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 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+ ··· + anxn ∈ R[x; α, δ] and c, r ∈ R. Then f(x)r = c if and only if a0r = c and air = 0 for each 1 ≤ i ≤ n.
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) = c ∈ R. 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/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 ᾱ-derivation.
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 2-primal ring and ℘ a minimal prime ideal of R. If R is an (α, δ)-compatible ring, then ℘ is 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 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).
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 + ··· + bmxm ∈ R[x; α, δ] such that g(x)f(x) = c is a constant, then there exist non-zero elements r, a ∈ R such that rf(x) = ac. In particular r = abp for some p, 0 ≤ p ≤ m, 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 ≤ 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 bja0 = 0 for each 0 ≤ j ≤ m. Since g(x) ≠ 0, there exists 0 ≤ s ≤ m 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 ≤ 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:
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Case 1: Let b0 = b1 = ··· = bm−1 = 0. Since c = g(x)f(x) = bmxmf(x) we have bmai = 0 for each 0 ≤ i ≤ n, by Lemma 2.1, and so c = 0. Thus r = bm and a = 1 are desired elements.
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Case 2: Assume that there exits 0 ≤ j ≤ m − 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 ≤ k ≤ t, where 0 ≤ i1, it ≤ n. 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, a1 ∈ R 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 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 b̄jāi = 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 b̄0āi = 0̄ from left by d̄0 we get āi = 0̄ for each i ≥ 1. This yields that ai ∈ 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) = a0 + a1x + ···+anxn be a non-zero element of R[x; α, δ]. If there is a non-zero element g(x) = b0 + b1x + ··· + bmxm ∈ R[x; α, δ] such that f(x)g(x) = c is a constant, then there exist non-zero elements r, a ∈ R such that f(x)r = ca. In particular r = bpa for some p, 0 ≤ p ≤ m, 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 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 α-derivation of R. For integers 0 ≤ i ≤ j, let us write fij 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, fjj={αj}, f0j={δj}and fj-1j={αj-1δ,αj-2δα,…,δαj-1}. Clearly each element of fijis an additive map on R. Also for each a ∈ R and each integers 0 ≤ i ≤ j, we write fij(a)={β(a)∣β∈fij}.
We say also that a set S ⊆ R 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 + ··· + amxm ∈ Niℓ (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 ≤ 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.
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 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 amαm(am)α2m …α(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 am ∈ 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 + ··· + am−1xm−1)t + h(x), where h(x) ∈ Niℓ(R)[x; α, δ]. Then (a0 +a1x+···+am−1xm−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, …, am−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), …, fk(x)} be a subset of L − rad(R)[x; α, δ]. Write fi(x) = ai0 + ai1x + ··· + aimxm, where aij is in L − rad(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 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.
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 ᾱ-rigid, by [9, Lemma 2.2]. For the forward direction, let f(x) = a0 + a1x + ··· + anxn ∈ 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, …, an are nilpotent, by Theorem 2.5.
For the reverse implication, let a0 is a unit and a1, …, an are nilpotent. Then a1x+···+anxn ∈ 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 + ··· + 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 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.
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 b ∈ R) implies that a + br ∈ U(R) for some r ∈ R. If Sr(a) = 1 for all a ∈ R, 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 a ∈ R 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−1 ∈ Niℓ(R). By a similar argument, one can show that a0, a1, …, an ∈ 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 Sr(R[x; α, δ]) > 1, as desired.