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

We call the ring A⋈fJ 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 A⋈I of A×A consisting of all pairs (x,y)∈A×A with x−y∈I. 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, A⋈I 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 A⋈I 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:A→B 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 A⋈fJ is a ZPI ring if and only if A is a ZPI ring and f(A)+J is Noetherian.

In this paper we study the ZPI properties on amalgamated duplication of A along an ideal I, A⋈I. First let us recall the following notions. Let A be a commutative ring and I be an ideal of A. Let A⋈I be the subring of A×A consisting of the elements (a,a+i) for a∈A and i∈I. Then the ring A⋈I 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=ℤ[​[X]​]. We will show that A is not a ZPI ring. Indeed, let M=(X,2)ℤ[​[X]​] and I=(X2,2X,2)ℤ[​[X]​]. Since 2∈I\M2, then M2⊂I. Moreover, I⊂M, because X∈M\I. Thus M2⊂I⊂M, 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=P1⋯Pk where P_i is a prime ideal of A for each 1≤i≤k. Thus J=P1⋯Pk/I=P1/I⋯Pk/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=ℤ[​[X]​]. Then by Example 2.1, A is not a ZPI ring. Let I=Xℤ[​[X]​]. Since A/I≃ℤ, then A/I is a ZPI ring.

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

We call the ring A⋈fJ amalgamation of A with B along J with respect to f. Let p_A and p_B be the restrictions to A⋈fJ of A×B onto A and B, respectively. Let π:B→B/J be the canonical projection and f^=π°f. Then A⋈fJ is the pullback f^×B/Jπ of f^ and π:

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

Proof. By [2, Proposition 5.1] A⋈fJ(0,J)≃A and A⋈fJ(f−1(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 P′f={(p,f(p)+j),∣p∈P and j∈J} and Q¯f={(a,f(a)+j)∣a∈A,j∈J and f(a)+j∈Q}. According to [1, Proposition 2.6], the set of maximal ideals of A⋈fJ is Max(A⋈fJ)={P′ f∣P∈ Max(A)}∪{Q¯f∣Q∈ Max(B)∖V(J)}, where V(J)={Q∈ Spec(B)∣J⊆Q}. Note that when A=B, f=id_A and J=I, then we obtain A⋈fJ =A⋈I 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)∣q∈Q,i∈I and I⊆Q}, 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 a∈A,∈I and a+i∈Q} for some maximal ideal Q of A such that I⊈Q. Since a+i∈Q, there exists q∈ Q such that a=q-i. Thus M={(q−i,q)∣i∈I and q∈Q}. This implies that M={(q+i,q)∣q∈Q,i∈I and I⊆Q}.

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 M2⊆J⊆M. By Remark 2.5, M=N⋈ I for some maximal ideal N of A or M={(q+i,q)∣q∈Q,i∈I} for some maximal ideal Q of A such that I⊆Q.

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,...,βn∈I such that a=α1β1+⋯+αnβn. Thus

Now, since M2⊆J⊆M, then N2⊆PA(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=N⋈I. It suffices to show that N⋈I⊆J. Let (a,a+i)∈N⋈I. Then a∈PA(J); so there exists j ∈ J such that (a,a+j)∈J. We have (a,a+i)=(a,a+i+j−j)=(a,a+j)+(0,i−j). By claim above, (0,I)⊆J; so (a,a+i)∈J. Thus J=N⋈I.

• 2. PA(J)=N2. We will prove that J=M2=(N⋈I)2. It suffices to show that J⊆(N⋈I)2. Let (a,a+i)∈J. Then a∈N2; so a=α1β1+⋯+αnβn for some α1,...αn,β1,...,βn∈N. Thus

Since (0,I)⊆(N⋈I)2, then (a,a+i)∈(N⋈I)2. This implies that J⊆(N⋈I)2. Hence J=(N⋈I)2.

Second case: M={(q+i,q)∣q∈Q,i∈I} for some maximal ideal Q of A such that I⊆Q.

Claim (I,0)⊆M2⊆J.

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,βk∈I. As for all 1≤k≤n,(αk,0)∈M, then (a,0)∈M2. This implies that (I,0)⊆M⊆J.

We set the projection:

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 q∈Q=H(J), there exista∈A,i′∈I 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+i∈Q2 which implies that a+i=α1β1+⋯+αnβn for some α1,...αn,β1,...,βn∈Q. We have

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

  For all 1≤k≤n,(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. A⋈0={(a,a)∣a∈A} 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:A→B a ring homomorphism and J an ideal of B. If J is included in the radical of Jacobson of B, then Max(A⋈fJ)={P′ f∣P∈Max(A)}.

Proof. Since J⊆∩ Q∈Max(B)Q, then for all Q∈ Max(B), J⊆Q; so {Q¯f, Q∈ Max(B)∖V(J)}=∅.

Let A and B be commutative rings with identity, f:A→B a ring homomorphism and J an ideal of B. According to [6, Proposition 3.2], A⋈fJ 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:A→B 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. A⋈fJ 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 A⋈fJ is Noetherian. It suffices to prove that for all maximal ideal M of A⋈fJ there is no ideal properly contained between M and M². Assume that there exists an ideal I of A⋈fJ such that M2⊆I⊆M, for some maximal ideal M of A⋈fJ. Since J is included in the radical of Jacobson of B, then by Proposition 2.9, Max(A⋈fJ)={P′ f∣P∈Max(A)}; so M=P⋈fJ, 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,...,βn∈J such that a=α1β1+⋯+αnβn. Thus

Now, since M2⊆I⊆M, then P2⊆PA(I)⊆P. This implies that PA(I)=P2 or PA(I)=P, because A is a ZPI ring.

First case: P_A(I)=P. We will prove that I=M=P⋈fJ. It suffices to show that P⋈fJ⊆I. Let (a,f(a)+j)∈P⋈fJ. Then a∈PA(I); so there exists an i ∈ J such that (a,f(a)+i)∈I. We have (a,f(a)+j)=(a,f(a)+j+i−i)=(a,f(a)+i)+(0,j−i). Since (0,J)⊆I,(a,f(a)+j)∈I. Thus I=P⋈fJ.

Second case: PA(I)=P2. We will prove that I=M2=(P⋈fJ)2. It suffices to show that I⊆(P⋈fJ)2. Let (a,f(a)+j)∈I. Then a∈P2; so a=α1β1+⋯+αnβn for some α1,...αn,β1,...,βn∈P. Thus

Since (0,J)⊆(P⋈fJ)2, then (a,f(a)+j)∈(P⋈fJ)2. This implies that I⊆(P⋈fJ)2. Hence I=(P⋈fJ)2.

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

Example 2.11. Let A and B be commutative rings with identity, f:A→Ba 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.

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