Rings whose non-units are square-nil clean

Throughout (the rest of) the present paper, all rings are supposed to be associative with identity and all modules are unitary. We denote by U(R), Nil(R) and Id(R) the set of invertible elements, the set of nilpotent elements and the set of idempotent elements in R, respectively. Likewise, J(R) denotes the Jacobson radical of R, and Z(R) denotes the centre of R. The ring of n × n matrices over R and the ring of n × n upper triangular matrices over R are, respectively, denoted by M_n(R) and T_n(R).

Mimicking Nicholson [1, 2], a ring is called clean if every its element can be written as a sum of a unit element and an idempotent element. Moreover, a ring R is termed exchange if, for each a ∈ R, there exists an idempotent e ∈ aR such that 1 − e ∈ (1 − a)R. Every clean ring is known to be exchange, but the converse implication is manifestly not valid in general; however, it is true in the abelian case (see, e.g. [1, Proposition 1.8]).

In [2], an element c of a ring R is defined to be strongly clean if there is an idempotent e ∈ R, which commutes with c, such that c − e is invertible in R. Analogously, a strongly clean ring is the one in which every element is strongly clean. Furthermore, it is shown that a strongly clean element of a ring satisfies a generalized version of the classical Fitting's Lemma. This yields (cf. [3, Section 10]) that each strongly π-regular element of a ring is strongly clean. (We also refer to Ref. [4] for an excellent overview of the relationship between the clean property and other classical ring theory notions.)

Next, imitating Diesl [5], a ring R is said to be (strongly) nil clean if, for each r ∈ R, there are q ∈ Nil(R) and e ∈ Id(R) such that r = q + e (in addition, qe = eq for strong nil cleanness). Many fundamental properties of (strongly) nil clean rings were proven in this source as well as their general theory was developed there.

Following Danchev et al. [6], a ring R is called generalized strongly nil clean, and briefly abbreviated by GSNC, if any non-invertible element of R is strongly nil clean.

Now, we say that an element a ∈ R is said to be square-idempotent, provided a² = a⁴. Motivated by the above discussion, we introduce and study the notion of NUS-nil clean rings. So, a ring R is called (strongly) square-nil clean if, for each a ∈ R, there are a square-idempotent e ∈ R and an element n ∈ Nil(R) such that a = e + n (in addition, en = ne for strong square-nil cleanness). If any non-unit (i.e. any non-invertible element) in a ring R is (strongly) square-nil clean, we say R is (strongly) NUS-nil clean. The class of strongly NUS-nil clean rings lies strictly between the classes of strongly nil clean rings and strongly clean rings. Specifically, we establish that: a ring R is strongly NUS-nil clean if, and only if, a⁴ − a² ∈ Nil(R) for all a∉U(R). We, moreover, prove that this property passes to corner rings and is not Morita invariant. Also, the direct product of strongly NUS-nil clean rings is too a strongly NUS-nil clean ring. In particular, strongly NUS-nil clean rings are always strongly π-regular. Likewise, the Jacobson radical of a strongly NUS-nil clean ring is a nil-ideal. Finally, a ring R with only trivial idempotents is strongly NUS-nil clean if, and only if, R is a local ring with nil Jacobson radical. Certain matrix constructions and group ring extensions will reach us with some new types of examples of NUS-nil clean rings.

Our strategic achievements are, respectively, Theorems 2.6, 2.28, 2.30 and 2.37 stated and proved in the sequel.

In this section, we officially introduce the concept of strongly NUS-nil clean rings and investigate their elementary but useful properties.

We begin our work with the next series of preliminary technicalities.

Proof. Assume a = e + n, where a is a non-unit, e is a square-idempotent and n is a nilpotent element of a ring R that commutes with e. It is now easily seen that a = 1 − e² + e − 1 + e² + n and e − 1 + e² + n ∈ U(R), as needed. □

Our immediate consequence is the following.

In other words, the class of strongly NUS-nil clean rings lies in a proper way between the classes of strongly nil clean rings and strongly clean rings. Indeed, it is worthwhile noting that $Z3×Z3$⁠, $M2(Z3)$ and $M2(Z2)$ are all strongly NUS-nil clean rings that are not strongly nil clean, while $Z2[[x]]$ and $Z3[[x]]$ are strongly clean rings that are not strongly NUS-nil clean.

Proof. The sufficiency being clear, we are focussing on necessity. To that aim, letting a_i ∈ R_i for some index i, whence

So, (0, …, 0, a_i, 0, …, 0) is strongly square-nil clean, and hence a_i is strongly square-nil clean. That is why, any R_i is strongly square-nil clean, as required. □

Recall the pivotal fact that, in view of [7, Lemma 2.6], if 2 ∈ U(R) and a³ − a is a nilpotent, then there exists a polynomial $α(t)∈Z[t]$ such that α(a)³ = α(a) and a − α(a) is a nilpotent.

We now manage to proceed by proving the following helpful necessary and sufficient condition.

Proof. Assume that R is strongly NUS-nil clean. Then, for all a∉U(R), there are e⁴ = e² ∈ R and n ∈ Nil(R) such that a = e + n and en = ne. Thus, a² = e² + n(2e + n). So,

Therefore, clearly a⁴ − a² ∈ Nil(R), as desired.

Conversely, assume that a⁴ − a² ∈ Nil(R) for every a∉U(R). It follows that a³ − a ∈ Nil(R) for each a∉U(R). If, for a moment, 2 ∈ U(R), then the aforementioned [7, Lemma 2.6] applies to get that there exists $α(a)∈Z[a]$ such that α(a)³ = α(a) and a − α(a) ∈ Nil(R). So, there is m ∈ Nil(R) such that a = α(a) + m. It is now easy to see that α(a)⁴ = α(a)² and mα(a) = α(a)m, so that we are done.

Next, if 2 ∉ U(R), then, by hypothesis, 12 = 2⁴–2² ∈ Nil(R). So, 12 = 3 × 2 × 2 ∈ Nil(R) insuring that 6 ∈ Nil(R). On the other hand, since, for every a∉U(R), it must be that a³ − a ∈ Nil(R), we get $(a3−a)6∈Nil(R)$ and hence

But, as 6 ∈ Nil(R), we perceive a¹⁸ + 3a¹⁴ + 3a¹⁰ + a⁶ ∈ Nil(R), and since a⁴ − a² ∈ Nil(R), we obtain that a⁶ − a⁴ ∈ Nil(R) and so a⁶ − a² ∈ Nil(R). Now, from a⁸(a⁶ − a²) ∈ Nil(R), we may write a¹⁴ − a¹⁰ ∈ Nil(R) and thus 3a¹⁴ − 3a¹⁰ ∈ Nil(R). Further, as a¹⁸ + 3a¹⁴ + 3a¹⁰ + a⁶ ∈ Nil(R), we have a¹⁸ + 6a¹⁴ + a⁶ ∈ Nil(R), and since 6 ∈ Nil(R), we deduce that a¹⁸ + a⁶ ∈ Nil(R) giving a¹³ + a ∈ Nil(R). However, from a³ − a ∈ Nil(R), we can see that a⁵ − a³ ∈ Nil(R) and so a⁵ − a³ + a³ − a ∈ Nil(R). Arguing in a similar way, one can see that a⁷ − a ∈ Nil(R) and a¹³ − a ∈ Nil(R). From this and a¹³ + a ∈ Nil(R), we derive 2a ∈ Nil(R) for all a∉U(R).

Furthermore, as a⁷ − a ∈ Nil(R), we deduce a(a⁶ − 1) ∈ Nil(R). Since a∉U(R), we can observe that −a⁴ + a∉U(R) whence −2a⁴ + 2a ∈ Nil(R). Thus, a(a⁶ − 1 − 2a³ + 2) ∈ Nil(R) guaranteeing that a(a⁶ − 2a³ + 1) ∈ Nil(R). It now follows from 6 ∈ Nil(R) that −6a⁶ − 6a² + 30a³ − 18a⁴ ∈ Nil(R) which forces that a(a⁶ − 20a³ + 1 − 6a⁵ − 6a + 30a²) ∈ Nil(R). Combining this and the relation 15a⁵ − 15a³ ∈ Nil(R), we arrive at

which means that

Therefore, a(a − 1)⁶ ∈ Nil(R) so that there is k ∈ N such that a^k(a − 1)^6k = 0 and thus a^6k(a − 1)^6k = 0. So, one infers that a(a − 1) ∈ Nil(R).

Furthermore, owing now to [8, Lemma 3.5], there exists $β(a)∈Z[a]$ such that β(a)² = β(a) and a − β(a) ∈ Nil(R) (note that the cited lemma works even without the assumption that 2 is not a unit in R). Finally, it follows that a = β(a) + k, where k ∈ Nil(R) and kβ(a) = β(a)k, as wanted. □

A ring R is known to be strongly π-regular, provided that, for any a ∈ R, there exists $n∈N$ such that aⁿ ∈ aⁿ⁺¹R (cf. [9]).

We now come to our critical statement.

Proof. Let R be a strongly NUS-nil clean ring and a ∈ R. If a ∈ U(R), then a is obviously a strongly π-regular element. If, however, a∉U(R), then, in virtue of Theorem 2.6, there exists $m∈N$ such that $(a4−a2)m=0$⁠. It, thereby, follows at once that a^2m = a^2m+1r for some r ∈ R. Consequently, R is a strongly π-regular ring, as expected. □

We next continue with further crucial claims.

Proof. Let a ∈ eRe such that a∉U(eRe). We write a = ea = ae = eae. If a ∈ U(R), then there exists b ∈ R such that ab = ba = 1, which assures a(ebe) = (ebe)a = e, a contradiction. Therefore, a∉U(R). Hence, according to Theorem 2.6, we have

Proof. Let j ∈ J(R). Thus, j ∉ U(R). Now, using Theorem 2.6, we detect j⁴ − j² ∈ Nil(R). It apparently follows that j²(j² − 1) ∈ Nil(R). So, $j2t(j2−1)t=0$ for some positive integer t. However, as j² − 1 ∈ U(R), we get j^2t = 0 and so j ∈ Nil(R), as pursued. □

• (i) For any nil-ideal I ⊆ R, R is strongly NUS-nil clean if, and only if, R/I is strongly NUS-nil clean.

• (ii) A ring R is strongly NUS-nil clean if, and only if, J(R) is nil and R/J(R) is strongly NUS-nil clean.

Proof. (i) Suppose R is a strongly NUS-nil clean ring and put $R¯≔R/I$⁠. If $a¯∉U(R¯)$⁠, then a∉U(R), which ensures with the aid of Theorem 2.6 that a⁴ − a² ∈ Nil(R), so that $a¯4−a¯2∈Nil(R¯)$⁠.

Conversely, suppose $R¯$ is a strongly NUS-nil clean ring. If a∉U(R), then $a¯∉U(R¯)$⁠, and thus Theorem 2.6 guarantees that $a¯4−a¯2∈Nil(R¯)$⁠. Therefore, there exists $k∈N$ such that $(a4−a2)k∈I⊆Nil(R)$⁠.

(ii) Utilizing Lemma 2.9 and part (i) of the proof, the arguments are complete. □

• (i) R/I is strongly NUS-nil clean.

• (ii) R/Iⁿ is strongly NUS-nil clean for all $n∈N$⁠.

• (iii) R/Iⁿ is strongly NUS-nil clean for some $n∈N$⁠.

Proof. (i) ⇒ (ii). For any $n∈N$⁠, we have $R/InI/In≅R/I$⁠. Since I/Iⁿ is a nil-ideal of R/Iⁿ and R/I is strongly NUS-nil clean, then Proposition 2.10 gives that R/Iⁿ is a strongly NUS-nil clean ring.

(ii) ⇒ (iii). This implication is trivial.

(iii) ⇒ (i). For any ideal I of R, we have $R/InI/In≅R/I$⁠, so that, via Proposition 2.10, we can conclude that R/I is strongly NUS-nil clean, as stated. □

• (i) R is strongly square-nil clean.

• (ii) T_n(R) is strongly NUS-nil clean for all $n∈N$⁠.

• (iii) T_n(R) is strongly NUS-nil clean for some n ≥ 2.

Proof. (ii) ⇒ (iii). This is trivial.

(iii) ⇒ (i). Let a ∈ R. Then, it must be that

It is evident that A is non-invertible in T₂(R). By hypothesis, we can find a square idempotent $E=e11e120e22$ and a nilpotent $Q=q11q120q22$ such that A = E + Q and EQ = QE. It now follows by plain inspection that a = e₁₁ + q₁₁ and e₁₁q₁₁ = q₁₁e₁₁. Hence, $e114=e112$⁠, and q₁₁ is a nilpotent in R. Thus, a has strongly NUS-nil clean decomposition. Therefore, point (i) is valid.

(i) ⇒ (ii). It suffices with Proposition 2.10 (ii) at hand to prove only that the quotient T_n(R)/J(T_n(R)) is strongly NUS-nil clean. To this target, observe that

So, we need just to show that $⊓i=1nR/J(R)$ is strongly NUS-nil clean. In fact, Lemma 2.4 is a guarantor that $⊓i=1nR/J(R)$ is strongly NUS-nil clean if, and only if, R/J(R) is strongly square-nil clean. It is now easy to see that, if R is strongly square-nil clean, then R/J(R) is strongly-square nil clean, as asked for. □

Let R be a ring and M a bi-module over R. The trivial extension of R and M is stated as

with addition defined component-wise and multiplication defined by

The next two consequences arrived quite naturally.

• (i) T(R, M) is a strongly NUS-nil clean ring.

• (ii) R is a strongly NUS-nil clean ring.

Proof. Set A ≔ T(R, M) and consider I ≔ T(0, M). It is not too hard to verify that I is a nil-ideal of A such that $AI≅R$⁠. So, the result follows directly from Proposition 2.10. □

Let α be an endomorphism of R and n a positive integer. Consider the skew triangular matrix ring

with addition point-wise and multiplication given by:

where

We, hereafter, denote the elements of T_n(R, α) by (a₀, a₁, …, a_n−1). If α is the identity endomorphism, then T_n(R, α) is a subring of the upper triangular matrix ring T_n(R).

• (i) T_n(R, α) is a strongly NUS-nil clean ring.

• (ii) R is a strongly NUS-nil clean ring.

Proof. Choose

Then, one easily checks that Iⁿ = (0) and $Tn(R,α)I≅R$⁠. Consequently, Proposition 2.10 employs to get the desired result. □

Furthermore, one mentions that Wang et al. introduced in Ref. [10] the matrix ring S_n,m(R) as follows: supposing R is a ring, consider the matrix ring S_n,m(R) = 

Likewise, let T_n,m(R) be

and let we state

Thereby, we have the following.

• (i) S_n,m(R) is a NUS-nil clean ring.

• (ii) T_n,m(R) is a NUS-nil clean ring.

• (iii) U_n(R) is a NUS-nil clean ring.

• (iv) R is a NUS-nil clean ring.

Let α be an endomorphism of R. We denote by R[x, α] the skew polynomial ring whose elements are the polynomials over R; the addition is defined as usual, and the multiplication is defined by the equality xr = α(r)x for any r ∈ R. So, there is a ring isomorphism

given by

with a_i ∈ R, 0 ≤ i ≤ n − 1. Thus, one deduces that $Tn(R,α)≅R[x,α]〈xn〉$⁠, where 〈xⁿ〉 is the ideal generated by xⁿ.

Besides, R[[x, α]] denotes the ring of skew formal power series over R; that is, all formal power series of x having coefficients from R with multiplication defined by xr = α(r)x for all r ∈ R. On the other hand, we know that the isomorphism $R[x,α]〈xn〉≅R[[x,α]]〈xn〉$ is fulfilled.

We, thus, extract the following three consequences.

• (i) $R[x,α]〈xn〉$ is a strongly NUS-nil clean ring.

• (ii) $R[[x,α]]〈xn〉$ is a strongly NUS-nil clean ring.

• (iii) R is a strongly NUS-nil clean ring.

• (i) $R[x]〈xn〉$ is a strongly NUS-nil clean ring.

• (ii) $R[[x]]〈xn〉$ is a strongly NUS-nil clean ring.

• (iii) R is a strongly NUS-nil clean ring.

$Sn(R)≔(aij)∈Tn(R)|a11=a22=⋯=ann$⁠.

Then, the following statements are equivalent:

• (i) S_n(R) is a strongly NUS-nil clean ring.

• (ii) R is a strongly NUS-nil clean ring.

Proof. Assuming I≔{(a_ij) ∈ S_n(R): a₁₁ = 0}, it is rather evident that I is a nil-ideal of S_n(R) such that S_n(R)/I ≅ R holds, whence Proposition 2.10 (i) works, as inspected. □

The following example demonstrates that the full matrix constructions are rather more complicated than we anticipate in comparison to the triangular matrix rings. Specifically, we exhibit a concrete construction of a strongly NUS-nil clean ring which is not strongly square-nil clean.

Thus, R is a strongly NUS-nil clean ring, but it is not strongly square-nil clean. Indeed, by contrary, assume that $M2(Z2)$ is strongly square-nil clean. Then, we choose $A=1110∈M2(Z2)$⁠, and so there are $E4=E2∈M2(Z2)$ and $N∈Nil(M2(Z2))$ such that A = E + N and EN = NE. It follows now by a plain calculation that $A4−A2=1001∈Nil(M2(Z2))$⁠, which is an obvious contradiction as this is exactly the identity matrix. Therefore, $M2(Z2)$ is really not strongly square-nil clean, as claimed.

The next assertion arises logically asking what happens for sizes greater than 2.

Proof. It suffices to establish that M₃(R) is not a strongly NUS-nil clean ring having in mind Lemma 2.8. To this goal, consider the matrix

Then, we have

Consequently, Theorem 2.6 is applicable to get that R cannot be a strongly NUS-nil clean ring, as asserted. □

An immediate consequence is the following one.

Proof. Assume on the contrary that there exists 0 ≠ e ∈ Id(R) such that eRe ≅ M_n(S) for some non-zero ring S. Since R is strongly NUS-nil clean, it follows from Lemma 2.8 that eRe has to be strongly NUS-nil clean too, and so M_n(S) is also strongly NUS-nil clean, implying a contradiction with Proposition 2.20, as expected. □

The following affirmation is somewhat surprising.

Proof. Let a ∈ R. Then, one finds that

Thus, Lemma 2.6 works to get that A⁴ − A² ∈ Nil(M₂(R)). So, a⁴ − a² ∈ Nil(R) whence a³ − a ∈ Nil(R). Therefore, bearing in mind [7, Theorem 2.12], one writes that a = e + n, where e³ = e, n ∈ Nil(R) and en = ne. As e⁴ = e², e is a square-idempotent and so R is a strongly square-nil clean ring, as formulated. □

We now intend to examine some structural characterizations.

Proof. Let a ∈ R and a∉U(R). Since R is local, a ∈ J(R) and hence a ∈ Nil(R). So, a is a nilpotent element, and thus it is a strongly NUS-nil clean element, as required. □

We now can extract the following criterion.

Proof. Assume that R is a strongly NUS-nil clean ring, so J(R) is nil in accordance with Lemma 2.9. If a∉U(R), then we have a = q + u, where qu = uq, q ∈ Nil(R) and either u² = 1 or u² = 0. Since a is not a unit, it must be that a = u + q, where qu = uq and u² = 0 giving a ∈ Nil(R). Thus, exploiting [9, Proposition 19.3], R must be a local ring.

Oppositely, suppose R is a local ring with a nil Jacobson radical J(R). So, for each a∉U(R), we have a ∈ J(R) ⊆Nil(R), whence a is a nil clean element, as requested. □

Proof. If 2 ∉ Nil(R), then Theorem 2.6 can be applied to derive that 2⁴–2² ∈ Nil(R). Hence, we get 12 ∈ Nil(R) and so 6 ∈ Nil(R). □

• (i) R is a strongly NUS-nil clean ring.

• (ii) R is a GSNC ring.

Proof. (ii) ⇒ (i). It is straightforward.

(i) ⇒ (ii). It is sufficient to prove only that a² − a ∈ Nil(R) for each a ∈ R \ U(R). To that end, choose a ∈ R \ U(R). Thus, Theorem 2.6 allows us to infer that a⁴ − a² ∈ Nil(R), and so a²(1 − a²) ∈ Nil(R). But, as 2 ∈ Nil(R), we get a²(1 − a² − 2a + 2a²) ∈ Nil(R) and, therefore, a²(1 − a)² ∈ Nil(R). Thus, a² − a ∈ Nil(R), as needed. □

• (i) R is a strongly NUS-nil clean ring.

• (ii) Either R is a GSNC ring or R is a strongly square-nil clean ring.

Proof. (ii) ⇒ (i). This is routine.

(i) ⇒ (ii). Applying Lemma 2.25, we have that either 2 ∈ Nil(R) or 6 ∈ Nil(R). If, for a moment, 2 ∈ Nil(R), then R is a GSNC ring using Lemma 2.26. However, if 6 ∈ Nil(R), we may decompose with the help of the Chinese Remainder Theorem that R ≅ R₁ ⊕ R₂, where 2 ∈ Nil(R₁) and 3 ∈ Nil(R₂). Now, Lemma 2.4 tells us that R₁ and R₂ are both strongly square-nil clean rings. Moreover, since 2 ∈ Nil(R₁), the ring R₁ is even strongly nil clean. It thus follows at once that R is a strongly square-nil clean ring, as promised. □

The following necessary and sufficient condition is key providing us with a satisfactory close transversal between the two new notions of strong square-nil cleanness and strong NUS-nil cleanness as defined above.

• (i) R is a strongly square-nil clean ring.

• (ii) R is a strongly NUS-nil clean ring and, for every u ∈ U(R), u² = 1 + n, where n ∈ Nil(R).

Proof. (i) ⇒ (ii). It is readily to see that R is strongly NUS-nil clean. Let u ∈ U(R). Thus, u = e + m, where e² = e⁴ and m ∈ Nil(R) with em = me. It follows that u² = e² + m(2e + m) and u⁴ = e⁴ + m(2e + m)(2e² + m(2e + m)) = e² + m(2e + m)(2e² + m(2e + m)). Hence, u⁴ − u² = m(2e + m)(2e² + m(2e + m) − 1) ∈ Nil(R). So, 1 − u² ∈ Nil(R). Consequently, there exists n ∈ Nil(R) such that u² = 1 + n, as desired.

(ii) ⇒ (i). Assume that a ∈ R. If a∉U(R), then a = f + q, where f² = f⁴ and q ∈ Nil(R) with qf = fq and we are done.

If, however, a ∈ U(R), then a² = 1 + n, where n ∈ Nil(R). It follows that a³ = a²a = a + na and, similarly, a³ = aa² = a + an. So, an = na and hence a − a³ = −an ∈ Nil(R). Now, we will distinguish two possible cases for the element 2 ∈ R.

If, firstly, 2 ∈ U(R), then [7, Lemma 2.6] applies to get that there exists $α(a)∈Z[a]$ such that α(a)³ = α(a) and a − α(a) ∈ Nil(R). So, there is m ∈ Nil(R) such that a = α(a) + m. It is now easy to see that α(a)⁴ = α(a)² and mα(a) = α(a)m, so that we are over.

However, if 2 ∉ U(R), then we claim that 2 ∈ Nil(R). Assuming on the contrary 2 ∉ Nil(R), then by point (ii) and using Theorem 2.6, we write 2⁴–2² ∈ Nil(R). Consequently, 6 ∈ Nil(R). Since, for every a ∈ U(R), we know a³ − a ∈ Nil(R), we perceive $(a3−a)6∈Nil(R)$ and hence

But, 6 ∈ Nil(R) enables us that a¹⁸ + 3a¹⁴ + 3a¹⁰ + a⁶ ∈ Nil(R), and since a⁴ − a² ∈ Nil(R), we obtain that a⁶ − a⁴ ∈ Nil(R) so that a⁶ − a² ∈ Nil(R). Now, from a⁸(a⁶ − a²) ∈ Nil(R), we write a¹⁴ − a¹⁰ ∈ Nil(R) and thus 3a¹⁴ − 3a¹⁰ ∈ Nil(R). But, as a¹⁸ + 3a¹⁴ + 3a¹⁰ + a⁶ ∈ Nil(R), we have a¹⁸ + 6a¹⁴ + a⁶ ∈ Nil(R), and since 6 ∈ Nil(R), we deduce that a¹⁸ + a⁶ ∈ Nil(R) leading to a¹³ + a ∈ Nil(R). Further, as a³ − a ∈ Nil(R), we can see that a⁵ − a³ ∈ Nil(R) and so a⁵ − a³ + a³ − a ∈ Nil(R). Utilizing analogous arguments, one can see that a¹³ − a ∈ Nil(R). Combining this and a¹³ + a ∈ Nil(R), we obtain 2a ∈ Nil(R) for all a ∈ U(R). It now follows that 2 ∈ Nil(R), as claimed.

Next, we assert that 1 − a∉U(R). Acting on contrary, we assume that 1 − a ∈ U(R) and then, by assumption, there is m ∈ Nil(R) such that (1 − a)² = 1 + m and m(1 − a) = (1 − a)m. It, thus, follows that 1 − 2a + a² = 1 + m. Since a² = 1 + n, we have that 1 − 2a + 1 + n = 1 + m and so 1 − 2a = m − n. As ma = am and n = a² − 1, we find mn = nm, thus showing that m − n ∈ Nil(R). So, 1 − 2a ∈ Nil(R), and since 2 ∈ Nil(R), we detect 1 ∈ Nil(R), being an absurd. Therefore, 1 − a∉U(R), as asserted, and, via (ii), we inspect that

It, thereby, follows that

Furthermore, as a³ − a ∈ Nil(R), we can write (1 − a)(a³ − a) ∈ Nil(R). Consequently, (1 − a)(−3a + 3a²) ∈ Nil(R). But, since 2 ∈ Nil(R), we arrive at (1 − a)(2a − 2a²) ∈ Nil(R). It, therefore, follows that

Thus, a(1 − a) = (1 − a)a ∈ Nil(R) for all a ∈ U(R) meaning that 1 − a ∈ Nil(R) with a ∈ 1 + Nil(R), because obviously a commutes with the existing element from Nil(R).

Finally, in either case, the element a ∈ U(R) has a strongly square-nil clean decomposition, as wanted. □

The next comments are well-positioned.

We are now prepared to prove our principal result that sounds somewhat curious and is a partial converse of Lemma 2.22.

Proof. Firstly, we show that 6 ∈ Nil(R). If, for a moment, 2 ∉ U(R), then Lemma 2.25 employs to get either 2 ∈ Nil(R) or 6 ∈ Nil(R). So, in this case, we have 6 ∈ Nil(R), because 2 ∈ Nil(R) would imply that 6 ∈ Nil(R).

If 2 ∈ U(R), then, by hypothesis, there exist e² = e⁴ ∈ R and n ∈ Nil(R) such that 2 = e + n and en = ne. So,

Therefore, 6 ∈ Nil(R) in either case.

Thus, one can decompose with the aid of the Chinese Remainder Theorem that R ≃ R₁ × R₂, where 2 ∈ Nil(R₁) and 3 ∈ Nil(R₂). But, since R is local, we have either $R/J(R)≅Z2$ or $R/J(R)≅Z3$⁠. So, M₂(R/J(R)) is a strongly NUS-nil clean ring. However, we know that

as well as that J(R) is a nilpotent ideal of R, because R is Noetherian. Consequently, one knows that M₂(J(R)) is a nilpotent ideal of M₂(R). Finally, Proposition 2.10 leads to the fact that M₂(R) is a strongly NUS-nil clean ring, as asked. □

The next assertion also sounds somewhat surprisingly.

Proof. For all u, v ∈ U(R), we have u² = v² = (uv)² = 1. Therefore, uv = (uv)⁻¹ = v⁻¹u⁻¹ = vu. Hence, the invertible elements commute with each other.

Now, we will illustrate that R is abelian. In fact, for each idempotent e in R and a ∈ R, we know 2e − 1 ∈ U(R) and (1 + ea(1 − e)) ∈ U(R). Since, by what we have shown above, the invertible elements commute with each other, we have 2ea(1 − e) = 0. Since 2 ∈ U(R), we get ea(1 − e) = 0, which means ea = eae.

On the other hand, since 2e − 1 ∈ U(R) and (1 + (1 − e)ae) ∈ U(R), we obtain 2(1 − e)ae = 0. So, ae = eae. Therefore, R is abelian, as claimed.

On the other side, since 1 + Nil(R) ⊆ U(R) and all invertible elements commute with each other, the nilpotent elements also commute with each other.

On the other hand, since 2 ∈ U(R) and, for every u ∈ U(R), u² = 1, we receive 2² = 1. It now follows that 3 ∈ Nil(R). Next, we demonstrate that a³ − a ∈ Nil(R) for any a ∈ R. To that purpose, choose a ∈ R. If a ∈ U(R), then a² = 1 and so a³ − a = 0.

If, however, a∉U(R), then Theorem 2.6 informs us that a⁴ − a² ∈ Nil(R). So, a³ − a ∈ Nil(R).

Now, given x, y ∈ R. Consulting with [7, Proposition 2.8], one writes that x = e₁ + e₂ + m and y = f₁ + f₂ + n, where e₁, e₂, f₁, f₂ ∈ Id(R) and m, n ∈ Nil(R). As all idempotents are known by the established above to be central in R, and nilpotent elements are commuting with each other, we finally conclude that xy = yx. Thus, R is a commutative ring, indeed. □

Let A, B be two rings and let M, N be (A, B)-bi-module and (B, A)-bi-module, respectively. Moreover, we consider the bilinear maps ϕ: M ⊗_BN → A and ψ: N ⊗_AM → B that apply to the following properties:

For m ∈ M and n ∈ N, define mn≔ϕ(m ⊗ n) and nm≔ψ(n ⊗ m). Now, the 4-tuple $R=AMNB$ becomes to an associative ring with obvious matrix operations that is called a Morita context ring. Designate the two-sided ideals Imϕ and Imψ to MN and NM, respectively, that are called the trace ideals of the Morita context.

We now have at our disposal all the instruments necessary to prove the following criterion.

Proof. Since, MN and NM are nilpotent ideals of A and B, respectively, one can says that MN ⊆ J(A) and NM ⊆ J(B). Therefore, adapting [12], we argue that

and hence the isomorphism

is fulfilled. Notice that R is a strongly NUS-nil clean ring if, and only if, so is the factor-ring R/J(R).

Furthermore, thanking to Lemma 2.4, the quotient R/J(R) is strongly NUS-nil clean ring if, and only if, $AJ(A)$ and $BJ(B)$ are both strongly square-nil clean rings.

Now, if R is strongly NUS-nil clean, then by what we have established so far J(R) is a nil-ideal and so J(A) and J(B) are nil as well. Observe also that Proposition 2.10 can be applied to get that both A and B are strongly square-nil clean rings, as stated.

Conversely, if A and B are strongly square-nil clean rings, then both J(A) and J(B) are nil, and besides $AJ(A)$ and $BJ(B)$ are strongly square-nil clean rings yielding that R/J(R) is a strongly NUS-nil clean ring. But, as J(A) and J(B) are nil, we elementarily deduce that J(R) is a nil-ideal of R. It now follows that R is a strongly NUS-nil clean ring, as formulated. □

Next, let R, S be two rings, and let M be an (R, S)-bi-module such that the operation (rm)s = r(ms) is valid for all r ∈ R, m ∈ M and s ∈ S. Given such a bi-module M, we consider the formal triangular matrix ring

where it obviously forms a ring with the usual matrix operations. Regarding Proposition 2.32, if we set N = {0}, then we will obtain the following immediate consequence.

We now deal with group ring extensions of the NUS-nil clean property as follows: let R be a ring, G a group, and we traditionally denote by RG the group ring of G over R.

In this section, we intend to establish a suitable criterion for a group ring to be strongly NUS-nil clean under some sensible circumstances on the former group and ring objects. Concretely, we are able to achieve this, provided the whole group is locally finite. Recall that a group G is a p-group if every element of G has order which is a power of the prime number p. Also, we recollect that a group is locally finite if each its finitely generated subgroup is finite.

Suppose now that G is an arbitrary group and R is an arbitrary ring. Standardly, in RG there is a homomorphism ɛ: RG → R, defined by

which is called the augmentation map of RG and its kernel, denoted by Δ(RG), is called the augmentation ideal of RG.

We now proceed by showing the following assertions.

Proof. We know that RG/Δ(RG) ≅ R. Thus, it follows at once that R must be a strongly NUS-nil clean ring as being an epimorphic image of RG. □

Proof. Knowing [13, Proposition 16], we see that Δ(RG) is a nil-ideal. Thus, since RG/Δ(RG) ≅ R, referring to Proposition 2.10 we infer that RG is a strongly NUS-nil clean ring. □

We also record the following interesting fact.

Having in hand the preceding three statements, we now have at our disposal all the machinery needed to establish the following major assertion.

We finish off our research work with the following claim.

Proof. Seeing that RG = Δ(RG) + R, because Δ(RG) ⊆ J(RG), we write RG = J(RG) + R. Therefore, the isomorphism

is true, and since R is strongly NUS-nil clean, one finds that R/(J(RG) ∩ R) is strongly NUS-nil clean too. Consequently, we conclude that RG/J(RG) is a strongly NUS-nil clean ring, as requested.

The authors express their sincere gratitude to the expert referee for the very careful reading of the initially submitted version and the numerous competent suggestions made, which led to a substantial improvement of the exposition.