A note on lower nil M-Armendariz rings

Abstract

In this article, we prove some results for lower nil $M$-Armendariz ring. Let $M$ be a strictly totally ordered monoid and $I$ be a semicommutative ideal of $R$. If $\frac{R}{I}$ is a lower nil $M$-Armendariz ring, then $R$ is lower nil $M$-Armendariz. Similarly, for above $M$, if $I$ is 2-primal with $N_{\ast }(R)\subseteq I$ and $R/I$ is $M$-Armendariz, then $R$ is a lower nil $M$-Armendariz ring. Further, we find that if $M$ is a monoid and $N$ a u.p.-monoid where $R$ is a $2$-primal $M$-Armendariz ring, then $R[N]$ is a lower nil $M$-Armendariz ring.

Keywords:

• Armendariz ring
• 2-primal ring
• Semicommutative ring
• Skew triangular matrix ring
• u.p.-monoid
• M-Armendariz ring
• Lower nil M-Armendariz ring

JEL classification:

• 16S36
• 16N60
• 16Y99

Public Interest Statement

In this manuscript, we proved some results for lower nil $M$- Armendariz rings which is introduced by Alhevaz and Hashemi in 2016. In fact, they discussed in their work about Upper nilradicals and Lower Nilradicals of unique product monoid rings in connection with the famous question of Amitsur of whether or not a polynomial ring over a nil coefficient ring is nil. In general, $N_{\ast }(R)[M]\ne N_{\ast }(R[M])$, but for lower nil $M$-Armendariz ring with semicommutative ring $R$, we proved that $R[M]$ is 2-primal ring. As the consequences, when ring $R$ is a lower nil $M$-Armendariz, then ring of triangular matrices and skew upper triangular matrices are lower nil $M$-Armendariz rings. The logical relationship among above-mentioned notions and other significant classes of Armendariz like rings can be helpful to provide the appropriate setting for obtaining results on radicals of the monoid rings of unique product monoids and also can be used to construct new classes of nil-Armendariz rings.

1. Introduction

Throughout this article, $R$ denotes an associative ring with identity unless otherwise stated. For a ring $R$, $N_{\ast }(R)$, $N^{\ast }(R)$ and $N(R)$ denote the prime radical (lower nilradical), upper nilradical and the set of nilpotent elements of $R$, respectively. It is known that $N_{\ast }(R)\subseteq N^{\ast }(R)\subseteq N(R)$. Here, $R[x]$ denotes the polynomial ring with an indeterminate $x$ over $R$ and $C_{f(x)}$ for the set of all coefficients of $f(x)\in R[x]$. $M_n(R)$ and $T_n(R)$ represent the full matrix ring and upper triangular matrix ring of order $n$ over the ring $R$, respectively.

A ring $R$ is said to be Armendariz ring if two polynomials $f(x),g(x)\in R[x]$, such that $f(x)g(x)=0$ implies $ab=0$ for each $a\in C_{f(x)}$, $b\in C_{g(x)}$. A ring $R$ is reduced if it has no nonzero nilpotent element. Armendariz (1974) himself proved that every reduced ring is satisfying above condition. Later, the term Armendariz ring was coined by Rege and Chhawchharia (1997). A ring $R$ is said to be semicommutative by Narbornne (1982), whenever $ab=0$ implies $aRb=0$ for $a,b\in R$ and it is 2-primal if $N_{\ast }(R)=N(R)$. Birkenmeier et al. (1993, Proposition 2.2) showed that the class of 2-primal rings is closed under subrings. A ring $R$ is called $NI$, by Marks (2001), if $N^{\ast }(R)=N(R)$.

Let $M$ be a monoid and $e$ the identity element of $M$. Then, $R[M]$ denotes the monoid ring of $M$ over the ring $R$. Due to Liu (2005), a ring $R$ is called $M$-Armendariz ring, whenever elements $\alpha =a_1{g}_1+a_2{g}_2+\cdots +a_n{g}_n$, $\beta =b_1{h}_1+b_2{h}_2+\cdots +b_m{h}_m\in R[M]$ satisfy $\alpha \beta =0$, then $a_i{b}_j=0$ for each $i$ and $j$. Clearly, every ring is an $M$-Armendariz ring, where $M=\left\{e\right\}$. But, if $S$ is a semigroup with trivial multiplication $st=0$ for all $s,t\in S$ and $M=S^1$, the semigroup $S$ with identity, then none of the ring is an $M$-Armendariz. If $M=\mathbb{N}\cup \left\{0\right\}$, then $R$ is $M$-Armendariz if and only if $R$ is Armendariz ring. Also, a monoid $M$ with $\left|M\right|\ge 2$ and $R$ is $M$-Armendariz, then $R$ is a p.p.-ring if and only if $R[M]$ is a p.p.-ring.

Alhevaz and Moussavi (2014) said a ring $R$ to be a nil $M$-Armendariz if $\alpha =a_1{g}_1+a_2{g}_2+\cdots +a_n{g}_n$, $\beta =b_1{h}_1+b_2{h}_2+\cdots +b_m{h}_m\in R[M]$ such that $\alpha \beta \in N(R)[M]$ implies $a_i{b}_j\in N(R)$ for each $i$ and $j$. Later, Alhevaz and Hashemi (2017) introduced the concept of upper and lower nil $M$-Armendariz ring relative to a monoid. A ring $R$ is upper $(\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r})$ nil $M$-Armendariz ring if whenever elements $\alpha =a_1{g}_1+a_2{g}_2+\cdots +a_n{g}_n$, $\beta =b_1{h}_1+b_2{h}_2+\cdots +b_m{h}_m\in R[M]$ such that $\alpha \beta \in N^{\ast }(R)[M]$ $(N_{\ast }(R)[M])$, then $a_i{b}_j\in N^{\ast }(R)$ $(a_i{b}_j\in N_{\ast }(R))$ for each $i$ and $j$.

2. Lower nil m-armendariz rings

Recall that a monoid $M$ is said to be u.p. monoid (unique product monoid) if for any two nonempty finite subsets $P,Q$ of $M$, there exists an element $s\in M$ uniquely presented in the form of $pq$ where $p\in P$ and $q\in Q$. The class of unique product monoids is quite large and important. For example, this class contains the right or left ordered monoids, torsion-free nilpotent groups, submonoids of a free group. Usefulness of unique product monoids and groups are extensively studied relating to the zero divisor problems. The ring theoretical property of unique product monoid has been established by many researchers in past, we refer Cheon and Kim (2012); Liu (2005); Okninski (1991); Passman (1977).

Proposition 2.1. For a unique product monoid $M$, every $2$-primal ring is a lower nil $M$-Armendariz ring.

Proof. Let $M$ be a unique product monoid. Take $\alpha =a_1{g}_1+a_2{g}_2+\dots +a_n{g}_n$, $\beta =b_1{h}_1+b_2{h}_2+\dots +b_m{h}_m\in R[M]$ such that $\alpha \beta \in N_{\ast }(R)[M]$. Then $\bar{\alpha }\bar{\beta }=\bar{0}$ in $R/N_{\ast }(R)[M]$. Since $R/N_{\ast }(R)$ is reduced, therefore by Liu (2005, Proposition 1.1), $R/N_{\ast }(R)$ is $M$-Armendariz ring. This implies $a_i{b}_j\in N_{\ast }(R)$ for each $i$ and $j$. Thus, $R$ is a lower nil $M$-Armendariz ring.□

Corollary 2.1. For any unique product monoid $M$, semicommutative ring is a lower nil $M$-Armendariz ring.

A monoid $M$ equipped with an order “$\le$” is said to be an ordered monoid if for any $r_1,r_2,s\in M$, $r_1\le r_2$ implies $r_{1}s\le r_{2}s$ and $s{r}_1\le s{r}_2$. Moreover, if $r_1<r_2$ implies $r_{1}s<r_{2}s$ and $s{r}_1<s{r}_2$, then $M$ is said to be strictly totally ordered monoid.

Since each strictly totally ordered monoid is a u.p. monoid, hence by Proposition 2.1, we have the following result.

Corollary 2.2 Let $M$ be a strictly totally ordered monoid. Then every $2$-primal ring is a lower nil $M$-Armendariz ring.

The following example shows that the condition u.p. monoid in Proposition 2.1 is not superfluous.

Example 2.1. Let $M=\left\{0,I,E_{11},E_{12},E_{21},E_{22}\right\}$, where $E_{ij}$ is the unit matrix of $M_2(\mathbb{Z})$ for each $1\le i,j\le 2$. Then $M$ is a monoid but it is not a u.p. monoid. Let $\alpha =1.E_{22}$ and $\beta =1.E_{11}-1.E_{12}$. Then $\alpha \beta =0\in N_{\ast }(\mathbb{Z})[M]$ but $1.1\notin N_{\ast }(\mathbb{Z})$. Hence, $\mathbb{Z}$ is a not a lower nil $M$-Armendariz ring.

Proposition 2.2. Let $M$ be a u.p. monoid, $R$ a lower nil $M$-Armendariz ring and $S$, a subring of $R$.

(1) If $N_{\ast }(S)\subseteq N_{\ast }(R)$, then $S$ is a lower nil $M$-Armendariz ring.

(2) If $R$ is an NI ring, then $S$ is a lower nil $M$-Armendariz ring.

Proof. (1) Let $\alpha \beta \in N_{\ast }(S)[M]$, where $\alpha ,\beta \in R[M]$. Then by assumption, $\alpha \beta \in N_{\ast }(R)[M]$. Since $R$ is a lower nil $M$-Armendariz ring, therefore $a_i{b}_j\in N_{\ast }(R)$ for each $i,j$. Also, $N_{\ast }(S)\subseteq N_{\ast }(R)$, so $N_{\ast }(S)=S\cap N_{\ast }(R)$ and hence $a_i{b}_j\in N_{\ast }(S)$ for each $i,j$.

(2) Since $R$ is NI, $N^{\ast }(R)=N(R)$. Also, for lower nil $M$-Armendariz ring, we have $N_{\ast }(R)=N^{\ast }(R)$. Therefore, $N^{\ast }(R)=N_{\ast }(R)=N(R)$. Thus, $S$ is a lower nil $M$-Armendariz ring.□

Proposition 2.3. For a u.p. monoid $M$, every lower nil $M$-Armendariz ring is nil $M$-Armendariz ring.

Proof. We have, $R$ is a lower nil $M$-Armendariz ring if and only if $R/N_{\ast }(R)$ is a $M$-Armendariz ring. Then, by using Alhevaz & Hashemi, 2017, Lemma 3.1(c)) and Hashemi (2013, Theorem 2.23), we obtain the result.□

But converse is not true. In this regard, we have an example.

Example 2.2. Suppose $M$ is a u.p. monoid and consider the ring given in Hwang, Jeon, and Lee (2006, Example 1.2). Let $S$ be a reduced ring, $n$ a positive integer and $R_n=U_{2^n}(S)$. Each $R_n$ is an NI ring by Hwang et al. (2006, Proposition 4.1(1)). Define a map $\sigma :R_n\to R_{n+1}$ by $A\mapsto \left(\begin{array}{cc}A& 0\\ 0& A\end{array}\right)$, then $R_n$ can be considered as subring of $R_{n+1}$ via $\sigma$ $(i.e.,A=\sigma (A)forA\in R_n)$. Notice that $D=\left\{R_n,{\sigma }_{nm}\right\}$, with ${\sigma }_{nm}={\sigma }^{m-n}$ whenever $n\le m$, is a direct system over $I=\left\{1,2,\dots \right\}$. Set $R=\hspace{0.17em}{R}_n$ be the direct limit of $D$. Then $R={\cup }_{n=1}^{\mathrm{\infty }}{R}_n$, and $R$ is an NI by Hwang et al. (2006, Proposition 1.1). By Alhevaz and Moussavi (2014, Theorem 2.2), NI ring is nil $M$-Armendariz ring. By Jeon, Kim, Lee, and Yoon (2009, Theorem 2.2(1)), $R$ is a semiprime ring, hence $N_{\ast }(R)=0$. Here $N^{\ast }(R)=\{m=(m_{ij})\in R|m_{ii}=0foralli\}$. Thus, by Alhevaz & Hashemi, 2017, Lemma 3.1(c)), $R$ is not a lower nil $M$-Armendariz ring.

Recall that a monoid $M$ is cancellative if for all $m,g,h\in M$, $g\ne h$ implies $mg\ne mh$ and $gm\ne hm$.

Proposition 2.4. Let $N$ be an ideal of a cancellative monoid $M$. If $R$ is a lower nil $N$-Armendariz ring, then $R$ is a lower nil $M$-Armendariz ring.

Proof. Suppose $\alpha =a_1{g}_1+a_2{g}_2+\dots +a_n{g}_n$, $\beta =b_1{h}_1+b_2{h}_2+\dots +b_m{h}_m\in R[M]$ such that $\alpha \beta \in N_{\ast }(R)[M]$. Take $k\in N$, then $k{g}_1,k{g}_2,\dots k{g}_n,h_{1}k,h_{2}k,\dots h_{m}k\in N$ and $k{g}_i\ne k{g}_j$, $h_{i}k\ne h_{j}k$ when $i\ne j$. Now,

(2.1) $\left(\sum _{i=1}^n{a}_i(k{g}_i)\right)\left(\sum _{j=1}^m{b}_j(h_{j}k)\right)\in N_{\ast }(R)[N].$(2.1)

Since $R$ is a lower nil $N$-Armendariz ring, therefore $a_i{b}_j\in N_{\ast }(R)$ for each $i,j$. Thus, $R$ is a lower nil $M$-Armendariz ring.□

Definition 2.1. Let $R$ be a ring and $S^{-1}R=\{u^{-1}a|u\in S,a\in R\}$ for a multiplicative closed subset $S$ consisting of all central regular elements of $R$. Then $S^{-1}R$ is a ring.

Proposition 2.5. Let $M$ be a monoid and $S,$ a multiplicatively closed subset of the ring $R$ consisting of central regular elements. Then $R$ is a lower nil $M$-Armendariz ring if and only if so is $S^{-1}R$.

Proof. Let $R$ be a lower nil $M$-Armendariz ring and $\alpha \beta \in N_{\ast }(S^{-1}R)[M]$ where $\alpha =p_1{g}_1+p_2{g}_2+\dots +p_n{g}_n$ and $\beta =q_1{h}_1+q_2{h}_2+\dots +q_m{h}_m\in (S^{-1}R)[M]$. Here, we consider $p_i=a_i{u}^{-1}$ and $q_j=b_j{v}^{-1}$, where $a_i,b_j\in R$ for each $i,j$ and $u,v\in S$. From $\alpha \beta \in N_{\ast }(S^{-1}R)[M]$, we have $(a_1{g}_1+a_2{g}_2+\dots +a_n{g}_n)(b_1{h}_1+b_2{h}_2+\dots +b_m{h}_m)\in N_{\ast }(R)[M]$. Since $R$ is the lower nil $M$-Armendariz ring, therefore $a_i{b}_j\in N_{\ast }(R)$. Moreover, $N_{\ast }(S^{-1}(R))=S^{-1}{N}_{\ast }(R)$, so $p_i{q}_j=a_i{u}^{-1}{b}_j{v}^{-1}=a_i{b}_j(uv{)}^{-1}\in N_{\ast }(S^{-1}(R))$. Thus, $S^{-1}R$ is a lower nil $M$-Armendariz ring.□

Converse is obvious, since $N_{\ast }(R)\subseteq N_{\ast }(S^{-1}R)$ and $R$ is a subring of $S^{-1}R$. Therefore, $R$ is a lower nil $M$-Armendariz ring.

Lemma 2.1. Let $M$ be a monoid with at least one non trivial element of finite order and $R$ be a ring such that $0\ne 1$. Then $R$ is not a lower nil $M$-Armendariz ring.

Proof. Suppose $e\ne g\in M$ has order $n$ and take $\alpha =1e+1g+\dots +1g^{n-1}$, $\beta =1e+(-1)g$, then $\alpha \beta =0$ i.e. $\alpha \beta \in N_{\ast }(R)[M]$ but $1\notin N_{\ast }(R)$. Hence, $R$ is not a lower nil $M$-Armendariz ring.□

Lemma 2.2. Let $N$ be a submonoid of the monoid $M$. If $R$ is a lower nil $M$-Armendariz ring, then $R$ is a lower nil $N$-Armendariz ring.

Definition 2.2. If $T(G)$ contains elements of finite order in an abelian group $G$, then $T(G)$ is a fully invariant subgroup of $G$. The group $G$ is said to be a torsion-free group if $T(G)=\left\{e\right\}$.

Theorem 2.1. Let $G$ be a finitely generated abelian group. Then the following conditions are equivalent:

(1) $G$ is torsion-free.

(2) There exists a ring $R$ with $\left|R\right|\ge 2$ such that $R$ is a lower nil $G$-Armendariz ring.

Proof. $(\mathbf{1})\Rightarrow (\mathbf{2}):$ If $G$ is a finitely generated abelian group with $T(G)=\left\{e\right\}$. Then $G\cong \mathbb{Z}\times \mathbb{Z}\times \dots \times \mathbb{Z}$. By Liu (2005, Lemma 1.13), $G$ is a u.p.-monoid. Also, by Corollary 2.1, $R$ is a lower nil $G$-Armendariz, for any semicommutative ring $R$ with $\left|R\right|\ge 2$.□

$(\mathbf{2})\Rightarrow (\mathbf{1}):$ Let $g\in T(G)$ and $g\ne e$. Then $H=⟨g⟩$ is a cyclic group of finite order. If a ring $R\ne \left\{0\right\}$ is a lower nil $G$-Armendariz ring, then $R$ is lower nil $H$-Armendariz ring by Lemma 2.2, which contradicts Lemma 2.1. Thus, every ring $R\ne \left\{0\right\}$ is not a lower nil $G$-Armendariz ring. Hence, $g=e$ and $T(G)=e$.

Proposition 2.6. For a monoid $M$, if $R$ is a semicommutative lower nil $M$-Armendariz ring, then $N_{\ast }(R)[M]=N_{\ast }(R[M])$.

Proof. Let $\alpha =a_1{g}_1+a_2{g}_2+\cdots +a_m{g}_m\in N_{\ast }(R)[M]$, where $a_i\in N_{\ast }(R)$ for each $1\le i\le m$. Now, $R{a}_{i}R\subseteq N(R)$ for each $1\le i\le m$. Then there exists $k_i$, a positive integer, such that $(R{a}_{i}R{)}^{k_i}=0$ for each fixed $a_i$, where $1\le i\le m$. Let $k>max\left\{k_i:1\le i\le m\right\}$. Let ${\beta }_1=b_1{h}_1+b_2{h}_2+\cdots +b_l{h}_l$ and ${\beta }_2=c_1{h}_1^{\prime }+c_2{h}_2^{\prime }+\cdots +c_n{h}_n^{\prime }\in R[M]$. Then ${\beta }_1\alpha {\beta }_2=(b_1{a}_1{c}_1{h}_1{g}_1{h}_{1\prime }+b_1{a}_1{c}_2{h}_1{g}_2{h}_{2\prime }+\cdots +b_1{a}_1{c}_n{h}_1{g}_1{h}_{n\prime })$ $+(b_1{a}_2{c}_1{h}_1{g}_2{h}_{1\prime }+\cdots +b_1{a}_2{c}_n{h}_1{g}_2{h}_{n\prime })+\cdots +(b_1{a}_m{c}_1{h}_1{g}_m{h}_{1^{\prime }}+\cdots +b_1{a}_m{c}_n{h}_1{g}_m{h}_{n^{\prime }})+(b_2{a}_1{c}_1{h}_2{g}_1{h}_{1^{\prime }}+\cdots +b_2{a}_m{c}_n{h}_2$ $g_1{h}_{n^{\prime }})+\cdots +(b_2{a}_m{c}_1{h}_2{g}_m{h}_{1\prime }+\cdots +b_2{a}_m{c}_n{h}_2{g}_m{h}_{n\prime })+\cdots +(b_l{a}_1{c}_1{h}_l{g}_1{h}_{1\prime }+\cdots +b_l{a}_1{c}_n{h}_l{g}_1{h}_{n\prime })$ $+\cdots +(b_l{a}_m{c}_1{h}_l{g}_m{h}_{1\prime }+\cdots +b_l{a}_m{c}_n{h}_l{g}_m{h}_{n\prime })$ . For brevity of notation, let ${\beta }_1\alpha {\beta }_2=A_1{t}_1+A_2{t}_2+\cdots +A_{lmn}{t}_{lmn}$. Then $({\beta }_1\alpha {\beta }_2{)}^{klmn}=(A_1{t}_1+A_2{t}_2+\cdots +A_{lmn}{t}_{lmn}{)}^{klmn}$ $=\sum _u(\sum _{t_{s_1}{t}_{s_2\dots t_{s_{klmn}}}=u}{A}_{s_1}{A}_{s_2}\dots A_{s_{klmn}})u$. Here, in $({\beta }_1\alpha {\beta }_2{)}^{klmn}$, each term of $A_{s_1}{A}_{s_2}\dots A_{s_{klmn}}$, there exist some $A_j$ for $1\le j\le lmn$, at least $k$ times. Therefore, we can replace $A_{s_1}{A}_{s_2}\dots A_{s_{klmn}}$ by $B_1{A}_j^{v_1}{B}_2{A}_j^{v_2}\dots B_w{A}_j^{v_p}$, where $v_1+v_2+\cdots +v_p>k$ and $B_q$ is a product of some elements from the set $\left\{A_1,A_2,\dots ,A_{lmn}\right\}$. Since $A^{v_1+v_2+\cdots +v_p}=0$ and $R$ is a semicommutative ring, so $B_1{A}_j^{v_1}{B}_2{A}_j^{v_2}\dots B_w{A}_j^{v_p}=0$. Therefore, $A_{s_1}{A}_{s_2}\dots A_{s_{klmn}}=0$ and hence $({\beta }_1\alpha {\beta }_2{)}^{klmn}=0$. Thus, $N_{\ast }(R)[M]\subseteq N_{\ast }(R[M])$.□

Conversely, let $\alpha =a_1{g}_1+a_2{g}_2+\cdots +a_m{g}_m\in N_{\ast }(R[M])$. Then, there exists a positive integer $s$ such that $({\beta }_1\alpha {\beta }_2{)}^s=0$, for each ${\beta }_1,{\beta }_2\in R[M].$ This implies, $({\beta }_1\alpha {\beta }_2{)}^s=(A_1{t}_1+A_2{t}_2+\cdots +A_{lmn}{t}_{lmn}{)}^s=0,$ i.e., $({\beta }_1\alpha {\beta }_2{)}^s\in N_{\ast }(R)[M]$. Since $R$ is a lower nil $M$-Armendariz ring, therefore, $A_j^s\in N_{\ast }(R)$ implies $A_j\in N_{\ast }(R)$. Hence, $a_i\in N_{\ast }(R)$ for each $1\le i\le m$ and this implies $N_{\ast }(R[M])\subseteq N_{\ast }(R)[M]$. Thus, $N_{\ast }(R[M])=N_{\ast }(R)[M]$.

Proposition 2.7. Let $M$ be a monoid and $N$ a u.p. monoid. If $R$ is a semicommutative lower nil $M$-Armendariz ring, then $R[M]$ is a lower nil $N$-Armendariz ring.

Proof. Here, every lower nil $M$-Armendariz ring is a nil $M$-Armendariz ring. Also, by Alhevaz and Moussavi (2014, Proposition 2.12), $N(R)[M]=N(R[M])$ and from Proposition 2.6, $N_{\ast }(R)[M]=N_{\ast }(R[M])$. Again, by semicommutative of $R$, we get $N(R[M])=N(R)[M]=N_{\ast }(R)[M]=N_{\ast }(R[M])$. Hence, $R[M]$ is a $2$-primal ring. Finally, by Proposition 2.1, we conclude that $R[M]$ is a lower nil $N$-Armendariz ring.□

Theorem 2.2. Let $M$ be a monoid and $N$ a u.p. monoid. If $R$ is a semicommutative and lower nil $M$-Armendariz ring, then $R[N]$ is a lower nil $M$-Armendariz ring.

Proof. First note that $\mathrm{\Phi }:R[N][M]\to R[M][N]$ defined by $\mathrm{\Phi }({\sum }_i({\sum }_j{a}_{i_j}{n}_j)m_i)={\sum }_j({\sum }_i{a}_{i_j}{m}_i)n_j$ is a ring isomorphism. Now, suppose $({\sum }_i{\alpha }_i{m}_i)({\sum }_j{\beta }_j{m}_j^{\prime })\in N_{\ast }(R[N])[M]$, where $m_i$, $m_j^{\prime }\in M$ and ${\alpha }_i,{\beta }_j\in R[N]$ for each $i,j$. In order to prove ${\alpha }_i{\beta }_j\in N_{\ast }(R[N])$ for each $i,j$, let ${\alpha }_i={\sum }_r{a}_{i_r}{n}_r$ and ${\beta }_j={\sum }_s{a}_{j_s}{n}_s^{\prime }\in R[N]$. From Proposition 2.6, we have $N_{\ast }(R)[M]=N_{\ast }(R[M])$. Further, $R$ is a semicommutative ring and $N$ a u.p. monoid, so by Corollary 2.1, $R$ is a lower nil $N$-Armendariz ring. Again, by Cheon and Kim (2012, Theorem 3), we have $N_{\ast }(R)[N]=N_{\ast }(R[N])$. Hence, $({\sum }_r({\sum }_i{a}_{i_r}{m}_i)n_r)({\sum }_s({\sum }_j{a}_{j_s}{m}_j^{\prime })n_s^{\prime })\in N_{\ast }(R[M])[N]$. Therefore, by Proposition 2.6, $({\sum }_i{a}_{i_r}{m}_i)({\sum }_j{a}_{j_s}{m}_j^{\prime })\in N_{\ast }(R[M])=N_{\ast }(R)[M]$, for each $r$ and $s$. Since $R$ is a lower nil $M$-Armendariz ring, therefore $a_{i_r}{b}_{j_s}\in N_{\ast }(R)$ for each $i,j,r,s$, this implies ${\alpha }_i{\beta }_j\in N_{\ast }(R)[N]=N_{\ast }(R[N])$. Thus, $R[N]$ is a lower nil $M$-Armendariz ring.

A ring $R$ is said to be a Dedekind finite $(vonNeumann$ - finite) if $ab=1$ implies $ba=1$ for each $a,b\in R$.

Proposition 2.8. Let $M$ be a cyclic group of order $n\ge 2$. Then each lower nil $M$-Armendariz ring is a Dedekind finite.

Proof. Let $R$ be a lower nil $M$-Armendariz ring and assume on the contrary $R$ is not a Dedekind finite. Then by Goodearl (1979, Proposition 5.5), $R$ contains an infinite set of matrix units, say $\left\{E_{11},E_{12},E_{13},\dots ,E_{21},E_{22},E_{23},\dots \right\}$. Consider the elements $\alpha =E_{11}e+E_{12}g$ and $\beta =(E_{22}-E_{11})e+(E_{11}-E_{12})g$ of $R[M]$. Then $\alpha \beta =0$ but $(E_{11}-E_{12})$ is not a strongly nilpotent, which contradicts the assumption. Hence, $R$ is a Dedekind finite.□

Theorem 2.3. Let $M$ be a monoid and $N$ a u.p. monoid. If $R$ is a semicommutative lower nil $M$-Armendariz ring, then $R$ is a lower nil $M\times N$-Armendariz ring.

Proof. By Liu (2005, Theorem 2.3), we have $R[M\times N]\cong R[M][N]$ and by Proposition 2.6, $N_{\ast }(R)[M]=N_{\ast }(R[M])$. Rest part of the proof follows Proposition 2.7.

Let $M_i$, $i\in I$, be monoids for index set $I$. Denote ${\coprod }_{i\in I}{M}_i=\{(h_i{)}_{i\in I}$ there exist only finite $i\prime s$ such that $h_i\ne e_i,$ the identity of $M_i\}$. Then ${\coprod }_{i\in I}{M}_i$ is a monoid under the binary operation $(h_i{)}_{i\in I}(h_i^{\prime }{)}_{i\in I}=(h_i{h}_i^{\prime }{)}_{i\in I}$.□

Corollary 2.3. Let $M_i,i\in I$ be the u.p. monoids and $R$ a semicommutative ring. If $R$ is a lower nil $M_{i_0}$-Armendariz ring for some $i_0\in I$, then $R$ is a lower nil $({\coprod }_{i\in I}{M}_i)$-Armendariz ring.

Proof. Let $\alpha ={\sum }_i{a}_i{g}_i$, $\beta ={\sum }_j{b}_j{h}_j\in R[{\coprod }_{i\in I}{M}_i]$ such that $\alpha \beta \in N_{\ast }(R)[{\coprod }_{i\in I}{M}_i]$. Then $\alpha ,\beta \in R[M_{i_0}\times M_1\times M_2\times \cdots \times M_n]$ for some finite subset $\left\{M_1,M_2,\dots ,M_n\right\}\subseteq \left\{M_i:i\in I\right\}$. From Theorem 2.3 and by applying induction, the ring $R$ is a lower nil $(M_{i_0}\times M_1\times M_2\times \cdots \times M_n)$-Armendariz ring, therefore $a_i{b}_j\in N_{\ast }(R)$. Hence, $R$ is a lower nil $({\coprod }_{i\in I}{M}_i)$-Armendariz ring.□

Theorem 2.4. Let $M$ be a strictly totally ordered monoid and $I$ an ideal of the ring $R$. If $I$ is semicommutative and $R/I$ is a lower nil $M$-Armendariz ring, then $R$ is a lower nil $M$-Armendariz.

Proof. Let $\alpha ={\sum }_{i=1}^m{a}_i{g}_i$ and $\beta ={\sum }_{j=1}^n{b}_j{h}_j$ $\in R[M]$ with $\alpha \beta \in N_{\ast }(R)[M]$, where $g_1<g_2<g_3<\cdots <g_m;h_1<h_2<h_3<\cdots <h_n$. Then ${\sum }_{g_i{h}_j=w}{a}_i{b}_j\in N_{\ast }(R)$. Now, we use transfinite induction on the strictly totally ordered set $(M,\le )$ to prove $a_i{b}_j\in N_{\ast }(R)$, for each $i,j$. Clearly, $\bar{\alpha }\bar{\beta }\in N_{\ast }(R/I)[M]$. Since $R/I$ is a lower nil $M$-Armendariz ring, so there exists a positive integer $s$ such that $(r_1{a}_i{b}_j{r}_2{)}^s\in N(I)$ for each $i,j$ and $r_1,r_2\in R$. Also, $g_1{h}_1<g_i{h}_j$ if $i\ne 1$ or $j\ne 1$. This implies that $a_1{b}_1\in N_{\ast }(R)$. Now, suppose $a_i{b}_j\in N_{\ast }(R)$, for any $g_i$ and $h_j$ with $g_i{h}_j<w$. In order to prove $a_i{b}_j\in N_{\ast }(R)$, for each $i$, $j$ and $g_i{h}_j=w$, consider $X=\{(g_i,h_j)|g_i{h}_j=w\}$. Then $X$ is a finite set. We write $X$ as $\{(g_{i_t},h_{j_t})|t=1,2,\dots ,k\}$ such that $g_{i_1}<g_{i_2}<\cdots <g_{i_k}$. Since $M$ is a cancellative monoid, $g_{i_1}=g_{i_2}$ and $g_{i_1}{h}_{j_1}=g_{i_2}{h}_{j_2}=w$ imply that $h_{j_1}=h_{j_2}$. Also, since $(M,\le )$ is a strictly totally ordered monoid, $g_{i_1}<g_{i_2}$ and $g_{i_1}{h}_{j_1}=g_{i_2}{h}_{j_2}=w$ imply $h_{j_2}<h_{j_1}$. So, we have $h_{j_k}<\cdots <h_{j_2}<h_{j_1}$. Now,

$\sum _{g_i{h}_j=w}{a}_i{b}_j=\sum _{(g_i,h_j)\in X}{a}_i{b}_j=\sum _{t=1}^k{a}_{i_t}{b}_{j_t}\in N_{\ast }(R)\subseteq N(R).$

(2.2) $\sum _{g_i{h}_j=w}{r}_1{a}_i{b}_j{r}_2=\sum _{(g_i,h_j)\in X}{r}_1{a}_i{b}_j{r}_2=\sum _{t=1}^k{r}_1{a}_{i_t}{b}_{j_t}{r}_2\in N(R)forany{r}_1,r_2\in R.$(2.2)

Note that, $g_{i_1}{h}_{j_t}<g_{i_t}{h}_{j_t}=w$ for any $t\ge 2$ and by induction hypothesis, we have $a_{i_1}{b}_{j_2}\in N_{\ast }(R)$. Consider $(a_{i_1}{b}_{j_2}{)}^p=0$ for some positive integer $p$, then $(b_{j_2}{a}_{i_1}{)}^{p+1}=0$. Therefore,

$((r_1{a}_{i_2}{b}_{j_2}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}{r}_1{a}_{i_2})(b_{j_2}{a}_{i_1}{)}^{p+1}(b_{j_2}{r}_2(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1})=0.$

$((r_1{a}_{i_2}{b}_{j_2}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}{r}_1{a}_{i_2})(b_{j_2}{a}_{i_1})(b_{j_2}{a}_{i_1}{)}^p(b_{j_2}{r}_2(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1})=0.$

$((r_1{a}_{i_2}{b}_{j_2}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}{r}_1{a}_{i_2})(b_{j_2})(r_2(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^s{r}_1)(a_{i_1})(b_{j_1}{r}_2(r_1)a_{i_2}{b}_{j_2}{r}_2)$

$(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}{r}_1{a}_{i_2})(b_{j_2}{a}_{i_1}{)}^p(b_{j_2}{r}_2(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1})=0.$

This implies

$[(r_1{a}_{i_2}{b}_{j_2}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}{]}^2[(r_1{a}_{i_2}{b}_{j_2}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}{r}_1{a}_{i_2}](b_{j_2})(r_2(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^s{r}_1)(a_{i_1})$

$(b_{j_1}{r}_2(r_1)a_{i_2}{b}_{j_2}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}{r}_1{a}_{i_2})(b_{j_2}{a}_{i_1}{)}^{p-1}(b_{j_2}{r}_2(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1})=0.$

Continuing the above procedure, we get

$[(r_1{a}_{i_2}{b}_{j_2}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}){]}^{2p+4}=0.$

This implies that $(r_1{a}_{i_2}{b}_{j_2}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}\in N(I)$. Similarly, we can see that $(r_1{a}_{i_t}{b}_{j_t}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}\in N(I)$ for $3\le t\le k$. Since $I$ is a semicommutative ideal, so $N(I)$ is an ideal of $I$, therefore ${\sum }_{t=2}^k(r_1{a}_{i_t}{b}_{j_t}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}\in N(I)$. On the other hand, multiplying by $(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}$ from right in Equation (2.2), we have ${\sum }_{t=1}^k(r_1{a}_{i_t}{b}_{j_t}{r}_2)(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+1}\in N(I)$, so $(r_1{a}_{i_1}{b}_{j_1}{r}_2{)}^{s+2}\in N(I)\subseteq N(R)$ and hence, $a_{i_1}{b}_{j_1}\in N_{\ast }(R)$. Continuing this process, we get $a_{i_t}{b}_{j_t}\in N_{\ast }(R)$ for $1\le t\le k$. So, $a_i{b}_j\in N_{\ast }(R)$ for each $i,j$ with $g_i{h}_j=w$. Thus, $a_i{b}_j\in N_{\ast }(R)$ for each $i,j$.□

Proposition 2.9. Let $M$ be a monoid with $\left|M\right|\ge 2$. Then the following conditions are equivalent:

(1) $R$ is a lower nil $M$-Armendariz ring.

(2) $H_3(R)=\left\{\left(\begin{array}{ccc}{a}_{11}& 0& 0\\ a_{21}& a_{22}& a_{23}\\ 0& 0& a_{33}\end{array}\right):a_{ij}\in R\right\}$ is a lower nil $M$-Armendariz ring.

Proof. First, we claim

$N_{\ast }(H_3(R))=\left(\begin{array}{ccc}{N}_{\ast }(R)& 0& 0\\ R& N_{\ast }(R)& R\\ 0& 0& N_{\ast }(R)\end{array}\right).$

For this, we consider $\left(\begin{array}{ccc}{a}_{11}& 0& 0\\ a_{21}& a_{22}& a_{23}\\ 0& 0& a_{33}\end{array}\right)\in \left(\begin{array}{ccc}{N}_{\ast }(R)& 0& 0\\ R& N_{\ast }(R)& R\\ 0& 0& N_{\ast }(R)\end{array}\right)$,

then $a_{ii}\in N_{\ast }(R)$ for each $1\le i\le 3$. So, $R{a}_{ii}R\subseteq N(R)$ for each $i$, $1\le i\le 3$. Then there exists a positive integer $n$ corresponding to nilpotency of all elements of $R{a}_{ii}R$ such that ${\left(\begin{array}{ccc}{a}_{11}& 0& 0\\ a_{21}& a_{22}& a_{23}\\ 0& 0& a_{33}\end{array}\right)}^{2n}=0\in N_{\ast }(H_3(R))$. Hence, $\left(\begin{array}{ccc}{N}_{\ast }(R)& 0& 0\\ R& N_{\ast }(R)& R\\ 0& 0& N_{\ast }(R)\end{array}\right)\subseteq N_{\ast }(H_3(R))$.□

Conversely, $\left(\begin{array}{ccc}{a}_{11}& 0& 0\\ a_{21}& a_{22}& a_{23}\\ 0& 0& a_{33}\end{array}\right)=C\in N_{\ast }(H_3(R))$. Then $(H_3(R))C(H_3(R))\subseteq N(H_3(R))$. This implies $R{a}_{ii}R\subseteq N(R)$ and hence $a_{ii}\in N_{\ast }(R)$ for each $1\le i\le 3$. Therefore

$N_{\ast }(H_3(R))\subseteq \left(\begin{array}{ccc}{N}_{\ast }(R)& 0& 0\\ R& N_{\ast }(R)& R\\ 0& 0& N_{\ast }(R)\end{array}\right).$

Next, we prove $H_3(R)$ is a lower nil $M$-Armendariz ring. Since $\mathrm{\Phi }:H_3(R)[M]\mapsto H_3(R[M])$ defined by

$\sum _{i=1}^n\left(\begin{array}{ccc}{a}_{11}^i& 0& 0\\ \\ a_{21}^i& a_{22}^i& a_{23}^i\\ \\ 0& 0& a_{33}^i\end{array}\right)g_i\mapsto \left(\begin{array}{ccc}\sum _{i=1}^n{a}_{11}^i{g}_i& 0& 0\\ \sum _{i=1}^n{a}_{21}^i{g}_i& \sum _{i=1}^n{a}_{22}^i{g}_i& \sum _{i=1}^n{a}_{23}^i{g}_i\\ 0& 0& \sum _{i=1}^n{a}_{33}^i{g}_i\end{array}\right)$

is an isomorphism. Let $\alpha =A_1{g}_1+A_2{g}_2+\cdots +A_n{g}_n$, $\beta =B_1{h}_1+B_2{h}_2+\cdots +B_m{h}_m\in H_3(R)[M]$ such that $\alpha \beta \in N_{\ast }(H_3(R))[M]$, where $A_i,B_j\in H_3(R)$. Also, let ${\alpha }_p={\sum }_{i=1}^n{a}_{pp}^i{g}_i$, ${\beta }_p={\sum }_{j=1}^m{b}_{pp}^j{h}_j\in R[M]$, for $p=1,2,3.$ Then by above isomorphism, it is easy to see that $\alpha \beta \in N_{\ast }(H_3(R)[M])$. So ${\alpha }_p{\beta }_p\in N_{\ast }(R)[M]$ and this implies $a_{pp}^i{b}_{pp}^j\in N_{\ast }(R)$ for each $i,j,p$. Therefore, $A_i{B}_j\in N_{\ast }(H_3(R))$. Thus, $H_3(R)$ is a lower nil $M$-Armendariz ring.

Proposition 2.10. Let $M$ be a monoid with $\left|M\right|\ge 2$. Then the following condition are equivalent:

(1) $R$ is a lower nil $M$-Armendariz ring.

(2) $T_n(R)$ is a lower nil $M$-Armendariz ring.

Proof. $(\mathbf{1})\Rightarrow (\mathbf{2}):$ It is easy to see that there exists an isomorphism between rings $T_n(R)[M]$ and $T_n(R[M])$, defined by

${\sum }_{i=1}^n\left(\begin{array}{cccc}{a}_{11}^i& a_{12}^i& \dots & a_{1n}^i\\ \\ 0& a_{22}^i& \dots & a_{2n}^i\\ \\ ⋮& ⋮& \ddots & ⋮\\ \\ 0& 0& \dots & a_{nn}^i\end{array}\right)g_i\mapsto \left(\begin{array}{cccc}{\mathrm{\Sigma }}_{i=1}^n{a}_{11}^i{g}_i& {\mathrm{\Sigma }}_{i=1}^n{a}_{12}^i{g}_i& \dots & {\mathrm{\Sigma }}_{i=1}^n{a}_{1n}^i{g}_i\\ \\ 0& {\mathrm{\Sigma }}_{i=1}^n{a}_{22}^i{g}_i& \dots & {\mathrm{\Sigma }}_{i=1}^n{a}_{2n}^i{g}_i\\ \\ ⋮& ⋮& \ddots & ⋮\\ \\ 0& 0& \dots & {\mathrm{\Sigma }}_{i=1}^n{a}_{nn}^i{g}_i\end{array}\right).$

Let $\alpha =A_1{g}_1+A_2{g}_2+\dots +A_m{g}_m$ and $\beta =B_1{h}_1+B_2{h}_2+\dots +B_n{h}_n\in T_n(R)[M]$ such that $\alpha \beta \in N_{\ast }(T_n(R))[M]$, where

$A_i=\left(\begin{array}{cccc}{a}_{11}^i& a_{12}^i& \dots & a_{1n}^i\\ \\ 0& a_{22}^i& \dots & a_{2n}^i\\ \\ ⋮& ⋮& \ddots & ⋮\\ \\ 0& 0& \dots & a_{nn}^i\end{array}\right),B_j=\left(\begin{array}{cccc}{b}_{11}^i& b_{12}^i& \dots & b_{1n}^i\\ \\ 0& b_{22}^i& \dots & b_{2n}^i\\ \\ ⋮& ⋮& \ddots & ⋮\\ \\ 0& 0& \dots & b_{nn}^i\end{array}\right)\in T_n(R).$

□

We have $N_{\ast }(T_n(R))=\{m=(m_{ij})\in T_n(R)|m_{ii}\in N_{\ast }(R)\}$. Therefore, from $\alpha \beta \in N_{\ast }(T_n(R))[M]$, we get $({\sum }_{i=0}^r{a}_{pp}^i{g}_i)({\sum }_{j=0}^s{b}_{pp}^j{h}_j)\in N_{\ast }(R)[M]$, for $p=1,2\dots n$. Since $R$ is a lower nil $M$-Armendariz ring, therefore $a_{pp}^i{b}_{pp}^j\in N_{\ast }(R)$, for each $p$ and all $i$, $j$ and hence $A_i{B}_j\in N_{\ast }(T_n(R))$ for each $i,j$. Thus, $T_n(R)$ is a lower nil M-Armendariz ring.

$(\mathbf{2})\Rightarrow (\mathbf{1}):$ Note that $R$ is isomorphic to $\left\{\left(\begin{array}{cccc}a& a& \dots & a\\ 0& a& \dots & a\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & a\end{array}\right):a\in R\right\},$ subring of $T_n(R)$. Clearly, $R$ is a lower nil $M$-Armendariz ring.

Now, following example makes it clear that full matrix ring $M_2(R)$ over a ring $R$ is not a lower nil $M$-Armendariz ring.

Example 2.3. Let $M$ be a monoid with $\left|M\right|\ge 2$ and $R$ is a ring. Take $e\ne g\in M$ and let $\alpha =\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)e-\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)g$ and $\beta =\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)e+\left(\begin{array}{cc}0& 0\\ 1& 1\end{array}\right)g$. Then $\alpha \beta \in N_{\ast }(M_2(R))[M]$ but $\left(\begin{array}{cc}1& 1\\ 0& 0\\ \hspace{0.17em}& \hspace{0.17em}\end{array}\right)\notin N_{\ast }(M_2(R))$. Therefore, $M_2(R)$ is not a lower nil $M$-Armendariz ring.

Let $R$ be a ring with an endomorphism $\alpha$ such that $\alpha (1)=1$. Chen, Yang, and Zhou (2006), considered the skew upper triangular matrix ring as a set of all upper triangular matrices with operations usual addition of matrices and multiplication subjected to the condition $E_{ij}r={\alpha }^{j-i}(r)E_{ij}$, i.e. for any two matrices $(a_{ij})$ and $(b_{ij})$, we have $(a_{ij})(b_{ij})=(c_{ij})$, where $c_{ij}=a_{ii}{b}_{ij}+a_{i,i+1}\alpha (b_{i+1,j})+\cdots +a_{ij}{\alpha }^{j-i}(b_{jj})$ for each $i\le j$ and it is denoted by $T_n(R,\alpha )$.

It is noted that $N_{\ast }(T_n(R,\alpha ))=(N_{\ast }(R),R,\dots ,R)$

The subring of the skew triangular matrices with constant main diagonal is denoted by $S(R,n,\alpha )$. Also, the subring of skew triangular matrices with constant diagonals is denoted by $T(R,n,\alpha )$. We can denote $A=(a_{ij})\in T(R,n,\alpha )$ by $(a_{11},a_{12},\dots ,a_{1n})$. Then $T(R,n,\alpha )$ is a ring with addition is pointwise and multiplication defined by

$(a_0,a_1,\dots ,a_{n-1})(b_0,b_1,\dots ,b_{n-1})=(a_0{b}_0,a_0\ast b_1+a_1\ast b_0,\dots ,a_0\ast b_{n-1}+\cdots +a_{n-1}\ast b_0)$

where $a_i\ast b_j=a_i{\alpha }^i(b_j)$ for each $i,j$. On the other hand, there exists a ring isomorphism $\mathrm{\Phi }:R[x;\alpha ]/(x^n)\to T(R,n,\alpha )$, defined by $\mathrm{\Phi }({\sum }_{i=0}^{n-1}{a}_i{x}^i)=(a_0,a_1,a_2,\dots ,a_{n-1})$, with $a_i\in R,$ $0\le i\le n-1$. So, $T(R,n,\alpha )\cong R[x;\alpha ]/(x^n)$, where $R[x;\alpha ]$ is the skew polynomial ring with multiplication subject to the condition $xr=\alpha (r)x$ for each $r\in R$.

Also, we consider the following subrings of $S(R,n,\alpha )$

$A(R,n,\alpha )=\left\{\sum _{j=1}^{⌊\frac{n}{2}⌋}\sum _{i=1}^{n-j+1}{a}_j{E}_{i,i+j-1}+\sum _{j=⌊\frac{n}{2}⌋+1}^n\sum _{i=1}^{n-j+1}{a}_j{E}_{i,i+j-1}\left|a_j,a_{i,k}\in R\right.\right\},$

$B(R,n,\alpha )=\{A+r{E}_{1k}|A\in A(R,n,\alpha ),r\in Randn=2k\ge 4\}.$

Let $S$ be a monoid and $f$ be an identity of $S$. Suppose $F\cup \left\{0\right\}$ is a free monoid generated by $W=\left\{w_1,w_2,\dots ,w_t\right\}$ and $S$ is a factor of $F$. Setting certain monomial in $W$ to $0$, it is enough to show that for some $n$, ${\delta }^n=0$ for any $\delta \ne f$. Let $R$ be a ring with an endomorphism $\alpha$. Then we can form the skew monoid ring $R[S;\alpha ]$, by taking its elements to be finite formal combinations ${\sum }_{v\in S}{r}_{v}v$, with multiplication subject to the relation $w_{i}r=\alpha (r)w_i$, for each $1\le i\le t$. It is easily seen that $N_{\ast }(R[S;\alpha ])=\{{\sum }_{v\in S}{r}_{v}v|r_v\in N_{\ast }(R)\}$ and also the ring $S(R,n,\alpha )$ and $T(R,n,\alpha )$ fit naturally into $R[S;\alpha ]$ with $W=\left\{E_{12},\dots ,E_{n-1,n}\right\}$ and $W=\left\{E_{12}+\cdots +E_{n-1,n}\right\}$, respectively.

Theorem 2.5. Let $M$ and $S$ be two monoids and $\alpha$ an endomorphism of the ring $R$ with $\alpha (1)=1$. Then the following hold:

(1) $R$ is lower nil $M$-Armendariz if and only if so is $S(R,n,\alpha )$;

(2) $R$ is lower nil $M$-Armendariz if and only if so is $T(R,n,\alpha )$;

(3) $R$ is lower nil $M$-Armendariz if and only if so is $A(R,n,\alpha )$;

(4) $R$ is lower nil $M$-Armendariz if and only if so is $B(R,n,\alpha )$;

(5) $R$ is lower nil $M$-Armendariz if and only if so is $T_n(R,\alpha )$;

(6) $R$ is lower nil $M$-Armendariz if and only if so is $R(S;\alpha ]$.

Proof. (1) Suppose that $R$ is a lower nil $M$-Armendariz ring. Let $\alpha ={\sum }_{i=1}^m{A}_i{g}_i$ and $\beta ={\sum }_{j=1}^n{B}_j{h}_j\in S(R,n,\alpha )[M]$ such that $\alpha \beta \in N_{\ast }(S(R,n,\alpha ))[M]$, where $A_i=(a_{st}^{(i)}),$ $B_j=(b_{st}^{(j)})$ for $1\le i\le m$ and $1\le j\le n$. Then ${\alpha }_0{\beta }_0\in N_{\ast }(R)[M]$, where ${\alpha }_0={\sum }_{i=1}^m{a}_{11}^{(i)}{g}_i$ and ${\beta }_0=b_{11}^{(j)}{h}_j\in R[M]$. Since $R$ is a lower nil $M$-Armendariz ring, therefore, $a_{11}^{(i)}{b}_{11}^{(j)}\in N_{\ast }(R)$ for each $i,j$ and hence $A_i{B}_j\in N_{\ast }(S(R,n,\alpha ))$ for each $i,j$. Thus, $S(R,n,\alpha )$ is a lower nil $M$-Armendariz ring.□

Rest results can be easily proved by following above arguments.

Proposition 2.11 (1) Direct product of lower nil $M$-Armendariz rings is a lower nil $M$-Armendariz ring.

(2) Direct sum of lower nil $M$-Armendariz rings is a lower nil $M$-Armendariz ring.

Proof. (1) It is well known that $N_{\ast }({\prod }_{R_i})=\prod (N_{\ast }(R_i))$. Let $\alpha ={\sum }_{i=0}^m{a}_{i_{\delta }}{g}_i\in R[M]$ and $\beta ={\sum }_{j=0}^n{b}_{j_{\delta }}{h}_j\in R[M]$ such that $\alpha \beta \in N_{\ast }(R)[M]$. Then $\alpha =({\alpha }_{\delta })$ and $\beta =({\beta }_{\delta })\in R[M]$, where ${\alpha }_{\delta }={\sum }_{i=0}^m{a}_{i_{\delta }}{g}_i$ and ${\beta }_{\delta }={\sum }_{j=0}^n{b}_{j_{\delta }}{g}_j$. So, ${\alpha }_{\delta }{\beta }_{\delta }\in N_{\ast }(R_{\delta })$ where $\delta \in I$. Since $R_{\delta }$ is the lower nil $M$-Armendariz ring, therefore $a_{i_{\delta }}{b}_{j_{\delta }}\in N_{\ast }(R_{\delta })$. Thus, $R$ is a lower nil $M$-Armendariz ring.

(2) It is easily seen that $N_{\ast }(\oplus R_i)=\oplus (N_{\ast }(R_i))$. As above, direct sum of lower nil $M$-Armendariz ring is a lower nil $M$-Armendariz ring.□

Theorem 2.6. The classes of lower nil $M$-Armendariz rings are closed under direct limit.

Proof. Let $D=\left\{R_i,{\beta }_{ij}\right\}$ be a direct system of lower nil $M$-Armendariz rings $R_i$ for $i\in I$ and ring homomorphisms ${\beta }_{ij}:R_i\to R_j$ for each $i\le j$ satisfying ${\beta }_{ij}(1)=1$, where $I$ is a directed partially ordered set. Let $R=R_i$ be the direct limit of $D$ with $l_i:R_i\to R$ and $l_j{\beta }_{ij}=l_i$. Now, we will prove $R$ is lower nil $M$-Armendariz ring. If we take $a,b\in R$, then $a=l_i(a_i)$, $b=l_j(b_j)$ for some $i,j\in I$ and there is $s\in I$ such that $i\le s$, $j\le n$. Define

$a+b=l_s({\beta }_{is}(a_i)+{\beta }_{js}(b_j))\hspace{0.17em}and\hspace{0.17em}ab=l_s({\beta }_{is}(a_i){\beta }_{js}(b_j))$

where ${\beta }_{is}(a_i)$, ${\beta }_{js}(b_j)\in R_s$. Then $R$ forms a ring with $l_i(0)=0$ and $l_i(1)=1$. Also, let $\alpha \delta \in N_{\ast }(R)[M]$ for $\alpha ={\sum }_{p=1}^m{a}_p{g}_p$ and $\delta ={\sum }_{q=1}^n{b}_q{h}_q\in R[M]$. Then there are $i_p,j_q,s\in I$ such that $a_p=l_{i_p}(a_{i_p})$, $b_q=l_{i_q}(b_{j_q})$, $i_p\le s$, $j_q\le s$ and hence $\alpha \delta \in N_{\ast }(R_s)[M]$. Since $R_s$ is a lower nil $M$-Armendariz, $l_s({\beta }_{i_{p}s}(a_i){\beta }_{j_{q}s}(b_j))\in N_{\ast }(R_s)$ and therefore, $a_p{b}_q\in N_{\ast }(R)$. Thus, $R$ is a lower nil $M$-Armendariz ring.

Weak Annihilator: Let $R$ be a ring. Then for a subset $X$ of $R$, the weak annihilator of $X$ in $R$ is $N_R(X):=\left\{a\in R:xa\in N(R)forallx\in X\right\}$. If $X$ is singleton, say $X=\left\{r\right\}$, then we use $N_R(r)$. Also, for the ring $R$, we define

$NAn{n}_R(2^R):=\left\{N_R(U):U\subseteq R\right\}$

$NAn{n}_{R[M]}(2^{R[M]}):=\left\{N_{R[M]}(V):V\subseteq R[M]\right\}.$

For an element $\beta \in R[M],$ $C_{\beta }$ denotes the set of all coefficients of $\beta$ and for a subset $W$ of $R[M]$, $C_W$ denotes the set ${\cup }_{\beta \in W}{C}_{\beta }$.□

Now, we present the results of Alhevaz and Hashemi (2017) and Cheon and Kim (2012), which are helpful forour next result.

Theorem 2.7. Alhevaz and Hashemi (2017, Theorem 3.2) Let $M$ be a u.p. monoid with nontrivial center. If $R$ is a lower nil $M$-Armendariz ring, then $N(R)[M]=N(R[M])$.

Theorem 2.8. Cheon and Kim (2012, Theorem 3) Let $R$ be a ring and $M$ a u.p. monoid. Then $N_{\ast }(R)[M]=N_{\ast }(R[M])$.

Theorem 2.9. Cheon and Kim (2012, Theorem 5) Let $R$ be a ring and $M$ a u.p. monoid. Then $N_{\ast }(R)=N(R)$ if and only if $N_{\ast }(R[M])=N(R[M])$.

Remark 2.1. Now, one can easily conclude that if $R$ is $2$-primal ring and $M$ a u.p. monoid, then $N(R)[M]=N(R[M])$.

Based on the above discussion, we have the following results:

Theorem 2.10. Let $M$ be a u.p. monoid with nontrivial center and $R$ a lower nil $M$-Armendariz. Then

$\mathrm{\Phi }:NAn{n}_R(2^R)\to NAn{n}_{R[M]}(2^{R[M]})$

defined by $\mathrm{\Phi }(I)=I[M]$ for every $I\in NAn{n}_R(2^R)$ is bijective.

Proof. By Alhevaz and Hashemi (2017, Theorem 3.2), we have $N(R)[M]=N(R[M])$ and also from Proposition 2.3, $R$ is a nil $M$-Armendariz ring. Therefore, by Lunqun and Jinwang (2013, Theorem 3.1), result is true.

A ring $R$ is said to be nilpotent p.p. ring if for any $q\notin N(R)$, the $N_R(q)$ is a right ideal generated by a nilpotent element of $R$.□

Theorem 2.11. Let $M$ be a monoid and $R$, a semicommutative lower nil $M$-Armendariz ring. If $R$ is nilpotent p.p. ring, then so is $R[M]$.

Proof. Let $\alpha =a_1{g}_1+a_2{g}_2+\cdots +a_n{g}_n\notin N(R[M])$ and $\beta =b_1{h}_1+b_2{h}_2+\cdots +b_m{h}_m\in N_{R[M]}(\alpha )$. Then $\alpha \beta \in N(R[M]$. Also, by Alhevaz and Moussavi (2014, Proposition 2.12), $N(R)[M]=N(R[M])$, therefore $\alpha \beta \in N(R)[M]$ and hence $a_i{b}_j\in N(R)$ for each $1\le i\le n,1\le j\le m$. Since $\alpha \notin N(R[M])=N(R)[M]$, so there exist at least one $i$, $1\le i\le n$ such that $a_i\notin N(R)$. Therefore, $R$ being nilpotent $p.p.$ ring, there exist some $s\in N(R)$ such that $N_R(a_i)=sR$.□

Now, we claim $N_{R[M]}(\alpha )=se.R[M]$. Since $b_j\in N_R(a_i)$ for each $1\le j\le m$, so $b_j=s{r}_j$ for some $r_j\in R$. Therefore, $\beta =se(r_1{h}_1+r_2{h}_2+\cdots +r_m{h}_m)\in se.R[M]$, hence $N_{R[M]}(\alpha )\subseteq se.R[M]$.

Also, for any $\delta =w_1{e}_1+w_2{e}_2+\cdots +w_q{e}_q\in R[M]$, since $s\in N(R)$ and $R$ semicommutative ring, we have $a_{i}s{w}_j\in N(R)$ for each $i,j$. So $\alpha .se.\delta \in N(R)[M]=N(R[M])$. Hence, $se.R[M]\subseteq N_{R[M]}(\alpha )$. Thus, $N_{R[M]}(\alpha )=se.R[M]$, where $se\in N(R[M])$.

3. Lower nil $M$-Armendarz rings from $M$-Armendariz rings

Theorem 3.1. Let $M$ be a strictly totally ordered monoid and $I$ a proper ideal of $R$ with $N_{\ast }(R)\subseteq I.$ If $R/I$ is $M$-Armendariz and $I$, the 2-primal ring, then $R$ is a lower nil $M$-Armendariz ring.

Proof. Suppose $\bar{R}=R/I$ is an $M$-Armendariz ring and $I$ is a 2-primal ring. Since, $N_{\ast }(R)\subseteq I$, so $N_{\ast }(R)\subseteq N_{\ast }(I)$. Also, $I$ is an ideal of $R$, so $N_{\ast }(I)\subseteq N_{\ast }(R)$ and hence $N_{\ast }(I)=N_{\ast }(R)$. Since $I$ is a 2-primal, $I/N_{\ast }(R)=I/N_{\ast }(I)\cong I/N(I)$, therefore, it is reduced.□

Let $\alpha =a_1{g}_1+a_2{g}_2+\dots +a_n{g}_n$ and $\beta =b_1{h}_1+b_2{h}_2+\dots +b_m{h}_n$ such that $\alpha \beta \in N_{\ast }(R)[M]$ with $g_1<g_2<\dots <g_n;h_1<h_2<\dots <h_m$, where $g_1,g_2,\dots ,g_n;h_1,h_2,\dots ,h_n\in M$. Now, we use transfinite induction on strictly totally ordered set $(M,\le )$ to show $a_i{b}_j\in N_{\ast }(R)$. Since $N_{\ast }(R)\subseteq I$, therefore $\alpha \beta \in I[M]$. Also, $R/I$ is $M$-Armendariz, therefore $a_i{b}_j\in I$ for each $i$ and $j$. If there exist $1\le i\le n$ and $1\le j\le m$ such that $g_i{h}_j=g_1{h}_1$, then $g_1\le g_i$ and $h_1\le h_j$. If $g_1<g_i$, then $g_1{h}_1<g_i{h}_1\le g_i{h}_j=g_1{h}_1$ implies $g_1{h}_1<g_1{h}_1$, a contradiction. Hence, $g_1=g_i$, similarly $h_1=h_j$. Thus, $a_1{b}_1\in N_{\ast }(R)$.

Now, let $w\in M$ be such that for any $g_i,h_j\in M$, $g_i{h}_j<w$, $a_i{b}_j\in N_{\ast }(R)$. To prove $a_i{b}_j\in N_{\ast }(R)$ for any $g_i$ and $h_j$ with $g_i{h}_j=w$, Consider $X=\left\{(g_i,h_j):g_i{h}_j=w\right\}$. Then $X$ is a finite set. So, we put $X=\left\{(g_{i_t},h_{j_t}):t=1,2,\dots k\right\}$ such that $g_{i_1}<g_{i_2}<\dots <g_{i_k}$. Here, $M$ is cancellative, $g_{i_1}=g_{i_2}$ and $g_{i_1}{h}_{j_1}=g_{i_2}{h}_{j_2}=w$ imply $h_{j_1}=h_{j_2}$. Since $(M,\le )$ strictly totally ordered monoid, $g_{i_1}<g_{i_2}$ and $g_{i_1}{h}_{j_1}=g_{i_2}{h}_{j_2}=w$, therefore, $h_{j_2}<h_{j_1}$. Thus, we have $h_{j_k}<\dots <h_{j_2}<h_{j_1}$. Now

(3.3) $\sum _{g_i{h}_j=w}{a}_i{b}_j=\sum _{t=1}^{t=k}{a}_{i_t}{b}_{j_t}\in N_{\ast }(R).$(3.3)

For any $t\ge 2$, $g_{i_1}{h}_{j_t}<g_{i_t}{h}_{j_t}=w$ and by induction hypothesis $a_{i_1}{b}_{j_t}\in N_{\ast }(R)$. Take $p=b_{j_t}(a_{i_1}{b}_{j_1})a_{i_1}$, then $p\in I$, since $a_i{b}_j\in I$ for each $i$ and $j$. Also, $a_{i_1}{b}_{j_t}\in N_{\ast }(R)$, so $p^2=b_{j_t}(a_{i_1}{b}_{j_1})a_{i_1}{b}_{j_t}(a_{i_1}{b}_{j_1})a_{i_1}\in N_{\ast }(R)$. Moreover, it is known that $I/N_{\ast }(R)$ is reduced, therefore, $p\in N_{\ast }(R)$. Again,

(3.4) $(a_{i_t}{b}_{j_t})(a_{i_1}{b}_{j_1}{)}^2=(a_{i_t}(b_{j_t}(a_{i_1}{b}_{j_1})a_{i_1})b_{j_1})\in N_{\ast }(R)$(3.4)

Multiplying (3.3) by $(a_{i_1}{b}_{j_1}{)}^2$ from right, we get

$(a_{i_1}{b}_{j_1}{)}^3+{\sum }_{t=2}^k(a_{i_t}{b}_{j_t})(a_{i_1}{b}_{j_1}{)}^2\in N_{\ast }(R)$. This implies $(a_{i_1}{b}_{j_1}{)}^3\in N_{\ast }(R)$. Since $I/N_{\ast }(R)$ is reduced, therefore, $a_{i_1}{b}_{j_1}\in N_{\ast }(R)$.

Again, by Equation (3.3), we have

(3.5) $a_{i_2}{b}_{j_2}+\dots +a_{i_k}{b}_{j_k}\in N_{\ast }(R)$(3.5)

so, $(a_{i_t}{b}_{j_t})(a_{i_2}{b}_{j_2}{)}^2=(a_{i_t}(b_{j_t}(a_{i_2}{b}_{j_2})a_{i_2})b_{j_2})\in N_{\ast }(R)$. Multiplying Equation (3.5) by $(a_{i_2}{b}_{j_2}{)}^2$ from right side, we get $(a_{i_2}{b}_{j_2}{)}^3\in N_{\ast }(R)$. This implies $a_{i_2}{b}_{j_2}\in N_{\ast }(R)$, because $I/N_{\ast }(R)$ is reduced. Continuing the procedure, we get $a_{i_t}{b}_{j_t}\in N_{\ast }(R)$ for $t=1,2,\dots k$ with $g_i{h}_j=w$. Therefore, by transfinite induction, $a_i{b}_j\in N_{\ast }(R)$, for each $i$ and $j$. Thus, $R$ is a lower nil $M$-Armendariz ring.

Recall that a monoid $M$ is torsion-free if $g,h\in M$ and $k\ge 1$ are such that $g^k=h^k$ implies $g=h$.

Corollary 3.1 Let $M$ be a commutative, cancellative and torsion-free monoid with $\left|M\right|\ge 2$. If either one of the following conditions holds, then $R$ is a lower nil $M$-Armendariz ring.

(1) $R$ is a 2-primal ring.

(2) $R/I$ is an $M$-Armendariz ring for some ideal $I$ of $R$ and $I$ is $2$-primal with $N_{\ast }(R)\subseteq I$.

Proof. If $M$ is commutative, cancellative and torsion-free monoid, then by Ribenboim (1992, Result 3.3), there exists a compatible strict total order monoid “≤” on $M$. Then, by Corollary 2.2 and Theorem 3.1, results hold.

A ring $R$ is a right (resp. left) uniserial ring if its lattice of right (resp. left) ideals is totally ordered by inclusion. Right uniserial rings are also called right chain rings or right valuation rings because they are obvious generalization of commutative valuation domains. Like commutative valuation domains, right uniserial rings have a rich theory and they offer remarkable examples, we refer Marks, Mazurek, and Ziembowski (2010).□

Proposition 3.1. (Alhevaz & Hashemi, 2017, Proposition 3.1(a)) For any u.p. monoid $M$, every $M$-Armendariz ring is a lower nil $M$-Armendariz ring.

Proposition 3.2. For any u.p. monoid $M$, every right or left uniserial ring is a lower nil $M$-Armendariz ring.

Proof. Let $R$ be a uniserial ring and $M$ a u.p. monoid. By Marks et al. (2010, Corollary 6.2), $R$ is $M$-Armendariz ring. Therefore, by Proposition 3.1, $R$ is a lower nil $M$-Armendariz ring.□

Proposition 3.3. Let $M$ be a monoid and $N$ a u.p.-monoid. If $R$ is semicommutative ring as well as $M$-Armendariz ring, then $R[M]$ is a lower nil $N$-Armendariz ring.

Proof. Let $\alpha =a_1{g}_1+a_2{g}_2+\dots +a_n{g}_n$ and $\beta =b_1{h}_1+b_2{h}_2+\dots +b_m{h}_m\in R[M]$ such that $\alpha \beta =0$. This implies $a_i{b}_j=0$ for each $i$ and $j$, since $R$ is $M$-Armendariz ring. Also $R$ is semicommutative, therefore $a_{i}R{b}_j=0$ for each $i,j$. Again, we can easily see that $\alpha R[M]\beta =0$. So from Corollary 2.1, $R[M]$ is a lower nil $N$-Armendariz ring.□

Lemma 3.1. Let $M$ be a monoid. If $R$ is a $2$-primal $M$-Armendariz ring, then $R[M]$ is $2$-primal ring and $R$ is a lower nil $M$-Armendariz ring with $N(R)[M]=N(R[M])=N_{\ast }(R)[M]=N_{\ast }(R[M])$.

Proof. By Lunqun and Jinwang (2013, Theorem 2.3), we have $R$ is a nil $M$-Armendariz ring. Since $R$ is $2$-primal, therefore $R$ is a lower nil $M$-Armendariz ring. Also, by Lunqun and Jinwang (2013, Lemma 2.2, Theorem 2.3), we have $N(R)[M]=N(R[M])$ and $R$ is $2$-primal, so $R[M]$ is $2$-primal. Hence, $N(R)[M]=N(R[M])=N_{\ast }(R)[M]=N_{\ast }(R[M])$.□

Proposition 3.4. Let $M$ be a monoid and $N,$ a u.p.-monoid. If $R$ is a $2$-primal $M$-Armendariz ring, then $R[N]$ is a lower nil $M$-Armendariz ring.

Proof. Since $N$ is a u.p.-monoid, therefore $N_{\ast }(R)[N]=N_{\ast }(R[N])$, by Cheon and Kim (2012, Theorem 3). Now, from Lemma 3.1, $N_{\ast }(R)[M]=N_{\ast }(R[M])$. Next, from Lemma 3.1, $R[M]$ is a lower nil $N$-Armendariz ring. Further, rings $R[N][M]$ and $R[M][N]$ are isomorphic under the map

$\sum _d\left(\sum _i{a}_{id}{n}_i\right)m_d\mapsto \sum _i\left(\sum _d{a}_{id}{m}_d\right)n_i.$

Now, let

$\left(\sum _i{\alpha }_i{g}_i\right)\left(\sum _j{\beta }_j{h}_j\right)\in N_{\ast }(R[N])[M],$

where ${\alpha }_i,{\beta }_j\in R[N]$ and $g_i,h_j\in M$. We claim that ${\alpha }_i{\beta }_j\in N_{\ast }(R[N])$ for each $i$ and $j$. Let

${\alpha }_i=\sum _d{a}_{id}{n}_d,{\beta }_j=\sum _e{b}_{je}{n}_e^{\prime }\in R[N].$

Then

$\left(\sum _i\left(\sum _d{a}_{id}{n}_d\right)g_i\right)\left(\sum _j\left(\sum _e{b}_{je}{n}_e^{\prime }\right)h_j\right)\in N_{\ast }(R[N])[M]=N_{\ast }(R)[N][M].$

Threfore,

$\left(\sum _d\left(\sum _i{a}_{id}{g}_i\right)n_d\right)\left(\sum _e\left(\sum _j{b}_{je}{h}_j\right)n_e^{\prime }\right)\in N_{\ast }(R)[M][N]=N_{\ast }(R[M])[N].$

By Lemma 3.1, $R[M]$ is a lower nil $N$-Armendariz ring, so

$\left(\sum _i{a}_{id}{g}_i\right)\left(\sum _j{b}_{je}{h}_j\right)\in N_{\ast }(R[M])=N_{\ast }(R)[M]$

for all $d,e$. Also, from Lemma 3.1, $R$ is a lower nil $M$-Armendariz ring, hence $a_{id}{b}_{je}\in N_{\ast }(R)$ for each $i,j,d,e$. Therefore,

${\alpha }_i{\beta }_j=\left(\sum _d{a}_{id}{n}_i\right)\left(\sum _e{b}_{je}{n}_j^{\prime }\right)\in N_{\ast }(R)[N]=N_{\ast }(R[N])$

for each $i,j$. Thus, $R[N]$ is a lower nil $M$-Armendariz ring.□

Proposition 3.5. Let $M$ be a monoid and $N$ a u.p.-monoid. If $R$ is $2$-primal $M$-Armendariz ring, then $R$ is a lower nil $M\times N$-Armendariz ring.

Proof. Construction of this proof is based on the proof of Liu (2005, Theorem 2.3).□

Let ${\sum }_{i=1}^t{a}_i(m_i,n_i)\in R[M\times N]$. Without loss of generality, we assume $\left\{n_1,n_2,\dots ,n_t\right\}=\left\{n_1,n_2,\dots ,n_s\right\}$ with $n_i\ne n_j$, where $1\le i\ne j\le s$.

$Let\hspace{0.17em}{A}_g=\left\{i:1\le i\le t,n_i=n_g\right\},for1\le g\le s.$

Then

$\sum _{i=1}^s\left(\sum _{i\in A_g}{a}_i{m}_i\right)n_g\in R[M][N].$

Note that $m_i\ne m_{i^{\prime }}$ for any $i,i^{\prime }\in A_g$, when $i\ne i^{\prime }$. It is easy to see that there exists an isomorphism between rings $R[M\times N]$ and $R[M][N]$ defined by

$\sum _{i=1}^t{a}_i(m_i,n_i)\mapsto \sum _{g=1}^s\left(\sum _{i\in A_g}{a}_i{m}_i\right)n_g.$

Assume

$\left(\sum _{i=1}^t{a}_i(m_i,n_i)\right)\left(\sum _{j=1}^{t\prime }{b}_j(m_j^{\prime },n_j^{\prime })\right)\in N_{\ast }(R[M\times N]).$

Then from the above isomorphism, we have

$\left(\sum _{g=1}^s\left(\sum _{i\in A_g}{a}_i{m}_i\right)n_g\right)\left(\sum _{h=1}^{s\prime }\left(\sum _{j\in B_h}{b}_j{m}_j^{\prime }\right)n_h^{\prime }\right)\in N_{\ast }(R[M][N]).$

By Lemma 3.1, $R[M]$ is a lower nil $N$-Armendariz ring. Therefore,

$\left(\sum _{i\in A_g}{a}_i{m}_i\right)\left(\sum _{j\in B_h}{b}_j{m}_j^{\prime }\right)\in N_{\ast }(R[M])=N_{\ast }(R)[M].$

Also, by Lemma 3.1, $R$ is a lower nil $M$-Armendariz ring. Therefore, $a_i{b}_j\in N_{\ast }(R)$ for each $i\in A_g$ and $j\in B_h$. Moreover, $a_i{b}_j\in N_{\ast }(R)$ for each $i$ and $j$, where $1\le i\le t$ and $1\le j\le t^{\prime }$. Hence, $R$ is a lower nil $M\times N$-Armendariz ring.

Example 3.1. Let $R={\mathbb{Z}}_2\oplus {\mathbb{Z}}_2$ be a ring and $M=\mathbb{N}\cup \left\{0\right\}$ a monoid. Then $R$ is a $2$-primal $M$-Armendariz ring. Here, $N_{\ast }(R[M])=N(R[M])$. Let $N$ be a monoid generated by elements $m_1,m_2,m_3,n_1,n_2,n_3$ by the following defining relations:

$m_1{n}_1=m_2{n}_3,m_1{n}_2=m_3{n}_1,m_1{n}_3=m_2{n}_2,m_3{n}_2=m_2{n}_1.$

Then by Okninski (1991, Example 13 of Chapter 10), due to Krempa, the monoid $M$ is a u.p.-monoid. Therefore, by Proposition 2.1, $R[M]$ is a lower nil $N$-Armendariz ring. Also, by Liu (2005, Theorem 2.3), we have $R[M][N]\cong R[M\times N]$, hence $R$ is a lower nil $M\times N$-Armendariz ring.

Acknowledgements

The authors are thankful to Indian Institute of Technology Patna for providing the research facilities and to anonymous referees and the editor of this journal for their valuable suggestions to improve the presentation of the manuscript.

Additional information

Funding

The authors received no direct funding for this research.

Notes on contributors

Sushma Singh

Sushma Singh completed her MSc in Mathematics from the Department of Mathematics, Banaras Hindu University, Varanasi, India, in 2010. Currently, she is a Ph. D. Research Scholar at the Department of Mathematics, Indian Institute of Technology, Patna, and her research area is Rings and Modules.

Om Prakash

Om Prakash is an associate professor at the Department of Mathematics, Indian Institute of Technology, Patna. He completed his MSc from Patna University, Patna, in 1999 and PhD from Banasthali University, Rajasthan, in 2010. He has 18 years of teaching experience in reputed institutions. His main research interest includes Rings and Modules, Algebraic Number Theory, Algebraic Coding Theory and Algebraic Graph Theory. He has the credit of published more than 30 research papers in international journals.