Strongly Nil-*-Clean Rings

A *-ring $R$ is called a strongly nil-*-clean ring if every element of $R$ is the sum of a projection and a nilpotent element that commute with each other. In this article, we show that $R$ is a strongly nil-*-clean ring if and only if every idempotent in $R$ is a projection, $R$ is periodic, and $R/J(R)$ is Boolean. For any commutative *-ring $R$, we prove that the algebraic extension $R[i]$ where $i^2=\mu i+\eta$ for some $\mu,\eta\in R$ is strongly nil-*-clean if and only if $R$ is strongly nil-*-clean and $\mu\eta$ is nilpotent. The relationships between Boolean *-rings and strongly nil-*-clean rings are also obtained.


Introduction
Let R be an associative ring with unity. A ring R is called strongly nil clean if every element of R is the sum of an idempotent and a nilpotent that commute. These rings are first discovered by Hirano-Tominaga-Yakub [11] and were refered to as [E-N]representable rings. In [8] and [9], Diesl refers to this class as strongly nil clean and studies their properties. Studying strongly nil cleanness is also relevent for Lie algebra. The decomposition of matrices as in the definition of strongly nil cleanness over a field must be the Jordan-Chevalley decomposition in Lie theory. An involution of a ring R is an operation * : R → R such that (x + y) * = x * + y * , (xy) * = y * x * and (x * ) * = x for all x, y ∈ R. A ring R with involution * is called a * -ring. An element p in a * -ring R is called a projection if p 2 = p = p * (see [2]). Recently the concept of strongly clean rings are considered for any * -ring. Vaš [15] calls a * -ring R strongly * -clean if each of its elements is the sum of a projection and a unit that commute with each other (see also [14]).
In this paper, we adapt strongly nil cleanness to * -rings. We call a * -ring R strongly nil * -clean if every element of R is the sum of a projection and a nilpotent element that commute. The paper consists of three parts. In Section 2, we characterize the class of strongly nil * -clean rings on several different ways. For example, we show that a ring R is a strongly nil * -clean ring if and only if every idempotent in R is a projection, R is periodic, and R/J(R) is Boolean. In Section 3, we prove a result related to the strongly nil * -cleanness of a commutative * -ring and its algebraic extension. For a commutative * -ring R with µ * = µ, η * = η ∈ R, R[i] = {a + bi | a, b ∈ R, i 2 = µi + η } is strongly nil * -clean if and only if R is strongly nil * -clean and µη is nilpotent. Foster [10] introduced the concept of Boolean-like rings as a generalization of Boolean rings in the commutative case. We adopt the concept of Boolean-like rings to rings with involution and prove that a * -ring R is * -Boolean-like if and only if R is strongly nil *clean and αβ = 0 for all nilpotent elements α, β in R. In the last section, we investigate submaximal ideals ( [12]) of strongly nil * -clean rings; and also define * -Boolean rings as * -rings over which every element is a projection and characterize them in terms of strongly nil * -cleanness.
Throughout this paper all rings are associative with unity (unless otherwise noted). We write J(R), N (R) and U (R) for the Jacobson radical of a ring R, the set of all nilpotent elements in R and the set of all units in R, respectively. The ring of all polynomials in one variable over R is denoted by R[x].

Characterization Theorems
The main purpose of this section is to explore the structure of strongly nil * -clean rings. A ring R is called uniquely nil clean if, for any x ∈ R, there exists a unique idempotent e ∈ R such that x − e ∈ N (R) [8]. If, in addition, x and e commute, R is called uniquely strongly nil clean [11]. Strongly nil cleanness and uniquely strongly nil cleanness are equivalent by [11,Theorem 3]. Analogously, for a * -ring, we define uniquely strongly nil * -clean rings by replacing "idempotent" with "projection" in the definition of uniquely strongly nil clean rings.
We will use the following lemma frequently. (i) R is strongly nil * -clean; (ii) R is uniquely nil clean and every idempotent in R is a projection; (iii) R is uniquely strongly nil * -clean.
Proof If R is strongly nil * -clean, then R is strongly * -clean. For, if x ∈ R, then there exist a projection e ∈ R and w ∈ N (R) such that 2 − x = e + w and ew = we. This gives that x = (1 − e) + (1 − w) where 1 − e is a projection and 1 − w ∈ U (R). If R is strongly * -clean, then every idempotent in R is a projection by [14,Theorem 2.2]. By [11,Theorem 3], the proof is completed.
We note that the condition "every idempotent in R is a projection" in Proposition 2.2 is necessary as the following example shows.
. Then R is a commutative * -ring with the usual matrix addition and multiplication. In fact, R is Boolean. Thus, for any x ∈ R, there exists a unique idempotent e ∈ R such that x − e ∈ R is nilpotent. But it is not strongly nil * -clean because the only projections are the trivial projections and there does not exist a projection e in R such that On the other hand, in [11,Theorem 3], it is proved that R is strongly nil clean if and only if N (R) is an ideal and R/N (R) is Boolean. Also, R is uniquely nil clean if and only if R is abelian, N (R) is an ideal and R/N (R) is Boolean [11,Theorem 4]. So if we adopt these results to rings with involution, immediately we have the following theorem by using Proposition 2.2. But we give a new proof to the necessity. (1) Every idempotent in R is a projection; (2) N (R) forms an ideal; Proof Assume that R is strongly nil * -clean. In view of Lemma 2.1, for any x ∈ R, there exist an idempotent g ∈ R and a nilpotent element v ∈ R such that x = g + v and gv = vg. Thus, Write (x − x 2 ) m = 0, and so x m ∈ x m+1 R. This shows that R is strongly π-regular. According to [1,Theorem 3], N (R) forms an ideal of R. Further, x − x 2 ∈ N (R), and so R/N (R) is Boolean. The converse is obvious by [11,Theorem 3].
A ring R is called strongly J- * -clean if for any x ∈ R there exists a unique projection e ∈ R such that x − e ∈ J(R) [7]. (1) R is strongly J- * -clean; (2) J(R) is nil.
Proof Suppose that R is strongly nil * -clean. In view of Theorem 2.4, N (R) forms an ideal of R, and this gives that N (R) ⊆ J(R). On the other hand, for any x ∈ J(R), there exists a projection e ∈ R such that x − e ∈ N (R). Then e = x − (x − e) ∈ J(R). This shows that e = 0, and so x is nilpotent. That is J(R) is nil, and so N (R) = J(R). In view of Proposition 2.2, we can see that there exists a unique projection e ∈ R such that x − e ∈ J(R). Hence R is strongly J- * -clean by [7,Theorem 3.3].
Conversely, assume that (1) and (2) hold. In view of [7, Proposition 2.1], R is strongly * -clean. Thus, R is abelian. Let x ∈ R. By virtue of [7,Theorem 3.3], there exist a projection e ∈ R and a w ∈ J(R) such that x = e + w and xe = ex. As J(R) is nil, w ∈ R is nilpotent. Therefore R is strongly nil * -clean.

From Lemma 2.5 and [7, Proposition 2.1], it follows that
The first inclusion is strict, because, for example, the power series ring Z 2 [[x]] is strongly J- * -clean but not strongly nil * -clean where * is the identity involution by [4, Example 2.5 (5)]. The second inclusion is also strict by [7, Example 2.2(2)].
We should note that a strongly nil clean ring may not be strongly J-clean (see [4, Example on p. 3799]). Hence strongly nil clean and strongly nil * -clean classes have different behavior when compared to classes of strongly J-clean and strongly J- * -clean classes respectively. Lemma 2.6 Let R be a * -ring. Then R is strongly nil * -clean if and only if (1) Every idempotent in R is a projection; (2) J(R) is nil; Proof Assume that (1), (2) and (3) hold. For any x ∈ R, x + J(R) = x 2 + J(R). As J(R) is nil, every idempotent in R lifts modulo J(R). Thus, we can find an idempotent e ∈ R such that x − e ∈ J(R) ⊆ N (R). By Lemma 2.1, xe = ex, and so the result follows. The converse is by Theorem 2.4 and Lemma 2.5.
Recall that a ring R is periodic if for any x ∈ R, there exist distinct m, n ∈ N such that x m = x n . With this information we can now prove the following. Theorem 2.7 Let R be a * -ring. Then R is strongly nil * -clean if and only if (1) Every idempotent in R is a projection; (2) R is periodic; Proof Suppose that R is strongly nil * -clean. By virtue of Lemma 2.6, every idempotent in R is a projection and Therefore the proof is completed by Lemma 2.6.

Proposition 2.8 A * -ring R is strongly nil * -clean if and only if
(1) R is strongly * -clean; Proof Suppose that R is strongly nil * -clean. By the proof of Lemma 2.5, Conversely, assume that (1) and (2) hold. Let a ∈ R. Then we can find a projection e ∈ R such that (a − 1) − e ∈ U (R) and e(a − 1) = (a − 1)e. That is, (1 − a) + e ∈ U (R). As 1 − (a − e) ∈ U (R), by hypothesis, a − e ∈ N (R). In addition, ea = ae. Accordingly, R is strongly nil * -clean.

Corollary 2.9
Let R be a * -ring. Then R is strongly nil * -clean if and only if so is Proof One direction is obvious. Conversely, assume that R is strongly nil * -clean.
, a 1 , · · · , a n−1 ∈ R}. Also note that R is abelian. Thus, it can be easily seen that every element in R[x]/(x n ) can be written as the sum of a projection and a nilpotent element that commute.

Algebraic Extensions
Let R be a commutative * -ring, and let µ, η ∈ R with µ * = µ and η * = η. Let The aim of this section is to explore the algebraic extensions of a strongly nil * -clean ring.
Since R is strongly nil * -clean, it follows from Theorem 2.4 that 2 − 2 2 ∈ N (R), and so 2 ∈ N (R). For any a + bi ∈ R[i], it is easy to verify that This shows that is Boolean. According to Theorem 2.4, we complete the proof.
As an immediate consequence, we deduce that a commutative * -ring R is strongly nil * -clean if and only if so is R[i] where i 2 = −1.
We now consider a subclass of strongly nil * -clean rings consisting of rings which we call * -Boolean-like. First recall that a ring R is called Boolean-like [10] if it is commutative with unit and is of characteristic 2 with ab(1 + a)(1 + b) = 0 for every a, b ∈ R. Any Boolean ring is clearly a Boolean-like ring but not conversely (see [10]). Any Boolean-like ring is uniquely nil clean by [  Since R is abelian, it follows that xy = yx. Hence R is commutative.
The following is an example of a * -Boolean-like ring.
Proof If R[i] is * -Boolean-like, then R is * -Boolean-like. Also µη ∈ R is nilpotent by Proposition 3.1 and Theorem 3.4. Since µ is unit and N (R) is an ideal, η is nilpotent.
We end this section with an example showing that strongly nil clean ring need not be strongly nil * -clean.
Example 3.7 Consider the ring Then for any x, y ∈ R, (x − x 2 )(y − y 2 ) = 0. Obviously, R is not commutative. This implies that R is not a * -Boolean-like ring for any involution * . Accordingly, R is not strongly nil * -clean for any involution * ; otherwise, every idempotent in R is a projection, a contradiction (see Lemma 2.1). We can also consider the involution

Submaximal Ideals and * -Boolean Rings
An ideal I of a ring R is called a submaximal ideal if I is covered by a maximal ideal of R. That is, there exists a maximal ideal J of R such that I J R and for any ideal K of R such that I ⊆ K ⊆ J then we have I = K or K = J. This concept was firstly introduced to study Boolean-like rings (cf. [12]).
A * -ring R is called a * -Boolean ring if every element of R is a projection. The purpose of this section is to characterize submaximal ideals of strongly nil *clean rings, and * -Boolean rings by means of strongly nil * -cleanness. We begin with the following lemma. (2) For any a ∈ R, n ≥ 1, a n ∈ M implies that a ∈ M .
Proof Suppose that M is maximal. Obviously, M is prime. Let a ∈ R and a n ∈ M . If a ∈ M , RaR + M = R. Thus, RaR = R where R = R/M and a = a + M . Clearly, R is an abelian clean ring, and so it is an exchange ring by [5,Theorem 17.2.2]. This implies that R/M is an abelian exchange ring. As in the proof of [5, Proposition 17.1.9], there exists an idempotent e + M ∈ R/M such that R(e + M )R = R and e + M ∈ aR. Thus, 1 − e ∈ M . Hence, 1 − ar ∈ M for some r ∈ R. This implies that a n−1 − a n r ∈ M , and so a n−1 ∈ M . By iteration of this process, we see that a ∈ M , as required.
Conversely, assume that (1) and (2) hold. Assume that M is not maximal. Then we can find a maximal ideal I of R such that M I R. Choose a ∈ I while a ∈ M . By hypothesis, there exist an idempotent e ∈ R and a nilpotent u ∈ R such that a = e + u. Write u m = 0. Then u m ∈ M . By hypothesis, u ∈ M . This shows that e ∈ M . Clearly, R is abelian. Thus eR(1 − e) ⊆ M . As M is prime, we deduce that 1 − e ∈ M . As a result, 1 − a = (1 − e) − u ∈ M , and so 1 = (1 − a) + a ∈ I. This gives a contradiction. Therefore M is maximal.
Let R be a strongly nil * -clean ring, and let x ∈ R. Then there exists a unique projection e ∈ R such that x − e ∈ N (R). We denote e by x P and x − e by x N . Lemma 4.2 Let I be an ideal of a strongly nil * -clean ring R, and let x ∈ R be such that x ∈ I. If x P ∈ I, then there exists a maximal ideal J of R such that I ⊆ J and x ∈ J.
Then Q is an ideal of R. If Q ∈ Ω, then x P ∈ Q, and so x P ∈ K i for some i. This gives a contradiction. Thus, Ω is inductive. By using Zorn's Lemma, there exists an ideal J of R which is maximal in Ω. Let a, b ∈ R such that a, b ∈ J. By the maximality of J, we see that RaR + J, RbR + J ∈ Ω. This shows that x P ∈ RaR + J RbR + J . Hence, x P = x 2 P ∈ RaRbR + J. This yields that aRb ∈ J; otherwise, x P ∈ J, a contradiction. Hence, J is prime. Assume that J is not maximal. Then we can find a maximal ideal M of R such that J M R. Clearly, R is abelian. By the maximality, we see that x P ∈ M , and so 1 − x P ∈ M . This implies that 1 − x P ∈ J. As x P R(1 − x P ) = 0 ⊆ J, we have that x P ∈ J, a contradiction. Therefore J is a maximal ideal, as asserted. Proposition 4.3 Let R be strongly nil * -clean. Then the intersection of two maximal ideals is submaximal and it is covered by each of these two maximal ideals. Further, there is no other maximal ideals containing it.
Proof Let I 1 and I 2 be two distinct maximal ideals of R. Then I 1 I 2 I 1 . Suppose I 1 I 2 ⊆ J I 1 . Then we can find some x ∈ I 1 while x ∈ J. Write x n N = 0. Then x n N ∈ I 1 . In light of Lemma 4.1, x N ∈ I 1 . Likewise, x N ∈ I 2 . Thus, x N ∈ I 1 I 2 ⊆ J. This shows that x P ∈ J. By virtue of Lemma 4.2, there exists a maximal ideal M of R such that J ⊆ M and x ∈ M . Hence, I 1 I 2 ⊆ M and I 1 = M . If I 2 = M , then I 2 + M = R. Write t + y = 1 with t ∈ I 2 , y ∈ M . Then for any z ∈ I 1 , z = zt + zy ∈ I 1 I 2 + M = M , and so I 1 = M . This gives a contradiction. Thus I 2 = M , and then J ⊆ M ⊆ I 2 . As a result, J ⊆ I 1 I 2 , and so I 1 I 2 = J. Therefore I 1 I 2 is a submaximal ideal of R. We claim that I 1 I 2 is semiprime. If K 2 ⊆ I 1 I 2 , then for any a ∈ K, we see that a 2 ∈ I 1 I 2 . In view of Lemma 4.1, a ∈ I 1 I 2 . This implies that K ⊆ I 1 I 2 . Hence, I 1 I 2 is semiprime. Therefore I 1 I 2 is the intersection of maximal ideals containing I 1 I 2 . Assume that K is a maximal ideal of R such that I 1 I 2 ⊆ K. If K = I 1 , I 2 , then I 1 + K = I 2 + K = R. This implies that I 1 I 2 + K = R, and so K = R, a contradiction. Thus, K = I 1 or K = I 2 , and so the proof is completed.
We call a local ring R absolutely local provided that for any 0 = x ∈ J(R), J(R) = RxR.
Corollary 4.4 Let R be strongly nil * -clean, and let I be an ideal of R. Then I is a submaximal ideal if and only if R/I is Boolean with four elements or R/I is absolutely local.
Proof Let I be a submaximal ideal of R.
Case I. I is contained in more than a maximal ideal. Then I is contained in two distinct maximal ideals of R. Since I is submaximal, there exists a maximal ideal J of R such that I is covered by J. Thus, we have a maximal ideal J ′ such that J ′ = J and I J ′ . Hence, I ⊆ J J ′ ⊆ J. Clearly, J J ′ = J as J + J ′ = R, and so I = J J ′ . In view of Proposition 4.3, there is no maximal ideal containing I except for J and J ′ . This shows that R/I has only two maximal ideals covering {0 + I}. For any a ∈ R, it follows from Theorem 2.4 that a − a 2 ∈ R is nilpotent. Write (a − a 2 ) n = 0. Then (a − a 2 ) n ∈ J. According to Lemma 4.1, a − a 2 ∈ J. Likewise, a − a 2 ∈ J ′ . Thus, a − a 2 ∈ J J ′ , and so a − a 2 ∈ I. This shows that R/I is Boolean. Therefore R/I is Boolean with four elements.
Case II. Suppose that I is contained in only one maximal ideal J of R. Then R/I has only one maximal ideal J/I. Clearly, R is an abelian exchange ring, and then so is R/I. Let e ∈ R/I be a nontrivial idempotent. Then I ⊆ I + ReR ⊆ J or I + ReR = R. Likewise, I ⊆ I + R(1− e)R ⊆ J or I + R(1− e)R = R. This shows that I + ReR = R or I + R(1 − e)R = R Thus, (R/I)(e + I)(R/I) = R/I or (R/I)(1 − e + I)(R/I) = R/I, a contradiction. Therefore all idempotents in R/I are trivial. It follows from [5, Lemma 17.2.1] that R/I is local. For any 0 = x ∈ J/I, we see that 0 = I ⊆ RxR ⊆ J. As I is submaximal, we deduce that J = RxR. Therefore R is absolutely local.
Conversely, assume that R/I is Boolean with four elements. Then R/I has precisely two maximal ideals covering {0+I}, and so R has precisely two maximal ideals covering I. Thus, we have a maximal ideal J such that I J. If I ⊆ K ⊆ J. Then K = I or K is maximal, and so K = J. Consequently, I is submaximal. Assume that R/I is absolutely local. Then R/I has a uniquely maximal ideal J/I. Hence, J is a maximal ideal of R such that I J. Assume that I K ⊆ J. Choose a ∈ K while a ∈ I. Then J = RaR ⊆ K, and so K = J. Therefore I is submaximal, as required.
Corollary 4.5 Let R be strongly nil * -clean. If I 1 and I 2 are distinct maximal ideals of R, then R/(I 1 I 2 ) is Boolean.
Proof Since I 1 /(I 1 I 2 ) and I 2 /(I 1 I 2 ) are distinct maximal ideals, R/(I 1 I 2 ) is not local. In view of Proposition 4.3, I 1 I 2 is a submaximal ideal of R. Therefore we complete the proof from Corollary 4.4.
Recall that an ideal I of a commutative ring R is primary provided that for any x, y ∈ R, xy ∈ I implies that x ∈ I or y n ∈ I for some n ∈ N. Clearly, every maximal ideal of a commutative ring is primary. We end this article by giving the relation between strongly nil * -clean rings and * -Boolean rings.
Lemma 4.6 Let R be a commutative strongly nil * -clean ring. Then the intersection of all primary ideals of R is zero.
Proof Let a be in the intersection of all primary ideal of R. Assume that a = 0. Let Ω = {I | I is an ideal of R such that a ∈ I}. Then Ω = ∅ as 0 ∈ Ω. Given any ideals Then M ∈ Ω. Thus, Ω is inductive. By using Zorn's Lemma, we can find an ideal Q which is maximal in Ω. It will suffice to show that Q is primary. If not, we can find some x, y ∈ R such that xy ∈ Q, but x ∈ Q and y n ∈ Q for any n ∈ N. This shows that a ∈ Q + (x), and so a = b + cx for some b ∈ Q, c ∈ R. Since R is strongly nil * -clean, it follows from Theorem 2.7 that there are some distinct k, l ∈ N such that y k = y l . Say k > l. Then y l = y k = y l+1 y k−l−1 = y l yy k−l−1 = y l+2 y 2(k−l−1) = · · · = y 2l y l(k−l−1) . Hence, y l(k−l) = y l (y l(k−l−1) ) = y 2l y 2l(k−l−1) = y l(k−l) 2 . Choose s = l(k − l). Then y s is an idempotent. Write y = y P + y N . Then y s − y P = (y P + y N ) s − y P = y N sy P + · · · + y s−1 N ∈ N (R). As R is a commutative ring, we see that (y s − y P ) 3 = y s − y P . This implies that y s = y P . Since xy ∈ Q, we show that xy s ∈ Q, and so xy P ∈ Q. It follows from a = b + cx that ay P = by P + cxy P ∈ Q. Clearly, y s ∈ Q, and so a ∈ Q + (y P ). Write a = d + ry P for some d ∈ Q, r ∈ R. We see that ay P = dy P + ry P , and so ry P ∈ Q. This implies that a ∈ Q, a contradiction. Therefore Q is primary, a contradiction. Consequently, the intersection of all primary ideal of R is zero.  (2) Every primary ideal of R is maximal; (3) R is strongly nil * -clean.
Proof Suppose that R is a * -Boolean ring. Clearly, R is a commutative strongly nil * -clean ring. Let I be a primary ideal of R. If I is not maximal, then there exists a maximal ideal M such that I M R. Choose x ∈ M while x ∈ I. As x is an idempotent, we see that xR(1 − x) ⊆ I, and so (1 − x) m ∈ I ⊂ M for some m ∈ N. Thus, 1 − x ∈ M . This implies that 1 = x + (1 − x) ∈ M , a contradiction. Therefore I is maximal, as required.
Conversely, assume that (1), (2) and (3) hold. Clearly, every maximal ideal of R is primary, and so J(R) = {P | P is primary}. In view of Lemma 4.6, J(R) = 0. Hence every element is a projection i.e. R is * -Boolean.

Corollary 4.8 A ring R is a Boolean ring if and only if
(1) R is commutative; (2) Every primary ideal of R is maximal; (3) R is strongly nil clean.
Proof Choose the involution as the identity. Then the result follows from Theorem 4.7.