P-CLEAN RINGS

Throughout this paper R denotes an associative ring with identity and all modules are unitary. We use the symbol U(R) to denote the group of units of R and Id(R) the set of idempotents of R, Un(R) the set of elements which are the sum of n units of R, UΣ(R) the set of elements each of which is the sum of finitely many units in R, RE(R) (URE(R)) the set of regular (unit regular) elements of R, and Peri(R) the set of periodic elements of R. The Jacobson radical and the prime radical of R are denoted by J(R) and Nil∗(R), respectively. Following Han and Nicholson [4], an element x of a ring R is called clean if x = e+u where e ∈ Id(R) and u∈U(R). A ring R is clean if every element of R is clean. This notion was first introduced by Nicholson [5] as early as 1977 in his study of lifting idempotents and exchange rings. Since then, a great deal is known about clean rings and their generalizations (cf. [1–9]). According to Ye [9], a ring R is called semiclean if every element of R has the form x = f + u, where u ∈ U(R) and f is periodic, that is, f p = f q for two different positive integers p and q . In [8], an element x of a ring R is called n-clean if x = e+u1 + ···+un where e ∈ Id(R), ui ∈ U(R), and n is a positive integer. The ring R is called n-clean if every element of R is n-clean for some fixed positive integer n. While R is called Σ-clean, if the n is a positive integer depending on x. Also Zhang and Tong in [10] defined R to be G-clean, if each x ∈ R has the form x = a+u where a is unit regular and u∈U(R). Motivated by the results of Han and Nicholson [4] on clean rings, Ye [9] on semiclean rings, Xiao and Tong [8] on n-clean rings and Σ-clean rings, and Zhang and Tong [10] on G-clean rings, in this paper we unify the structures of various clean rings by introducing


Introduction
Throughout this paper R denotes an associative ring with identity and all modules are unitary.We use the symbol U(R) to denote the group of units of R and Id(R) the set of idempotents of R, U n (R) the set of elements which are the sum of n units of R, U Σ (R) the set of elements each of which is the sum of finitely many units in R, RE(R) (URE(R)) the set of regular (unit regular) elements of R, and Peri(R) the set of periodic elements of R. The Jacobson radical and the prime radical of R are denoted by J(R) and Nil * (R), respectively.
Following Han and Nicholson [4], an element x of a ring R is called clean if x = e + u where e ∈ Id(R) and u ∈ U(R).A ring R is clean if every element of R is clean.This notion was first introduced by Nicholson [5] as early as 1977 in his study of lifting idempotents and exchange rings.Since then, a great deal is known about clean rings and their generalizations (cf.[1][2][3][4][5][6][7][8][9]).
According to Ye [9], a ring R is called semiclean if every element of R has the form x = f + u, where u ∈ U(R) and f is periodic, that is, f p = f q for two different positive integers p and q .In [8], an element x of a ring R is called n-clean if x = e + u 1 + ••• + u n where e ∈ Id(R), u i ∈ U(R), and n is a positive integer.The ring R is called n-clean if every element of R is n-clean for some fixed positive integer n.While R is called Σ-clean, if the n is a positive integer depending on x.Also Zhang and Tong in [10] defined R to be G-clean, if each x ∈ R has the form x = a + u where a is unit regular and u ∈ U(R).
Motivated by the results of Han and Nicholson [4] on clean rings, Ye [9] on semiclean rings, Xiao and Tong [8] on n-clean rings and Σ-clean rings, and Zhang and Tong [10] on G-clean rings, in this paper we unify the structures of various clean rings by introducing the notion of P-clean rings and the common properties of those rings.By the way, we answer a question of Xiao and Tong [8] in the negative and extend some known results of [8,9].

P-clean rings
We start this section by the following definitions.
For two subsets A and B of a ring R, the sum of A and B is defined as follows: The sum of more than two subsets of an R can be defined inductively.
Let P be a property which is meaningful for elements of a ring.For any ring R, let P(R) be the subset {a ∈ R | a has property P} of R. Definition 2.1.Property P will be called admissible if the following conditions are satisfied.
(3) For any e ∈ Id(R), P(eRe For convenience, an element of P(R) is called a P-element in R. In this paper P will always be an admissible property.
(2) We prove this by using induction on n.In fact, the case n = 2 is condition (3) of Definition 2.1.Assume (2) holds for n − 1.Let e 1 + e 2 + ••• + e n−1 = f .Then multiplied by e i on the two sides of the above equation, we have e i f = f e i = e i , which gives e i = f e i f and so e i ∈ Id( f R f ).Note that f R f is a ring with identity f .It yields that P(e 1 Re 1 ) + P(e 2 Re 2 ) + Lemma 2.4.Let R be a ring and e ∈ Id(R).Then the following hold. ( Proof.We only prove ( 3) and ( 8), the others are very similar. ( Then there exist positive integers m > n and p > q such that f m = f n and g p = g q .By Ye [9, Lemma 5.2], f n(m−n) and g q(p−q) are both idempotents.Set t = 2n(m − n) q(p − q).Then f 2t = f t and g 2t = g t .Since It is easy to show that an n-clean element is m-clean whenever n ≤ m, since for any e ∈ Id(R), e = (1 − e) + (2e − 1) where 1 − e ∈ Id(R) and (2e − 1) 2 = 1.So without loss of generality, we can assume n = m.Using (1), f + g ∈ Id(R).And from (6) Using Lemma 2.4, it is easy to check that for any ring R, 0, R, are all subsets of R defined by a suitable admissible property P.
From the above arguments, the following proposition is immediate.Proposition 2.5.Let R be a ring.Then the following conclusions hold.
(2) Pri(R)-clean rings are precisely semiclean rings.Note that here an (S,2)-ring is a ring in which every element can be expressed as a sum of two units of R. While in some literature it referred to a ring in which every element can be written as a sum of no more than two units.Proposition 2.6.Any homomorphic image of a P-clean ring is P-clean.
It should be noted that Proposition 2.7 is not true for an infinite direct product of rings R i .For example, the ring Z of integers is a Σ-clean ring, but R = ∞ i=1 Z is not Σclean since (1,2,...,n,...) is obviously not Σ-clean.

Proposition 2.8. The ring R is P-clean if and only if the ring R[[x]] of formal power series over R is
The following corollary extends [8, Proposition 2.5] which states that for a commutative ring [4] that if e is an idempotent in a ring R such that eRe and (1 − e)R(1 − e) are both clean rings, then R is clean.Hence the ring of n × n matrices over R is clean.Similar results hold for semiclean rings, n-clean rings, and Σ-clean rings.We now extend these results to P-clean rings.

It has been proved by Han and Nicholson in
Note that p + q ∈ P(eRe) + P(ēRē) ⊆ P(R) by Definition 2.1, the proof is complete.
Using Lemma 2.10, an inductive argument gives immediately.The following two results are direct consequences of Theorem 2.11 Corollary 2.12.If R is a P-clean ring, so also is the matrix ring M n (R).
Since any homomorphic image of a P-clean ring is again P-clean, with Theorem 2.11, this gives the following.
Corollary 2.14.If A and B are rings and V = A V B is a bimodule, the split-null extension R is P-clean if and only if both A and B are P-clean, where In particular, induction shows that for each n ≥ 1, a ring R is P-clean if and only if the ring of all n × n upper triangular matrices over R is P-clean.
Let R be a ring and let I be an ideal of R. We say P-elements in R/I lift modulo I, if for any p ∈ P(R/I) there exists a ∈ P(R) such that π(a) = p where π is the canonical ring homomorphism from R onto R/I.
We close this section with the following proposition whose proof is very easy.
Proposition 2.15.Let R be a ring and let I be an ideal contained in J(R).If R/I is a P-clean ring and P-elements lift modulo I, then R is a P-clean ring.

Some remarks
It is known by [4,Proposition 6] that a ring R is clean if and only if R/I is clean for any ideal I ⊆ J(R) and idempotents lift modulo I. Xiao and Tong [8] naturally claimed that they do not know whether for any n-clean ring R, idempotents of R/I lift modulo I where I is any ideal of R contained in J(R).The following counterexample shows that the answer is negative.

Example 3.1.
There is a 4-clean ring R in which idempotents of R/J(R) cannot be lifted to R.
Proof.Let R be the subring of rational numbers Q given by R . Then R has only two maximal ideals: 2R and 3R, so J(R) = 6R.Denote the ring of integers modulo n by Z n , then R/J(R) ∼ = Z 2 × Z 3 , which has four idempotents.But R has only two idempotents.This shows that idempotents of R/J(R) cannot be lifted to R modulo J(R).But it can be shown that R is a 4-clean ring.
The following two results are obtained by Xiao and Tong [8] for commutative rings, now we extend them to 2-primal rings (rings whose prime radical coincides with the set of nilpotent elements).Proposition 3.2.For any 2-primal ring R, the polynomial ring R[x] is not Σ-clean.Proof.Assume the contrary, then x = e(x) Since R is 2-primal, a polynomial over R is invertible if and only if its constant term is in U(R) and the other coefficients are in Nil * (R) by [3,Theorem 2.4], so u i j ∈ Nil * (R) for each j ≥ 1. Hence x = e(x) + u 1 (x) + ••• + u n (x) gives a 1 + u 11 + ••• + u n1 = 1, so a 1 is a unit in R, and a 2 + u 12 + ••• + u n2 = 0 implies a 2 ∈ Nil * (R).On the other hand, e(x) 2 = e(x) implies e 2 0 = e 0 , and e(x) 2 = e 0 + (e 0 a 1 + a 1 e 0 ) x + (e 0 a 2 + a 2 1 + a 2 e 0 ) x 2 + ••• + a 2 m x 2m .So a 2 = e 0 a 2 + a 2 1 + a 2 e 0 by comparing the coefficient of x 2 in e(x) 2 = e(x).Note that the sum of a unit and a nilpotent element must be a unit and e 0 a 2 + a 2 e 0 ∈ Nil * (R).It follows that a 2 ∈ U(R).This is a contradiction, and the proof is complete.From Proposition 3.2, the following corollary is immediate.Corollary 3.3.For any 2-primal ring R, the polynomial ring R[x] is not n-clean.
We conclude this paper with the following proposition.Proposition 3.4.For any 2-primal ring R, the polynomial ring R[x] is not semiclean.
Proof.Assume the contrary, then x = p(x) + u(x) where p(x) is a periodic element and u(x) is a unit.Let p(x) = p 0 + p 1 x + ••• + p n x n and u(x) = u 0 + u 1 x + ••• + u n x n .Since R is 2-primal, u i ∈ Nil * (R) for each i ≥ 1 by [3,Theorem 2.4].By comparing the coefficient of x = p(x) + u(x), we have p 1 + u 1 = 1, which implies p 1 is a unit in R, and p i + u i = 0 gives p i ∈ Nil * (R) for each i ≥ 2. Clearly we can assume that p(x) s = p(x) t for positive integers s > t ≥ 2. Then a routine calculation shows that the coefficient of x s in p(x) s is i1+i2+•••+is=s p i1 p i2 ... p is = p s 1 + a for some a ∈ Nil * (R).Comparing the coefficients of x s on two sides of p(x) s = p(x) t , we have p 1 ∈ Nil * (R), which is a contradiction.
The above result is obtained by Ye [9] only for a commutative ring.

Lemma 2 . 10 .
Let e ∈ Id(R) be such that eRe and (1 − e)R(1 − e) are both P-clean rings.Then R is a P-clean ring.Proof.For convenience, write r = 1 − r for each r ∈ R. We use the Pierce decomposition of the ring R: R = eRe + eRē + ēRe + ēRē.(2.1)Let x = a + b + c + d where a ∈ eRe, b ∈ eRē, c ∈ ēRe, and d ∈ ēRē.By hypothesis, write a = p + u where p ∈ P(eRe) and u ∈ U(eRe) with inverse u 1 .Then
Re 1 ) + P(e 2 Re 2 ) + ••• + P(e n−1 Re n−1 ) + P(e n Re n Theorem 2.11.If 1 = e 1 + e 2 + ••• + e n in a ring R where e i are orthogonal idempotents and each e i Re i is P-clean, then R is P-clean.