EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 3, 2023, 1675-1684 ISSN 1307-5543 – ejpam.com Published by New York Business Global Strongly 2-Nil Clean Rings with Units of Order Two Renas T. M.Salim1, Nazar H. Shuker2,∗ 1 Department of Mathematics, University of Zakho, Faculty of Science, Zakho, Iraq 2 Department of Math. College of Computer Science and Math. Mosul University, Mosul, Iraq Abstract. A ring R is considered a strongly 2-nil clean ring, or (strongly 2-NC ring for short), if each element in R can be expressed as the sum of a nilpotent and two idempotents that commute with each other. In this paper, further properties of strongly 2-NC rings are given. Furthermore, we introduce and explore a special type of strongly 2-NC ring where every unit is of order 2, which we refer to as a strongly 2-NC rings with U(R) = 2. It was proved that the Jacobson radical over a strongly 2-NC ring is a nil ideal, here, we demonstrated that the Jacobson radical over strongly 2-NC ring with U(R) = 2 is a nil ideal of characteristic 4. We compare this ring with other rings, since every SNC ring is strongly 2-NC, but not every unit of order 2, and if R is a strongly 2-NC with U(R) = 2, then R need not be SNC ring. In order to get Nil(R) = 0, we added one more condition involving this ring. 2020 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Clean, Nil clean, Strongly 2-nil clean, Tripotent 1. Introduction In [1] W.K. Nicholson defined a clean ring as having an Σ = Σ2 and a unit u with a = Σ + u. In [2], an element a ∈ R is said to be strongly clean if a = Σ + u with u ∈ U(R),Σ ∈ Id(R) and uΣ = Σu. While the ring R is strongly clean if every element of R is strongly clean. Clearly, Z9 is a strongly clean ring. A nil-clean ring is defined as a ring with each element is the sum of an idempotent and a nilpotent was first proposed by Diesl in [3], R is considered a strongly nil clean (SNC for short) if the idempotent and nilpotent commute [4]. The structure of SNC rings and related topics was given for example in [5] and [6]. Clearly, Z8 is an SNC ring. A strongly 2-NC ring was defined by Chen and Sheibani in [7] as a ring R with each element is a sum of two idempotents and a nilpotent that commute with each other. Many authors have been working on these topics see for example [8] a ring R is called strongly ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i3.4797 Email addresses: renas.salim@uoz.edu.krd (R. T. M.Salim), nazarh 2013@yahoo.com (N. H. Shuker) https://www.ejpam.com 1675 © 2023 EJPAM All rights reserved. R. T. M.Salim, N. H. Shuker / Eur. J. Pure Appl. Math, 16 (3) (2023), 1675-1684 1676 2-nil-*-clean if every element in R is the sum of two projections and a nilpotent that commute, [9] if every element in R is the sum of an idempotent and two nilpotents , then R is called 2-nil-clean and [10] a ring R is defined to be 2-nil-good if every element in R is the sum of two units and a nilpotent. The purpose of this paper is to present new properties of strongly 2-NC rings, and their connection with other related rings. We prove that if R is a strongly 2-NC ring, with n2 + 2n = 0 for every n ∈ Nil(R). Then R is of characteristic 48 with every unit is of order 4. Additionally, we introduce and investigate a strongly 2-NC rings with U(R) = 2, providing their fundamental properties and their connection with tripotent rings and other related rings. Among other results we prove that: If R is a strongly 2-NC ring with 2 ∈ U(R). Then 24 = 0, and the Jacobson radical over a strongly 2-NC ring is a nil ideal of characteristic 4. In addition, we show that if R is a strongly 2-NC ring with U(R) = 2 and 2 ∈ U(R), then Nil(R) = 0. In this paper, we define R as an associative ring containing an identity element. Finally, it is worth mentioning that ring theory has several applications in many field, see for example [11], [12] and [13]. To represent the set of units, idempotents and nilpotents in R, we will use the symbols U(R), Id(R) and Nil(R), respectively. Additionally, we will use J(R) to denote the Jacobson radical and Zn for the ring of integers modulo n. Recall that: Definition 1. [14]. A ring R is considered to be n-good if each element is a sum of n units. Definition 2. [15]. If t = t3 is referred to as a tripotent. R is called a tripotent ring if every element of R is tripotent. Clearly, Z6 is a tripotent ring. Definition 3. For any a ∈ R, we define Ann(a) = {b ∈ R : ab = ba = 0}. Theorem 1. [7]. Let R be a ring. Then the following are equivalent: 1. R is strongly 2-NC. 2. For all a ∈ R, a− a3 ∈ Nil(R). 3. For all a ∈ R, a2 is SNC element. Theorem 2. [7]. A ring R is strongly 2-NC if and only if 1. J(R) is nil. 2. R/J(R) is tripotent. Theorem 3. [16] The following are equivalent for a ring R: 1. Every element of R is a sum of a nilpotent and two tripotents that commute with one another. 2. a5 − a is nilpotent for all a ∈ R. R. T. M.Salim, N. H. Shuker / Eur. J. Pure Appl. Math, 16 (3) (2023), 1675-1684 1677 2. Fundamental properties of strongly 2-NC rings This section presents new properties of strongly 2-NC rings, and we provide a condition for strongly 2-NC rings to be tripotent rings. Example 1. Consider the ring Z18. Note that: Nil(Z18) = {0, 6, 12}, and Id(Z18) = {0, 1, 9, 10}. By direct calculation, we may find that Z18 is a strongly 2-NC. Chen and Sheibani in [7] proved that: Lemma 1. The following two issues are equivalent: 1. R is a strongly 2-NC ring. 2. a = Σ1 − Σ2 + n, for each a ∈ R, and some Σ1,Σ2 ∈ Id(R), n ∈ Nil(R), that commute. Next, we shall record the following two lemmas, that will be used extensively through- out our current work. Lemma 2. [17]. If u ∈ U(R) and n ∈ Nil(R), and if un = nu, then 1 + n and u+ n are units. Lemma 3. Suppose that Σ1 and Σ2 are two commuting idempotents. Then: 1. (Σ1 − Σ2) 2 is an idempotent. 2. (Σ1 − Σ2) 3 is tripotent. 3. (Σ1 − Σ2) 2 + (Σ1 − Σ2)− 1 is a unit of order 2. 4. 2(Σ1 − Σ2) 2 − 1 is a unit of order 2. Proof. 1. (Σ1 − Σ2) 4 = Σ4 1 − 4Σ3 1Σ2 + 6Σ2 1Σ 2 2 − 4Σ1Σ 3 2 +Σ4 2 = Σ1 − 4Σ1Σ2 + 6Σ1Σ2 − 4Σ1Σ2 +Σ2 = (Σ1 − Σ2) 2. 2. (Σ1 − Σ2) 3 = Σ3 1 − 3Σ2 1Σ2 + 3Σ1Σ 2 2 − Σ3 2 = Σ1 − 3Σ1Σ2 + 3Σ1Σ2 − Σ2 = (Σ1 − Σ2). 3. ((Σ1 − Σ2) 2 + (Σ1 − Σ2)− 1)((Σ1 − Σ2) 2 + (Σ1 − Σ2)− 1) = (Σ1 − Σ2) 4 + (Σ1 − Σ2) 3 − (Σ1 − Σ2) 2 + (Σ1 − Σ2) 3 + (Σ1 − Σ2) 2 − (Σ1 − Σ2)− (Σ1 − Σ2) 2 − (Σ1 − Σ2) + 1 = (Σ1 − Σ2) 2 + (Σ1 − Σ2)− (Σ1 − Σ2) 2 + (Σ1 − Σ2) + (Σ1 − Σ2) 2 − (Σ1 − Σ2)− (Σ1 − Σ2) 2 − (Σ1 − Σ2) + 1 = 1. R. T. M.Salim, N. H. Shuker / Eur. J. Pure Appl. Math, 16 (3) (2023), 1675-1684 1678 4. (2(Σ1 − Σ2) 2 − 1)(2(Σ1 − Σ2) 2 − 1) = 4(Σ1 − Σ2) 4 − 2(Σ1 − Σ2) 2 − 2(Σ1 − Σ2) 2 + 1 = 4(Σ1 − Σ2) 2 − 2(Σ1 − Σ2) 2 − 2(Σ1 − Σ2) 2 + 1 = 1. Next, we shall give the following results. Proposition 1. Let R be a strongly 2-NC ring, then for any a ∈ R we have: 1. a2 is an SNC. 2. a2 is the sum of a tripotent and a nilpotent that commute. 3. a2 is the sum of an idempotent, a unit of order 2, and a nilpotent that commutes. Proof. 1. Given a ∈ R, there existing some Σ1,Σ2 ∈ Id(R) and n ∈ Nil(R), that commute with one another, such that a = Σ1−Σ2+n. Thus, a2 = (Σ1−Σ2) 2+2(Σ1−Σ2)n+n2. But 2(Σ1 − Σ2)n + n2 = n1 ∈ Nil(R), so a2 = (Σ1 − Σ2) 2 + n1. According to Lemma 3(1) (Σ1 − Σ2) 2 is an idempotent. Yielding a2 is an SNC element. 2. Follows from Lemma 3(2). 3. By (1) a2 is a SNC element, then a2 = Σ + n where Σ ∈ Id(R), n ∈ Nil(R) that commute, we may write a2 = (1 − Σ) + (2Σ − 1) + n. Clearly, (1 − Σ)2 = 1−Σ, (2Σ− 1)2 = 1. Thus, a2 is the sum of an idempotent, a unit of order 2, and a nilpotent. Proposition 2. Suppose R is a ring, and let a ∈ R. Then: 1. If a2 is a strongly 2-NC, then a and −a are strongly clean. 2. If a2 is a strongly 2-NC, then a is the sum of two tripotents and a nilpotent commute one another. Proof. 1. Take a2 = Σ1 − Σ2 + n, by Proposition 1(1), a4 is a SNC element, so a4 = Σ + n where Σ ∈ Id(R), n ∈ Nil(R) that commute. Write a4 = (1 − Σ) + (2Σ − 1) + n. But (2Σ − 1)2 = 1, then (2Σ − 1) + n = u1 ∈ U(R). So a4 = (1 − Σ) + u1, implies a4 − (1 − Σ) = u1, but (1 − Σ)4 = 1 − Σ, yields (a2 − (1 − Σ))(a2 + (1 − Σ)) = u1. and hence, (a− (1−Σ))(a+ (1−Σ))(a2 + (1−Σ)) = u1. Thus, a− (1−Σ) ∈ U(R) and −a− (1− Σ) ∈ U(R). 2. Let a in R. Applying Theorem 1, (a2)3 − a2 ∈ Nil(R). Hence a(a5 − a) ∈ Nil(R), so (a4 − 1)a(a5 − a) = (a5 − a)2 ∈ Nil(R). Using Theorem 3, a is a sum of two tripotents and a nilpotent that commute. R. T. M.Salim, N. H. Shuker / Eur. J. Pure Appl. Math, 16 (3) (2023), 1675-1684 1679 Proposition 3. Suppose R is a strongly 2-NC ring, and a = Σ1 − Σ2 + n for any a ∈ R. Then: 1. Ann(a) ∩ (Σ1 − Σ2)R = 0. 2. If 2 ∈ U(R), then a is 3-good element. 3. If a ∈ U(R), then (Σ1 − Σ2) 2 = 1. 4. If a is a non-zero divisor, then a ∈ U(R). Proof. 1. Let c ∈ Ann(a)∩(Σ1−Σ2)R. Then ac = ca = 0 and c = (Σ1−Σ2)r, for some r ∈ R. Hence a(Σ1−Σ2)r = 0, so (Σ1−Σ2+n)(Σ1−Σ2)r = 0, (Σ1−Σ2) 2+n(Σ1−Σ2) = 0. Applying Lemma 3, we get ((Σ1 −Σ2) 2 + n(Σ1 −Σ2) 3)r = 0, (Σ1 −Σ2) 2(1+ n(Σ1 − Σ2))r = 0. But 1 + n(Σ1 − Σ2) ∈ U(R), say u, then we have (Σ1 − Σ2) 2ur = 0, so (Σ1 − Σ2) 2r = 0. Multiply by (Σ1 − Σ2), we have (Σ1 − Σ2)r = c = 0. Therefore, Ann(a) ∩ (Σ1 − Σ2)R = 0. 2. We may write a = Σ1+1+Σ2+1+n−2. Consider (Σ1+1)(2−Σ1) = 2Σ1−Σ1+2− Σ1 = 2. Since 2 ∈ U(R), then Σ1 + 1 = u1 ∈ U(R). Similarly Σ2 + 1 = u2 ∈ U(R). Furthermore, n− 2 ∈ U(R), say u3. Thus, a = u1 + u2 + u3. 3. Let a = (Σ1−Σ2)+n, and let a ∈ U(R). Then a−n = (Σ1−Σ2) ∈ U(R). Applying Lemma 3(2), then (Σ1 − Σ2) = (Σ1 − Σ2) 3. Thus (Σ1 − Σ2) 2 = 1. 4. Let a be a non-zero divisor element. Applying Theorem 1, a3 − a ∈ Nil(R), this gives a(a2 − 1) ∈ Nil(R), thus, ar(a2 − 1)r = 0, for some positive integer r. Since ar is a non-zero divisor, then (a2 − 1)r = 0, so a2 − 1 = n1 ∈ Nil(R), implies a2 = 1 + n1 ∈ U(R), then a ∈ U(R). It was proved in [18], that. Proposition 4. [18, Proposition 1]. Assume R is a nil clean ring with every nilpotent is the difference between two commuting idempotents, then R is a Boolean ring. We here extend this result as follows: Theorem 4. Suppose R is a strongly 2-NC ring, with any nilpotent is the difference between two commuting idempotents. Then R is a tripotent ring. Proof. Let a in R, then a = Σ1−Σ2+n for some existing Σ1,Σ2 ∈ Id(R), n ∈ Nil(R), that commute which each other. Then n = Σ3 −Σ4 for some Σ3,Σ4 ∈ Id(R) and Σ3Σ4 = Σ4Σ3. So n+Σ4 = Σ3, this implies (n+Σ4) 2 = (n+Σ4), then n2 +2nΣ4 +Σ2 4 = n+Σ4, this gives n2 + 2nΣ4 − n = 0, so n2 + n(2Σ4 − 1) = 0, but (2Σ4 − 1)2 = 1, then we have n = −n2(2Σ4 − 1)−1. As n is nilpotent, then n = 0. Thus, a = Σ1 − Σ2. Applying Lemma 3(2), (Σ1 − Σ2) 3 = Σ1 − Σ2. Hence, a = a3 therefore, R is a tripotent ring. R. T. M.Salim, N. H. Shuker / Eur. J. Pure Appl. Math, 16 (3) (2023), 1675-1684 1680 3. Strongly 2-NC rings with units of order two In this section, we introduce and investigate a strongly 2-NC rings with every unit is of order 2, we refer to this type of ring as strongly 2-NC rings with U(R) = 2. Definition 4. A ring R is called strongly 2-NC with U(R) = 2 if for every a ∈ R, existing two idempotents Σ1,Σ2 and a nilpotent n, that commute and every unit is of order 2, such that a = Σ1 +Σ2 + n. Example 2. The rings Z4, Z6, Z8, Z12, Z24 are all strongly 2-NC with U(R) = 2, while the ring Z9 is not strongly 2-NC with U(R) = 2. We start this section with some fundamental properties of a strongly 2-NC ring with U(R) = 2. Proposition 5. Homomorphic images of strongly 2-NC ring with U(R) = 2 is again strongly 2-NC ring with every unit is of order 2. Proof. Let f : R → R′ be a homomorphism from a strongly 2-NC ring R with U(R) = 2 onto R′. Then for any b ∈ R′, there exists a ∈ R, such that b = f(a), a = Σ1 +Σ2 +n and U(R) = 2, where Σ1,Σ2 ∈ Id(R), n ∈ Nil(R) that commute of with one another. Now, b = f(a) = f(Σ1 + Σ2 + n) = f(Σ1) + f(Σ2) + f(n). Clearly, f(Σ1), f(Σ2) ∈ Id(R′) and f(n) ∈ Nil(R′). On the other hand for any u ∈ (R), where u is a unit, (f(u))2 = f(u2) = f(1), this shows that f(u) is a unit of order 2. Therefore R′ is a strongly 2-NC ring with U(R′) = 2. Proposition 6. If R is a strongly 2-NC ring with U(R) = 2. Then 24 = 0. Proof. Assume that a in R, then existing two idempotents Σ1,Σ2 and a nilpotent n that commute with one another, such that a = Σ1 − Σ2 + n. By Theorem 1, a3 − a ∈ Nil(R), this gives 23 − 2 = 6 ∈ Nil(R). Since every unit is of order 2, and since 6 is nilpotent, then 6− 1 = 5 ∈ U(R). This gives 52 = 1, so 24 = 0. Example 3. Consider the ring Z24. Clearly, Z24 is a strongly 2-NC, with U(Z24) = {1, 5, 7, 11, 13, 17, 19, 23}. Observe that 12 = 52 = 72 = 112 = 132 = 172 = 192 = 232 = 1. Observe that every SNC ring is strongly 2-NC, but not every unit of order 2. Example 4. The ring Z16 is an SNC which is strongly 2-NC, but Z16 is not strongly 2-NC ring with U(R) = 2, since the units 3, 5, 11, 13 are not of order 2. Note that: If R is a strongly 2-NC with U(R) = 2, then R need not to be SNC ring. Example 5. In the ring Z12. Then U(Z12) = {1, 5, 7, 11} and 12 = 52 = 72 = 112 = 1. Clearly, Z12 is a strongly 2-NC with U(Z12) = 2, but (Z12) is not SNC ring. Since 2 is not SNC element. R. T. M.Salim, N. H. Shuker / Eur. J. Pure Appl. Math, 16 (3) (2023), 1675-1684 1681 Proposition 7. If a ring R is a strongly 2-NC ring with U(R) = 2, for which 3 ∈ U(R), then R is SNC ring of characteristic 8. Proof. Assume R is a strongly 2-NC ring with U(R) = 2. Then By Proposition 1(1), a2 is a SNC element for every a ∈ R. Applying Proposition 2(1), a is strongly clean. Then a may be written a − 1 = Σ + u, where Σ ∈ Id(R) and u2 = 1. Then a = Σ + u + 1. Since 3 ∈ U(R), then 32 = 1, gives 8 = 0 thus, 2 ∈ Nil(R). So (u + 1)2 = u2 + 2u + 1 = 2(u+ 1) ∈ Nil(R). Thus, u+ 1 ∈ Nil(R). Therefore R is an SNC ring. Example 6. Consider the ring Z8. Then U(Z8) = {1, 3, 5, 7}. So 12 = 32 = 52 = 72 = 1. Clearly, Z8 is a strongly 2-NC with U(Z8) = 2. Observe that 3 ∈ U(Z8). Then Z8 is an SNC ring. It was proved in Theorem 2, if a ring R is a strongly 2-NC, then J(R) is nil. In the next result, we consider J(R) over a strongly 2-NC ring with U(R) = 2. Theorem 5. If R is a strongly 2-NC ring with U(R) = 2, then J(R) is nil of characteristic 4. Proof. Given a ∈ J(R), then a = Σ1 −Σ2 + n, where Σ1,Σ2 ∈ Id(R) and n ∈ Nil(R), that commute with one another. Write a = 1−(Σ1−Σ2) 2+(Σ1−Σ2) 2+(Σ1−Σ2)−1+n. According to Lemma 3(3), (Σ1 − Σ2) 2 + (Σ1 − Σ2) − 1 = u1 is a unit of order 2, then a = 1−(Σ1−Σ2) 2+u1+n implies a = 1−(Σ1−Σ2) 2+u2, where u2 = u1+n. Since a ∈ J(R), so a − u2 ∈ U(R), applying to Proposition 3(3), we conclude that 1 − (Σ1 − Σ2) 2 = 1, gives (Σ1 − Σ2) 2 = 0, whence it follows that a = 1 + u2, with u22 = 1. Now consider a2 = (1 + u2) 2 = 2(1 + u2) = 2a, and a3 = 22(1 + u2) = 4a. Choose a = 2b, then (2b)3 = 4(2b), so 8b3 = 8b, implies 8b(1 − b2) = 0, but b ∈ J(R), gives 1 − b2 ∈ U(R), gives 8b = 0. Thus 4a = a3 = 0. Example 7. Consider the ring Z24. Then Z24 is a strongly 2-NC with U(Z24) = 2. Now J(Z24) = {0, 6, 12, 18}. So J(Z24) is a nil ideal of characteristic 4. Corollary 1. If R is a strongly 2-NC ring with U(R) = 2 and if 2 ∈ U(R), then J(R) = 0. Proof. Let a ∈ J(R), then by Theorem 5, 4a = 0, since 2 ∈ U(R), then a = 0. Proposition 8. If R is a strongly 2-NC ring with U(R) = 2, and if 2 ∈ U(R), then Nil(R) = 0. Proof. Given a ∈ R, then a − 1 = Σ1 − Σ2 + n, so a = Σ1 − Σ2 + n + 1, but n+1 ∈ U(R), say u, then a = Σ1−Σ2+u. Let n ∈ Nil(R), then n = Σ1−Σ2+u, implies Σ1 − Σ2 = n − u, since n − u ∈ U(R), according to Proposition 3(3), (Σ1 − Σ2) 2 = 1. Furthermore, n2 = (Σ1−Σ2) 2+2(Σ1−Σ2)u+u2 = 1+2(Σ1−Σ2)u+1 = 2(1+(Σ1−Σ2)u). Observe that nu = (Σ1 − Σ2)u + 1. Thus, n2 = 2nu, so n(n − 2u) = 0. since 2 ∈ U(R), by assumption then n− 2u ∈ U(R). Whence it follows that n = 0. Next, we shall explore the relationship between strongly 2-NC ring with U(R) = 2 and a tripotent ring. R. T. M.Salim, N. H. Shuker / Eur. J. Pure Appl. Math, 16 (3) (2023), 1675-1684 1682 Theorem 6. A ring R with 2 ∈ U(R) is strongly 2-NC with U(R) = 2 if and only if R is a tripotent. Proof. Let R be a strongly 2-NC ring with U(R) = 2, and let a ∈ R, then a = Σ1 −Σ2 + n, where Σ1,Σ2 ∈ Id(R), n ∈ Nil(R), that commute with one another. Accord- ing to Proposition 8, n = 0. Thus, a = Σ1 − Σ2 = (Σ1 − Σ2) 3 = a3. Conversely, assume that R is a tripotent ring, and t = t3 ∈ R, since 2 ∈ U(R), then t may be written as t = t2+t 2 − t2−t 2 . Note that: ( t 2+t 2 )2 = t2+2t+t2 4 = t2+t 2 , and ( t 2−t 2 )2 = t2−2t+t2 4 = t2−t 2 , so ( t 2+t 2 ), ( t 2−t 2 ) ∈ Id(R). Observe that for any unit u, u3 = u thus, u2 = 1. Therefore, R is a strongly 2-NC ring with U(R) = 2. To end this section, we consider a strongly 2-NC ring, with every unit is of order 4. Proposition 9. Suppose R is a strongly 2-NC ring, and if n2+2n = 0 for every nilpotent n. Then every unit of R is of order 4, and 48 = 0. Proof. Given a ∈ R, then by Proposition 1(1), a2 is an SNC element. Write a2 = Σ+n, where Σ ∈ Id(R), n ∈ Nil(R) and Σn = nΣ. Let u ∈ U(R), then u2 = Σ + n, implies Σ = u2−n = v ∈ U(R). Thus, Σ = 1. Hence u2 = 1+n, implies u4 = (1+n)2 = 1+2n+n2. By assumption n2 + 2n = 0, then u4 = 1. On the other hand 6 ∈ Nil(R) Theorem 1. Thus, 62 + 2(6) = 0, gives 48 = 0. Example 8. In the ring Z48. Then U(Z48) = {1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47}, Nil(Z48) = {0, 6, 12, 18, 24, 30, 36, 42}, Id(Z48) = {0, 1, 16, 33}. By direct calculation, one easily check that Z48 is a strongly 2-NC ring, with every unit is of order 4. 4. Conclusion In this article, new properties of a strongly 2-NC rings are given. Additionally, we added certain conditions for strongly 2-NC ring with each unit must be present of order four. We also introduce and investigated a strongly 2-NC ring with every unit of order two. We discuss some of the fundamental properties and present several examples. It was proved that the Jacobson radical over a strongly 2-NC ring is a nil ideal, here, we demonstrated that the Jacobson radical over strongly 2-NC ring with U(R) = 2 is a nil ideal of characteristic 4. 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