Semicommutative and Armendariz Matrix Rings

Abstract

In this paper, we construct some interesting high-order upper triangular matrix rings, which have semicommutative and Armendariz properties. Also, the relatively maximality of these rings as subrings of certain matrix rings is considered.

1. Introduction

Throughout this work, R is always an associative ring with identity. According to Bell [1], a ring R satisfies the insertion-of-factors-property (IFP) if for any $a,b\in R$, the condition $ab=0$ implies $aRb=0$. The same property appears in the literature under different names: semicommutative in [2] and SI in [3]. We will consistently use the term semicommutative herein. It is noteworthy that, by (Lemma 1.2 [3]), a ring R is semicommutative if and only if for every $a\in R$, the set $\{x\in R\mid ax=0\}$ is an ideal of R. Owing to its importance, semicommutativity has been extensively studied in noncommutative ring theory; for the theory of semicommutative rings, we refer to [4,5,6], and for its generalizations, we refer to [7,8,9].

Let n be a positive integer. We denote the $n\times n$ full matrix ring over R by $M_n\left(R\right)$ and the $n\times n$ upper triangular matrix ring by $T_n\left(R\right)$; the identity matrix is denoted by $I_n$. A natural question is whether $M_n\left(R\right)$ or $T_n\left(R\right)$ remains semicommutative when R is semicommutative. However, it is known that for any ring R with identity and $n\ge 2$, the ring $T_n\left(R\right)$ (and consequently $M_n\left(R\right)$) is not semicommutative (Example 5 [4]). This fact motivates the interesting problem of identifying semicommutative subrings within $T_n\left(R\right)$.

Define a subring $S_n\left(R\right)$ of $T_n\left(R\right)$ by

$$S_n\left(R\right)=\left\{\left(\begin{array}{ccccc}r& r_{12}& r_{13}& \cdots & r_{1n}\\ 0& r& r_{23}& \cdots & r_{2n}\\ 0& 0& r& \cdots & r_{3n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 0& r\end{array}\right)|r,r_{ij}\in R\right\}.$$

Kim and Lee [5] proved that, for a reduced ring R, $S_3\left(R\right)$ is semicommutative, whereas $S_n\left(R\right)$ fails to be semicommutative for $n\ge 4$ (see Proposition 1.2 and Example 1.3 [5]). Motivated by this discrepancy, it is natural to investigate what other semicommutative subrings exist within the upper triangular matrix ring $T_n\left(R\right)$ for $n\ge 4$. Building on the work of Liu [10], Wang [11], Yang [12], and others, we continue to investigate this area.

Let R be a ring and $R\left[X\right]$ its polynomial ring in an indeterminate X. Following [13], we say R is an Armendariz ring if for any polynomials

$$f\left(X\right)=a_0+a_{1}X+\cdots +a_m{X}^m,g\left(X\right)=b_0+b_{1}X+\cdots +b_n{X}^n\in R\left[X\right]$$

with $f\left(X\right)g\left(X\right)=0$; it follows that $a_i{b}_j=0$ for all $0\le i\le m$ and $0\le j\le n$. The terminology is due to the observation by Armendariz (Lemma 1 [14]) that reduced rings always have this property. It is well documented that the class of Armendariz rings exhibits a rich structure (see [4,15,16,17] for a detailed account).

This paper also studies the Armendariz property in the context of specific subrings of the upper triangular matrix ring $T_n\left(R\right)$ that we introduce.

Notations

Let $m,n$ be two nonnegative integers that are not both zero. Define the $m\times m$ matrix $V_m$ by $V_m=\left(\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& 0& 0& \cdots & 1\\ 0& 0& 0& \cdots & 0\end{array}\right)$ and let O denote a zero matrix of appropriate dimensions (not necessarily square).

We define the set $U_n^m\left(R\right)$ to consist of all $(m+n)\times (m+n)$ matrices over R satisfying the following conditions:

(I)

$a_{ij}=0$ for all $i>j$;

(II)

$a_{ii}=a_{jj}$ for all $i,j$;

(III)

For all $1\le i,j,k,l\le m$ with $i<j$ and $k<l$, $a_{ij}=a_{kl}$ whenever $j-i=l-k$;

(IV)

$a_{ij}=0$ for all $m+1\le i<j\le m+n$ with $i<j$.

In the block matrix form, we can express $U_n^m\left(R\right)$ as

$$U_n^m\left(R\right)=\left\{\left(\begin{array}{cc}{\displaystyle a_1{I}_m+\sum _{k=1}^{m-1}{a}_{k+1}{V}_m^k}& M\\ ineO& a_1{I}_n\end{array}\right)|a_1,a_2,\cdots ,a_m\in R\right\},$$

where O is the $n\times m$ zero matrix and M ranges over all $m\times n$ matrices over R. Also, a matrix of $U_n^m\left(R\right)$ can be expressed explicitly in the following form:

$$\left(\begin{array}{cccccccccc}{a}_1& a_2& \cdots & a_{m-1}& a_m& a_{11}& a_{12}& \cdots & a_{1,n-1}& a_{1n}\\ 0& a_1& \ddots & \ddots & a_{m-1}& a_{21}& a_{22}& \cdots & a_{2,n-1}& a_{2n}\\ ⋮& ⋮& \ddots & \ddots & ⋮& ⋮& ⋮& \cdots & ⋮& ⋮\\ 0& \cdots & 0& a_1& a_2& a_{m-1,1}& a_{m-1,2}& \cdots & a_{m-1,n-1}& a_{m-1,n}\\ 0& \cdots & 0& 0& a_1& a_{m1}& a_{m2}& \cdots & a_{m,n-1}& a_{mn}\\ ine0& \cdots & 0& 0& 0& a_1& 0& \cdots & 0& 0\\ 0& \cdots & 0& 0& 0& 0& a_1& \cdots & 0& 0\\ ⋮& \cdots & ⋮& ⋮& ⋮& ⋮& \ddots & \ddots & ⋮& ⋮\\ 0& \cdots & 0& 0& 0& 0& 0& \ddots & a_1& 0\\ 0& \cdots & 0& 0& 0& 0& 0& \cdots & 0& a_1\end{array}\right)$$

with all entries in R. It is easily verified that $U_n^m\left(R\right)$ forms a subring of the upper triangular matrix ring $T_{m+n}\left(R\right)$. We will show that if R is reduced, then $U_n^m\left(R\right)$ is semicommutative. In fact, $U_n^m\left(R\right)$ turns out to be a ring with simple 0 multiplication and also an Armendariz ring.

Symmetrically, we define the subring $W_n^m\left(R\right)$ of the upper triangular matrix ring to consist of all $(m+n)\times (m+n)$ matrices over R, satisfying the following:

(I′)

$a_{ij}=0$ for all $i>j$;

(II′)

$a_{ii}=a_{jj}$ for all $i,j$;

(III′)

$a_{ij}=0$ for all $1\le i<j\le m$;

(IV′)

For all $m+1\le i,j,k,l\le m+n$ with $i<j$ and $k<l$, $a_{ij}=a_{kl}$ whenever $j-i=l-k$.

In the block matrix form,

$$W_n^m\left(R\right)=\left\{\left(\begin{array}{cc}{a}_1{I}_m& M\\ ineO& {\displaystyle a_1{I}_n+\sum _{k=1}^{n-1}{a}_{k+1}{V}_n^k}\end{array}\right)|a_1,a_2,\cdots ,a_n\in R\right\},$$

where M ranges over all $m\times n$ matrices over R. Also, a matrix of $W_n^m\left(R\right)$ can be expressed explicitly in the following form:

$$\left(\begin{array}{cccccccccc}{a}_1& 0& \cdots & 0& 0& a_{11}& a_{12}& \cdots & a_{1,n-1}& a_{1n}\\ 0& a_1& \ddots & \ddots & 0& a_{21}& a_{22}& \cdots & a_{2,n-1}& a_{2n}\\ ⋮& ⋮& \ddots & \ddots & ⋮& ⋮& ⋮& \cdots & ⋮& ⋮\\ 0& \cdots & 0& a_1& 0& a_{m-1,1}& a_{m-1,2}& \cdots & a_{m-1,n-1}& a_{m-1,n}\\ 0& \cdots & 0& 0& a_1& a_{m1}& a_{m2}& \cdots & a_{m,n-1}& a_{mn}\\ ine0& \cdots & 0& 0& 0& a_1& a_2& \cdots & a_{n-1}& a_n\\ 0& \cdots & 0& 0& 0& 0& a_1& \ddots & \ddots & a_{n-1}\\ ⋮& \cdots & ⋮& ⋮& ⋮& ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& 0& 0& 0& 0& \cdots & a_1& a_2\\ 0& \cdots & 0& 0& 0& 0& 0& \cdots & 0& a_1\end{array}\right)$$

with all entries in R. Similarly, one can show that if R is reduced, then $W_n^m\left(R\right)$ is semicommutative. In fact, $W_n^m\left(R\right)$ is also a ring with simple 0 multiplication and an Armendariz ring.

Definition 1 

([17]). Let $M_n\left(R\right)$ be an $n\times n$ matrix ring over R, and let S be a subring of $M_n\left(R\right)$. The ring S is said to have simple 0 multiplication if for any $A=\left(a_{ij}\right),B=\left(b_{ij}\right)\in S$ with $AB=0$, we have $a_{ik}{b}_{kj}=0$ for all $1\le i,j,k\le n$.

2. Main Results

A ring R is reduced if it has no nonzero nilpotent elements. We will use the following well-known properties of reduced rings (see Lemma 2.2 [17]):

Lemma 1. 

If we let R be a reduced ring, then the following holds:

(1) 

If $a_1{a}_2{a}_3=0$ for some $a_1,a_2,a_3\in R$, then $a_{\sigma \left(1\right)}{a}_{\sigma \left(2\right)}{a}_{\sigma \left(3\right)}=0$ for any permutation σ of $\{1,2,3\}$.

(2) 

If $a^{2}b=0$ or $a{b}^2=0$ for some $a,b\in R$, then $ab=0$.

(3) 

The polynomial ring $R\left[X\right]$ is reduced.

Corollary 1. 

If we let R be a reduced ring, and if $ab=0$ for some $a,b\in R$, then $acb=0$ for all $c\in R$.

Proof. 

If we suppose $ab=0$, then for any $c\in R$, we have $\left(ab\right)c=0$, i.e., $abc=0$. By Lemma 1 (1), it follows that $acb=0$. □

Lemma 2. 

If we let R be a reduced ring, and let $m,n$ be nonnegative integers that are not both zero, then the subring $U_n^m\left(R\right)$ of the upper triangular matrix ring $T_{m+n}\left(R\right)$ has simple 0 multiplication.

Proof. 

If we assume that $A=\left(\begin{array}{cc}\overline{A}& G\\ ineO& a_1{I}_n\end{array}\right)$ and $B=\left(\begin{array}{cc}\overline{B}& H\\ ineO& b_1{I}_n\end{array}\right)$ are in $U_n^m\left(R\right)$, satisfying

$$AB=\left(\begin{array}{cc}\overline{A}\overline{B}& \overline{A}H+G{b}_1\\ ineO& a_1{b}_1{I}_n\end{array}\right)=O,$$

where ${\displaystyle \overline{A}=a_1{I}_m+\sum _{k=1}^{m-1}{a}_{k+1}{V}_m^k}$ and ${\displaystyle \overline{B}=b_1{I}_m+\sum _{k=1}^{m-1}{b}_{k+1}{V}_m^k}$ are $m\times m$ matrices, $G={\left(a_{ij}\right)}_{m\times n}$ and $H={\left(b_{ij}\right)}_{m\times n}$ are $m\times n$ matrices, and $I_n$ is the n-by-n identity matrix, then we infer that $\overline{A}\overline{B}=O$, $\overline{A}H+G{b}_1=O$, and $a_1{b}_1=0$.

To show that $T_{m+n}\left(R\right)$ has simple 0 multiplication, we only need to verify the three systems of equations—$(\star )$, $(\ast )$, and $(\ast \ast )$:

(⋆)

$a_i{b}_j=0,\forall i+j\in \{2,3,\cdots ,m,m+1\}.$

(∗)

$a_{ij}{b}_1=0,\forall (i,j)\in \{1,2,\cdots ,m\}\times \{1,2,\cdots ,n\}.$

(∗∗)

$a_i{b}_{jk}=0,\forall 1\le i\le j\le m;\mathrm{and}\forall 1\le k\le n.$

It follows from $\overline{A}\overline{B}=O$ that the following equations hold:

$$\left\{\begin{array}{cc}{a}_1{b}_1=0,\hfill & \left(1\right)\hfill \\ a_1{b}_2+a_2{b}_1=0,\hfill & \left(2\right)\hfill \\ a_1{b}_3+a_2{b}_2+a_3{b}_1=0,\hfill & \left(3\right)\hfill \\ \cdots ,\hfill & \\ a_1{b}_{m-1}+a_2{b}_{m-2}+\cdots +a_{m-2}{b}_2+a_{m-1}{b}_1=0,\hfill & \left(4\right)\hfill \\ a_1{b}_m+a_2{b}_{m-1}+\cdots +a_{m-1}{b}_2+a_m{b}_1=0,\hfill & \left(5\right)\hfill \end{array}\right.$$

It follows from Corollary 1 that Equation (1), $a_1{b}_1=0$, implies $a_{1}c{b}_1=0$ for any $c\in R$ as R is reduced. By multiplying Equation (2) on the right-hand side by $b_1$, we have $a_1{b}_2{b}_1+a_2{b}_1^2=0$. It follows that $a_2{b}_1^2=0$ as $a_1{b}_2{b}_1=0$, and so $a_2{b}_1=0$ (see Lemma 1), which implies $a_1{b}_2=0$. That is, we have shown that $a_1{b}_2=0$ and $a_2{b}_1=0$ in Equation (2). Now, assume that we have shown $a_i{b}_j=0$ for $i+j=2,3,\cdots ,m$. In the next section, we will show that $a_i{b}_j=0$ for $i+j=m+1$. By multiplying Equation (5) on the right-hand side by $b_1$, we have

$$a_1{b}_m{b}_1+a_2{b}_{m-1}{b}_1+\cdots +a_{m-1}{b}_2{b}_1+a_m{b}_1^2=0.$$

By our assumption that $a_i{b}_j=0$ for $i+j=2,3,\cdots ,m$, from Corollary 1, we obtain that $a_{i}x{b}_1=0$ for any $i=1,2,3,\cdots ,m-1$ and any element $x\in R$. Thus, we have $a_1{b}_m{b}_1+a_2{b}_{m-1}{b}_1+\cdots +a_{m-1}{b}_2{b}_1=0$, which yields $a_m{b}_1^2=0$, and hence, $a_m{b}_1=0$ by Lemma 1. Now, Equation (5) can be simplified as

$$a_1{b}_m+a_2{b}_{m-1}+\cdots +a_{m-2}{b}_3+a_{m-1}{b}_2=0.$$

By multiplying the equation above on the right-hand side by $b_2$, we have $a_1{b}_m{b}_2+a_2{b}_{m-1}{b}_2+\cdots +a_{m-2}{b}_3{b}_2+a_{m-1}{b}_2^2=0.$ By our assumption that $a_i{b}_j=0$ for $i+j=2,3,\cdots ,m$, from Corollary 1, we obtain that $a_{i}x{b}_2=0$ for any $i=1,2,3,\cdots ,m-2$ and any element $x\in R$. Thus, we have $a_1{b}_m{b}_2+a_2{b}_{m-1}{b}_2+\cdots +a_{m-2}{b}_3{b}_2=0$, which yields $a_{m-1}{b}_2^2=0$, and hence, $a_{m-1}{b}_2=0$ by Lemma 1. Inductively, we can show that $a_{m+1-j}{b}_j=0$ for $j=1,2,3,\cdots ,m$. Thus, we have shown that

$$(\star )a_i{b}_j=0,\forall i+j\in \{2,3,\cdots ,m,m+1\}.$$

By multiplying equation $\overline{A}H+G{b}_1=O$ of matrices on the right-hand side by $b_1$, we have $\overline{A}H{b}_1+G{b}_1^2=O$. It should be noted that $a_i{b}_1=0$ for all $i\in \{1,2,3,\cdots ,m\}$, which we have already implied with $\overline{A}{b}_1=0$. Since R is reduced, it follows easily that $\overline{A}H{b}_1=O$, and so the matrix $G{b}_1^2=O$, which obviously implies that the following holds:

$$(\ast )a_{ij}{b}_1=0,\forall (i,j)\in \{1,2,\cdots ,m\}\times \{1,2,\cdots ,n\}.$$

That is, $G{b}_1=O$, and consequently, $\overline{A}H=O$. We then obtain that the first column entries in the matrix $\overline{A}H$ satisfy the following system of equations:

$$\left\{\begin{array}{cc}{a}_1{b}_{11}+a_2{b}_{21}+\cdots +a_{m-1}{b}_{m-1,1}+a_m{b}_{m1}=0,\hfill & \left(6\right)\hfill \\ a_1{b}_{21}+\cdots +a_{m-2}{b}_{m-1,1}+a_{m-1}{b}_{m1}=0,\hfill & \left(7\right)\hfill \\ \cdots ,\hfill & \\ a_1{b}_{m-1,1}+a_2{b}_{m1}=0,\hfill & \left(8\right)\hfill \\ a_1{b}_{m1}=0,\hfill & \left(9\right)\hfill \end{array}\right.$$

It follows from Corollary 1 that Equation (9) implies that $a_{1}c{b}_{m1}=0$ for any $c\in R$ as R is reduced. By multiplying Equation (8) on the right-hand side by $b_{m1}$, we have $a_1{b}_{m-1,1}{b}_{m1}+a_2{b}_{m1}^2=0$. It follows that $a_2{b}_{m1}^2=0$, and so $a_2{b}_{m1}=0$ (see Lemma 1), which then implies that $a_1{b}_{m-1,1}=0$. That is, we have shown all the addition items, $a_1{b}_{m-1,1}=0$ and $a_2{b}_{m1}=0$, in Equation (8), and $a_1{b}_{m1}=0$ in Equation (9). Now, it is assumed that we have shown all the addition items of $a_i{b}_{j1}=0$ in Equations (7), …, and (8) for $1\le i<j\le m$. In the next section, we will show all the addition items of $a_i{b}_{i1}=0$ in Equation (6) for $i=1,2,\cdots ,m$. By multiplying Equation (6) on the right-hand side by $b_{m1}$, we have

$$(a_1{b}_{11}{b}_{m1}+a_2{b}_{21}{b}_{m1}+\cdots +a_{m-1}{b}_{m-1,1}{b}_{m1})+a_m{b}_{m1}^2=0.$$

By our assumption that $a_i{b}_{m1}=0$ for $i=1,2,3,\cdots ,m-1$, from Corollary 1, we obtain that $a_{i}x{b}_{m1}=0$ for any $i=1,2,3,\cdots ,m-1$ and any element $x\in R$. Thus, we have $a_1{b}_{11}{b}_{m1}+a_2{b}_{21}{b}_{m1}+\cdots +a_{m-1}{b}_{m-1,1}{b}_{m1}=0$, which yields $a_m{b}_{m1}^2=0$, and hence, $a_m{b}_{m1}=0$ by Lemma 1. Now, Equation (6) can be simplified as

$$a_1{b}_{11}+a_2{b}_{21}+\cdots +a_{m-1}{b}_{m-1,1}=0.$$

By multiplying the equation above on the right-hand side by $b_{m-1,1}$, we have

$$(a_1{b}_{11}{b}_{m-1,1}+a_2{b}_{21}{b}_{m-1,1}+\cdots +a_{m-2}{b}_{m-2,1}{b}_{m-1,1})+a_{m-1}{b}_{m-1,1}^2=0.$$

By our assumption that $a_i{b}_{m-1,1}=0$ for $i=1,2,3,\cdots ,m-2$, from Corollary 1, we obtain that $a_{i}x{b}_{m-1,1}=0$ for any $i=1,2,3,\cdots ,m-2$ and any element $x\in R$. Thus, we have $a_1{b}_{11}{b}_{m-1,1}+a_2{b}_{21}{b}_{m-1,1}+\cdots +a_{m-2}{b}_{m-2,1}{b}_{m-1,1}=0$, which yields $a_{m-1}{b}_{m-1,1}^2=0$, and hence, $a_{m-1}{b}_{m-1,1}=0$ by Lemma 1. Inductively, we can show that $a_i{b}_{i1}=0$ for $i=1,2,3,\cdots ,m-2$. Thus, we have shown that $a_i{b}_{i1}=0$ in Equation (6) for $i=1,2,\cdots ,m$. This proves all the addition items

$$a_i{b}_{j1}=0,\forall 1\le i\le j\le m,$$

which lie in each entry of the first column of the matrix $\overline{A}H$.

Continuing this process, we then obtain all the addition items of

$$(\ast \ast )a_i{b}_{jk}=0,\forall 1\le i\le j\le m;\mathrm{and}\forall 1\le k\le n,$$

which lie in each entry of the matrix $\overline{A}H$.

Combining the three systems of equations—$(\star )$, $(\ast )$, and $(\ast \ast )$—we conclude that the ring $U_n^m\left(R\right)$ has simple 0 multiplication. □

Now, we present one of the main results of this paper.

Theorem 1. 

If we let R be a reduced ring, and let $m,n$ be two nonnegative integers that are not both zero, then $U_n^m\left(R\right)$ is a semicommutative ring.

Proof. 

Let $A=\left(\begin{array}{cc}\overline{A}& G\\ ineO& a_1{I}_n\end{array}\right),B=\left(\begin{array}{cc}\overline{B}& H\\ ineO& b_1{I}_n\end{array}\right)\in U_n^m\left(R\right)$ be two matrices as in Lemma 2 such that $AB=\left(\begin{array}{cc}\overline{A}\overline{B}& \overline{A}H+G{b}_1\\ ineO& a_1{b}_1{I}_n\end{array}\right)=O$. Now, for any matrix

$$C=\left(\begin{array}{cc}\overline{C}& K\\ ineO& c_1{I}_n\end{array}\right)\in U_n^m\left(R\right),$$

where $\overline{C}=c_1{I}_m+{\sum }_{k=1}^{m-1}{b}_{k+1}{V}_m^k$ is an m-by-m matrix, and $K=\left(k_{ij}\right)$ is an m-by-n matrix, it is easily seen that

$$\begin{array}{cc}\hfill ACB& =\left(\begin{array}{cc}\overline{A}& G\\ ineO& a_1{I}_n\end{array}\right)\left(\begin{array}{cc}\overline{C}& K\\ ineO& c_1{I}_n\end{array}\right)\left(\begin{array}{cc}\overline{B}& H\\ ineO& b_1{I}_n\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}\overline{A}\overline{C}& \overline{A}K+G{c}_1\\ ineO& a_1{c}_1{I}_n\end{array}\right)\left(\begin{array}{cc}\overline{B}& H\\ ineO& b_1{I}_n\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}\overline{A}\overline{C}\overline{B}& \overline{A}\overline{C}H+(\overline{A}K+G{c}_1)b_1\\ ineO& a_1{c}_1{b}_1{I}_n\end{array}\right)\hfill \\ \hfill & =O.\hfill \end{array}$$

The last equation follows from Lemma 2. In fact, since R is reduced, this implies that $acb=0$ for any $c\in R$ when $ab=0$ for some $a,b\in R$. We readily infer that $\overline{A}K{b}_1=O$ and $G{c}_1{b}_1=O$ as $\overline{A}{b}_1=O$ and $G{b}_1=O$. By the proof of $\overline{A}H=O$ in Lemma 2, we obtain that $\overline{A}\overline{C}H=O$. It follows easily that $\overline{A}\overline{C}\overline{B}=O$ and $a_1{c}_1{b}_1=0$ under the conditions. Hence, the ring $U_n^m\left(R\right)$ is semicommutative. □

Notation

Let $m,n$ be two nonnegative integers that are not both zero. In [11,17], the authors define the subring $T_{m,n}\left(R\right)$ of the upper triangular matrix ring to consist of all the following $(m+n)\times (m+n)$ matrices over R:

$$\left(\begin{array}{cccccccccc}{a}_1& a_2& \cdots & a_{m-1}& a_m& a_{11}& a_{12}& \cdots & a_{1,n-1}& a_{1n}\\ 0& a_1& \ddots & \ddots & a_{m-1}& a_{21}& a_{22}& \cdots & a_{2,n-1}& a_{2n}\\ ⋮& ⋮& \ddots & \ddots & ⋮& ⋮& ⋮& \cdots & ⋮& ⋮\\ 0& \cdots & 0& a_1& a_2& a_{m-1,1}& a_{m-1,2}& \cdots & a_{m-1,n-1}& a_{m-1,n}\\ 0& \cdots & 0& 0& a_1& \begin{array}{c}\hfill a_{m1}\end{array}& \begin{array}{c}\hfill a_{m2}\end{array}& \begin{array}{c}\hfill \cdots \end{array}& \begin{array}{c}\hfill a_{m,n-1}\end{array}& \begin{array}{c}\hfill a_{mn}\end{array}\\ ine0& \cdots & 0& 0& 0& a_1& \begin{array}{c}\hfill a_{m1}\end{array}& \cdots & \begin{array}{c}\hfill a_{m,n-2}\end{array}& \begin{array}{c}\hfill a_{m,n-1}\end{array}\\ 0& \cdots & 0& 0& 0& 0& a_1& \ddots & \ddots & \begin{array}{c}\hfill a_{m,n-2}\end{array}\\ ⋮& \cdots & ⋮& ⋮& ⋮& ⋮& \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& 0& 0& 0& 0& \ddots & a_1& \begin{array}{c}\hfill a_{m1}\end{array}\\ 0& \cdots & 0& 0& 0& 0& 0& \cdots & 0& a_1\end{array}\right)$$

One will see that the following result is of little independent interest in this paper.

Remark 1. 

Wang (Theorem 2.3 [11]) proved that $T_{m,n}\left(R\right)$ is semicommutative for any reduced ring R.

The following results can be proven similarly to Lemma 2 and Theorem 1.

Lemma 3. 

If we let R be a reduced ring, and let $m,n$ be two nonnegative integers that are not both zero, then the ring $W_n^m\left(R\right)$ has simple 0 multiplication.

We now present the second main result of this paper.

Theorem 2. 

If we let R be a reduced ring, and let $m,n$ be nonnegative integers that are not both zero, then $W_n^m\left(R\right)$ is semicommutative.

Following arguments similar to those in Theorems 1 and 2, we obtain the following result on Armendariz rings.

Proposition 1. 

If we let R be a reduced ring, and let $m,n$ be two nonnegative integers that are not both zero, then $U_n^m\left(R\right)$ and $W_n^m\left(R\right)$ are Armendariz rings.

Proof. 

We only need to show that $U_n^m\left(R\right)$ is Armendariz since the other result can be proved similarly. Since R is reduced, it follows from Lemma 1 that $R\left[X\right]$ is reduced. The map $\theta :U_n^m\left(R\right)\left[X\right]\to U_n^m\left(R\left[X\right]\right)$ is defined by

$$\theta (A_0+A_{1}X+A_2{X}^2+\cdots +A_m{X}^m)=\left(f_{ij}\left(X\right)\right),$$

where $f_{ij}\left(X\right)={\sum }_{k=0}^m{a}_{ij}^{\left(k\right)}{X}^k$ and $a_{ij}^{\left(k\right)}$ denotes the $(i,j)$ entry of the matrix $A_k$ in $U_n^m\left(R\right)$. One may verify that $\theta$ is an isomorphism of rings, and so $\theta \left(\left(U_n^m\left(R\right)\right)\left[X\right]\right)=U_n^m\left(R\left[X\right]\right)$. In view of Lemma 2, we get that $U_n^m\left(R\left[X\right]\right)$ has simple 0 multiplication. Then, we infer from Theorem (Theorem 2.1 (ii) [17]) that $U_n^m\left(R\right)$ is an Armendariz ring, as desired. □

Remark 2. 

It is proven by Wang, Puczyłowski, and Li (Theorem 2.3 (3) [17]) that the ring $T_{m,n}\left(R\right)$ over a reduced ring R is Armendariz.

Corollary 2. 

If we let R be a reduced ring and $m\ge 2$ be an integer, then

$$U_0^m=W_m^0=\left\{\left(\begin{array}{ccccc}{a}_1& a_2& \cdots & a_{m-1}& a_m\\ 0& a_1& \ddots & \ddots & a_{m-1}\\ ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& a_1& a_2\\ 0& \cdots & 0& 0& a_1\end{array}\right)|a_i\in R\right\},$$

$$U_1^m=\left\{\left(\begin{array}{cccccc}{a}_1& a_2& \cdots & a_{m-1}& a_m& a_{11}\\ 0& a_1& \ddots & \ddots & a_{m-1}& a_{21}\\ ⋮& ⋮& \ddots & \ddots & ⋮& ⋮\\ 0& \cdots & 0& a_1& a_2& a_{m-1,1}\\ 0& \cdots & 0& 0& a_1& a_{m1}\\ ine0& \cdots & 0& 0& 0& a_1\end{array}\right)|a_i,a_{i1}\in R\right\}$$

and

$$W_m^1=\left\{\left(\begin{array}{cccccc}{a}_1& a_{11}& a_{12}& \cdots & a_{1,m-1}& a_{1m}\\ ine0& a_1& a_2& \cdots & a_{m-1}& a_m\\ 0& 0& a_1& \ddots & \ddots & a_{m-1}\\ ⋮& ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& 0& 0& \cdots & a_1& a_2\\ 0& 0& 0& \cdots & 0& a_1\end{array}\right)|a_i,a_{1i}\in R\right\}$$

are semicommutative and Armendariz rings.

Proof. 

The results follow directly from Theorems 1 and 2 and Proposition 1. □

It should be noted that every subring of a semicommutative ring is semicommutative, and every subring of an Armendariz ring is Armendariz. Now, the following set is considered:

$${\Lambda }_1=\left\{\left(\begin{array}{cc}{\displaystyle a_1{I}_m+\sum _{k=1}^{m-1}{a}_{k+1}{V}_m^k}& O\\ ineO& a_1{I}_n\end{array}\right)|a_1,a_2,\cdots ,a_m\in R\right\},$$

which forms a subring of $U_n^m\left(R\right)$ for $n\ge 1$. It follows that ${\Lambda }_1$ is semicommutative (resp. Armendariz). One can readily check that, in fact, there is a ring isomorphism ${\Lambda }_1\cong U_0^m\left(R\right)$. We also remark that the ring $U_0^m\left(R\right)$ precisely consists of all upper triangular Toeplitz matrices over R.

It follows from Wang (Theorem 2.6 [11]) (resp. Corollary 2.6 [17]) that $U_1^m\left(R\right)$ and $W_m^1\left(R\right)$ are maximal semicommutative (resp. Armendariz) subrings of $T_{m+1}\left(R\right)$ for every $m\ge 2$ over a reduced ring R.

In what follows, we aim to identify certain “relatively maximal” semicommutative (resp. Armendariz) subrings of matrix rings over R. Here, the term “relatively maximal" refers to a subring that is maximal within a particular family of subrings of R, though it need not be a maximal subring of R itself. Lee and Zhou have also studied this topic on Armendariz before (see Propositions 1.8 and 1.9 [16]). To this end, for integers $m\ge 2$, $n\ge 2$, and $2\le k\le n$, we define the set

$$U_{n,k}^m\left(R\right)=\left\{\left(\begin{array}{cc}{\displaystyle a_1{I}_m+\sum _{j=1}^{m-1}{a}_{j+1}{V}_m^j}& M\\ ineO& a_1{I}_n+a{E}_{1,k}\end{array}\right)|a,a_1,a_2,\cdots ,a_m\in R\right\},$$

where $E_{1,k}$ denotes the matrix unit in $M_n\left(R\right)$ and M ranges over all $m\times n$ matrices over R. One may verify that $U_{n,k}^m\left(R\right)$ is a subring of $T_{m+n}\left(R\right)$. It should be noted that $U_{n,k}^m\left(R\right)=U_n^m\left(R\right)+R{E}_{m+1,m+k}$, where $E_{m+1,m+k}$ is the matrix unit in $M_{m+n}\left(R\right)$; hence, $U_n^m\left(R\right)$ is a subring of $U_{n,k}^m\left(R\right)$.

Theorem 3. 

If we let R be a reduced ring, and let $m\ge 2$ and $n\ge 2$ be two integers, then $U_n^m\left(R\right)$ is a maximal semicommutative subring of $U_{n,k}^m\left(R\right)$ for any $2\le k\le n$.

Proof. 

If we suppose that $U_n^m\left(R\right)$ is not a maximal semicommutative subring of $U_{n,k}^m\left(R\right)$ for some $2\le k\le n$, then there is a semicommutative subring $U\subseteq U_{n,k}^m\left(R\right)$ such that U properly contains $U_n^m\left(R\right)$, and so there is an element $0\ne a\in R$ such that $a{E}_{m+1,m+k}\in U$. Now, if we let $A=E_{1,m}\in U_n^m\left(R\right)\subseteq U$ and $B=a{E}_{m+1,m+k}\in U$, then $AB=0$, but $ACB=a{E}_{1,m+k}\ne 0$, where $C=E_{m,m+1}\in U_n^m\left(R\right)\subseteq U$, and $E_{i,j}$ denotes the matrix units of $M_{m+n}\left(R\right)$. Thus, U is not semicommutative, and so it is a contradiction. This completes the proof. □

Theorem 4. 

If we let R be a reduced ring, and let $m\ge 2$ and $n\ge 2$ be two integers, then $U_n^m\left(R\right)$ is a maximal Armendariz subring of $U_{n,k}^m\left(R\right)$ for any $2\le k\le n$.

Proof. 

If we suppose that $U_n^m\left(R\right)$ is not a maximal Armendariz subring of $U_{n,k}^m\left(R\right)$ for some $2\le k\le n$, then there is an Armendariz subring $U\subseteq U_{n,k}^m\left(R\right)$ such that U properly contains $U_n^m\left(R\right)$, and so there is an element $0\ne a\in R$ such that $a{E}_{m+1,m+k}\in U$. Now, if we let $f\left(X\right)=E_{1,m}-E_{1,m+1}X\in U_n^m\left(R\right)\left[X\right]\subseteq U\left[X\right]$ and $g\left(X\right)=a{E}_{m+1,m+k}+a{E}_{m,m+k}X\in U\left[x\right]$, then $f\left(X\right)g\left(X\right)=0$, but $E_{1,m}·a{E}_{m,m+k}=a{E}_{1,m+k}\ne 0$, where $E_{i,j}$ denotes the matrix units of $M_{m+n}\left(R\right)$. Thus, U is not Armendariz, and so it is a contradiction. This completes the proof. □

In the following, we define a set

$$P_{m+n}\left(R\right)=\left\{\left(\begin{array}{cc}{\displaystyle a_1{I}_m+\sum _{k=1}^{m-1}{a}_{k+1}{V}_m^k}& M\\ ineO& {\displaystyle a_1{I}_n+\sum _{k=1}^{n-1}{b}_{k+1}{V}_n^k}\end{array}\right)|a_i,b_j\in R\right\}$$

of $T_{m+n}\left(R\right)$ with $m,n\ge 2$, where M runs over all m-by-n matrices over R. It is readily checked that $P_{m+n}\left(R\right)$ forms a ring, and so it is a subring of $T_{m+n}\left(R\right)$.

Remark 3. 

If we let $m\ge 2$ and $n\ge 2$ be two integers, then as subrings of $T_{m+n}\left(R\right)$, we have $T_{m,n}\left(R\right)\subset P_{m+n}\left(R\right)$. It is proven in Theorem 2.6 [11] (resp. Corollary 2.6 [17]) that $T_{m,n}\left(R\right)$ is a maximal semicommutative (resp. maximal Armendariz) subring of $T_{m+n}\left(R\right)$, and so it is a maximal semicommutative (resp. maximal Armendariz) subring of $P_{m+n}\left(R\right)$.

It follows from the following result that there is another maximal semicommutative (resp. maximal Armendariz) subring of $P_{m+n}\left(R\right)$ other than $T_{m,n}\left(R\right)$.

Theorem 5. 

If we let R be a reduced ring, and let $m\ge 2$ and $n\ge 2$ be two integers, then $U_n^m\left(R\right)$ is a maximal semicommutative (Armendariz) subring of $P_{m+n}\left(R\right)$.

Proof. 

It is easy to see that $U_n^m\left(R\right)$ is a subring of $P_{m+n}\left(R\right)$. We only need to show that any subring satisfying $U_n^m\left(R\right)\subset P\subseteq P_{m+n}\left(R\right)$ is neither semicommutative nor Armendariz.

If we suppose that $U_n^m\left(R\right)\subset P\subseteq P_{m+n}\left(R\right)$, and P is a ring, then there is an element

$$\left(\begin{array}{cc}{\displaystyle a_1{I}_m+\sum _{k=1}^{m-1}{a}_{k+1}{V}_m^k}& G\\ ineO& {\displaystyle a_1{I}_n+\sum _{k=1}^{n-1}{c}_{k+1}{V}_n^k}\end{array}\right)\in P$$

such that not all the elements $c_2,c_3,\cdots c_n\in R$ are zero. We infer that

$$\begin{array}{cc}\hfill B:& =\left(\begin{array}{cc}O& O\\ ineO& {\displaystyle \sum _{k=1}^{n-1}{c}_{k+1}{V}_n^k}\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{\displaystyle a_1{I}_m+\sum _{k=1}^{m-1}{a}_{k+1}{V}_m^k}& G\\ ineO& {\displaystyle a_1{I}_n+\sum _{k=1}^{n-1}{c}_{k+1}{V}_n^k}\end{array}\right)-\left(\begin{array}{cc}{\displaystyle a_1{I}_m+\sum _{k=1}^{m-1}{a}_{k+1}{V}_m^k}& G\\ ineO& \begin{array}{c}\\ a_1{I}_n\\ \end{array}\end{array}\right)\hfill \end{array}$$

is in P as $U_n^m\left(R\right)\subseteq P$ is a subring of P and $B\ne O$. Consequently,

$$B=\left(\begin{array}{cc}O& O\\ ineO& {\displaystyle \sum _{k=l}^n{c}_k{V}_n^{k-1}}\end{array}\right)\in P,$$

where $l=\inf \left\{k\right|c_k\ne 0\}$.

We use $E_{i,j}$ to denote the matrix units of $M_{m+n}\left(R\right)$.

If we let $A=E_{1,m}\in U_n^m\left(R\right)\subseteq P$, then $AB=O$, but $ACB={\sum }_{k=l}^n{c}_k{E}_{1,m+k}\ne O$, where $C=E_{m,m+1}\in U_n^m\left(R\right)\subseteq P$. Thus, P is not semicommutative, and so it follows from Theorem 1 that $U_n^m\left(R\right)$ is a maximal semicommutative subring of $P_{m+n}\left(R\right)$.

Now, if we let $f\left(X\right)=E_{1,m}-E_{1,m+1}x\in U_n^m\left(R\right)\left[X\right]\subseteq P\left[X\right]$ and $g\left(x\right)=B+CX\in P\left[X\right]$, where $C={\sum }_{k=l}^n{c}_k{E}_{m,m+k}\in U_n^m\left(R\right)$, then $f\left(X\right)g\left(X\right)=O$, but $E_{1,m}·C={\sum }_{k=l}^n{c}_k{E}_{1,m+k}\ne O$. Thus, P is not Armendariz, and so it follows from Proposition 1 that $U_n^m\left(R\right)$ is a maximal Armendariz subring of $P_{m+n}\left(R\right)$. This completes the proof. □

In the next section, we define a set

$$W_n^{m,k}\left(R\right)=\left\{\left(\begin{array}{cc}{a}_1{I}_m+a{E}_{k,m}& G\\ ineO& {\displaystyle a_1{I}_n+\sum _{j=1}^{n-1}{a}_{j+1}{V}_n^j}\end{array}\right)|a,a_j\in R\right\}$$

of $T_{m+n}\left(R\right)$ with $m\ge 2$, $n\ge 2$, and $1\le k\le m-1$, where $E_{k,m}$ denotes the m-by-m matrix units, and $I_n$ is the n-by-n identity matrix. It is readily checked that $W_n^{m,k}\left(R\right)$ forms a ring, and so it is a subring of $T_{m+n}\left(R\right)$.

By an analogous proof to Theorems 3, 4, and 5, we immediately obtain the result below without giving the proof.

Theorem 6. 

If we let R be a reduced ring, and let $m\ge 2$ and $n\ge 2$ be two integers, then the following statements hold:

(1) 

The ring $W_n^m\left(R\right)$ is a maximal semicommutative (Armendariz) subring of $W_n^{m,k}\left(R\right)$.

(2) 

The ring $W_n^m\left(R\right)$ is a maximal semicommutative (Armendariz) subring of $P_{m+n}\left(R\right)$.

At the end of this section, we provide an example below.

Example 1. 

If we let R be a reduced ring, then

$$U_2^2=\left\{\left(\begin{array}{cccc}{a}_1& a_{12}& a_{13}& a_{14}\\ 0& a_1& a_{23}& a_{24}\\ 0& 0& a_1& \begin{array}{c}\hfill 0\end{array}\\ 0& 0& 0& a_1\end{array}\right)|a_1,a_{ij}\in R\right\},$$

$$S_4\left(R\right)=\left\{\left(\begin{array}{cccc}{a}_1& a_{12}& a_{13}& a_{14}\\ 0& a_1& a_{23}& a_{24}\\ 0& 0& a_1& a_{34}\\ 0& 0& 0& a_1\end{array}\right)|a_1,a_{ij}\in R\right\},$$

$$T_{2,2}=\left\{\left(\begin{array}{cccc}{a}_1& a_{12}& a_{13}& a_{14}\\ 0& a_1& \begin{array}{c}\hfill a_2\end{array}& a_{24}\\ 0& 0& a_1& \begin{array}{c}\hfill a_2\end{array}\\ 0& 0& 0& a_1\end{array}\right)|a_1,a_2,a_{ij}\in R\right\}.$$

By Theorems 3 and 4, it follows that $U_2^2\left(R\right)$ is a both maximal semicommutaive and maximal Armendariz subring of $S_4\left(R\right)$ as $S_4\left(R\right)=U_{2,2}^2$. By Remark 3, $T_{2,2}\left(R\right)$ is a maximal semicommutative (resp. maximal Armendariz) subring of $S_4\left(R\right)$ as $S_4\left(R\right)=P_{m+n}\left(R\right)$.

3. Applications

If we let M be a left R-module and recall from [18,19] that M is called a semicommutative left R-module if for any $a\in R$ and $m\in M$, $am=0$ implies $arm=0$ for any $r\in R$, as well as recall from [20] that M is called an Armendariz left R-module if whenever polynomials $f\left(X\right)=a_0+a_{1}X+a_2{X}^2+\cdots +a_s{X}^s\in R\left[X\right]$ and $g\left(X\right)=m_0+m_{1}X+m_2{X}^2+\cdots +m_t{X}^t\in M\left[X\right]$ satisfy $f\left(X\right)g\left(X\right)=0$, then $a_i{m}_j=0$ for each $0\le i\le s$ and $0\le j\le t$, then, similarly, one can define a semicommutative (resp. an Armendariz) right R-module.

Given a ring R and two positive integers $m,n$, the set of all m-by-n matrices over R is denoted by $M_{m\times n}\left(R\right)$. Theorems 1 and 2 and Proposition 1 can be used to give some new semicommutative and Armendariz R-modules.

Proposition 2. 

If we let R be a reduced ring, and $m,n$ be two positive integers, then the following statements hold:

(1) 

$M_{m\times n}\left(R\right)$ is a semicommutative (resp. an Armendariz) left $U_0^m\left(R\right)$-module.

(2) 

$M_{m\times n}\left(R\right)$ is a semicommutative (resp. an Armendariz) right $W_n^0\left(R\right)$-module.

Proof. 

It suffices to prove (1), and (2) can be proved similarly. Let

$$A=a_1{I}_m+\sum _{k=1}^{m-1}{a}_{k+1}{V}_m^k\in U_0^m\left(R\right)\mathrm{and}M\in M_{m\times n}\left(R\right)$$

satisfy $AM=O$, where $V_m=\left(\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& \ddots & \ddots & ⋮\\ 0& 0& 0& \cdots & 1\\ 0& 0& 0& \cdots & 0\end{array}\right)$ is a square matrix of order m. Then, we have

$$\left(\begin{array}{cc}A& O\\ ineO& a_1{I}_n\end{array}\right)\left(\begin{array}{cc}O& M\\ ineO& O\end{array}\right)=O$$

in $U_n^m\left(R\right)$. For any $B=b_1{I}_m+{\sum }_{k=1}^{m-1}{b}_{k+1}{V}_m^k\in U_0^m\left(R\right)$, it follows that

$$\left(\begin{array}{cc}A& O\\ ineO& a_1{I}_n\end{array}\right)\left(\begin{array}{cc}B& O\\ ineO& b_1{I}_n\end{array}\right)\left(\begin{array}{cc}O& M\\ ineO& O\end{array}\right)=O$$

since $U_n^m\left(R\right)$ is a semicommutative ring by Theorem 1. This implies that

$$ABM=O,$$

and so $M_{m\times n}\left(R\right)$ is a semicommutative left $U_0^m\left(R\right)$-module.

To show that $M_{m\times n}\left(R\right)$ is an Armendariz left $U_0^m\left(R\right)$-module, we let

$$f\left(X\right)=A_0+A_{1}X+A_2{X}^2+\cdots +A_s{X}^s\in U_0^m\left(R\right)\left[X\right]$$

and

$$g\left(X\right)=M_0+M_{1}X+M_2{X}^2+\cdots +M_t{X}^t\in M_{m\times n}\left(R\right)\left[X\right],$$

satisfying $f\left(X\right)g\left(X\right)=O$; then, it is easy to check that

$$F\left(X\right)G\left(X\right)=0,$$

where

$$\begin{array}{cc}\hfill F\left(X\right)& =\left(\begin{array}{cc}f\left(X\right)& O\\ ineO& a_0{I}_n+a_{1}X{I}_n+\cdots +a_s{X}^s{I}_n\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{A}_0& O\\ ineO& a_0{I}_n\end{array}\right)+\left(\begin{array}{cc}{A}_1& O\\ ineO& a_1{I}_n\end{array}\right)X+\cdots +\left(\begin{array}{cc}{A}_s& O\\ ineO& a_s{I}_n\end{array}\right)X^s\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill G\left(X\right)& =\left(\begin{array}{cc}O& g\left(X\right)\\ ineO& O\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}O& M_0\\ ineO& O\end{array}\right)+\left(\begin{array}{cc}O& M_1\\ ineO& O\end{array}\right)X+\cdots +\left(\begin{array}{cc}O& M_t\\ ineO& O\end{array}\right)X^t\hfill \end{array}$$

are both in $U_n^m\left(R\right)\left[X\right]$, and $a_i$ is the diagonal element of $A_i\in U_0^m\left(R\right)$. By Proposition 1, the ring $U_n^m\left(R\right)$ is Armendariz. This implies that

$$\left(\begin{array}{cc}{A}_i& O\\ ineO& a_i{I}_n\end{array}\right)\left(\begin{array}{cc}O& M_j\\ ineO& O\end{array}\right)=\left(\begin{array}{cc}O& A_i{M}_j\\ ineO& O\end{array}\right)=O$$

for each $0\le i\le s$ and $0\le j\le t$, and so we have $A_i{M}_j=O$ for each $0\le i\le s$ and $0\le j\le t$, as desired. □

Remark 4. 

In fact, $M_{m\times 1}\left(R\right)$ is a free left $U_0^m\left(R\right)$-module, and so is $M_{m\times n}\left(R\right)$. Thus, it is easily seen that if R is reduced, then any free left $U_0^m\left(R\right)$-module $M_{m\times 1}{\left(R\right)}^{(\Lambda )}$ is a semicommutative (resp. an Armendariz) left $U_0^m\left(R\right)$-module, where $\Lambda$ is an index set.

Remark 5. 

If we let $\mathbb{Z}\left[\omega \right]=\left\{a+b\omega |a,b\in \mathbb{Z}\right\}$ be the ring of Eisenstein integers, which is actually a two-dimensional algebraic equation over $\mathbb{Z}$, where $\omega =-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{3}}{2}}i$ is a primitive third root of unity, since $\mathbb{Z}\left[\omega \right]$ is a Euclidean domain, it is reduced. Thus, the matrix rings $U_n^m\left(R\right)$ and $W_n^m\left(R\right)$ preserve semicommutativity and Armendariz properties when R is taken to be the Eisenstein ring $\mathbb{Z}\left[\omega \right]$. Eisenstein integers possess significant applications in the fields of coding, communications, and signal processing (see [21]).