PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 102(116) (2017), 195–202 DOI: https://doi.org/10.2298/PIM1716195C

TOPOLOGICALLY BOOLEAN AND g(x) -CLEAN RINGS

Angelina Yan Mui Chin and Kiat Tat Qua

Abstract. Let R be a ring with identity and let g(x) be a polynomial in Z(R)[x] where Z(R) denotes the center of R. An element r ∈ R is called g(x)-clean if r = u + s for some u, s ∈ R such that u is a unit and g(s) = 0. The ring R is g(x)-clean if every element of R is g(x)-clean. We consider g(x) = x(x − c) where c is a unit in R such that every root of g(x) is central in R. We show, via set-theoretic topology, that among conditions equivalent to R being g(x)-clean, is that R is right (left) c-topologically boolean.

1. Introduction

Let R be a ring with identity and let g(x) be a polynomial in Z(R)[x] where Z(R) denotes the center of R. Let Id(R) and U(R) denote the set of idempotents and the set of units in R, respectively. The notion of g(x)-clean rings first appeared in a 2002 paper of Camillo and Simón [1], where an element r ∈ R is called g(x)- clean if r = u + s for some u ∈ U(R) and s ∈ R such that g(s) = 0. The ring R is g(x)-clean if every element of R is g(x)-clean. Note that if r ∈ R is g(x)-clean and g(x) is a factor of a polynomial h(x) ∈ Z(R)[x], then r is also h(x)-clean.

Clearly, if g(x) = x^{2} − x, then g(x)-clean rings are clean. However, in general, g(x)-clean rings are not necessarily clean. A well-known example is the group ring ℤ_{(7)}C_{3} where ℤ_{(7)} = {m/n | m, n ∈ ℤ, gcd(7, n) = 1} and C_{3} is the cyclic group of order 3. By [7, Example 2.7], ℤ_{(7)}C_{3} is (x^{4}−1)-clean. However, Han and Nicholson [4] have shown that ℤ_{(7)}C_{3} is not clean.

Conversely, for a clean ring R, there may exist a g(x) ∈ Z(R)[x] such that R is not g(x)-clean (see [3, Example 2.3]). Indeed, let R be a Boolean ring containing more than two elements. Let c ∈ R where 0 ≠ c ≠ 1 and let g(x) = x^{2}+(1+c)x+c = (x + 1)(x + c). Since R is Boolean, so it is clean. Suppose that R is g(x)-clean. Then c = u + s for some u ∈ U(R) and s ∈ R such that g(s) = 0. Note that u = 1 since R is Boolean. Therefore, s = c + 1. However, g(c + 1) = c ≠ 0 which contradicts the assumption that g(s) = 0. Hence, it follows that R is clean but not g(x)-clean.

2010 Mathematics Subject Classification: Primary 16U99.

Key words and phrases: g(x)-clean, n-clean, topologically boolean.

Communicated by Zoran Petrović.

In [2], a ring R (not necessarily commutative) is said to be right (left) topologically boolean, or a right (left) tb-ring for short, if for every pair of distinct maximal right (left) ideals of R, there is a nontrivial idempotent in exactly one of them. The case where R is commutative has been considered earlier in [5]. Now let g_{c}(x) = x(x−c) ∈ Z(R)[x]. Here, we define a ring R to be right (left) c-topologically boolean, or a right (left) c-tb ring for short, if for every pair of distinct maximal right (left) ideals of R, there is a root of g_{c}(x) in exactly one of them. We say that R is a c-tb ring if it is both right and left c-tb. Clearly, when c = 1, a right (left) c-tb ring is just a right (left) tb-ring.

In this paper we consider g(x) = x(x − c) ∈ Z(R)[x] where c is a unit in R such that every root of g(x) is central in R and show via set-theoretic topology that among conditions equivalent to R being g(x)-clean is that R is right (left) c-tb. Throughout this paper, all rings are assumed to be associative with identity.

2. Some preliminaries

Let n be a positive integer. For a ring R and polynomial g(x) ∈ Z(R)[x], an element r ∈ R is said to be (n, g(x))-clean if r = u_{1} + ⋯+ u_{n} + s for some u_{1}, …, u_{n} ∈ U(R) and s ∈ R such that g(s) = 0. The ring R is (n, g(x))-clean if all of its elements are (n, g(x))-clean. Clearly, a (1, g(x))-clean ring is g(x)-clean. In [8], an element r ∈ R is said to be n-clean if r = e + u_{1} + ⋯+ u_{n} for some e ∈ Id(R) and u_{1}, …, u_{n} ∈ U(R). The ring R is n-clean if all of its elements are n-clean.

In [7, Theorem 2.1], Wang and Chen showed that if g(x) = (x − a)(x − b) ∈ Z(R)[x] with b − a ∈ U(R), then R is g(x)-clean if and only if R is clean. In [3, Theorem 3.2], Fan and Yang gave another proof of the same result. In the following, we give an extension to n-clean rings as follows:

Theorem 2.1. Let R be a ring and let g(x) = (x − a)(x − b)h(x) ∈ Z(R)[x]

such that b − a ∈ U(R). If R is n-clean, then R is (n, g(x))-clean (n ∈ ℕ).

Proof. Let r ∈ R. Since R is n-clean, then (r −a)(b−a)^{−1} = e+u_{1} +⋯ +u_{n}

for some e ∈ Id(R) and u_{i} ∈ U(R) (i = 1, …, n). Thus, r = (e(b − a) + a) + u_{1}(b − a) + ⋯+ u_{n}(b − a), where u_{i}(b − a) ∈ U(R) (i = 1, …, n). Note that g((e(b−a)+a) = e(b−a)(e(b−a)−(b−a))h(e(b−a)+a) = 0· h(e(b−a)+a) = 0. Hence, e(b − a) + a is a root of g(x). It follows that R is (n, g(x))-clean. □

By Theorem 2.1 and the fact that clean rings are n-clean for any integer n ⩾ 1 (by [9, Lemma 2.1]), we obtain the following:

Corollary 2.1. Let R be a ring and let g(x) = (x − a)(x − b) ∈ Z(R)[x] such that b − a ∈ U(R). Then R is g(x)-clean if and only if R is n-clean for all positive integers n.

TOPOLOGICALLY BOOLEAN AND g(x)-CLEAN RINGS 197

Let R be a ring and let g(x) ∈ Z(R)[x]. An element r ∈ R is called weakly g(x)-clean if r = u + s or r = u − s for some u ∈ U(R) and s ∈ R such that g(s) = 0. We say that R is weakly g(x)-clean if every element in R is weakly g(x)-clean. Clearly, a g(x)-clean ring is weakly g(x)-clean. It is also clear that if R is a weakly g(x)-clean ring and g(x) is a factor of a polynomial h(x) ∈ Z(R)[x], then R is also a weakly h(x)-clean ring.

In the following we obtain some results which generalise parts of Theorem 3.5 in [3].

Proposition 2.1. Let R be a ring which is weakly x(x − c)-clean where c ∈ Z(R). Then c ∈ U(R).

Proof. Let g(x) = x(x−c) ∈ Z(R)[x]. Since R is weakly g(x)-clean, c = u+s or c = u−s for some u ∈ U(R) and s ∈ R such that g(s) = 0. For the case c = u+s, we have that s = −u + c and hence, s^{2} = (−u + c)^{2} = u^{2} + cr for some r ∈ R. Since 0 = g(s) = s(s − c), we also have s^{2} = cs. Thus, c(s − r) = u^{2} ∈ U(R). This implies that c ∈ U(R). For the case c = u − s, we have that s = u − c and hence, s^{2} = (u − c)^{2} = u^{2} − cr for some r ∈ R. Since 0 = g(s) = s(s − c), we also have s^{2} = cs. Thus, c(s + r) = u^{2} ∈ U(R) which implies that c ∈ U(R). □

Lemma 2.1. Let R be a ring and let g(x) = ax^{m} − bx^{n}, h(x) = ax^{m} + bx^{n} ∈ Z(R)[x] where m, n are positive integers of different parity. Then R is g(x)-clean if and only if R is h(x)-clean.

Proof. (⇒): Assume that R is a g(x)-clean ring. Then for any r ∈ R, −r = u+s where u ∈ U(R) and s ∈ R such that g(s) = 0. It follows that r = (−u)+(−s). Note that

h(−s) = a(−s)^{m} + b(−s)^{n} = (−1)^{m}as^{m} + (−1)^{n}bs^{n}
= \{ \begin{matrix}as^{m} − bs^{n}, & if m is even, n is odd \\ −(as^{m} − bs^{n}), & if m is odd, n is even \end{matrix}
= 0.

It follows that r is h(x)-clean.

(⇐): Suppose that R is h(x)-clean. Then for any r ∈ R, −r = u + s where u ∈ U(R) and s ∈ R such that h(s) = 0. It follows that r = (−u) + (−s). Then since

g(−s) = a(−s)^{m} − b(−s)^{n} = (−1)^{m}as^{m} − (−1)^{n}bs^{n}
= \{ \begin{matrix}as^{m} + bs^{n}, & if m is even, n is odd \\ −(as^{m} + bs^{n}), & if m is odd, n is even \end{matrix}
= 0,

we have that r is g(x)-clean. □

Theorem 2.2. Let R be a ring and let c ∈ Z(R). Then the following are equivalent:

(a) R is x(x − c)-clean;

(b) R is x(x + c)-clean;

(c) R is n-clean for all positive integers n and c ∈ U(R).

Proof. (a)⇔ (b): This follows readily by Lemma 2.1. (a)⇒ (c): Assume (a). By Proposition 2.1, we have c ∈ U(R). It follows by Corollary 2.1 that R is n-clean for all positive integers n. (c)⇒ (a): This follows readily by Theorem 2.1 (take n = 1). □

Lemma 2.2. Let R be a ring, let c ∈ U(R) and let all roots of g(x) = x(x − c) in R be central. For any a, b ∈ R, if ab = c, then ba = c.

Proof. Let a, b ∈ R such that ab = c. Since c is a root of g(x), we have that c is central and therefore, ba(ba−c) = baba−c(ba) = b(ab)a−c(ba) = c(ba)−c(ba) = 0. Thus, ba is a root of g(x) and hence, ba is also central. Then ca = (ab)a = a(ba) = baa and it follows that c^{2} = c(ab) = (ca)b = (baa)b = bac. Since c ∈ U(R) (by the hypothesis), it follows that c = ba. □

3. Some equivalent conditions for x(x − c)-clean rings

Let R be a ring. A proper right (left) ideal P of R is said to be prime if aRb ⊆ P with a, b ∈ R implies that a ∈ P or b ∈ P. Given a ring R, let Spec_{r}(R) be the set of all proper right ideals of R which are prime. It has been shown in [10, Corollary 2.8] that if R is not a right quasi-duo ring, then Spec_{r}(R) is a topological space with the weak Zariski topology but not with the Zariski topology. For a right ideal I of R, let 𝒰_{r}(I) = {P ∈ Spec_{r}(R) | P ⊉ I} and 𝒱_{r}(I) = Spec_{r}(R) ∖ 𝒰_{r}(I). Let τ = {𝒰_{r}(I) | I is a right ideal of R}. Then τ contains the empty set and Spec_{r}(R). In general, τ is just a subbase of the weak Zariski topology on Spec_{r}(R). For any element a ∈ R, let 𝒰_{r}(a) = 𝒰_{r}(aR) and 𝒱_{r}(a) = 𝒱_{r}(aR). Then 𝒰_{r}(a) = {P ∈ Spec_{r}(R) | a ∉ P } and 𝒱_{r}(a) = {P ∈ Spec_{r}(R) | a ∈ P }. The left prime spectrum Spec_{l}(R) and the weak Zariski topology associated with it are defined analogously. Let Max_{r}(R) (Max_{l}(R)) be the set of all maximal right (left) ideals of R. Since maximal right (left) ideals are prime right (left) ideals (see [6]), Max_{r}(R) (Max_{l}(R)) inherits the weak Zariski topology on Spec_{r}(R) (Spec_{l}(R)). Let U_{r}(I) = Max_{r}(R) ∩ 𝒰_{r}(I) and V_{r}(I) = Max_{r}(R) ∩ 𝒱_{r}(I) for any right ideal I of R. Then, in particular, U_{r}(a) = Max_{r}(R) ∩ 𝒰_{r}(a) and V_{r}(a) = Max_{r}(R) ∩ 𝒱_{r}(a) for any a ∈ R.

Recall that a clopen set in a topological space is a set which is both open and closed. A topological space is said to be zero-dimensional if it has a base consisting of clopen sets.

We begin with the following lemmas.

Lemma 3.1. Let R be a ring, let g(x) = x(x − c) ∈ Z(R)[x] where c ∈ U(R) and let s ∈ R be a central root of g(x). Let N be a maximal right ideal of R. If s ∉ N, then c − s ∈ N.

Proof. Since g(s) = 0, we have that s(s − c) = 0 ∈ P for any prime right ideal P of R. Then since s is central, it follows that every prime right ideal of R contains either s or s − c. Now since c = s + (c − s) and c ∈ U(R), we have that 1 = sc^{−1} + (c − s)c^{−1} . Hence, every prime right ideal of R contains either s or c − s but not both. Since maximal right ideals are prime right ideals (by [6]), it follows that if s ∉ N, then c − s ∈ N. □

Lemma 3.2. Let R be a ring and let g(x) = x(x−c) ∈ Z(R)[x] where c ∈ U(R). Let s, t ∈ R be central roots of g(x). Then c^{−1} st, s + t − c^{−1} st and c − s are also roots of g(x).

Proof. We first note that since s(s − c) = 0 and t(t − c) = 0, we thus have s = c^{−1} s^{2} and t = c^{−1} t^{2}. Then

g(c^{−1} st) = c^{−1} st(c^{−1} st − c) = c^{−2} (st)^{2} − st
= c^{−2} (st)^{2} − c^{−1} s^{2}t = c^{−2} (s^{2}t)(t − c) = 0.

We also have that

g(s + t − c^{−1} st) = (s + t − c^{−1} st)(s + t − c^{−1} st − c)
= s(s − c) + s(t − c^{−1} st) + t(s − c^{−1} st)
+ t(t − c) − c^{−1} st(s − c^{−1} st) − c^{−1} st(t − c) = 0.

Finally, we note that g(c − s) = (c − s)((c − s) − c) = (s − c)s = g(s) = 0. □

Let R be a ring and let g(x) = x(x − c) ∈ Z(R)[x] where c ∈ U(R). Let ξ = {U_{r}(s) | s ∈ R is a central root of g(x) = x(x − c)}. By Lemma 3.2 and the following lemma, we may deduce that ξ is closed under intersection and union.

Lemma 3.3. Let R be a ring and let g(x) = x(x − c) ∈ Z(R)[x] with c ∈ U(R) such that every root of g(x) is central in R. If s, t ∈ R are roots of g(x), then the following hold.

(a) U_{r}(s) ∩ U_{r}(t) = U_{r}(c^{−1} st);

(b) U_{r}(s) ∪ U_{r}(t) = U_{r}(s + t − c^{−1} st);

(c) U_{r}(s) = V_{r}(c − s). In particular, every set in ξ is clopen.

Proof. (a) Let P ∈ 𝒰_{r}(s) ∩ 𝒰_{r}(t). Then P ∈ Spec_{r}(R) with s, t ∉ P. Note that c ∉ P. Since c, s, t are central in R and P is a prime right ideal of R, it follows that c^{−1} st ∉ P. Hence, P ∈ 𝒰_{r}(c^{−1} st) and therefore, 𝒰_{r}(s) ∩ 𝒰_{r}(t) ⊆ 𝒰_{r}(c^{−1} st). Conversely, suppose that P ∈ 𝒰_{r}(c^{−1} st). If s or t belongs to P, then since s, t are central in R and P is a right ideal of R, it follows that c^{−1} st ∈ P; a contradiction. Thus s and t do not belong to P, that is, P ∈ 𝒰_{r}(s) ∩ 𝒰_{r}(t). Hence, 𝒰_{r}(c^{−1} st) ⊆ 𝒰_{r}(s) ∩ 𝒰_{r}(t). The equality 𝒰_{r}(s) ∩ 𝒰_{r}(t) = 𝒰_{r}(c^{−1} st) thus follows. Then U_{r}(s)∩ U_{r}(t) = 𝒰_{r}(s)∩ 𝒰_{r}(t)∩Max_{r}(R) = 𝒰_{r}(c^{−1} st)∩Max_{r}(R) = U_{r}(c^{−1} st).

(b) Let P ∈ 𝒰_{r}(s) ∪ 𝒰_{r}(t). Then s ∉ P or t ∉ P. Without loss of generality, suppose that s ∉ P. Since s(s − c) = 0 ∈ P and s ∉ P with s central in R, it follows that s−c ∈ P. Then (1−c^{−1} s)t = −c^{−1} (s−c)t ∈ P. If s+(1−c^{−1} s)t ∈ P, then it will follow that s ∈ P; a contradiction. Thus, s + (1 − c^{−1} s)t ∉ P and hence, P ∈ 𝒰_{r}(s + (1 − c^{−1} s)t). The inclusion 𝒰_{r}(s) ∪ 𝒰_{r}(t) ⊆ 𝒰_{r}(s + (1 − c^{−1} s)t) therefore holds. For the reverse inclusion, suppose that P ∈ 𝒰_{r}(s + (1 − c^{−1} s)t). Then s + (1 − c^{−1} s)t ∉ P. If s and t both belong to P, then s + (1 − c^{−1} s)t ∈ P; a contradiction. Hence, either s ∉ P or t ∉ P, that is, P ∈ 𝒰_{r}(s) or P ∈ 𝒰_{r}(t). Therefore, P ∈ 𝒰_{r}(s) ∪ 𝒰_{r}(t) and the inclusion 𝒰_{r}(s + (1 − c^{−1} s)t) ⊆ 𝒰_{r}(s) ∪ 𝒰_{r}(t) follows. Hence, 𝒰_{r}(s) ∪ 𝒰_{r}(t) = 𝒰_{r}(s + (1 − c^{−1} s)t). It follows that

U_{r}(s) ∪ U_{r}(t) = (𝒰_{r}(s) ∩ Max_{r}(R)) ∪ (𝒰_{r}(t) ∩ Max_{r}(R))
= (𝒰_{r}(s) ∪ 𝒰_{r}(t)) ∩ Max_{r}(R)
= 𝒰_{r}(s + (1 − c^{−1} s)t) ∩ Max_{r}(R) = U_{r}(s + (1 − c^{−1} s)t).

(c) By using Lemma 3.1, we have U_{r}(s) = Max_{r}(R) ∖ U_{r}(c − s) = V_{r}(c − s). It follows that every set in ξ is clopen. □

Next, we extend Proposition 2.4 in [2] as follows:

Proposition 3.1. Let R be an x(x − c)-clean ring with c ∈ Z(R) such that every root of x(x − c) is central in R. Then R is a right c-tb ring.

Proof. By Proposition 2.1, c ∈ U(R). Let M and N be distinct maximal right ideals of R. Then there exists a ∈ M ∖ N and N + aR = R. Hence, 1 − ar ∈ N for some r ∈ R. Since N is a right ideal of R, c − arc = (1 − ar)c ∈ N. Let y = arc. Then c − y ∈ N and y ∈ M ∖ N. Since R is x(x − c)-clean, there exist a unit u ∈ R and a root s ∈ R of x(x − c) such that y = u + s. If s ∈ M, then u = y − s ∈ M from which it follows that M = R; a contradiction since M is a maximal right ideal of R. Thus, s ∉ M. If s ∉ N, then c − s ∈ N (by Lemma 3.1) and hence, u = y − s = (y − c) + (c − s) ∈ N. It follows that N = R which is also not possible since N is a maximal right ideal of R. We thus have that s is a root of x(x − c) belonging to N only. Hence, R is a right c-tb ring. □

Proposition 3.2. Let R be a ring and let g(x) = x(x − c) ∈ Z(R)[x] with c ∈ U(R) such that every root of g(x) in R is central. If R is a right c-tb ring, then ξ forms a base for the weak Zariski topology on Max_{r}(R). In particular, Max_{r}(R) is a compact, zero-dimensional Hausdorff space.

Proof. Note that if M_{1} and M_{2} are two distinct maximal right ideals of R, then since R is a right c-tb ring, there exists a root s ∈ R of g(x) such that s ∉ M_{1}, s ∈ M_{2} (that is, M_{1} ∈ U_{r}(s), M_{2} ∉ U_{r}(s)). The points in Max_{r}(R) can therefore be separated by disjoint clopen sets belonging to ξ. Hence, Max_{r}(R) is Hausdorff. By [2, Lemma 2.1] we have that Max_{r}(R) is compact.

To show that ξ forms a base for the weak Zariski topology on Max_{r}(R), let K ⊆ Max_{r}(R) be a closed subset and take M ∉ K. For each N ∈ K, since N ≠ M, there exists a clopen set U_{r}(s_{N}) ∈ ξ separating M and N, say N ∈ U_{r}(s_{N}). The collection {U_{r}(s_{N}) | N ∈ K} is therefore an open cover of the set K. Since K is compact, it has a finite subcover, that is, K is contained in a finite cover of sets of the form U_{r}(s_{N}) with N ∈ K. By Lemma 3.3, there exists a clopen set C ∈ ξ separating M from K. Hence, ξ forms a base for the weak Zariski topology on Max_{r}(R). Since every set in ξ is clopen (by Lemma 3.3), it follows that Max_{r}(R) is zero-dimensional. □

Proposition 3.3. Let R be a ring and let g(x) = x(x − c) ∈ Z(R)[x] with c ∈ U(R) such that every root of g(x) in R is central. If ξ forms a base for the weak Zariski topology on Max_{r}(R), then for any a ∈ R, there exists a root s of g(x) such that s ∉ M for every M ∈ V_{r}(a) and s ∈ N for every N ∈ V_{r}(a − c).

Proof. Consider the disjoint closed sets V_{r}(a) and V_{r}(a − c). Since ξ forms a base for the weak Zariski topology on Max_{r}(R) and Max_{r}(R) is compact, there is a clopen set U_{r}(s) ∈ ξ separating the sets V_{r}(a) and V_{r}(a − c). Without loss of generality, assume that V_{r}(a) ⊆ U_{r}(s) and V_{r}(a − c) ⊆ V_{r}(s). Then it follows that s ∉ M for every M ∈ V_{r}(a) and s ∈ N for every N ∈ V_{r}(a − c). □

Proposition 3.4. Let R be a ring and let g(x) = x(x − c) ∈ Z(R)[x] with c ∈ U(R) such that every root of g(x) in R is central. If for every a ∈ R there exists a root s ∈ Z(R) of g(x) such that V_{r}(a) ⊆ U_{r}(s) and V_{r}(a − c) ⊆ V_{r}(s), then R is g(x)-clean.

Proof. Let a ∈ R. By the hypothesis, there exists a root s ∈ Z(R) of g(x) such that V_{r}(a) ⊆ U_{r}(s) and V_{r}(a − c) ⊆ V_{r}(s). We claim that a − s is a unit. Let M be a maximal right ideal of R. Note that if a ∈ M, then a−s ∉ M, since s ∉ M. Next, suppose that a ∉ M. If a − s ∈ M, then s ∉ M, and hence, c − s ∈ M (by Lemma 3.1). Then since (a−c)+(c−s) = a−s ∈ M, it follows that a−c ∈ M and hence, s ∈ M (because V_{r}(a − c) ⊆ V_{r}(s)); a contradiction. Thus, a − s ∉ M. We have therefore shown that a − s ∉ M for any maximal right ideal M of R. Hence, a − s has a right inverse, that is, (a − s)v = 1 for some v ∈ R. Then (a − s)(vc) = c and by Lemma 2.2, we have that (vc)(a − s) = c. Since c ∈ U(R) ∩ Z(R), we can conclude that a − s is a unit in R. Hence, a is the sum of a unit and a root of g(x) in R. Since a is arbitrary in R, it follows that R is g(x)-clean. □

We are now ready for the main result.

Theorem 3.1. Let R be a ring and let x(x − c) ∈ Z(R)[x] with c ∈ U(R). If every root of x(x − c) is central in R, then the following conditions are equivalent.

(a) R is x(x − c)-clean;

(b) R is x(x + c)-clean;

(c) R is n-clean for all positive integers n;

(d) R is a right c-tb ring;

(e) The collection ξ = {U_{r}(s) | s ∈ R is a root of x(x − c)} forms a base for the weak Zariski topology on Max_{r}(R);

(f) For every a ∈ R, there exists a root s ∈ Z(R) of x(x − c) such that V_{r}(a) ⊆ U_{r}(s) and V_{r}(a − c) ⊆ V_{r}(s);

(g) R is a left c-tb ring;

(h) The collection ξ = {U_{l}(s) | s ∈ R is a root of x(x − c)} forms a base for the weak Zariski topology on Max_{l}(R).

Proof. By Theorem 2.2, it follows readily that (a) ⇔ (b) ⇔ (c). By Proposition 3.1, we readily have (a) ⇒ (d). The implications (d) ⇒ (e) ⇒ (f) follow by Propositions 3.2 and 3.3, respectively. The implication (f) ⇒ (a) is straightforward by using Proposition 3.4. By using the left analogue of the arguments in the proofs of (a) ⇒ (d) ⇒ (e) ⇒ (f) ⇒ (a), we obtain the equivalence (a) ⇔ (g) ⇔ (h). □

A ring R is said to be strongly clean if every element of R is the sum of an idempotent and a unit which commute with one another. A strongly clean ring is therefore clean and hence, x(x − 1)-clean. On the other hand, an abelian x(x − 1)- clean ring is clearly strongly clean. We thus have the following as a consequence of Theorem 3.1:

Corollary 3.1. Let R be an abelian ring. The following conditions are equivalent:

(a) R is clean;

(b) R is strongly clean;

(c) R is x(x + 1)-clean;

(d) R is n-clean for all positive integers n;

(e) R is a right tb-ring;

(f) The collection ξ = {U_{r}(s) | s ∈ Id(R)} forms a base for the weak Zariski topology on Max_{r}(R);

(g) For every a ∈ R, there exists s ∈ Id(R) such that V_{r}(a) ⊆ U_{r}(s) and V_{r}(a − 1) ⊆ V_{r}(s);

(h) R is a left tb-ring;

(i) The collection ξ = {U_{l}(s) | s ∈ Id(R)} forms a base for the weak Zariski topology on Max_{l}(R).

References

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2. A. Y. M. Chin, Clean elements in abelian rings, Proc. Indian Acad. Sci., Math. Sci. 119 (2009), 145–148.

3. L. Fan, X. Yang, On rings whose elements are the sum of a unit and a root of a fixed polynomial, Commun. Algebra 36 (2008), 269–278.

4. J. Han, W. K. Nicholson, Extensions of clean rings, Commun. Algebra 29 (2001), 2589–2595.

5. W. Wm. McGovern, Neat rings, J. Pure Appl. Algebra 205 (2006), 243–265.

6. S. H. Sun, Rings in which every prime ideal is contained in a unique maximal right ideal, J. Pure Appl. Algebra 78 (1992), 183–194.

7. Z. Wang, J. Chen, A note on clean rings, Algebra Colloq. 14 (2007), 537–540.

8. G. Xiao, W. Tong, n-clean rings and weakly unit stable range rings, Commun. Algebra 33 (2005), 1501–1517.

9. , n-clean rings, Algebra Colloq. 13 (2006), 599–606.

10. G. Zhang, W. Tong, F. Wang, Spectra of maximal 1-sided ideals and primitive ideals, Commun. Algebra 34 (2006), 2879–2896.

Institute of Mathematical Sciences (Received 20 05 2015) University of Malaya (Revised 11 08 2017) Kuala Lumpur Malaysia

acym@um.edu.my

Department of Mathematical and Actuarial Sciences Universiti Tunku Abdul Rahman Kajang, Selangor Malaysia

quakt@utar.edu.my