In this section, we consider some types of strongly (*x*^{2} + *cx* + *d*)-nil clean rings and we will investigate the relation between strongly *g*(*x*)-nil clean rings and strongly nil clean rings.

#### Theorem 3.1

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

#### Proof

⇒) : Suppose *R* is strongly nil clean with *b* − *a* ∈ *N*(*R*) and let *r* ∈ *R*. Then
$\frac{r-(a+1)}{(b-a)-1}=e+n,$
where *e*^{2} = *e*, *n* ∈ *N*(*R*) and *en* = *ne*. Therefore, *r* = [*e*((*b* − *a*) − 1) + (*a* + 1)] + *n*((*b* − *a*) − 1). Write *s* = [*e*((*b* − *a*) − 1) + (*a* + 1)] and *v* = *n*((*b* − *a*) − 1). Then *r* = *s* + *v* where clearly *v* ∈ *N*(*R*) and *sv* = *vs*. Moreover, *s* is a root of (*x* − (*a* + 1))(*x* − *b*) since
$$\begin{array}{}\phantom{\rule{1em}{0ex}}[e((b-a)-1)+(a+1)-(a+1)][e((b-a)-1)+(a+1)-b]\\ ={e}^{2}((b-a)-1{)}^{2}-e((b-a)-1)((b-a)-1)=0\end{array}$$

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

⇐) : Suppose *R* is a strongly (*x* − (*a* + 1))(*x* − *b*)-nil clean. Write *a* = *s* + *n* where *n* ∈ *N*(*R*), (*s* − (*a* + 1))(*s* − *b*) = 0 and *sn* = *ns*. Then, *s* − *a* ∈ *N*(*R*) and so *s* − (*a* + 1) ∈ *U*(*R*). Therefore, *s* = *b* and *b* − *a* ∈ *N*(*R*). Now, let *r* ∈ *R*. By the assumption, we can write *r*((*b* − *a*) − 1) + (*a* + 1) = *s* + *n* where *n* ∈ *N*(*R*), *s* is a root of (*x* − (*a* + 1))(*x* − *b*) and *sn* = *ns*. Thus, *r* = *t* + *m* where *t* = (*s* − (*a* + 1))((*b* − *a*) − 1)^{−1} and *m* = *n*((*b* − *a*) − 1)^{− 1}. Since *a*, *b* ∈ *C*(*R*), it is clear that *m* ∈ *N*(*R*) and *tm* = *mt*. Moreover,
$$\begin{array}{}{t}^{2}=((s-(a+1))((b-a)-1{)}^{-1}{)}^{2}=(s-(a+1))(s-b+b-(a+1))((b-a)-1{)}^{-2}\\ \phantom{\rule{1em}{0ex}}=(s-(a+1))(s-b)+(s-(a+1))(b-(a+1))((b-a)-1{)}^{-2}\\ \phantom{\rule{1em}{0ex}}=(s-(a+1))((b-a)-1{)}^{-1}=t\end{array}$$

Therefore, *R* is strongly nil clean. □

#### Lemma 3.2

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

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

#### Corollary 3.3

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

*R* *is strongly nil clean*.

*R* *is strongly* (*x*^{2} − (2*b* + 1)*x* + *b*(*b* + l))-*nil clean*.

*R* *is strongly* (*x*^{2} − (2*b* + 1 − 2*n*)*x* + (*b*^{2} + *b*(1 − 2*n*))-*nil clean*.

*R* *is strongly* (*x*^{2} − (2*b* + 1 + 2*n*) *x* + (*b*^{2} + *b*(1 + 2*n*))-*nil clean*.

#### Proof

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

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

#### Definition 3.5

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

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

#### Example 3.6

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

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

#### Example 3.7

*Let* *g*(*x*) = *x*^{n} + *a*_{n − 1}*x*^{n − 1} + … + *a*_{1}*x* + *a*_{0} ∈ *C*(*R*)[*x*] *where* *R* *is a non zero ring and* *n* ∈ ℕ. *If* *a*_{0} ∈ *U*(*R*), *then* *R* *is not strongly* *g*(*x*)-*nil clean. In particular*, *R* *is not strongly* (*x*^{n} ± 1)-*nil clean*.

#### Proof

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

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

#### Proposition 3.8

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

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

#### Proposition 3.9

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

#### Proof

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

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

#### Proposition 3.10

*Let* *R* *be a ring and* *g*(*x*) = *x*^{n} + *a*_{n−1}*x*^{n − 1} + … + *a*_{1}*x* + *a*_{0} ∈ *C*(*R*)[*x*] *where* *n* ∈ ℕ *and* *a*_{0} ∈ *U*(*R*). *If* *R* *is strongly* *g*(*x*)-nil^{*} *clean, then* *R* = *U*_{2}(*R*). *In particular if* *R* *is strongly* (*x*^{n − 2} + *x*^{n − 3} + … + *x* + 1)-*nil*^{*} *clean, then* *R* = *U*_{2}(*R*) *is strongly* (*x*^{n} − *x*)-*nil clean*.

#### Proof

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

By choosing *n* = 4 in Proposition 3.10, we conclude that if *R* is strongly (*x*^{2} + *x* + 1)-nil^{*}clean, then *R* = *U*_{2}(*R*) is strongly (*x*^{4}− *x*)-nil clean. Also, we have *R* = *U*_{2}(*R*) if *R* is strongly (*x*^{n} ± 1)-nil^{*}clean.

#### Lemma 3.11

*Let R be a Boolean ring and G any cyclic group of order p (prime). Then all elements of the group ring RG are roots of g(x)* = *x*^{2p−1} − *x*.

By taking for example *p* = 3 in Lemma 3.11, we see that the group ring ℤ_{2}(*C*_{3}) is (strongly) (*x*^{4} − *x*)-nil clean which is not strongly nil clean. Moreover, the field ℤ_{3} is (strongly) (*x*^{3} − *x*)-nil clean which is not strongly nil clean. Therefore, strongly (*x*^{n} − *x*)-nil cleanness of a ring *R* does not imply strongly nil cleanness of *R* whether *n* is odd or even.

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

#### Proposition 3.12

*Let R be a ring*, *n* ∈ ℕ *and a*, *b* ∈ *R*. *Then R is strongly* (*ax*^{2n} − *bx*)-*nil clean if and only if R is strongly* (*ax*^{2n} + *bx*)-*nil clean*.

By Proposition 3.12 and Lemma 3.11, we conclude that ℤ_{2}(*C*_{p}) is also (*x*^{2p−1} + *x*)-nil clean for any prime *p*. However, one can see that Proposition 3.12 need not be true for odd degrees. For example, the field ℤ_{3} is strongly (*x*^{3} − *x*)-nil clean which is not strongly (*x*^{3} + *x*)-nil clean.

Recall that an element *p* in a ring *R* is called periodic if *p*^{k} = *p*^{m} for some distinct positive integers *k* and *m*. Equivalently, *p* is periodic if *p* is a root of *g*(*x*) = *x*^{2n} − *x*^{n} for some positive integer *n*. *R* is called periodic if every element in *R* is periodic. In the following Lemma, we can see a sufficient condition for a ring to be periodic.

#### Lemma 3.13

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

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

#### Theorem 3.14

*The following are equivalent for a ring R*.

*R is periodic*.

*R is a strongly* (*x*^{n} − *x*)-*nil clean rings for some n* ∈ ℕ.

*For every r* ∈ *R*, *there exists k* ∈ ℕ *and a* ∈ 〈*r*〉 *such that r*^{k} = *r*^{k+1}*a*.

#### Proof

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

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

(3) ⇒ (1) : Lemma 3.13. □

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