Recall that a ring R is said to be indecomposable, if R cannot be written as R = R1 × R2, where R1 and R2 are both nonzero rings. It is well known, and not difficult to prove, that a ring R is indecomposable if and only if 1 is the only nonzero idempotent of R.
Lemma 2.1
The following statements are equivalent for an ideal I in R.
The ideal I can be written as I = J1 ∩ J2 ∩ … ∩ Jn, where J1, J2,…,Jnare ideals in R such that each of the R/Jiis indecomposable.
The ideal I can be written as I = K1K2…Km = K1∩K2∩…∩Km, where K1, K2,…,Kmare pairwise comaximal ideals in R such that each of the R/Kiis indecomposable.
R/I has only finitely many idempotents.
Proof(1) ⇒ (2) Let I can be written as I = J1∩J2∩…∩Jn, where J1, J2,…,Jn are ideals in R such that each of the R/Ji is indecomposable. Let two of the Ji’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/Jn are indecomposable. Repeating this argument, we see that the ideal I can be written as I = K1 ∩K2 ∩ … ∩Km, where K1, K2,…,Km are pairwise comaximal ideals of R such that each of the R/Ki is indecomposable. Now since K1,K2,…,Km are pairwise comaximal, we have K1K2…Km = K1 ∩ K2 ∩ … ∩ Km. (2) ⇒ (3) Let the ideal I can be written as I = K1K2…Km = K1 ∩ K2 ∩ … ∩ Km, where K1, K2,…,Km are pairwise comaximal ideals of R such that each of the R/Ki is indecomposable. By the Chinese Remainder Theorem, R/I≅⊕i=1mR/Ki. Now since each of the R/Ki 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).
Theorem 2.2
Let R be a ring such that Max(R) is a Noetherian topological space as a subspace of Spec(R) with respect to the Zariski topology. Then every proper ideal I of R can be written as I = J1 ∩ J2 ∩ … ∩ Jn, where J1, J2,…,Jnare ideals of R such that each of the R/Jiis indecomposable.
ProofLet 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∩…∩Jn, where J1, J2,…,Jn are ideals of R such that each of the R/Ji is indecomposable. By Lemma 2.1, R has infinitely many distinct idempotents, say α1, α2,…,αn,…, and so f1 = α1, f2 = α1α2,…, fn = α1α2…αn,…, are infinitely many distinct idempotents in R with fifi+1 = fi+1. Set ei = fi − fi+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 − ei ≠ 1, there exists ∈ Max(R) such that 1 − ei ∈ .
Set Jn = ∩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 Jk = Jk+1. Therefore, Jk+1 = ∩i∈ℕ{1,2,…,k+1} ⊆ Jk = ∩i∈ℕ{1,2,…,k} ⊆ . Now for all i ≠ k + 1, ek+1 = ek+1 − 0 = (1 − ei)ek+1 ∈ . Thus, ek+1 ∈ Jk+1 ⊆ . Therefore, ek+1 ∈ and 1 − ek+1 ∈ , a contradiction.
We will need the following well known fact about indecomposable rings which is a consequence of [1, Proposition 27.1].
Lemma 2.3
Let I be a proper ideal of R. Then R/I is indecomposable if and only if R/I is indecomposable.
An ideal I of a ring R is called semi-primary if I is a prime ideal. In [3] and [2], Gilmer considered rings whose semi-primary ideals are primary. Before proceeding, we state some useful results.
Theorem 2.4. ([2, Theorem 2])
Let R be a ring whose semi-primary ideals are primary. If Q is a P-primary ideal of R where P is a nonmaximal prime ideal of R, then Q = P.
By Lemma 2.3, if I is a semi-primary ideal, then R/I is indecomposable. The next result is a consequence of Theorem 2.4.
Corollary 2.5
Let R be a ring with the property that if R/I is indecomposable for an ideal I of R, then I is primary. Then if Q is a P-primary ideal of R where P is a nonmaximal prime ideal of R, then Q = P.
Lemma 2.6
Let R be a one-dimensional ring such that for any chain P ⊂ of prime ideals of R and p ∈ P, there exists m ∈ such that p = pm. Then if P is a nonmaximal prime ideal of R and I is an ideal of R with I=P, then I = P.
Corollary 2.7. ([3, Corollay 2.2])
Let R be a ring whose semi-primary ideals are primary. Then R has dimension less than two.
Remark 2.8
It is easily seen that if an ideal I of a ring R can be generated by a set of idempotents, then every element of I is a multiple of an idempotent of I.
Lemma 2.9
Let I be a ideal of R such that R/I is indecomposable, and let I′ = 〈{e ∈ I | e2 = e}〉. Then R/I′is also indecomposable.
ProofLet 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.
Lemma 2.10
Let R be a zero-dimensional ring. If {Pα}α∈Λis a family of prime ideals of R such that R/∩α∈ΛPαis indecomposable, then | Λ |= 1.
ProofLet {Pα}α∈Λ be a family of prime ideals of R such that R/∩α∈ΛPα is indecomposable. Since ∩α∈ΛPα is a radical ideal, R/∩α∈ΛPα is a zero-dimensional reduced ring, and so R/∩α∈ΛPα is a V on Neumann regular ring. Suppose r + ∩α∈ΛPα is a nonzero element of R/∩α∈ΛPα, if r + ∩α∈ΛPα is nonunit, then there exists s+∩α∈ΛPα in R/∩α∈ΛPα such that r+∩α∈ΛPα = rsr+∩α∈ΛPα. It is easily seen that sr+∩α∈ΛPα is a nontrivial idempotent of R/∩α∈ΛPα. So R/∩α∈ΛPα is not indecomposable, a contradiction. Thus, every nonzero element of R/∩α∈ΛPα is unit, and so R/∩α∈ΛPα is a field. Thus, | Λ |= 1.
Proposition 2.11
The following statements are equivalent for a ring R.
For an ideal I of R if R/I is indecomposable, then I is primary.
R is a zero-dimensional ring or R is a one-dimensional ring such that every nonmaximal prime ideal of R can be generated by its idempotents.
Proof(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.
Case 1If ∩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.
Case 2If ∩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.
Theorem 2.12
The following statements are equivalent for a ring R.
ProofBy 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…Km = K1 ∩K2 ∩ … ∩Km, where K1, K2,…,Km are pairwise comaximal ideals in R such that each of the R/Ki is indecomposable. Proposition 2.11 implies that each of the Ki 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.
Theorem 2.13
Let D be an integral domain which is not a field, then the following statements are equivalent:
ProofBy 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 ∩ … ∩ Jn, where J1, J2,…,Jn are ideals of D such that each of the D/Ji is indecomposable. By Proposition 2.11, each of the Ji 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 ∩ … ∩ Jn, such that each of the Ji is a power of a maximal ideal, say Ji=miri for some ri ∈ ℕ. Hence, I=J1∩J2∩…∩Jn=m1r1∩m2r2∩…∩mnrn=m1r1m2r2…mnrn. Therefore, D is a Dedekind domain.
Theorem 2.14
Let D be an integral domain which is not a field, then the following statements are equivalent:
D is an almost Dedekind domain and each nonzero element of D is contained in only finitely many maximal ideals of D.
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.
ProofBy Theorem 2.13 and [4, Theorem 37.2].