Ore Extensions Over Right Strongly Hopfian Rings

Abstract

An associative ring is said to be right strongly Hopfian if the chain of right annihilators $r_R(a)\subseteq r_R(a^2)\subseteq \cdots$ stabilizes for each $a\in R$. In this article, we are interested in the class of right strongly Hopfian rings and the transfer of this property from an associative ring R to the Ore extension $R[x;\alpha ,\delta ]$ and the monoid ring R[M]. It is proved that if R is $(\alpha ,\delta )$-compatible and $R[x;\alpha ,\delta ]$ is reversible, then the Ore extension $R[x;\alpha ,\delta ]$ is right strongly Hopfian if and only if R is right strongly Hopfian, and it is also shown that if M is a strictly totally ordered monoid and R[M] is a reversible ring, then the monoid ring R[M] is right strongly Hopfian if and only if R is right strongly Hopfian. Consequently, several known results regarding strongly Hopfian rings are extended to a more general setting.

1 Introduction

Throughout this paper, all rings are associative with identity. For a nonempty subset X of a ring R, $l_R(X)=\{a\in R\mid aX=0\}$ and $r_R(X)=\{a\in R\mid Xa=0\}$ denote the left and the right annihilators of X in R, respectively. Following A. Hmaimou et al [5], a ring R is left strongly Hopfian if for every endomorphism f of R, the chain ker $f\subseteq$ ker $f^2\subseteq \cdots$ stabilizes. Equivalently, R is left strongly Hopfian if the chain of left annihilators $l_R(a)\subseteq l_R(a^2)\subseteq \cdots$ stabilizes for each $a\in R$. The class of left strongly Hopfian rings is very large. It contains Noetherian rings, Laskerian rings, rings satisfying acc on d-annihilators, those satisfying acc on d-colons, and so on [4]. If R is a commutative ring, then a left strongly Hopfian ring is also called a strongly Hopfian ring. A. Hmaimou et al [5] showed that for a commutative ring R, the ring R is strongly Hopfian if and only if the polynomial ring R[x] is strongly Hopfian if and only if the Laurent polynomial ring $R[x;x^{-1}]$ is strongly Hopfian. Let R be a commutative ring. In [4], S. Hizem provided an example of a strongly Hopfian ring R such that the power series ring R[[x]] is not necessarily strongly Hopfian, and also gave some necessary and sufficient conditions for R[[x]] to be strongly Hopfian. For more details and properties of left strongly Hopfian rings, see [2, 4, 5, 7, 8].

Let $\alpha$ be an endomorphism, and $\delta$ an $\alpha$-derivation of R, that is, $\delta$ is an additive map such that $\delta (ab)=\delta (a)b+\alpha (a)\delta (b)$, for a, $b\in R$. According to Annin [1], a ring R is said to be $\alpha$-compatible if for each $a, b \in R, ab=0 \Leftrightarrow a\alpha (b)=0$. Clearly, this may only happen when the endomorphism $\alpha$ is injective. Moreover, R is said to be $\delta$-compatible if for each $a, b \in R, ab=0 \Rightarrow a\delta (b)=0$. A ring R is $(\alpha , \delta )$-compatible if it is both $\alpha$-compatible and $\delta$-compatible. Recall that a ring R is reversible if $ab=0\Rightarrow ba=0$ for all a, $b\in R$, and a ring R is semicommutative if $ab=0$ implies $aRb=0$ for any a, $b\in R$. Clearly, any subring of a reversible ring is also reversible, and if R is a reversible ring, then for any $n\in \mathbb {N}$ and any permutation $\sigma \in S_n$, $x_1x_2\cdots x_n=0$ implies $x_{\sigma (1)}Rx_{\sigma (2)}R\cdots x_{\sigma (n)}R=0$ for any $x_i\in R$, $1\le i\le n$. Reversible rings are semicommutative, but the reverse is not true in general [6, Example 1.5]. Let $f(x)=a_0+a_1x+\cdots +a_nx^n\in R[x;\alpha ,\delta ]$, $\{a_0, a_1, \ldots , a_n\}$ denote the subset of R that comprised the coefficients of f(x).

In this article, we are interested in the class of right strongly Hopfian rings and the transfer of this property from an associative ring R to the Ore extension $R[x;\alpha ,\delta ]$ and the monoid ring R[M]. We first provide some examples of right strongly Hopfian rings. We next show that (1) if R is $(\alpha ,\delta )$-compatible and $R[x;\alpha ,\delta ]$ is reversible, then the Ore extension $R[x;\alpha ,\delta ]$ is right strongly Hopfian if and only if R is right strongly Hopfian; (2) if M is a strictly totally ordered monoid and R[M] a reversible ring, then the monoid ring R[M] is right strongly Hopfian if and only if R is right strongly Hopfian.

2 Extensions of Right Strongly Hopfian Rings

Definition 2.1

A ring R is right strongly Hopfian if the chain of right annihilators $r_R(a)\subseteq r_R(a^2)\subseteq \cdots$ stabilizes for each $a\in R$.

The next Lemma is known and very useful. We leave the proof for the reader.

Lemma 2.2

Let $a\in R$. Then the chain $r_R(a)\subseteq r_R(a^2)\subseteq \cdots$ stabilizes if and only if there exists $n>m$ such that $r_R(a^n)=r_R(a^m)$.

Lemma 2.3

Let $A\subset B$ be an extension of rings. If B is right strongly Hopfian, then so is A.

Proof

Let $a\in A$. Then $r_A(a)=r_B(a)\cap A$.

Proposition 2.4

Let $T_n(R)$ denote the $n\times n$ upper triangular matrix ring over a ring R. Then the following conditions are equivalent:

• (1) R is right strongly Hopfian;

• (2) $T_n(R)$ is right strongly Hopfian.

Proof

$(1)\Rightarrow (2)$. Suppose R is right strongly Hopfian and let

$$\begin{aligned} A=\left( \begin{array}{cccc} a_{11}&{}\quad a_{12}&{}\quad \cdots &{}\quad a_{1n}\\ 0&{}\quad a_{22}&{}\quad \cdots &{}\quad a_{2n}\\ \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \cdots \\ 0&{}\quad 0&{}\quad \cdots &{}\quad a_{nn}\\ \end{array}\right) \in T_n(R). \end{aligned}$$

We proceed by induction on n to show that $T_n(R)$ is right strongly Hopfian. Let $n=2$. Put $\alpha =\left( \begin{array}{ll}a&{}b\\ 0&{}c\\ \end{array}\right) \in T_2(R)$. Since R is right strongly Hopfian, there exists $m\in \mathbb {N}$ such that for any $n>m$, $r_R(a^n)=r_R(a^m)$ and $r_R(c^n)=r_R(c^m)$. Now we show that $r_{T_2(R)}(\alpha ^{2m+1})=r_{T_2(R)}(\alpha ^{2m})$. If $\beta =\left( \begin{array}{ll}x&{}y\\ 0&{}z\\ \end{array}\right) \in r_{T_2(R)}(\alpha ^{2m+1}),$ then

$$\begin{aligned}&\alpha ^{2m+1}\beta \\&\quad ={\left( \begin{array}{ll}a^{2m+1}&{}\quad a^{2m}b+a^{2m-1}bc+\cdots +a^mbc^m+\cdots +bc^{2m}\\ 0&{}\quad c^{2m+1}\\ \end{array}\right) \left( \begin{array}{ll}x&{}y\\ 0&{}z\\ \end{array}\right) }\\&\quad \!=\!{\left( \begin{array}{ll}a^{2m+1}x&{}\quad a^{2m+1}y+(a^{2m}b+a^{2m-1}bc+\cdots +a^mbc^m+\cdots +bc^{2m})z\\ 0&{} \quad c^{2m+1}z\\ \end{array}\right) \!=\!0.}\\ \end{aligned}$$

Thus $x\in r_R(a^{2m+1})=r_R(a^{2m})$ and $z\in r_R(c^{2m+1})=r_R(c^{2m})=\cdots =r_R(c^m)$. Hence the equation

$$\begin{aligned} a^{2m+1}y+(a^{2m}b+a^{2m-1}bc+\cdots +a^mbc^m+\cdots +bc^{2m})z=0 \end{aligned}$$

becomes

$$\begin{aligned}&a^{2m+1}y+(a^{2m}b+a^{2m-1}bc+\cdots +a^{m+1}bc^{m-1})z\\&\quad =a^{m+1}(a^my+(a^{m-1}b+a^{m-2}bc+\cdots +bc^{m-1})z)=0.\\ \end{aligned}$$

Then

$$\begin{aligned} a^my+(a^{m-1}b+a^{m-2}bc+\cdots +bc^{m-1})\quad z\in r_R(a^{m+1}), \end{aligned}$$

and so

$$\begin{aligned} a^my+(a^{m-1}b+a^{m-2}bc+\cdots +bc^{m-1})z\in r_R(a^m). \end{aligned}$$

Hence

$$\begin{aligned}&a^m(a^my+(a^{m-1}b+a^{m-2}bc+\cdots +bc^{m-1})z)\\&\quad =a^{2m}y+(a^{2m-1}b+a^{2m-2}bc+\cdots +a^mbc^{m-1})z=0. \end{aligned}$$

Then

$$\begin{aligned}&a^{2m}\beta \\&\quad ={\left( \begin{array}{ll}a^{2m}&{} \quad a^{2m-1}b+a^{2m-2}bc+\cdots +abc^{2m-2}+bc^{2m-1}\\ 0&{} \quad c^{2m}\\ \end{array}\right) \left( \begin{array}{ll}x&{}y\\ 0&{}z\\ \end{array}\right) }\\&\quad ={\left( \begin{array}{ll}a^{2m}x&{} \quad a^{2m}y+(a^{2m-1}b+a^{2m-2}bc+\cdots +a^mbc^{m-1}+\cdots +bc^{2m-1})z\\ 0&{} \quad c^{2m}z\\ \end{array}\right) }\\&\quad ={\left( \begin{array}{ll}0&{} \quad a^{2m}y+(a^{2m-1}b+a^{2m-2}bc+\cdots +a^mbc^{m-1})z\\ 0&{} \quad 0\\ \end{array}\right) =0.} \end{aligned}$$

Hence $r_{T_2(R)}(\alpha ^{2m+1})\subseteq r_{T_2(R)}(\alpha ^{2m})$ and so $r_{T_2(R)}(\alpha ^{2m+1})= r_{T_2(R)}(\alpha ^{2m}).$ Therefore $T_2(R)$ is right strongly Hopfian.

Next, we assume that the result is true for $n-1$, $n>2$, and let

$$\begin{aligned} A=\left( \begin{array}{cccc} a_{11}&{}\quad a_{12}&{}\quad \cdots &{}\quad a_{1n}\\ 0&{}\quad a_{22}&{}\quad \cdots &{}\quad a_{2n}\\ \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \cdots \\ 0&{}\quad 0\quad &{}\cdots &{}\quad a_{nn}\\ \end{array}\right) \in T_n(R). \end{aligned}$$

We show that $r_{T_n(R)}(A)\subseteq r_{T_n(R)}(A^2)\subseteq \cdots$ stabilizes. Put

$$\begin{aligned} A=\left( \begin{array}{cccc} a_{11}&{}\quad a_{12}&{}\quad \cdots &{}\quad a_{1n}\\ 0&{}\quad a_{22}&{}\quad \cdots &{}\quad a_{2n}\\ \cdots &{}\quad \cdots &{}\quad \cdots &{}\quad \cdots \\ 0&{}\quad 0&{}\quad \cdots &{}\quad a_{nn}\\ \end{array}\right) =\left( \begin{array}{ll}A_{n-1}&{} B\\ 0&{}a_{nn}\\ \end{array}\right) . \end{aligned}$$

By the induction hypothesis, we can find $m\in \mathbb {N}$ such that for any $s>m$, $r_{T_{n-1}(R)}(A_{n-1}^s)=r_{T_{n-1}(R)}(A_{n-1}^m)$ and $r_R(a_{nn}^s)=r_R(a_{nn}^m)$. Then using the same way as above, we can show that $r_{T_n(R)}(A^{2m+1})=r_{T_n(R)}(A^{2m})$ and so $T_n(R)$ is right strongly Hopfian by induction.

$(2)\Rightarrow (1)$ This follows easily from Lemma 2.3.

Corollary 2.5

Let $L_n(R)$ denote the lower triangular matrix ring over R. Then the following conditions are equivalent:

• (1) R is right strongly Hopfian;

• (2) $L_n(R)$ is right strongly Hopfian.

Let

$$\begin{aligned} S_n(R)= & {} \left. \left\{ \left( \begin{array}{llll} a&{}a_{12}&{}\cdots &{} a_{1n}\\ 0&{}a&{}\cdots &{} a_{2n}\\ \cdots &{}\cdots &{}\cdots &{}\cdots \\ 0&{}0&{}\cdots &{}a\\ \end{array} \right) \right| a, a_{ij}\in R\right\} ,\\ G_n(R)= & {} \left. \left\{ \left( \begin{array}{llll} a_1&{}a_{2}&{}\cdots &{} a_{n}\\ 0&{}a_1&{}\cdots &{} a_{n-1}\\ \cdots &{}\cdots &{}\cdots &{}\cdots \\ 0&{}0&{}\cdots &{}a_1\\ \end{array} \right) \right| a_{i}\in R, 1\le i\le n \right\} , \end{aligned}$$

and let $R\bowtie R$ denote the trivial extension of R by R.

Corollary 2.6

The following conditions are equivalent:

• (1) R is right strongly Hopfian;

• (2) $S_n(R)$ is right strongly Hopfian;

• (3) $G_n(R)$ is right strongly Hopfian;

• (4) $R[x]/(x^n)$ is right strongly Hopfian;

• (5) $R\bowtie R$ is right strongly Hopfian.

Proof

Note that $R[x]/(x^n)\cong G_n(R)$ and $R\bowtie R\cong G_2(R).$

Let R be a ring. Immediately, we deduce that the lower triangular matrix ring over R is right strongly Hopfian if and only if the upper triangular matrix ring is right strongly Hopfian. Let R be a ring, and let $W(R)=\left. \left\{ \left( \begin{array}{lll}a_{11}&{}0&{}0\\ a_{21}&{}a_{22}&{}a_{23}\\ 0&{}0&{}a_{33}\\ \end{array}\right) \right| a_{ij}\in R\right\}$. Then W(R) is a $3\times 3$ subring of $M_3(R)$ under usual matrix addition and multiplication. A natural problem asks if the right strongly Hopfian property of such a ring coincides with that of R. This inspires us to consider the right strongly Hopfian property of W(R).

Proposition 2.7

Let R be a ring. Then W(R) is right strongly Hopfian if and only if R is right strongly Hopfian.

Proof

Suppose R is right strongly Hopfian and let

$$\begin{aligned} \alpha =\left( \begin{array}{lll}a&{}\quad 0&{}\quad 0\\ x&{} \quad b&{} \quad y\\ 0&{}\quad 0&{} \quad c\\ \end{array}\right) \in W(R). \end{aligned}$$

Then there exists $m\in \mathbb {N}$ such that for any $n>m$, $r_R(a^n)=r_R(a^m)$, $r_R(b^n)=r_R(b^m)$, and $r_R(c^n)=r_R(c^m)$. Now we show that $r_{W(R)}(\alpha ^{2m+1})=r_{W(R)}(\alpha ^{2m})$. If

$$\begin{aligned} \beta =\left( \begin{array}{lll}d&{} \quad 0&{} \quad 0\\ s&{}\quad e&{} \quad t\\ 0&{}\quad 0&{} \quad f\\ \end{array}\right) \in r_{W(R)}(\alpha ^{2m+1}), \end{aligned}$$

then

$$\begin{aligned} \alpha ^{2m+1}\beta =\left( \begin{array}{lll}a^{2m+1}d&{}\quad 0&{} \quad 0\\ ud+b^{2m+1}s&{}\quad b^{2m+1}e&{} \quad b^{2m+1}t+vf\\ 0&{}\quad 0&{} \quad c^{2m+1}f\\ \end{array}\right) =0, \end{aligned}$$

where

$$\begin{aligned} u=xa^{2m}+bxa^{2m-1}+\cdots +b^mxa^m+b^{m+1}xa^{m-1}+\cdots +b^{2m-1}xa+b^{2m}x, \end{aligned}$$

and

$$\begin{aligned} v=b^{2m}y+b^{2m-1}yc+b^{2m-2}yc^2+\cdots +byc^{2m-1}+yc^{2m}. \end{aligned}$$

Hence

$$\begin{aligned} d\in r_R(a^{2m+1})= & {} r_R(a^{2m})=\cdots =r_R(a^m),\\ e\in r_R(b^{2m+1})= & {} r_R(b^{2m})=\cdots =r_R(b^m), \end{aligned}$$

and

$$\begin{aligned} f\in r_R(c^{2m+1})=r_R(c^{2m})=\cdots =r_R(c^m). \end{aligned}$$

Then

$$\begin{aligned} 0= & {} ud+b^{2m+1}s\\= & {} (xa^{2m}+bxa^{2m-1}+\cdots +b^mxa^m+b^{m+1}xa^{m-1}+\cdots +b^{2m}x)d+b^{2m+1}s\\= & {} (b^{m+1}xa^{m-1}+b^{m+2}xa^{m-2}+\cdots +b^{2m}x)d+b^{2m+1}s\\= & {} b^{m+1}((xa^{m-1}+bxa^{m-2}+\cdots +b^{m-1}x)d+b^ms), \end{aligned}$$

and

$$\begin{aligned} 0= & {} b^{2m+1}t+vf\\= & {} b^{2m+1}t+(b^{2m}y+b^{2m-1}yc+\cdots +b^{m+1}yc^{m-1}+b^myc^m+\cdots +yc^{2m})f\\= & {} b^{2m+1}t+(b^{2m}y+b^{2m-1}yc+\cdots +b^{m+1}yc^{m-1})f\\= & {} b^{m+1}(b^mt+(b^{m-1}y+b^{m-2}yc+\cdots +yc^{m-1})f). \end{aligned}$$

Hence

$$\begin{aligned} (xa^{m-1}+bxa^{m-2}+\cdots +b^{m-1}x)d+b^ms\in r_R(b^{m+1})=r_R(b^m) \end{aligned}$$

and

$$\begin{aligned} b^mt+(b^{m-1}y+b^{m-2}yc+\cdots +yc^{m-1})f\in r_R((b^{m+1})=r_R(b^m). \end{aligned}$$

So

$$\begin{aligned}&b^m((xa^{m-1}+bxa^{m-2}+\cdots +b^{m-1}x)d+b^ms)\\&\quad =b^mxa^{m-1}+b^{m+1}xa^{m-2}+\cdots +b^{2m-1}x)d+b^{2m}s=0, \end{aligned}$$

and

$$\begin{aligned}&b^m(b^mt+(b^{m-1}y+b^{m-2}yc+\cdots +yc^{m-1})f)\\&\quad =b^{2m}t+(b^{2m-1}y+b^{2m-2}yc+\cdots +b^myc^{m-1})f=0. \end{aligned}$$

Then by routine computations, we can show that $\alpha ^{2m}\beta =0$ and so $\beta \in r_{W(R)}(\alpha ^{2m})$. Hence $r_{W(R)}(\alpha ^{2m+1})=r_{W(R)}(\alpha ^{2m}).$ Therefore W(R) is right strongly Hopfian.

Conversely, if W(R) is right strongly Hopfian, then by Lemma 2.3 R is right strongly Hopfian.

Let $\alpha$ be an endomorphism and $\delta$ an $\alpha$-derivation of R. We denote by $R[x;\alpha , \delta ]$ the Ore extension whose elements are the polynomials over R, the addition is defined as usual, and the multiplication is subject to the relation $xa=\alpha (a)x+\delta (a)$ for any $a\in R$. From this rule, an inductive argument can be made in order to calculate an expression for $x^ja$, for all $j\in \mathbb {N}$ and $a\in R$. To recall this result, we shall use some convenient notation introduced in [9].

Notation 2.8

Let $\delta$ be an $\alpha$-derivation of R. For integers i, j with $0\le i\le j$, $f^j_i\in End(R, + )$ will denote the map which is the sum of all possible words in $\alpha , \delta$ built with i letters $\alpha$ and $j-i$ letters $\delta$. For instance, $f_0^0=1, f^j_j=\alpha ^j, f^j_0=\delta ^j$, and $f^j_{j-1}=\alpha ^{j-1}\delta +\alpha ^{j-2}\delta \alpha +\cdots +\delta \alpha ^{j-1}.$ Using recursive formulas for the $f_i^j$’s and induction, as done in [9], one can show with a routine computation that

$$\begin{aligned} x^ja=\sum \limits _{i=0}^jf^j_i(a)x^i. \end{aligned}$$

The following Lemma is well known and we omit the proof (see [3, Lemma 2.1]).

Lemma 2.9

Let R be an $(\alpha ,\delta )$-compatible ring. Then we have the following:

• (1) If $ab=0$, then $a\alpha ^n(b) =\alpha ^n(a)b=0$ for all positive integers n.

• (2) If $\alpha ^k(a)b=0$ for some positive integer k, then $ab=0$.

• (3) If $ab=0$, then $\alpha ^n(a)\delta ^m(b)=0=\delta ^ m(a)\alpha ^n(b)$ for all positive integers m, n.

• (4) If $ab=0$, then $af^j_i(b)=0$ and $f_i^j(a)b=0$ for all i, j.

Lemma 2.10

Let R be an $\alpha$-compatible ring. If $\alpha ^{k_1}(a_1)\alpha ^{k_2}(a_2)\cdots \alpha ^{k_n}(a_n)=0$ for some positive integers, then $a_1a_2\cdots a_n=0$.

Proof

Using induction, for $n=1$, the result is true by the injectivity of $\alpha$. Now suppose $\alpha ^{k_1}(a_1)\alpha ^{k_2}(a_2)\cdots \alpha ^{k_n}(a_n)=0$. Then $\alpha ^{k_1}(a_1)\alpha ^{k_2}(a_2)\cdots \alpha ^{k_{n-1}}(a_{n-1})a_n=0$, and so $\alpha ^{k_1}(a_1)\alpha ^{k_2}(a_2)\cdots \alpha ^{k_{n-1}}(a_{n-1}a_n)=0$. Then $a_1a_2\cdots a_n=0$.

Lemma 2.11

Let R be an $(\alpha ,\delta )$-compatible ring, $f(x)=a_0+a_1x+\cdots +a_nx^n$ and $g(x)=b_0+b_1x+\cdots +b_mx^m$ be two polynomials in $R[x;\alpha ,\delta ].$ Then we have the following:

• (1) If for all $0\le i\le n$ and $0\le j\le m$, $a_ib_j=0,$ then $f(x)g(x)=0$.

• (2) If R is semicommutative and $c\in R$ is such that for all $0\le j\le m$, $cb_j=0$, then $cf(x)g(x)=0$.

Proof

(1) We have

$$\begin{aligned} f(x)g(x)= & {} (a_0+a_1x+\cdots +a_nx^n)(b_0+b_1x+\cdots +b_mx^m)\\= & {} \sum \limits _{l=0}^{m+n}\left( \sum \limits _{s+t=l}\left( \sum _{i=s}^na_if^i_s(b_t)\right) \right) x^l. \end{aligned}$$

By Lemma 2.9, $a_ib_t=0$ implies $a_if^i_s(b_t)=0$. Thus it is easy to see that $f(x)g(x)=0$.

(2) Since R is semicommutative, for all $0\le i\le n$ and $0\le j\le m$, $cb_j=0$ implies $ca_ib_j=0$. Thus by (1) we complete the proof.

For two polynomials f(x) and g(x) in $R[x;\alpha ,\delta ]$, in order to calculate an expression for $(f(x)+g(x))^n$, for all $n\in \mathbb {N}$, we denote by $[Q^n_if(x)g(x)]$ the polynomial which is the sum of all possible terms, with each term being a product of i polynomials f(x) and $n-i$ polynomials g(x). Using this convenient notation, we have $(f(x)+g(x))^n=f(x)^n+[Q^n_{n-1}f(x)g(x)]+[Q^n_{n-2}f(x)g(x)]+\cdots +[Q^n_{1}f(x)g(x)]+g(x)^n$.

Lemma 2.12

Let R be an $(\alpha ,\delta )$-compatible semicommutative ring, $ax^r$, $f(x)=b_0+b_1x+\cdots +b_mx^m$, $g(x)=c_0+c_1x+\cdots +c_qx^q$ be three polynomials in $R[x;\alpha ,\delta ]$. If $c_j\in r_R(a^n)$ for all $0\le j\le q$, then for any $p>n$, $[Q_n^p(ax^r)f(x)]g(x)=0$.

Proof

It is easy to check that the coefficients of $[Q^p_n(ax^r)f(x)$] can be written as sums of monomials of length p in $f^t_s(a)$ and $f^v_u(b_j)$, where $b_j\in \{ b_0, b_1,\ldots , b_m\}$ and $t\ge s\ge 0$, $v\ge u\ge 0$ are nonnegative positive integers. Consider each monomial $f^{t_1}_{s_1}(v_1)f^{t_2}_{s_2}(v_2)\cdots f^{t_p}_{s_p}(v_p)$ where $v_1$, $v_2$, ${\ldots }$, . It would contain n letters a. Suppose $v_{r_1}=v_{r_2}=\cdots =v_{r_n}=a$ for some $1\le r_1< r_2<\cdots <r_n\le p$. Then we write the monomial $f^{t_1}_{s_1}(v_1)f^{t_2}_{s_2}(v_2)\cdots f^{t_p}_{s_p}(v_p)$ as

$$\begin{aligned} f^{t_1}_{s_1}(v_1)\cdots f^{t_{r_1}}_{s_{r_1}}(a) f^{t_{r_1+1}}_{s_{r_1+1}}(v_{r_1+1})\cdots f^{t_{r_n-1}}_{s_{r_n-1}}(v_{r_n-1})f^{t_{r_n}}_{s_{r_n}}(a)f^{t_{r_n+1}}_{s_{r_n+1}}(v_{r_n+1})\cdots f^{t_p}_{s_p}(v_p), \end{aligned}$$

where $v_s\in \{b_0, b_1, \ldots , b_m\}$ if $s\not \in \{r_1, r_2,\ldots , r_n\}$. For each $0\le j\le q$, since R is $(\alpha ,\delta )$-compatible and semicommutative, $a^nc_j=aa\cdots ac_j=0$ implies

$$\begin{aligned} f^{t_{r_1}}_{s_{r_1}}(a)f^{t_{r_2}}_{s_{r_2}}(a)\cdots f^{t_{r_n}}_{s_{r_n}}(a)c_j=0, \end{aligned}$$

and so

$$\begin{aligned}&f^{t_1}_{s_1}(v_1)\cdots f^{t_{r_1}}_{s_{r_1}}(a) f^{t_{r_1+1}}_{s_{r_1+1}}(v_{r_1+1})\cdots \\&f^{t_{r_n-1}}_{s_{r_n-1}}(v_{r_n-1})f^{t_{r_n}}_{s_{r_n}}(a)f^{t_{r_n+1}}_{s_{r_n+1}}(v_{r_n+1})\cdots f^{t_p}_{s_p}(v_p)c_j=0. \end{aligned}$$

Thus by Lemma 2.11 we complete the proof.

The same idea can be used to prove the following.

Corollary 2.13

Let R be an $(\alpha ,\delta )$-compatible semicommutative ring, $ax^r$, $f(x)=b_0+b_1x+\cdots +b_mx^m$, $g(x)=c_0+c_1x+\cdots +c_qx^q$ be three polynomials in $R[x;\alpha ,\delta ]$. If $c_j\in r_R(a^n)$ for all $0\le j\le q$, Then we have the following:

• (1) For any $p>n+l$, $[Q_{n+l}^p(ax^r)f(x)]g(x)=0$.

• (2) $R[x;\alpha ,\delta ](a^{i_1}x^{n_1})R[x;\alpha ,\delta ](a^{i_2}x^{n_2})R[x;\alpha ,\delta ]\cdots (a^{i_k}x^{n_k})R[x;\alpha ,\delta ]g(x)=0$ if $i_1+i_2+\cdots +i_k\ge n$.

Proposition 2.14

Let R be $(\alpha ,\delta )$-compatible and $R[x;\alpha ,\delta ]$ be reversible. Then the following conditions are equivalent:

• (1) R is right strongly Hopfian;

• (2) $R[x;\alpha ,\delta ]$ is right strongly Hopfian.

Proof

$(1)\Rightarrow (2)$ Let $f(x)=a_0+a_1x+\cdots +a_nx^n\in R[x;\alpha ,\delta ]$. Since R is right strongly Hopfian, there exists $k\in \mathbb {N}$ such that for all $l>k$ and all $0\le i\le n$, $r_R(a_i^l)=r_R(a_i^k)$. Now we show that $r_{R[x;\alpha ,\delta ]}(f(x)^{(n+1)k+1})=r_{R[x;\alpha ,\delta ]}(f(x)^{(n+1)k}).$ If

$$\begin{aligned} g(x)=b_0+b_1x+\cdots +b_mx^m\in r_{R[x;\alpha ,\delta ]}(f(x)^{(n+1)k+1}), \end{aligned}$$

then

$$\begin{aligned} 0= & {} f(x)^{(n+1)k+1}g(x)=(a_0+a_1x+\cdots +a_nx^n)^{(n+1)k+1}(b_0+b_1x+\cdots +b_mx^m)\\= & {} a_n\alpha ^n(a_n)\alpha ^{2n}(a_n)\cdots \alpha ^{(n+1)kn}(a_n)\alpha ^{[(n+1)k\!+\!1]n}(b_m)x^{[(n+1)k+1]n+m}\!+\! {\mathrm{lower}}\,\, {\mathrm{terms}}. \end{aligned}$$

Hence

$$\begin{aligned} a_n\alpha ^n(a_n)\alpha ^{2n}(a_n)\cdots \alpha ^{(n+1)kn}(a_n)\alpha ^{[(n+1)k+1]n}(b_m)=0. \end{aligned}$$

By Lemma 2.10, we obtain $a_n^{(n+1)k+1}b_m=0.$ Hence

$$\begin{aligned} b_m\in r_R\left( a_n^{(n+1)k+1}\right) =r_R\left( a_n^k\right) . \end{aligned}$$

From $f(x)^{(n+1)k+1}g(x)=0$, we have $a_n^kf(x)^{(n+1)k+1}g(x)=0$. Then by Lemma 2.11 we obtain

$$\begin{aligned} 0= & {} a_n^kf(x)^{(n+1)k+1}g(x)\\= & {} a_n^k(a_0+a_1x+\cdots +a_nx^n)^{(n+1)k+1} (b_0+b_1x+\cdots +b_mx^m)\\= & {} a_n^k(a_0+a_1x+\cdots +a_nx^n)^{(n+1)k+1}(b_0+b_1x+\cdots +b_{m-1}x^{m-1})\\&+a_n^k(a_0+a_1x+\cdots +a_nx^n)^{(n+1)k+1}b_mx^m\\= & {} a_n^k(a_0+a_1x+\cdots +a_nx^n)^{(n+1)k+1}(b_0+b_1x+\cdots +b_{m-1}x^{m-1})\\= & {} a_n^ka_n\alpha ^n(a_n)\cdots \alpha ^{(n+1)kn}(a_n)\alpha ^{[(n+1)k+1]n}(b_{m-1})x^{[(n+1)k+1]n+m-1}+ {\mathrm{lower}}\,\, {\mathrm{terms}}. \end{aligned}$$

Hence

$$\begin{aligned} a_n^{k+1}\alpha ^n(a_n)\alpha ^{2n}(a_n)\cdots \alpha ^{(n+1)kn}(a_n)\alpha ^{[(n+1)k+1]n}(b_{m-1})=0 \end{aligned}$$

and so

$$\begin{aligned} b_{m-1}\in r_R\left( a_n^{(n+2)k+1}\right) =r_R\left( a_n^k\right) . \end{aligned}$$

Using the same method repeatedly, we obtain

$b_j\in r_R(a_n^{(n+1)k+1})=r_R(a_n^k)$ for all $0\le j\le m$.

Consider the polynomial f(x) as the sum of two polynomials $a_nx^n$ and $h(x)=a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots +a_0.$ Then by Corollary 2.13 we obtain

$$\begin{aligned} 0= & {} f(x)^{(n+1)k+1}g(x)=\left( a_nx^n+h(x)\right) ^{(n+1)k+1}g(x)\nonumber \\= & {} (a_nx^n)^{(n+1)k+1}g(x)+\left[ Q_{(n+1)k}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)+\cdots \nonumber \\&+ \left[ Q_{k}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x) +\left[ Q_{k-1}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)\nonumber \\&+\cdots +\left[ Q_{1}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x) +h(x)^{(n+1)k+1}g(x)\nonumber \\= & {} \left[ Q_{k-1}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)+\left[ Q_{k-2}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)+\cdots \nonumber \\&+\left[ Q_{1}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)+ h(x)^{(n+1)k+1}g(x). \end{aligned}$$

(1)

Multiplying Eq. (1) on the left side by $(a_nx^n)^{k-1}$, then by Lemma 2.12 and Corollary 2.13, we obtain

$$\begin{aligned} (a_nx^n)^{k-1}h(x)^{(n+1)k+1}g(x)=0. \end{aligned}$$

Since $R[x;\alpha ,\delta ]$ is reversible, this implies

$$\begin{aligned} \left[ Q_{k-1}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)h(x)^{k-1}=0. \end{aligned}$$

(2)

Multiplying Eq. (1) on the right side by $h(x)^{k-1}$, we obtain

$$\begin{aligned}&\left[ Q_{k-2}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)h(x)^{k-1} +\left[ Q_{k-3}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)h(x)^{k-1}\nonumber \\&\quad +\cdots +h(x)^{(n+1)k+1}g(x)h(x)^{k-1}=0. \end{aligned}$$

(3)

Multiplying Eq. (3) on the left side by $(a_nx^n)^{k-2}$, we obtain

$$\begin{aligned}&(a_nx^n)^{k-2}\left[ Q_{1}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)h(x)^{k-1}\nonumber \\&\quad +(a_nx^n)^{k-2}h(x)^{(n+1)k+1}g(x) h(x)^{k-1}=0. \end{aligned}$$

(4)

By equation $(a_nx^n)^{k-1}h(x)^{(n+1)k+1}g(x)=0$ and because $R[x;\alpha ,\delta ]$ is reversible, it is easy to see that $(a_nx^n)^{k-2}\left[ Q_{1}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)h(x)^{k-1}=0$. Hence Eq. (4) becomes

$$\begin{aligned} (a_nx^n)^{k-2}h(x)^{(n+1)k+1}g(x) h(x)^{k-1}=0. \end{aligned}$$

Since $R[x;\alpha ,\delta ]$ is reversible, this implies

$$\begin{aligned} \left[ Q_{k-2}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)h(x)^{k-1}h(x)^{k-2}=0. \end{aligned}$$

Multiplying Eq. (3) on the right side by $h(x)^{k-2}$, we obtain

$$\begin{aligned}&\left[ Q_{k-3}^{(n+1)k+1}(a_nx^n)h(x)\right] g(x)h(x)^{k-1}h(x)^{k-2}\\&\quad + \cdots +h(x)^{(n+1)k+1}g(x)h(x)^{k-1}h(x)^{k-2}=0. \end{aligned}$$

Continuing this process yields that

$$\begin{aligned} h(x)^{(n+1)k+1}g(x)h(x)^{k-1}h(x)^{k-2}\cdots h(x)=0, \end{aligned}$$

and so

$$\begin{aligned}&h(x)^{(n+1)k+1+\frac{k(k-1)}{2}}g(x)\\&\quad =(a_0+a_1x+\cdots +a_{n-1}x^{n-1})^{(n+1)k+1+\frac{k(k-1)}{2}}(b_0+ b_1x+\cdots b_mx^m)=0, \end{aligned}$$

since $R[x;\alpha ,\delta ]$ is reversible. Now by the same way as above, we obtain

$$\begin{aligned} b_j\in r_R\left( a_{n-1}^{(n+1)k+1+\frac{k(k-1)}{2}}\right) =r_R\left( a_{n-1}^{k}\right) \end{aligned}$$

for all $0\le j\le m$. Using induction on n, we obtain

$$\begin{aligned} b_j\in r_R\left( a_i^{(n+1)k+1}\right) =r_R\left( a_i^{k}\right) \end{aligned}$$

for all $0\le j\le m$ and $0\le i\le n$. It is now easy to check that $f(x)^{(n+1)k}g(x)=0$. Hence $r_{R[x;\alpha ,\delta ]}(f(x)^{(n+1)k+1})=r_{R[x;\alpha ,\delta ]}(f(x)^{(n+1)k})$. Therefore $R[x;\alpha ,\delta ]$ is right strongly Hopfian.

$(2)\Rightarrow (1)$ It is trivial.

Corollary 2.15

We have the following:

• (1) If R is $\alpha$-compatible and the skew polynomial ring $R[x;\alpha ]$ is reversible, then the skew polynomial ring $R[x;\alpha ]$ is right strongly Hopfian if and only if R is right strongly Hopfian.

• (2) If R is $\delta$-compatible and the differential polynomial ring $R[x;\delta ]$ is reversible, then the differential polynomial ring $R[x;\delta ]$ is right strongly Hopfian if and only if R is right strongly Hopfian.

Proof

It is an immediate consequence of Proposition 2.14.

Corollary 2.16

([5, Theorem 5.1]).   Let R be a commutative strongly Hopfian ring, then the polynomial ring R[x] is strongly Hopfian.

Let M be a multiplicative monoid. In the following, e will always stand for the identity of M. Then R[M] will denote the monoid ring over R consisting of all elements of the form $\sum _{i=1}^n r_ig_i$ with $r_i\in R$, $g_i\in M$, $i=1, 2,\ldots ,n$, where the addition is given naturally and the multiplication is given by

$$\begin{aligned} \left( \sum _{i=1}^n r_ig_i\right) \left( \sum _{j=1}^m s_jh_j\right) =\sum _{i=1}^n\sum _{j=1}^m(r_is_j)(g_ih_j). \end{aligned}$$

Recall that the ordered monoid $(M,\le )$ is a strictly ordered monoid if for any g, $g'$, $h\in M$, $g<g'$ implies that $gh<g'h$ and $hg<hg'$.

For two elements $\alpha$ and $\beta$ in R[M], in order to calculate an expression for $(\alpha +\beta )^n$, for all $n\in \mathbb {N}$, we denote by $[Q_i^n\alpha \beta ]$ the sum of all possible terms with each term being a product of i elements $\alpha$ and $n-i$ elements $\beta$. Using this convenient notation, we have $(\alpha +\beta )^n=\alpha ^n+[Q_{n-1}^n\alpha \beta ]+[Q_{n-2}^n\alpha \beta ]+\cdots + [Q_1^n\alpha \beta ]+\beta ^n$.

Lemma 2.17

Let $(M,\le )$ be a strictly totally ordered monoid and R a semicommutative ring, $\alpha =ag$, $\beta =b_1h_1+b_2h_2+\cdots +b_nh_n$ and $\gamma =c_1v_1+c_2v_2+\cdots +c_mv_m$ be three elements in R[M]. If there exists a positive integer $n\in \mathbb {Z}$ such that $c_j\in r_R(a^n)$ for all $1\le j\le m$, then for any $p>n$, $[Q_n^p\alpha \beta ]\gamma =0$.

Proof

The coefficients of $[Q_n^p\alpha \beta ]$ can be written as sums of monomials of length p in a and $b_j$, where $b_j\in \{b_1, b_2,\ldots , b_n\}$. Consider one of such monomials, $d_1d_2\cdots d_p$, where $d_i\in \{a, b_1, b_2,\ldots , b_n\}$, $0\le i\le p$. It would contain n letters a. Suppose $d_{r_1}=d_{r_2}=d_{r_n}=a$ for some $1\le r_1<r_2<\cdots <r_n\le p$. Then we can write the monomial as $d_1d_2\cdots d_{r_1-1}ad_{r_1+1}\cdots d_{r_n-1}ad_{r_n+1}\cdots d_p$. Since R is semicommutative and $c_j\in r_R(a^n)$ for all $1\le j\le m$, $a^nc_j=aa\cdots ac_j=0$ implies

$$\begin{aligned} d_1d_2\cdots d_{r_1-1}ad_{r_1+1}\cdots d_{r_n-1}ad_{r_n+1}\cdots d_pc_j=0 \end{aligned}$$

for all $1\le j\le m$. Hence each monomial appearing in $[Q_n^p\alpha \beta ]\gamma$ is equal to 0. Therefore $[Q_n^p\alpha \beta ]\gamma =0$.

The same idea can be used to prove the following.

Corollary 2.18

Let $(M,\le )$ be a strictly totally ordered monoid and R a semicommutative ring, $\alpha =ag$, $\beta =b_1h_1+b_2h_2+\cdots +b_nh_n$ and $\gamma =c_1v_1+c_2v_2+\cdots +c_mv_m$ be three elements in R[M]. If there exists a positive integer $n\in \mathbb {Z}$ such that $c_j\in r_R(a^n)$ for all $1\le j\le m$, then for any $p>n +l$, we have the following:

• (1) $[Q_{n+l}^p\alpha \beta ]\gamma =0$.

• (2) $R[M](ag)^{n_1}R[M](ag)^{n_2}R[M]\cdots (ag)^{n_k}R[M]\gamma R[M]=0$ if $n_1+n_2+\cdots +n_k\ge n$.

Proposition 2.19

Let M be a strictly totally ordered monoid and R[M] a reversible ring. Then the following conditions are equivalent:

• (1) R is a right strongly Hopfian ring;

• (2) R[M] is a right strongly Hopfian ring.

Proof

$(1)\Rightarrow (2)$. Let $\alpha =a_1g_1+a_2g_2+\cdots +a_ng_n\in R[M]$ with $g_i<g_j$ if $i<j$. Since R is right strongly Hopfian, there exists $k\in \mathbb {N}$ such that for all $l>k$ and all $1\le i\le n$, $r_R(a_i^l)=r_R(a_i^k).$ Now we show that $r_{R[M]}(\alpha ^{nk+1})=r_{R[M]}(\alpha ^{nk})$. If $\beta =b_1h_1+b_2h_2+\cdots b_mh_m\in r_{R[M]}(\alpha ^{nk+1})$ with $h_s<h_t$ if $s<t$, then

$$\begin{aligned} 0=\alpha ^{nk+1}\beta =(a_1g_1+a_2g_2+\cdots +a_ng_n)^{nk+1}(b_1h_1+b_2h_2+\cdots b_mh_m). \end{aligned}$$

Considering the coefficient of the largest element $g_n^{nk+1}h_m$ in $\alpha ^{nk+1}\beta$, we obtain $a_n^{nk+1}b_m=0$. Hence $b_m\in r_R(a_n^{nk+1})=r_R(a_n^k)$. From $\alpha ^{nk+1}\beta =0$, we have

$$\begin{aligned} 0= & {} (a_n^ke)\alpha ^{nk+1}\beta =(a_n^ke)(a_1g_1+a_2g_2+\cdots +a_ng_n)^{nk+1}(b_1h_1+b_2h_2+\cdots b_mh_m)\\= & {} (a_n^ke)(a_1g_1+a_2g_2+\cdots +a_ng_n)^{nk+1}(b_1h_1+b_2h_2+\cdots b_{m-1}h_{m-1})\\&+(a_n^ke)(a_1g_1+a_2g_2+\cdots +a_ng_n)^{nk+1}b_mh_m\\= & {} (a_n^ke)(a_1g_1+a_2g_2+\cdots +a_ng_n)^{nk+1}(b_1h_1+b_2h_2+\cdots b_{m-1}h_{m-1})\\= & {} (a_n^ke)\alpha ^{nk+1}(\beta -b_mh_m). \end{aligned}$$

Considering the coefficient of the largest element $g_n^{nk+1}h_{m-1}$ in $(a_n^ke)\alpha ^{nk+1}(\beta -b_mh_m)$, we obtain

$$\begin{aligned} b_{m-1}\in r_R\left( a_n^{(n+1)k+1}\right) =r_R\left( a_n^k\right) . \end{aligned}$$

Continuing this process yields that $b_j\in r_R(a_n^{nk+1})=r_R(a_n^k)$ for all $1\le j\le m$. Consider the element $\alpha$ as the sum of two elements $a_ng_n$ and $\gamma =a_1g_1+a_2g_2+\cdots +a_{n-1}g_{n-1}$. Then by Lemma 2.17 and Corollary 2.18 we obtain

$$\begin{aligned} 0= & {} \alpha ^{nk+1}\beta =(a_ng_n+\gamma )^{nk+1}\beta \nonumber \\= & {} (a_ng_n)^{nk+1}\beta +\left[ Q_{nk}^{nk+1}(a_ng_n)\gamma \right] \beta + \cdots +\left[ Q_{k-1}^{nk+1}(a_ng_n)\gamma \right] \beta +\cdots +\gamma ^{nk+1}\beta \nonumber \\= & {} \left[ Q_{k-1}^{nk+1}(a_ng_n)\gamma \right] \beta +\left[ Q_{k-2}^{nk+1}(a_ng_n)\gamma \right] \beta +\cdots +\gamma ^{nk+1}\beta . \end{aligned}$$

(5)

Multiplying Eq. (5) on the left side by $(a_ng_n)^{k-1}$, and by Corollary 2.18, we obtain $(a_ng_n)^{k-1}\gamma ^{nk+1}\beta =0$ and so $[Q_{k-1}^{nk+1}(a_ng_n)\gamma ]\beta \gamma ^{k-1}=0$ since R[M] is reversible. Multiplying Eq. (5) on the right side by $\gamma ^{k-1}$, we obtain

$$\begin{aligned} \left[ Q_{k-2}^{nk+1}(a_ng_n)\gamma \right] \beta \gamma ^{k-1}+\left[ Q_{k-3}^{nk+1}(a_ng_n)\gamma \right] \beta \gamma ^{k-1} +\cdots +\gamma ^{nk+1}\beta \gamma ^{k-1}=0.\nonumber \\ \end{aligned}$$

(6)

Multiplying Eq. (6) on the left side by $(a_ng_n)^{k-2}$, we obtain

$$\begin{aligned} (a_ng_n)^{k-2}\left[ Q_1^{nk+1}(a_ng_n)\gamma \right] \beta \gamma ^{k-1}+(a_ng_n)^{k-2}\gamma ^{nk+1}\beta \gamma ^{k-1}=0. \end{aligned}$$

Since R[M] is reversible, $(a_ng_n)^{k-1}\gamma ^{nk+1}\beta =0$ implies

$$\begin{aligned} (a_ng_n)^{k-2}\left[ Q_1^{nk+1}(a_ng_n)\gamma \right] \beta \gamma ^{k-1}=0. \end{aligned}$$

Hence we obtain $(a_ng_n)^{k-2}\gamma ^{nk+1}\beta \gamma ^{k-1}=0$, and so $[Q_{k-2}^{nk+1}(a_ng_n)\gamma ]\beta \gamma ^{k-1}\gamma ^{k-2}=0$ since R[M] is reversible. Then multiplying Eq. (6) on the right side by $\gamma ^{k-2}$, we obtain

$$\begin{aligned}&\left[ Q_{k-3}^{nk+1}(a_ng_n)\gamma \right] \beta \gamma ^{k-1}\gamma ^{k-2} +\left[ Q_{k-4}^{nk+1}(a_ng_n)\gamma \right] \beta \gamma ^{k-1}\gamma ^{k-2}\\&\quad +\cdots +\gamma ^{nk+1}\beta \gamma ^{k-1}\gamma ^{k-2}=0. \end{aligned}$$

Continuing this process, we obtain $\gamma ^{nk+1}\beta \gamma ^{\frac{k(k-1)}{2}}=0$ and so $\gamma ^{nk+1+\frac{k(k-1)}{2}}\beta =0$. Using the same way as above, we can show that

$$\begin{aligned} b_j\in r_R(a_{n-1}^{nk+1+\frac{k(k-1)}{2}})=r_R(a_{n-1}^k) \end{aligned}$$

for all $1\le j\le m$. Using induction on n, we obtain

$$\begin{aligned} b_j\in r_R\left( a_i^{nk+1}\right) =r_R\left( a_i^k\right) \end{aligned}$$

for all $1\le j\le m$ and $1\le i\le n$. Then it is easy to check that $\alpha ^{nk}\beta =0$. Hence $\beta \in r_{R[M]}(\alpha ^{nk})$, and so $r_{R[M]}(\alpha ^{nk+1})=r_{R[M]}(\alpha ^{nk})$. Therefore R[M] is right strongly Hopfian.

$(2)\Rightarrow (1)$ It is trivial.

Corollary 2.20

([5, Corollary 5.4]) Let R be a commutative strongly Hopfian ring, then $R[x,x^{-1}]$ is strongly Hopfian.