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.
© Kyungpook Mathematical Journal. All rights reserved.

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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.

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)iI=(αi(ai))iI.

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}. 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=0mbjxjR[x] be such that f(x)g(x)=0=s=0n+mi+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=0rckxkR[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=0mbjxjR[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=0rckxkR[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} 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=0majxj and g1(x)=j=0mbjxj. 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)xkR[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,dR}

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. ForA=((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,dR}

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(ajbjcj0ajdj00aj)xj

andh(x)=k=0r(akbkck0akdk00ak)xkS3(R),

where f0(x)=i=0naixi,f1(x)=i=0nbixi,f2(x)=i=0ncixi,f3(x)=i=0ndixi,g0(x)=j=0majxj,g1(x)=j=0mbjxj,g2(x)=j=0mcjxj,g3(x)=j=0mdjxj. 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 aiakα(aj)=0,aiakα(bj)=0,biakα(aj)=0,biakα(dj)=0,aiakα(cj)=0,aibkα(dj)=0,biakα(dj)=0,aickα(aj)=0,bidkα(aj)=0,ciakα(aj)=0,aiakα(dj)=0,aidkα(aj)=0,diakα(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)={(aa12a13a1n0aa23a2n00aa3n000a)a,aijR}.

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,aijR}.

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=(a1a2a3an-2ab0a1a2an-3an-2c00a1an-4an-3an-2000a1a2a30000a1a200000a1)ai,a,b,cR}.

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

(a1a2a3an-2a1,n-1a1n0a1a2an-3an-2a2n00a1an-4an-3an-2000a1a2a30000a1a200000a1),(b1b2b3bn-2b1,n-1b1n0b1b2bn-3bn-2b2n00b1bn-4bn-3bn-2000b1b2b30000b1b200000b1)

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)=0f1(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(a1ia2ia3ian-2ia1,n-1ia1ni0a1ia2ian-3ian-2ia2ni00a1ian-4ian-3ian-2i000a1ia2ia3i0000a1ia2i00000a1i)xi,g(x)=i=0n(b1jb2jb3jbn-2jb1,n-1jb1nj0b1jb2jbn-3jbn-2jb2nj00b1jbn-4jbn-3jbn-2j000b1jb2jb3j0000b1jb2j00000b1j)xiand h(x)=k=0r(c1kc2kc3kcn-2kc1,n-1kc1nk0c1kc2kcn-3kcn-2kc2nk00c1kcn-4kcn-3kcn-2k000c1kc2kc3k0000c1kc2k00000c1k)xkVn(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 0A(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++hryrR[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)tiΣi=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)tjA(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)tkA(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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