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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2022; 62(2): 205-211

Published online June 30, 2022

### ZPI Property In Amalgamated Duplication Ring

Ahmed Hamed and Achraf Malek*

Department of Mathematics, Faculty of Sciences, Monastir, Tunisia
e-mail : hamed.ahmed@hotmail.fr and achraf_malek@yahoo.fr

Received: January 23, 2021; Revised: September 14, 2021; Accepted: December 6, 2021

### Abstract

Let A be a commutative ring. We say that A is a ZPI ring if every proper ideal of A is a finite product of prime ideals [5]. In this paper, we study when the amalgamated duplication of A along an ideal I, A⋈ I to be a ZPI ring. We show that if I is an idempotent ideal of A, then A is a ZPI ring if and only if A⋈ I is a ZPI ring.

Keywords: ZPI ring, Amalgamated duplication ring, Dedekind domain

### 1. Introduction

All rings considered in this paper are commutative and unitary. Let A and B be commutative rings with identity, f:AB a ring homomorphism and J an ideal of B. Then the subring AfJ of A×B is defined as follows:

AfJ={(a,f(a)+j)|aAandjJ}

We call the ring AfJ the amalgamation of A with B along J with respect to f. This construction was introduced and studied by D'Anna, Finacchiaro and Fontana [1, 2]. The study of the amalgamation ring widespread and can improve early studies on classical constructions like A+XB[X], A+XB[[X]] and D+M which are in fact, special cases of amalgamated algebra rings. However, we will be mostly interested in the amalgamated duplication ring which is a particular case of the amalgamated algebra ring. Let A be a commutative ring and I an ideal of A. The following ring construction called the amalgamated duplication of A along I was introduced by D'Anna in [3]. It is the subring AI of A×A consisting of all pairs (x,y)A×A with xyI. Motivations and additional applications of the amalgamated duplication are discussed in detail in [3, 4]. Recall that a commutative ring A is called a ZPI ring if every proper ideal of A is a finite product of prime ideals [5, 8, 9].

In this paper we study the ZPI property and we give a necessary and sufficient condition for the amalgamated duplication of A along an ideal I, AI to be a ZPI ring, where I is an idempotent ideal (i.e., I2=I). We show that if A is a ZPI ring, then A/I is a ZPI ring, and we prove that the reverse is not true in general. Let I be an idempotent ideal of A. We show that A is a ZPI ring if and only if AI is a ZPI ring. We end this paper by a sufficient condition for the amalgamated algebra along an ideal to be a ZPI ring. Let A and B be commutative rings with identity, f:AB a ring homomorphism and J an idempotent ideal of B. We show that if J is included in the radical of Jacobson of B, then AfJ is a ZPI ring if and only if A is a ZPI ring and f(A)+J is Noetherian.

### 2. Main Results

In this paper we study the ZPI properties on amalgamated duplication of A along an ideal I, AI. First let us recall the following notions. Let A be a commutative ring and I be an ideal of A. Let AI be the subring of A×A consisting of the elements (a,a+i) for aA and iI. Then the ring AI is called the amalgamated duplication of A along an ideal I. Recall that a commutative ring A is said to be ZPI; if every proper ideal of A is a finite product of prime ideals of A. It was shown in [7][Theorem 9.10] that A is a ZPI ring if and only if A is Noetherian and for all maximal ideal M of A, there is no ideal properly contained between M2 and M.

Example 2.1. Let A=[[X]]. We will show that A is not a ZPI ring. Indeed, let M=(X,2)[[X]] and I=(X2,2X,2)[[X]]. Since 2I\M2, then M2I. Moreover, IM, because XM\I. Thus M2IM, and hence A is not a ZPI ring.

Lemma 2.2. Let A be a ZPI ring and I an ideal of A. Then A/I is a ZPI ring.

Proof. Let J be an ideal of A/I. Then J=B/I is such that B is an ideal of A containing I. Since A is a ZPI ring, B=P1Pk where Pi is a prime ideal of A for each 1ik. Thus J=P1Pk/I=P1/IPk/I and therefore A/I is a ZPI ring.

The following example proves that the reverse of the previous lemma is not true in general.

Example 2.3. A=[[X]]. Then by Example 2.1, A is not a ZPI ring. Let I=X[[X]]. Since A/I, then A/I is a ZPI ring.

Let A and B be commutative rings with identity, f:AB a ring homomorphism and J an ideal of B. Then the subring AfJ of A×B is defined as follows:

AfJ={(a,f(a)+j)|aAandjJ}.

We call the ring AfJ amalgamation of A with B along J with respect to f. Let pA and pB be the restrictions to AfJ of A×B onto A and B, respectively. Let π:BB/J be the canonical projection and f^=π°f. Then AfJ is the pullback f^×B/Jπ of f^andπ:

Proposition 2.4. Let A and B be commutative rings with identity, f:AB a ring homomorphism and J an ideal of B. If AfJ is a ZPI ring, then A and f(A)+J are ZPI rings.

Proof. By [2, Proposition 5.1] AfJ(0,J)A and AfJ(f1(J),0)f(A)+J. Then by Lemma 2.2, A and f(A)+J are ZPI rings.

Let P be a prime ideal of A and Q be a prime ideal of B. We note Pf={(p,f(p)+j),pP and jJ} and Q¯f={(a,f(a)+j)aA,jJ and f(a)+jQ}. According to [1, Proposition 2.6], the set of maximal ideals of AfJ is Max(AfJ)={P fP Max(A)}{Q¯fQ Max(B)V(J)}, where V(J)={Q Spec(B)JQ}. Note that when A=B, f=idA and J=I, then we obtain AfJ =AI the amalgamated duplication of A along an ideal I.

Remark 2.5. Let I be an ideal of a commutative ring A. Then the maximal ideals of A ⋈ I are:

• 1. N ⋈ I, where N is a maximal ideal of A.

• 2. {(q+i,q)qQ,iI and IQ}, where Q is a maximal ideal of A.

Proof. Let M be a maximal ideal of A ⋈ I. Then M=N ⋈ I where N is a maximal ideal of A or M={(a,a+i), with aA,I and a+iQ} for some maximal ideal Q of A such that IQ. Since a+iQ, there exists q∈ Q such that a=q-i. Thus M={(qi,q)iI and qQ}. This implies that M={(q+i,q)qQ,iI and IQ}.

Recall that an ideal I of a commutative ring A is said to be idempotent if I2=I.

Theorem 2.6. Let I be an idempotent ideal of A. Then the following assertions are equivalent:

• 1. A is a ZPI ring.

• 2. A ⋈ I is a ZPIring.

Proof. (2) (1). Follows from Proposition 2.4.

(1) (2). Let M be a maximal ideal of A ⋈ I. Suppose that there exists an ideal J of A ⋈ I such that M2JM. By Remark 2.5, M=N⋈ I for some maximal ideal N of A or M={(q+i,q)qQ,iI} for some maximal ideal Q of A such that IQ.

First case: M=N ⋈ I, where N is a maximal ideal of A.

Claim (0,I)J.

Proof of claim. Let (0,a)(0,I). Since I is an idempotent ideal of A, there exist α1,...αn,β1,...,βnI such that a=α1β1++αnβn. Thus

(0,a)=(0,α1)(0,β1)++(0,αn)(0,βn)(NI)2J.

Now, since M2JM, then N2PA(J)N. This implies that PA(J)=N2 or PA(J)=N, because A is a ZPI ring.

• 1. PA(J)=N.We will prove that J=M=NI. It suffices to show that NIJ. Let (a,a+i)NI. Then aPA(J); so there exists j ∈ J such that (a,a+j)J. We have (a,a+i)=(a,a+i+jj)=(a,a+j)+(0,ij). By claim above, (0,I)J; so (a,a+i)J. Thus J=NI.

• 2. PA(J)=N2. We will prove that J=M2=(NI)2. It suffices to show that J(NI)2. Let (a,a+i)J. Then aN2; so a=α1β1++αnβn for some α1,...αn,β1,...,βnN. Thus

(a,a+i)=(α1,α1)(β1,β1)++(αn,αn)(βn,βn)+(0,i).

Since (0,I)(NI)2, then (a,a+i)(NI)2. This implies that J(NI)2. Hence J=(NI)2.

Second case: M={(q+i,q)qQ,iI} for some maximal ideal Q of A such that IQ.

Claim (I,0)M2J.

Proof of claim. Let (a,0)(I,0). Since I is an idempotent ideal, (a,0)=(α1,0)(β1,0)++(αn,0)(βn,0) for some αk,βkI. As for all 1kn,(αk,0)M, then (a,0)M2. This implies that (I,0)MJ.

We set the projection:

H:AIA(a,a+i)a+i.

Now, we have Q2=H(M2)H(J)H(M)=Q. Since A is a ZPI ring, then H(J)=Q or H(J)=Q2.

• 1. H(J)=Q. We will show that M=J. Let (q+i,q) be an element of M. Since qQ=H(J), there existaA,iI such that q=a+i' with (a,a+i)J. By the claim above (q+i,q)=(a+i+i,a+i)=(a,a+i)+(i+i,0)J. Hence M=J.

• 2. If H(J)=Q2. We show that J=M2. Let (a,a+i)J. Then a+iQ2 which implies that a+i=α1β1++αnβn for some α1,...αn,β1,...,βnQ. We have

(a,a+i)=(α1β1++αnβni,α1β1++αnβn)  =(α1β1,α1β1)++(αnβn,αnβn)+(i,0)  =(α1,α1)(β1,β1)++(αn,αn)(βn,βn)+(i,0).

For all 1kn,(αk,αk)(βk,βk)M2, then the claim above (I,0)M2. This implies that (a,a+i)M2, and M2=J.

Hence A ⋈ I is a ZPI ring.

Question 2.7. Is the property I idempotent in Theorem 2.6 is necessary?

Recall that an integral domain is said to be a Dedekind domain if every proper ideal of A is a finite product of prime ideals. Note that A ⋈ I is an integral domain if and only if A is an integral domain and I=(0).

Corollary 2.8. Let A be an integral domain. Then the following assertions are equivalent:

• 1. A is a Dedekind domain.

• 2. A0={(a,a)aA} is a Dedekind domain.

Proof. (1) (2) Let I=(0). Then I is an idempotent ideal of A. Since A is a ZPI ring, then by Theorem 2.6, A ⋈ I is a ZPI ring. Moreover, A is an integral domain and I=(0), then A ⋈ I is an integral domain. Hence A ⋈ I is a Dedekind domain.

(2) (1) Since A ⋈ 0 is a ZPI ring, then by Theorem 2.6, A is a ZPI ring. As A ⋈ 0 is an integral domain, then A is an integral domain. Hence A is a Dedekind domain.

Proposition 2.9. Let A and B be commutative rings with identity, f:AB a ring homomorphism and J an ideal of B. If J is included in the radical of Jacobson of B, then Max(AfJ)={P fPMax(A)}.

Proof. Since J QMax(B)Q, then for all Q∈ Max(B), JQ; so {Q¯f,Q Max(B)V(J)}=.

Let A and B be commutative rings with identity, f:AB a ring homomorphism and J an ideal of B. According to [6, Proposition 3.2], AfJ is a Noetherian ring if and only if A and f(A)+J are Noetherian.

Proposition 2.10. Let A and B be commutative rings with identity, f:AB a ring homomorphism and J an idempotent ideal of B. Assume that J is included in the radical of Jacobson of B. Then the following assertions are equivalent:

• 1. AfJ is a ZPI ring.

• 2. A is a ZPI ring and f(A)+J is a Noetherian ring.

Proof. (1) (2) Follows from Proposition 2.4.

(2) (1) Since A and f(A)+J are Noetherian, then AfJ is Noetherian. It suffices to prove that for all maximal ideal M of AfJ there is no ideal properly contained between M and M2. Assume that there exists an ideal I of AfJ such that M2IM, for some maximal ideal M of AfJ. Since J is included in the radical of Jacobson of B, then by Proposition 2.9, Max(AfJ)={P fPMax(A)}; so M=PfJ, for some maximal ideal P of A. We will show that (0,J)I. Let (0,a)(0,J). Since J is an idempotent ideal of B, there exist α1,...αn,β1,...,βnJ such that a=α1β1++αnβn. Thus

(0,a)=(0,α1)(0,β1)++(0,αn)(0,βn)(PfJ)2I.

Now, since M2IM, then P2PA(I)P. This implies that PA(I)=P2 or PA(I)=P, because A is a ZPI ring.

First case: PA(I)=P. We will prove that I=M=PfJ. It suffices to show that PfJI. Let (a,f(a)+j)PfJ. Then aPA(I); so there exists an i ∈ J such that (a,f(a)+i)I. We have (a,f(a)+j)=(a,f(a)+j+ii)=(a,f(a)+i)+(0,ji). Since (0,J)I,(a,f(a)+j)I. Thus I=PfJ.

Second case: PA(I)=P2. We will prove that I=M2=(PfJ)2. It suffices to show that I(PfJ)2. Let (a,f(a)+j)I. Then aP2; so a=α1β1++αnβn for some α1,...αn,β1,...,βnP. Thus

(a,f(a)+j)=(α1,f(α1))(β1,f(β1))++(αn,f(αn))(βn,f(βn))+(0,j).

Since (0,J)(PfJ)2, then (a,f(a)+j)(PfJ)2. This implies that I(PfJ)2. Hence I=(PfJ)2.

Let A be a commutative ring. We denote by Γ(A):={(a,f(a))aA}the Graph of A.

Example 2.11. Let A and B be commutative rings with identity, f:ABa ring homomorphism and J=(0). It is easy to see that J is an idempotent ideal of B included in the radical of Jacobson of B. By Proposition 2.10, Γ(A) is a ZPI ring if and only if A is a ZPI ring and f(A) is Noetherian.

### Acknowledgments.

The authors would like to thank the referee for his/her insightful suggestions towards the improvement of the paper.

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