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KYUNGPOOK Math. J. 2019; 59(4): 603-615
Ore Extension Rings with Constant Products of Elements
Ebrahim Hashemi∗ and Abdollah Alhevaz
Faculty of Mathematical Sciences, Shahrood University of Technology, Shahrood, P. O. Box 316-3619995161, Iran
e-mail : eb_hashemi@yahoo.com, eb_hashemi@shahroodut.ac.ir and a.alhevaz@gmail.com, a.alhevaz@shahroodut.ac.ir
* Corresponding Author.
Received: April 7, 2018; Accepted: November 20, 2018; Published online: December 23, 2019.
© Kyungpook Mathematical Journal. All rights reserved.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

Let R be an associative unital ring with an endomorphism α and α-derivation δ. The constant products of elements in Ore extension rings, when the coefficient ring is reversible, is investigated. We show that if$f(x)=∑i=0naixi$ and $g(x)=∑j=0mbjxj$ be non-zero elements in Ore extension ring R[x; α, δ] such that g(x)f(x) = cR, then there exist non-zero elements r, aR such that rf(x) = ac, when R is an (α, δ)-compatible ring which is reversible. Among applications, we give an exact characterization of the unit elements in R[x; α, δ], when the coeficient ring R is (α, δ)-compatible. Furthermore, it is shown that if R is a weakly 2-primal ring which is (α, δ)-compatible, then J(R[x; α, δ]) = Niℓ(R)[x; α, δ]. Some other applications and examples of rings with this property are given, with an emphasis on certain classes of NI rings. As a consequence we obtain generalizations of the many results in the literature. As the final part of the paper we construct examples of rings that explain the limitations of the results obtained and support our main results.

Keywords: Ore extensions, 2-primal rings, constant products, reversible rings, stable range one.
1. Introduction and Preliminary Definitions

Throughout, unless mentioned otherwise, R denotes an associative ring with unity. Letting R be a ring, α be an endomorphism of R and δ be an α-derivation of R (so δ is an additive map satisfying δ(ab) = δ(a)b + α(a)δ(b)), the general (left) Ore extension R[x; α, δ] is the ring of polynomials over R in the variable x, with termwise addition and with coefficients written on the left of x, subject to the skew-multiplication rule xr = α(r)x + δ(r) for rR. If α is an identity map on R or δ = 0, then we denote R[x; α, δ] by R[x; δ] and R[x; α], respectively. We use Niℓ*(R), Niℓ* (R), L-rad(R), Niℓ (R) and J(R) to denote the prime radical, upper nil radical, Levitzki radical, the set of all nilpotent elements of R and the Jacobson radical of R, respectively. Given a polynomial f(x) over a ring R, we denote by deg(f(x)) the degree of f(x).

Recall that a ring R is reduced if it has no non-zero nilpotent element. According to Krempa [14], an endomorphism α of a ring R is called rigid if (a) = 0 implies a = 0 for aR. R is called an α-rigid ring [11] if there exists a rigid endomorphism α of R. Note that any rigid endomorphism of a ring is a monomorphism and α-rigid rings are reduced by Hong et al. [11]. Properties of α-rigid rings have been studied in Krempa [14], Hirano [10] and Hong et al. [11, 12].

According to Hong et al. [12], for an endomorphism α of a ring R, an α-ideal I is called to be an α-rigid ideal if (a) ∈ I implies aI for aR. Hong et al. [12] studied connections between the α-rigid ideals of R and the related ideals of some ring extensions.

In [9], the authors defined α-compatible rings, which are a generalization of α- rigid rings. A ring R is called α-compatible if for each a, bR, ab = 0 ⇔ (b) = 0. Moreover, R is said to be δ-compatible if for each a, bR, ab = 0 ⇒ (b) = 0. If R is both α-compatible and δ-compatible, we say that R is (α, δ)-compatible. In [9, Lemma 2.2], the authors showed that R is α-rigid if and only if R is α-compatible and reduced. Thus the α-compatible ring is a generalization of an α-rigid ring to the more general case where R is not assumed to be reduced. In [7], the author defined α-compatible ideals, which are a generalization of α-rigid ideals. An ideal I is called an α-compatible ideal if for each a, bR, abI(b) ∈ I. Moreover, I is said to be a δ-compatible ideal if for each a, bR, abI(b) ∈ I. If I is both α-compatible and δ-compatible, we say that I is an (α, δ)-compatible ideal. In [7, Proposition 2.4], the author showed that an ideal I is α-rigid if and only if I is α-compatible and completely semiprime(i.e., if a2I, then aI).

In [16], Nasr-Isfahani studied Ore extensions of 2-primal rings. He showed that if R is an (α, δ)-compatible ring, then R is 2-primal if and only if R[x; α, δ] is 2- primal if and only if Niℓ(R)[x; α, δ] = Niℓ*(R[x; α, δ]) if and only if every minimal (α, δ)-prime ideal of R is completely prime.

Following [1], a ring R is called reversible if ab = 0 implies ba = 0 for a, bR. Also, following [2], a ring R is called 2-primal if Niℓ*(R) = Niℓ(R). Shin in [18, Proposition 1.11] showed that a ring R is 2-primal if and only if every minimal prime ideal P of R is completely prime (i.e. R/P is a domain). Moreover, a ring R is called weakly 2-primal if Niℓ(R) =L − rad(R). If Niℓ(R) = Niℓ*(R), then R is called NI. It is known that the following implications holds between the mentioned classes of rings:

$reduced⇒reversible⇒2-primal⇒weakly 2-primal⇒NI.$

But the converses does not hold (see [4, 13]). Moreover, a ring is right (resp., left) duo if every right (resp., left) ideal is an ideal. The importance of the study of these classes of rings in noncommutative ring theory is because of the famous Köthe’s problem which ask whether every one-sided nil ideal of any associative ring is contained in a two-sided nil ideal of the ring. As observed by Bell [1], these rings fulfill the requirements of the Köthe’s Conjecture.

In [3] Chen proved that if R is an α-compatible ring and $f(x)=∑i=0naixi$and $g(x)=∑j=0mbjxj$ are non-zero polynomials in R[x; α] such that g(x)f(x) = cR, then b0a0 = c and there exist non-zero elements r, aR such that rf(x) = ac. As an application of the above result, he showed that if R is an α-compatible and weakly 2-primal ring, then a polynomial f(x) ∈ R[x; α] is unit if and only if its constant term is unit and other coefficients are all nilpotent. In [4] Chen and Cui proved that if R is an α-compatible and weakly 2-primal ring, then R[x; α] is weakly 2-primal.

In this paper we prove some results which concern the constant products of elements in Ore extension rings over reversible ring. Roughly speaking, our main theorems give a characterization of the unit elements in Ore extension ring. We will also pay a particular attention to stable range one property of Ore extension rings. As the final part of the paper we construct examples of rings that support our main results and explain the limitations of the results obtained in former sections.

2. Ore Extension of Reversible Rings with Constant Products of Elements

In this section we prove some results which concern the constant products of elements in Ore extension rings over reversible ring. Note that the methods used for the “unmixed” Ore extensions do not apply to the general case. We also note that in the investigation of Ore extension rings R[x; α, δ], our results based on the twist property in multiplication of polynomials. We start this section by the following lemma, which will be useful in the sequel.

### Lemma 2.1

Let R be an (α, δ)-compatible ring and a, bR. Then we have the following:

(1) If ab = 0, then aαn(b) = 0 = αn(a)b for any non-negative integer n.

(2) If αk(a)b = 0 for some non-negative integer k, then ab = 0.

(3) If ab = 0, then αn(a)δm(b) = 0 = δm(a)αn(b) for any non-negative integers m, n.

(4) If ab = 0, then α(a)α(b) = 0 = δ(a)δ(b) .

(5) If an = 0, then (α(a))n = 0 = (δ(a))n for any positive integer n.

(6) If ab = 0 then axmb = 0 for each m ≥ 0.

If axmb = 0 in R[x; α, δ], for some m ≥ 0, then ab = 0.

Proof

(1), (2) and (3) are proved in [9, Lemma 2.1].

(4) Since ab = 0, then by (1) and (2) we have α(a)b = 0 = δ(a)b. Hence α(a)α(b) = 0 = δ(a)δ(b), since R is (α, δ)-compatible.

(5), (6) and (7) follow from (4), (3) and (1), respectively.

### Lemma 2.2.([8, Lemma 2.2])

Let R be an (α, δ)-compatible ring, f(x) = a0+a1x+ ··· + anxnR[x; α, δ] and c, rR. Then f(x)r = c if and only if a0r = c and air = 0 for each 1 ≤ in.

### Lemma 2.3.([8, Lemma 2.3])

Let R be an α-rigid ring and also f(x) = a0 +a1x+ ···+anxn and g(x) = b0 +b1x+···+bmxm be non-zero elements of R[x; α, δ] such that f(x)g(x) = cR. Then a0b0 = c and aibj = 0, for each i, j with i + j ≥ 1

Recall that an ideal I of R is called an α-ideal if α(I) ⊆ I; I is called an α- invariant if α−1(I) = I; I is called a δ-ideal if δ(I) ⊆ I; I is called an (α, δ)-ideal if it is both an α- and a δ-ideal. Clearly, each α-compatible ideal is an α-invariant ideal, and each δ-compatible ideal is δ-ideal.

If I is an (α, δ)-ideal, then ᾱ : R/IR/I defined by ᾱ(a) = α(a) + I is an endomorphism and δ̄ : R/IR/I defined by δ̄(a) = δ(a) + I is an ᾱ-derivation.

Recall also that, an ideal ℘ of R is completely prime if ab ∈ ℘ implies a ∈ ℘ or b ∈ ℘ for a, bR.

### Lemma 2.4

Let R be a 2-primal ring anda minimal prime ideal of R. If R is an (α, δ)-compatible ring, thenis an α-invariant and a δ-ideal of R. Moreover, ℘ is an (α, δ)-compatible ideal of R.

Proof

Since R is 2-primal, Niℓ(R) = Niℓ*(R). Then = R/Niℓ(R) is a reduced ring. By Lemma 2.1, Niℓ(R) is an (α, δ)-compatible ideal of R, and so is (ᾱ, δ̄)- compatible, by [7, Proposition 2.1]. Clearly ℘/Niℓ(R) is a minimal prime ideal of . Then, by [14, Lemma 1.5], ℘/Niℓ(R) is a collection of some right annihilators of subsets of . Thus ℘/Niℓ(R) is an ᾱ-invariant and a δ̄-ideal of , since is (ᾱ, δ̄)-compatible. Therefore ℘ is an α-invariant and a δ-ideal of R.

Now, since ℘ is completely prime, α-invariant and a δ-ideal of R, one can easily show that ℘ is an (α, δ)-compatible ideal of R.

Now we are in a position to give one of our main theorems in this paper. Recall that following [17], an associative ring R with unity is called left McCoy when the equation g(x)f(x) = 0 implies rf(x) = 0 for some non-zero element rR, where f(x) and g(x) are non-zero polynomials in R[x]. Right McCoy rings are defined dually and they satisfy dual properties. A ring R is called McCoy if it is both left and right McCoy. This name for them was chosen by Nielsen in [17] in recognition of McCoy’s proof in [15, Theorem 2] that commutative rings satisfy this condition. McCoy rings are unified generalization of a reversible and right duo rings. These rings, though may look a bit specific, were studied by many authors and are related to important ring theory problems. Systematic studies of these rings were started in [17] and next continued in a number of papers, generalizing the McCoy condition in many different ways. The following two theorems are generalizations of [3, Theorems 2.4 and 2.5]. When Niℓ(R) forms an ideal of a ring R, we say an element rR is unit modulo Niℓ(R) if r + Niℓ(R) is unit in R/Niℓ(R).

### Theorem 2.5

Let R be a reversible (α, δ)-compatible ring and f(x) = a0 + a1x + ···+anxn be a non-zero element of R[x; α, δ]. If there is a non-zero element g(x) = b0 + b1x + ··· + bmxmR[x; α, δ] such that g(x)f(x) = c is a constant, then there exist non-zero elements r, aR such that rf(x) = ac. In particular r = abp for some p, 0 ≤ pm, and a is either one or a product of at most m coefficients from f(x). Furthermore, if b0is a unit element modulo Niℓ(R), then a1, a2, …, an are all nilpotent.

Proof

Let deg(f(x)) = 0. Then f(x) = a0. Hence bja0 = 0 for each 1 ≤ jm and b0a0 = c, by Lemma 2.2. If c ≠ 0, then b0 ≠ 0 and so r = b0 and a = 1 are desired non-zero elements. If c = 0, then bja0 = 0 for each 0 ≤ jm. Since g(x) ≠ 0, there exists 0 ≤ sm such that bs ≠ 0. Thus r = bs and a = 1 are desired non-zero elements.

Now let deg(f(x)) = n ≥ 1. We proceed by induction on degree of g(x). If deg(g(x)) = m = 0, then g(x) = b0 ≠ 0. Since g(x)f(x) = c, we have b0a0 = c and b0ai = 0 for each 1 ≤ in. Thus r = b0 and a = 1 are desired elements. Assume that the conclusion is true for all polynomials of degree less that m. Let g(x)f(x) = c and deg(g(x)) = m. We proceed by dividing the proof into two cases:

• Case 1: Let b0 = b1 = ··· = bm−1 = 0. Since c = g(x)f(x) = bmxmf(x) we have bmai = 0 for each 0 ≤ in, by Lemma 2.1, and so c = 0. Thus r = bm and a = 1 are desired elements.

• Case 2: Assume that there exits 0 ≤ jm − 1 such that bj ≠ 0. If bmf(x) = 0, then bmxmf(x) = 0, by Lemma 2.1, and so c = g(x)f(x) = (b0 + b1x + ··· + bm−1xm−1)f(x). Since g1(x) = b0 + b1x + ··· + bm−1xm−1 ≠ 0, hence deg(g1(x)) ≤ m − 1 and g1(x)f(x) = c, so the result follows from induction hypothesis. Now assume that bmf(x) ≠ 0. Then bmxmf(x) ≠ 0, by Lemma 2.1. Hence c = g(x)f(x) = g(x)(ai1xi1 + ··· + aitxit ) such that g(x)aik ≠ 0 for each 1 ≤ kt, where 0 ≤ i1, itn. Thus bmαm(ait) = 0 and so bm ait = 0, by Lemma 2.1. Hence aitbm = 0, since R is reversible. Thus aitc = (aitg(x))f(x). Since aitg(x) ≠ 0, deg(aitg(x)) ≤ m − 1 and (aitg(x))f(x) = aitc, hence by induction hypothesis there exist elements 0 ≠ r1, a1R such that r1f(x) = a1aitc. Thus r = r1 and a = a1aitare desired elements.

Now we will show that the elements a1, a2, …, an are all nilpotent, when b0 is unit modulo Niℓ(R). Since R is reversible and hence Niℓ(R) is an ideal of R, then = R/Niℓ(R) is reduced. On the other hand, since Niℓ(R) is an (α, δ)- compatible ideal of R, hence by [7, Proposition 2.1], = R/Niℓ(R) is (ᾱ, δ̄)- compatible, and therefore R is ᾱ-rigid, by [9, Lemma 2.2]. Now from g(x)f(x) = c, we get (x) (x) = , and hence by Lemma 2.3 we have 0ā0 = and jāi = 0̄ for each i, j with i + j ≥ 1. Now since b0 is a unit element of R, then d0b0 = 1 for some d0R. Multiplying 0āi = 0̄ from left by 0 we get āi = 0̄ for each i ≥ 1. This yields that aiNiℓ(R) for each i ≥ 1, completing the proof.

The same idea as that in the proof of Theorem 2.5, can be used to prove the following theorem.

### Theorem 2.6

Let R be a reversible (α, δ)-compatible ring and f(x) = a0 + a1x + ···+anxn be a non-zero element of R[x; α, δ]. If there is a non-zero element g(x) = b0 + b1x + ··· + bmxmR[x; α, δ] such that f(x)g(x) = c is a constant, then there exist non-zero elements r, aR such that f(x)r = ca. In particular r = bpa for some p, 0 ≤ pm, and a is either one or a product of at most m coefficients from f(x). Furthermore, if b0is a unit element modulo Niℓ(R), then a1, …, an are nilpotent.

Theorems 2.5 and 2.6, have the following immediate corollary.

### Corollary 2.7

Let R be a reversible (α, δ)-compatible ring. Then f(x)R[x; α, δ] is a right or left zero-divisor, if and only if there exists a non-zero constant rR such that rf(x) = 0 = f(x)r.

Proof

It follows from Theorems 2.5, 2.6 and Lemma 2.2.

In particular, taking c = 0, α to be identity and δ equal to zero, we obtain the following main result of [17], as another corollary of Theorem 2.5.

### Corollary 2.8.([17, Theorem 2])

Let R be a reversible ring. Then R is right and left McCoy ring.

Let δ be an α-derivation of R. For integers 0 ≤ ij, let us write $fij$ for the set of all “words” in α and δ in which there are i factors of α and ji factors of δ. For instance, $fjj={αj}, f0j={δj}$and $fj-1j={αj-1δ,αj-2δα,…,δαj-1}$. Clearly each element of $fij$is an additive map on R. Also for each aR and each integers 0 ≤ ij, we write $fij(a)={β(a)∣β∈fij}$.

We say also that a set SR is locally nilpotent if for any subset {s1, …, sn} ⊆ S, there exists an integer t, such that any product of t elements from {s1, …, sn} is zero.

### Lemma 2.9

Let R be an (α, δ)-compatible ring and Niℓ(R) a locally nilpotent ideal of R. Then Niℓ(R)[x; α, δ] is a nil ideal of R.

Proof

Since R is (α, δ)-compatible ideal of R, Niℓ(R) is an (α, δ)-compatible ideal of R, by Lemma 2.1. Hence Niℓ (R)[x; α, δ] is an ideal of R[x; α, δ]. Let f(x) = a0+ a1x + ··· + amxmNiℓ (R)[x; α, δ]. Assume that M = {a0, a1, …, am} ⊆ Niℓ (R). Since Niℓ (R) is locally nilpotent, there exists a positive integer t such that any product of t elements from M is zero. Let $N=∪fij(ar)$, where 0 ≤ rm and 0 ≤ ij are integers. Then any product of t elements of N is zero, by Lemma 2.1. Thus (f(x))t = 0, since each coefficient of (f(x))t is a finite sum of the product of t elements from N. Therefore Niℓ(R)[x; α, δ] is a nil ideal of R[x; α, δ].

We continue by proving the second main result of this paper, which investigates the equivalences of the weakly 2-primal property of Ore extension ring with the coefficent ring R, when R is an (α, δ)-compatible ring.

### Theorem 2.10

Let R be an (α, δ)-compatible ring. Then R is a weakly 2-primal ring if and only if R[x; α, δ] is a weakly 2-primal ring. In this case R[x; α, δ] is a weakly semicommutative ring.

Proof

For the forward direction, since R is weakly 2-primal, we have Lrad(R) = Niℓ(R). We will show that Lrad(R[x; α, δ]) = Niℓ(R[x; α, δ]). It is enough to show that Niℓ(R[x; α, δ]) ⊆ Lrad(R[x; α, δ]), since the reverse inclusion is obvious. First we show that Niℓ(R[x; α, δ]) = Niℓ(R)[x; α, δ] = Lrad(R)[x; α, δ]. By Lemma 2.9, we have Niℓ(R)[x; α, δ] is a nil ideal of R[x; α, δ], and hence Niℓ(R)[x; α, δ] ⊆ Niℓ(R[x; α, δ]). For the reverse inclusion, assume that $f(x)=∑i=0maixi$ is a nilpotent element with the nilpotency index t. Therefore we get amαm(am)α2mα(t−1)m(am) = 0, since it is the leading coefficient of (f(x))t = 0. Then $amt=0$, by Lemma 2.1, and so amNiℓ(R). Since Niℓ(R) is an (α, δ)-compatible ideal, $fij(am)⊆Niℓ(R)$ for each integers 0 ≤ ij. Thus 0 = (f(x))t = (a0 + a1x + ··· + am−1xm−1)t + h(x), where h(x) ∈ Niℓ(R)[x; α, δ]. Then (a0 +a1x+···+am−1xm−1)tNiℓ(R)[x; α, δ]. Then f(x) ∈ Niℓ(R[x; α, δ]), since Niℓ(R)[x; α, δ]) ⊆ Niℓ(R[x; α, δ]). Now by induction hypothesis on degree of f(x) we conclude that a0, a1, …, am−1Niℓ(R), which implies that f(x) ∈ Niℓ(R)[x; α, δ]. Therefore Niℓ(R[x; α, δ]) = Niℓ(R)[x; α, δ] = Lrad(R)[x; α, δ].

Next we show that Lrad(R)[x; α, δ] is a locally nilpotent ideal of R[x; α, δ]. Assume that M = {f1(x), …, fk(x)} be a subset of Lrad(R)[x; α, δ]. Write fi(x) = ai0 + ai1x + ··· + aimxm, where aij is in Lrad(R) for all i = 1, 2, ···, k and j = 0, 1, …, n. Let N = {ai0, ai1, …, aim | i = 1, 2, …, k}. Then N is a finite subset of Lrad(R) and since Lrad(R) is locally nilpotent, there exists a positive integer t such that any product of t elements from N is zero. Assume that $W=∪fij(ars)$, where 1 ≤ rk, 0 ≤ sm and 0 ≤ ij are non-negative integers. Then by Lemma 2.1, any product of t elements from W is zero, which implies that any product of t elements from M is zero. Therefore Lrad(R)[x; α, δ] is a locally nilpotent ideal of R[x; α, δ]. The backward direction is clear, since each subring of a weakly 2-primal ring is weakly 2-primal.

Moreover it is clear that the ring R[x; α, δ] is weakly semicommutative, since R[x; α, δ]/Niℓ(R[x; α, δ]) is a reduced ring, and the proof is complete.

### Corollary 2.11

Let R be a reversible ring which is (α, δ)-compatible. Then a polynomial f(x) ∈ R[x; α, δ] is unit if and only if its constant term is a unit and the other coefficionts are nilpotent.

Proof

Let = R/Niℓ(R). Then is ᾱ-rigid, by [9, Lemma 2.2]. For the forward direction, let f(x) = a0 + a1x + ··· + anxnR[x; α, δ] is a unit. Then there exists g(x) ∈ R[x; α, δ] such that f(x)g(x) = 1 = g(x)f(x). Hence (x) (x) = 1. Then ā0 is a unit element of , by Lemma 2.3, and so a1, …, an are nilpotent, by Theorem 2.5.

For the reverse implication, let a0 is a unit and a1, …, an are nilpotent. Then a1x+···+anxnniℓ(R)[x; α, δ] = Lrad(R[x; α, δ]) ⊆ J(R[x; α, δ]), by Theorem 2.10. Thus f(x) = a0 + a1x + ··· + anxn is a unit of R[x; α, δ].

We denote the unit group of R by U(R).

### Corollary 2.12

Let R be an α-rigid ring. Then U(R[x; α, δ]) = U(R).

### Corollary 2.13

Let R be a NI ring. If Niℓ(R) is an (α, δ)-compatible ideal of R, then U(R[x; α, δ]) ⊆ U(R) + Niℓ(R)[x; α, δ].

Proof

Since R is NI, hence Niℓ(R) is an ideal of R and then = R/Niℓ(R) is reduced. Since Niℓ(R) is an (α, δ)-compatible ideal of R, then the ring is (ᾱ, δ̄)- compatible, by [7, Proposition 2.1]. Hence is ᾱ-rigid, by [9, Lemma 2.2]. Now, the result follows from Corollary 2.12.

### Corollary 2.14

Let R be a weakly 2-primal ring which is (α, δ)-compatible. Then U(R[x; α, δ]) = U(R) + Niℓ(R)[x; α, δ] = U(R) + Niℓ(R[x; α, δ]).

Proof

Since R is (α, δ)-compatible, Niℓ(R) is (α, δ)-compatible, by Lemma 2.1. Since a weakly 2-primal ring is NI, we have U(R[x; α, δ]) ⊆ U(R)+Niℓ(R)[x; α, δ], by Corollary 2.13. Since R is weakly 2-primal, Niℓ(R)[x; α, δ] = Niℓ(R[x; α, δ]). Thus U(R) + Niℓ(R)[x; α, δ] ⊆ U(R[x; α, δ]). Therefore U(R[x; α, δ]) = U(R) + Niℓ(R)[x; α, δ].

### Corollary 2.15

Let R be a weakly 2-primal ring which is (α, δ)-compatible. Then J(R[x; α, δ]) = Niℓ(R)[x; α, δ].

Proof

Following Theorem 2.10, we get

$Niℓ(R)[x;α,δ]=Niℓ(R[x;α,δ]=L-rad(R)[x;α,δ]=L-rad(R[x;α,δ]).$

Thus the result follows from Corollary 2.14.

Our next result concerns with stable range 1 property of Ore extension rings. Recall that an element a in any ring R is said to have (right) stable range 1 (written Sr(a) = 1) if aR + bR = R (for any bR) implies that a + brU(R) for some rR. If Sr(a) = 1 for all aR, then a ring R is said to have stable range 1, written Sr(R) = 1. It is well known that this property is left-right symmetric.

### Proposition 2.16

If Niℓ(R) is an (α, δ)-compatible ideal of R, then we have Sr(R[x; α, δ]) > 1.

Proof

We will begin by assuming, for a contradiction, that Sr(R[x; α, δ]) = 1. Now, since x(−x)+1+x2 = 1, then there exists f(x) ∈ R[x; α, δ] such that x+(1+x2)f(x) is a unit in R[x; α, δ]. For each aR we have x2a = δ2(a) + [αδ(a) + δα(a)]x + α2(a)x2. Write f(x) = a0 + a1x + ··· + anxn. Then

$x+(1+x2)f(x) =[δ2(a0)+a0] +[δ2(a1)+αδ(a0)+δα(a0)+a1+1]x +[δ2(a2)+αδ(a1)+δα(a1)+α2(a0)+a2]x2 +[δ2(a3)+αδ(a2)+δα(a2)+α2(a1)+a3]x3 ⋮ +[δ2(an-1)+αδ(an-2)+δα(an-2)+α2(an-3)+an-1]xn-1 +[δ2(an)+αδ(an-1)+δα(an-1)+α2(an-2)+an]xn +[αδ(an)+δα(an)+α2(an-1)]xn-1 +[α2(an)]xn+2.$

Since Niℓ(R) is an ideal of R, hence R is weakly 2-primal and so by Corollary 2.14, the constant term of x+(1+x2)f(x) is unit and other coefficients are all nilpotent. Then nilpotency of α2(an) and Lemma 2.1 imply that an is nilpotent. Since Niℓ(R) is an α-invariant and a δ-ideal, we have αδ(an)+δα(an) ∈ Niℓ(R). Now nilpotency of [αδ(an)+δα(an)+α2(an−1)] and Lemma 2.1 implies that an−1Niℓ(R). By a similar argument, one can show that a0, a1, …, anNiℓ(R). Then [δ2(a0) + a0] ∈ Niℓ(R), which is a contradiction, since it is the constant term of x + (1 + x2)f(x). Therefore Sr(R[x; α, δ]) > 1, as desired.

3. The Examples

Below we construct examples of rings which help us to support our main results and also explain the limitations of the obtained results. The class of NI rings which are (α, δ)-compatible are quite large and important. For some other classes of examples, we direct the reader to see [6]. As a first example, we present some classes of rings with this property.

### Example 3.1

Let S be any reduced ring and consider the reversible and hence NI ring R = {(a, b) | a, bS} with addition pointwise and multiplication given by (a, b)(c, d) = (ac, ad + bc). Let α : RR be an automorphism defined by α((a, b)) = (a,−b) and δ : RR be an α-derivation defined by δ((a, b)) = (a, b) − α((a, b)) = (0,−b). We will show that R is (α, δ)-compatible. To see this, let (a, b)(c, d) = 0. Thus ac = ad + bc = 0. Therefore ca = 0, since S is reduced. Now, multiplying ad+bc = 0 from left by c, we get cad+cbc = 0 and then cbc = 0. Hence (bc)2 = bcbc = 0 and so bc = ad = 0, since S is reduced. Then (a, b)α((c, d)) = (ac, bcad) = (0, 0). Similarly, one can see that (a, b)α((c, d)) = (0, 0) implies that (a, b)(c, d) = (0, 0). Thus R is α-compatible. On the other hand, let (a, b)(c, d) = 0. Then similar computations as above show that (a, b)δ((c, d)) = 0. Therefore R is (α, δ)-compatible, as desired.

Let R be a ring and σ denotes an endomorphism of R with σ(1) = 1. In [5], the authors introduced skew triangular matrix ring as a set of all triangular matrices with addition point-wise and a new multiplication subject to the condition Eijr = σji(r)Eij. So (aij)(bij) = (cij ), where cij = aiibij + ai,i+1σ(bi+1,j) + ··· + aijσji(bjj), for each ij and denoted it by Tn(R, σ).

The subring of the skew triangular matrices with constant main diagonal is denoted by S(R, n, σ); and the subring of the skew triangular matrices with constant diagonals is denoted by T(R, n, σ). We can denote A = (aij) ∈ T(R, n, σ) by (a11, …, a1n). Then T(R, n, σ) is a ring with addition point-wise and multiplication given by:

$(a0,…,an-1)(b0,…,bn-1)=(a0b0,a0*b1+a1*b0,…,a0*bn-1+⋯+an-1*b0),$

with ai * bj = aiσi(bj ), for each i and j. Therefore, clearly one can see that T(R, n, σ) ≅ R[x; σ]/(xn), where (xn) is the ideal generated by xn in R[x; σ].

We consider the following two subrings of S(R, n, σ), as follows:

$A(R,n,σ)={∑j=1⌊n2⌋∑i=1n-j+1ajEi,i+j-1+∑j=⌊n2⌋+1n∑i=1n-j+1ai,i+j-1Ei,i+j-1};B(R,n,σ)={A+rE1k∣A∈A(R,n,σ) and r∈R} n=2k≥4.$

Let α and σ be endomorphisms of a ring R and δ is an α-derivation, with ασ = σα and δσ = σδ. The endomorphism α of R is extended to the endomorphism ᾱ : Tn(R, σ) → Tn(R, σ) defined by ᾱ((aij)) = (α(aij)) and the α-derivation δ of R is also extended to δ̄ : Tn(R, σ) → Tn(R, σ) defined by δ̄((aij)) = (δ(aij )).

### Example 3.2

α be a rigid endomorphism of a ring R, δ and α-derivation and σ an endomorphism of R such that ασ = σα and σδ = δσ. If R is α-rigid, then the ring T(R, n, σ)[x; ᾱ, δ̄] is a reversible ring.

Proof

We break the proof in to some parts.

Claim 1

If R is a (α, δ) compatible and (α, δ)-skew Armendariz, then R is a reversible ring if and only if R[x; α, δ] is reversible.

### Proof

We only need to prove the necessity. For this, let $f(x)=∑i=0maixi$ and $g(x)=∑j=0nbjxj∈R[x;α,δ]$ such that f(x)g(x) = 0. Then aixibjxj = 0, for each i and j, by (α, δ)-skew Armendariz property of R. Then by Lemma 2.1, we have aibj = 0, for each i and j. Since R is reversible bjai = 0 and hence bjαk(ai) = 0, for each k ≥ 0 and bjδl(ai) = 0, for each l ≥ 0, since R is (α, δ) compatible. Then g(x)f(x) = 0 and therefore R[x; α, δ] is reversible.

Claim 2

Let R be a σ-rigid ring and A = (aij ), B = (bij) ∈ A(R, n, σ) such that AB = 0, then aikbkj = 0, for each 1 ≤ i, j, kn.

### Proof

We proceed by induction on n. The case n = 1 is clear and hence the base case of our induction is established. So, we may assume n > 1 and that the claim is true for all smaller values by inductive assumption. Let A = (aij) and B = (bij) in A(R, n, σ) such that AB = 0. Consider the following elements in $A(R,n,σ): A′=(aij′),A″=(aij″),B′=(bij′),B″=(bij″)$, where $aij′=aij,aij″=ai+1,j+1,bij′=bij,bij″=bi+1,j+1$, for all 1 ≤ i, jn−1. Now from AB = 0, we get AB′ = 0 = AB″. Thus by induction hypothesis, we have

$aikbkj=0, alnbnn=0, alkbkn=0,$

for all 1 ≤ i, j, kn−1 and 2 ≤ ln. Now it is sufficient to show that a1kbkn = 0, for each k. From the entry 1n in AB = 0, we have that

$a11b1n+a12σ(b2n)+⋯+a1nσn-1(bnn)=0.$

Since A,BA(R, n, σ), we have that aij = ai+1,j+1, for $1≤i≤j≤⌊n2⌋$ and bkn = bk−1,n−1 for $⌈n2⌉≤k≤n$, where ⌊x⌋ denotes the largest integer less than or equal to x and also ⌈x⌉ denotes the smallest integer greater than or equal to x.

Since R is σ-rigid, we have σj−1(bjn)a11 = σj−1(bjn)ajj = 0, for all 1 < jn. Thus multiplying (3.1) by a11 on the right, we obtain a11b1na11 = 0. Since R is reduced, we get a11b1n = 0. Therefore

$a12σ(b2n)+a13σ2(b3n)+⋯+a1nσn-1(bnn)=0.$

Since R is σ-rigid, σj−1(bjn)a12 = σj−1(bjn)aj−1,j = 0, for all 2 < jn if $2≤⌊n2⌋$. Thus, multiplying (3.2) by a12 on the right, we obtain a12σ(b2n)a12 = 0. Since R is reduced, we get a12σ(b2n) = 0. Therefore

$a13σ2(b3n)+⋯+a1nσn-1(bnn)=0,$

and also, by σ-compatibility of R, we get a12b2n = 0. Repeating in this way, we get a1kbkn = 0, for $1≤k≤⌊n2⌋$ and also

$a1,⌊n2⌋+1σ⌊n2⌋(b⌊n2⌋+1,n)+⋯+a1nσn-1(bnn)=0.$

Since R is σ-rigid, we get σn−1(bnn)a1k = σn−1(bkk)a1k = 0, for all 1 ≤ k < n. Thus, multiplying (3.4) by σn−1(bnn) on the left, we obtain σn−1(bnn)a1nσn−1(bnn) = 0. Using the reduced property of a ring R, we get a1nσn−1(bnn) = 0. Therefore

$a1,⌊n2⌋+1σ⌊n2⌋(b⌊n2⌋+1,n)+⋯+a1n-1σn-2(bn-1,n)=0$

and also using the σ-compatibility of R, we get a1nbnn = 0. By continuing in this way, we can show that a1kbkn = 0, for all $⌈n2⌉≤k≤n$. Thus if n is even, we are done. Also, if n is odd, we obtain that

$a1,⌊n2⌋+1σ⌊n2⌋(b⌊n2⌋+1,n)=0,$

and by σ-compatibility of R, we get

$a1,⌊n2⌋+1b⌊n2⌋+1,n=0.$

Therefore aikbkj = 0, for all 1 ≤ i, j, kn, as desired.

Now, by Claim 2, one can see easily that T(R, n, σ) is reversible and (ᾱ, δ̄)- compatible. Therefore by Claim 1, it is sufficient to prove that T(R, n, σ) is (ᾱ, δ̄)-skew Armendariz ring. For, consider the map ϕ : T(R, n, σ)[x; ᾱ, δ̄] → T(R[x; α, δ], n, σ̄) defined by $ϕ(∑k=0mAkxk)=(fij)$, where $Ak=(aij(k))$ and $fij=∑k=0maij(k)xk$, for each m and 1 ≤ i, jn. It is easy to show that ϕ is an isomorphism. Let $f=∑k=0rAkxk$ and $g=∑l=0sBlxl∈T(R,n,σ)[x;α¯,δ¯]$ and fg = 0, where $Ak=(aij(k))$ and $Bl=(bij(l))$. So we have (fij)(gij) = 0, where $fij=∑k=0raij(k)xk$ and $gij=∑l=0sbij(l)xl$. Since R is σ-rigid, R[x; α, δ] is σ̄-rigid. This is because, if $f(x)=∑i=0maixi∈T[x;α,δ]$ and f(xσ̄(f(x)) = 0, then we have amαm(σ(am)) = 0 and consequently amσ(αm(am)) = 0, since ασ = σα. Thus amαm(am) = 0, since R is σ-rigid and so am = 0, since R is α-rigid. Hence f(x) = 0 and hence R[x; α, δ] is σ̄-rigid. Thus fitgtj = 0, for each 1 ≤ i, j, tn and easy calculations show that AkxkBlxl = 0, for each k and l, showing that T(R, n, σ) is (ᾱ, δ̄)-skew Armendariz ring, and the result follows.

The following example shows that in Theorems 2.5 and 2.6, the hypothesis that “R is (α, δ)-compatible” can not be dropped.

### Example 3.3

(1) Let R0 be any reduced ring. Then for a reversible ring R = R0[x], consider the endomorphism α : RR given by α(f(x)) = f(0) and α-derivation δ : RR given by δ(f(x)) = xf(x) − α(f(x))x. Then one can see easily that the ring R is δ-compatible but R is not α-compatible. Now considering the elements F(y) = a0 + a1y and G(y) = b0 + b1y in R[y; α, δ], where a0 = b0 = x, a1 = 0 and b1 = −x, we get G(y)F(y) = x2x3 = cR but b0a0 = x2c.

(2) Let S and T be any reduced rings. Suppose R = ST with the usual addition and multiplication. Then for reversible ring R, let α : RR be an endomorphism defined by α((a, b)) = (b, a) and δ : RR be an α-derivation defined by δ((a, b)) = (ab, 0). Then it can be easily seen that the ring R is neither α-compatible nor δ-compatible. Let f(x) = (−1, 0) + (1, 0)x and g(x) = (1, 0) + (−1, 0)x be non-zero elements in R[x; α, δ]. Then f(x)g(x) = (0, 0), but (−1, 0)(1, 0) = (−1, 0) ≠ (0, 0).

It is well known that when R is an NI ring and (α, δ)-compatible, then Niℓ(R) is an (α, δ)-compatible ideal of R. But, the following example shows that the converse is not true in general.

### Example 3.4

Let ℤ4 be the ring of integers modulo 4. Consider the reversible and hence NI ring R = {(a, b) | a, b ∈ ℤ4} with addition pointwise and multiplication given by (a, b)(c, d) = (ac, ad + bc). Let α : RR be an endomorphism defined by α((a, b)) = (a, 2b). We will show that R is not α-compatible ring. To see this, let r = (2, 0) and s = (0, 1) in R, then rs = (0, 2) ≠ (0, 0) whereas (s) = (2, 0)(0, 2) = (0, 0). Thus R is not α-compatible. On the other hand, since Niℓ(R) = {(0, b), (2, b)| b ∈ ℤ4}, easy calculations show that Niℓ(R) is an α-compatible ideal of R.

Acknowledgements

This research was in part supported by a grant from Shahrood University of Technology.

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# Let f(x) = x{eq}^2 {/eq} + 3x - 5. Find f'(0), f'(3), f'(-1), and f'(-2).

## Question:

Let f(x) = x{eq}^2 {/eq} + 3x - 5. Find f'(0), f'(3), f'(-1), and f'(-2).

## Signs of the First Derivative:

The first derivative of a real variable function can change sign at values where this derivative is zero or at values where it is not defined. The sign in each interval is precisely defined by evaluating in the first derivative a point that belongs to the interval and verifying the sign of the result.

## Answer and Explanation: 1

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Differentiate the given function with respect to {eq}x{/eq}.

\begin{align} f(x) &= x^2+3x-5\\[0.3cm] f'(x) &=...

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