring, and using Theorem 2.14, that Iis an A-module along itself
Corollary 2.17. Let Rbe a ring and Ian ideal of R. Assume that I6⊆ ZR(I). Then Iis
an A-module along itself if and only if Iis an A-module.
3 Amalgamated duplication and SA-property
The main purpose of this section is to characterize when is the amalgamated duplication
R I of a ring Ralong an ideal Ian SA-ring.
On Property (A) of the amalgamated duplication of a ring along an ideal 8
If Ris a ring and Ian ideal of R, for easiness, we adopt the following notation: for any
subset Eof R, we consider the next two subsets of R I :
E I ={(e, e +i)∈R I :e∈Eand i∈I}
I E := {(e+i, e) : e∈Eand i∈I}
and the subset of R:
E+I:= {e+i:e∈Eand i∈I}.
We begin by stating the main theorem of this section.
Theorem 3.1. Let Rbe a ring and Ian ideal of R. Then the following statements are
equivalent:
1) R I is an SA-ring;
2) Ris an SA-ring and I⊆Z(R).
To prove this theorem, we need the following preparatory results.
Lemma 3.2. Let Rbe a ring and Ian ideal of R. Then
1) Z(R) I ⊆Z(R I).
2) I ZR(I)⊆Z(R I ).
Proof. 1) Let a∈Z(R)and e∈I. Then two cases arise.
Case 1:annR(a)∩annR(e)6= (0). Then let b∈annR(a)∩annR(e)such that b6= 0. Hence
(b, b)(a, a +e) = (0,0), so that (a, a +e)∈Z(R I).
Case 2: annR(a)∩annR(e) = (0). Let b∈annR(a)such that b6= 0. Then b /∈annR(e)
and thus be ∈I\ {0}. Hence (be, 0)(a, a +e) = (0,0) and (be, 0) 6= (0,0). Therefore
(a, a +e)∈Z(R I).
It follows from the above cases that Z(R) I ⊆Z(R I).
2) Let a∈ZR(I)and i∈I. Then there exists e∈I\ {0}such that ae = 0. Hence
(a+i, a)(0, e) = (0,0) and thus (a+i, a)∈Z(R I). It follows that I ZR(I)⊆Z(R I),
as desired.
Lemma 3.3. Let Rbe a ring and Ian ideal of R. Let abe a regular element of Rand e∈I.
Then
(a, a +e)∈Z(R I)⇔a+e∈ZR(I).
Proof. ⇐)Assume that a+e∈ZR(I). Then there exists i∈I\ {0}such that (a+e)i= 0.
Hence (a, a +e)(0, i) = (0,0) and (0, i)6= (0,0). It follows that (a, a +e)∈Z(R I).
⇒)Suppose that (a, a +e)∈Z(R I). Then there exists (b, b +j)∈R I \ {(0,0)}such
that (b, b +j)(a, a +e) = (0,0). Hence ba = 0 and (b+j)(a+e) = 0. Since a /∈Z(R), we get
b= 0. Therefore j(a+e) = 0. It follows that a+e∈ZR(I), as desired.
We recall that S=R\Z(R)stands for the set of all regular elements of R.
On Property (A) of the amalgamated duplication of a ring along an ideal 9
Corollary 3.4. Let Rbe a ring and Ian ideal of R. Then
Z(R I) = (Z(R) I )G(S I)\(R×ZR(I)) .
Proof. It follows from Lemma 3.2 and Lemma 3.3.
Lemma 3.5. Let Rbe a ring and Ian ideal of R. If Ris an SA-ring and I⊆Z(R), then
Z(R)\(S+I) = ∅.
Proof. Assume that Ris an SA-ring and that I⊆Z(R). Then, in particular, Z(R)is an
ideal of R. Suppose that (S+I)TZ(R)6=∅. Then there exists b∈Sand i∈Isuch that
b+i∈Z(R). Since i∈I⊆Z(R),b+i∈Z(R)and Z(R)is an ideal of R, we get b∈Z(R)
which is absurd. Hence (S+I)TZ(R) = ∅, as desired.
Proof of Theorem 3.1. 1) ⇒2) Let J= (a1, a2,· · · , an)be a finitely generated ideal of R
such that ai∈Z(R)for each i∈ {1, ..., n}. By Lemma 3.2, we have (ak, ak)∈Z(R
I)for each k∈ {1,2,· · · , n}. Now, since Z(R I )is an ideal of R I, we get K=
( (a1, a1), ..., (an, an) )R I ⊆Z(R I). Therefore, as R I is an S A-ring, there exists
(a, a +e)∈(R I)\ {(0,0)}such that (a, a +e)K= (0,0). If a6= 0, then aak= 0 for each
k∈ {1,· · · , n}, so that, aJ = 0 and thus ann(J)6= (0). If a= 0, then e6= 0 and eJ = 0,
that is, ann(J)6= (0). It follows that Ris an SA-ring. On the other hand, we claim that
(S I)\(R×ZR(I)) = ∅.
In fact, assume that (S I)∩(R×ZR(I)) 6=∅. Then there exists b∈Sand e∈Isuch
that b+e∈ZR(I). Note that, by Corollary 3.4, (b, b +e)∈Z(R I)and (0, e)∈Z(R I ).
Since Z(R I)is an ideal, we get (b, b)∈Z(R I)which is absurd as b /∈Z(R). Hence
(S I)∩(R×ZR(I)) = ∅proving our claim. Now, assume that I*Z(R). Then there
exists j∈Isuch that j /∈Z(R). Hence (j, 0) ∈(S I)∩(R×ZR(I)) which leads to a
contradiction in view of the above claim. It follows that I⊆Z(R).
2) ⇒1) Suppose that Ris an SA-ring and I⊆Z(R). Then, by Lemma 3.5, we have
(S+I)∩Z(R) = ∅. Therefore (S I)∩(R×ZR(I)) = ∅. It follows, by corollary 3.4,
that Z(R I) = Z(R) I . Now, let (a1, a1+e1),· · · ,(an, an+en)∈Z(R I ). Then
a1, a2,· · · , an∈Z(R). Therefore, as Ris an SA-ring, there exists a∈R\ {0}such that
aak= 0 for each k∈ {1,· · · , n}. If aek= 0 for each k∈ {1,· · · , n}, then, for each
k∈ {1,· · · , n},
(a, a)(ak, ak+ek) = (0,0) and (a, a)∈R I \ {(0,0)}.
Assume that there exits t∈ {1,2,· · · , n}such that aet6= 0. Hence (aet,0)(ak, ak+ek) = (0,0)
for each k∈ {1,2,· · · , n}and (aet,0) ∈R I \ {(0,0)}. It follows from both cases
that annR./I ((a1, a1+e1), ..., (an, an+en)) 6={(0,0)}. Consequently, R I is an SA-ring
completing the proof of the theorem .
On Property (A) of the amalgamated duplication of a ring along an ideal 10
4 Amalgamated duplication and A-property
The aim of this section is to totally characterize when an amalgmated duplication R I of
a ring Ralong an ideal Iis an A-ring. This characterization involves the notion of an ideal
which is an A-module along itself defined in Section 2.
Let Rbe a ring and Man R-module. We recall that Spec(Z(R)) (resp., Max(Z(R))) de-
notes the set of prime ideals (resp., maximal ideals) of Rcontained in Z(R)and Spec(ZR(M))
(resp., Max(ZR(M))) denotes the set of prime ideals (resp., maximal ideals) of Rcontained
in ZR(M). Also, given an ideal Iof R, we denote by ZI(R)the subset of Rdefined by
ZI(R) := {a∈R:a+I⊆Z(R)}.
Note that ZI(R)⊆Z(R)and put SI=R\ZI(R). If I⊆Z(R), then we denote by MaxI(Z(R))
the set of the maximal primes of Rcontaining I, that is,
MaxI(Z(R)) = {m∈Max(Z(R)) : I⊆m}.
The next theorems determine the set ZI(R)when Rhas only finite number of maximal
primes.
Theorem 4.1. Let Rbe a ring and Ian ideal of R. Then
1) S
m∈MaxI(Z(R))
m⊆ZI(R).
2) If Radmits a finite number of maximal primes, then
ZI(R) = [
m∈MaxI(Z(R))
m.
Proof. 1) It is clear.
2) Assume that Max(Z(R)) = {P1, P2,· · · , Pn}is a finite set. Then Z(R) = P1∪P2∪ · · ·∪ Pn.
Let MaxI(Z(R)) = {P1, P2,· · · , Pt}. Then I⊆P1∩P2∩ · · · ∩ Ptand I*Pt+1 ∪ · · · ∪ Pn.
Suppose that ZI(R)*P1∪ · · ·∪ Ptand let x∈ZI(R)\P1∪· · · ∪ Pt. Let i∈I\Pt+1 ∪ · · ·∪ Pn.
Without loss of generality, let x∈Pt+1 ∩ · · · ∩ Pr\Pr+1 ∪ · · · ∪ Pnfor some r∈ {t+ 1,· · · , n}.
Note that, if Pr+1 ∩ · · · ∩ Pn= (0), then, in particular, Pr+1 ∩ · · · ∩ Pn⊆P1and thus there
exists k∈ {r+ 1,· · · , n}such that Pk⊆P1so that Pk=P1which is absurd as I⊆P1while
I*Pk. Hence Pr+1 ∩ · · · ∩ Pn6= (0). Let y∈Pr+1 ∩ · · · ∩ Pn\Pt+1 ∪ · · · ∪ Pr. Consider
z:= x+yi. Then z∈Z(R)as x∈ZI(R). If z∈P1∪P2∪ · · · ∪ Pt, then x∈P1∪ · · · ∪ Ptas
I⊆P1∩ · · · ∩ Pt. This leads to a contradiction since x6∈ P1∪ · · · ∪ Pt. If z∈Pt+1 ∪ · · · ∪ Pr,
then yi ∈Pt+1 ∪ · · · ∪ Pras x∈Pt+1 ∩ · · · ∩ Prwhich is absurd since i, y 6∈ Pt+1 ∪ · ·· ∪ Pr.
If z∈Pr+1 ∪ · ·· ∪ Pn, then x∈Pr+1 ∪ · · · ∪ Pnsince y∈Pr+1 ∩ · · · ∩ Pn. This is absurd as
x6∈ Pr+1 ∪ · · · ∪ Pn. It follows that ZI(R)⊆P1∪ · · · ∪ Ptand thus the desired equality holds
comlpleting the proof.
Recall that if a ring Ris Noetherian, then it admits a finite number of maximal primes
[18, Theorem 80]. Thus the following corollary is a direct consequence of Theorem 4.1.
Corollary 4.2. Let Rbe a Noetherian ring and Ian ideal of R. Let MaxI(Z(R)) =
{m1, m2,· · · , mr}denote the set of maximal primes of Rcontaining I. Then
ZI(R) = m1∪m2∪ · · · ∪ mr.
On Property (A) of the amalgamated duplication of a ring along an ideal 11
The following theorem proves that the sets ZI(R)and ZR(I)shares the set of prime ideals
of a zero-dimensional ring R.
Theorem 4.3. Let Rbe a commutative ring such that dim(R)=0and let Ibe an ideal of
R. Let pbe a prime ideal of R. Then either p⊆ZI(R)or p⊆ZR(I). More precisely, if
I⊆p, then p⊆ZI(R)and if I*p, then p⊆ZR(I).
Proof. Let pbe a prime ideal of R. First, note that p⊆Z(R). If I⊆p, then p+I=p⊆Z(R)
and thus p⊆ZI(R). Now, assume that I*pand let i∈I\p. Let y∈p. Then, as pis a
minimal prime ideal of R, there exists s∈R\psuch that syn= 0 for some integer n≥1. Note
that as s6∈ pand i6∈ p, then, in particular, si 6= 0. Let r= max{1≤j≤n−1:(si)yj6= 0}.
Then (si)yr6= 0 and (si)yr+1 = 0. Let a:= (si)yr. Then a∈I\ {0}and ay = 0, that is,
y∈ZR(I). It follows that p⊆ZR(I)completing the proof.
Next, we announce the main theorem of this section. It totally characterizes when the
amalgamated duplication of a ring Ralong an ideal Isatisfies the Property (A).
Theorem 4.4. Let Rbe a ring and Ian ideal of R. Then R I is an A-ring if and only
if the following assertions hold:
1) Ris an A-ring.
2) Iis an A-module along itself.
3) If p∈Spec(Z(R)) such that p⊆ZR(I)∪ZI(R), then either p⊆ZR(I)or p⊆ZI(R).
In the next remark, we record various cases of rings Rwithin which the condition (3) of
Theorem 4.4 holds such as the case where Ris Noetherian or zero-dimensional.
Remark 4.5. Let Rbe a ring and Ian ideal of R. Then
1) If I⊆J(R)and R=Q(R), then, as S
m∈MaxI(Z(R))
m=S
m∈Max(Z(R))
m⊆ZI(R), we get
ZI(R) = Z(R)and the condition (3) of Theorem 4.4 holds.
2) If either ZI(R) = Z(R)or ZR(I) = Z(R), then it is easy to see that the condition (3) of
Theorem 4.4 holds.
3) If Ris Noetherian, then ZR(I)is a finite union of prime ideals and, by Corollary 4.2, ZI(R)
is a finite union of the maximal primes of R, then the condition (3) of Theorem 4.4 holds.
4) If Ris a zero-dimensional ring, then, by Theorem 4.3, the condtion (3) of Theorem 4.4
holds.
The proof of Theorem 4.4 requires the following lemmas. The first one describes the
prime spectrum of R I in terms of the prime spectrum of R.
Lemma 4.6. [10, Proposition 2.5] Let Rbe a ring and Ian ideal of R. Let pbe a prime
ideal of R. Then
1) If I⊆p, then P:= p I is the unique prime ideal of R I such that P∩R=p.
2) If I*p, then p I and I p are distinct and are the only prime ideals of R I
contracting to pover R.
The following lemma characterizes the prime ideals of R I contained in Z(R I).
On Property (A) of the amalgamated duplication of a ring along an ideal 12
Lemma 4.7. Let Rbe a ring and Ian ideal of R. Let p∈Spec(R). Then
1) p I ⊆Z(R I )if and only if p⊆Z(R).
2) I p ⊆Z(R I )if and only if p⊆ZR(I)∪ZI(R).
Proof. 1) Assume that p I ⊆Z(R I ). Suppose, by way of contradiction, that there exists
a∈psuch that a6∈ Z(R). Then, by Lemma 3.3, a+I⊆ZR(I), that is, a∈ZI(R)⊆Z(R)
which is absurd. Hence p⊆Z(R). The converse is clear applying Lemma 3.2.
2) Assume that I p ⊆Z(R I). Let a∈p. Then (a+i, a)∈Z(R I)for each i∈I.
Assume that a6∈ ZR(I). Then, by Lemma 3.3, a+I⊆Z(R)and thus a∈ZI(R). It follows
that a∈ZR(I)∪ZI(R)for each a∈pand thus p⊆ZR(I)∪ZI(R). Conversely, assume
that p⊆ZR(I)∪ZI(R). Let (a+i, a)∈I p with a∈pand i∈I. If a∈ZR(I), then,
by Lemma 3.2, (a+i, a)∈Z(R I). Suppose that a6∈ ZR(I). Then, by Corollary 3.4,
a∈ZI(R), that is, a+I⊆Z(R). In particular, we get a+i∈Z(R). Hence, by Lemma 3.2,
(a+i, a)∈Z(R I). It follows that I p ⊆Z(R I)establishing the desired equivalence
and completing the proof.
Lemma 4.8. Let Rbe a ring and Ian ideal. Then the following assertion are equivalent:
1) If pis a prime ideal of Rsuch that p⊆ZR(I)∪ZI(R), then either p⊆ZR(I)or
p⊆ZI(R).
2) For any ideal Hof R I contained in Z(R I ), either H⊆Z(R) I or H⊆I
ZR(I).
Proof. 1) ⇒2) Let Hbe an ideal of R I such that H⊆Z(R I). Then there exists a
prime ideal Pof R I such that H⊆Pand P⊆Z(R I ). Then, by Lemma 4.6, either
P=p I or P=I p for p=P∩R. Assume that P=p I. Then, by Lemma 4.7,
p⊆Z(R)and thus P⊆Z(R) I. Hence H⊆Z(R) I. Now, suppose that P=I p.
Then, by Lemma 4.7, p⊆ZR(I)∪ZI(R). Hence, using (1), we get either p⊆ZR(I)or
p⊆ZI(R). If p⊆ZR(I), then P=I p ⊆I ZR(I)and thus H⊆I ZR(I). If
p⊆ZI(R), then
P={(a+i, a) : a∈pand i∈I} ⊆ Z(R) I
so that H⊆Z(R) I. This proves (2).
2) ⇒1) Let pbe a prime ideal of Rsuch that p⊆ZR(I)∪ZI(R). Then, by Lemma 4.7,
I p ⊆Z(R I ). Hence, using (2), we get either I p ⊆Z(R) I or I p ⊆I ZR(I).
If I p ⊆Z(R) I , then p+I⊆Z(R), that is, p⊆ZI(R). If I p ⊆I ZR(I), then
p⊆ZR(I). It follows that either p⊆ZR(I)or p⊆ZI(R)proving (1) and completing the
proof of the lemma.
Proof of Theorem 3.1. Assume that R I is an A-ring. Next, we prove that the three
conditions 1), 2) and 3) hold:
1) Let K= (a1, a2,· · · , an)be a finitely generated ideal of Rsuch that K⊆Z(R). Let
z∈((a1, a1),· · · ,(an, an))R I. Then
z=
n
X
t=1
(bt, bt+it)(at, at)=(b1a1+· · · +bnan, b1a1+· · · +bnan+i),
On Property (A) of the amalgamated duplication of a ring along an ideal 13
with i=
n
X
t=1
itat∈I. As
n
X
t=1
btat∈K⊆Z(R), we get, by Lemma 3.2, that z∈Z(R I).
Hence ((a1, a1),· · · ,(an, an))R I ⊆Z(R I). Since R I is an A-ring, there exists
(a, a +j)∈R I \ {(0,0)}such that (a, a +j)(ar, ar) = (0,0) for each r∈ {1,2,· · · , n}.
Then aar= 0 and (a+j)ar= 0 for each r∈ {1,2,· · · , n}. Thus aar=jar= 0 for each
r∈ {1,2,· · · , n}. If a6= 0, then a∈annR(K). If not, that is, if a= 0, then j∈annR(K)
and j∈I\ {0}. Therefore annR(K)6= (0). It follows that Ris an A-ring.
2) Let Jbe a finitely generated ideal of Rsuch that J⊆ZR(I)and J∩(S+I)6=∅. Put
J= (b1,· · · , bn)and let b∈Sand j∈Isuch that b+j∈J. Consider the finitely generated
ideal H= ((b1, b1),· · · ,(bn, bn),(b, b +j)) of R I. Let t∈H. Then
t=
n
X
r=1
(αr, αr+ir)(br, br)+(α, α +i)(b, b +j)
=n
X
r=1
αrbr+αb,
n
X
r=1
(αr+ir)br+ (α+i)(b+j)
for (αr, αr+ir),(α, α +i)∈R I for each r∈ {1,2,· · · , n}. Note that
n
X
r=1
(αr+ir)br+ (α+i)(b+j)∈J.
Since J⊆ZR(I), we obtain
n
X
r=1
(αr+ir)br+ (α+i)(b+j)∈ZR(I). Hence, there exists
ω∈I\ {0}such that
n
X
r=1
(αr+ir)br+ (α+i)(b+j)ω= 0.
Therefore t(0, ω)=0and (0, ω)6= (0,0), and thus t∈Z(R I). It follows that H⊆Z(R
I)and thus, since R I is an A-ring, there exists (a, a +e)∈R I \ {(0,0)}such that
H(a, a +e) = (0,0). In particular, (b, b +j)(a, a +e) = (0,0), so that ba = 0 = (b+j)(a+e).
As b∈S:= R\Z(R), we get a= 0. It follows that e∈I\ {0}and Je = (0) since
H(0, e) = (0). Consequently, annI(J)6= (0) and thus Jis an A-module along itself.
3) Let Hbe an ideal R I such that H⊆Z(R I). Assume that H*Z(R) I
and H*I ZR(I). Then there exists (a, a +i),(b, b +j)∈Hsuch that a /∈Z(R)and
b+j /∈ZR(I)with i, j ∈I. Let K= ((a, a +i),(b, b +j))R I ⊆H⊆Z(R I). As R I
is an A-ring, there exists (c, c +k)∈R I \ {(0,0)}with k∈Isuch that
(c, c +k)(a, a +i) = (0,0) and (c, c +k)(b, b +j) = (0,0).
We obtain ac = 0 and thus c= 0 as a /∈Z(R). Therefore k(b+j) = 0 with k∈I. It follows
that k= 0 as b+j /∈ZR(I)yielding (c, c +k) = (0,0) which is absurd. Consequently, either
H⊆Z(R) I or H⊆I ZR(I). This proves (3) by virtue of Lemma 4.8.
Conversely, assume that the assertions (1), (2) and (3) are satisfied. Let H= ((a1, a1+
i1),· · · ,(an, an+in))R I be a finitely generated ideal of R I such that H⊆Z(R I).
Let K= (a1,· · · , an)Rand J= (a1+i1,· · · , an+in)R. Using (3) and Lemma 4.8, we get
On Property (A) of the amalgamated duplication of a ring along an ideal 14
either H⊆Z(R) I or H⊆I ZR(I), that is, K⊆Z(R)or J⊆ZR(I). The following
two cases arise.
Case 1. K⊆Z(R). Then, as Ris an A-ring, there exists r∈R\ {0}such that rK = 0,
that is, rai= 0,∀i∈ {1,· · · , n}. If rit= 0, for each t∈ {1,· · · , n}, then (r, r)H= (0,0)
so that annR ./ I(H)6= (0,0). Assume that there exists t∈ {1,· · · , n}such that rit6= 0.
Then (rit,0)(ak, ak+ik) = (0,0) for each k∈ {1,2,· · · , n}and (rit,0) 6= (0,0). Therefore
annR ./I (H)6= (0,0).
Case 2. K*Z(R). Then J⊆ZR(I)and there exits b=α1a1+· · · +αnan∈K\Z(R)
with the αj∈R. Let i=α1i1+α2i2+· · · +αnin∈I. Then b+i=α1(a1+i1) + α2(a2+
i2) + · · · +αn(an+in)∈J∩(S+I)so that J∩(S+I)6=∅. It follows, using (2), that
annI(J)6= (0). Therefore there exits k∈I\ {0}such that kJ = 0. Hence (0, k)∈annR./I(H)
and (0, k)6= (0,0). Then annR./I (H)6= (0,0).
It follows from the above two cases that R I is an A-ring completing the proof of the
theorem.
Next, we list consequences of Theorem 4.4.
Corollary 4.9. Let Rbe a Noetherian ring or a zero-dimensional ring. Let Ibe an ideal of
R. Then R I is an A-ring if and only if Ris an A-ring.
Proof. First, note that Ris an A-ring. By Proposition 2.2, Iis an A-module along itself.
Also, by Remark 4.5, the condition (3) of Theorem 4.4 holds. Then the result easily follows
from Theorem 4.4.
We recall that Rad(R)stands for the nilradical of R.
Corollary 4.10. Let Rbe a ring and Ian ideal of Rsuch that I⊆Rad(R). Then the
following assertions are equivalent:
1) R I is an A-ring;
2) Ris an A-ring.
The proof of Corollary 4.10 follows from Theorem 4.4 and the next lemma. Note that by
Remark 4.5(2), if ZI(R) = Z(R), then the condition (3) of Theorem 4.4 holds.
Lemma 4.11. Let Rbe a ring and Ian ideal of R. Assume that I⊆Rad(R). Then
1) Iis an A-module along itself.
2) Z(R) = ZI(R).
Proof. 1) It is direct by Proposition 2.4.
2) Note that ZI(R)⊆Z(R)always holds. Let a∈Z(R)and b6= 0 be an element of Rsuch
that ab = 0. Let i∈I\ {0}. Then there exists n≥1such that in= 0. Consider the positive
integer r:= max{m∈N:bim6= 0}. Note that 0≤r≤n−1,bir6= 0 and bir+1 = 0. Then
it is easily verified that (a+i)bir= 0 and thus (a+i)∈Z(R). It follows that a+I⊆Z(R),
that is, a∈ZI(R). Therefore Z(R) = ZI(R), as desired.
On Property (A) of the amalgamated duplication of a ring along an ideal 15
Corollary 4.12. Let Rbe a ring and Ian ideal of Rsuch that In= (0) for some integer
n≥1. Then the following assertions are equivalent:
1) R I is an A-ring;
2) Ris an A-ring.
Corollary 4.13. Let Rbe a ring and Ian ideal of Rsuch that ZR(I) = Z(R). Then R I
is an A-ring if and only if Ris an A-ring.
Proof. It follows from the combination of Theorem 4.4, Proposition 2.8 and Remark 4.5(2).
Corollary 4.14. Let Rbe a ring and Ian ideal of R. Assume that I*Z(R). Then the
following assertions are equivalent:
1) R I is an A-ring;
2) Ris an A-ring.
3) Iis an A-module;
4) Iis an A-module along itself.
Proof. It follows from the combination of Theorem 4.4, Theorem 2.14, Lemma 2.15 and
Corollary 4.13.
Corollary 4.15. Let Rbe a ring and Ian ideal of R. Assume that Z(R)is an ideal of R.
The R I is an A-ring if and only if Ris an A-ring and Iis an A-module along itself.
Moreover, if I⊆J(ZR(I)), then R I is an A-ring if and only if Ris an A-ring.
Proof. Note that, as Z(R)is an ideal, ZI(R) = Z(R). Then, by Remark 4.5(2), the condition
(3) of Theorem 4.4 holds and thus Theorem 4.4 establishes the first equivalence. Now, if
I⊆J(ZR(I)), then, by Proposition 2.3, Iis an A-module along itself, and thus the first step
allows to deduce the second equivalence completing the proof.
Acknowledgement. The authors would thank the referee for his careful reading of the
paper and for many comments and corrections.
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