(1) ⇒ (2) Let I can be written as I = J₁∩J₂∩…∩J_n, where J₁, J₂,…,J_n are ideals in R such that each of the R/J_i is indecomposable. Let two of the J_i’s, say J₁ and J₂, are contained in a maximal ideal of R. If e+(J₁∩J₂) is an idempotent in R/(J₁ ∩J₂), then e+J₁ and e+J₂ are idempotents in R/J₁ and R/J₂, respectively. Now since R/J₁ and R/J₂ are indecomposable and J₁ ⊆ and J₂ ⊆ , we have e + (J₁ ∩ J₂) = 0+(J₁ ∩ J₂) or e + (J₁ ∩ J₂) = 1+(J₁ ∩ J₂). Hence, R/(J₁ ∩ J₂) is also indecomposable. Set I1′=(J1∩J2). Hence, I=I1′∩I3∩…∩Jn such that R/I1′, R/J₃, …, R/J_n are indecomposable. Repeating this argument, we see that the ideal I can be written as I = K₁ ∩K₂ ∩ … ∩K_m, where K₁, K₂,…,K_m are pairwise comaximal ideals of R such that each of the R/K_i is indecomposable. Now since K₁,K₂,…,K_m are pairwise comaximal, we have K₁K₂…K_m = K₁ ∩ K₂ ∩ … ∩ K_m. (2) ⇒ (3) Let the ideal I can be written as I = K₁K₂…K_m = K₁ ∩ K₂ ∩ … ∩ K_m, where K₁, K₂,…,K_m are pairwise comaximal ideals of R such that each of the R/K_i is indecomposable. By the Chinese Remainder Theorem, R/I≅⊕i=1mR/Ki. Now since each of the R/K_i has no nontrivial idempotents, R/I has only finitely many idempotents.

(3) ⇒ (1) If R/I is indecomposable, there is nothing to prove. Suppose that R/I is not indecomposable. Then there exists a nontrivial idempotent r + I in R/I. Thus, {0 + I} = I/I = (〈I, r〉/I) ∩ (〈I, r − 1〉/I) = (〈I, r〉/I)(〈I, r − 1 〉/I). Hence, I = 〈I, r〉 ∩ 〈I, r − 1〉 and R/I ≅ R/〈I, r〉 ⊕ R/〈I, r − 1〉. Now if R/〈I, r〉 and R/〈I, r − 1〉 are indecomposable, the proof is complete. Otherwise, R/〈I, r〉 or R/〈I, r−1〉 can be written as a direct sum of nonzero rings as above. Since R/I has finitely many idempotents, this process terminates after finite steps. This completes the proof.

For a ring R, let Spec(R) and Max(R) denote the set of all prime ideals and all maximal ideals of R, respectively. The Zariski topology on Spec(R) is the topology obtained by taking the collection of sets of the form (I) = {P ∈ Spec(R) | I ⊈ P} (resp. (I) = {P ∈ Spec(R) | I ⊆ P}), for every ideal I of R, as the open (resp. closed) sets. When considering as a subspace of Spec(R), Max(R) is called Max−Spectrum of R. So, its closed and open subsets are D(I) = (I)∩Max(R) = { ∈ Max(R) | I ⊈ } and V(I) = (I) ∩ Max(R) = { ∈ Max(R) | I ⊆ }, respectively.

A topological space X is called Noetherian if every nonempty set of closed subsets of X, ordered by inclusion, has a minimal element. An ideal I of R is called a J-radical ideal if it is an intersection of maximal ideals. Clearly, J-radical ideals of R correspond to closed subsets of Max(R), and Max-Spectrum of R is Noetherian if and only if R satisfies the ascending chain condition for J-radical ideals (See [7] for more details).

Let I be a proper ideal of R. Since Max(R) is Noetherian, Max(R/I) is also Noetherian. Thus, it is sufficient to show that the result is true for the zero ideal. Suppose, on the contrary, that the zero ideal cannot be written as I = J₁∩J₂∩…∩J_n, where J₁, J₂,…,J_n are ideals of R such that each of the R/J_i is indecomposable. By Lemma 2.1, R has infinitely many distinct idempotents, say α₁, α₂,…,α_n,…, and so f₁ = α₁, f₂ = α₁α₂,…, f_n = α₁α₂…α_n,…, are infinitely many distinct idempotents in R with f_if_i+1 = f_i+1. Set e_i = f_i − f_i+1 for each i ∈ ℕ. It is easily seen that {ei}i=1∞ is an infinite set of nonzero orthogonal idempotents. For each i ∈ ℕ, as 1 − e_i ≠ 1, there exists ∈ Max(R) such that 1 − e_i ∈ .

Set J_n = ∩_{i∈ℕ{1,2,…,n}} for each n ∈ ℕ. Thus J₁ ⊆ J₂ ⊆ J₃ ⊆ …, is an ascending chain of J-radical ideals of R. By hypothesis, there exists k ∈ ℕ such that J_k = J_k+1. Therefore, J_k+1 = ∩_{i∈ℕ{1,2,…,k+1}} ⊆ J_k = ∩_{i∈ℕ{1,2,…,k}} ⊆ . Now for all i ≠ k + 1, e_k+1 = e_k+1 − 0 = (1 − e_i)e_k+1 ∈ . Thus, e_k+1 ∈ J_k+1 ⊆ . Therefore, e_k+1 ∈ and 1 − e_k+1 ∈ , a contradiction.

We will need the following well known fact about indecomposable rings which is a consequence of [1, Proposition 27.1].

Let P be a nonmaximal prime ideal of R, and let I be an ideal of R with I=P. Consider the ring R̄ = R/I, and write “bar” for the quotient map. Since I=P, P̄ is the unique nonmaximal prime ideal of R̄. Let p̄ ∈ P̄. If is a maximal ideal of R̄, then P̄ ⊂ . By hypothesis, there exists m̄ ∈ such that p¯=pm¯. Now let χ̄ be an element of R̄ such that χ̄ ∉ . Then (x¯-xm¯)p¯=0¯. Since x¯-xm¯∉m¯, the annihilator of p̄ must be R̄. Hence, p¯=pm¯=0¯. Therefore, I = P.

Let x²+I′ = x+I′ for some x ∈ R. Thus, x²−x ∈ I′ ⊆ I and so x²+I = x+I. Since R/I is indecomposable and x + I is an idempotent element in R/I, we have x ∈ I or x−1 ∈ I. Suppose that x ∈ I. Nowsince x²−x ∈ I′, by Remark 2.8, there exists e² = e ∈ I′ such that x² − x = re for some r ∈ R. Thus, x² − x = (x² − x)e. Hence, (1−e)x² = (1−e)x. Thus, ((1−e)x)² = (1−e)²x² = (1−e)x² = (1−e)x. This shows that (1 − e)x is an idempotent element in I, hence (1 − e)x ∈ I′. Now since e ∈ I′, we have x ∈ I′. A similar argument works when x − 1 ∈ I. Therefore, R/I′ has no nontrivial idempotents, and so R/I′ is indecomposable.

Let {P_α}_α∈Λ be a family of prime ideals of R such that R/∩_α∈ΛP_α is indecomposable. Since ∩_α∈ΛP_α is a radical ideal, R/∩_α∈ΛP_α is a zero-dimensional reduced ring, and so R/∩_α∈ΛP_α is a V on Neumann regular ring. Suppose r + ∩_α∈ΛP_α is a nonzero element of R/∩_α∈ΛP_α, if r + ∩_α∈ΛP_α is nonunit, then there exists s+∩_α∈ΛP_α in R/∩_α∈ΛP_α such that r+∩_α∈ΛP_α = rsr+∩_α∈ΛP_α. It is easily seen that sr+∩_α∈ΛP_α is a nontrivial idempotent of R/∩_α∈ΛP_α. So R/∩_α∈ΛP_α is not indecomposable, a contradiction. Thus, every nonzero element of R/∩_α∈ΛP_α is unit, and so R/∩_α∈ΛP_α is a field. Thus, | Λ |= 1.

(1) ⇒ (2) By Corollary 2.7 and the fact that R/I is indecomposable for every semi-primary ideal I, R has dimension less than two. Let R be a one-dimensional ring, and let P be a nonmaximal prime ideal of R. By Lemma 2.9, R/P′ is indecomposable, where P′ = 〈{e ∈ P | e² = e}〉. By hypothesis, P′ is primary. Since P is a minimal prime ideal over P′, P′ is P-primary ideal. Thus P′ = P by Corollary 2.5. Therefore, every nonmaximal prime ideal of R can be generated by its idempotents. (2) ⇒ (1) Let R be a zero-dimensional ring, and let R/I be indecomposable for an ideal I of R. By Lemma 2.3, R/I=R/∩I⊆P∈Spec(R)P is indecomposable. Thus, by Lemma 2.10, I=∩I⊆P∈Spec(R)P must be a maximal ideal of R. Hence, I is primary in this case.

Now let R be one-dimensional, and let R/I be indecomposable for an ideal I of R. By Lemma 2.3, R/I=R/∩I⊆P∈Spec(R)P is indecomposable. If R/∩ _{I⊆P∈Spec(R)} is a zero-dimension ring, as above, I=∩I⊆P∈Spec(R)P is a maximal ideal of R, and so I is primary.

Now let R/∩ _{I⊆P∈Spec(R)}P be a one-dimension ring. We now consider the cases ∩_{I⊆P∈Spec(R)}P is a prime ideal and ∩_{I⊆P∈Spec(R)}P is not a prime ideal.

If ∩_{I⊆P∈Spec(R)}P is a prime ideal of R, then ∩_{I⊆P∈Spec(R)}P is a nonmaximal prime ideal of R. If p ∈ ∩_{I⊆P∈Spec(R)}P, by hypothesis and Remark 2.8, there exists an idempotent e ∈ ∩_{I⊆P∈Spec(R)}P such that p = re for some r ∈ R. Hence, p = peⁿ for each n ∈ ℕ. Since I=∩I⊆P∈Spec(R)P, thus a power of e is in I. Hence, p ∈ I, and so ∩_{I⊆P∈Spec(R)}P = I. Therefore, I is primary.

If ∩_{I⊆P∈Spec(R)}P is not a prime ideal, there exists a non-maximal prime ideal P₁ of a ring R containing ∩_{I⊆P∈Spec(R)}P. By hypothesis, there exists a non-trivial idempotent e ∈ P₁∩_{I⊆P∈Spec(R)}P. Thus the ring R/∩_{I⊆P∈Spec(R)}P has a nontrivial idempotent, and so R/∩_{I⊆P∈Spec(R)}P is not indecomposable, a contradiction.

Now we can state the main results of this paper.

By Theorem 1.3 and the definition of ZPI-rings, it is sufficient to prove (2) ⇒ (1). By Theorem 2.2 and Lemma 2.1, every proper ideal I of R can be written as I = K₁K₂…K_m = K₁ ∩K₂ ∩ … ∩K_m, where K₁, K₂,…,K_m are pairwise comaximal ideals in R such that each of the R/K_i is indecomposable. Proposition 2.11 implies that each of the K_i is primary, and so every proper ideal of R can be written as a product of primary ideals. By hypothesis, each primary ideal of R is a power of its radical. Thus, every ideal of R can be written as a product of prime ideals of R. Therefore, R is a ZPI-ring.

By Theorem 1.1 and 1.2, it is sufficient to prove (2) ⇒ (1). Suppose that D is an almost Dedekind domain such that Max(D) is a Noetherian topological space as a subspace of Spec(D) with respect to the Zariski topology. Thus, by Theorem 2.2, every proper ideal I of D can be written as I = J₁ ∩ J₂ ∩ … ∩ J_n, where J₁, J₂,…,J_n are ideals of D such that each of the D/J_i is indecomposable. By Proposition 2.11, each of the J_i is primary, and so every proper ideal of D can be written as an intersection of primary ideals. Since D is an almost Dedekind domain, each nonzero primary ideal of D is a power of its radical. Thus, every nonzero proper ideal I of D can be written as I = J₁ ∩ J₂ ∩ … ∩ J_n, such that each of the J_i is a power of a maximal ideal, say Ji=miri for some r_i ∈ ℕ. Hence, I=J1∩J2∩…∩Jn=m1r1∩m2r2∩…∩mnrn=m1r1m2r2…mnrn. Therefore, D is a Dedekind domain.

By Theorem 2.13 and [4, Theorem 37.2].