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, b ∈ R. Then we have the following:
(1) If ab = 0, then aα^{n}(b) = 0 = α^{n}(a)b for any nonnegative integer n.
(2) If α^{k}(a)b = 0 for some nonnegative integer k, then ab = 0.
(3) If ab = 0, then α^{n}(a)δ^{m}(b) = 0 = δ^{m}(a)α^{n}(b) for any nonnegative integers m, n.
(4) If ab = 0, then α(a)α(b) = 0 = δ(a)δ(b) .
(5) If a^{n} = 0, then (α(a))^{n} = 0 = (δ(a))^{n} for any positive integer n.
(6) If ab = 0 then ax^{m}b = 0 for each m ≥ 0.
If ax^{m}b = 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) = a_{0}+a_{1}x+ ··· + a_{n}x^{n} ∈ R[x; α, δ] and c, r ∈ R. Then f(x)r = c if and only if a_{0}r = c and a_{i}r = 0 for each 1 ≤ i ≤ n.
Lemma 2.3.([8, Lemma 2.3])
Let R be an αrigid ring and also f(x) = a_{0} +a_{1}x+ ···+a_{n}x^{n} and g(x) = b_{0} +b_{1}x+···+b_{m}x^{m} be nonzero elements of R[x; α, δ] such that f(x)g(x) = c ∈ R. Then a_{0}b_{0} = c and a_{i}b_{j} = 0, for each i, j with i + j ≥ 1
Recall that an ideal I of R is called an αideal if α(I) ⊆ I; I is called an α invariant if α^{−1}(I) = I; I is called a δideal if δ(I) ⊆ I; I is called an (α, δ)ideal if it is both an α and a δideal. Clearly, each αcompatible ideal is an αinvariant ideal, and each δcompatible ideal is δideal.
If I is an (α, δ)ideal, then ᾱ : R/I → R/I defined by ᾱ(a) = α(a) + I is an endomorphism and δ̄ : R/I → R/I defined by δ̄(a) = δ(a) + I is an ᾱderivation.
Recall also that, an ideal ℘ of R is completely prime if ab ∈ ℘ implies a ∈ ℘ or b ∈ ℘ for a, b ∈ R.
Lemma 2.4
Let R be a 2primal ring and ℘ a minimal prime ideal of R. If R is an (α, δ)compatible ring, then ℘ is an αinvariant and a δideal of R. Moreover, ℘ is an (α, δ)compatible ideal of R.
Proof
Since R is 2primal, Niℓ(R) = Niℓ_{*}(R). Then R̄ = R/Niℓ(R) is a reduced ring. By Lemma 2.1, Niℓ(R) is an (α, δ)compatible ideal of R, and so R̄ is (ᾱ, δ̄) compatible, by [7, Proposition 2.1]. Clearly ℘/Niℓ(R) is a minimal prime ideal of R̄. Then, by [14, Lemma 1.5], ℘/Niℓ(R) is a collection of some right annihilators of subsets of R̄. Thus ℘/Niℓ(R) is an ᾱinvariant and a δ̄ideal of R̄, since R̄ is (ᾱ, δ̄)compatible. Therefore ℘ is an αinvariant and a δideal of R.
Now, since ℘ is completely prime, αinvariant and a δideal 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 nonzero element r ∈ R, where f(x) and g(x) are nonzero 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 r ∈ R 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) = a_{0} + a_{1}x + ···+a_{n}x^{n} be a nonzero element of R[x; α, δ]. If there is a nonzero element g(x) = b_{0} + b_{1}x + ··· + b_{m}x^{m} ∈ R[x; α, δ] such that g(x)f(x) = c is a constant, then there exist nonzero elements r, a ∈ R such that rf(x) = ac. In particular r = ab_{p} for some p, 0 ≤ p ≤ m, and a is either one or a product of at most m coefficients from f(x). Furthermore, if b_{0}is a unit element modulo Niℓ(R), then a_{1}, a_{2}, …, a_{n} are all nilpotent.
Proof
Let deg(f(x)) = 0. Then f(x) = a_{0}. Hence b_{j}a_{0} = 0 for each 1 ≤ j ≤ m and b_{0}a_{0} = c, by Lemma 2.2. If c ≠ 0, then b_{0} ≠ 0 and so r = b_{0} and a = 1 are desired nonzero elements. If c = 0, then b_{j}a_{0} = 0 for each 0 ≤ j ≤ m. Since g(x) ≠ 0, there exists 0 ≤ s ≤ m such that b_{s} ≠ 0. Thus r = b_{s} and a = 1 are desired nonzero 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) = b_{0} ≠ 0. Since g(x)f(x) = c, we have b_{0}a_{0} = c and b_{0}a_{i} = 0 for each 1 ≤ i ≤ n. Thus r = b_{0} 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 b_{0} = b_{1} = ··· = b_{m}_{−1} = 0. Since c = g(x)f(x) = b_{m}x^{m}f(x) we have b_{m}a_{i} = 0 for each 0 ≤ i ≤ n, by Lemma 2.1, and so c = 0. Thus r = b_{m} and a = 1 are desired elements.

Case 2: Assume that there exits 0 ≤ j ≤ m − 1 such that b_{j} ≠ 0. If b_{m}f(x) = 0, then b_{m}x^{m}f(x) = 0, by Lemma 2.1, and so c = g(x)f(x) = (b_{0} + b_{1}x + ··· + b_{m}_{−1}x^{m}^{−1})f(x). Since g_{1}(x) = b_{0} + b_{1}x + ··· + b_{m}_{−1}x^{m}^{−1} ≠ 0, hence deg(g_{1}(x)) ≤ m − 1 and g_{1}(x)f(x) = c, so the result follows from induction hypothesis. Now assume that b_{m}f(x) ≠ 0. Then b_{m}x^{m}f(x) ≠ 0, by Lemma 2.1. Hence c = g(x)f(x) = g(x)(a_{i}_{1}x^{i}^{1} + ··· + a_{i}_{t}x^{it} ) such that g(x)a_{i}_{k} ≠ 0 for each 1 ≤ k ≤ t, where 0 ≤ i_{1}, i_{t} ≤ n. Thus b_{m}α_{m}(a_{i}_{t}) = 0 and so b_{m} a_{i}_{t} = 0, by Lemma 2.1. Hence a_{i}_{t}b_{m} = 0, since R is reversible. Thus a_{i}_{t}c = (a_{i}_{t}g(x))f(x). Since a_{i}_{t}g(x) ≠ 0, deg(a_{i}_{t}g(x)) ≤ m − 1 and (a_{i}_{t}g(x))f(x) = a_{i}_{t}c, hence by induction hypothesis there exist elements 0 ≠ r_{1}, a_{1} ∈ R such that r_{1}f(x) = a_{1}a_{i}_{t}c. Thus r = r_{1} and a = a_{1}a_{i}_{t}are desired elements.
Now we will show that the elements a_{1}, a_{2}, …, a_{n} are all nilpotent, when b_{0} is unit modulo Niℓ(R). Since R is reversible and hence Niℓ(R) is an ideal of R, then R̄ = 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], R̄ = R/Niℓ(R) is (ᾱ, δ̄) compatible, and therefore R is ᾱrigid, by [9, Lemma 2.2]. Now from g(x)f(x) = c, we get ḡ (x) f̄ (x) = c̄ ∈ R̄, and hence by Lemma 2.3 we have b̄_{0}ā_{0} = c̄ and b̄_{j}ā_{i} = 0̄ for each i, j with i + j ≥ 1. Now since b_{0} is a unit element of R, then d_{0}b_{0} = 1 for some d_{0} ∈ R. Multiplying b̄_{0}ā_{i} = 0̄ from left by d̄_{0} we get ā_{i} = 0̄ for each i ≥ 1. This yields that a_{i} ∈ Niℓ(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) = a_{0} + a_{1}x + ···+a_{n}x^{n} be a nonzero element of R[x; α, δ]. If there is a nonzero element g(x) = b_{0} + b_{1}x + ··· + b_{m}x^{m} ∈ R[x; α, δ] such that f(x)g(x) = c is a constant, then there exist nonzero elements r, a ∈ R such that f(x)r = ca. In particular r = b_{p}a for some p, 0 ≤ p ≤ m, and a is either one or a product of at most m coefficients from f(x). Furthermore, if b_{0}is a unit element modulo Niℓ(R), then a_{1}, …, a_{n} 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 zerodivisor, if and only if there exists a nonzero constant r ∈ R 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 αderivation of R. For integers 0 ≤ i ≤ j, let us write ${f}_{i}^{j}$ for the set of all “words” in α and δ in which there are i factors of α and j −i factors of δ. For instance, ${f}_{j}^{j}=\{{\alpha}^{j}\},\hspace{0.17em}{f}_{0}^{j}=\{{\delta}^{j}\}$and ${f}_{j1}^{j}=\{{\alpha}^{j1}\delta ,{\alpha}^{j2}\delta \alpha ,\dots ,\delta {\alpha}^{j1}\}$. Clearly each element of ${f}_{i}^{j}$is an additive map on R. Also for each a ∈ R and each integers 0 ≤ i ≤ j, we write ${f}_{i}^{j}(a)=\{\beta (a)\mid \beta \in {f}_{i}^{j}\}$.
We say also that a set S ⊆ R is locally nilpotent if for any subset {s_{1}, …, s_{n}} ⊆ S, there exists an integer t, such that any product of t elements from {s_{1}, …, s_{n}} 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) = a_{0}+ a_{1}x + ··· + a_{m}x^{m} ∈ Niℓ (R)[x; α, δ]. Assume that M = {a_{0}, a_{1}, …, a_{m}} ⊆ 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 $N=\cup {f}_{i}^{j}({a}_{r})$, where 0 ≤ r ≤ m and 0 ≤ i ≤ j 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 2primal 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 2primal ring if and only if R[x; α, δ] is a weakly 2primal ring. In this case R[x; α, δ] is a weakly semicommutative ring.
Proof
For the forward direction, since R is weakly 2primal, we have L−rad(R) = Niℓ(R). We will show that L − rad(R[x; α, δ]) = Niℓ(R[x; α, δ]). It is enough to show that Niℓ(R[x; α, δ]) ⊆ L − rad(R[x; α, δ]), since the reverse inclusion is obvious. First we show that Niℓ(R[x; α, δ]) = Niℓ(R)[x; α, δ] = L − rad(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 $f(x)={\sum}_{i=0}^{m}{a}_{i}{x}^{i}$ is a nilpotent element with the nilpotency index t. Therefore we get a_{m}α^{m}(a_{m})α^{2}^{m} …α^{(}^{t}^{−1)}^{m}(a_{m}) = 0, since it is the leading coefficient of (f(x))^{t} = 0. Then ${a}_{m}^{t}=0$, by Lemma 2.1, and so a_{m} ∈ Niℓ(R). Since Niℓ(R) is an (α, δ)compatible ideal, ${f}_{i}^{j}({a}_{m})\subseteq Ni\ell (R)$ for each integers 0 ≤ i ≤ j. Thus 0 = (f(x))^{t} = (a_{0} + a_{1}x + ··· + a_{m}_{−1}x^{m}^{−1})^{t} + h(x), where h(x) ∈ Niℓ(R)[x; α, δ]. Then (a_{0} +a_{1}x+···+a_{m}_{−1}x^{m}^{−1})^{t} ∈ Niℓ(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 a_{0}, a_{1}, …, a_{m}_{−1} ∈ Niℓ(R), which implies that f(x) ∈ Niℓ(R)[x; α, δ]. Therefore Niℓ(R[x; α, δ]) = Niℓ(R)[x; α, δ] = L − rad(R)[x; α, δ].
Next we show that L − rad(R)[x; α, δ] is a locally nilpotent ideal of R[x; α, δ]. Assume that M = {f_{1}(x), …, f_{k}(x)} be a subset of L − rad(R)[x; α, δ]. Write f_{i}(x) = a_{i}_{0} + a_{i}_{1}x + ··· + a_{im}x^{m}, where a_{ij} is in L − rad(R) for all i = 1, 2, ···, k and j = 0, 1, …, n. Let N = {a_{i}_{0}, a_{i}_{1}, …, a_{im}  i = 1, 2, …, k}. Then N is a finite subset of L − rad(R) and since L − rad(R) is locally nilpotent, there exists a positive integer t such that any product of t elements from N is zero. Assume that $W=\cup {f}_{i}^{j}({a}_{rs})$, where 1 ≤ r ≤ k, 0 ≤ s ≤ m and 0 ≤ i ≤ j are nonnegative 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 L−rad(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 2primal ring is weakly 2primal.
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 R̄ = R/Niℓ(R). Then R̄ is ᾱrigid, by [9, Lemma 2.2]. For the forward direction, let f(x) = a_{0} + a_{1}x + ··· + a_{n}x^{n} ∈ R[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 f̄ (x) ḡ (x) = 1. Then ā_{0} is a unit element of R̄, by Lemma 2.3, and so a_{1}, …, a_{n} are nilpotent, by Theorem 2.5.
For the reverse implication, let a_{0} is a unit and a_{1}, …, a_{n} are nilpotent. Then a_{1}x+···+a_{n}x^{n} ∈ niℓ(R)[x; α, δ] = L−rad(R[x; α, δ]) ⊆ J(R[x; α, δ]), by Theorem 2.10. Thus f(x) = a_{0} + a_{1}x + ··· + a_{n}x^{n} 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 αrigid 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 R̄ = R/Niℓ(R) is reduced. Since Niℓ(R) is an (α, δ)compatible ideal of R, then the ring R̄ is (ᾱ, δ̄) compatible, by [7, Proposition 2.1]. Hence R̄ is ᾱrigid, by [9, Lemma 2.2]. Now, the result follows from Corollary 2.12.
Corollary 2.14
Let R be a weakly 2primal 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 2primal 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 2primal, 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 2primal ring which is (α, δ)compatible. Then J(R[x; α, δ]) = Niℓ(R)[x; α, δ].
Proof
Following Theorem 2.10, we get
$$Ni\ell (R)[x;\alpha ,\delta ]=Ni\ell (R[x;\alpha ,\delta ]=Lrad(R)[x;\alpha ,\delta ]=Lrad(R[x;\alpha ,\delta ]).$$
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 S_{r}(a) = 1) if aR + bR = R (for any b ∈ R) implies that a + br ∈ U(R) for some r ∈ R. If S_{r}(a) = 1 for all a ∈ R, then a ring R is said to have stable range 1, written S_{r}(R) = 1. It is well known that this property is leftright symmetric.
Proposition 2.16
If Niℓ(R) is an (α, δ)compatible ideal of R, then we have S_{r}(R[x; α, δ]) > 1.
Proof
We will begin by assuming, for a contradiction, that S_{r}(R[x; α, δ]) = 1. Now, since x(−x)+1+x^{2} = 1, then there exists f(x) ∈ R[x; α, δ] such that x+(1+x^{2})f(x) is a unit in R[x; α, δ]. For each a ∈ R we have x^{2}a = δ^{2}(a) + [αδ(a) + δα(a)]x + α^{2}(a)x^{2}. Write f(x) = a_{0} + a_{1}x + ··· + a_{n}x^{n}. Then
$$\begin{array}{ll}x+(1+{x}^{2})f(x)\hspace{0.17em}\hfill & =[{\delta}^{2}({a}_{0})+{a}_{0}]\hfill \\ \hspace{0.17em}\hfill & +[{\delta}^{2}({a}_{1})+\alpha \delta ({a}_{0})+\delta \alpha ({a}_{0})+{a}_{1}+1]x\hfill \\ \hspace{0.17em}\hfill & +[{\delta}^{2}({a}_{2})+\alpha \delta ({a}_{1})+\delta \alpha ({a}_{1})+{\alpha}^{2}({a}_{0})+{a}_{2}]{x}^{2}\hfill \\ \hspace{0.17em}\hfill & +[{\delta}^{2}({a}_{3})+\alpha \delta ({a}_{2})+\delta \alpha ({a}_{2})+{\alpha}^{2}({a}_{1})+{a}_{3}]{x}^{3}\hfill \\ \hspace{0.17em}\hfill & \vdots \hfill \\ \hspace{0.17em}\hfill & +[{\delta}^{2}({a}_{n1})+\alpha \delta ({a}_{n2})+\delta \alpha ({a}_{n2})+{\alpha}^{2}({a}_{n3})+{a}_{n1}]{x}^{n1}\hfill \\ \hspace{0.17em}\hfill & +[{\delta}^{2}({a}_{n})+\alpha \delta ({a}_{n1})+\delta \alpha ({a}_{n1})+{\alpha}^{2}({a}_{n2})+{a}_{n}]{x}^{n}\hfill \\ \hspace{0.17em}\hfill & +[\alpha \delta ({a}_{n})+\delta \alpha ({a}_{n})+{\alpha}^{2}({a}_{n1})]{x}^{n1}\hfill \\ \hspace{0.17em}\hfill & +[{\alpha}^{2}({a}_{n})]{x}^{n+2}.\hfill \end{array}$$
Since Niℓ(R) is an ideal of R, hence R is weakly 2primal and so by Corollary 2.14, the constant term of x+(1+x^{2})f(x) is unit and other coefficients are all nilpotent. Then nilpotency of α^{2}(a_{n}) and Lemma 2.1 imply that a_{n} is nilpotent. Since Niℓ(R) is an αinvariant and a δideal, we have αδ(a_{n})+δα(a_{n}) ∈ Niℓ(R). Now nilpotency of [αδ(a_{n})+δα(a_{n})+α^{2}(a_{n}_{−1})] and Lemma 2.1 implies that a_{n}_{−1} ∈ Niℓ(R). By a similar argument, one can show that a_{0}, a_{1}, …, a_{n} ∈ Niℓ(R). Then [δ^{2}(a_{0}) + a_{0}] ∈ Niℓ(R), which is a contradiction, since it is the constant term of x + (1 + x^{2})f(x). Therefore S_{r}(R[x; α, δ]) > 1, as desired.