Home Rings in which elements are the sum of a nilpotent and a root of a fixed polynomial that commute
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Rings in which elements are the sum of a nilpotent and a root of a fixed polynomial that commute

  • Ali H. Handam EMAIL logo and Hani A. Khashan
Published/Copyright: April 17, 2017
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Open Mathematics
From the journal Open Mathematics Volume 15 Issue 1

Abstract

An element in a ring R with identity is said to be strongly nil clean if it is the sum of an idempotent and a nilpotent that commute, R is said to be strongly nil clean if every element of R is strongly nil clean. Let C(R) be the center of a ring R and g(x) be a fixed polynomial in C(R)[x]. Then R is said to be strongly g(x)-nil clean if every element in R is a sum of a nilpotent and a root of g(x) that commute. In this paper, we give some relations between strongly nil clean rings and strongly g(x)-nil clean rings. Various basic properties of strongly g(x) -nil cleans are proved and many examples are given.

Keywords: Nil clean ring; Strogly nil clean ring; g(x)-nil clean ring; Strongly g(x)-nil clean ring
MSC 2010: 16N40; 16U99

1 Introduction

Let R be an associative ring with identity. Following [10], we define an element r of a ring R to be clean if there is an idempotent e ∈ R and a unit u ∈ R such that r = u + e. A clean ring is defined to be one in which every element is clean. Clean rings were first introduced by Nicholson [10] as a class of exchange rings. In [11], an element r of a ring R is defined to be strongly clean provided that there exist an idempotent e ∈ R and a unit u ∈ R such that r = e + u and eu = ue. Analogously, a strongly clean ring is one in which every element is strongly clean. In [7], Diesl introduced the concept of strongly nil cleanness. An element r ∈ R is strongly nil clean provided that there exist an idempotent e ∈ R and a nilpotent element n ∈ R such that r = n + e and ne = en. Recently, the concept of nil clean rings was generalized to g(x)-nil clean rings (see Khashan and Handam [9]). We refer the reader to [1, 4–6, 14] for a survey on nil clean rings.

Let C(R) denote the center of a ring R and g(x) be a polynomial in C(R)[x]. Following Camillo and Simón [2], R is called g(x)-clean if for each r ∈ R, there exist a unit u ∈ R and s ∈ R such that r = s + u and g(s) = 0. Of course (x2 − x)-clean rings are clean rings.

Nicholson and Zhou [12] proved that if g(x) ∈ (x − a)(x − b)C(R)[x] where a, b ∈ C(R) are such that b and b − a are both units in R and RM is a semisimple left R-module, then End(RM) is g(x)-clean. In 2008, Fan and Yang [8], studied more properties of g(x)-clean rings. Among many conclusions, they proved that if g(x) ∈ (x − a)(x − b)C(R)[x] where a, b ∈ C(R) with b − a a unit in R, then R is a clean ring if and only if R is (x − a)(x − b)-clean. Recently, Khashan and Handam [9] completely determined the relation between nil clean rings and g(x)-nil clean rings.

In this paper, we define and study strongly g(x)-nil clean rings. For a ring R and g(x) ∈ C(R)[x], We say that r ∈ R is strongly g(x)-nil clean provided that there exist a nilpotent b ∈ R and s ∈ R such that r = s + b, g(s) = 0 and sb = bs. Moreover, R is called strongly g(x)-nil clean if every element in R is strongly g(x)-nil clean.

In section 2, we define the concept of strongly g(x)-nil clean and some general properties of strongly g(x)-nil clean rings are given. In section 3, we determine the relation between strongly nil clean rings and strongly g(x)-nil clean rings and we give many characterizations for a strongly nil clean ring R in terms of some strongly g(x)-nil clean rings.

Throughout this article, U(R), Id(R) and N(R) will denote, respectively, the group of units, the set of idempotents and the set of nilpotent elements in R. We write Mn(R) and Tn(R) for the rings of all n × n matrices and all upper triangular matrices over R, respectively. The ring of integers modulo n is denoted by ℤn.

2 Strongly g(x)-nil clean rings

We study in this section the basic properties of strongly g(x)-nil clean rings, and some examples are given.

Definition 2.1

Let R be a ring and g(x) be a fixed polynomial in C(R)[x]. An element r ∈ R is called strongly g(x)-nil clean if r = n + s where n ∈ N(R), g(s) = 0 and ns = sn. A ring R is called strongly g(x)-nil clean if each of its elements is strongly g(x)-nil clean.

Clearly, every strongly g(x)-nil clean ring is g(x)-nil clean. Moreover, the two concepts coincide in the commutative case. However, the converse need not be true in non commutative rings. For example, one can see that the ring M2(ℤ2) is (x4 + x2)-nil clean which is not strongly (x4 + x2)-nil clean.

Proposition 2.2

Every strongly g(x)-nil clean ring is strongly g(x)-clean.

Proof

Suppose R is a strongly g(x)-nil clean ring and let r ∈ R. Write r − 1 = n + s where n ∈ N(R), g(s) = 0 and ns = sn. Then r = (n + 1) + s where n + 1 ∈ U(R) and (n + 1)s = s(n + 1) . Hence, R is strongly g(x)-clean.     □

The converse of Proposition 2.2 is not true in general. For example, simple computations show that the non strongly (x4 + x2)-nil clean ring M2(ℤ2) is strongly (x4 + x2)-clean.

For any two rings R and S, consider the ring homomorphism η : C(R) → C(S) with η(1R) = 1S. If g(x)= ∑i=0naixi∈C(R)[x], we let gη(x):=∑i=0nη(ai)xi∈C(S)[x]. In particular, if g(x) ∈ ℤ[x], then gη(x) = g(x).

Next, we give some properties of the class of strongly g(x)-nil clean rings. The proof of the following proposition is straightforward.

Proposition 2.3

Let R and S be two rings, η : R → S be a ring epimorphism and g(x) ∈ C(R)[x]. If R is strongly g(x)-nil clean, then S is strongly gη(x)-nil clean.

Corollary 2.4

Let R be a ring.

  1. If R is a strongly g(x)-nil clean ring and I is an ideal of R, then R/I is strongly gη(x)-nil clean where η : R → R/I is defined by η(r) = r + I.

  2. If the formal power series ring R[[t]] is strongly g(x)-nil clean, then R is strongly gη(x)-nil clean where

    η : R[[t]] → R is defined by η(∑i=0∞aiti)=a0.

Corollary 2.5

Let R be a ring and 1 < n ∈ ℕ. If Tn(R) is strongly g(x)-nil clean, then R is strongly gη(x)-nil clean.

Proof

Define η : Tn(R) → R by η(A) = a11 where A = (aij) ∈ Tn(R). Then clearly η is a ring epimorphism and R is strongly gη(x)-nil clean.     □

Proposition 2.6

Let R1, R2, Rk be rings and g(x) ∈ ℤ[x]. Then R=∏i=1kRi is strongly g(x)-nil clean if and only if Ri is strongly g(x)-nil clean for all i ∈ {1, 2, …, k}.

Proof

⇒) : For each i ∈ {1, 2, …, k}, Ri is a homomorphic image of ∏i=1kRi under the projection homomorphism. Hence, Ri is strongly g(x)-nil clean by Proposition 2.3.

⇐) : Let r=(x1,x2,...,xk)∈∏i=1kRi. For each i, write xi = ni + si where ni ∈ N(Ri), g(si) = 0 and nisi = sini. Let n = (n1, n2, nk) and s = (s1, s2, sk). Then r = n + s where clearly n ∈ N(R), g(s) = 0 and ns = sn. Therefore, R is strongly g(x)-nil clean.     □

Proposition 2.7

If R is a commutative ring, then the ring of polynomials R[t] is not strongly nil clean (and hence not strongly (x2 − x)-nil clean).

Proof

It is well known that N(R[t]) = {a0 + a1t + … + aktk : ai ∈ N(R)} and Id(R[t]) = Id(R). Suppose t = (a0 + a1t + … + aktk) + e where ai ∈ N(R) and e2 = e. Then a1 = 1 ∈ N(R) which is a contradiction. Therefore, R[t] is not strongly nil clean.     □

The converse of (1) in Corollary 2.4 is not true in general. For example, the field ℤ2 ≃ ℤ2[x]/〈x〉 is strongly (x2 − x)-nil clean but ℤ2[x] is not strongly (x2 − x)-nil clean by Proposition 2.7.

Let A and B be rings, BMA be a bimodule and consider the formal triangular matrix T=A0MB. If g(x)= a00m0b0+a10m1b1x+...+an0mnbnxn∈C(T)[x], then clearly gA(x) = a0 + a1x + … + anxn ∈ C(A)[x] and gB(x) = b0 + b1x + … + bnxn ∈ C(B)[x].

Proposition 2.8

Let A and B be rings and let BMA be a bimodule. If T=A0MB is strongly g(x)-nil clean for some g(x) ∈ C(T)[x], then A is strongly gA(x)-nil clean and B is strongly gB(x)-nil clean.

Proof

Suppose that T=A0MB is strongly g(x)-nil clean. For every a ∈ A, b ∈ B, m ∈ M, write a0mb=n10n2n3+s10s2s3wheren10n2n3∈N(T),g(s10s2s3)=0andn10n2n3s10s2s3=s10s2s3n10n2n3. Then a = n1 + s1 and b = n3 + s3. It is easy to see that n1 ∈ N(A), n3 ∈ N(B), gA(s1) = gB(s3) = 0, n1s1 = s1n1 and n3s3 = s3n3. Therefore, A is a strongly gA(x)-nil clean and B is a strongly gB(x)-nil clean.     □

3 Strongly (x2 + cx + d)-nil clean rings

In this section, we consider some types of strongly (x2 + cx + d)-nil clean rings and we will investigate the relation between strongly g(x)-nil clean rings and strongly nil clean rings.

Theorem 3.1

Let R be a ring and a, b ∈ C(R). Then R is strongly nil clean with b − a ∈ N(R) if and only if R is strongly (x − (a + 1))(x − b)-nil clean.

Proof

⇒) : Suppose R is strongly nil clean with b − a ∈ N(R) and let r ∈ R. Then r−(a+1)(b−a)−1=e+n, where e2 = e, n ∈ N(R) and en = ne. Therefore, r = [e((b − a) − 1) + (a + 1)] + n((b − a) − 1). Write s = [e((b − a) − 1) + (a + 1)] and v = n((b − a) − 1). Then r = s + v where clearly v ∈ N(R) and sv = vs. Moreover, s is a root of (x − (a + 1))(x − b) since

[e((b−a)−1)+(a+1)−(a+1)][e((b−a)−1)+(a+1)−b]=e2((b−a)−1)2−e((b−a)−1)((b−a)−1)=0

Hence, R is strongly (x − (a + 1))(x − b)-nil clean.

⇐) : Suppose R is a strongly (x − (a + 1))(x − b)-nil clean. Write a = s + n where n ∈ N(R), (s − (a + 1))(s − b) = 0 and sn = ns. Then, s − a ∈ N(R) and so s − (a + 1) ∈ U(R). Therefore, s = b and b − a ∈ N(R). Now, let r ∈ R. By the assumption, we can write r((b − a) − 1) + (a + 1) = s + n where n ∈ N(R), s is a root of (x − (a + 1))(x − b) and sn = ns. Thus, r = t + m where t = (s − (a + 1))((b − a) − 1)−1 and m = n((b − a) − 1)− 1. Since a, b ∈ C(R), it is clear that m ∈ N(R) and tm = mt. Moreover,

t2=((s−(a+1))((b−a)−1)−1)2=(s−(a+1))(s−b+b−(a+1))((b−a)−1)−2=(s−(a+1))(s−b)+(s−(a+1))(b−(a+1))((b−a)−1)−2=(s−(a+1))((b−a)−1)−1=t

Therefore, R is strongly nil clean.     □

Lemma 3.2

([7]). If a ring R is nil clean, then 2 is a (central) nilpotent element in R.

As 2 is a central nilpotent in any (strongly) nil clean ring R, then 2n ∈ N(R) for any integer n. On the other hand, a non zero strongly g(x)-nil clean ring can’t have a nilpotent of the form 2n + 1 since otherwise 1 ∈ N(R) which is a contradiction. Next, we give some special cases of Theorem 3.1.

Corollary 3.3

Let R be a ring and n ∈ ℤ. For any b ∈ C(R), the following conditions are equivalent

  1. R is strongly nil clean.

  2. R is strongly (x2 − (2b + 1)x + b(b + l))-nil clean.

  3. R is strongly (x2 − (2b + 1 − 2n)x + (b2 + b(1 − 2n))-nil clean.

  4. R is strongly (x2 − (2b + 1 + 2n) x + (b2 + b(1 + 2n))-nil clean.

Proof

In Theorem 3.1, we take a = b to get (1) ⇔ (2), a = b − 2n to get (1) ⇔(3) and a = b + 2n to get (1) ⇔(4).     □

For example, strongly nil clean, strongly (x2 − 3x + 2)-nil clean, strongly (x2 − 5x + 6)-nil clean, strongly (x2 + x)-nil clean, strongly (x2 + 3x)-nil clean and strongly (x2 − 3x)-nil clean rings are all equivalent.

Remark 3.4

The equivalence in Theorem 3.1 of strongly nil clean rings and (x − (a + 1))(x − b)-nil clean rings is a global property. That is, it holds for a ring R but it may fail for a single element. For example, 1 ∈ ℤ10 is (strongly) nil clean but it is not (strongly) (x2 + x)-nil clean in ℤ10.

Let R be a strongly g(x)-nil clean ring and suppose 0 = s + n where n ∈ N(R), g(s) = 0 and ns = sn. If s is a unit, then clearly R = {0}. In particular, if g(x) = xn − 1 or g(x) = xn + 1, then the only strongly g(x)-nil clean ring is the zero ring. Therefore, we give the following definition.

Definition 3.5

A ring R is called strongly g(x)-nil* clean if for every 0 ≠ r ∈ R, r = s + n where n ∈ N(R), g(s) = 0 and ns = sn.

It is clear that every strongly g(x)-nil clean ring is strongly g(x)-nil*clean. Moreover, the converse is true if there is a non unit root s of g(x) such that 0 = s + b for some b ∈ N(R). In particular, strongly g(x)-nil clean rings and strongly g(x)-nil* clean rings coincide if all roots of g(x) are non unit. However, there are examples of strongly g(x)-nil*clean rings which are not strongly g(x)-nil clean.

Example 3.6

For any prime integer p, the field ℤp is strongly (xp − 1 − 1)-nil* clean which is not strongly (xp − 1 − 1)-nil clean. In fact by Fermat’s Theorem, any non zero element in ℤp is a root of xp − 1−1 and so ℤp is strongly (xp − 1−1)-nil* clean. On the other hand, the element 0 can’t be written as a sum of a nilpotent and a root of xp − 1 − 1. Therefore, ℤp is not (xp − 1 − 1)-nil clean.

In general, we prove in the next example that for any polynomial g(x) ∈ C(R)[x] with a unit constant term, R is not strongly g(x)-nil clean.

Example 3.7

Let g(x) = xn + an − 1xn − 1 + … + a1x + a0 ∈ C(R)[x] where R is a non zero ring and n ∈ ℕ. If a0 ∈ U(R), then R is not strongly g(x)-nil clean. In particular, R is not strongly (xn ± 1)-nil clean.

Proof

Suppose R is strongly g(x)-nil clean. If we write 0 = s + r where r ∈ N(R), g(s) = 0 and rs = sr, then s(sn − 1 + an − 1sn−2 + … + a1) = −a0 ∈ U(R) and so s ∈ U(R) . But, also s = −r ∈ N(R), so 1 = 0 and R = {0}, a contradiction.     □

Next, for any n ∈ ℕ, we give in the following proposition a characterization of strongly (xn − 1)-nil*clean rings. The proof is trivial.

Proposition 3.8

Let R be a ring and n ∈ ℕ. Then R is strongly (xn − 1)-nil* clean if and only if for every 0 ≠ r ∈ R, r = v + b where b ∈ N(R), vn = 1 and vb = bv.

Since in any ring, the sum of a unit and a nilpotent that commute is again a unit, then we conclude by Proposition 3.8 that any strongly (xn − 1)-nil* clean ring is a division ring.

Proposition 3.9

Let R be a ring and 2 ≤ n ∈ ℕ. If R is strongly (xn − 1 − 1)-nil* clean, then R is strongly (xn − x)- nil clean. Moreover, the converse is true if for every 0 ≠ a ∈ R, (a + 1)R (or\ R(a + 1)) does not contain non trivial idempotents.

Proof

Suppose R is strongly (xn − 1 − 1)-nil*clean. If r = 0, then clearly r is a strongly (xn − x)-nil clean element. Suppose 0 ≠ r ∈ R. Then r = v + a where a ∈ N(R) and vn − 1 = 1 and so v is a root of xn − x. Therefore, R is strongly (xn − x)-nil clean. Now, suppose that for every a ∈ R, (a + 1)R contains no non trivial idempotents and R is strongly (xn − x)-nil clean. Let 0 ≠ r ∈ R and write r = s + a where a ∈ N(R), sn = s and sa = as. Then rsn − 1 = s + asn − 1 and so r (1 − sn − 1) = a(1 − sn − 1). If we set y = 1 + a, then y ∈ U(R) and (r + 1)(1 − sn − 1) = (a + 1)(1 − sn − 1) = y (1 − sn − 1). Now, clearly y(1 − sn − 1)y− 1 is an idempotent with y(1 − sn − 1)y− 1 = (r + 1)(1 − sn − 1)y− 1 ∈ (r + 1)R. By assumption, y(1 − sn − 1)y− 1 = 0 and so 1 − sn − 1 = 0. Therefore, s is a root of xn − 1 − 1 and R is strongly (xn − 1 − 1)-nil*clean.     □

For a ring R and n ∈ ℕ, let Un (R) denotes the set of elements in R that can be written as a sum of no more than n units.

Proposition 3.10

Let R be a ring and g(x) = xn + an−1xn − 1 + … + a1x + a0 ∈ C(R)[x] where n ∈ ℕ and a0 ∈ U(R). If R is strongly g(x)-nil* clean, then R = U2(R). In particular if R is strongly (xn − 2 + xn − 3 + … + x + 1)-nil* clean, then R = U2(R) is strongly (xn − x)-nil clean.

Proof

Let 1 ≠ r ∈ R and write r − 1 = s + b where b ∈ N(R), sn + an − 1 sn − 1 + … + a1s + a0 = 0 and sb = bs. Then r = s + (b + 1) where b + 1 ∈ U(R). Since also, s(sn − 1 + an − 1 sn − 2 + … +a1) = −a0 ∈ U(R), then s ∈ U(R). Thus, r ∈ U2(R). Since also 1 ∈ U2(R), then R = U2(R). Now, suppose R is strongly (xn − 2 + xn − 3 + … + x + 1)-nil* clean. Then R = U2(R) by taking a0=1 ∈ U(R). If r = 0, then r is clearly a strongly (xn − x)-nil clean element. Let 0 ≠ r ∈ R and write r = s + b where b ∈ N(R), sn − 2 + sn − 3 + … + s + 1 = 0 and sb = bs. Then sn − s = s(s − 1)(sn − 2 + sn − 3 + … + s + 1) = 0 and so R is strongly (xn − x)-nil clean.     □

By choosing n = 4 in Proposition 3.10, we conclude that if R is strongly (x2 + x + 1)-nil*clean, then R = U2(R) is strongly (x4− x)-nil clean. Also, we have R = U2(R) if R is strongly (xn ± 1)-nil*clean.

Lemma 3.11

Let R be a Boolean ring and G any cyclic group of order p (prime). Then all elements of the group ring RG are roots of g(x) = x2p−1 − x.

By taking for example p = 3 in Lemma 3.11, we see that the group ring ℤ2(C3) is (strongly) (x4 − x)-nil clean which is not strongly nil clean. Moreover, the field ℤ3 is (strongly) (x3 − x)-nil clean which is not strongly nil clean. Therefore, strongly (xn − x)-nil cleanness of a ring R does not imply strongly nil cleanness of R whether n is odd or even.

In the following proposition, we can see that (xn − x)-nil clean and (xn + x)-nil clean rings coincide for any even integer n. The proof can be easily verified.

Proposition 3.12

Let R be a ring, n ∈ ℕ and a, b ∈ R. Then R is strongly (ax2n − bx)-nil clean if and only if R is strongly (ax2n + bx)-nil clean.

By Proposition 3.12 and Lemma 3.11, we conclude that ℤ2(Cp) is also (x2p−1 + x)-nil clean for any prime p. However, one can see that Proposition 3.12 need not be true for odd degrees. For example, the field ℤ3 is strongly (x3 − x)-nil clean which is not strongly (x3 + x)-nil clean.

Recall that an element p in a ring R is called periodic if pk = pm for some distinct positive integers k and m. Equivalently, p is periodic if p is a root of g(x) = x2n − xn for some positive integer n. R is called periodic if every element in R is periodic. In the following Lemma, we can see a sufficient condition for a ring to be periodic.

Lemma 3.13

([3]). Let R be a ring. If for each x ∈ R, there is n ∈ ℕ and p(x) ∈ ℤ[x] such that xn = xn+1p(x), then R is periodic.

In the following Theorem, periodic rings are characterized as strongly (xn − x)-nil clean rings for some n ∈ ℕ. This Theorem is essentially proved in [13]. However, for the sake of completeness, we shall give the proof.

Theorem 3.14

The following are equivalent for a ring R.

  1. R is periodic.

  2. R is a strongly (xn − x)-nil clean rings for some n ∈ ℕ.

  3. For every r ∈ R, there exists k ∈ ℕ and a ∈ 〈r〉 such that rk = rk+1a.

Proof

(1) ⇒ (2) : Suppose R is periodic and let r ∈ R. Then there exists k ∈ ℕ such that r2k = rk. Let s = rk+1 and b = r − rk+1. Then r = s + b where clearly s is a root of g(x) = xk+1 − x, bk = 0 and sb = bs. Therefore, R is a strongly (xk+1 − x)-nil clean.

(2)⇒ (3) : Suppose R is strongly (xn − x)-nil clean for some n ∈ ℕ. Let r ∈ R and write r = s + b where sn − s = 0, b ∈ N(R) and sb = bs. Now, r − rn = (s + b) − (s + b)n = b + bt where t ∈ 〈s, b〉 and so r − rn ∈ N(R). By simple computations, there exists k ∈ ℕ such that rk = rk+1 a where a ∈ 〈r〉.

(3) ⇒ (1) : Lemma 3.13. □

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Received: 2016-1-12
Accepted: 2017-2-2
Published Online: 2017-4-17

© 2017 Handam and Khashan

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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https://doi.org/10.1515/math-2017-0031
Keywords for this article
Nil clean ring; Strogly nil clean ring; g(x)-nil clean ring; Strongly g(x)-nil clean ring
Creative Commons
BY-NC-ND 3.0

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