PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 102(116) (2017), 133–148 DOI https://doi.org/10.2298/PIM1716133C

RINGS IN WHICH THE POWER OF EVERY ELEMENT IS THE SUM OF AN IDEMPOTENT AND A UNIT

Huanyin Chen and Marjan Sheibani

Abstract. A ring R is uniquely π-clean if the power of every element can be uniquely written as the sum of an idempotent and a unit. We prove that a ring R is uniquely π-clean if and only if for any a ∈ R, there exists an integer m and a central idempotent e ∈ R such that a^{m} − e ∈ J(R), if and only if R is Abelian; idempotents lift modulo J(R); and R/P is torsion for all prime ideals P ⊇ J(R). Finally, we completely determine when a uniquely π-clean ring has nil Jacobson radical.

1. Introduction

An attractive problem in ring theory is to determine when a ring is generated additively by idempotents and units. An element of a ring is uniquely clean if it can be uniquely written as the sum of an idempotent and a unit. A ring R is uniquely clean if every element in R is uniquely clean. Many results on such rings can be found in [3, 5, 6]. Following Zhou [6], a ring R is uniquely π-clean if some power of every element in R is uniquely clean. This is a natural generalization of uniquely clean rings. The motivation of this paper is to develop explicit characterizations of such rings.

In Section 2, we explore the structures of uniquely π-clean rings, and prove that a ring R is uniquely π-clean if and only if for any a ∈ R, there exists an m ∈ ℕ and a central idempotent e ∈ R such that a^{m} − e ∈ J(R), if and only if for any a ∈ R, there exists an m ∈ ℕ and a unique e ∈ R such that a^{m} − e ∈ J(R), and J(R) = {x ∈ R | x^{m} − 1 ∈ U(R) f or all m ∈ ℕ}. This extends Lee and Zhou’s theorem as well.

In Section 3, we characterize uniquely π-cleanness by means of certain prime ideals. It is shown that a ring R is uniquely π-clean if and only if R is Abelian; every idempotent lifts modulo J(R); and R/P is torsion for all prime ideals P containing the Jacobson radical J(R). Furthermore, we consider a type of radical-like ideal J^{∗}(R), and characterize uniquely π-clean ring R by using such a special one.

2010 Mathematics Subject Classification: 16S34, 16U60, 16U99, 16E50.

Key words and phrases: idempotent, unit, Jacobson radical, uniquely clean ring, π-uniquely clean ring.

Communicated by Zoran Petrović.

Finally, we completely determine when a uniquely π-clean ring has nil Jacobson radical. Recall that an element a ∈ R is uniquely nil-clean provided that there exists a unique idempotent e ∈ R such that a − e ∈ N(R) [3]. We say that a ∈ R is uniquely π-nil-clean provided that a^{n} ∈ R is uniquely nil-clean for some n ∈ ℕ. A ring R is uniquely π-nil-clean if every element in R is uniquely π-nil-clean. A ring R is periodic if for any a ∈ R there exist distinct m, n ∈ ℕ such that a^{m} = a^{n}. In the last section, we characterize uniquely π-nil-clean rings. We prove that a ring R is uniquely π-nil-clean if and only if R is uniquely π-clean and J(R) is nil, if and only if R is an Abelian periodic ring, if and only if for any a ∈ R there exists some m ∈ ℕ such that a^{m} ∈ R is uniquely nil clean, if and only if for any a ∈ R, there exists some m ∈ ℕ and a unique idempotent e ∈ R such that a^{m} − e ∈ P (R), where P (R) is the prime radical of R. Here, an element a ∈ R is uniquely nil clean if there exists a unique idempotent e ∈ R such that a − e ∈ R is nilpotent [3, 5].

Throughout the paper, all rings are associative with an identity. We use J(R) and P (R) to denote the Jacobson radical and prime radical of a ring R. N(R) stands for the set of all nilpotent elements in R.

2. Structure Theorems

The aim of this is to explore the structures of uniquely π-clean rings. Recall that a ring R is an exchange ring if for any a ∈ R there exists an idempotent e ∈ aR such that 1 − e ∈ (1 − a)R. A ring R is an exchange ring if and only if, for every right R-module A and any two decompositions A = M ⊕ N = ⨁ _{i∈I} A_{i}, where M_{R} ≅ R and the index set I is finite, there exist submodules A′_{i} ⊆ A_{i} such that A = M ⊕ ( ⨁ _{i∈I} A′_{i}). The class of exchange rings is very large. For instances, regular rings, π-regular rings, strongly π-regular rings, semiperfect rings, left or right continuous rings, clean rings and unit C^{∗}-algebras of real rank zero, etc. We begin with

Lemma 2.1. Every uniquely π-clean ring is an Abelian exchange ring.

Proof. Let R be uniquely π-clean, let e ∈ R be an idempotent, and let r ∈ R. Then x := 1 − (e + er(1 − e)) ∈ R is an idempotent. By hypothesis, x ∈ R is uniquely clean. One easily checks that

x = e + (1 − 2e − er(1 − e)) = (e + er(1 − e)) + (1 − 2(e + er(1 − e))).

Further,

e = e^{2} ∈ R, (1 − 2e − er(1 − e))^{−1} = (1 − er(1 − e))(1 − 2e),
(e + er(1 − e)) = (e + er(1 − e))^{2}, (1 − 2(e + er(1 − e)))^{2} = 1.

By the uniqueness, we get e = e + er(1 − e), and then er = ere. Likewise, re = ere. Thus, er = re, and therefore R is Abelian.

For any a ∈ R, then we can find some m ∈ ℕ such that a^{m} ∈ R is clean. Write a^{m} = f + v, where f = f^{2}, v ∈ U(R). Then a^{m} − f^{m} = v, and so a − f ∈ U(R).

This implies that R is strongly clean. In view of [9, Theorem 30.2], every clean ring is an exchange ring. Therefore R is an exchange ring, as asserted. □

A ring R is strongly clean if for any a ∈ R there exists an idempotent e ∈ R such that a − e ∈ U(R) and ea = ae. As a consequence of Lemma 2.1, every uniquely π-clean ring is strongly clean. A ring R is uniquely clean provided that every element in R can be uniquely written as the sum of an idempotent and a unit. It is easy to verify that ℤ/3ℤ is not uniquely clean as 2 = 0 + 2 = 1 + 1, while ℤ/3ℤ is uniquely π-clean. Let R = ⨁ _{p is prime} ℤ/(p+1)ℤ. Then R is strongly clean. For any 1 ⩽ m ⩽ [log_{2}p], 2^{m} ∈ ℤ/(p + 1)ℤ is not uniquely clean. Thus, R is not uniquely π-clean. Therefore, we conclude that {uniquely clean rings} ⊊ {uniquely π-clean rings} ⊊ {strongly clean rings}.

Theorem 2.1. Let R be a ring. Then R is uniquely π-clean if and only if

(1) R is Abelian;

(2) Every idempotent lifts modulo J(R);

(3) R/J(R) is uniquely π-clean.

Proof. Suppose R is uniquely π-clean. In view of Lemma 2.1, R is an Abelian exchange ring. This proves (1) and (2), in terms of [9, Theorem 30.2]. For any ā ∈ R/J(R), then a ∈ R is uniquely π-clean. Thus, we have some n ∈ ℕ such that a^{n} ∈ R is uniquely clean. This implies that a^{n} = e + u, e = e^{2} ∈ R, u ∈ U(R). Hence, ā^{n} = \overline{e} + \overline{u}. Write ā^{n} = \overline{f} + \overline{v}, \overline{f} = \overline{f} ^{2} ∈ R/J(R), \overline{v} ∈ U(R/J(R)). Clearly, every unit lifts modulo J(R). So we may assume that f = f^{2} ∈ R, v ∈ U(R). As a result, there exists some r ∈ J(R) such that a^{n} = e + u = f + (v + r). By the uniqueness, we get e = f. Therefore R/J(R) is uniquely π-clean.

Conversely, assume that (1)–(3) hold. For any a ∈ R, we have ā ∈ R/J(R), and so there exists some n ∈ ℕ such that ā^{n} ∈ R is uniquely clean. By hypothesis, idempotents lift modulo J(R). In addition, units lift modulo J(R). Thus, a^{n} = e + u, e = e^{2} ∈ R, u ∈ U(R). Write a^{n} = f + v, f = f^{2}, v ∈ U(R). Then ā^{n} = \overline{f} + \overline{v}. By the uniqueness, we get \overline{e} = \overline{f} , i.e., e − f ∈ J(R). This infers that f(1 − e) = (e − f)(e − 1) ∈ J(R). As every idempotent in R is central, f(1 − e) ∈ R is an idempotent, thus, f(1 − e) = 0. It follows that f = fe. Likewise, e = ef. Consequently, e = f, and therefore R is uniquely π-clean. □

Corollary 2.1. Every corner of a uniquely π-clean ring is uniquely π-clean.

Proof. Let R be uniquely π-clean, and let e = e^{2} ∈ R. In light of Theorem 2.1, e ∈ R is central. For any eae ∈ eRe, then eae + 1 − e ∈ R is uniquely π-clean. So we have some n ∈ ℕ such that (eae + 1 − e)^{n} ∈ R is uniquely clean. Thus, (eae + 1 − e)^{n} = f + u, f = f^{2} ∈ R, u ∈ U(R), and so (eae)^{n} = efe + eue is clean in eRe. Write (eae)^{n} = g + v, g = g^{2} ∈ eRe, v ∈ U(eRe). Then (eae + 1 − e)^{n} = (eae)^{n} + 1 − e = g + (v + 1 − e), where g = g^{2} ∈ R. Write vw = wv = e. Then (v + 1 − e)^{−1} = w + 1 − e, and so v + 1 − e ∈ U(R). Thus, g = f = ege = efe, as required. □

Lemma 2.1 shows that every uniquely π-clean ring is an Abelian exchange ring. We now exhibit an exchange-like property of such rings.

Theorem 2.2. Let R be a ring. Then R is uniquely π-clean if and only if

(1) R is Abelian;

(2) For any a ∈ R, there exists an n ∈ ℕ and a unique idempotent e ∈ a^{n}R such that 1 − e ∈ (1 − a^{n})R.

Proof. Suppose that R is uniquely π-clean. In view of Lemma 2.1, every idempotent in R is central. For any a ∈ R, there exists some n ∈ ℕ such that a^{n} ∈ R is uniquely clean. Write a^{n} = f + v, where f = f^{2}, v ∈ U(R). Set g = 1 − f. Then g = g^{2} ∈ R. Obviously, we get

(a^{n} − g)v = (f + v − v(1 − f)v^{−1} )v = v^{2} + fv − v + vf = a^{2n} − a^{n}.

Thus g − a^{n} ∈ (a^{n} − a^{2n})R, and so g ∈ a^{n}R and 1 − g ∈ (1 − a^{n})R.

If there exists an idempotent h ∈ a^{n}R such that 1 − h ∈ (1 − a^{n})R. Write h = a^{n}x, xh = x. Then xa^{n}x = x. It is easy to verify that xa^{n} = x(a^{n}x)a^{n} = a^{n}x(xa^{n}) = a^{n}(xa^{n})x = a^{n}x. Write 1 − h = (1 − a^{n})y, y(1 − h) = y. Likewise, y(1 − a^{n}) = (1 − a^{n})y. One directly checks that (a^{n} − (1 − h))^{−1} = x − y, i.e., a^{n} − (1 − h) ∈ U(R), By the uniqueness, we get 1 − h = f. Hence, g = 1 − f = h, as desired.

Conversely, assume that (1) and (2) hold. For any a ∈ R, there exists an n ∈ ℕ and a unique idempotent e ∈ a^{n}R such that 1 − e ∈ (1 − a^{n})R. As in the preceding discussion, we get a^{n} − (1 − e) ∈ U(R). Write a^{n} = f + v, where f = f^{2}, v ∈ U(R). Set g = 1 − f. Then g = g^{2} ∈ R. Further, we have g ∈ a^{n}R and 1 − g ∈ (1 − a^{n})R. By the uniqueness, we obtain g = e. Thus, f = 1 − e, hence the result. □

Corollary 2.2. Let R be a ring. Then R is uniquely π-clean if and only if

(1) Every idempotent in R is central.

(2) For any a ∈ R, there exists an n ∈ ℕ and a unique idempotent e ∈ Ra^{n} such that 1 − e ∈ R(1 − a^{n}).

Proof. Obviously, a ring R is uniquely π-clean if and only if so is the opposite ring R^{op}. Applying Theorem 2.2 to R^{op}, we complete the proof. □

A ring R is local if it has only one maximal right ideal. A ring R is potent if for any a ∈ R there exists some n ∈ ℕ such that a^{n} = a. We note that every potent ring is commutative.

Lemma 2.2. Let R be a local ring. If R is uniquely π-clean, then R/J(R) is potent.

Proof. Suppose that there exists some a ∈ R such that a^{n} − a ∉ J(R) for all n ⩾ 2. Then a(a^{n−1} − 1) ∈ U(R) as R is a local ring. This implies that a ∈ U(R) and a^{n−1} − 1 ∈ U(R) for all n ⩾ 2. Since R is uniquely π-clean, we have an m ∈ ℕ such that a^{m} ∈ R is uniquely clean. But a^{m} = 0 + a^{m} = 1 + (a^{m} − 1), a contradiction. Therefore, for any a ∈ R, there exists some integer n ⩾ 2 such that a^{n} − a ∈ J(R). That is, R/J(R) is potent. □

Lemma 2.3. [6, Theorem 3.1] Let R be a ring. Then R is uniquely π-clean if and only if

(1) R is Abelian;

(2) Every idempotent lifts modulo J(R);

(3) R/J(R) is potent.

We have accumlated all the information necessary to prove the following.

Theorem 2.3. Let R be a ring. Then the following statements are equivalent:

(1) R is uniquely π-clean.

(2) For any a ∈ R, there exist an m ∈ ℕ and a central idempotent e ∈ R such that a^{m} − e ∈ J(R).

Proof. (1) ⇒ (2) In view of Lemma 2.3, R/J(R) is potent. For any a ∈ R, ā ∈ R/J(R) is potent, and so ā^{m} ∈ R/J(R) is an idempotent for some m ∈ ℕ. By using Lemma 2.3 again, we can find a central idempotent e ∈ R such that ā^{m} = \overline{e}, and so a^{m} − e ∈ J(R).

(2) ⇒ (1) If e ∈ R is an idempotent, then we have a central idempotent f ∈ R such that e − f ∈ J(R). As (e − f)^{3} = e − f, we deduce that e = f; hence, every idempotent in R is central. If e − e^{2} ∈ J(R), then we can find a central idempotent f ∈ R such that e^{m} − f ∈ J(R) for some m ∈ ℕ. As e − e^{2} ∈ J(R), if m ⩾ 3, we see that e − e^{m} = (e − e^{2}) + (e − e^{2})e + ⋯+ (e − e^{2})e^{m−2} ∈ J(R). Thus e − f ∈ J(R), and then idempotents lift modulo J(R).

For any a ∈ R, there exists m ∈ ℕ such that a^{m} − e ∈ J(R) for a central idempotent. Hence, ā^{m} = \overline{e} in R/J(R). Thus, S := R/J(R) is periodic. Thus, S is an Abelian exchange ring. If x^{2} = 0 and x ≠ 0 in S, then x ∉ J(S). For any r ∈ S, there exists some idempotent g ∈ Srx such that 1 − g ∈ S(1 − rx). Write g = crx for a c ∈ S. Then g = g^{2} = (crx)g = (cr)gx = (cr)(crx)x = (cr)^{2}x^{2} = 0, as S is Abelian. Thus, 1 − rx ∈ S is left invertible. Since S is Abelian, it is easy to check that 1 − rx ∈ U(S). This shows that x ∈ J(S); hence, x = 0. This gives a contradiction. Therefore S is reduced.

Let a ∈ R; there exist m, n (m > n) such that ā^{m} = ā^{n} in S. Choose k = n(m − n). It is easy to verify that p = ā^{k+1} is potent and w = ā − ā^{k+1} ∈ N(S). Further, ā = p + w = p is potent, and so S is potent. Applying Lemma 2.3, we complete the proof. □

Corollary 2.3. Let R be a ring. Then R is uniquely clean if and only if

(1) R is uniquely π-clean;

(2) J(R) = {x ∈ R | x − 1 ∈ U(R)}.

Proof. Obviously, J(R) ⊆ {x ∈ R | 1 − x ∈ U(R)}. Suppose that 1 − x ∈ U(R). Then we have an idempotent e ∈ R and an element u ∈ J(R) such that x = e + u and ex = xe by [10, Theorem 20]. Thus, 1 − e = (1 − x) + u ∈ U(R), and so 1 − e = 1. This implies that e = 0, whence x = u ∈ J(R). Therefore J(R) = {x ∈ R | 1 − x ∈ U(R)}.

Conversely, assume that (1) and (2) hold. In view of Lemma 2.3, R/J(R) is potent. It follows from J(R) = {x ∈ R | x − 1 ∈ U(R)} that U(R/J(R)) = {¯1}.

Write p = p^{n}(n ⩾ 2) in R/J(R). Then (1 − p^{n−1} + p)^{−1} = 1 − p^{n−1} + p^{n−2} . Hence, p = p^{n−1} , and so p^{2} = p^{n} = p. This implies that R/J(R) is Boolean. Therefore we complete the proof by Lemma 2.1 and [10, Theorem 20]. □

Theorem 2.4. Let R be a ring. Then R is uniquely π-clean if and only if

(1) For any a ∈ R, there exists an m ∈ ℕ and a unique e ∈ R such that a^{m} − e ∈ J(R).

(2) J(R) = {x ∈ R | x^{m} − 1 ∈ U(R) for all m ∈ ℕ}.

Proof. Suppose that R is uniquely π-clean. Let a ∈ R. In view of Theorem 2.3, there exist an m ∈ ℕ and a central idempotent g ∈ R such that a^{m} − g ∈ J(R). If there exists an idempotent f ∈ R such that a^{m} − f ∈ J(R), then g − f = (a^{m} − f) − (a^{m} − g) ∈ J(R). Clearly, (g − f)^{3} = g − f, and so (g − f)(1 − (g − f)^{2}) = 0. Thus, g = f, i.e., the uniqueness is verified.

Clearly, J(R) ⊆ {x ∈ R | x^{m} − 1 ∈ U(R) for all m ∈ ℕ}. If x ∉ J(R), then 0 ≠ xR ⊈ J(R). In view of Lemma 2.1, R is an exchange ring, and so there exists an idempotent 0 ≠ e ∈ xR. Write e = xr for an r ∈ R. Choose a = exe + 1 − e. Then we can find some m ∈ ℕ such that a^{m} ∈ R is uniquely clean. In addition, R is Abelian by Lemma 2.1. Obviously, a^{m} = 0+(ex^{m}e+1−e) = e+(e(x^{m} −1)e+1−e). If x^{m} − 1 ∈ U(R), then 0 = e, a contradiction. This implies that x^{m} − 1 ∉ U(R). That is, x ∉ {x ∈ R | x^{m} − 1 ∈ U(R) for all m ∈ ℕ}. Therefore {x ∈ R | x^{m} − 1 ∈ U(R) for all m ∈ ℕ} ⊆ J(R), as required.

Conversely, assume that (1) and (2) hold. Let x ∈ N(R). Then x^{m} − 1 ∈ U(R) for all m ∈ ℕ. By hypothesis, we get x ∈ J(R). Therefore, every nilpotent element in R is contained in J(R). Let e ∈ R be an idempotent, and let r ∈ R. Then e + er(1 − e) ∈ R is an idempotent. Hence, there exists a unique f ∈ R such that (e + er(1 − e)) − f ∈ J(R). By the preceding discussion, (e + er(1 − e)) − e = er(1 − e) ∈ J(R). The uniqueness forces e = f. But (e+er(1−e))−(e+er(1−e)) ∈ J(R), and so e + er(1 − e) = f = e. This shows that er = ere. Likewise, re = ere. That is, er = re, and then R is Abelian. For any a ∈ R, there exist an m ∈ ℕ and a unique e ∈ R such that w := a^{m} − e ∈ J(R). Then a^{m} = (1 − e) + (2e − 1 + w). As (2e − 1)^{2} = 1, we see that 2e − 1 + w ∈ U(R). If there exists an idempotent f ∈ R such that a^{m} − f ∈ U(R), then e − f = (a^{m} − f) − (a^{m} − e) ∈ U(R). One easily checks that (e + f − 1)(e − f)^{2} = 0, and therefore e + f − 1 = 0. Thus, f = 1 − e, hence the result. □

Corollary 2.4. Let R be a ring. Then R is uniquely π-clean if and only if

(1) For any a ∈ R, there exist an m ∈ ℕ and a unique e ∈ R such that a^{m} − e ∈ J(R).

(2) N(R) ⊆ J(R).

Proof. Suppose that R is uniquely π-clean. (1) is obvious by Theorem 2.4. Let a ∈ N(R). Then 1 − a^{m} ∈ U(R) for all m ∈ ℕ. It follows by Theorem 2.4 that a ∈ J(R). Therefore N(R) ⊆ J(R).

Conversely, assume that (1) and (2) hold. Let e ∈ R, and let x ∈ R. Then ex(1 − e) ∈ J(R). By hypothesis, we have some m ∈ ℕ such that the expressions (e + ex(1 − e))^{m} = (e + ex(1 − e)) + 0 = e + ex(1 − e) are unique. This implies that ex(1 − e) = 0, and so ex = exe. Likewise, xe = exe. Therefore R is Abelian. This yields the result by Theorem 2.3. □

Corollary 2.5. Let R be a local ring. Then the following statements are equivalent:

(1) R is uniquely π-clean.

(2) U(R) = {x ∈ R | There is an m ∈ ℕ such that x^{m} − 1 ∈ J(R)}.

(3) J(R) = {x ∈ R | x^{m} − 1 ∈ U(R) for all m ∈ ℕ}.

Proof. (1) ⇒ (3) is clear from Theorem 2.4.

(3) ⇒ (2) Obviously, {x ∈ R | there is an m ∈ ℕ such that x^{m} − 1 ∈ J(R)} ⊆ U(R). For any x ∈ U(R), x ∉ J(R). By hypothesis, there exists some m ∈ ℕ such that x^{m} − 1 ∉ U(R). As R is local, x^{m} − 1 ∈ J(R). This implies that U(R) ⊆ {x ∈ R | there is an m ∈ ℕ such that x^{m} − 1 ∈ J(R)}, as required.

(2) ⇒ (1) For any x ∈ R, we see that either x ∈ J(R) or x ∈ U(R). This implies that x̄ = ¯0 or x̄^{m} = ¯1 in R/J(R). Thus R/J(R) is potent. In light of Lemma 2.3, R is uniquely π-clean. □

3. Factors of Prime Ideals

The aim of this section is to characterize uniquely π-clean rings by means of prime ideals containing the Jacobson radicals. We use J-spec(R) to denote the set {P ∈ Spec(R) | J(R) ⊆ P }. Obviously, every maximal ideal is contained in J-spec(R). Set

J^{∗}(R) = ⋂ {P | P is a maximal ideal of R}.

We will see that J(R) ⊆ J^{∗}(R). In general, they are not the same. For instance, J(R) = 0 and J^{∗}(R) = {x ∈ R | dim_{F} (xV) < ∞}, where R = End_{F} (V) and V is an infinite-dimensional vector space over a field F. Furthermore, we characterize a uniquely π-clean ring R by means of the radical-like ideal J^{∗}(R). A ring R is strongly π-regular if, for any a ∈ R there exists n ∈ ℕ such that a^{n} ∈ a^{n+1} R. We have

Lemma 3.1. [7, Corollary 2.8] Let R be a commutative ring. Then the following statements are equivalent:

(1) R is strongly π-regular.

(2) R is an exchange ring in which every prime ideal of R is maximal.

Lemma 3.2. Let R be an Abelian exchange ring, and let x ∈ R. Then RxR = R if and only if x ∈ U(R).

Proof. If x ∈ U(R), then RxR = R. Conversely, assume that RxR = R. As in the proof of [4, Proposition 17.1.9], there exists an idempotent e ∈ R such that e ∈ xR such that ReR = R. This implies that e = 1. Write xy = 1. Then yx = y(xy)x = (yx)^{2}. Hence, yx = y(yx)x. Therefore 1 = x(yx)y = xy(yx)xy = yx, and so x ∈ U(R). This completes the proof. □

Herstein’s theorem says that a ring R is periodic if and only if for any a ∈ R, there exists n ∈ ℕ such that a^{n} = a^{n+1} f(a) for some f(t) ∈ ℤ[t]. We recall that a ring R is torsion, provided that for any nonzero a ∈ R there exists m ∈ ℕ such that a^{m} = 1. With this information we now derive

Theorem 3.1. Let R be a ring. Then R is uniquely π-clean if and only if

(1) R is Abelian;

(2) Every idempotent lifts modulo J(R);

(3) R/P is torsion for all P ∈ J-spec(R).

Proof. Suppose R is uniquely π-clean. In view of Lemmas 2.1 and 2.3, R is an Abelian exchange ring, and R/J(R) is potent. Let P ∈ J-spec(R). Then R/J(R)/P/J(R) ≅ R/P is prime; hence, P/J(R) is a prime ideal of R/J(R). As every potent ring is commutative, R/J(R) is a commutative π-regular ring. It follows from Lemma 3.1 that P/J(R) is a maximal ideal of R/J(R). We infer that P is a maximal ideal of R.

Clearly, \overline{R} := R/P is an Abelian exchange ring. Since P is maximal, R/P is simple. For any 0 ≠ x ∈ \overline{R}, we have \overline{R}x\overline{R} = \overline{R}. By Lemma 3.2, x ∈ U(R/P). Hence, R/P is a division ring. On the other hand, R/P ≅ R/J(R)/P/J(R) is potent. Thus, we have some m ∈ ℕ such that x^{m+1} = x, and so x^{m} = 1. This implies that R/P is torsion, as required.

Conversely, assume that (1)–(3) hold. Assume that R is not uniquely π-clean. Set S = R/J(R). In view of Theorem 2.3, S is not periodic. By using Herstein’s theorem, there exists some a ∈ S such that a^{m} ≠ a^{m+1} f(a) for any m ∈ ℕ and any f(x) ∈ ℤ[x]. Let Ω = {I ◁ S | ā^{m} ≠ ā^{m+1} f(ā) in S/I for any m ∈ ℕ and any f(x) ∈ ℤ[x]}. Then Ω is an nonempty inductive. By using Zorn’s lemma, there exists an ideal Q of S which is maximal in Ω. If Q is not prime, then there exist two ideals K and L of R such that K, L ⊈ Q and KL ⊆ Q. By the maximality of Q, we can find some s, t ∈ ℕ and some f(x), g(x) ∈ ℤ[x] such that ā^{s} = ā^{s+1} f(ā) in R/(K + Q) and ā^{t} = ā^{t+1} g(ā) in R/(L + Q). Thus, a^{s} − a^{s+1} f(a) ∈ K + Q and a^{t} − a^{t+1} g(a) ∈ L + Q, and so (a^{s} − a^{s+1} f(a))(a^{t} − a^{t+1} g(a)) ∈ (K + Q)(L + Q) ⊆ KL + Q ⊆ Q. In S/Q, we have ā^{s+t} = ā^{s+t+1} h(ā) for some h(x) ∈ ℤ[x]. This contradicts the choice of Q. Hence, Q ∈ J-spec(R). By hypothesis, R/Q is torsion, and so R/Q is periodic, which is impossible. Therefore R is uniquely π-clean. □

Corollary 3.1. A ring R is uniquely clean if and only if

(1) R is uniquely π-clean.

(2) R/M ≅ ℤ_{2} for all maximal ideals M of R.

Proof. Suppose R is uniquely clean. Then R is uniquely π-clean. (2) is proved by [3, Theorem 2.1].

Conversely, assume that (1) and (2) hold. For all maximal ideals M of R, 1_{R/M} is not the sum of two units in R/M. In view of Lemma 2.1, R is an Abelian exchange ring, and so it is clean. Let x ∈ R. Write x = e_{1} + u_{1} = e_{2} + u_{2}, e_{1} = e_{1}^{2}, e_{2} = e_{2}^{2} and u_{1}, u_{2} ∈ U(R). If R(1 − e_{2}(1 − e_{1}))R ≠ R, then there exists a maximal ideal M of R such that R(1 − e_{2}(1 − e_{1}))R ⊆ M. Clearly, J(R) ⊆ M. Hence, x̄ = \overline{e_{1}} + \overline{u_{1}} = \overline{e_{2}} + \overline{u_{2}} in R/M. By Theorem 3.1, R/M is a division ring. This implies that \overline{e_{i}} are ¯0 or ¯1. If \overline{e_{1}} ≠ \overline{e_{2}}, then 1_{R/M} is the sum of two units, a contradiction. Therefore we get e_{1} − e_{2} ∈ M. This implies that e_{2}(1 − e_{1}) = (e_{1} − e_{2})(e_{1} − 1) ∈ M, and so 1 = e_{2}(1 − e_{1}) + (1 − e_{2}(1 − e_{1})) ∈ M, a contradiction. As a result, R(1 − e_{2}(1 − e_{1}))R = R. As e_{2}(1 − e_{1}) ∈ R is an idempotent, we get e_{2}(1 − e_{1}) = 0, and so e_{2} = e_{2}e_{1}. Likewise, e_{1} = e_{1}e_{2}. Consequently, e_{1} = e_{2}, and then u_{1} = u_{2}. Therefore R is uniquely clean. □

Let S(R) be the nonempty set of all ideals of a ring R generated by central idempotents. By Zorn’s lemma, S(R) contains maximal elements. As usually, we say that R/P is a Pierce stalk if P is a maximal element of the set S(R), and that P is a Pierce ideal. Let Pier(R) be the set of all Pierce ideals of R.

Proposition 3.1. Every uniquely π-clean ring is the subdirect product of rings R_{i}, where each R_{i}/J(R_{i}) is torsion.

Proof. Let R be a uniquely π-clean ring. [9, Remark 11.2] says that the intersection of all Pierce ideals of R is zero, i.e., ⋂ { P | P ∈ Pier(R)} = 0. Let φ_{P} : R → R/P be the natural epimorphism. Then ⋂ _{P ∈Pier (R)} kerφ_{P} = ⋂ _{P ∈Pier (R)} P = 0. Hence, R is the subdirect product of all R/P, where P ∈ Pier(R). In view of Lemma 2.1, R is an Abelian exchange ring. Let P ∈ Pier(R). Then R/P is an exchange ring. As R is indecomposable, we see that R/P is a local ring. By an argument in [6], R/P is uniquely π-clean, and so R/P/J(R/P) is potent from Lemma 2.3, as needed. □

Lemma 3.3. Let R be an Abelian exchange ring. Then J^{∗}(R) = J(R).

Proof. Let M be a maximal ideal of R. If J(R) ⊈ M, then J(R) + M = R. Write x + y = 1 with x ∈ J(R), y ∈ M. Then y = 1 − x ∈ U(R), an absurd. Hence, J(R) ⊆ M. This implies that J(R) ⊆ J^{∗}(R). Let x ∈ J^{∗}(R), and let r ∈ R. If R(1 − xr)R ≠ R, then we can find a maximal ideal M of R such that R(1 − xr)R ⊆ M, and so 1 − xr ∈ M. It follows that 1 = xr + (1 − xr) ∈ M, which is imposable. Therefore R(1 − xr)R = R. In light of Lemma 3.2, 1 − xr ∈ U(R), and then x ∈ J(R). This completes the proof. □

Theorem 3.2. Let R be a ring. Then R is uniquely π-clean if and only if

(1) R is an exchange ring;

(2) R/J^{∗}(R) is potent and every idempotent uniquely lifts modulo J^{∗}(R).

Proof. Suppose R is uniquely π-clean. Then R is an Abelian exchange ring by Lemma 2.1. In view of Lemma 3.3, J^{∗}(R) = J(R). It follows from Lemma 2.3 that R/J^{∗}(R) is potent. Let e−e^{2} ∈ J(R). Then we can find an idempotent f ∈ R such that e − f ∈ J(R). Since (e − f)^{2}(1 − (e − f)) = 0, we get e = f, as desired.

Conversely, assume that (1) and (2) hold. Let e ∈ R be an idempotent, and let r ∈ R. Then \overline{er(1 − e)} ∈ R/J^{∗}(R) is potent. This implies that \overline{er(1 − e)} = ¯0, and so er(1 − e) ∈ J^{∗}(R). Since e − e, e − (e + er(1 − e)) ∈ J^{∗}(R), by the uniqueness, we get e = e + er(1 − e), and so er = ere. Likewise, re = ere; hence that er = re.

Thus, R is Abelian. In light of Lemma 3.3, J^{∗}(R) = J(R). Therefore we complete the proof, in terms of Lemma 2.3. □

Corollary 3.2. Let R be a ring which has finitely many maximal ideals. Then R is uniquely π-clean if and only if

(1) R is an exchange ring;

(2) R/J^{∗}(R) is the direct sum of finitely many torsion rings and every idempotent uniquely lifts modulo J^{∗}(R).

Proof. ⇒: Clearly, R is an exchange ring. Let M be a maximal ideal of R. As in the proof of Lemma 3.3, we see that J(R) ⊆ M. This shows that M ∈ J-spec(R). Therefore R/M is torsion by Theorem 3.1. Since R has finitely many maximal ideals M_{1}, …, M_{n}, we see that R/J^{∗}(R) ≅ R/M_{1} ⊕ ⋯⊕ R/M_{n}. Therefore R/J^{∗}(R) is the direct sum of finitely many torsion rings, as desired.

⇐: As every torsion ring is potent, we see that R/J^{∗}(R) is potent. Therefore we complete the proof, by Theorem 3.2. □

Theorem 3.3. Let R be a ring. Then R is uniquely π-clean if and only if

(1) For any a ∈ R, there exist an m ∈ ℕ and a unique e ∈ R such that a^{m} − e ∈ J^{∗}(R).

(2) J^{∗}(R) = {x ∈ R | x^{m} − 1 ∈ U(R) for all m ∈ ℕ}.

Proof. One direction is obvious by Lemma 3.3 and Theorem 2.4.

Conversely, assume that (1) and (2) hold. Let x ∈ N(R). Then x^{m} − 1 ∈ U(R) for all m ∈ ℕ. By hypothesis, we have x ∈ J^{∗}(R), and so N(R) ⊆ J^{∗}(R). Let e ∈ R be an idempotent, and let r ∈ R. Then e + er(1 − e) + 0 = e + er(1 − e) with 0, er(1 − e) ∈ J^{∗}(R). By the uniqueness, we get er = ere. Similarly, we have re = ere. That is, er = re. We infer that R is Abelian. For any a ∈ R, there exist an m ∈ ℕ and a unique e ∈ R such that w := a^{m} − e ∈ J^{∗}(R). Then a^{m} = (1 − e) + (2e − 1 + w). But 2e − 1 + w = (1 − 2e)((1 − 2e)w − 1) ∈ U(R), by (2). If there exists an idempotent f ∈ R such that a^{m} − f ∈ U(R), then e − f = (a^{m} − f) − (a^{m} − e) = (a^{m} − f)(1 − (a^{m} − f)^{−1} (a^{m} − e)) ∈ U(R). It follows from (e + f − 1)(e − f)^{2} = 0 that f = 1 − e, and we are through. □

Let P (R) be the intersection of all prime ideals of R, i.e., P (R) is the prime radical of R. As is well known, P (R) is the intersection of all minimal prime ideals of R.

Corollary 3.3. Let R be a uniquely π-clean in which every prime ideal is maximal. Then

P (R) = {x ∈ R | x^{m} − 1 ∈ U(R) for all m ∈ ℕ}.

Proof. As every maximal ideal is prime, we deduce that J^{∗}(R) = P (R), and therefore we complete the proof by Theorem 3.3. □

4. Certain Classes

In this section we investigate certain classes of uniquely π-clean rings. We now recall the concept of ideal-extensions. Let R be a ring with an identity and S be a ring (not necessary unitary), and let S be an R-R-bimodule in which (s_{1}s_{2})r = s_{1}(s_{2}r), r(s_{1}s_{2}) = (rs_{1})s_{2} and (s_{1}r)s_{2} = s_{1}(rs_{2}) for all s_{1}, s_{2} ∈ S, r ∈ R. The ideal extension I(R; S) is defined to be the additive Abelian group R ⊕ S with multiplication (r_{1}, s_{1})(r_{2}, s_{2}) = (r_{1}r_{2}, s_{1}s_{2} + r_{1}s_{2} + s_{1}r_{2}) (see [10]). We start this section by examining when an ideal extension is uniquely π-clean.

Theorem 4.1. The ideal-extension I(R; S) is uniquely π-clean and S is idempotent-free if and only if

(1) R is uniquely π-clean;

(2) If e = e^{2} ∈ R, then es = se for all s ∈ S;

(3) If s ∈ S, then there exists an s′ ∈ S such that ss′ = s′s and s+s′+ss′ = 0.

Proof. Assume that (1)–(3) hold. Let e ∈ S be an idempotent. Then (−e) + s′+ (−e)s′ = 0 for some s′ ∈ S. Hence, (1 − e)(1 + s′) = 1, and so e = 0. That is, S is idempotent-free. Let (a, s) ∈ I(R; S). Then a ∈ R is uniquely π-clean. Thus, we have some n ∈ ℕ such that a^{n} ∈ R is uniquely clean. Write a^{n} = e + u, e = e^{2} ∈ R, u ∈ U(R). Hence, (a, s)^{n} = (a^{n}, x) = (e, 0) + (u, x) for some x ∈ S. Clearly, (e, 0)^{2} = (e, 0). As x ∈ S, we see that u^{−1} x ∈ S, and so we have some t ∈ S such that u^{−1} x + t + u^{−1} xt = 0 and u^{−1} xt = tu^{−1} x. This implies that 1 + u^{−1} x = (1+t)^{−1} ∈ U(R). One easily checks that (u, x)^{−1} = (u^{−1} , −(1+u^{−1} x)^{−1} u^{−1} xu^{−1} ); hence, (u, x) ∈ U(I(R; S)). Write (a, s)^{n} = (f, y) + (v, w), (f, y)^{2} = (f, y) and (v, w) ∈ U(I(R; S)). Then f = f^{2} ∈ R, y = 0 and v ∈ U(R). Clearly, a^{n} = f + v. Further, x = y + w = w. This implies that f = e, v = u, and so (f, y) = (e, 0), (v, w) = (u, x). As a result, (a, s) ∈ I(R; S) is uniquely π-clean, and so I(R; S) is uniquely π-clean.

Assume that I(R; S) is uniquely π-clean and S is idempotent-free. Then R is uniquely π-clean. Let e = e^{2} ∈ R and s ∈ S. In view of Lemma 2.1, (e, 0) = (e, 0)^{2} ∈ I(R; S) is central. Hence, (e, 0)(0, s) = (0, s)(e, 0), and so es = se. For any s ∈ S, there exists some n ∈ ℕ such that (1, s)^{n} ∈ I(R; S) is uniquely clean. Write (1, s)^{n} = (1, x) = (f, y)+(u, v) where x ∈ S, (f, y) ∈ I(R; S) is an idempotent and (u, v) ∈ I(R; S) is a unit. Clearly, f = 0, and so y = 0. This implies that x = y + v = v; hence, (1, x) ∈ I(R; S) is a unit. Further, (1, s) ∈ I(R; S) is a unit. Write (1, s)^{−1} = (1, s′) for a s′ ∈ S. Then ss′ = s′s and s + s′+ ss′ = 0, hence the result. □

Corollary 4.1. Let R be uniquely π-clean. Then S = {(a_{ij} ) ∈ T_{n}(R) | a_{11} =

⋯= a_{nn} } is uniquely π-clean.

Proof. Let T = {(a_{ij} ) ∈ T_{n}(R) | a_{11} = ⋯= a_{nn} = 0}. Then S ≅ I(R; T). Then the result follows by Theorem 4.1. □

A ring R is called potently J-clean if for any a ∈ R there exists a potent p ∈ R such that a − p ∈ J(R). We shall show that such rings form a subclass of uniquely π-clean rings. A ring R is an exchange ring if and only if R/J(R) is an exchange ring, and every idempotent lifts modulo J(R). We have

Lemma 4.1. Every potently J-clean ring is an exchange ring.

Proof. Let R be a potently J-clean ring. Then R/J(R) is potent, and so it is an exchange ring. Let \overline{e} ∈ R/J(R) be an idempotent. Then we have a potent p ∈ R such that w := e − p ∈ J(R). Write p = p^{n} for some n ⩾ 2. Then p^{n−1} ∈ R is an idempotent. Moreover, e = p + w, and so e^{n−1} = p^{n−1} + v for some v ∈ J(R). But e − e^{n−1} ∈ J(R). Hence, e − p^{n−1} = (e − e^{n−1} ) + (e^{n−1} − p^{n−1} ) ∈ J(R). So idempotents can be lifted modulo J(R). Therefore R is an exchange ring. □

Theorem 4.2. Every Abelian potently J-clean ring is uniquely π-clean.

Proof. Let R be an Abelian potently J-clean ring. Then R is an exchange ring by Lemma 4.1. Thus, every idempotent in R lifts modulo J(R). For any a ∈ R, there exists a potent p ∈ R such that a − p ∈ J(R). This implies that ā ∈ R/J(R) is potent, and so R/J(R) is potent. According to Lemma 2.3, R is uniquely π-clean. □

Corollary 4.2. Let R be Abelian. If for any sequence of elements {a_{i}} ⊆ R there exists a k ∈ ℕ and n_{1}, ⋯, n_{k} ⩾ 2 such that (a_{1} − a_{1}^{n_{1}} ) ⋯ (a_{k} − a_{k}^{n_{k}} ) = 0, then R is uniquely π-clean.

Proof. For any a ∈ R, we have a k ∈ ℕ and n_{1}, ⋯, n_{k} ⩾ 2 such that (a − a^{n_{1}} ) ⋯ (a − a^{n_{k}} ) = 0. This implies that a^{k} = a^{k+1} f(a) for some f(t) ∈ ℤ[t]. By Herstein’s theorem, R is periodic. Therefore every element in R is the sum of a potent element and a nilpotent element.

Clearly, R/J(R) is isomorphic to a subdirect product of some primitive rings R_{i}. Case 1. There exists a subring S_{i} of R_{i} which admits an epimorphism ϕ_{i} :

S_{i} → M_{2}(D_{i}) where D_{i} is a division ring.

Case 2. R_{i} ≅ M_{m_{i}} (D_{i}) for a division ring D_{i}. Clearly, the hypothesis is inherited by all subrings, all homomorphic images and all corners of R, we claim that, for any sequence of elements {a_{i}} ⊆ M_{2}(D_{i}) there exists s ∈ ℕ and m_{1}, …, m_{s} ⩾ 2 such that (a_{1} − a_{1}^{m_{1}} ) ⋯ (a_{s} − a_{s}^{m_{s}} ) = 0. Choose a_{i} = e_{12} if i is odd and a_{i} = e_{21} if i is even. Then (a_{1} −a_{1}^{m_{1}} )(a_{2} −a_{2}^{m_{2}} ) ⋯ (a_{s} −a_{s}^{m_{s}} ) = a_{1}a_{2} ⋯ a_{s} ≠ 0, a contradiction. This forces m_{i} = 1 for all i. We infer that each R_{i} is reduced, and then so is R/J(R). If a ∈ N(R), we have some n ∈ ℕ such that a^{n} = 0, and thus ā^{n} = 0 is R/J(R). Hence, ā ∈ J(R/J(R)) = 0. This implies that a ∈ J(R), and so N(R) ⊆ J(R). Therefore R is potently J-clean, hence the result by Theorem 4.2. □

5. Uniquely π -nil Clean Rings

In this section, we explore uniquely π-nil-clean rings, and completely determine when a ring is uniquely π-nil-clean ring.

Lemma 5.1. Let R be a ring. Then the following statements are equivalent:

(1) R is uniquely π-nil-clean.

(2) R is an Abelian periodic ring.

Proof. (1) ⇒ (2). Let e ∈ R be an idempotent and r ∈ R. Choose a = e + er(1 − e). Then we can find some m ∈ ℕ such that a^{m} ∈ R is uniquely nil clean. As a = a^{m} = e + er(1 − e) = (e + er(1 − e)) + 0, by the uniqueness, we get er(1 − e) = 0, and so er = ere. Likewise, re = ere, and so er = re. Therefore R is Abelian. Let a ∈ R. Then there exists some n ∈ ℕ such that a^{n} = f + u, where f = f^{2} ∈ R and u ∈ N(R). Hence, a^{2n} = f + v for a v ∈ N(R) and uv = vu. This shows that a^{n} − a^{2n} = u − v ∈ N(R). Thus, we have a k ∈ ℕ such that a^{nk} = a^{nk+1} f(a) for some f(t) ∈ ℤ[t]. In light of Herstein’s theorem, R is periodic.

(2) ⇒ (1) Let a ∈ R. Since R is periodic, there exists some m ∈ ℕ such that a^{m} ∈ R is an idempotent. Write a^{m} = e + w where e = e^{2} ∈ R and w ∈ N(R). Then a^{m} − e = w ∈ N(R). As R is Abelian, we see that (a^{m} − e)^{3} = a^{m} − e. Thus, (a^{m} − e)(1 − (a^{m} − e)^{2}) = 1, and so a^{m} = e, as required. □

As every finite ring is periodic, it follows from Lemma 5.1 that every finite commutative ring is uniquely π-nil-clean, e.g., ℤ_{n}[α] = {a + bα | a, b ∈ ℤ_{n}, α = −\frac{1}{2} + \frac{\sqrt{3}}{2}i, i^{2} = −1\}.

The above observation leads us to the following result alluded to earlier.

Theorem 5.1. Let R be a ring. Then the following are equivalent:

(1) R is uniquely π-nil-clean.

(2) R is uniquely π-clean and J(R) is nil.

(3) For any a ∈ R, there exist some m ∈ ℕ and a unique idempotent e ∈ R such that a^{m} − e ∈ P (R).

Proof. (1) ⇒ (3). By virtue of Lemma 5.1, R is an Abelian periodic ring. In view of [2, Theorem 2], N(R) forms an ideal of R, and so N(R) = P (R). For any a ∈ R, there exists some m ∈ ℕ such that a^{m} is uniquely nil clean. Write a^{m} = e + w with e = e^{2} and w ∈ N(R). Therefore a^{m} − e ∈ P (R), as required.

(3) ⇒ (2). Let e ∈ R be an idempotent, and let r ∈ R. Then we have an idempotent f ∈ R such that er(1 − e) = f + w for a w ∈ P (R). Hence, 1 − f = 1 − er(1 − e) + w = (1 − er(1 − e))(1 + (1 + er(1 − e))w) ∈ U(R). We infer that f = 0, and so er(1 − e) = w ∈ P (R). But we have a unique expression e + er(1 − e) = e + er(1 − e) + 0 where er(1 − e), 0 ∈ P (R). By the uniqueness, we get e = e + er(1 − e), and so er = ere. Similarly, re = ere. Therefore er = re, i.e., R is Abelian.

Let x ∈ J(R). Write x = h + v with h = h^{2} ∈ R, v ∈ P (R). Then h = x − v ∈ J(R); hence that h = 0. It follows that J(R) = P (R). Accordingly, for any a ∈ R, there exist some m ∈ ℕ and a unique idempotent e ∈ R such that a^{m} − e ∈ J(R).

If x ∈ N(R), then we have an idempotent g ∈ R and a u ∈ P (R) such that x = g + u, and so g = x − u. As R is Abelian, we see that xu = ux, and then g ∈ N(R). This shows that g = 0. Consequently, x = u ∈ P (R) ⊆ J(R). We infer that N(R) ⊆ J(R). In light of Corollary 2.4, R is uniquely π-clean, as desired.

(2) ⇒ (1). In view of Lemma 2.1, R is Abelian. In view of Lemma 2.3, R/J(R) is potent. Let a ∈ R. Then ā = \overline{a^{m}} (m ⩾ 2), and so a − a^{m} ∈ J(R). As J(R) is nil, every idempotent lifts modulo J(R). Hence, we can find some n ∈ ℕ such that (a − a^{m})^{n} = 0, and so a^{n} = a^{n+1} f(a) for some f(t) ∈ ℤ[t]. In terms of Herstein’s theorem, R is periodic. This completes the proof, by Lemma 5.1. □

Corollary 5.1. Let R be a ring. Then the following statements are equivalent:

(1) R is uniquely π-nil-clean.

(2) R/J(R) is potent, R is Abelian and J(R) is nil.

(3) For any a ∈ R, there exists some m ∈ ℕ and a central idempotent e ∈ R such that a^{m} − e ∈ P (R).

Proof. (1) ⇔ (2) is proved by Theorem 5.1 and Lemma 2.3.

(1) ⇒ (3) This is obvious, in view of Lemma 5.1 and Theorem 5.1.

(3) ⇒ (1). For any a ∈ R, there exist some m ∈ ℕ and a central idempotent e ∈ R such that a^{m} − e ∈ P (R). Write a^{m} − f ∈ P (R) for an idempotent f ∈ R. Then e − f = (a^{m} − f) − (a^{m} − e) ∈ P (R). As (e − f)^{3} = e − f, we conclude that e = f, and we are through by Theorem 5.1. □

Let n ⩾ 2 be a fixed integer. Following Yaqub [8], a ring R is said to be generalized n-like provided that for any a, b ∈ R, (ab)^{n} − ab^{n} − a^{n}b + ab = 0.

Proposition 5.1. Every generalized n-like ring is uniquely π-nil-clean.

Proof. Let a ∈ R. Then a^{2n} − 2a^{n+1} + a^{2} = 0, and so (a − a^{n})^{2} = 0. Thus, a − a^{n} ∈ N(R). Hence, a^{m} = a^{m+1} f(a) for some f(t) ∈ ℤ[t]. Accordingly, R is periodic by Herstein’s theorem.

Let e, f ∈ R. Since R is a generalized n-like ring, we have

((1 − e)f)^{n}e = ((1 − e)fe)^{n} − (1 − e)fe + (1 − e)fe = 0;
((1 − e)f)^{n} = (1 − e)f + (1 − e)f − (1 − e)f = (1 − e)f.

Reiterating in the last, we get (1−e)f = ((1−e)f)^{2n}, and so (1−e)fe = 0. Hence, fe = efe. Likewise, ef = efe. Therefore ef = fe. We infer that R is Abelian. Therefore we conclude that R is uniquely π-nil-clean, in terms of Lemma 5.1. □

Let R = \{( \begin{matrix}x & y & z \\ 0 & x^{2} & 0 \\ 0 & 0 & x \end{matrix} ) | x, y, z ∈ GF (4)\}. It is easy to check that for each a ∈ R,

a^{7} = a or a^{7} = a^{2} = 0. Therefore R is a generalized 7-like ring. By Proposition 5.1, R is uniquely π-clean which is a noncommutative periodic ring.

An element a ∈ R is uniquely weakly nil-clean provided that a or −a is uniquely nil-clean. A ring R is uniquely weakly nil-clean ring provided that every element in R is uniquely weakly nil-clean [5].

Lemma 5.2. Every uniquely weakly nil-clean ring is uniquely π-nil-clean.

Proof. Let R be a uniquely weakly nil-clean ring. In view of [5, Theorem 12], R is Abelian. Let a ∈ R. Then there exists an idempotent e ∈ R such that a − e ∈ N(R) or −a − e ∈ N(R). If a − e ∈ N(R), then a − a^{2} ∈ N(R). If −a − e ∈ N(R), then a + a^{2} ∈ N(R). In any case, a^{n} = a^{n+1} f(a) for some f(t) ∈ ℤ[t]. In view of Herstein’s theorem, R is periodic. Therefore R is uniquely π-nil-clean, in terms of Lemma 5.1. □

[5, Theorem 12] says that a ring R is a uniquely weakly nil-clean ring if and only if R is Abelian, J(R) is nil and R/J(R) is a Boolean ring, ℤ_{3} or the product of two such rings. We have

Theorem 5.2. A ring R is a uniquely weakly nil-clean ring if and only if

(1) R is uniquely π-nil-clean;

(2) R/J(R) is a Boolean ring, ℤ_{3} or the product of two such rings.

Proof. ⇒: In view of Lemma 5.2, proving (1). Further, proving (2) in terms of [4, Theorem 18].

⇐: As R is uniquely π-nil-clean, in view of Corollary 5.1, R is Abelian and J(R) is nil, then by (2) and in light of [5, Theorem 18], R is a uniquely weakly nil-clean ring. □

Corollary 5.2. A ring R is a uniquely weakly nil-clean ring if and only if for any a ∈ R, there exists a central idempotent e ∈ R such that a − e ∈ P (R) or a + e ∈ P (R).

Proof. ⇒: In view of Corollary 5.1, R is uniquely π-nil-clean. For any a ∈ R, by hypothesis, we see that ā or −ā is an idempotent in R/J(R). By virtue of [1, Theorem 1.12], R/J(R) is a Boolean ring, ℤ_{3} or the product of two such rings.

⇐: Let a ∈ R. By (2), there exists a central idempotent e ∈ R such that a − e ∈ P (R) or a + e ∈ P (R). Hence, a^{2} − e = (a − e)(a + e) ∈ P (R). Thus, R is uniquely π-nil-clean, by Theorem 5.1. Let x ∈ J(R). Then there exists a central idempotent f ∈ R such that x − f or x + f is in P (R). If x − f ∈ P (R), then f ∈ J(R), and so f = 0. This implies that x ∈ P (R). If x + f ∈ J(R), similarly, x ∈ P (R). Hence, J(R) ⊆ P (R). We infer that J(R) = P (R). Thus, R/J(R) is a Boolean ring, ℤ_{3} or the product of two such rings, by [1, Theorem 1.12]. In light of Theorem 5.2, the result follows. □

A ring R is uniquely nil-clean provided that every element in R is uniquely nil-clean. [5, Corollary 13] says that R is a uniquely nil-clean ring if and only if R is a uniquely weakly nil-clean ring and 2 ∈ J(R). Further, we derive

Corollary 5.3. A ring R is a uniquely nil-clean ring if and only if

(1) R is uniquely π-nil-clean;

(2) R/J(R) is a Boolean ring.

Proof. ⇒: Clearly, R is uniquely π-nil-clean. In view of [3, Theorem 4.5], R/J(R) is Boolean.

⇐: By virtue of Theorem 5.2, R is a uniquely weakly nil-clean ring. As \overline{2}^{2} = \overline{2}, we see that 2 ∈ J(R). Therefore R is a uniquely nil-clean ring, in terms of [5, Corollary 13]. □

Acknowledgements. The authors are grateful to the referee for his/her helpful comments and suggestions which led to much improved paper. Huanyin Chen was supported by the Natural Science Foundation of Zhejiang Province, China (No. LY17A010018).

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Department of Mathematics (Received 31 05 2015) Hangzhou Normal University (Revised 09 01 2016) Hangzhou China

   huanyinchen@aliyun.com

Women’s University of Semnan (Farzanegan)

Semnan Iran

   m.sheibani1@gmail.com