Journal of Pure and Applied Algebra 220 (2016) 633–646

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Journal of Pure and Applied Algebra

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Nil-clean and strongly nil-clean rings

Tamer Koşan a, Zhou Wang b, Yiqiang Zhou c,∗ a Department of Mathematics, Gebze Technical University, Gebze/Kocaeli, Turkey b Department of Mathematics, Southeast University, Nanjing 210096, PR China c Department of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, Nfld A1C 5S7, Canada a r t i c l e i n f o a b s t r a c t

Article history: An element a of a ring R is nil-clean if a = e + b where e2 = e ∈ R and b is a Received 28 August 2014 nilpotent; if further eb = be, the element a is called strongly nil-clean. The ring R Received in revised form 8 May 2015 is called nil-clean (resp., strongly nil-clean) if each of its elements is nil-clean (resp., Available online 5 August 2015 strongly nil-clean). It is proved that an element a is strongly nil-clean iff a is a sum Communicated by S. Iyengar of an idempotent and a unit that commute and a −a2 is a nilpotent, and that a ring MSC: R is strongly nil-clean iff R/J(R)is boolean and J(R)is nil, where J(R) denotes the Primary: 16U99; secondary: 16E50; Jacobson radical of R. The strong nil-cleanness of Morita contexts, formal matrix 16N40; 16S34; 16S50; 16S70 rings and group rings is discussed in details. A necessary and sufficient condition is obtained for an ideal I of R to have the property that R/I strongly nil-clean implies R is strongly nil-clean. Finally, responding to the question of when a matrix ring is nil-clean, we prove that the matrix ring over a 2-primal ring R is nil-clean iff R/J(R)is boolean and J(R)is nil, i.e., R is strongly nil-clean. © 2015 Elsevier B.V. All rights reserved.

1. Introduction

There has been considerable interest in the structure of the rings whose elements are sums of certain special elements. For instance, a ring is called 2-good if every element is a sum of two units, while a ring is called (strongly) clean if every element is a sum of an idempotent and a unit (that commute with each other). Clean and strongly clean rings, and 2-good rings are active subjects, which can be traced back to Nicholson’s work [25] in 1977 and Zelinsky’s paper [31] in 1954, respectively. Not to mention, there are other important examples in the literature. This paper is concerned with two interesting variants of the clean property of rings, introduced by Diesl in [10]. Following Diesl [10], an element of a ring is called (strongly) nil-clean if it is a sum of an idempotent and a nilpotent (that commute with each other), and a ring is called (strongly) nil-clean if every element is (strongly) nil-clean. It comes as no surprise that nil-clean and strongly nil-clean rings are naturally connected to clean and strongly clean rings. Besides, the

* Corresponding author. E-mail address: [email protected] (Y. Zhou). http://dx.doi.org/10.1016/j.jpaa.2015.07.009 0022-4049/© 2015 Elsevier B.V. All rights reserved. 634 T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646 study of (strongly) nil-clean rings finds their close connections to strongly π-regular rings, boolean rings, and uniquely strongly clean rings. Furthermore, the nil-cleanness of a matrix ring is tightly linked to the famous Köthe Conjecture (see Section 6). The reader is referred to the papers [10,5,16,2] for the background and current stage of the study of nil-clean and strongly nil-clean rings. We continue the study of nil-clean and strongly nil-clean rings with the focus on the structure and construction of strongly nil-clean rings and the question of when a matrix ring is nil-clean. We start by proving that strongly nil-clean elements are exactly those strongly clean elements a with a − a2 nilpotent, with which some useful equivalent conditions of a strongly nil-clean element are obtained. These equivalent conditions are then used to prove the structure of a strongly nil-clean ring. This structure theorem, improving several results in [10], is utilized to conduct a detailed discussion regarding the strong nil-cleanness of Morita contexts, formal matrix rings and group rings in Sections 3 and 4, and the results obtained give various new families of strongly nil-clean rings. In Section 5, we go further to show the structure of a non-unital strongly nil-clean ring, and prove an extension theorem as an application. In Section 6, we prove that the matrix ring over a 2-primal ring R is nil-clean iff R/J(R)is boolean and J(R)is nil, i.e., R is a strongly nil-ring. This seems to be the best answer, so far, to the question of when a matrix ring is nil-clean. Throughout, R is an associative ring with identity, and C(R), J(R), U(R)and Nil(R)denote the center, the Jacobson radical, the unit group and the set of all nilpotent elements of R, respectively.

2. Strongly nil-clean rings

Let R be a ring and a ∈ R. If a = e + b where e2 = e ∈ R, b ∈ R is a nilpotent and eb = be, then this expression is called a strongly nil-clean decomposition of a in R. Similarly, we define clean, nil-clean and strongly clean decompositions of an element. In this section, we give important equivalent conditions of a strongly nil-clean element, and prove the structure theorem of a strongly nil-clean ring. We also show that the so-called Jacobson Lemma holds for strongly nil-clean elements.

Theorem 2.1. An element a ∈ R is strongly nil-clean iff a is strongly clean in R and a − a2 is a nilpotent.

Proof. (⇒). Let a = e + b be a strongly nil-clean decomposition in R. Then a =(1 − e) +(2e − 1 + b)is a strongly clean decomposition in R. Moreover, a2 = e +2eb + b2, so a − a2 =(1 − 2e − b)b is a nilpotent. (⇐). Let a = e +u be a strongly clean decomposition in R and a −a2 be a nilpotent. Then a2 = e +2eu +u2 and so a − a2 =(1 − 2e − u)u. It follows that 1 − 2e − u is a nilpotent. So a =(1 − e) +(−1 +2e + u)is a strongly nil-clean decomposition in R. 2

Corollary 2.2. (See [10, Corollary 3.10].) A unit u of a ring R is strongly nil-clean iff 1 − u is a nilpotent.

A uniquely strongly clean element of R is an element having a unique strongly clean decomposition in R. An element a ∈ R is called strongly π-regular if an ∈ Ran+1 ∩ an+1R for some positive integer n.

Lemma 2.3. Let a be a strongly nil-clean element of R. Then:

(1) a has a unique strongly nil-clean decomposition in R. (2) a is a strongly π-regular element of R. (3) a is a uniquely strongly clean element of R.

Proof. (1) is [10, Corollary 3.8]; (2) is by [10, p. 203, Remark]. (3) By Theorem 2.1, we know that a is strongly clean and a2−a is a nilpotent. By the proof of Theorem 2.1, two different idempotents which give strongly clean decompositions of a must yield two different idempotents T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646 635 which give strongly nil-clean decompositions of a. But this is impossible. Thus a is uniquely strongly clean. 2

By [10, Proposition 2.5], an element a ∈ R is strongly π-regular iff there is an idempotent e ∈ R and a unit u ∈ R such that a = e + u, ae = ea and eae is nilpotent; this elementwise decomposition of a strongly π-regular element is referred to as a strongly π-regular decomposition. By [10, Proposition 2.6], a strongly π-regular element has only one such decomposition. Strongly nil-clean elements, uniquely strongly clean elements and strongly π-regular elements share the following interesting property. For a ∈ R, the commutant of a in R is denoted by C(a), i.e., C(a) = {x ∈ R : ax = xa}.

Proposition 2.4. Let a ∈ R. Then C(a) ⊆ C(e) holds in any of the following cases:

(1) a is a strongly nil-clean element of R with a strongly nil-clean decomposition a = e + b. (2) a is a strongly π-regular element of R with a strongly π-regular decomposition a = e + u. (3) a is a uniquely strongly clean element of R with a strongly clean decomposition a = e + u. eRe eR(1 − e) Proof. Let x ∈ C(a). Consider the Peirce decomposition R = with respect (1 − e)Re (1 − e)R(1 − e) ea 0 to the idempotent e. Then a = . 0(1−e)a e 0 eb 0 eex(1 − e) eb −ex(1 − e) e 0 (1) Since a = + = + = + 00 0(1− e)b 00 0(1− e)b (1 − e)xe 0 eb 0 are all strongly nil-clean decompositions of a in R, it follows that ex(1 − e) = −(1 − e)xe (1 − e)b (1 − e)xe =0. So x ∈ C(e). e 0 eu 0 eex(1 − e) eu −ex(1 − e) e 0 (2) Since a = + = + = + 00 0(1− e)u 00 0(1− e)u (1 − e)xe 0 eu 0 are all strongly π-regular decompositions of a in R, it follows that ex(1 − e) = −(1 − e)xe (1 − e)u (1 − e)xe =0. So x ∈ C(e). (3) It is similar to (2). 2

An element of a ring is called uniquely clean if it has only one clean decomposition in the ring. For a uniquely clean element a ∈ R with a clean decomposition a = e + u, C(a) need not be contained in C(e). In fact, a and e may not commute (see [15]). The next corollary follows from Theorem 2.1 and Lemma 2.3.

Corollary 2.5. Let a ∈ R. The following are equivalent:

(1) a is strongly nil-clean. (2) a is strongly π-regular and a − a2 is a nilpotent. (3) a is uniquely strongly clean and a − a2 is a nilpotent.

Corollary 2.6. A strongly nil-clean ring is uniquely strongly clean, i.e., every element is uniquely strongly clean.

A uniquely strongly clean element need not be strongly nil-clean. In fact, a uniquely strongly clean ring need not be strongly nil-clean: Z2[[t]] is a uniquely strongly clean ring by [8, Theorem 20], but its Jacobson radical is not nil, so it is not strongly nil-clean by Theorem 2.7 below. 636 T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646

Diesl [10, Corollary 3.11] obtained a characterization of a strongly nil-clean ring: A ring R is strongly nil-clean iff R is a strongly π-regular ring and U(R) =1 +Nil(R). We now present the structure of a strongly nil-clean ring.

Theorem 2.7. A ring R is strongly nil-clean iff R/J(R) is boolean and J(R) is nil.

Proof. (⇐). Let a ∈ R. Then a − a2 ∈ J(R). As J(R)is nil, there exists a polynomial f(t)over Z such that e := f(a)is an idempotent of R and a − e ∈ J(R). Hence a = e +(a − e)is a strongly nil-clean decomposition. (⇒). By Corollary 2.6, R is uniquely strongly clean, so R/J(R)is boolean by [8, Corollary 18]. Let a ∈ J(R), and let a = e + b be a strongly nil-clean decomposition. As R/J(R)is boolean and b is a nilpotent, it follows that b ∈ J(R). Thus, e = a − b ∈ J(R), so e =0. Hence a = b is a nilpotent. 2

Corollary 2.8. (See [10, Corollary 3.20].) Let R be a commutative ring. Then R is nil-clean iff R/J(R) is boolean and J(R) is nil.

It is well-known that, for a, b ∈ R, 1 −ab ∈ U(R) ⇔ 1 −ba ∈ U(R). This result is also known as Jacobson’s Lemma. In [19], the authors proved that Jacobson Lemma holds for Drazin invertible elements, π-regular elements, strongly clean elements, and uniquely strongly clean elements. Here we show that Jacobson Lemma holds for strongly nil-clean elements.

Theorem 2.9. Let R be a ring and let a, b ∈ R. If ab is strongly nil-clean, then so is ba.

Proof. Since it is known that ab is strongly clean iff ba is strongly clean, we need only show that (ab)2 −ab is nilpotent iff (ba)2 − ba is nilpotent. However, this is clear because [(xy)2 − xy]n+1 = x[(yx)2 − yx]n(yxy − y) for any x, y ∈ R. 2

Corollary 2.10. Let R be a ring and let a, b ∈ R. If 1 − ab is strongly nil-clean, then so is 1 − ba.

Proof. Note that x ∈ R is strongly nil-clean iff 1 − x is strongly nil-clean. Suppose 1 − ba is strongly nil-clean. Then ba is strongly nil-clean, and so ab is strongly nil-clean by Theorem 2.9. Hence 1 − ab is strongly nil-clean. 2

Question 2.11. Does Jacobson Lemma hold for nil-clean elements?

3. Morita contexts and formal matrix rings AM In [10], the author showed that a formal matrix ring is strongly nil-clean iff A, B are strongly 0 B nil-clean, and that a proper matrix ring is never strongly nil-clean. This suggests that strongly nil-clean rings may be constructed through Morita contexts. Here we consider when a Morita context is a strongly nil-clean ring, and determine when a formal matrix ring is strongly nil-clean. AM A Morita context is a 4-tuple , where A, B are rings, M and N are bimodules, and NB A B B A there exist context products M ×N → A and N × M → B written multiplicatively as (w, z) → wz and AM (z, w) → zw, such that is an associative ring with the obvious matrix operations. NB The next result (in fact, a more general result) can be found in [28]. T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646 637 AM J(A) M Lemma 3.1. (See [28].) Let R := be a Morita context. Then J(R) = 0 , where NB N0 J(B) M0 = x ∈ M : xN ⊆ J(A) and N0 = y ∈ N : yM ⊆ J(B) . Canonically, M/M0 is an A/J (A), B/J(B) -bimodule and N/N0 is a B/J(B), A/J(A) -bimodule, and A/J(A) M/M this induces a Morita context 0 where the context products are given by N/N0 B/J(B)

(x + M0)(y + N0)=xy + J(A), (y + N0)(x + M0)=yx + J(B) for all x ∈ M and y ∈ N. AM Lemma 3.2. (See [29, Proposition 2.6].) Let R := be a Morita context, and let NB A/J(A) M/M0 N/N0 B/J(B) ∼ A/J(A) M/M0 be defined above. Then R/J(R) = . N/N0 B/J(B) AM Lemma 3.3. Let R := be a Morita context. Then R/J(R) is boolean iff A/J(A), B/J(B) are NB boolean, MN ⊆ J(A) and NM ⊆ J(B). ∼ A/J(A) M/M0 Proof. By Lemma 3.2, R/J(R) = . If R/J(R)is boolean, then it must be that N/N0 B/J(B) M/M0 =0and N/N0 =0, i.e., M = M0 and N = N0. It follows that MN ⊆ J(A)and NM ⊆ J(B)and ∼ that R/J(R) = A/J(A) × B/J(B). So A/J(A), B/J(B)are boolean. ∼ Conversely, MN ⊆ J(A)and NM ⊆ J(B)gives M0 = M and N0 = N. So, it follows that R/J(R) = A/J(A) M/M0 ∼ = A/J(A) ×B/J(B). Thus, A/J(A)and B/J(B)being boolean implies that R/J(R) N/N0 B/J(B) is boolean. 2 AM Theorem 3.4. Let R := be a Morita context. If R is a strongly nil-clean ring, then A, B are NB strongly nil-clean rings, MN ⊆ J(A) and NM ⊆ J(B); the converse holds if MN, NM are nilpotent ideals of A and B respectively.

Proof. Suppose R is strongly nil-clean. Then A, B are strongly nil-clean by [10, Corollary 3.26]. Since R/J(R)is boolean, one infers MN ⊆ J(A)and NM ⊆ J(B)by Lemma 3.3. For the converse, we see that R/J(R)is boolean by Lemma 3.3. Moreover, A, B are strongly π-regular rings by [10, Corollary 3.7]. With the assumption that MN, NM are nilpotent ideals of A and B respectively, we infer that R is strongly π-regular by [29, Theorem 3.5]. Hence J(R)is nil, and so R is strongly nil-clean by Theorem 2.7. 2

We do not know whether the condition that MN, NM are nilpotent is necessary in Theorem 3.4. RR Given a ring R and a central element s of R, the 4-tuple becomes a ring with addition defined RR componentwise and with multiplication defined by a x a x a a + sx y a x + x b 1 1 2 2 = 1 2 1 2 1 2 1 2 . y1 b1 y2 b2 y1a2 + b1y2 sy1x2 + b1b2 638 T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646 AM This ring is denoted by K (R). A Morita context with A = B = M = N = R is called a s NB generalized matrix ring over R. It was observed by Krylov in [17] that a ring S is a generalized matrix ring over R iff S = Ks(R)for some s ∈ C(R). Here MN = NM = sR, so ‘MN ⊆ J(A) ⇔ s ∈ J(R)’, ‘NM ⊆ J(B) ⇔ s ∈ J(R)’, and ‘MN, NM are nilpotent ⇔ s is a nilpotent’. Thus, Theorem 3.4 has the following quick consequence.

Corollary 3.5. Let R be a ring with s ∈ C(R). Then Ks(R) is strongly nil-clean iff R is strongly nil-clean and s ∈ J(R).

Corollary 3.5 clearly implies that Ks2 (R)is strongly nil-clean iff R is strongly nil-clean and s ∈ J(R). As explained below, Ks2 = M2(R; s), a formal matrix ring defined by s. Following [30], for n ≥ 2and for s ∈ C(R), the n × n formal matrix ring over R defined by s, denoted as Mn(R; s), is the set of all n × n matrices over R with usual addition of matrices and with multiplication defined below: for (aij)and (bij)in Mn(R; s),

n δikj (aij)(bij )=(cij), where cij = s aikbkj. k=1

Here δijk =1 + δik − δij − δjk with δik, δij, δjk the Kronecker delta symbols. Note that a proper matrix ring is never strongly nil-clean (see [10, p. 208]).

Theorem 3.6. Let R be a ring with s ∈ C(R) and let n ≥ 2. Then Mn(R; s) is strongly nil-clean iff R is strongly nil-clean and s ∈ J(R).

Proof. The claim holds for n =2as noted above. Suppose n > 2and assume that the claim⎛ holds for⎞ M1n AM ⎜ . ⎟ M − (R; s). Let A = M − (R; s). Then M (R; s) = is a Morita context, where M = ⎝ . ⎠ n 1 n 1 n NR . ⎛ Mn−1,n⎞ x1n ⎜ ⎟ ··· − . ∈ and N =(Mn1 Mn,n−1 )with Min = Mni = R for all i =1, ..., n 1. Moreover, for x = ⎝ . ⎠

xn−1,n M and y =(yn1 ··· yn,n−1 ) ∈ N, we have ⎛ ⎞ 2 s x1nyn1 sx1nyn2 ··· sx1nyn,n−1 ⎜ 2 ⎟ ⎜ sx2nyn1 s x2nyn2 ··· sx2nyn,n−1 ⎟ xy = ⎜ . . . ⎟ ∈ A (3.1) ⎝ . . . ⎠ 2 sxn−1,nyn1 sxn−1,nyn2 ··· s xn−1,nyn,n−1 and

2 2 2 yx = s yn1x1n + s yn2x2n + ···+ s yn,n−1xn−1,n ∈ R. (3.2)

If Mn(R; s)is strongly nil-clean, then, by Theorem 3.4, R is strongly nil-clean and NM ⊆ J(R). Thus, by (3.2), s2R = NM ⊆ J(R), so s2 ∈ J(R). As R/J(R)is boolean, s ∈ J(R). Conversely, suppose that R is strongly nil-clean and s ∈ J(R). Then s is a nilpotent and A is strongly nil-clean by induction hypothesis. So NM = s2R is nilpotent in R. But, by (3.1), MN ⊆ sA. As sA is nilpotent in A, MN is nilpotent in A. Hence Mn(R; s)is strongly nil-clean by Theorem 3.4. 2 T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646 639

4. Group rings

Strongly nil-clean rings can be constructed using group rings. In [22], the authors characterized the commutative ring R and the abelian group G such that the group ring of G over R is (strongly) nil-clean. Here we will show that the group ring of a locally finite 2-group over a strongly nil-clean ring is strongly nil-clean. If R is a ring and G is a group, RG denotes the group ring of the group G over R. The ring homomorphism

ω : RG → R, Σrgg → Σrg is called the augmentation map, and ker(ω)is called the augmentation ideal of the group ring RG and is denoted by (RG). That is (RG) = {Σrgg :Σrg =0}.

Lemma 4.1. If RG/J(RG) is boolean, then (RG) ⊆ J(RG). And if in addition J(R) =0, then (RG) = J(RG).

Proof. The booleanness of RG/J(RG) implies that 1 −g ∈ J(RG)for all g ∈ G. Now if x =Σrgg ∈ (RG), then 0 = ω(x) =Σrg, and thus 1 − x =1 +Σrg − Σrgg =1 +Σrg(1 − g) ∈ U(RG). Hence (RG) ⊆ J(RG). Because ω : RG → R is a ring homomorphism, we have ω(J(RG)) ⊆ J(R). So, for J(R) =0, we get J(RG) ⊆ (RG), and hence (RG) = J(RG). 2

A ring R is called semiboolean if R/J(R)is boolean and idempotents lift modulo J(R)(see [27]).

Lemma 4.2. The following are equivalent for a group G:

∼ (1) Z2G/J(Z2G) = Z2. (2) Z2G is a uniquely clean ring. (3) Z2G is a semiboolean ring.

Proof. (1) ⇒ (2). This is by [26, Theorem 15]. (2) ⇒ (3). This is obvious.

(3) ⇒ (1). If Z2G is semiboolean, then Z2G/J(Z2G)is boolean. By Lemma 4.1, J(Z2G) = (RG)(as ∼ J(Z2) =0). Thus, Z2G/J(Z2G) = Z2G/ (Z2G) = Z2. 2

Lemma 4.3. (See [9, Proposition 9].) If R is a ring and G is a locally finite group then J(R) = J(RG) ∩ R. In particular, J(R)(RG) ⊆ J(RG).

Theorem 4.4. Let R be a ring and G be a group.

(1) If RG is a semiboolean ring, then R is a semiboolean ring and G is a 2-group. (2) If R is a semiboolean ring and G is a locally finite 2-group, then RG is a semiboolean ring.

Proof. (1) If RG is semiboolean, then R is semiboolean (an image of R) and so Z2 is an image of R. Thus, Z2G is semiboolean (an image of RG). Hence Z2G is uniquely clean by Lemma 4.2. So G is a 2-group by [7, Theorem 5]. (2) By [27, Lemma 24], it suffices to show that for any x ∈ RG, there exists y2 = y ∈ RG such that x −y ∈ J(RG). Since ω(x) ∈ R and since R is semiboolean, there exists e2 = e ∈ R such that ω(x) −e ∈ J(R). So ω(x − e) = ω(x) − ω(e) = ω(x) − e ∈ J(R). Thus, by Lemma 4.3, ω(x − e) ∈ J(RG). Because R/J(R)is boolean and because G is a locally finite 2-group, we have J(RG) = {x ∈ RG : ω(x) ∈ J(R)} by [7, Lemma 10]. Hence x − e ∈ J(RG). So RG is semiboolean. 2 640 T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646

Corollary 4.5. Let R be a ring and G be a locally finite group. Then RG is semiboolean iff R is semiboolean and G is a 2-group.

Corollary 4.6. Let R be a ring and G be a group. Then RG is semiboolean iff R is semiboolean and (RG) ⊆ J(RG).

Proof. (⇒). R is an image of RG, so it is semiboolean. Since RG/J(RG)is boolean, (RG) ⊆ J(RG)by Lemma 4.1. (⇐). If (RG) ⊆ J(RG), then J RG/ (RG) = J(RG)/ (RG) =[J(RG) + (RG)]/ (RG), and (RG)is a radical ring, so it is a semiboolean general ring without 1(see [27, p. 307]). Moreover, ∼ RG/ (RG) = R is semiboolean. Now if x2 − x ∈ (RG), then ω(x)2 − ω(x) = ω(x2 − x) =0. So ω(x)2 = ω(x) ∈ R ⊆ RG. Because ω(x − ω(x)) =0, x − ω(x) ∈ (RG). So idempotents lift modulo (RG) in RG. By [27, Theorem 29], RG is semiboolean. 2

Theorem 4.7. Let R be a ring and G be a group.

(1) If RG is strongly nil-clean, then R is strongly nil-clean and G is a 2-group. (2) If R is strongly nil-clean and G is a locally finite 2-group, then RG is strongly nil-clean.

Proof. (1) If RG is strongly nil-clean, then its image R is strongly nil-clean. As RG is semiboolean, G is a 2-group by Theorem 4.4. (2) Assume that R is strongly nil-clean and G is a locally finite 2-group. Then R is semiboolean, so RG is semiboolean by Theorem 4.4. Thus, by Theorem 2.7, it suffices to show that each x ∈ J(RG)is nil. As G is a locally finite 2-group, there exists a finite 2-subgroup H of G such that x ∈ GH. By [9, Proposition 9], J(RH) ⊇ RH ∩ J(RG). So x ∈ J(RH). Therefore, without loss of generality, we may assume that G is a finite 2-group. Since R/J(R)is boolean and J(R)is nil (see Theorem 2.7), 2 ∈ J(R)is a nilpotent. So, by [9, Theorem 9], (RG)is nilpotent. As ω : RG → R is a ring homomorphism, ω(J(RG)) ⊆ J(R), so ω(x) ∈ J(R). Hence, there exists k>0such that ω(x)k =0, i.e., ω(xk) =0. This shows that xk ∈ (RG). As (RG)is nilpotent, it follows that x is a nilpotent. 2

5. Non-unital rings and the extension problem

In [10, Corollary 3.22], the author showed that, for any nilpotent ideal I of R, R is strongly nil-clean iff R/I is strongly nil-clean. This motivates us to ask which ideals I of a ring R ensure that R/I being strongly nil-clean implies that R is strongly nil-clean. This is actually the so-called extension problem for strongly nil-clean rings. An answer to the question helps construct examples of strongly nil-clean rings. First we recall some notations. A non-unital ring is an associative ring with or without identity. Through- out this section, I is a non-unital ring. We still use J(I)and Nil(I)to denote the Jacobson radical and the set of all nilpotent elements of I, respectively. If S is a ring and I is a non-unital ring such that I = SIS is a bimodule, then the additive group S ⊕ I with multiplication (r, a)(s, b) =(rs, rb + as + ab) becomes an associative ring if and only if the conditions s(ab) =(sa)b, a(sb) =(as)b and (ab)s = a(bs)are satisfied for all s ∈ S and a, b ∈ I. In this case, this ring is called the ideal extension of S by I and is denoted by I(S; I). It is well known that any non-unital ring I is an ideal of its Dorroh extension I(Z; I). We now define (strong) nil-cleanness of a non-unital ring.

Definition 5.1. (See [10, p. 202].) Let I be a non-unital ring. An element a ∈ I is called nil-clean if there is an idempotent e ∈ I and a nilpotent b ∈ I such that a = e + b. The element a is further called strongly nil-clean if such an idempotent and a nilpotent can be chosen such that be = eb. A non-unital ring is called nil-clean (respectively, strongly nil-clean) if each of its elements is nil-clean (respectively, strongly nil-clean). T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646 641

Lemma 5.2. Let I be an ideal of a ring R and a ∈ I. Then a is a strongly nil-clean element of I iff a is a strongly nil-clean element of R.

Proof. (⇒). It is obvious. (⇐). Let a = e +b be a strongly nil-clean decomposition in R. Then a =(1 −e) +(2e −1 +b)is a strongly clean decomposition in R. So ae =(2e − 1 + b)e, and hence e =(2e − 1 + b)−1ae ∈ I as I is an ideal of R. So a = e + b is a strongly nil-clean decomposition in I. 2

Let Q(I) = {q ∈ I : q + p − qp = p + q − pq =0 for some p ∈ I}. Note that q ∈ Q(R) ⇔ 1 − q ∈ U(R) and that, for any ideal I ¡ R, Q(I) = I ∩ Q(R).

Lemma 5.3. An element a ∈ Q(I) is strongly nil-clean in I iff a is a nilpotent.

Proof. There exists a ring R such that I is an ideal of R. Let a ∈ Q(I), and let a = e + b be a strongly nil-clean decomposition in I. Then 1 − e =(1 − a) + b in R. As 1 − a ∈ U(R)and (1 − a)b = b(1 − a), b being a nilpotent implies that (1 − a) + b is a unit, so 1 − e =1. Hence e =0and so a = b is a nilpotent. 2

Lemma 5.4. If I is strongly nil-clean and J(I) =0, then 2I =0.

Proof. Assume 2I =0. As I is clean, every nonzero right ideal contains a nonzero idempotent by [27, Corollary 6]. So, there exists a nonzero idempotent e in 2I. Write e =2a with a ∈ I. Then ea = ae. As argued in the proof of [8, Lemma 16], we see that (1 − 3e)(1 − 3ae) =1 − 3ae − 3e +9eae =1 − 3ae − 3e +9ae = 1 − 3e +6ae =1 − 3e +3(2a)e =1. Similarly, (1 − 3ae)(1 − 3e) =1. So 1 − 3e ∈ U(R). Moreover, 1 − 2e ∈ U(R). That is, 2e, 3e ∈ Q(R) ∩ I = Q(I). Then, by Lemma 5.3, 2e and 3e are (commuting) nilpotents. So, e =3e − 2e is a nilpotent, a contradiction. 2

Lemma 5.5. Let I be strongly nil-clean with J(I) =0. Then I is boolean.

Proof. By Lemma 5.4, R := I(Z2, I)is well-defined. To show I is boolean, it suffices to show that R is boolean. First we show that R is strongly nil-clean. Let x := (m, a) ∈ R. If m =0, let a = e + b be a strongly nil-clean decomposition in I and so x =(0, a) =(0, e) +(0, b)is a strongly nil-clean decomposition in R. If m =1, let −a = e + b be a strongly nil-clean decomposition in I and so x =(1, a) =(1, −e) +(0, −b)is a strongly nil-clean decomposition in R. Hence, R is strongly nil-clean. Next we show that, for any 0 = x =(m, a) ∈ R, xR contains a nonzero idempotent. If m =0, then xR =(0, aZ2 +aI). As a nonzero right ideal of I, aZ2+aI contains a nonzero idempotent by [27, Corollary 6]. Thus, we can assume x =(1, a). If a + a2 =0, then x2 = x; otherwise x(0, a) =(1, a)(0, a) =(0, a + a2) =0, and x(0, a)R (and hence xR) contains a nonzero idempotent as argued above. Hence we infer J(R) =0, so R is boolean by Theorem 2.7. 2

We have the non-unital analogue of Theorem 2.7.

Theorem 5.6. A non-unital ring I is strongly nil-clean iff I/J(I) is boolean and J(I) is nil.

Proof. (⇒). If I is strongly nil-clean, then I/J(I)is certainly strongly nil-clean and J(I/J(I)) =0, so I/J(I)is boolean by Lemma 5.5. By Lemma 5.3, J(I)is nil. (⇐). Let a ∈ I. Then a − a2 ∈ J(I)since I/J(I)is boolean. So a − a2 is a nilpotent. By a standard argument (see [1, p. 301]), there exists a polynomial f(t)over Z such that e := f(a)is an idempotent and a − e ∈ J(I). As ea = ae, a = e +(a − e)is a strongly nil-clean decomposition in I. 2 642 T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646

Corollary 5.7. Let I be a non-unital ring and A ¡ I. If I is strongly nil-clean, then A is strongly nil-clean.

∼ Proof. We have J(A) = A ∩ J(I) ⊆ J(I)and A/J(A) = A/(A ∩ J(I)) = (A + J(I))/J(I) ⊆ I/J(I). By Theorem 5.6, I/J(I)is boolean and J(I)is nil. It follows that A/J(A)is boolean and J(A)is nil. So A is strongly nil-clean. 2

Theorem 5.8. Let I be a non-unital ring and A ¡ I. Then I is strongly nil-clean iff the following conditions hold:

(1) A and I/A are strongly nil-clean. (2) Every idempotent of I/A can be lifted to an idempotent of I. (3)A J(I/ ) =(A + J(I))/A.

Proof. (⇒). Suppose I is strongly nil-clean. Then clearly I/A is strongly nil-clean and, by Corollary 5.7, A is strongly nil-clean. As I is a clean ring, idempotents lift modulo A. Finally, as I/J(I)is boolean, every image of I/J(I)has zero Jacobson radical. Hence, by [27, Lemma 20], J(I/K) =(K + J(I))/K for all K ¡ I. In particular, J(I/A) =(A + J(I))/A. (⇐). We first show that J(I)is nil. Let x ∈ J(I). As J(I/A) =(A + J(I))/A is nil, x + A ∈ J(I/A)is a nilpotent, so xn ∈ A for some n > 0. As A is strongly nil-clean, xn is strongly nil-clean. But since xn ∈ J(I), it is a nilpotent by Lemma 5.3. Thus x is a nilpotent. So, by Theorem 5.6, it suffices to show that I/J(I)is boolean. However, in view of Theorem 5.6, this follows from [27, Theorem 29]. Hence I is strongly nil-clean by Theorem 5.6. 2

Corollary 5.9. Let I be a nil ideal of R. Then R is strongly nil-clean iff R/I is strongly nil-clean.

Corollary 5.10. Let S = I(R; V ) be an ideal extension, where R is a ring and V = RVR is a bimodule. Then S is strongly nil-clean iff the following conditions hold:

(1) R and V are strongly nil-clean. r(2) Fo all a ∈ J(R), there exists x ∈ V such that (a, x) ∈ J(S).

∼ ∼ Proof. Let A = {(0, x) : x ∈ V }. Then A  S, S/A = R, and A = V . We can easily verify the following: (1) holds iff A and S/A are strongly nil-clean; (2) holds iff J(S/A) =(A + J(S))/A. Moreover, note that idempotents of S/A can always be lifted to idempotents of S. So the claim follows from Theorem 5.8. 2

If R is a boolean ring and RVR is a bimodule such that I(R; V )is a well-defined ring, then I(R; V ) is strongly nil-clean iff V is strongly nil-clean. For instance, if R is boolean then I(R; Tn(R)) is strongly nil-clean.

6. Nil-cleanness of matrix rings

Matrix rings over a clean ring are all clean [12]. Motivated by this, Diesl [10] asks whether a matrix ring over any nil-clean ring is again nil-clean. One can ask a slightly different question of when a matrix ring

Mn(R)is nil-clean. By [10, Corollary 3.17], Mn(R)is nil-clean iff M(R/J(R)) is nil-clean and Mn(J(R)) is nil. The Köthe conjecture claims that the sum of two nil left ideals is nil in any ring, or equivalently the matrix ring over a nil ring is again nil. But, since the conjecture is still unsettled, it would be difficult to characterize the arbitrary ring R such that Mn(R)is nil-clean. Here, we work on this question focusing on some classes of rings for which the Köthe conjecture holds true. Let us recall some known progress: For a T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646 643

∼ field or a division ring D, Mn(D)is nil-clean iff D = F2 (see [5,16]); for a strongly regular ring R, Mn(R)is nil-clean iff R is boolean (see [16]). In this section, we will provide some further improving answers to the aforementioned question. Recall that the prime radical N(R)of a ring R is defined to be the intersection of all the prime ideals in R. By Levitzki [20], the prime radical N(R)coincides with the lower radical of R as defined by Baer [4] (or the lower nilradical denoted by Nil∗(R)). A 2-primal ring is a ring R such that N(R) =Nil(R). It is known that a ring R is 2-primal iff R/I is a domain for every minimal prime ideal I of R. Examples of 2-primal rings include commutative rings and reduced rings (i.e., rings without nonzero nilpotents). Also recall that a semipotent ring is a ring for which any one-sided ideal not contained in the Jacobson radical contains a nonzero idempotent. Note that every clean ring is semipotent.

Theorem 6.1. Let R be a 2-primal ring and n ≥ 1. Then Mn(R) is nil-clean iff R/J(R) is boolean and J(R) is nil, i.e., R is strongly nil-clean.

∼ Proof. (⇐). As R/J(R)is boolean, Mn(R)/J(Mn(R)) = Mn(R/J(R)) is nil-clean by [5, Corollary 6]. So to M M show n(R)is nil-clean, it suffices to show that n(J(R)) is nil by [10, Corollary 3.17]. As R is a 2-primal ring and J(R)is nil, we see J(R) =Nil∗(R). So Mn(J(R)) = Mn Nil∗(R) =Nil∗ Mn(R) (see [18, p. 172]), which is nil. ∼ (⇒). Since Mn(R)is nil-clean, Mn(R)/J(Mn(R)) = Mn(R/J(R)) is nil-clean and J(Mn(R)) = Mn(J(R)) is nil by [10, Corollary 3.17]. It follows that J(R)is nil. Since R is 2-primal, it follows that J(R) = N(R), and hence R/J(R)is a reduced ring. Thus, by [3], R/J(R)is a subdirect product of a family of domains

{Sα : α ∈ Γ}. As an image of Mn(R/J(R)), Mn(Sα)is also nil-clean, and hence it is clean (see [10, Proposition 3.4]). Thus, Mn(Sα)is semipotent. By [24, Corollary 1.7], Sα is semipotent. Being a domain, ∼ Z ∈ Sα must be a division ring. Now it follows from [16, Theorem 3] that Sα = 2 for all α Γ. Hence, we see that R/J(R)is a subring of the boolean ring Sα. So R/J(R)is a boolean ring. 2

Corollary 6.2. Let R be a commutative ring and n ≥ 1. Then Mn(R) is nil-clean iff R/J(R) is boolean and J(R) is nil.

Corollary 6.3. Let R be a reduced ring. Then Mn(R) is nil-clean iff R is boolean.

Corollary 6.4. Let R be a right self-injective, strongly π-regular ring with J(R) =0. Then Mn(R) is nil-clean iff R is a finite direct product of matrix rings over boolean rings.

Proof. The sufficiency is clear. For the necessity, we note that R is a finite direct product of matrix rings over strongly regular rings by [14]. Since Mn(R)is nil-clean, it follows that some matrix ring over each of the strongly regular rings is nil-clean, so each of the strongly regular rings is boolean by Corollary 6.3. 2

Corollary 6.5. (See [6].) Let R be an abelian ring. Then Mn(R) is nil-clean iff R/J(R) is boolean and Mn(J(R)) is nil. ∼ Proof. (⇐). Since R/J(R)is boolean, Mn(R)/J(Mn(R)) = Mn(R/J(R)) is nil-clean. As Mn(J(R)) is nil, Mn(R)is nil-clean by [10, Corollary 3.17]. (⇒). Since Mn(R)is nil-clean, Mn(J(R)) is nil by [10, Corollary 3.17]. Moreover, Mn(R/J(R)) is nil-clean. So R/J(R)is semipotent (see the proof of Theorem 6.1). As R/J(R)is semiprimitive and abelian, it must be reduced by [21, Theorem 2.1]. So R/J(R)is boolean by Corollary 6.3. 2

A ring R is said to be of bounded index (of nilpotence) if there is a positive integer n such that an =0 for all nilpotent elements a of R. The least such integer is called the index of R. 644 T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646

Lemma 6.6. Let R be of bounded index. If J(R) is nil, then Mn(J(R)) is nil for all n ≥ 1.

Proof. It is clear that N(R) ⊆ J(R). Since R is of bounded index, it follows that R/N(R)is of bounded index. As J(R)is nil, J(R)/N (R)is nil. So there exists a fixed positive integer k such that a¯k = 0for¯ all a¯ ∈ J(R)/N (R). Since R/N(R)is semiprime, it follows from [13, Lemma 1.1] that J(R)/N (R) = 0,¯ i.e.,

J(R) = N(R). As argued in the proof of Theorem 6.1, we can deduce that Mn(J(R)) is nil. 2

Corollary 6.7. Let R be a right continuous ring of bounded index. Then Mn(R) is nil-clean iff J(R) is nil and R/J(R) is isomorphic to a finite direct product of matrix rings over boolean rings.

Proof. (⇐). Since R is of bounded index, J(R)nil implies that J(Mn(R)) = Mn(J(R)) is nil by Lemma 6.6. Moreover, R/J(R)being isomorphic to a finite direct product of matrix rings over boolean rings clearly implies that Mn(R)/J(Mn(R)) is nil-clean. Hence Mn(R)is nil-clean by [10, Corollary 3.17]. (⇒). Since Mn(R)is nil-clean, J(Mn(R)) = Mn(J(R)) is nil by [10, Corollary 3.17], and it follows that J(R)is nil. Thus, as R is of bounded index, R/J(R)is also of bounded index. Since R is right continuous, by [23, Theorem 3.11, Corollary 3.13], R/J(R) = S × T , where S is a regular, right self-injective ring and T is a reduced regular ring. Since S is of bounded index, S is a finite direct product of matrix rings over reduced ∼ M ⊕ M ⊕···⊕M regular rings by [11, Theorem 7.20]. Thus, R/J(R) = n1 (A1) n2 (A2) nt (At), where each Ai M ∼ M M is a reduced regular ring. Since nin(Ai) = n ni (Ai) is also nil-clean, Ai is boolean by Corollary 6.3. So R/J(R)is a finite direct product of matrix rings over boolean rings. 2

Corollary 6.8. If R is a strongly nil-clean ring of bounded index, then Mn(R) is nil-clean for all n ≥ 1.

Proof. Since R is strongly nil-clean, J(R)is nil and R/J(R)is boolean by Theorem 2.7, so Mn(J(R)) is nil by Lemma 6.6,and Mn(R/J(R)) is nil-clean. So Mn(R)is nil-clean by [10, Corollary 3.17]. 2

We let eij be the matrix with (i, j)-entry 1and the other entries 0.

Lemma 6.9. Let S = Mn(R) where R is a ring and n ≥ 2. For 1 ≤ i ≤ n − 1, let bi = e12 + e23 + ···+ ··· n−1 ei−1,i + ei+1,i+2 + + en−1,n + en1. Then bi = ei+1,i.

Proof. The verification is straightforward. 2

Lemma 6.10. Let e be a non-central idempotent of S. If eSe = Mn(R) where R is a ring and n ≥ 2, then S contains a nilpotent of index n +1.

−  −  Proof. Since e is non-central, either eS(1 e) =0or (1 e)Se =0. Without loss of generality, we may −   ∈ −  M n M assume eS(1 e) =0. Take 0 = x eS(1 e), so ex =0. We note that n(R) = i=1 n(R)eii = Mn(R)e21 + Mn(R)e32 + ···+ Mn(R)en,n−1 + Mn(R)e1n. By Lemma 6.9, there exist nilpotents bi of index n−1 − ··· n such that bi = ei+1,i for i =1, ..., n 1. Moreover, if bn = e12 + e23 + + en−1,n, then bn is n−1  M  n−1  a nilpotent of index n and bn = e1n. Since ex =0, n(R)x =0, so bi x =0for some i. But n n n−1 n−1  n+1 n−1 2 (bi + x) = bi + bi x = bi x =0, and (bi + x) =(bi + x)bi x =0.

Theorem 6.11. Let R be of bounded index such that R/J(R) is an indecomposable ring. Then Mk(R) is ∼ nil-clean iff J(R) is nil and R/J(R) = Mn(Z2) for some n ≥ 1.

Proof. (⇐). Since R is of bounded index, J(R)nil implies that J(Mk(R)) = Mk(J(R)) is nil by Lemma 6.6. ∼ ∼ ∼ Moreover, if R/J(R) = Mn(Z2), then Mk(R)/J(Mk(R)) = Mk(R/J(R)) = Mnk(Z2)is nil-clean. So Mk(R) is nil-clean by [10, Corollary 3.17]. T. Koşan et al. / Journal of Pure and Applied Algebra 220 (2016) 633–646 645

(⇒). First the hypothesis implies that Mk(R/J(R)) is nil-clean and J(R)is nil, and hence it follows that R/J(R)is also of bounded index. Therefore, without loss of generality, we may assume that R is an indecomposable ring of bounded index such that J(R) =0. Let us assume that R has bounded index n. ∼ We next show that R = Mn(Z2). The claim is true for n =1by Corollary 6.3. So we assume n ≥ 2. Being nil-clean, Mk(R)is semipotent, so R is semipotent by [24, Corollary 1.7]. As R is of bounded index n, there exists r ∈ R such that rn = 0 but rn−1 =0. Since R is semipotent and J(R) =0, RrR contains a system 2 2 ∼ of n -matrix units by [21]. Thus, there exists e = e ∈ RrR such that eRe = Mn(A)for some nontrivial ring A. Since R is of bounded index n, e is central by Lemma 6.10. Hence e =1as R is indecomposable. ∼ Thus, R = Mn(A). Clearly, A is of bounded index m for some 1 ≤ m ≤ n. Next we show m =1. Assume m ≥ 2and let a ∈ A with am = 0 but am−1 =0. As A is also semipotent and semiprimitive, there exists 2 ∼ f = f ∈ AaA such that fAf = Mm(T )for a nontrivial ring T , and f must be central by Lemma 6.10. As ∼ ∼ A is also indecomposable, f =1A, so A = Mm(T ). This shows that R = Mnm(T ), and this clearly shows that R has bounded index greater than n. Hence, m =1, i.e., A is a reduced ring. Since A is indecomposable ∼ ∼ ∼ and Mk(R) = Mnk(A)is nil-clean, A = Z2 by Corollary 6.3. Hence R = Mn(Z2). 2

Acknowledgements

The authors are grateful to the referee whose valuable comments helped formulate Proposition 2.4 and simplify several proofs. Yiqiang Zhou acknowledges gratefully the support from TUBITAK of Turkey for his visit to the Gebze Technical University, and the hospitality received from the host university. The research of the second author was supported by the National Natural Science Foundation of China (No. 11201064) and the Fundamental Research Funds for the Central Universities of China (No. 2242014R30008), and the third author by a Discovery Grant from NSERC of Canada (No. 194196).

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