Some remarks on multiplicatively closed sets

Rings considered in this article are commutative with identity. A subring of a ring is assumed to contain the identity element of the ring. Let S be a multiplicatively closed subset of a ring R satisfying the following property (P): whenever \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${ab \in S}$$\end{document} with at least one of a, b is in S, then both of them are in S. The (P)-closure of any multiplicatively closed subset of R is discussed; and saturated multiplicatively closed subsets of the form 1 + I, where I is an ideal of R are also considered. We investigate the question of determining rings admitting only a finite number of such multiplicatively closed sets. In addition, we discuss the problem of whether a ring R admitting only a finite number of saturated multiplicatively closed subsets of the form 1 + I, where I is an ideal of R, extends to the polynomial ring R[x] in one variable x over R.

Observe that P ∈ if and only if 1 / ∈ P + I if and only if P + I = R. It bas been pointed to me by the referee that if 1 = {M ∈ |M is a maximal ideal of R}, then S = R\(∪ M∈ 1 M). Since 1 ⊆ , it is clear that ∪ M∈ 1 M ⊆ ∪ P∈ P. Let P ∈ . Then P + I = R. Hence, there exists a maximal ideal M of R such that P + I ⊆ M. Now, M is a prime ideal of R, M + I = R and so M ∈ . Thus, M ∈ 1 and as P ⊆ M, it follows that ∪ P∈ P ⊆ ∪ M∈ 1 M. This proves that ∪ P∈ P = ∪ M∈ 1 M and so S = R\(∪ M∈ 1 M). However, if I is any ideal of R, then 1 + I has the following property: whenever ab ∈ 1 + I for some a, b ∈ R and if either a or b belongs to 1 + I , then both a and b belong to 1 + I (for a proof see Example 2.1(i)).
The observations mentioned in the previous paragraph motivate the following definition. Let R be a ring and let S be a multiplicatively closed subset of R. We say that S satisfies property (P) if the following holds: whenever ab ∈ S for some a, b ∈ R and if either a or b belongs to S, then both of them belong to S. In such a case, we simply say that S satisfies (P).
Let R be a ring. In Sect. 2 of this article, we provide some examples of multiplicatively closed subsets S of R that satisfy (P). It is noted in Remark 2.2(ii) that the multiplicatively closed subset Z (R) ∪ {1} satisfies (P) if and only if Z (R) ∪ {1} = R, where Z (R) denotes the set of all zero-divisors of R. In the rest of Sect. 2, we attempt to characterize rings R such that Z (R) ∪ {1} = R. Though I do not know any characterization of a ring R with the property Z (R) ∪ {1} = R, however, under some additional hypotheses on R, we characterize R such that Z (R) ∪ {1} = R (see Propositions 2.7 and 2.11). We provide in Proposition 2.12, a necessary and sufficient condition in order that a multiplicatively closed subset S of a ring R to satisfy (P).
Let S be a multiplicatively closed subset of a ring R. Note that S 1 = ∩ T ⊇S T , where T varies over all multiplicatively closed subsets of R, T ⊇ S, and T satisfies (P) is the smallest multiplicatively closed subset of R such that S 1 ⊇ S and S 1 satisfies (P). We call S 1 , the (P)-closure of S and we denote it by (P) − cl(S). In Proposition 3.2, we characterize (P) − cl(S) in terms of its elements. Let T be a ring and let R be a subring of T . Let S be a multiplicatively closed subset of R. We call the smallest multiplicatively closed subset of T containing S and which satisfies (P), the (P)-closure of S in T and denote it by (P) − cl T (S). Under the hypothesis that T is an integral domain, in the rest of Sect. 3, we determine (P) − cl T (S) for some multiplicatively closed subsets S of R.
Let I be an ideal of a ring R. In Sect. 4, we focus our study on ideals I such that the multiplicatively closed subset 1 + I of R is saturated. We prove in Proposition 4.1 that if I is an ideal of Z , then 1 + I is saturated if and only if either I = Z or I = 2Z. Motivated by the assertion of Proposition 4.1, we attempt in this section to provide classes of rings R such that R admits only finitely many saturated multiplicatively closed subsets of the form 1 + I , where I is an ideal of R (see Propositions 4.3(i) and 4.5). We also provide some examples to illustrate the results proved in Sect. 4.
For any ring R, let R[x] (respectively R [[x]]) denote the polynomial (respectively the power series) ring in one variable x over R. We denote by R, the collection of all rings R with the property that R admits only a finite number of saturated multiplicatively closed subsets of the form 1+I , where I is an ideal of R. The main problem discussed in Sect. 5 is to determine whether a ring R ∈ R implies that R[x] ∈ R(respectively R [[x]] ∈ R). We prove in Proposition 5.3 that R ∈ R if and only if R [[x]] ∈ R. I do not know whether R ∈ R implies that R[x] ∈ R. However, we prove under some additional hypotheses on R that if R ∈ R, then R[x] ∈ R.

Examples of multiplicatively closed subsets with property ( P)
In this section, we give some examples of multiplicatively closed subsets of a ring R that satisfy (P). We begin with the following example: Example 2.1 Let R be a ring.
(i) If I is any ideal of R, then the multiplicatively closed subset S = 1 + I of R satisfies (P).
(iv) Let R[x] denote the polynomial ring in one variable over R.
Then S is a multiplicatively closed subset of R[x], S satisfies (P) but S = 1 + I for any ideal I of R [x].
The assertions (ii), (iii), and (iv) can be easily verified and hence their verifications are omitted.
We mention in Example 2.3, a multiplicatively closed subset of a ring which does not satisfy the property (P) with the help of the following remark. Proof The proof of this remark is obvious and hence is omitted.

Remark 2.4
Let R be an integral domain which is not a field. Let I be a nonzero ideal of R with I = R. Let S = (I \{0}) ∪ {1}. Then S is a multiplicatively closed subset of R and S does not satisfy (P). In particular, if n ∈ Z with n ≥ 2, then S = (nZ\{0}) ∪ {1} is a multiplicatively closed subset of Z and S does not satisfy (P).
Proof The proof of this remark is quite obvious and hence is omitted.
It is noted in Remark 2.2(ii) that the multiplicatively closed subset S = Z (R) ∪ {1} of a ring R satisfies (P) if and only if Z (R) ∪ {1} = R. It is natural to consider the following question.
Q. What are the necessary and sufficient conditions on a ring R in order that R satisfies Z (R) ∪ {1} = R?
In the following remarks, we attempt to answer the above question. Note that Remark 2.5 Let R be a ring. If U (R) = {1}, then the following hold: Proof Note that −1 ∈ U (R) = {1} and so −1 = 1. Hence 1 + 1 = 0 and so r + r = 0 for all r ∈ R.
(ii) Let a be a nilpotent element of R. Then 1 + a ∈ U (R). Since U (R) = {1}, it follows that 1 + a = 1 and this implies that a = 0. This proves that R is reduced.
The following remark determines rings R with the following properties: Z (R) is an ideal of R and Z (R) ∪ {1} = R. Remark 2.6 Let R be a ring. Suppose that Z (R) is an ideal of R. Then Z (R) ∪ {1} = R if and only if R is isomorphic to Z/2Z as rings.
Conversely, if R is isomorphic to Z/2Z as rings, then it is clear that Let R be a ring. Note that S = R\Z (R) is a saturated multiplicatively closed subset of R. Hence, we obtain from [1, Problem 7(i), p. 44] that Z (R) = R\S is a union of prime ideals of R. Moreover, if the zero ideal of R admits a primary decomposition, then Z (R) is the union of a finite number of prime ideals of R [1,Proposition 4.7]. In Proposition 2.7, we determine rings R such that Z (R) ∪ {1} = R under the assumption that Z (R) is the union of a finite number of prime ideals of R.
Conversely, suppose that R ∼ = T = F 1 × · · · × F m as rings for some m ≥ 1, Proof Let {M α } α∈ denote the set of all maximal ideals of R. Since Z (R) ∪ {1} = R, it follows that U (R) = {1} and hence from Remark 2.5(iii), we obtain that ∩ α∈ M α = (0). Observe that the mapping φ : R → α∈ R/M α given by φ(r ) = {r + M α } α∈ is a homomorphism of rings and is injective. Let T = α∈ R/M α . Since R/M α is a field for each α ∈ , it follows that T is von Neumann regular. Note that R ∼ = φ(R) as rings and φ(R) is a subring of T .
We determine in Proposition 2.11 when a von Neumann regular ring R satisfies Z (R) ∪ {1} = R. We make use of the following lemma in the proof of Proposition 2.11 and some other results of Sect. 4 of this article. For any ideal I of a ring R and an element r ∈ R, we denote r + I , the residue class of r modulo I by r . Proof Suppose that the multiplicatively closed subset S = 1+ I of R is saturated. It is clear that {1} ⊆ U (R/I ). Let a ∈ R be such that a ∈ U (R/I ). Then there exists b ∈ R such that ab ∈ S = 1 + I . Since S is saturated, a, b ∈ S. Hence a = 1 + i for some i ∈ I . Therefore, we obtain that a = 1. This proves that U (R/I ) ⊆ {1} and so we obtain that U (R/I ) = {1}.
Conversely, suppose that U (R/I ) = {1}. Let x, y ∈ R be such that x y ∈ S. Hence x y = 1 + c for some c ∈ I . This implies that x y = 1 and so x, y ∈ U (R/I ). Since U (R/I ) = {1}, it follows that x = y = 1. Hence x, y ∈ 1 + I = S. This proves that S = 1 + I is saturated.
We next mention the following corollary to Lemma 2.9, which is also used in the proof of Proposition 2.11. Corollary 2.10 Let R be a Boolean ring (that is, r = r 2 for each r ∈ R). Then for any ideal I of R, the multiplicatively closed subset S = 1 + I of R is saturated.
Proof If I = R, then it is clear that S = 1 + I = R and hence is saturated. Suppose that I = R. In view of Lemma 2.9, to prove that the multiplicatively closed subset S = 1 + I of R is saturated, it is enough to show that U (R/I ) = {1}. Note that R/I is also a Boolean ring. We complete the proof of this corollary by showing that for any Boolean ring T, U (T ) = {1}. Let t ∈ T . Suppose that t = 1. Now from t = t 2 , it follows that t (1 − t) = 0 and hence t ∈ Z (T ). Therefore, t cannot be a unit in T . This proves that U (T ) = {1}, and moreover, Z (T ) ∪ {1} = T .
The following proposition determines when a von Neumann regular ring R satisfies Z (R) ∪ {1} = R and also shows that the converse of Corollary 2.10 is true.

Proposition 2.11
Let R be a ring. The following statements are equivalent:

is a von Neumann regular ring and Z (R) ∪ {1} = R.
(ii) R is a Boolean ring (that is, r = r 2 for each r ∈ R). (iii) If S is any subring of R, then Z (S) ∪ {1} = S. (iv) If I is any ideal of R, then the multiplicatively closed subset S = 1 + I of R is saturated.
(v) R is a von Neumann regular ring and |R/M| = 2 for each maximal ideal M of R.
Since R is von Neumann regular, we obtain from [3, Exercise 29, p. 113] that r = ue where u is a unit in R and e is an idempotent element of R. It now follows from Z (R) ∪ {1} = R that 1 is the only unit of R. Thus, r = e = e 2 = r 2 .
(ii) ⇒ (iii) Let S be any subring of R. Then s = s 2 for each s ∈ S. It is shown in the proof of Corollary 2.10 that for any Boolean ring T , We next show that R is a von Neumann regular ring. In view of [3, Exercise 16, p. 111], it is enough to prove that R is reduced and dim R = 0. Since U (R) = {1}, we know from Remark 2.5(ii) that R is reduced. Moreover, we obtain from Remark 2.5(i) that 1 + 1 = 0. Hence R contains S = {0, 1} as a subring. Note that S ∼ = Z/2Z as rings. We claim that R is integral over S. Let r ∈ R. If r ∈ S, then it is clear that r is integral over S. Hence, we may assume that r / ∈ S. Consider the subring Hence r ∈ Z (S[r ]) and so there exists t ∈ S[r ]\{0} such that rt = 0. Let t = s 0 + s 1 r + · · · + s n r n for some n ≥ 1, s i ∈ S for i ∈ {0, 1 . . . , n} with s n = 0. Since S = {0, 1}, it follows that s n = 1 and from rt = 0, we obtain that r n+1 + s n−1 r n +· · ·+s 0 r = 0. This proves that r is integral over S and so R is integral over S. Now we obtain from [4,Theorem 48 , we obtain that R is a Boolean ring. Hence, we obtain from Corollary 2.10 that for any ideal I of R, the multiplicatively closed subset S = 1 + I of R is saturated.
(iv) ⇒ (i) Let I be any ideal of R. We claim that I = √ I . This is clear if I = R. Suppose that I = R. Since the multiplicatively closed subset S = 1 + I of R is saturated, we obtain from Lemma 2.9 that U (R/I ) = {1}. Hence it follows from Remark 2.5(ii) that R/I is a reduced ring. This implies that I = √ I . Let r ∈ R. Now Rr = √ Rr = √ Rr 2 = Rr 2 . Hence r = r 2 s for some s ∈ R. This proves that R is a von Neumann regular ring. As R is a von Neumann regular ring, it is clear that Hence we obtain that Z (R) ∪ {1} = R. This proves (iv) ⇒ (i).
(ii) ⇒ (v) Since R is a Boolean ring, R is von Neumann regular. Let M be any maximal ideal of R. Since each element of R is idempotent, it follows that each element of the field R/M is idempotent. Hence we obtain that R/M = {0 + M, 1 + M}. This proves that |R/M| = 2 for each maximal ideal M of R.
(v) ⇒ (ii) Let r ∈ R. Let M be any maximal ideal of R. Since |R/M| = 2, it follows that r + M = (r + M) 2 = r 2 + M. Hence, r − r 2 ∈ M for each maximal ideal M of R. As R is a von Neumann regular ring, R is reduced and dim R = 0 [3, Exercise 16(d), p. 111]. Thus, r − r 2 ∈ ∩ P∈Spec(R) P = the nilradical of R = (0). Hence r = r 2 for each r ∈ R.
This completes the proof of Proposition 2.11.
The following proposition determines when a multiplicatively closed subset of a ring satisfies (P).

Proposition 2.12 Let R be a ring and let S be a multiplicatively closed subset of R. Then S satisfies (P) if and only if R\S = ∪ a∈R\S Sa.
Proof Assume that S satisfies (P). First observe that if r is any element of R, then as 1 ∈ S, it follows that r = 1r ∈ Sr. Hence, we obtain that R\S ⊆ ∪ a∈R\S Sa. Let a ∈ R\S. As S satisfies (P), it is clear that Sa ⊆ R\S. Hence ∪ a∈R\S Sa ⊆ R\S. Therefore, we obtain that R\S = ∪ a∈R\S Sa. Conversely, suppose that the multiplicatively closed subset S of R satisfies R\S = ∪ a∈R\S Sa. Let x, y ∈ R be such that x, x y ∈ S. If y / ∈ S, then as Sy ⊆ R\S, it follows that x y / ∈ S. This is a contradiction. Hence y ∈ S. This shows that S satisfies (P).
Let S be a multiplicatively closed subset of R with U (R) ⊆ S. The following proposition determines when S is saturated.

Proposition 2.13 Let R be a ring and let S be a multiplicatively closed subset of R with U (R) ⊆ S. Then S is saturated if and only if (R\S) ∪ {1} is a multiplicatively closed subset of R and S satisfies (P).
Proof Suppose that S is saturated. Then it is clear that (R\S) ∪ {1} is a multiplicatively closed subset of R and S satisfies (P).
Conversely, suppose that (R\S) ∪ {1} is a multiplicatively closed subset of R and S satisfies (P). Let a, b ∈ R be such that ab ∈ S. We want to show that a, b ∈ S.
Since U (R) ⊆ S and S is multiplicatively closed, it follows that a, b ∈ S if either a or b is a unit in R. Hence, we may assume that a / ∈ U (R) and b / ∈ U (R). Suppose that a / ∈ S. Since ab ∈ S and S satisfies (P), it follows that b / ∈ S. Thus a, b ∈ (R\S) ∪ {1}. By assumption, (R\S) ∪ {1} is a multiplicatively closed subset of R. Hence ab ∈ (R\S) ∪ {1}. Since ab ∈ S, we obtain that ab = 1. This is impossible since a / ∈ U (R). Therefore, a ∈ S. Similarly, it follows that b ∈ S. This proves that S is saturated.

( P)-closure of multiplicatively closed sets
We begin with the following remark.
Remark 3.1 Let S be a multiplicatively closed subset of a ring R. Note that if W is the collection of all multiplicatively closed subsets T of R with the following properties: (i) S ⊆ T and (ii) T satisfies (P), then S 1 = ∩ T ∈W T ∈ W and S 1 is the smallest element of W with respect to the inclusion relation. That is, S 1 is the smallest multiplicatively closed subset of R such that S ⊆ S 1 and S 1 satisfies (P). We call S 1 , the (P)-closure of S and for convenience, we denote it by (P) − cl(S).
Let R, S be as in Remark 3.1. In the following proposition, we describe the elements of (P) − cl(S). For any subset A of R and an element b of R, we denote the set {r ∈ R|rb ∈ A} by (A : R b).

Proposition 3.2 Let S be a multiplicatively closed subset of a ring R. Then
Proof For convenience, let us denote the set ∪ s∈S (S : R s) by T . We first show that T is a multiplicatively closed subset of R. Note that since S is a multiplicatively closed subset of R, S ⊆ (S : R s) for each s ∈ S. Hence we obtain that S ⊆ T . As 1 ∈ S, it follows that 1 ∈ T . Let t 1 , t 2 ∈ T . Now there exist s 1 , s 2 ∈ S such that t 1 s 1 ∈ S and t 2 s 2 ∈ S. Since S is a multiplicatively closed subset of R, it follows that s 1 s 2 ∈ S and t 1 s 1 t 2 s 2 ∈ S. Hence, we obtain that t 1 t 2 ∈ (S : R s 1 s 2 ) ⊆ T . This proves that T is a multiplicatively closed subset of R.
We next verify that T satisfies (P). Let r 1 , r 2 ∈ R be such that r 1 r 2 , r 1 ∈ T . Now there exist s 1 , s ∈ S such that r 1 s 1 ∈ S and r 1 r 2 s ∈ S. Observe that r 1 s 1 s ∈ S and r 2 ∈ (S : R r 1 s 1 s) ⊆ T . This shows that T satisfies (P).
Let T 1 be any multiplicatively closed subset of R such that S ⊆ T 1 and T 1 satisfies (P). We assert that T ⊆ T 1 . Let t ∈ T . Then ts ∈ S for some s ∈ S. Now ts, s ∈ S ⊆ T 1 and as T 1 satisfies (P), it follows that t ∈ T 1 . Hence T ⊆ T 1 .
This proves that T is the (P)-closure of S. Thus, We next provide some examples to illustrate Proposition 3.2. We also state and prove some corollaries to Proposition 3.2.
It follows from the above paragraph that if n is any natural number with n > 1, then the multiplicatively Let T be a commutative ring with identity and let R be a subring of T . Let S be a multiplicatively closed subset of R. Observe that S can also be regarded as a multiplicatively closed subset of T . We call the smallest multiplicatively closed subset of T which contains S and satisfies (P), the (P)-closure of S in T and for convenience, we denote the (P)-closure of S in T by (P) − cl T (S). Thus, (P) − cl(S) = (P) − cl R (S). In Corollary 3.4, we determine (P) − cl T (S), under the hypothesis that T is an integral domain and S = R\{0}.

Corollary 3.4 Let R be a subring of an integral domain T . Let K be the quotient field of R. Let S = R\{0}.
Then the following hold: The following statements are equivalent: is any multiplicatively closed subset of R and if S 1 satisfies (P), then S 1 regarded as a multiplicatively closed subset of T satisfies (P).
Proof (a) Since R is an integral domain, S = R\{0} is a multiplicatively closed subset of R (indeed, S is a saturated multiplicatively closed subset of R). As T contains R as a subring, it is clear that S is a multiplicatively closed subset of T . We now verify that . Thus, t 1 , t 2 ∈ R and are such that t 1 t 2 , t 1 ∈ S 1 . Since S 1 regarded as a multiplicatively closed subset of R satisfies (P), we obtain that t 2 ∈ S 1 . This proves that S 1 regarded as a multiplicatively closed subset of T satisfies (P).
is a saturated multiplicatively closed subset of R. Hence S satisfies (P). Now we obtain from (iii) that S regarded as a multiplicatively closed subset of T satisfies (P).
We next provide the following remark which illustrates Corollary 3.4. Remark 3.5 (i) Let R be a subring of an integral domain T . Suppose that R is integrally closed and T is integral over R. Let S 1 ⊆ R\{0} be a multiplicatively closed subset of R and S 1 satisfies (P). Then S 1 regarded as a multiplicatively closed subset of T satisfies (P). (ii) Let K be any algebraic number field. Let R K denote the ring of algebraic integers in K . That is, R K = {α ∈ K |α is integral over Z}. Let S 1 ⊆ Z\{0} be a multiplicatively closed subset of Z and suppose that S 1 satisfies (P). Then S 1 regarded as a multiplicatively closed subset of R K satisfies (P). (iii) Let T be an integral domain. Let L denote the quotient field of T . Let R be a subring of T and let K be its quotient field. Let S = R\{0}. If L is algebraic over K , then the saturation of S regarded as a multiplicatively closed subset of T equals T \{0}. Thus S regarded as a multiplicatively closed subset of T is saturated if and only if T = R.
Proof (i) Let S 1 ⊆ R\{0} be a multiplicatively closed subset of R and suppose that S 1 satisfies (P). We assert that S 1 regarded as a multiplicatively closed subset of T satisfies (P). To prove this assertion, in view of Corollary 3.
Hence, S 1 regarded as a multiplicatively closed subset of T satisfies (P). (ii) This follows immediately from (i) since R K is integral over Z and Z is integrally closed. (iii) We know from [1, Exercise 7(ii), p. 44] that the saturation of S regarded as a multiplicatively closed for each α ∈ . Therefore, we obtain that the saturation of S regarded as a multiplicatively closed subset of T equals T \{0}. Now, S = R\{0} regarded as a multiplicatively closed subset of T is saturated if and only if S = the saturation of S regarded as a multiplicatively closed subset of T = T \{0} if and only if T = R.
We next have the following remark.
Remark 3.6 Let R be a subring of an integral domain T . Let a ∈ R\{0} be such that a is not a unit in R. Let S = {a n |n ≥ 0}. Then the following hold: Proof (i) If a is not a unit in T , then it follows from Example 2.1(ii) that S satisfies (P) when regarded as a multiplicatively closed subset of T . Hence we obtain that If n ≥ 0, then a n ∈ S and so a n ∈ (S : T : s) for each s ∈ S. If n < 0, then a −n ∈ S and a n a −n = 1 ∈ S. Thus a n ∈ (S : T a −n ). This proves that Let R be a subring of a ring T . In the following remark, we provide some sufficient condition under which the following holds: if S 1 ⊆ R\{0} is any multiplicatively closed subset of R and if S 1 satisfies (P), then S 1 satisfies (P) when regarded as a multiplicatively closed subset of T .
Let S 1 ⊆ R\{0} be a multiplicatively closed subset of R. If S 1 satisfies (P), then S 1 satisfies (P) when regarded as a multiplicatively closed subset of T .
. If S 1 ⊆ R\{0} is any multiplicatively closed subset of R which satisfies (P), then S 1 satisfies (P) when regarded as a multiplicatively closed subset of T .
Proof (i) Let K denote the quotient field of R. Let S 1 ⊆ R\{0} be any multiplicatively closed subset of R and suppose that S 1 satisfies (P). To prove that S 1 regarded as a multiplicatively closed subset of T satisfies (P), in view of Corollary 3.
If R is a subring of a ring T and if T regarded as an R-module is faithfully flat, then it is known that for any ideal I of R, I T ∩ R = I [5, Theorem 7.5(ii)]. (ii) Let S 1 ⊆ R\{0} be any multiplicatively closed subset of R and suppose that S 1 satisfies (P). Let . Since for any r ∈ R, r T ∩ R = r R, it follows from (i) that S 1 regarded as a multiplicatively closed subset of T satisfies (P).
In Example 3.9, we illustrate that Remark 3.7(ii) may fail to hold if R is not an integral domain. We make use of the following lemma in the verification of Example 3.9. Proof Since I T is an ideal of T , it follows from Example 2.1(i) that the multiplicatively closed subset 1 + I T of T satisfies (P). As S 1 = 1 + I ⊆ 1 + I T , we obtain that (P) − cl T (S 1 ) ⊆ 1 + I T . We next verify that 1 + I T ⊆ (P) − cl T (S 1 ). By hypothesis, ((0) : R I ) + I = R. Hence there exist b ∈ ((0) : R I ), a ∈ I such that a + b = 1. Thus b = 1 − a ∈ S 1 ∩ ((0) : R I ). Let t ∈ 1 + I T . Note that t = 1 + y for some y ∈ I T . As b ∈ ((0) : R I ) and y ∈ I T , it follows that by = 0. Hence tb = (1+ y)b = b ∈ S 1 . This shows that t ∈ (S 1 : T b). It follows from Proposition 3.2 that (S 1 : T b) ⊆ (P) − cl T (S 1 ). Hence we obtain that t ∈ (P) − cl T (S 1 ). This proves that 1 + I T ⊆ (P) − cl T (S 1 ) and so we obtain that (P) − cl T (S 1 ) = 1 + I T . (i) S 1 is a multiplicatively closed subset of R and S 1 satisfies (P).
Proof (i) This is obvious.
4 Some results on saturated multiplicatively closed sets of the form 1 + I Throughout Sects. 4 and 5, we use R to denote the collection of all commutative rings R with identity satisfying the property that R admits only a finite number of saturated multiplicatively closed sets of the form 1 + I where I is an ideal of R. If R is semiquasilocal, then it is shown in Proposition 4.3(i) that R ∈ R. Let R be an integral domain, dim R = 1 and if R has Noetherian spectrum, then we prove in Proposition 4.5 that R ∈ R. Moreover, we mention some examples to illustrate the results proved in this section. We begin with the following proposition. Proof Let I be an ideal of Z. Suppose that the multiplicatively closed subset S = 1 + I of Z is saturated and I = Z. If I = (0), then S = {1}. As (−1) 2 = 1 ∈ S, whereas −1 / ∈ S, it follows that I = (0). Note that I = nZ for some n ≥ 2. It is well known that |U (Z/nZ)| = φ(n), where φ is the Euler's phi-function. Since S is saturated, it follows from Lemma 2.9 that |U (Z/nZ)| = 1. Thus φ(n) = 1. As φ(n) ≥ 2 for all n ≥ 3, we obtain that n = 2. Hence either I = Z or I = 2Z.
Conversely, suppose that either I = Z or I = 2Z. If I = Z, then S = Z is clearly saturated. If I = 2Z, then as U (Z/2Z) = {1}, it follows from Lemma 2.9 that S = 1 + 2Z is saturated (this also follows from the fact that if the product of two integers is odd, then both of them must be odd).
We make use of the following fact in the proof of Proposition 4.3. Let R be a semiquasilocal ring (that is, R has only a finite number of maximal ideals). Proposition 4.3 determines ideals I of R such that the multiplicatively closed subset S = 1 + I of R is saturated. Let R be an integral domain. If R has Noetherian spectrum and if dim R = 1, then we prove in Proposition 4.5 that R ∈ R. First it is useful to recall the following.
Recall from [6] that a ring R has Noetherian spectrum, if the topological space Spec(R) (that is, the set of all prime ideals of R endowed with the Zariski topology) is Noetherian. It is well known that a ring R has Noetherian spectrum if and only if R satisfies ascending chain condition on radical ideals of R [6]. Moreover, a ring R has Noetherian spectrum if and only if the following conditions hold: (i) R satisfies ascending chain condition on prime ideals of R, (ii) each proper ideal of R admits only a finite number of prime ideals minimal over it [4,Theorem 88 and Exercise 25, p. 65].
We make use of the following lemma in the proof of Proposition 4.5.

Lemma 4.4 Let R be an integral domain. Suppose that R has Noetherian spectrum and dim R = 1. Let M denote the collection of all maximal ideals M of R such that the multiplicatively closed subset S = 1 + M of R is saturated. Then the collection M is finite.
Proof Suppose that M is infinite. Then there exist distinct maximal ideals M 1 , M 2 , M 3 , . . . of R such that M n ∈ M for each n ∈ N. Since the multiplicatively closed subset S n = 1 + M n of R is saturated for each n ∈ N, it follows from Lemma 2.9 that U (R/M n ) = {1}. As R/M n is a field for each n ∈ N, we obtain that R/M n = {0, 1}. Let a be a nonzero nonunit of R. Since R has Noetherian spectrum, the ideal Ra has only a finite number of prime ideals minimal over it. As R is an integral domain and dim R = 1, it follows that a can belong to only a finite number of maximal ideals of R. Hence, there exists m ∈ N such that a / ∈ M n for all n ≥ m. Therefore, we obtain that a + M n = 1 + M n for all n ≥ m. Thus a − 1 ∈ ∩ n≥m M n . Since any nonzero element of R can belong to only a finite number of maximal ideals, it follows that a − 1 = 0. Hence a = 1. This is in contradiction to the choice of a.
This proves that M is finite.
We next have the following proposition the proof of which is a consequence of Fact 4.2 and Lemma 4.4.

Proposition 4.5 Let R be an integral domain. Suppose that R has Noetherian spectrum and dim
Proof Let C denote the collection of all ideals I of R such that the multiplicatively closed subset 1 + I of R is saturated. We know from Lemma 4.4 that C can contain only a finite number of maximal ideals of R. Let I be a nonzero proper ideal of R such that I ∈ C. Let W be the collection of all maximal ideals M of R such that M ⊇ I . Since R is a one-dimensional integral domain and has Noetherian spectrum, it follows that W is finite. Let W = {M 1 , . . . , M n }. We know from Fact 4.2 that I = ∩ M∈W M = ∩ n i=1 M i . Now, we obtain from the Chinese remainder theorem [1, Proposition 1.10(ii)] that R/I ∼ = R/M 1 × · · · × R/M n as rings. Hence, U (R/I ) ∼ = U (R/M 1 × · · · × R/M n ) = U (R/M 1 ) × · · · × U (R/M n ) as groups. As U (R/I ) is trivial by Lemma 2.9, we obtain that U (R/M i ) is trivial for i = 1, . . . , n. Hence by Lemma 2.9, we obtain that M i ∈ C for i = 1, . . . , n. Since C contains only a finite number of maximal ideals, we obtain from the above discussion that the collection C is finite. This shows that R can admit only a finite number of ideals I such that the multiplicatively closed subset S = 1 + I of R is saturated, and hence we obtain that R ∈ R. We next mention an example to illustrate that in Proposition 4.5, the hypothesis that R has Noetherian spectrum cannot be omitted. Assume by induction that we have chosen algebraic integers t 1 , t 2 , . . . , t k such that the following conditions are satisfied: . . , k} and let D i denote the integral closure of Z in F i . With this notation, there are exactly It now follows by the principle of mathematical induction that there exist algebraic integers . . . , t i ) and D i is the integral closure of Z in F i , then conditions (a), (b), and (c) hold for each i ∈ N.
Observe that D is integral over Z and hence D is integral over D k for each k ∈ N. Hence by [4,Theorem 48], dim D = dim Z = 1 and so any nonzero prime ideal of D is a maximal ideal of D. We assert that there exist infinitely many maximal ideals of D which are lying over 2Z in Z. Let k ≥ 1. Since D is integral over D k , it follows from [1, Theorem 5.10] that there exists a prime ideal P j of D such that P j ∩ D k = M (k) t for any two distinct s, t ∈ {1, . . . , 2 k }, it follows that P s = P t . Moreover, as P j ∩ Z = M (k) j ∩ Z = 2Z for j = 1, . . . , 2 k , we obtain that for each k ≥ 1, there exist at least 2 k prime ideals of D which are lying over 2Z in Z. This shows that there are infinitely many maximal ideals of D which are lying over 2Z in Z. Since D is a one-dimensional integral domain and as 2 belongs to infinitely many maximal ideals of D, we obtain that D does not have Noetherian spectrum.
We now show that there are infinitely many maximal ideals M of D such that the multiplicatively closed subset S = 1 + M of D is saturated. Indeed, we show that if M is any maximal ideal of D with M ∩ Z = 2Z, . . , 2 k }} is the set of all prime ideals of D k which are lying over 2Z in Z, it follows that as fields for each j = 1, . . . , 2 k , we obtain that D/M ∼ = Z/2Z as fields. This proves that U (D/M) is trivial. Hence we obtain from Lemma 2.9 that the multiplicatively closed subset S = 1 + M of D is saturated. Since there are infinitely many maximal ideals of D which are lying over 2Z in Z, it follows that there are infinitely many maximal ideals M of D such that S = 1 + M is saturated. Hence D / ∈ R. The above example also shows that the hypothesis that R is semiquasilocal cannot be omitted in Proposition 4.3(i).
We mention in Example 4.8, a one-dimensional integral domain R such that R ∈ R but R does not have Noetherian spectrum. We make use of the following fact in the verification of Example 4.8.

Fact 4.7 Let S be a saturated multiplicatively closed subset of a ring T and let R be a subring of T . Then the following hold:
(i) S ∩ R is a saturated multiplicatively closed subset of R.
(ii) If for some ideal I of T , the multiplicatively closed subset S = 1 + I of T is saturated, then the multiplicatively closed subset 1 + (I ∩ R) of R is saturated.
, the proof of (ii) follows immediately from (i).
We next have the following example.  6] is such that p 1 belongs to infinitely many maximal ideals of Z * . Hence Z * does not have Noetherian spectrum. We verify below that Z * ∈ R. Let C denote the collection of ideals I of Z * such that the multiplicatively closed subset S = 1 + I of Z * is saturated. It is clear that Z * ∈ C. Let I be a proper ideal of Z * such that I ∈ C. Then it follows from Lemma 2.9 that U (Z * /I ) = {1}. Since {1, −1} ⊆ U (Z * ), it follows that I = (0). With the notations being as in [3,Example 42.6], we know that Z * = ∪ ∞ i=1 Z i , where Z i denotes the integral closure of Z in F i = Q(t 1 , . . . , t i ). Observe that I = ∪ ∞ k=1 (I ∩ Z k ). Now it follows from Fact 4.7(ii) that the multiplicatively closed subset S k = 1 + (I ∩ Z k ) of Z k is saturated for each k ∈ N and the multiplicatively closed subset 1 + (I ∩ Z) of Z is saturated. As Z * is integral over Z and I is a nonzero ideal of Z * , we obtain that I ∩ Z is a nonzero ideal of Z. Since I = Z * , it is clear that I ∩ Z = Z. Now it follows from Proposition 4.1 that I ∩ Z = 2Z. Hence (I ∩Z k )∩Z = I ∩Z = 2Z for k = 1, 2, . . .. Moreover, since the multiplicatively closed subset S k = 1+(I ∩Z k ) of Z k is saturated, it follows from Fact 4.2 that I ∩ Z k = ∩ M M, where M ranges through all the maximal ideals of Z k with M ⊇ I ∩ Z k for k = 1, 2, . . .. As (I ∩ Z k ) ∩ Z = 2Z, we obtain that M ∩ Z = 2Z for each maximal ideal M of Z k with M ⊇ I ∩ Z k . Let A k denote the set of all maximal ideals of Z k which are lying over 2Z in Z. Note that in the notation of [3, Example 42.6], where B k is some nonempty subcollection of A k . Now we obtain from the Chinese remainder theorem [1, Proposition 1.10(ii)] that Z k /I ∩ Z k ∼ = M∈B k Z k /M. We know from Lemma 2.9 that U (Z k /I ∩ Z k ) is trivial. Hence we obtain that Z k /M ∼ = Z/2Z as fields for each M ∈ B k . It follows from the construction of [3,Example 42.6 k+1 for any k ≥ 1. Hence we obtain that k+1 , then as Z k /M (k) k+1 ∼ = Z/2Z as fields for each k ≥ 1, it follows that Z * /I ∼ = Z/2Z as fields. Hence we obtain from Lemma 2.9 that the multiplicatively closed subset S = 1 + I of Z * is saturated.
This proves that there are exactly two ideals I of Z * such that the multiplicatively closed subset 1 + I of Z * is saturated and they are I = Z * and I = ∪ ∞ k=1 M (k) k+1 . Hence we obtain that Z * ∈ R. We next have the following proposition.

Proposition 4.9 Let K be an algebraic number field and let R K denote the ring of algebraic integers in K .
Let C denote the collection of all ideals J of R K such that the multiplicatively closed subset 1 + J of R K is saturated. Then the following statements hold: Proof (i) Let [K : Q] = n. We know from [2, Exercise 4.2.3], that R K is a free Z-module of rank n. Thus R K is a finite integral extension domain of Z. Hence dim R K = 1 and R K is Noetherian. Since any Noetherian ring has Noetherian spectrum, we obtain from Proposition 4.5 that C is finite. (ii) Let J ∈ C be such that J = R K . Let I = J ∩Z. We know from Fact 4.7(ii) that the multiplicatively closed subset 1 + I of Z is saturated. Since J = R K , it follows that I = Z. Now it follows from Proposition 4.1 that J ∩ Z = I = 2Z.
The following remark provides another proof of Proposition 4.9(i).

Remark 4.10
We remark here that Proposition 4.9(i) can also be proved as follows. Let J ∈ C be such that J = R K . From Proposition 4.9(ii), we obtain that J ∩ Z = 2Z and so J ⊇ 2R K . We know from [2, Exercise 4.4.3] that for any nonzero ideal A of R K , R K /A is a finite ring. Hence it follows that R K /2R K is a finite ring.
Hence we obtain that J = J i .This proves that C is finite.
We next provide some examples to illustrate Proposition 4.9.
Example 4.11 Let d be a square-free integer. Let K = Q( √ d). Let R K denote the ring of algebraic integers in K . Let C denote the collection of all ideals J of R K such that the multiplicatively closed subset 1 + J of R K is saturated. Then the following hold: Proof First note that it follows from Lemma 2.9 and Remark 2.5(ii) that any J ∈ C is a radical ideal of R K .
(i) We know that R K = Z[ √ d] [2,Example 4.3.4]. Suppose that J ∈ C and J = R K . We know from Proposition 4.9(ii) that be obtained by replacing the coefficients of f (x) by their respective residue classes modulo 2. We consider the following cases. Case (a). d ≡ 2(mod 4). Then f (x) = x 2 . Hence it follows from [2, Theorem 5.5.1] that P = 2R K + √ d R K is a prime ideal of R K and moreover, 2R K = P 2 . Furthermore, [R K /P : Z/2Z] = 1. Since any nonzero prime ideal of R K is maximal, it follows from 2R K = P 2 that P is the only radical ideal of R K with the property that P ∩ Z = 2Z. As R K /P ∼ = Z/2Z, it follows that U (R K /P) = {1} and so we obtain from Lemma 2.9 that P ∈ C. Now it is clear from the above discussion that P is the only element of C with P = R K . Hence we obtain that C = {R K , P}.
As any nonzero prime ideal of R K is maximal, it follows from 2R K = N 2 that N is the only radical ideal of R K such that N ∩ Z = 2Z. Since R K /N ∼ = Z/2Z, it follows that U (R K /N ) = {1}. Now we obtain from Lemma 2.9 that N ∈ C. From the above discussion, it is clear that denote the polynomial obtained on replacing the coefficients of f (x) by their respective residue classes modulo 2. We need to consider the following two cases. Case (I). d ≡ 1(mod 8). Then f (x) = x 2 − x = x(x + 1). Now it follows from [2, Theorem 5.5.1] that P 1 = 2R K + ((1 + √ d)/2)R K and P 2 = 2R K + (1 + (1 + √ d)/2)R K are distinct prime ideals of R K with 2R K = P 1 P 2 . Moreover, [R K /P 1 : Z/2Z] = 1 = [R K /P 2 : Z/2Z]. Observe that R K /P i ∼ = Z/2Z for i = 1, 2 and so U (R K /P i ) is trivial for i = 1, 2. As P 1 + P 2 = R K , it follows from the Chinese remainder theorem [1, Proposition 1.10(ii)] that R K /P 1 ∩ P 2 ∼ = R K /P 1 × R K /P 2 ∼ = Z/2Z × Z/2Z.
is also trivial. Now it follows from Lemma 2.9 that P 1 , P 2 , P 1 ∩ P 2 ∈ C. Since 2R K = P 1 P 2 , we obtain that P 1 , P 2 , P 1 ∩ P 2 are the only radical ideals J of R K with J ∩ Z = 2Z. Thus in this case we obtain that C = {R K , P 1 , P 2 , P 1 ∩ P 2 }. Case (II). d ≡ 5(mod 8). Then f (x) = x 2 + x + 1. Note that f (x) is irreducible over Z/2Z. Now it follows from [2, Theorem 5.5.1] that P = 2R K is a prime ideal of R K and moreover [R K /P : Z/2Z] = 2. Observe that P is the only proper radical ideal of R K with the property that P ∩ Z = 2Z. As [R K /P : Z/2Z] = 2, it follows that U (R K /P) is not trivial. Hence we obtain that P / ∈ C. Thus R K is the only element of C.
Example 4.8 illustrates that the hypothesis that R has Noetherian spectrum is not necessary in Proposition 4.5. In Example 4.13, we provide another example to illustrate it. We make use of the following fact in the verification of Example 4.13. Proof Let I be an ideal of T such that the multiplicatively closed subset S = 1 + I of T is saturated. Since R = Z[α] is a subring of T , it follows from Fact 4.7(ii) that the multiplicatively closed subset 1 + (I ∩ R) of R is saturated. Hence we obtain from Example 4.11(ii) case (II) that I ∩ R = R. Therefore, we obtain that I = T .

Example 4.13
Let T = {z ∈ C|z is an algebraic integer} be the ring of all algebraic integers. Let C denote the collection of all ideals I of T such that the multiplicatively closed subset S = 1 + I of T is saturated. We assert that T is the only element of C. It is clear that T ∈ C. Moreover, since Z[(1 + √ 5)/2] is a subring of T , it follows from Fact 4.12 that C = {T }. This shows that T ∈ R.
Since T is integral over Z, it follows from [4,Theorem 48] that dim T = dim Z = 1. It is known that any nonzero nonunit of T belongs to uncountably many maximal ideals of T [3, Proposition 42.8(1)]. Hence it follows that T does not have Noetherian spectrum.

When does R ∈ R imply that R[x] ∈ R?
Let R be as mentioned in the beginning of Sect. 4 Proof Let C denote the collection of all ideals I of R 1 such that the multiplicatively closed subset 1 + I of R 1 is saturated. Let D denote the collection of all ideals J of R 2 such that the multiplicatively closed subset 1 + J of R 2 is saturated. Let J ∈ D. We assert that J = φ(I ) for some I ∈ C. If J = R 2 , then J = R 2 = φ(R 1 ) and it is clear that R 1 ∈ C. Suppose that J = R 2 . Since φ : R 1 → R 2 is a surjective homomorphism of rings, it follows that R 1 /φ −1 (J ) ∼ = R 2 /J as rings. Hence we obtain that U (R 1 /φ −1 (J )) ∼ = U (R 2 /J ) as groups. Since J ∈ D and J = R 2 , it follows from Lemma 2.9 that U (R 2 /J ) = {1}, where 1 is the residue class of 1 modulo J . Therefore, we obtain that U (R 1 /φ −1 (J )) = {1} (here 1 is the residue class of 1 modulo φ −1 (J )). Now it follows from Lemma 2.9 that φ −1 (J ) ∈ C. Let I = φ −1 (J ). Since φ is onto, it is clear that Thus it is shown that D ⊆ {φ(I )|I ∈ C}. Since R 1 ∈ R, C must be a finite collection. Hence we obtain that the collection D is finite. This proves that R 2 ∈ R.
As an immediate consequence of Lemma 5.1, we have the following corollary.

Corollary 5.2 Let R be a ring and let T = R[x] or R[[x]
]. If T ∈ R, then R ∈ R.
Proof The mapping φ : T → R given by φ( f (x)) = f (0) for any f (x) ∈ T is a surjective homomorphism of rings. Hence if T ∈ R, then it follows from Lemma 5.1 that R ∈ R.
We next have the following proposition.

Proposition 5.3 Let R be a ring. Then R ∈ R if and only if R[[x]] ∈ R.
Proof Let C denote the collection of all ideals I of R such that the multiplicatively closed subset 1 + I of R is saturated. Let D denote the collection of all ideals J of R [[x]] such that the multiplicatively closed subset Hence the collection C is finite. We prove that the collection D is finite.
Let  We make use of the following lemma in the proof of Proposition 5.5.

Lemma 5.4
Let R i , i = 1, . . . , n be rings and let R = R 1 × · · · × R n be their direct product. Then R ∈ R if and only if R i ∈ R for i = 1, . . . , n.
Proof Suppose that R ∈ R. Since R i is a homomorphic image of R, it follows from Lemma 5.1 that R i ∈ R for i = 1, . . . , n. Conversely, assume that R i ∈ R for i = 1, . . . , n. We prove that R = R 1 × · · · × R n ∈ R. Let C(R) denote the collection of all ideals I of R such that the multiplicatively closed subset (1, . . . , 1) + I of R is saturated. Let C(R i ) denote the collection of all ideals I i of R i such that the multiplicatively closed subset 1 + I i of R i is saturated for i = 1, . . . , n. Since R i ∈ R, the collection C(R i ) is finite for i = 1, . . . , n. We prove that the collection C(R) is finite.
Let I ∈ C(R). Note that I = I 1 × · · · × I n for some ideal I i of R i for i = 1, . . . , n. Suppose that I = R. Then I i = R i for at least one i ∈ {1, . . . , n}. Let A = {i ∈ {1, . . . , n}|I i = R i }. Note that R/I ∼ = i∈A R i /I i as rings. Hence U (R/I ) ∼ = i∈A U (R i /I i ) as groups. Since I ∈ C (R) with I = R, we obtain from Lemma 2.9 that U (R/I ) is trivial and hence U (R i /I i ) is trivial for each i ∈ A. Therefore, it follows from Lemma 2.9 that I i ∈ C(R i ) for each i ∈ A. If i / ∈ A, then I i = R i which is clearly in C(R i ). Thus I i ∈ C(R i ) for i = 1, . . . , n and so I = I 1 × · · · × I n ∈ C(R 1 ) × · · · × C(R n ). This proves that C(R) ⊆ C(R 1 ) × · · · × C(R n ). Proceeding as above, with the help of Lemma 2.9, it is easy to show that C(R 1 ) × · · · × C(R n ) ⊆ C(R). Hence we obtain that C(R) = C(R 1 ) × · · · × C(R n ). Since C(R i ) is a finite collection for i = 1, . . . , n, it follows that the collection C(R) is finite. This proves that R ∈ R In Remark 5.6, we deduce with the help of Proposition 5.5 that for any semiquasilocal ring R, R[x] ∈ R. Proof We know from Proposition 4.3(i) that R ∈ R. Let C denote the collection of all ideals I of R such that the multiplicatively closed subset 1 + I of R is saturated. Since R is semiquasilocal, it follows from Fact 4.2 that any proper ideal I ∈ C is the intersection of a finite number of maximal ideals of R. Now it follows from Proposition 5.5 that R[x] ∈ R.

Proposition 5.5 Let
As an immediate consequence of Propositions 4.5 and 5.5, we next have the following proposition. Proof Let C denote the collection of all ideals I of R such that the multiplicatively closed subset 1 + I of R is saturated. By hypothesis, R is a one-dimensional integral domain with Noetherian spectrum. Hence it follows from Proposition 4.5 that R ∈ R. Let I ∈ C with I = R. Hence from Lemma 2.9, we obtain that U (R/I ) is trivial. Since char(R) = 2 by assumption, {1, −1} ⊆ U (R). Hence I = (0). As R is an integral domain, dim R = 1, and R has Noetherian spectrum, it follows that I can be contained in only a finite number of maximal ideals of R. Since I ∈ C, we obtain from Let R be a Boolean ring. We determine in the following proposition when R ∈ R. Proposition 5.8 Let R be a Boolean ring. Then the following statements are equivalent: (ii) R ∼ = F 1 × · · · × F m as rings for some m ≥ 1, where F i = Z/2Z for i = 1, . . . , m.
(iii) R[x] ∈ R, where R[x] is the polynomial ring in one variable over R.
Proof (i) ⇒ (ii) We know from (ii) ⇒ (iv) of Proposition 2.11 that for any ideal I of R, the multiplicatively closed subset 1 + I of R is saturated. By hypothesis, R ∈ R. Hence, we obtain that R can have only a finite number of ideals and so R can admit only a finite number of maximal ideals. Let {M 1 , . . . , M m } denote the set of all maximal ideals of R. Since R is a Boolean ring, R is reduced and dim R = 0. Hence it follows that ∩ m i=1 M i = (0). Now we obtain using the Chinese remainder theorem [1, Proposition 1.10(ii)] that R ∼ = R/M 1 × · · · × R/M m as rings. Since |R/M i | = 2 for i = 1, . . . , m, it is clear that R ∼ = F 1 × · · · × F m as rings, where F i = Z/2Z for i = 1, . . . , m.
(ii) ⇒ (iii) By (ii), R ∼ = F 1 × · · · × F m as rings, where F i = Z/2Z Let R be a von Neumann regular ring. As a consequence of Lemma 5.1, Propositions 5.5 and 5.8, we prove in the following corollary that R ∈ R if and only if R[x] ∈ R. Corollary 5.9 Let R be a von Neumann regular ring. Then the following statements are equivalent: (ii) R ∈ R and if I is a proper ideal of R such that the multiplicatively closed subset 1 + I of R is saturated, then I is the intersection of a finite number of maximal ideals of R.
Proof (i) ⇒ (ii) It is clear that R ∈ R. Let C denote the collection of ideals I of R such that the multiplicatively closed subset 1 + I of R is saturated. Let I ∈ C with I = R. We know from Lemma 2.9 that U (R/I ) = {1}. Moreover, we know from Fact 4.2 that I = ∩ M M, where M varies over all the maximal ideals M of R with M ⊇ I . Since R is a von Neumann regular ring, it follows that R/I is also von Neumann regular ring. Now it follows from [3, Exercise 29, p. 113] and U (R/I ) = {1} that each element of R/I is idempotent. Hence, we obtain that R/I is a Boolean ring. Since R ∈ R, it follows from Lemma 5.1 that R/I ∈ R. Thus, the Boolean ring R/I ∈ R and hence we obtain from the proof of (i) ⇒ (ii) of Proposition 5.8 that R/I can admit only a finite number of maximal ideals (indeed,it follows from (i) ⇒ (ii) of Proposition 5.8 that R/I ∼ = F 1 × · · · × ×F m as rings for some m ≥ 1, where F i = Z/2Z for i = 1, . . . , m). Hence, I is the intersection of a finite number of maximal ideals of R.