Kyungpook Mathematical Journal 2018; 58(2): 203-219
Extensions of Strongly α-semicommutative Rings
Ayoub Elshokry*, Eltiyeb Ali and Liu ZhongKui
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.

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

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.

Keywords: semicommutative rings; strongly semicommutative rings; α-semicommutative rings; strongly α-semicommutative rings
1. Introduction

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, bR, 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 aR, 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, bR, 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 IR-semicommutative, where IR 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 iI, let Ri be strongly αi-semicommutative where αi is an endomorphism of Ri. Set W = ΠiIRi. Define an endomorphism α of W as following:

$α(ai)i∈I=(αi(ai))i∈I.$

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]α2(g(x)) = 0. So, by induction hypothesis, we obtain f(x)R[x]αk(g(x)) = 0 and f(x)αk(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 ℤ2 be the ring of integers modulo 2 and consider the ring R = ℤ2⊕ℤ2, with the usual addition and multiplication. Then R is strongly semicommutative, since R is a commutative reduced ring. Now, let α : RR 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))α2(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 $R={(ab0a)∣a,b∈ℤ}$. Let α : RR be an endomorphism defined by $α((ab0a))=(a-b0a)$. 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 $f(x)=(f0(x)f1(x)0f0(x)),g(x)=(g0(x)g1(x)0g0(x))∈R[x]$. Then f0(x)g0(x) = 0 and f0(x)g1(x) + f1(x)g0(x) = 0. For $h(x)=(h0(x)h1(x)0h0(x))∈R[x]$, we have $(f0(x)f1(x)0f0(x)) (h0(x)h1(x)0h0(x))α((g0(x)g1(x)0g0(x)))=(f0(x)h0(x)g0(x)-f0(x)h0(x)g1(x)+f0(x)h1(x)g0(x)+f1(x)h0(x)g0(x)0f0(x)h0(x)g0(x))$. Since f0(x)g0(x) = 0, f0(x) = 0 or g0(x) = 0. If f0(x) = 0 then f1(x)g0(x) = 0. So f(x)R[x]α(g(x)) = 0. If g0(x) = 0 then f0(x)g1(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 (x) = 0 for 0 ≠ xR.

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, bR. 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 $f(x)=∑i=0naixi,g(x)=∑j=0mbjxj∈R[x]$ be such that $f(x)g(x)=0=∑s=0n+m∑i+j=saibjxs$. Since every reduced ring is an Armendariz ring, we obtain aibj = 0. Then α(bj)ai = 0 (by α-reversibility). Now for arbitrary element $h(x)=∑k=0rckxk∈R[x]$, we have α(bj)aick = 0 for each i, j, k, so aickα(bj) = 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) = a0+a1x+⋯+amxm, g(x) = b0+b1x+⋯+bnxnR[x] satisfy f(x)g(x) = 0, then aibj = 0 for each i and j. Hong et al. [7] called a ring R α-Armendariz if whenever f(x) = a0+a1x+⋯+amxm, g(x) = b0+b1x+⋯+bnxnR[x; α] satisfy f(x)g(x) = 0, then aibj = 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 $f(x)=∑i=0naixi,g(x)=∑j=0mbjxj∈R[x]$ satisfy f(x)g(x) = 0. Then, since R is Armendariz, each aibj is zero, additionally R is α-semicommutative, therefore aickα(bj) = 0 for any element ck in R for all i, j, k. Now it is easy to check that f(x)h(x)α(g(x)) = 0 for any $h(x)=∑k=0rckxk∈R[x]$.

### 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 rR. 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−1R.

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−1f(x) and G(x) = v−1g(x) ∈ (Δ−1R)[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−1f(x)v−1g(x) = (u−1v−1)f(x)g(x) = (uv)−1f(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−1R)[x]α(g(x)) = 0 for any regular element s. This implies F(x)(Δ−1R)[x]α(G(x)) = 0. Therefore Δ−1R 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 $∑i=knmixi$ with obvious addition and multiplication, where miR 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 Mn(R) and Tn(R) the n×n matrix ring and n×n upper triangular matrix ring over R, respectively.

Given a ring R and a bimodule RMR, the trivial extension of R by M is the ring T(R,M) = RM with the usual addition and the following multiplication (r1,m1)(r2,m2) = (r1r2, r1m2+m1r2). This is isomorphic to the ring of all matrices $(rm0r)$, where rR,mM 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 $α¯((ab0a))=(α(a)α(b)0α(a))$ 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 $R={(ab0a)∣a,b∈ℤ}$ with an endomorphism α defined by $α((ab0a))=(a-b0a)$ in Example 2.7. For

$A=((0100)(-110-1)(0000)(0100)),B=((0100)(1101)(0000)(0100))∈T(R,R)$

we have AB = 0. However, for

$C=((1001)(0000)(0000)(1001))∈T(R,R),$

we obtain

$0≠((0000)(0200)(0000)(0000))=ACα¯(B)∈AT(R,R)α¯(B).$

Thus, T(R,R) is not strongly ᾱ-semicommutative.

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) = (f0(x), f1(x)), g(x) = (g0(x), g1(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

$f0(x)g0(x)=0,$$f0(x)g1(x)+f1(x)g0(x)=0.$

Since R is reduced, R[x] is reduced. Therefore, (2.1) implies g0(x)f0(x) = 0. Multiplying (2.2) on the left side by g0(x) we get f1(x)g0(x) = 0, and so f0(x)g1(x) = 0. Let $f(x)=∑i=0n(ai,bi)xi,g(x)=∑j=0m(aj′,bj′)xj$, where $f0(x)=∑i=0naixi,f1(x)=∑i=0nbixi,g0(x)=∑j=0maj′xj$ and $g1(x)=∑j=0mbj′xj$. Since every reduced ring is an Armendariz ring, we obtain that $aiaj′=0,aibj′=0,biaj′=0$ for all i, j by the preceding results. With these facts and the fact that R is α-semicommutative, we have $aickα(aj′)=0,aickα(bj′)=0,aidkα(bj′)=0,bickα(aj′)=0$, for any elements ck, dk. Thus, f(x)h(x)α(g(x)) = 0, for any arbitrary $h(x)=∑k=0r(ck,dk)xk∈R[x]$. This implies that T(R,R) is strongly ᾱ-semicommutative.

The trivial extension T(R,R) of a ring R is extended to

$S3(R)={(abc0ad00a)∣a,b,c,d∈R}$

and an endomorphism α of a ring R is also extended to the endomorphism ᾱ of S3(R) defined by ᾱ((aij)) = (ᾱ(aij)). There exists a reduced ring R such that S3(R) is not strongly ᾱ-semicommutative by the following example.

### Example 2.20

We consider the commutative reduced ring R = ℤ2⊕ℤ2, and the automorphism α of R defined by α((a, b)) = (b, a), in Example 2.3. Then S3(R) is not strongly ᾱ-semicommutative. For$A=((1,0)(0,0)(0,0)(0,0)(1,0)(0,0)(0,0)(0,0)(1,0)),B=((0,1)(0,0)(0,0)(0,0)(0,1)(0,0)(0,0)(0,0)(0,1))∈S3(R)$

, then AB = 0, but AAᾱ(B) = A ≠ 0. Thus AS3(R) ᾱ(B) ≠ 0, and therefore S3(R) is not strongly ᾱ-semicommutative.

However, we obtain that S3(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

$S3(R)={(abc0ad00a)∣a,b,c,d∈R}$

is strongly ᾱ-semicommutative.

Proof

For

$(a1b1c10a1d100a1),(a2b2c20a2d200a2)∈S3(R),$

we can denote their addition and multiplication by

$(a1,b1,c1,d1) (a2,b2,c2,d2)=(a1+a2,b1+b2,c1+c2,d1+d2),(a1,b1,c1,d1) (a2,b2,c2,d2)=(a1a2,a1b2+b1a2,a1c2+b1d2+c1a2,a1d2+d1a2),$

, respectively. So every polynomial in S3[x] can be expressed in the form of (f0, f1, f2, f3) for some fis in R[x]. Let f(x) = (f0(x), f1(x), f2(x), f3(x)), g(x) = (g0(x), g1(x), g2(x), g3(x)) ∈ S3[x] with f(x)g(x) = 0. Then f(x)g(x) = (f0(x)g0(x), f0(x)g1(x)+f1(x)g0(x), f0(x)g2(x)+f1(x)g3(x)+f2(x)g0(x), f0(x)g3(x)+f3(x)g0(x)), we shall prove f(x)S3(R)[x]α(g(x)) = 0. So we have the following system of equations:

$f0(x)g0(x)=0,$$f0(x)g1(x)+f1(x)g0(x)=0,$$f0(x)g2(x)+f1(x)g3(x)+f2(x)g0(x)=0,$$f0(x)g3(x)+f3(x)g0(x)=0.$

Use the fact that R[x] is reduced. From Eq. (2.3), we get g0(x)f0(x) = 0. If we multiply Eq. (2.4), on the right side by g0(x), then $0=(f0(x)g1(x)+f1(x)g0(x))g0(x)=f1(x)g02(x)$, and so f1(x)g0(x) = 0 and f0(x)g1(x) = 0. Similarly, from Eq. (2.6), we have f3(x)g0(x) = 0, and f0(x)g3(x) = 0. Also, in Eq. (2.5), $0=(f0(x)g2(x)+f1(x)g3(x)+f2(x)g0(x))g0(x)=f2(x)g02(x)$ implies f2(x)g0(x) = 0 and

$f0(x)g2(x)+f1(x)g3(x)=0.$

Multiplying (2.7) on left side by f0(x) gives $0=f0(x)(f0(x)g2(x)+f1(x)g3(x))=f02(x)g2(x)$, and so f0(x)g2(x) = 0 hence f1(x)g3(x) = 0. Let

$f(x)=∑i=0n(aibici0aidi00ai) xi,g(x)=∑j=0m(aj′bj′cj′0aj′dj′00aj′)xj$

and$h(x)=∑k=0r(ak″bk″ck″0ak″dk″00ak″) xk∈S3(R)$,

where $f0(x)=∑i=0naixi,f1(x)=∑i=0nbixi,f2(x)=∑i=0ncixi,f3(x)=∑i=0ndixi,g0(x)=∑j=0maj′xj,g1(x)=∑j=0mbj′xj,g2(x)=∑j=0mcj′xj,g3(x)=∑j=0mdj′xj$. Since every reduced ring is an Armendariz ring, we obtain that $aiaj′=0,aibj′=0,biaj′=0,aicj′=0,bidj′=0,ciaj′=0,aidj′=0,diaj′=0$, for all i, j by the preceding results. With these facts and the fact that R is α-semicommutative ring, we have $aiak″α(aj′)=0,aiak″α(bj′)=0,biak″α(aj′)=0,biak″α(dj′)=0,aiak″α(cj′)=0,aibk″α(dj′)=0,biak″α(dj′)=0,aick″α(aj′)=0,bidk″α(aj′)=0,ciak″α(aj′)=0,aiak″α(dj′)=0,aidk″α(aj′)=0,diak″α(aj′)=0$. Consequently, we get the equation:

$f(x)h(x)α(g(x))=(f0(x),f1(x),f2(x),f3(x))S3(R)[x]α((g0(x),g1(x),g2(x),g3(x))=(f0(x)S3(R)[x]α(g0(x)),f0(x)S3(R)[x]α(g1(x))+f1(x)S3(R)[x]α(g0(x)),f0(x)S3(R)[x]α(g2(x))+f1(x)S3(R)[x]α(g3(x))+f2(x)S3(R)[x]α(g0(x)),f0(x)S3(R)[x]α(g3(x))+f3(x)S3(R)[x]α(g0(x)))=0.$

Therefore S3(R) is strongly ᾱ-semicommutative.

Let R be a ring. Define a subring Sn of the n-by-n full matrix ring Mn(R) over R as follows:

$Sn(R)={(aa12a13⋯a1n0aa23⋯a2n00a⋯a3n⋮⋮⋮⋱⋮000⋯a)∣a,aij∈R}.$

For an α-rigid ring R and n ≥ 2, by Proposition 2.21, we may suspect that Sn(R) may be strongly ᾱ-semicommutative ring for n ≥ 4. But the possibility is eliminated by the next example.

### Example 2.22

Let R be an α-rigid and

$S4={(aa12a13a140aa23a2400aa34000a)∣a,aij∈R}.$

Note that if R is an α-rigid ring, then α(e) = e, for e2 = eR by [6, Proposition 5]. In particular α(1) = 1. For $A=(01-10000000000000),B=(0000000100010000)∈S4(R)$, we obtain AB = 0. But we have $0≠(01-10000000000000)=ACα¯(B)∈S4(R)$, for $C=(0000001000000000)∈S4(R)$. Thus AC ᾱ(B) ≠ 0 and so S4(R) is not strongly ᾱ-semicommutative. Similarly, it can be proved that Sn(R) is not strongly ᾱ-semicommutative for n ≥ 5.

Let R be a ring and let

$Vn(R)={S=(a1a2a3⋯an-2ab0a1a2⋯an-3an-2c00a1⋯an-4an-3an-2⋮⋮⋮⋱⋮⋮⋮000⋯a1a2a3000⋯0a1a2000⋯00a1)∣ai,a,b,c∈R}.$

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 S3(R) is strongly ᾱ-semicommutative, but Sn(R) is not strongly ᾱ-semicommutative for n ≥ 4. In the next theorem we will show that a special subring Vn(R) of Tn(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 Vn(R) is strongly ᾱ-semicommutative.

Proof

Suppose that

$(a1a2a3⋯an-2a1,n-1a1n0a1a2⋯an-3an-2a2n00a1⋯an-4an-3an-2⋮⋮⋮⋱⋮⋮⋮000⋯a1a2a3000⋯0a1a2000⋯00a1),(b1b2b3⋯bn-2b1,n-1b1n0b1b2⋯bn-3bn-2b2n00b1⋯bn-4bn-3bn-2⋮⋮⋮⋱⋮⋮⋮000⋯b1b2b3000⋯0b1b2000⋯00b1)$

are in Vn(R). So every polynomial in Vn(R)[x] can be expressed in the form of (f1, f2,, fn−2, f1,n−1, f1n, f2n) for some fi’s in R[x]. Let f(x) = (f0(x), f1(x), ⋯, f2n(x)), g(x) = (g0(x), g1(x), ⋯, g2n(x)) ∈ Vn(R)[x] with f(x)g(x) = 0. We shall prove f(x)Vn(R)[x]α(g(x)) = 0. Now we have the following system of equations:

$f1(x)g1(x)=0,$$f1(x)g2(x)+f2(x)g1(x)=0,$$f1(x)g3(x)+f2(x)g2(x)+f3(x)g1(x)=0⋮f1(x)gn-2(x)+f2(x)gn-3(x)+⋯+fn-2(x)g1(x)=0,f1(x)g1,n-1(x)+f2(x)gn-2(x)+⋯+fn-2(x)g2(x)+f1,n-1(x)g1(x)=0,$$f1(x)g1n(x)+f2(x)g2n(x)+⋯+f1,n-1(x)g2(x)+f1n(x)g1(x)=0,$$f1(x)g2n(x)+f2(x)gn-2(x)+⋯+fn-2(x)g2(x)+f2n(x)g1(x)=0.$

Use the fact that R[x] is reduced. From Eq. (2.8), we get g1(x)f1(x) = 0. If we multiply Eq. (2.9) on the right side by f1(x), then f1(x)g2(x)f1(x) + f2(x)g1(x)f1(x) = 0. Thus f1(x)g2(x)f1(x) = 0 and hence f1(x)g2(x) = 0. From Eq. (2.9) it follows that f2(x)g1(x) = 0. Continuing in this manner, we can show that fi(x)gj(x) = 0 when i + j = 2,, n − 1. Hence gj(x)fi(x) = 0. Multiplying Eq. (2.10) on the right side by f1(x), we obtain 0 = f1(x)g1,n−1(x)f1(x) + f2(x)gn−2(x)f1(x) + ⋯ + fn−2(x)g2(x)f1(x) + f1,n−1(x)g1(x)f1(x) = f1(x)g1,n−1(x)f1(x). Thus f1(x)g1,n−1(x) = 0. Hence

$f2(x)gn-2(x)+⋯+fn-2(x)g2(x)+f1,n-1(x)g1(x)=0,$

Multiplying Eq. (2.13) on the right side by f2(x), we obtain

$0=f2(x)gn-2(x)f2(x)+⋯+fn-2(x)g2(x)f2(x)+f1,n-1(x)g1(x)f2(x)=f2(x)gn-2(x)f2(x).$

Thus f2(x)gn−2(x) = 0. Continuing in this manner, we can show that fi(x)gj(x) = 0 when i + j = n and f1(x)g1,n−1(x) = 0, f1,n−1(x)g1(x) = 0. Similarly, from Eq. (2.12), it follows that f1(x)g2n(x) = 0 and f2n(x)g1(x) = 0. Now multiplying Eq. (2.11) on the right side by f1(x), we have

0 = f1(x)g1n(x)f1(x)+f2(x)g2n(x)f1(x)+f3(x)gn−2(x)f1(x)+⋯+fn−2(x)g3(x) f1(x)+f1,n−1(x)g2(x)f1(x)+f1n(x)g1(x)f1(x) = f1(x)g1n(x)f1(x). Thus f1(x)g1n(x) = 0. Hence

$f2(x)g2n(x)+f3(x)gn-2(x)+⋯+f1,n-1(x)g2(x)+f1n(x)g1(x)=0,$

If we multiply Eq. (2.14) on the right side by f2(x), then 0 = f2(x)g2n(x)f2(x) + f3(x)gn−2(x)f2(x)+⋯+f1,n−1(x)g2(x)f2(x)+f1n(x)g1(x)f2(x) = f2(x)g2n(x)f2(x). Thus f2(x)g2n(x) = 0. Continuing in this manner, we can show that fi(x)gj(x) = 0 when i + j = n + 1, f1,n−1(x)g2(x) = 0 and f1n(x)g1(x) = 0. Let

$f(x)=∑i=0n(a1ia2ia3i⋯an-2ia1,n-1ia1ni0a1ia2i⋯an-3ian-2ia2ni00a1i⋯an-4ian-3ian-2i⋮⋮⋮⋱⋮⋮⋮000⋯a1ia2ia3i000⋯0a1ia2i000⋯00a1i)xi,g(x)=∑i=0n(b1jb2jb3j⋯bn-2jb1,n-1jb1nj0b1jb2j⋯bn-3jbn-2jb2nj00b1j⋯bn-4jbn-3jbn-2j⋮⋮⋮⋱⋮⋮⋮000⋯b1jb2jb3j000⋯0b1jb2j000⋯00b1j)xiand h(x)=∑k=0r(c1kc2kc3k⋯cn-2kc1,n-1kc1nk0c1kc2k⋯cn-3kcn-2kc2nk00c1k⋯cn-4kcn-3kcn-2k⋮⋮⋮⋱⋮⋮⋮000⋯c1kc2kc3k000⋯0c1kc2k000⋯00c1k)xk∈Vn(R)[x],$

where $f1(x)=∑i=0na1ixi,f2(x)=∑i=0na2ixi,⋯,fn-2(x)=∑i=0nan-2ixi,f1,n-1(x)=∑i=0na1,n-1ixi,f1n(x)=∑i=0na1nixi,f2n(x)=∑i=0na2nixi,g1(x)=∑j=0mb1jxj,g2(x)=∑j=0mb2jxj,⋯,gn-2(x)=∑j=0mbn-2jxj,g1,n-1(x)=∑j=0mb1,n-1jxj,g1n(x)=∑j=0mb1njxj,g2n(x)=∑j=0mb2njxj$. Since every reduced ring is an Armendariz ring, we obtain that $a1ib1j=0,a1ib2j=0,a2ib1j=0,a1ib3j=0,a2ib2j=0,a3ib1j=0,⋯,a1ibn-2j=0,a2ibn-3j=0,⋯,an-2ib1j=0,a1ib1,n-1j=0,a2ibn-2j=0,⋯,an-2ib2j=0,a1,n-1ib1j=0,a1ib1nj=0,a2ib2nj=0,a3ibn-1j=0,⋯,an-2ib3j=0,a1,n-1ib2j=0,a1nib1j=0,a1ib2nj=0,a2ibn-2j=0,⋯,an-2ib2j=0,a2nib1j=0$ for all i, j by the preceding results. With these facts and the fact that R is α-semicommutative ring, we have $a1ic1kα(b1j)=0,a1ic1kα(b2j)=0,a1ic2kα(b1j)=0,a2ic1kα(b1j)=0,a1ic1kα(b3j)=0,a1ic2kα(b2j)=0,a2ic1kα(b2j)=0,a1ic2kα(b1j)=0,a2ic2kα(b1j)=0,a3ic1kα(b1j)=0,⋯,a1ic2nkα(b1j)=0,a2icn-2kα(b1j)=0,⋯,an-2ic2kα(b1j)=0,a2nic1kα(b1j)=0$.

Therefore Vn(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 e2 = eR.

• (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 (r1, s1) + (r2, s2) = (r1 + r2, s1 + s2) and (r1, s1)(r2, s2) = (r1r2 + s1r2 + s2r1, s1s2), where riR and siS. 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) = (f1(x), f2(x)), g(x) = (g1(x), g2(x)) ∈ D(x) with (f1(x), f2(x))(g1(x), g2(x)) = 0. Then f1(x)g1(x) + f2(x)g2(x) + g2(x)f1(x) = 0 and f2(x)g2(x) = 0. Since S is a domain, we have f2(x) = 0 or g2(x) = 0. If f2(x) = 0, then 0 = f1(x)g1(x) + f2(x)g2(x) + g2(x)f1(x) = f1(x)g1(x) + g2(x)f1(x) and so f1(x)(g1(x) + g2(x)) = 0. Since R is strongly α-semicommutative with α(1) = 1, 0 = f1(x)(g1(x) + g2(x)) = f1(x)(g1(x)) + f1(x)tg2(x)), for all tR. This yields (f1(x), f2(x))(r, s)ᾱ(g1(x), g1(x)) = (f1(x)r + sf1(x))α(g1(x)) + (f1(x)r + sf1(x)g2(x), 0) = 0 for any (r, s) ∈ D, and hence (f1(x), f2(x))Dᾱ(g1(x), g2(x)) = 0. Now let g2(x) = 0. Then (f1(x) + f2(x))g1(x) = 0, and so 0 = (f1(x) + f2(x))(g1(x)) = 0. We similarly obtain (f1(x), f2(x))Dᾱ(g1(x), g2(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 = ℤ2⊕ℤ2, and let α: RR defined by α((a, b)) = (0, b). Consider the Dorroh extension D of R by the ring of integers ℤ2. 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/IR/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 $R=(FF0F)$ and an endomorphism α of R defined by $α ((ab0c))=(a-b0c)$. Then R is not strongly α-semicommutative. In fact, for $A=(1100),B=(0-101)∈R$, we have AB = 0, but $0≠A (1-101) α(B)∈ARα(B)$. Note that for the only nonzero proper ideals of R

$I=(FF00),J=(0F0F),K=(0F00),$

it can be easily checked that they are strongly α-semicommutative. Since R/IF and R/JF, 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)(g(x)) ⊆ I[x] and α(g(x))(f(x)) = 0, since α(g(x))(f(x)) ⊆ I[x], (α((g(x)(f(x)))2 = 0 and I[x] is reduced. Thus, (f(x)(g(x))I)2 = f(x)(g(x))If(x)(g(x))I = 0 and so f(x)(g(x))I = 0, thus f(x)(g(x))α(f(x)(g(x))) ⊆ f(x)(g(x))I = 0 since f(x)(g(x)) ⊆ I[x] and α(I) ⊆ I. Then f(x)(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

$f(y)=f0+f1y+⋯+fmym,g(y)=g0+g1y+⋯+gnyn,$

and

$h(y)=h0+h1y+⋯+hryr∈R[x][y].$

We also let fi = ai0 + ai1x + ⋯ + aiwxiw, gj = bj0 + bj1x + ⋯ + bjvxjv, hk = ck0 + ck1x + ⋯ + ckuxkuR[x] for each 0 ≤ im, 0 ≤ jn and 0 ≤ kr, where ai0, ai1, ⋯, aiw, bj0, bj1, ⋯, bjv, ck0, ck1, ⋯, ckuR. We claim that p(y)R[x]q(y) = 0. Take a positive integer k such that k ≥ max {deg(fi), deg(gj), deg(hk)}, for any 0 ≤ im, 0 ≤ jn, 0 ≤ kr, where the degree is as polynomials in R[x] and the degree of the zero polynomial is taken to be 0. Let f(xs) = f0 + f1xs + ⋯ + fnxms, g(xs) = g0 + g1xs + ⋯ + gnxns, h(xs) = h0 + h1xs + ⋯ + hrxrsR[x]. Then the set of coefficients of the fis, gjs (respectively, hks) is equal to the set of coefficients of f(xs), g(xs) (respectively, h(xs)). Since f(y)g(y) = 0, x commutes with elements of R in the polynomial ring R[x], we have f(xs)g(xs) = 0, in R[x]. Since R is strongly α-semicommutative, we have f(xs)(g(xs)) = 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 {xiaixi|aR, i ≥ 0} of the skew Laurent polynomial ring R[x, x−1; α], where α : RR 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 xibjxj, where bR and i, j are non-negative integers. Multiplication is subject to xa = α(a)x and ax−1 = x−1α(a) for all aR. Note that for each j ≥ 0, xiaixi = x−(i+j)αj(ai)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

$x-iaixi+x-jbjxj=x-(i+j)(αj(ai)+αi(bj))x(i+j)(x-iaixi)(x-jbjxj)=x-(i+j)(αj(ai)αi(bj))x(i+j)$

for a, bR 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

$Σi=0m(x-iaixi)ti→Σi=0m(x-iα(ai)xi)ti$

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 $f(t)=Σi=0m(x-iaixi)ti,g(t)=Σj=0n(x-jbjxj)tj∈A(R,α)[t]$ with f(t)g(t) = 0. Since A(R, α) is Armendariz, we have (xiaixi)(xjbjxj) = 0, and so x−(i+j)(αj(ai)αi(bj))x(i+j) = 0. This implies that αj(ai)αi(bj) = 0, and so αj+k(ai)αi+k(bj) = 0. Hence αj+k(ai)i+k+1(bj) = 0. Since R is α- semicommutative, for any $h(t)=Σk=0p(x-kckxk)tk∈A(R,α)[t]$, we have

$f(t)h(t)g(t)=(Σi=0m(x-iaixi)ti)(Σk=0p(x-kckxk)tk)α(Σj=0n(x-jbjxj)tj) =(Σi+k=0m+p(x-iaixi)(x-kckxk)ti+k)(Σj=0n(x-jα(bj)xj)tj) =(Σi+k=0m+p(x-(i+k)(αk(ai)αi(ck))xi+k)ti+k)(Σj=0n(x-jα(bj)xj)tj) =(Σi+j+k=0m+n+p(x-(i+k)(αk(ai)αi(ck))xi+k)(x-jα(bj)xj)ti+j+k =(Σi+j+k=0m+n+p(x-(i+j+k)(αj(αk(ai)αi(ck))αi+k)(α(bj))(xi+j+k)ti+j+k =(Σi+j+k=0m+n+p(x-(i+j+k)(αk+j(ai)αi+j(ck)αi+k+1(bj))(xi+j+k)ti+j+k.$

As (αk+j(ai)αi+j(ck)αi+k+1(bj) = 0, f(t)h(t)α(g(t)) = 0. So A(R, α) is strongly α-semicommutative.

### Corollary 2.34

Let A(R, α) be an Armendariz ring. If R is semicommutative, then A(R, α) is strongly semicommutative.

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