(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.

Since R is 2-primal, 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 ᾱ-invariant and a δ̄-ideal of R̄, since R̄ 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 r ∈ R, 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 r ∈ R is unit modulo Niℓ(R) if r + Niℓ(R) is unit in R/Niℓ(R).

Let deg(f(x)) = 0. Then f(x) = a₀. Hence b_ja₀ = 0 for each 1 ≤ j ≤ m 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 b_ja₀ = 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 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 b₀a_i = 0 for each 1 ≤ i ≤ n. 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 = ··· = b_m−1 = 0. Since c = g(x)f(x) = b_mx^mf(x) we have b_ma_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_mf(x) = 0, then b_mx^mf(x) = 0, by Lemma 2.1, and so c = g(x)f(x) = (b₀ + b₁x + ··· + b_m−1x^m−1)f(x). Since g₁(x) = b₀ + b₁x + ··· + b_m−1x^m−1 ≠ 0, hence deg(g₁(x)) ≤ m − 1 and g₁(x)f(x) = c, so the result follows from induction hypothesis. Now assume that b_mf(x) ≠ 0. Then b_mx^mf(x) ≠ 0, by Lemma 2.1. Hence c = g(x)f(x) = g(x)(a_i1x^i₁ + ··· + a_itx^i_t ) such that g(x)a_ik ≠ 0 for each 1 ≤ k ≤ t, where 0 ≤ i₁, i_t ≤ n. Thus b_mα_m(a_it) = 0 and so b_m a_it = 0, by Lemma 2.1. Hence a_itb_m = 0, since R is reversible. Thus a_itc = (a_itg(x))f(x). Since a_itg(x) ≠ 0, deg(a_itg(x)) ≤ m − 1 and (a_itg(x))f(x) = a_itc, hence by induction hypothesis there exist elements 0 ≠ r₁, a₁ ∈ R such that r₁f(x) = a₁a_itc. Thus r = r₁ and a = a₁a_itare desired elements.

Now we will show that the elements a₁, a₂, …, a_n are all nilpotent, when b₀ 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 ᾱ-rigid, 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 b̄₀ā₀ = c̄ and b̄_jā_i = 0̄ for each i, j with i + j ≥ 1. Now since b₀ is a unit element of R, then d₀b₀ = 1 for some d₀ ∈ R. Multiplying b̄₀ā_i = 0̄ from left by d̄₀ 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.

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.

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 + ··· + a_mx^m ∈ Niℓ (R)[x; α, δ]. Assume that M = {a₀, a₁, …, 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=∪fij(ar), 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 2-primal property of Ore extension ring with the coefficent ring R, when R is an (α, δ)-compatible ring.

For the forward direction, since R is weakly 2-primal, 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)=∑i=0maixi is a nilpotent element with the nilpotency index t. Therefore we get a_mα^m(a_m)α^2m …α^(t−1)m(a_m) = 0, since it is the leading coefficient of (f(x))^t = 0. Then amt=0, by Lemma 2.1, and so a_m ∈ Niℓ(R). Since Niℓ(R) is an (α, δ)-compatible ideal, fij(am)⊆Niℓ(R) for each integers 0 ≤ i ≤ j. Thus 0 = (f(x))^t = (a₀ + a₁x + ··· + a_m−1x^m−1)^t + h(x), where h(x) ∈ Niℓ(R)[x; α, δ]. Then (a₀ +a₁x+···+a_m−1x^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 a₀, a₁, …, 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 = {f₁(x), …, f_k(x)} be a subset of L − rad(R)[x; α, δ]. Write f_i(x) = a_i0 + a_i1x + ··· + a_imx^m, where a_ij is in L − rad(R) for all i = 1, 2, ···, k and j = 0, 1, …, n. Let N = {a_i0, a_i1, …, 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=∪fij(ars), where 1 ≤ r ≤ k, 0 ≤ s ≤ m and 0 ≤ i ≤ j 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 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 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.

Let R̄ = R/Niℓ(R). Then R̄ is ᾱ-rigid, by [9, Lemma 2.2]. For the forward direction, let f(x) = a₀ + a₁x + ··· + a_nxⁿ ∈ 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 ā₀ is a unit element of R̄, by Lemma 2.3, and so a₁, …, a_n are nilpotent, by Theorem 2.5.

For the reverse implication, let a₀ is a unit and a₁, …, a_n are nilpotent. Then a₁x+···+a_nxⁿ ∈ niℓ(R)[x; α, δ] = L−rad(R[x; α, δ]) ⊆ J(R[x; α, δ]), by Theorem 2.10. Thus f(x) = a₀ + a₁x + ··· + a_nxⁿ is a unit of R[x; α, δ].

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

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 ᾱ-rigid, by [9, Lemma 2.2]. Now, the result follows from Corollary 2.12.

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].

Following Theorem 2.10, we get

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 left-right symmetric.

We will begin by assuming, for a contradiction, that S_r(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 a ∈ R we have x²a = δ²(a) + [αδ(a) + δα(a)]x + α²(a)x². Write f(x) = a₀ + a₁x + ··· + a_nxⁿ. Then

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(a_n) and Lemma 2.1 imply that a_n is nilpotent. Since Niℓ(R) is an α-invariant and a δ-ideal, we have αδ(a_n)+δα(a_n) ∈ Niℓ(R). Now nilpotency of [αδ(a_n)+δα(a_n)+α²(a_n−1)] and Lemma 2.1 implies that a_n−1 ∈ Niℓ(R). By a similar argument, one can show that a₀, a₁, …, a_n ∈ Niℓ(R). Then [δ²(a₀) + a₀] ∈ Niℓ(R), which is a contradiction, since it is the constant term of x + (1 + x²)f(x). Therefore S_r(R[x; α, δ]) > 1, as desired.

We break the proof in to some parts.

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 a_ixⁱb_jx^j = 0, for each i and j, by (α, δ)-skew Armendariz property of R. Then by Lemma 2.1, we have a_ib_j = 0, for each i and j. Since R is reversible b_ja_i = 0 and hence b_jα^k(a_i) = 0, for each k ≥ 0 and b_jδ^l(a_i) = 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.

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 = (a_ij) and B = (b_ij) 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, j ≤ n−1. Now from AB = 0, we get A′B′ = 0 = A″B″. Thus by induction hypothesis, we have

for all 1 ≤ i, j, k ≤ n−1 and 2 ≤ l ≤ n. Now it is sufficient to show that a_1kb_kn = 0, for each k. From the entry 1n in AB = 0, we have that

Since A,B ∈ A(R, n, σ), we have that a_ij = a_i+1,j+1, for 1≤i≤j≤⌊n2⌋ and b_kn = b_k−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 σ^j−1(b_jn)a₁₁ = σ^j−1(b_jn)a_jj = 0, for all 1 < j ≤ n. Thus multiplying (3.1) by a₁₁ on the right, we obtain a₁₁b_1na₁₁ = 0. Since R is reduced, we get a₁₁b_1n = 0. Therefore

Since R is σ-rigid, σ^j−1(b_jn)a₁₂ = σ^j−1(b_jn)a_j−1,j = 0, for all 2 < j ≤ n if 2≤⌊n2⌋. Thus, multiplying (3.2) by a₁₂ on the right, we obtain a₁₂σ(b_2n)a₁₂ = 0. Since R is reduced, we get a₁₂σ(b_2n) = 0. Therefore

and also, by σ-compatibility of R, we get a₁₂b_2n = 0. Repeating in this way, we get a_1kb_kn = 0, for 1≤k≤⌊n2⌋ and also

Since R is σ-rigid, we get σⁿ⁻¹(b_nn)a_1k = σⁿ⁻¹(b_kk)a_1k = 0, for all 1 ≤ k < n. Thus, multiplying (3.4) by σⁿ⁻¹(b_nn) on the left, we obtain σⁿ⁻¹(b_nn)a_1nσⁿ⁻¹(b_nn) = 0. Using the reduced property of a ring R, we get a_1nσⁿ⁻¹(b_nn) = 0. Therefore

and also using the σ-compatibility of R, we get a_1nb_nn = 0. By continuing in this way, we can show that a_1kb_kn = 0, for all ⌈n2⌉≤k≤n. Thus if n is even, we are done. Also, if n is odd, we obtain that

and by σ-compatibility of R, we get

Therefore a_ikb_kj = 0, for all 1 ≤ i, j, k ≤ n, 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, j ≤ n. 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 (f_ij)(g_ij) = 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 a_mα^m(σ(a_m)) = 0 and consequently a_mσ(α^m(a_m)) = 0, since ασ = σα. Thus a_mα^m(a_m) = 0, since R is σ-rigid and so a_m = 0, since R is α-rigid. Hence f(x) = 0 and hence R[x; α, δ] is σ̄-rigid. Thus f_itg_tj = 0, for each 1 ≤ i, j, t ≤ n and easy calculations show that A_kx^kB_lx^l = 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 : R → R given by α(f(x)) = f(0) and α-derivation δ : R → R 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) = x² − x³ = c ∈ R but b₀a₀ = x₂ ≠ c.

(2) Let S and T be any reduced rings. Suppose R = S ⊕T with the usual addition and multiplication. Then for reversible ring R, let α : R → R be an endomorphism defined by α((a, b)) = (b, a) and δ : R → R be an α-derivation defined by δ((a, b)) = (a − b, 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 : R → R 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 rα(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.