Department of Mathematics, Northwest Normal University, Lanzhou, 730070, China. Department of Mathematics, Faculty of Education, University of Khartoum, Omdurman, Sudan, e-mail : ayou1975@yahoo.com and eltiyeb76@gmail.com, Department of Mathematics, Northwest Normal University, Lanzhou, 730070, China, e-mail : liuzk@nwnu.edu.cn
^{*}Corresponding Author.
Received: August 30, 2017; Accepted: March 9, 2018; Published online: June 23, 2018.
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Abstract
This paper is devoted to the study of strongly α-semicommutative rings, a generalization of strongly semicommutative and α-rigid rings. Although the n-by-n upper triangular matrix ring over any ring with identity is not strongly ᾱ-semicommutative for n ≥ 2, we show that a special subring of the upper triangular matrix ring over a reduced ring is strongly ᾱ-semicommutative under some additional conditions. Moreover, it is shown that if R is strongly α-semicommutative with α(1) = 1 and S is a domain, then the Dorroh extension D of R by S is strongly ᾱ-semicommutative.
Throughout this paper, R denotes an associative ring with identity and α denotes a nonzero and non-identity endomorphism, unless specified otherwise. A ring R is called semicommutative, if for all a, b ∈ R, ab = 0 implies aRb = 0. This is equivalent to the usual definition by [18, Lemma 1.2] or [8, Lemma 1]. Properties, examples and counterexamples of semicommutative rings were given in Huh, Lee and Smoktunowicz [8], Kim and Lee [10], Liu [13] and Yang [19]. One of generalizations of semicommutative rings was investigated by Liu and Zhao in [14].
Recall that an endomorphism α of a ring R is called rigid [11] if for a ∈ R, aα(a) = 0 implies a = 0, and R is called an α-rigid ring [6] 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 rings by [6, Proposition 5]. Due to [1], an endomorphism α of a ring R is called semicommutative if whenever ab = 0 for a, b ∈ R, aRα(b) = 0. A ring R is called α-semicommutative if there exists a semicommutative endomorphism α of R. Gang and Ruijuan [5] called a ring R strongly semicommutative, if whenever polynomials f(x), g(x) in R[x] satisfy f(x)g(x) = 0, then f(x)R[x]g(x) = 0. In general the polynomial rings over α-semicommutative rings need not be α-semicommutative. In this paper, we consider the α-semicommutative rings over which polynomial rings are also α-semicommutative and we call them strongly α-semicommutative rings, i.e., if α is an endomorphism of R, then α is called strongly semicommutative if whenever polynomials f(x), g(x) ∈ R[x] satisfy f(x)g(x) = 0, then f(x)R[x]α(g(x)) = 0. A ring R is called strongly α-semicommutative if there exists a strongly semicommutative endomorphism α of R. Clearly strongly α-semicommutative rings are α-semicommutative but not conversely. If R is Armendariz, then these two concepts coincide (see, Proposition 2.11). We characterize α-rigid rings by showing that a ring R is α-rigid if and only if R is a reduced strongly α-semicommutative ring and α is a monomorphism. It is also shown that a ring R is strongly α-semicommutative if and only if the polynomial ring R[x] over R is strongly α-semicommutative. Some extensions of α-semicommutative rings are considered.
2. Strongly α-semicommutative Rings
In this section we introduce the concept of a strongly α-semicommutative ring and study its properties. Observe that the notion of strongly α-semicommutative rings not only generalizes that of α-rigid rings, but also extends that of strongly semicommutative rings. We start by the following definition.
Definition 2.1
An endomorphism α of a ring R is called strongly semicommutative if whenever polynomials f(x), g(x) ∈ R[x] satisfy f(x)g(x) = 0, then f(x)R[x]α(g(x)) = 0. A ring R is called strongly α-semicommutative if there exists a strongly semicommutative endomorphism α of R.
It is clear that a ring R is strongly semicommutative, if R is strongly I_{R}-semicommutative, where I_{R} is the identity endomorphism of R. It is easy to see that every subring S with α(S) ⊆ S of a strongly α-semicommutative ring is also strongly α-semicommutative. For any i ∈ I, let R_{i} be strongly α_{i}-semicommutative where α_{i} is an endomorphism of R_{i}. Set W = Π_{i}_{∈}_{I}R_{i}. Define an endomorphism α of W as following:
Then it is easy to see that W is strongly α-semicommutative.
Remark 2.2
Let R be a strongly α-semicommutative ring with f(x)g(x) = 0 for f(x), g(x) ∈ R[x]. Then f(x)R[x]α(g(x)) = 0 and, in particular, f(x)α(g(x)) = 0. Since R is strongly α-semicommutative, we get f(x)R[x]α^{2}(g(x)) = 0. So, by induction hypothesis, we obtain f(x)R[x]α^{k}(g(x)) = 0 and f(x)α^{k}(g(x)) = 0, for any positive integer k.
The following example shows that there exists an endomorphism α of strongly semicommutative ring R such that R is not strongly α-semicommutative.
Example 2.3
Let ℤ_{2} be the ring of integers modulo 2 and consider the ring R = ℤ_{2}⊕ℤ_{2}, with the usual addition and multiplication. Then R is strongly semicommutative, since R is a commutative reduced ring. Now, let α : R → R be defined by α((a, b)) = (b, a). Then α is an automorphism of R. For f(x) = (1, 0)+(1, 0)x and g(x) = (0, 1)+(0, 1)x, it is clear that f(x)g(x) = 0. But (0, 0) ≠ ((1, 0) + (1, 0)x)(1, 1)x((1, 0) + (1, 0)x) ∈ f(x)R[x]α(g(x)). Thus R is not strongly α-semicommutative.
Lemma 2.4
R is a reduced ring if and only if so is R[x].
Lemma 2.5
A ring R is α-rigid if and only if R[x] is α-rigid.
Theorem 2.6
A ring R is α-rigid if and only if R is a reduced strongly α-semicommutative ring and α is a monomorphism.
Proof
(⇒) Let R be an α-rigid ring. Then R is reduced and α is a monomorphism by [6, p.218]. Assume that f(x)g(x) = 0, for f(x), g(x) ∈ R[x]. Let h(x) be an arbitrary polynomial of R[x]. Then g(x)f(x) = 0 since R[x] is reduced by Lemma 2.4. Thus f(x)h(x)α(g(x))α(f(x)h(x)α(g(x))) = f(x)h(x)α(g(x)f(x))α(h(x))α^{2}(g(x)) = 0. Since R is α-rigid, f(x)h(x)α(g(x)) = 0 by Lemma 2.5 so f(x)R[x]α(g(x)) = 0. Thus R is strongly α-semicommutative.
(⇐) Assume that f(x)α(f(x)) = 0 for f(x) ∈ R[x]. Since R is reduced and strongly α-semicommutative, α(f(x))f(x) = 0 and so α(f(x))R[x]α(f(x)) = 0. Hence α((f(x))^{2}) = 0 and so f(x) = 0, since α is a monomorphism and R is reduced. Therefore R is α-rigid.
The following examples show that the condition “R is reduced ring” and “α is a monomorphism” in Theorem 2.6 cannot be dropped respectively.
Example 2.7
Let ℤ be the ring of integers. Consider $R=\left\{\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill a\hfill \end{array}\right)\mid a,b\in \mathbb{Z}\right\}$. Let α : R → R be an endomorphism defined by $\alpha \left(\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill a\hfill \end{array}\right)\right)=\left(\begin{array}{cc}\hfill a\hfill & \hfill -b\hfill \\ \hfill 0\hfill & \hfill a\hfill \end{array}\right)$. Note that α is an automorphism. By [1, Example 2.5(1)] R is not reduced and hence R is not α-rigid. Thus R[x] is not α-rigid by Lemma 2.5.
Let f(x)g(x) = 0 for $f(x)=\left(\begin{array}{cc}{f}_{0}(x)& {f}_{1}(x)\\ 0& {f}_{0}(x)\end{array}\right),g(x)=\left(\begin{array}{cc}{g}_{0}(x)& {g}_{1}(x)\\ 0& {g}_{0}(x)\end{array}\right)\in R[x]$. Then f_{0}(x)g_{0}(x) = 0 and f_{0}(x)g_{1}(x) + f_{1}(x)g_{0}(x) = 0. For $h(x)=\left(\begin{array}{cc}{h}_{0}(x)& {h}_{1}(x)\\ 0& {h}_{0}(x)\end{array}\right)\in R[x]$, we have $\begin{array}{l}\left(\begin{array}{cc}{f}_{0}(x)& {f}_{1}(x)\\ 0& {f}_{0}(x)\end{array}\right)\mathrm{\hspace{0.17em}\u200a\u200a}\left(\begin{array}{cc}{h}_{0}(x)& {h}_{1}(x)\\ 0& {h}_{0}(x)\end{array}\right)\alpha \left(\left(\begin{array}{cc}{g}_{0}(x)& {g}_{1}(x)\\ 0& {g}_{0}(x)\end{array}\right)\right)\\ =\left(\begin{array}{cc}{f}_{0}(x){h}_{0}(x){g}_{0}(x)& -{f}_{0}(x){h}_{0}(x){g}_{1}(x)+{f}_{0}(x){h}_{1}(x){g}_{0}(x)+{f}_{1}(x){h}_{0}(x){g}_{0}(x)\\ 0& {f}_{0}(x){h}_{0}(x){g}_{0}(x)\end{array}\right)\end{array}$. Since f_{0}(x)g_{0}(x) = 0, f_{0}(x) = 0 or g_{0}(x) = 0. If f_{0}(x) = 0 then f_{1}(x)g_{0}(x) = 0. So f(x)R[x]α(g(x)) = 0. If g_{0}(x) = 0 then f_{0}(x)g_{1}(x) = 0. Again f(x)R[x]α(g(x)) = 0. Thus R is strongly α-semicommutative.
Example 2.8
Let F be a field and R = F[x] the polynomial ring over F. Define α : R[x] → R[x] by α(f(x)) = f(0) where f(x) ∈ R[x]. Then R[x] is a commutative domain (and so reduced) and α is not a monomorphism. If f(x)g(x) = 0 for f(x), g(x) ∈ R[x] then f(x) = 0 or g(x) = 0, and so f(x) = 0 or α(g(x)) = 0. Hence f(x)R[x]α(g(x)) = 0, and thus R is strongly α-semicommutative. Note that R is not α-rigid, since xα(x) = 0 for 0 ≠ x ∈ R.
Observe that if R is a domain then R is both strongly semicommutative and strongly α-semicommutative for any endomorphism α of R. Example 2.7 also shows that there exists a strongly α-semicommutative ring R which is not a domain. According to Cohn [4], a ring R is called reversible if ab = 0 implies ba = 0 for a, b ∈ R. Baser and et al. [2] called a ring R right (respectively, left) α-reversible if there exists a right (respectively, left) reversible endomorphism α of R. A ring is α-reversible if it is both left and right α-reversible.
Lemma 2.9
([16, Proposition 3]) A reduced α-reversible ring is α-semicommutative.
Proposition 2.10
Let R be a reduced and α-reversible ring. Then R is strongly α-semicommutative.
Proof
Let $f(x)={\sum}_{i=0}^{n}{a}_{i}{x}^{i},g(x)={\sum}_{j=0}^{m}{b}_{j}{x}^{j}\in R[x]$ be such that $f(x)g(x)=0={\sum}_{s=0}^{n+m}{\sum}_{i+j=s}{a}_{i}{b}_{j}{x}^{s}$. Since every reduced ring is an Armendariz ring, we obtain a_{i}b_{j} = 0. Then α(b_{j})a_{i} = 0 (by α-reversibility). Now for arbitrary element $h(x)={\sum}_{k=0}^{r}{c}_{k}{x}^{k}\in R[x]$, we have α(b_{j})a_{i}c_{k} = 0 for each i, j, k, so a_{i}c_{k}α(b_{j}) = 0 (by reducibility). Hence, f(x)h(x)α(g(x)) = 0. Therefore R is strongly α-semicommutative.
Rege and Chhawchharia [17] called a ring R an Armendariz ring if whenever polynomials f(x) = a_{0}+a_{1}x+⋯+a_{m}x^{m}, g(x) = b_{0}+b_{1}x+⋯+b_{n}x^{n} ∈ R[x] satisfy f(x)g(x) = 0, then a_{i}b_{j} = 0 for each i and j. Hong et al. [7] called a ring R α-Armendariz if whenever f(x) = a_{0}+a_{1}x+⋯+a_{m}x^{m}, g(x) = b_{0}+b_{1}x+⋯+b_{n}x^{n} ∈ R[x; α] satisfy f(x)g(x) = 0, then a_{i}b_{j} = 0 for each i and j.
Proposition 2.11
Let R be an Armendariz ring. If R is α-semicommutative, then R is strongly α-semicommutative.
Proof
Suppose that $f(x)={\sum}_{i=0}^{n}{a}_{i}{x}^{i},g(x)={\sum}_{j=0}^{m}{b}_{j}{x}^{j}\in R[x]$ satisfy f(x)g(x) = 0. Then, since R is Armendariz, each a_{i}b_{j} is zero, additionally R is α-semicommutative, therefore a_{i}c_{k}α(b_{j}) = 0 for any element c_{k} in R for all i, j, k. Now it is easy to check that f(x)h(x)α(g(x)) = 0 for any $h(x)={\sum}_{k=0}^{r}{c}_{k}{x}^{k}\in R[x]$.
Lemma 2.12
([10, Proposition 3.1(2)]) If R is a reversible α-Armendariz ring, then R is α-semicommutative.
Liu and Yang [20] called a ring R strongly reversible, if whenever polynomials f(x), g(x) ∈ R[x] satisfy f(x)g(x) = 0, then g(x)f(x) = 0.
Proposition 2.13
If R is a strongly reversible α-Armendariz ring, then R is strongly α-semicommutative.
Proof
Let f(x)g(x) = 0, for f(x), g(x) ∈ R[x]. Then g(x)f(x) = 0 since R is strongly reversible. By [7, Proposition 1.3(1)], we obtain α(g(x))f(x) = 0, and so α(g(x))f(x)h(x) = 0 for all h(x) ∈ R[x]. Hence, f(x)h(x)α(g(x)) = 0 for all h(x) ∈ R[x] since R is strongly reversible and f(x)R[x]α(g(x)) = 0. Therefore, R is strongly α-semicommutative.
Recall that an element u of a ring R is right regular if ur = 0 implies r = 0 for r ∈ R. Similarly, left regular elements can be defined. An element is regular if it is both left and right regular (and hence not a zero divisor).
Proposition 2.14
Let Δ be a multiplicatively closed subset of a ring R consisting of central regular elements. Then R is strongly α-semicommutative if and only if so is Δ^{−1}R.
Proof
It is enough to show that the necessity. Suppose that R is strongly α-semicommutative. Let F(x)G(x) = 0, for F(x) = u^{−1}f(x) and G(x) = v^{−1}g(x) ∈ (Δ^{−1}R)[x] where u, v are regular and f(x), g(x) ∈ R[x]. Since Δ is contained in the center of R we have 0 = F(x)G(x) = u^{−1}f(x)v^{−1}g(x) = (u^{−1}v^{−1})f(x)g(x) = (uv)^{−1}f(x)g(x) and so f(x)g(x) = 0. Since R is strongly α-semicommutative, f(x)R[x]α(g(x)) = 0 and f(x)(s^{−1}R)[x]α(g(x)) = 0 for any regular element s. This implies F(x)(Δ^{−1}R)[x]α(G(x)) = 0. Therefore Δ^{−1}R is strongly α-semicommutative.
The ring of Laurent polynomials in x with coefficients in a ring R, denoted by R[x; x^{−1}], consists of all formal sums ${\sum}_{i=k}^{n}{m}_{i}{x}^{i}$ with obvious addition and multiplication, where m_{i} ∈ R and k, n are (possibly negative) integers.
Corollary 2.15
Let R be a ring with α(1) = 1. Then R[x] is strongly α-semicommutative if and only if R[x; x^{−1}] is strongly α-semicommutative.
Corollary 2.16
Let R be an Armendariz ring. Then the following are equivalent:
(1) R is α-semicommutative.
(2) R is strongly α-semicommutative.
(3) R[x; x^{−1}] is strongly α-semicommutative.
Proposition 2.17
Let R be a ring, e a central idempotent of R, with α(e) = e. Then the following statements are equivalent:
(1) R is strongly α-semicommutative rings.
(2) eR and (1 − e)R are strongly α-semicommutative rings.
Proof
(1)⇔(2) This is straightforward since subrings and finite direct products of strongly α-semicommutative rings are strongly α-semicommutative.
We denote by M_{n}(R) and T_{n}(R) the n×n matrix ring and n×n upper triangular matrix ring over R, respectively.
Given a ring R and a bimodule _{R}M_{R}, the trivial extension of R by M is the ring T(R,M) = R⊕M with the usual addition and the following multiplication (r_{1},m_{1})(r_{2},m_{2}) = (r_{1}r_{2}, r_{1}m_{2}+m_{1}r_{2}). This is isomorphic to the ring of all matrices $\left(\begin{array}{cc}r& m\\ 0& r\end{array}\right)$, where r ∈ R,m ∈ M and the usual matrix operations are used.
For an endomorphism α of a ring R and the trivial extension T(R,R) of R, ᾱ : T(R,R) → T(R,R) defined by $\overline{\alpha}\left(\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)\right)=\left(\begin{array}{cc}\alpha (a)& \alpha (b)\\ 0& \alpha (a)\end{array}\right)$ is an endomorphism of T(R,R). Since T(R, 0) is isomorphic to R, we can identify the restriction of ᾱ by T(R, 0) to α. Notice that the trivial extension of a α-semicommutative ring is not ᾱ-semicommutative by [1, Example 2.9]. Now, we may ask whether the trivial extension T(R,R) is strongly ᾱ-semicommutative if R is strongly α-semicommutative. But the following example erases the possibility.
Example 2.18
Consider the strongly α-semicommutative ring $R=\left\{\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)\mid a,b\in \mathbb{Z}\right\}$ with an endomorphism α defined by $\alpha \left(\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)\right)=\left(\begin{array}{cc}a& -b\\ 0& a\end{array}\right)$ in Example 2.7. For
It was shown in [1, Proposition 2.10], that if R is a reduced α-semicommutative ring, then T(R,R) is an ᾱ-semicommutative. Here we have the following results.
Proposition 2.19
Let R be a reduced ring. If R is α-semicommutative, then T(R,R) is strongly ᾱ-semicommutative
Proof
Let f(x) = (f_{0}(x), f_{1}(x)), g(x) = (g_{0}(x), g_{1}(x)) ∈ T(R,R)[x] with f(x)g(x) = 0. We shall prove f(x)T(R,R)[x]α(g(x)) = 0. Now we have
Since R is reduced, R[x] is reduced. Therefore, (2.1) implies g_{0}(x)f_{0}(x) = 0. Multiplying (2.2) on the left side by g_{0}(x) we get f_{1}(x)g_{0}(x) = 0, and so f_{0}(x)g_{1}(x) = 0. Let $f(x)={\sum}_{i=0}^{n}({a}_{i},{b}_{i}){x}^{i},g(x)={\sum}_{j=0}^{m}({a}_{j}^{\prime},{b}_{j}^{\prime}){x}^{j}$, where ${f}_{0}(x)={\sum}_{i=0}^{n}{a}_{i}{x}^{i},{f}_{1}(x)={\sum}_{i=0}^{n}{b}_{i}{x}^{i},{g}_{0}(x)={\sum}_{j=0}^{m}{a}_{j}^{\prime}{x}^{j}$ and ${g}_{1}(x)={\sum}_{j=0}^{m}{b}_{j}^{\prime}{x}^{j}$. Since every reduced ring is an Armendariz ring, we obtain that ${a}_{i}{a}_{j}^{\prime}=0,{a}_{i}{b}_{j}^{\prime}=0,{b}_{i}{a}_{j}^{\prime}=0$ for all i, j by the preceding results. With these facts and the fact that R is α-semicommutative, we have ${a}_{i}{c}_{k}\alpha ({a}_{j}^{\prime})=0,{a}_{i}{c}_{k}\alpha ({b}_{j}^{\prime})=0,{a}_{i}{d}_{k}\alpha ({b}_{j}^{\prime})=0,{b}_{i}{c}_{k}\alpha ({a}_{j}^{\prime})=0$, for any elements c_{k}, d_{k}. Thus, f(x)h(x)α(g(x)) = 0, for any arbitrary $h(x)={\sum}_{k=0}^{r}({c}_{k},{d}_{k}){x}^{k}\in R[x]$. This implies that T(R,R) is strongly ᾱ-semicommutative.
The trivial extension T(R,R) of a ring R is extended to
and an endomorphism α of a ring R is also extended to the endomorphism ᾱ of S_{3}(R) defined by ᾱ((a_{ij})) = (ᾱ(a_{ij})). There exists a reduced ring R such that S_{3}(R) is not strongly ᾱ-semicommutative by the following example.
Example 2.20
We consider the commutative reduced ring R = ℤ_{2}⊕ℤ_{2}, and the automorphism α of R defined by α((a, b)) = (b, a), in Example 2.3. Then S_{3}(R) is not strongly ᾱ-semicommutative. For$A=\left(\begin{array}{ccc}(1,0)& (0,0)& (0,0)\\ (0,0)& (1,0)& (0,0)\\ (0,0)& (0,0)& (1,0)\end{array}\right),B=\left(\begin{array}{ccc}(0,1)& (0,0)& (0,0)\\ (0,0)& (0,1)& (0,0)\\ (0,0)& (0,0)& (0,1)\end{array}\right)\in {S}_{3}(R)$
, then AB = 0, but AAᾱ(B) = A ≠ 0. Thus AS_{3}(R) ᾱ(B) ≠ 0, and therefore S_{3}(R) is not strongly ᾱ-semicommutative.
However, we obtain that S_{3}(R) is strongly ᾱ-semicommutative for a reduced α-semicommutative ring R by the similar method to the proof of Proposition 2.19 as follows:
Proposition 2.21
Let R be a reduced ring. If R is α-semicommutative, then
, respectively. So every polynomial in S_{3}[x] can be expressed in the form of (f_{0}, f_{1}, f_{2}, f_{3}) for some f_{i}’s in R[x]. Let f(x) = (f_{0}(x), f_{1}(x), f_{2}(x), f_{3}(x)), g(x) = (g_{0}(x), g_{1}(x), g_{2}(x), g_{3}(x)) ∈ S_{3}[x] with f(x)g(x) = 0. Then f(x)g(x) = (f_{0}(x)g_{0}(x), f_{0}(x)g_{1}(x)+f_{1}(x)g_{0}(x), f_{0}(x)g_{2}(x)+f_{1}(x)g_{3}(x)+f_{2}(x)g_{0}(x), f_{0}(x)g_{3}(x)+f_{3}(x)g_{0}(x)), we shall prove f(x)S_{3}(R)[x]α(g(x)) = 0. So we have the following system of equations:
Use the fact that R[x] is reduced. From Eq. (2.3), we get g_{0}(x)f_{0}(x) = 0. If we multiply Eq. (2.4), on the right side by g_{0}(x), then $0=({f}_{0}(x){g}_{1}(x)+{f}_{1}(x){g}_{0}(x)){g}_{0}(x)={f}_{1}(x){g}_{0}^{2}(x)$, and so f_{1}(x)g_{0}(x) = 0 and f_{0}(x)g_{1}(x) = 0. Similarly, from Eq. (2.6), we have f_{3}(x)g_{0}(x) = 0, and f_{0}(x)g_{3}(x) = 0. Also, in Eq. (2.5), $0=({f}_{0}(x){g}_{2}(x)+{f}_{1}(x){g}_{3}(x)+{f}_{2}(x){g}_{0}(x)){g}_{0}(x)={f}_{2}(x){g}_{0}^{2}(x)$ implies f_{2}(x)g_{0}(x) = 0 and
$${f}_{0}(x){g}_{2}(x)+{f}_{1}(x){g}_{3}(x)=0.$$
Multiplying (2.7) on left side by f_{0}(x) gives $0={f}_{0}(x)({f}_{0}(x){g}_{2}(x)+{f}_{1}(x){g}_{3}(x))={f}_{0}^{2}(x){g}_{2}(x)$, and so f_{0}(x)g_{2}(x) = 0 hence f_{1}(x)g_{3}(x) = 0. Let
where $\begin{array}{l}{f}_{0}(x)={\sum}_{i=0}^{n}{a}_{i}{x}^{i},{f}_{1}(x)={\sum}_{i=0}^{n}{b}_{i}{x}^{i},{f}_{2}(x)={\sum}_{i=0}^{n}{c}_{i}{x}^{i},{f}_{3}(x)={\sum}_{i=0}^{n}{d}_{i}{x}^{i},{g}_{0}(x)\\ ={\sum}_{j=0}^{m}{a}_{j}^{\prime}{x}^{j},{g}_{1}(x)={\sum}_{j=0}^{m}{b}_{j}^{\prime}{x}^{j},{g}_{2}(x)={\sum}_{j=0}^{m}{c}_{j}^{\prime}{x}^{j},{g}_{3}(x)={\sum}_{j=0}^{m}{d}_{j}^{\prime}{x}^{j}\end{array}$. Since every reduced ring is an Armendariz ring, we obtain that ${a}_{i}{a}_{j}^{\prime}=0,{a}_{i}{b}_{j}^{\prime}=0,{b}_{i}{a}_{j}^{\prime}=0,{a}_{i}{c}_{j}^{\prime}=0,{b}_{i}{d}_{j}^{\prime}=0,{c}_{i}{a}_{j}^{\prime}=0,{a}_{i}{d}_{j}^{\prime}=0,{d}_{i}{a}_{j}^{\prime}=0$, for all i, j by the preceding results. With these facts and the fact that R is α-semicommutative ring, we have $\begin{array}{l}{a}_{i}{a}_{k}^{\u2033}\alpha ({a}_{j}^{\prime})=0,{a}_{i}{a}_{k}^{\u2033}\alpha ({b}_{j}^{\prime})=0,{b}_{i}{a}_{k}^{\u2033}\alpha ({a}_{j}^{\prime})=0,{b}_{i}{a}_{k}^{\u2033}\alpha ({d}_{j}^{\prime})=0,{a}_{i}{a}_{k}^{\u2033}\alpha ({c}_{j}^{\prime})=\\ 0,{a}_{i}{b}_{k}^{\u2033}\alpha ({d}_{j}^{\prime})=0,{b}_{i}{a}_{k}^{\u2033}\alpha ({d}_{j}^{\prime})=0,{a}_{i}{c}_{k}^{\u2033}\alpha ({a}_{j}^{\prime})=0,{b}_{i}{d}_{k}^{\u2033}\alpha ({a}_{j}^{\prime})=0,{c}_{i}{a}_{k}^{\u2033}\alpha ({a}_{j}^{\prime})=\\ 0,{a}_{i}{a}_{k}^{\u2033}\alpha ({d}_{j}^{\prime})=0,{a}_{i}{d}_{k}^{\u2033}\alpha ({a}_{j}^{\prime})=0,{d}_{i}{a}_{k}^{\u2033}\alpha ({a}_{j}^{\prime})=0\end{array}$. Consequently, we get the equation:
For an α-rigid ring R and n ≥ 2, by Proposition 2.21, we may suspect that S_{n}(R) may be strongly ᾱ-semicommutative ring for n ≥ 4. But the possibility is eliminated by the next example.
Note that if R is an α-rigid ring, then α(e) = e, for e^{2} = e ∈ R by [6, Proposition 5]. In particular α(1) = 1. For $A=\left(\begin{array}{cccc}0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right),B=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right)\in {S}_{4}(R)$, we obtain AB = 0. But we have $0\ne \left(\begin{array}{cccc}0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)=AC\overline{\alpha}(B)\in {S}_{4}(R)$, for $C=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\in {S}_{4}(R)$. Thus AC ᾱ(B) ≠ 0 and so S_{4}(R) is not strongly ᾱ-semicommutative. Similarly, it can be proved that S_{n}(R) is not strongly ᾱ-semicommutative for n ≥ 5.
Note that if a = c, then the matrix S is called an upper triangular Toeplitz matrix over R, see [15].
We proved in Proposition 2.21 and Example 2.22 that when R is a reduced ring and R is an α-semicommutative ring, then S_{3}(R) is strongly ᾱ-semicommutative, but S_{n}(R) is not strongly ᾱ-semicommutative for n ≥ 4. In the next theorem we will show that a special subring V_{n}(R) of T_{n}(R) for any positive integer n ≥ 2 is strongly ᾱ-semicommutative, where R is a reduced and α-semicommutativethe ring.
Theorem 2.23
Let R be a reduced ring. If R is α-semicommutative, then V_{n}(R) is strongly ᾱ-semicommutative.
are in V_{n}(R). So every polynomial in V_{n}(R)[x] can be expressed in the form of (f_{1}, f_{2}, ⋯, f_{n}_{−2}, f_{1}_{,n}_{−1}, f_{1}_{n}, f_{2}_{n}) for some f_{i’}s in R[x]. Let f(x) = (f_{0}(x), f_{1}(x), ⋯, f_{2}_{n}(x)), g(x) = (g_{0}(x), g_{1}(x), ⋯, g_{2}_{n}(x)) ∈ V_{n}(R)[x] with f(x)g(x) = 0. We shall prove f(x)V_{n}(R)[x]α(g(x)) = 0. Now we have the following system of equations:
Use the fact that R[x] is reduced. From Eq. (2.8), we get g_{1}(x)f_{1}(x) = 0. If we multiply Eq. (2.9) on the right side by f_{1}(x), then f_{1}(x)g_{2}(x)f_{1}(x) + f_{2}(x)g_{1}(x)f_{1}(x) = 0. Thus f_{1}(x)g_{2}(x)f_{1}(x) = 0 and hence f_{1}(x)g_{2}(x) = 0. From Eq. (2.9) it follows that f_{2}(x)g_{1}(x) = 0. Continuing in this manner, we can show that f_{i}(x)g_{j}(x) = 0 when i + j = 2, …, n − 1. Hence g_{j}(x)f_{i}(x) = 0. Multiplying Eq. (2.10) on the right side by f_{1}(x), we obtain 0 = f_{1}(x)g_{1}_{,n}_{−1}(x)f_{1}(x) + f_{2}(x)g_{n}_{−2}(x)f_{1}(x) + ⋯ + f_{n}_{−2}(x)g_{2}(x)f_{1}(x) + f_{1}_{,n}_{−1}(x)g_{1}(x)f_{1}(x) = f_{1}(x)g_{1}_{,n}_{−1}(x)f_{1}(x). Thus f_{1}(x)g_{1}_{,n}_{−1}(x) = 0. Hence
Thus f_{2}(x)g_{n}_{−2}(x) = 0. Continuing in this manner, we can show that f_{i}(x)g_{j}(x) = 0 when i + j = n and f_{1}(x)g_{1}_{,n}_{−1}(x) = 0, f_{1}_{,n}_{−1}(x)g_{1}(x) = 0. Similarly, from Eq. (2.12), it follows that f_{1}(x)g_{2}_{n}(x) = 0 and f_{2}_{n}(x)g_{1}(x) = 0. Now multiplying Eq. (2.11) on the right side by f_{1}(x), we have
If we multiply Eq. (2.14) on the right side by f_{2}(x), then 0 = f_{2}(x)g_{2}_{n}(x)f_{2}(x) + f_{3}(x)g_{n}_{−2}(x)f_{2}(x)+⋯+f_{1}_{,n}_{−1}(x)g_{2}(x)f_{2}(x)+f_{1}_{n}(x)g_{1}(x)f_{2}(x) = f_{2}(x)g_{2}_{n}(x)f_{2}(x). Thus f_{2}(x)g_{2}_{n}(x) = 0. Continuing in this manner, we can show that f_{i}(x)g_{j}(x) = 0 when i + j = n + 1, f_{1}_{,n}_{−1}(x)g_{2}(x) = 0 and f_{1}_{n}(x)g_{1}(x) = 0. Let
where $\begin{array}{l}{f}_{1}(x)={\sum}_{i=0}^{n}{a}_{1}^{i}{x}^{i},{f}_{2}(x)={\sum}_{i=0}^{n}{a}_{2}^{i}{x}^{i},\cdots ,{f}_{n-2}(x)={\sum}_{i=0}^{n}{a}_{n-2}^{i}{x}^{i},\\ {f}_{1,n-1}(x)={\sum}_{i=0}^{n}{a}_{1,n-1}^{i}{x}^{i},{f}_{1n}(x)={\sum}_{i=0}^{n}{a}_{1n}^{i}{x}^{i},{f}_{2n}(x)={\sum}_{i=0}^{n}{a}_{2n}^{i}{x}^{i},{g}_{1}(x)=\\ {\sum}_{j=0}^{m}{b}_{1}^{j}{x}^{j},{g}_{2}(x)={\sum}_{j=0}^{m}{b}_{2}^{j}{x}^{j},\cdots ,{g}_{n-2}(x)={\sum}_{j=0}^{m}{b}_{n-2}^{j}{x}^{j},{g}_{1,n-1}(x)={\sum}_{j=0}^{m}{b}_{1,n-1}^{j}{x}^{j},\\ {g}_{1n}(x)={\sum}_{j=0}^{m}{b}_{1n}^{j}{x}^{j},{g}_{2n}(x)={\sum}_{j=0}^{m}{b}_{2n}^{j}{x}^{j}\end{array}$. Since every reduced ring is an Armendariz ring, we obtain that $\begin{array}{l}{a}_{1}^{i}{b}_{1}^{j}=0,{a}_{1}^{i}{b}_{2}^{j}=0,{a}_{2}^{i}{b}_{1}^{j}=0,{a}_{1}^{i}{b}_{3}^{j}=0,{a}_{2}^{i}{b}_{2}^{j}=\\ 0,{a}_{3}^{i}{b}_{1}^{j}=0,\cdots ,{a}_{1}^{i}{b}_{n-2}^{j}=0,{a}_{2}^{i}{b}_{n-3}^{j}=0,\cdots ,{a}_{n-2}^{i}{b}_{1}^{j}=0,{a}_{1}^{i}{b}_{1,n-1}^{j}=0,{a}_{2}^{i}{b}_{n-2}^{j}=\\ 0,\cdots ,{a}_{n-2}^{i}{b}_{2}^{j}=0,{a}_{1,n-1}^{i}{b}_{1}^{j}=0,{a}_{1}^{i}{b}_{1n}^{j}=0,{a}_{2}^{i}{b}_{2n}^{j}=0,{a}_{3}^{i}{b}_{n-1}^{j}=0,\cdots ,{a}_{n-2}^{i}{b}_{3}^{j}=\\ 0,{a}_{1,n-1}^{i}{b}_{2}^{j}=0,{a}_{1n}^{i}{b}_{1}^{j}=0,{a}_{1}^{i}{b}_{2n}^{j}=0,{a}_{2}^{i}{b}_{n-2}^{j}=0,\cdots ,{a}_{n-2}^{i}{b}_{2}^{j}=0,{a}_{2n}^{i}{b}_{1}^{j}=0\end{array}$ for all i, j by the preceding results. With these facts and the fact that R is α-semicommutative ring, we have $\begin{array}{l}{a}_{1}^{i}{c}_{1}^{k}\alpha ({b}_{1}^{j})=0,{a}_{1}^{i}{c}_{1}^{k}\alpha ({b}_{2}^{j})=0,{a}_{1}^{i}{c}_{2}^{k}\alpha ({b}_{1}^{j})=0,\\ {a}_{2}^{i}{c}_{1}^{k}\alpha ({b}_{1}^{j})=0,{a}_{1}^{i}{c}_{1}^{k}\alpha ({b}_{3}^{j})=0,{a}_{1}^{i}{c}_{2}^{k}\alpha ({b}_{2}^{j})=0,{a}_{2}^{i}{c}_{1}^{k}\alpha ({b}_{2}^{j})=0,{a}_{1}^{i}{c}_{2}^{k}\alpha ({b}_{1}^{j})=\\ 0,{a}_{2}^{i}{c}_{2}^{k}\alpha ({b}_{1}^{j})=0,{a}_{3}^{i}{c}_{1}^{k}\alpha ({b}_{1}^{j})=0,\cdots ,{a}_{1}^{i}{c}_{2n}^{k}\alpha ({b}_{1}^{j})=0,{a}_{2}^{i}{c}_{n-2}^{k}\alpha ({b}_{1}^{j})=0,\cdots ,\\ {a}_{n-2}^{i}{c}_{2}^{k}\alpha ({b}_{1}^{j})=0,{a}_{2n}^{i}{c}_{1}^{k}\alpha ({b}_{1}^{j})=0\end{array}$.
Therefore V_{n}(R) is strongly ᾱ-semicommutative.
The next result can be proved by using the technique used in the proof of [3, Proposition 2.6]. A ring is called Abelian if every idempotent is central. Reduced rings are clearly Abelian.
Proposition 2.24
Let R be a strongly α-semicommutative ring. Then
(1) α(1) = 1, where 1 is the identity of R, if and only if α(e) = e for any e^{2} = e ∈ R.
(2) If α(1) = 1, then R is Abelian.
Let R be an algebra over a commutative ring S. Recall that the Dorroh extension of R by S is the ring D = R × S with operations (r_{1}, s_{1}) + (r_{2}, s_{2}) = (r_{1} + r_{2}, s_{1} + s_{2}) and (r_{1}, s_{1})(r_{2}, s_{2}) = (r_{1}r_{2} + s_{1}r_{2} + s_{2}r_{1}, s_{1}s_{2}), where r_{i} ∈ R and s_{i} ∈ S. For an endomorphism α of R, the S-endomorphism ᾱ of D defined by ᾱ(r, s) = (α(r), s) is an S-algebra homomorphism.
Proposition 2.25
If R is a strongly α-semicommutative ring with α(1) = 1 and S is a domain, then the Dorroh extension D of R by S is strongly ᾱ-semicommutative.
Proof
We apply the method in the proof of [3, Proposition 2.8.] Let f(x) = (f_{1}(x), f_{2}(x)), g(x) = (g_{1}(x), g_{2}(x)) ∈ D(x) with (f_{1}(x), f_{2}(x))(g_{1}(x), g_{2}(x)) = 0. Then f_{1}(x)g_{1}(x) + f_{2}(x)g_{2}(x) + g_{2}(x)f_{1}(x) = 0 and f_{2}(x)g_{2}(x) = 0. Since S is a domain, we have f_{2}(x) = 0 or g_{2}(x) = 0. If f_{2}(x) = 0, then 0 = f_{1}(x)g_{1}(x) + f_{2}(x)g_{2}(x) + g_{2}(x)f_{1}(x) = f_{1}(x)g_{1}(x) + g_{2}(x)f_{1}(x) and so f_{1}(x)(g_{1}(x) + g_{2}(x)) = 0. Since R is strongly α-semicommutative with α(1) = 1, 0 = f_{1}(x)tα(g_{1}(x) + g_{2}(x)) = f_{1}(x)tα(g_{1}(x)) + f_{1}(x)tg_{2}(x)), for all t ∈ R. This yields (f_{1}(x), f_{2}(x))(r, s)ᾱ(g_{1}(x), g_{1}(x)) = (f_{1}(x)r + sf_{1}(x))α(g_{1}(x)) + (f_{1}(x)r + sf_{1}(x)g_{2}(x), 0) = 0 for any (r, s) ∈ D, and hence (f_{1}(x), f_{2}(x))Dᾱ(g_{1}(x), g_{2}(x)) = 0. Now let g_{2}(x) = 0. Then (f_{1}(x) + f_{2}(x))g_{1}(x) = 0, and so 0 = (f_{1}(x) + f_{2}(x))Rα(g_{1}(x)) = 0. We similarly obtain (f_{1}(x), f_{2}(x))Dᾱ(g_{1}(x), g_{2}(x)) = 0, and thus the Dorroh extension D is strongly ᾱ-semicommutative.
Corollary 2.26
([17, Proposition 3.17(2)]) Let R be an algebra over a commutative domain S, and D be the Dorroh extension of R by S. Then R is strongly semicommutative if and only if D is strongly semicommutative.
Note that the condition α(1) = 1 in Proposition 2.25 cannot be dropped by the next example.
Example 2.27
Let R = ℤ_{2}⊕ℤ_{2}, and let α: R → R defined by α((a, b)) = (0, b). Consider the Dorroh extension D of R by the ring of integers ℤ_{2}. We clearly have ((1, 0), 0)((1, 0),−1) = 0, but ((1, 0), 0)((1, 0), 0)ᾱ((1, 0),−1) = ((1, 0),−1) ≠ 0 in D. Thus D is not strongly ᾱ-semicommutative.
For an ideal I of R, if α(I) ⊆ I, then ᾱ: R/I → R/I defined by ᾱ(a + I) = α(a) + I is an endomorphism of the factor ring R/I.
There exists a non-identity automorphism α of a ring R such that R/I is strongly ᾱ-semicommutative and I is strongly α-semicommutative for any nonzero proper ideal I of R, but R is not strongly α-semicommutative by the next example.
Example 2.28
Let F be a field. Consider the ring $R=\left(\begin{array}{cc}F& F\\ 0& F\end{array}\right)$ and an endomorphism α of R defined by $\alpha \hspace{0.17em}\left(\left(\begin{array}{cc}a& b\\ 0& c\end{array}\right)\right)=\left(\begin{array}{cc}a& -b\\ 0& c\end{array}\right)$. Then R is not strongly α-semicommutative. In fact, for $A=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right),B=\left(\begin{array}{cc}0& -1\\ 0& 1\end{array}\right)\in R$, we have AB = 0, but $0\ne A\hspace{0.17em}\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)\hspace{0.17em}\alpha (B)\in AR\alpha (B)$. Note that for the only nonzero proper ideals of R
it can be easily checked that they are strongly α-semicommutative. Since R/I ≅ F and R/J ≅ F, R/I and R/J are also strongly ᾱ-semicommutative. Finally, the factor ring R/K is reduced and ᾱ is an identity map on R/K. Thus, R/K is also strongly ᾱ-semicommutative.
Proposition 2.29
Let R be a ring with an endomorphism α, and I an ideal of R with α(I) ⊆ I. Suppose that R/I is a strongly ᾱ-semicommutative ring. If I is α-rigid as a ring without identity, then R is strongly α-semicommutative.
Proof
Let f(x)g(x) = 0 with f(x), g(x) ∈ R[x]. Then we have f(x)Rα(g(x)) ⊆ I[x] and α(g(x))Iα(f(x)) = 0, since α(g(x))Iα(f(x)) ⊆ I[x], (α((g(x)Iα(f(x)))^{2} = 0 and I[x] is reduced. Thus, (f(x)Rα(g(x))I)^{2} = f(x)Rα(g(x))If(x)Rα(g(x))I = 0 and so f(x)Rα(g(x))I = 0, thus f(x)Rα(g(x))α(f(x)Rα(g(x))) ⊆ f(x)Rα(g(x))I = 0 since f(x)Rα(g(x)) ⊆ I[x] and α(I) ⊆ I. Then f(x)Rα(g(x)) = 0 as I is α-rigid. Therefore, R is strongly α-semicommutative.
Theorem 2.30
Let α be an endomorphism of a ring R. Then R is strongly α-semicommutative if and only if R[x] is strongly α-semicommutative.
Proof
(⇐) The converse is obvious since R is a subring of R[x].
(⇒) Assume that R is strongly α-semicommutative. Let f(y), g(y) ∈ R[x][y] such that f(y)g(y) = 0. Let
We also let f_{i} = a_{i}_{0} + a_{i}_{1}x + ⋯ + a_{i}_{w}x^{i}^{w}, g_{j} = b_{j}_{0} + b_{j}_{1}x + ⋯ + b_{j}_{v}x^{j}^{v}, h_{k} = c_{k}_{0} + c_{k}_{1}x + ⋯ + c_{k}_{u}x^{k}^{u} ∈ R[x] for each 0 ≤ i ≤ m, 0 ≤ j ≤ n and 0 ≤ k ≤ r, where a_{i}_{0}, a_{i}_{1}, ⋯, a_{i}_{w}, b_{j}_{0}, b_{j}_{1}, ⋯, b_{j}_{v}, c_{k}_{0}, c_{k}_{1}, ⋯, c_{k}_{u} ∈ R. We claim that p(y)R[x]q(y) = 0. Take a positive integer k such that k ≥ max {deg(f_{i}), deg(g_{j}), deg(h_{k})}, for any 0 ≤ i ≤ m, 0 ≤ j ≤ n, 0 ≤ k ≤ r, where the degree is as polynomials in R[x] and the degree of the zero polynomial is taken to be 0. Let f(x^{s}) = f_{0} + f_{1}x^{s} + ⋯ + f_{n}x^{ms}, g(x^{s}) = g_{0} + g_{1}x^{s} + ⋯ + g_{n}x^{ns}, h(x^{s}) = h_{0} + h_{1}x^{s} + ⋯ + h_{r}x^{rs} ∈ R[x]. Then the set of coefficients of the f_{i}’s, g_{j}’s (respectively, h_{k}’s) is equal to the set of coefficients of f(x^{s}), g(x^{s}) (respectively, h(x^{s})). Since f(y)g(y) = 0, x commutes with elements of R in the polynomial ring R[x], we have f(x^{s})g(x^{s}) = 0, in R[x]. Since R is strongly α-semicommutative, we have f(x^{s})Rα(g(x^{s})) = 0. Hence f(y)R[x]α(g(y)) = 0, therefore R[x] is strongly α-semicommutative.
Corollary 2.31
Let R be a ring. Then R is strongly semicommutative if and only if R[x] is strongly semicommutative.
Corollary 2.32
Let α be an endomorphism of a ring R. Then the following are equivalent:
(1) R is strongly α-semicommutative.
(2) R[x] is strongly α-semicommutative.
(3) R[x; x^{−1}] is strongly α-semicommutative.
Let A(R, α) or A be the subset {x^{−}^{i}a_{i}x^{i}|a ∈ R, i ≥ 0} of the skew Laurent polynomial ring R[x, x^{−1}; α], where α : R → R is an injective ring endomorphism of a ring R (see [9] for more details). Elements of R[x, x^{−1}; α] are finite sums of elements of the form x^{−}^{i}b_{j}x^{j}, where b ∈ R and i, j are non-negative integers. Multiplication is subject to xa = α(a)x and ax^{−1} = x^{−1}α(a) for all a ∈ R. Note that for each j ≥ 0, x^{−}^{i}a_{i}x^{i} = x^{−(}^{i}^{+}^{j}^{)}α^{j}(a_{i})x^{(}^{i}^{+}^{j}^{)}. It follows that the set A(R, α) of all such elements forms a subring of R[x, x^{−1}; α] with
for a, b ∈ R and i, j ≥ 0. Note that α is actually an automorphism of A(R, α). Let A(R, α) be the ring defined above. Then for the endomorphism α in A(R, α), the map A(R, α)[t] → A(R, α)[t] defined by
is an endomorphism of the polynomial ring A(R, α)[t].
Proposition 2.33
Let A(R, α) be an Armendariz ring. If R is α-semicommutative, then A(R, α) is strongly α-semicommutative.
Proof
Let $f(t)={\mathrm{\Sigma}}_{i=0}^{m}({x}^{-i}{a}_{i}{x}^{i}){t}^{i},g(t)={\mathrm{\Sigma}}_{j=0}^{n}({x}^{-j}{b}_{j}{x}^{j}){t}^{j}\in A(R,\alpha )[t]$ with f(t)g(t) = 0. Since A(R, α) is Armendariz, we have (x^{−}^{i}a_{i}x^{i})(x^{−}^{j}b_{j}x^{j}) = 0, and so x^{−(}^{i}^{+}^{j}^{)}(α^{j}(a_{i})α^{i}(b_{j}))x^{(}^{i}^{+}^{j}^{)} = 0. This implies that α^{j}(a_{i})α^{i}(b_{j}) = 0, and so α^{j}^{+}^{k}(a_{i})α^{i}^{+}^{k}(b_{j}) = 0. Hence α^{j}^{+}^{k}(a_{i})Rα^{i}^{+}^{k}^{+1}(b_{j}) = 0. Since R is α- semicommutative, for any $h(t)={\mathrm{\Sigma}}_{k=0}^{p}({x}^{-k}{c}_{k}{x}^{k}){t}^{k}\in A(R,\alpha )[t]$, we have