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Abstract

In this paper we study the transfer of the property of Hopfian modules between the right ${\textstyle R}$-module ${\textstyle M_{R}}$ and some of its extension classes. Namely, under certain conditions, we show that: ${\textstyle M_{R}}$ is a Hopfian right ${\textstyle R}$-module if and only if the skew generalized power series module ${\textstyle [[M^{S,\leq }]]}$ is a Hopfian right ${\textstyle [[R^{S,\leq },\omega ]]}$-module.

Keywords

Hopfian module; Hopfian ring; Skew generalized power series module

2010 Mathematics Subject Classification

primary, 06F05, 16W60; secondary, 13E10

1. Introduction

Throughout this paper ${\textstyle R}$ denotes an associative ring not necessarily commutative with the identity and ${\textstyle M_{R}}$ a unitary right ${\textstyle R}$-module. As it has been noted by Hiremath  [3], the concept of Hopfian groups was introduced by Baumslag  [1]. In fact, the study of endomorphism rings of various rings and modules has been a topic of keen interest since the end of the nineteen sixties when injectivity and its variants began to flourish. In 1986, Hiremath introduced the concept of the Hopfian module as follows: A right ${\textstyle R}$-module ${\textstyle M_{R}}$ is called Hopfian   if any surjective endomorphism of ${\textstyle M_{R}}$ is an isomorphism. The term “Hopfian” is said to be in honor of Heinz Hopf and his use of the concept of the Hopfian group in his work on fundamental groups of surfaces. Any noetherian module is Hopfian and if ${\textstyle R}$ is a right noetherian ring, then every finitely generated ${\textstyle R}$-module is a Hopfian module. Also, a simple ring is Hopfian, since the kernel of any endomorphism is an ideal, which is necessarily zero in a simple ring.

The module ${\textstyle R_{R}}$ is Hopfian if and only if ${\textstyle R}$ is a directly finite ring. Symmetrically, these two are also equivalent to the left ${\textstyle R}$-module ${\textstyle _{R}R}$ being Hopfian. The full linear ring ${\textstyle End_{D}(V)}$ of a countable dimensional vector space is a Hopfian ring which is not Hopfian as a module, since it only has three ideals, but it is not directly finite.

Varadarajan  [5] and [6] showed that the right ${\textstyle R}$-module ${\textstyle M_{R}}$ is Hopfian if and only if the right ${\textstyle R[x]}$-module ${\textstyle M[x]}$ is Hopfian.

The motivation of this paper is to investigate how the property of Hopfian modules behaves under passage to the skew generalized power series modules.

2. Hopfian modules over skew generalized power series rings

In this section we extend the results of  [9] to the skew generalized power series modules.

Let ${\textstyle (S,\leq )}$ be an ordered commutative monoid. Unless stated otherwise, the operation of ${\textstyle S}$ will be denoted additively, and the identity element by 0. Recall that ${\textstyle (S,\leq )}$ is artinian if every strictly decreasing sequence of elements of ${\textstyle S}$ is finite and that ${\textstyle (S,\leq )}$ is narrow if every subset of pairwise order-incomparable elements of ${\textstyle S}$ is finite. The following construction is due to Zhongkui [10]:

Let ${\textstyle (S,\leq )}$ be a strictly ordered monoid (that is, if ${\textstyle s,s^{^{'}},t\in S}$ and ${\textstyle s<s^{^{'}}}$, then ${\textstyle s+t<s^{^{'}}+t}$), ${\textstyle R}$ a ring and ${\textstyle \omega :S\longrightarrow End(R)}$ a monoid homomorphism. Consider the set ${\textstyle A=[[R^{S,\leq },\omega ]]}$ of all maps ${\textstyle f:S\longrightarrow R}$ whose support (${\textstyle supp(f)=\lbrace s\in S\vert f(s)\not =0\rbrace }$) is artinian and narrow.

For every ${\textstyle s\in S}$ and ${\textstyle f,g\in A}$, let

${\displaystyle X_{s}(f,g)=\lbrace (u,v)\in S\times S\vert u+v=s;f(u)\not =0,g(v)\not =0\rbrace {\mbox{.}}}$

It follows from ([4], 4.1) that ${\textstyle X_{s}(f,g)}$ is a finite set.

This fact allows us to define the operation of multiplication (convolution) as follows:

${\displaystyle (fg)(s)=\sum _{(u,v)\in X_{s}(f,g)}f(u)\omega _{u}(g(v)){\mbox{,}}}$

and ${\textstyle (fg)(s)=0}$ if ${\textstyle X_{s}(f,g)=\phi }$. With this operation and pointwise addition ${\textstyle A=[[R^{S,\leq },\omega ]]}$ becomes a ring, which is called the  ring of skew generalized power series   with coefficients in ${\textstyle R}$ and exponents in ${\textstyle S}$.

In  [8], Zhao and Jiao generalized this construction to obtain the skew generalized power series modules over skew generalized power series rings, as follows:

Let ${\textstyle M_{R}}$ be a right ${\textstyle R}$-module, let ${\textstyle B}$ be the set of all maps ${\textstyle \varphi :S\longrightarrow M}$ such that supp${\textstyle (\varphi )=\lbrace s\in S\vert \varphi (s)\not =0\rbrace }$ is artinian and narrow. With pointwise addition, ${\textstyle B=[[M^{S,\leq }]]}$ is an abelian additive group. For each ${\textstyle f\in A=[[R^{S,\leq },\omega ]]}$ and ${\textstyle \varphi \in B}$, the set

${\displaystyle X_{s}(\varphi ,f)=\lbrace (u,v)\in S\times S\vert u+v=s;\varphi (u)\not =0,f(v)\not =0\rbrace }$

is finite (see  [9], Lemma 1). This allows us to define the scalar multiplication of the elements of ${\textstyle B}$ by scalars from ${\textstyle A}$ as follows:

${\displaystyle (\varphi f)(s)=\sum _{(u,v)\in X_{s}(\varphi ,f)}\varphi (u)\omega _{u}(f(v)){\mbox{,}}}$

and ${\textstyle (\varphi f)(s)=0}$ if ${\textstyle X_{s}(\varphi ,f)=\phi }$. With this operation and pointwise addition, one can easily show that ${\textstyle B}$ is a right ${\textstyle A}$-module, which is called the module of skew generalized power series   with coefficients in ${\textstyle M}$ and exponents in ${\textstyle S}$.

For every ${\textstyle s\in S}$ if we set ${\textstyle \omega (s)=Id_{R}\in Aut(R)\subset End(R)}$, the identity map of ${\textstyle R}$, then ${\textstyle A=[[R^{S,\leq },\omega ]]=[[R^{S,\leq }]]}$ is the ring of generalized power series in the sense of Ribenboim  [4] and ${\textstyle B=[[M^{S,\leq }]]}$ is the untwisted module of generalized power series in the sense of  [7].

For any ${\textstyle r\in R}$ we associated the map ${\textstyle c_{r}\in A}$ defined by:

${\displaystyle c_{r}(x)=\lbrace {\begin{array}{cc}r{\mbox{,}}&{\mbox{if}}x=0{\mbox{,}}\\0{\mbox{,}}&{\mbox{if}}x\not =0{\mbox{.}}\end{array}}}$

For any ${\textstyle m\in M}$ and ${\textstyle s\in S}$, we define a map ${\textstyle d_{m}^{s}\in B}$ by:

${\displaystyle d_{m}^{s}(x)=\lbrace {\begin{array}{cc}m{\mbox{,}}&{\mbox{if}}x=s{\mbox{,}}\\0{\mbox{,}}&{\mbox{if}}x\not =s{\mbox{.}}\end{array}}}$

If ${\textstyle (S,\leq )}$ is a strictly totally ordered monoid, then ${\textstyle supp(f)}$ is a nonempty well-ordered subset of ${\textstyle S}$, for every ${\textstyle 0\not =f\in A}$, and we denote by ${\textstyle \pi (f)}$ the smallest element of ${\textstyle supp(f)}$. Also, ${\textstyle supp(\varphi )}$ is a nonempty well-ordered subset of ${\textstyle S}$, for every ${\textstyle 0\not =\varphi \in B}$, and we denote by ${\textstyle \pi (\varphi )}$ the smallest element of ${\textstyle supp(\varphi )}$.

The following are required in the sequel.

Definition 1 [2].

A monoid ${\textstyle S}$ is called finitely generated if there exists a finite subset ${\textstyle \lbrace s_{1},\ldots ,s_{n}\rbrace }$ of ${\textstyle S}$ such that ${\textstyle S=\lbrace \sum _{i=1}^{n}k_{i}s_{i}\vert k_{i}\geq 0\rbrace }$.

Lemma 1 [2].

If  ${\textstyle S}$is a finitely generated monoid, then every ideal of  ${\textstyle S}$is finitely generated, so every strictly increasing sequence of ideals is finite.

The following lemma is crucial in developing the proof of the main result.

Lemma 2.

Let  ${\textstyle (S,\leq )}$be a strictly totally ordered monoid which is finitely generated and satisfies the condition that  ${\textstyle 0\leq s}$for every  ${\textstyle s\in S}$. Assume that  ${\textstyle \omega _{s}(1)=1}$for each  ${\textstyle s\in S}$. Then  ${\textstyle \pi (\alpha (\varphi ))\geq \pi (\varphi )}$where  ${\textstyle \alpha \in End_{A}(B)}$and  ${\textstyle 0\not =\varphi \in B}$.

Proof.

For each element ${\textstyle x\in S}$, we denote by ${\textstyle x^{+}}$ the subset ${\textstyle x^{+}=\lbrace x+y\vert {\mbox{   }}y\in S\rbrace }$. Set ${\textstyle s=\pi (\alpha (\varphi ))}$ and ${\textstyle t=\pi (\varphi )}$. Suppose that ${\textstyle s<t}$. Define ${\textstyle \psi _{1}\in B}$ via

${\displaystyle \psi _{1}(x)=\varphi (x+t){\mbox{.}}}$

Set ${\textstyle \varphi _{1}=\varphi -\psi _{1}e_{t}}$. If ${\textstyle \varphi _{1}=0}$, then ${\textstyle \varphi =\psi _{1}e_{t}}$. Thus

${\displaystyle \alpha (\varphi )=\alpha (\psi _{1}e_{t})=\alpha (\psi _{1})e_{t}{\mbox{.}}}$

For any ${\textstyle x\in S}$ with ${\textstyle x<t}$, we have

${\displaystyle {\begin{array}{l}\displaystyle (\alpha (\psi _{1})e_{t})(x)=\\\displaystyle =\sum _{(u,v)\in X_{x}(\alpha (\psi _{1}),e_{t})}\alpha (\psi _{1})(u)\omega _{u}(e_{t}(v)){\mbox{.}}\end{array}}}$

For each pair ${\textstyle (u,v)\in X_{x}(\alpha (\psi _{1}),e_{t})}$, we have ${\textstyle x=u+v<t}$. Then ${\textstyle u<t}$ and ${\textstyle v<t}$. Thus ${\textstyle e_{t}(v)=0}$ and ${\textstyle (\alpha (\psi _{1})e_{t})(x)=0}$. Hence ${\textstyle \pi (\alpha (\psi _{1})e_{t})\geq t}$. It follows that ${\textstyle s\geq t}$, a contradiction. So ${\textstyle \varphi _{1}\not =0}$. Since ${\textstyle \varphi =\varphi _{1}+\psi _{1}e_{t}}$, we have

${\displaystyle \alpha (\varphi )=\alpha (\varphi _{1}+\psi _{1}e_{t})=\alpha (\varphi _{1})+\alpha (\psi _{1})e_{t}{\mbox{.}}}$

It follows that

${\displaystyle supp(\alpha (\varphi ))\subseteq supp(\alpha (\varphi _{1}))\cup supp(\alpha (\psi _{1})e_{t}){\mbox{.}}}$

If ${\textstyle s\in supp(\alpha (\psi _{1})e_{t})}$, then ${\textstyle s\geq \pi (\alpha (\psi _{1})e_{t})\geq t}$, a contradiction with the assumption that ${\textstyle s<t}$. Thus ${\textstyle s\in supp(\alpha (\varphi _{1}))}$.

Denote ${\textstyle s_{1}=\pi (\alpha (\varphi _{1}))}$ and ${\textstyle t_{1}=\pi (\varphi _{1})}$. Then ${\textstyle s\geq s_{1}}$. Since

${\displaystyle {\begin{array}{l}\displaystyle \varphi _{1}(t)=(\varphi -\psi _{1}e_{t})(t)=\varphi (t)-(\psi _{1}e_{t})(t)=\varphi (t)-\\\displaystyle -\psi _{1}(0)\omega _{0}(e_{t}(t))=\varphi (t)-\psi _{1}(0)\omega _{0}(1)=\varphi (t)-\psi _{1}(0){\mbox{,}}\end{array}}}$

we have

${\displaystyle \varphi _{1}(t)=\varphi (t)-\varphi (0+t)=\varphi (t)-\varphi (t)=0{\mbox{.}}}$

It is clear that ${\textstyle t_{1}>t}$. If ${\textstyle t_{1}\in t^{+}=\lbrace t+u\vert u\in S\rbrace }$, then there exists ${\textstyle u\in S}$ such that ${\textstyle t_{1}=t+u}$. Thus

${\displaystyle {\begin{array}{l}\displaystyle 0\not =\varphi _{1}(t_{1})=(\varphi -\psi _{1}e_{t})(t_{1})=\varphi (t_{1})-(\psi _{1}e_{t})(t_{1})=\varphi (t_{1})-\\\displaystyle -\sum _{(y,z)\in X_{t_{1}}(\psi _{1},e_{t})}\psi _{1}(y)\omega _{y}(e_{t}(z))=\varphi (t_{1})-\\\displaystyle -\psi _{1}(u)\omega _{u}(e_{t}(t))=\varphi (t_{1})-\psi _{1}(u)\omega _{u}(1)=\varphi (t_{1})-\psi _{1}(u)=\\\displaystyle =\varphi (t_{1})-\varphi (u+t)=\varphi (t_{1})-\varphi (t_{1})=0{\mbox{,}}\end{array}}}$

a contradiction. Hence ${\textstyle t_{1}\not \in t^{+}}$ and ${\textstyle t_{1}+S\not \subseteq t+S}$. Then ${\textstyle t+S\subsetneqq (t+S)\cup (t_{1}+S)}$. Suppose that for a positive integer ${\textstyle n\geq 2}$, we have found ${\textstyle \varphi _{1},\psi _{1},\ldots ,\varphi _{n-1},\psi _{n-1}\in B}$ such that

${\displaystyle \psi _{i}(x)=\varphi _{i-1}(x+t_{i-1})\quad {\mbox{and}}\quad \varphi _{i}=\varphi _{i-1}-\psi _{i}e_{t_{i-1}}{\mbox{,}}}$

where

${\displaystyle t_{i}=\pi (\varphi _{i}),t<t_{1}<\cdots <t_{i-1}<t_{i}\quad {\mbox{and}}}$

We set ${\textstyle t_{0}=t}$ and ${\textstyle \varphi _{0}=\varphi }$. Define ${\textstyle \psi _{n}\in B}$ via

${\displaystyle \psi _{n}(x)=\varphi _{n-1}(x+t_{n-1}){\mbox{.}}}$

Let ${\textstyle \varphi _{n}=\varphi _{n-1}-\psi _{n}e_{t_{n-1}}}$. Hence

${\displaystyle \varphi =\psi _{1}e_{t}+\psi _{2}e_{t_{1}}+\cdots +\psi _{n}e_{t_{n-1}}+\varphi _{n}{\mbox{.}}}$

If ${\textstyle \varphi _{n}=0}$, then

${\displaystyle \alpha (\varphi )=\alpha (\psi _{1})e_{t}+\alpha (\psi _{2})e_{t_{1}}+\cdots +\alpha (\psi _{n})e_{t_{n-1}}{\mbox{.}}}$

Thus

${\displaystyle supp(\alpha (\varphi ))\subseteq {\underset {i=0}{\overset {n-1}{\bigcup }}}supp(\alpha (\psi _{i+1})e_{t_{i}}){\mbox{,}}}$

which implies that there exists ${\textstyle i}$ such that ${\textstyle s=\pi (\alpha (\varphi ))\in supp(\alpha (\psi _{i+1})e_{t_{i}})}$. Thus ${\textstyle s\geq \pi (\alpha (\psi _{i+1})e_{t_{i}})\geq t_{i}\geq t}$, a contradiction. Now, suppose that ${\textstyle \varphi _{n}\not =0}$. Denote ${\textstyle t_{n}=\pi (\varphi _{n})}$. Since ${\textstyle \varphi _{n}=\varphi _{n-1}-\psi _{n}e_{t_{n-1}}}$, we have

${\displaystyle {\begin{array}{l}\displaystyle \varphi _{n}(t_{n-1})=\varphi _{n-1}(t_{n-1})-(\psi _{n}e_{t_{n}-1})(t_{n-1})=\\\displaystyle =\varphi _{n-1}(t_{n-1})-\\\displaystyle -\sum _{(y,z)\in X_{t_{n}-1}(\psi _{n},e_{t_{n}-1})}\psi _{n}(y)\omega _{y}(e_{t_{n}-1}(z))=\\\displaystyle =\varphi _{n-1}(t_{n-1})-\psi _{n}(0)\omega _{0}(e_{t_{n}-1}(t_{n-1}))=\varphi _{n-1}(t_{n-1})-\\\displaystyle -\psi _{n}(0)\omega _{0}(1)=\varphi _{n-1}(t_{n-1})-\psi _{n}(0)=\varphi _{n-1}(t_{n-1})-\\\displaystyle -\varphi _{n-1}(0+t_{n-1})=0{\mbox{.}}\end{array}}}$

For any ${\textstyle x\in S}$ with ${\textstyle x<t_{n-1}}$ and for every ${\textstyle (y,z)\in X_{x}(\psi _{n},e_{t_{n-1}})}$, we have ${\textstyle x=y+z<t_{n-1}}$. Then ${\textstyle y<t_{n-1}}$ and ${\textstyle z<t_{n-1}}$, and hence ${\textstyle \omega _{y}(e_{t_{n-1}}(z))=\omega _{y}(0)=0}$. It follows that

${\displaystyle {\begin{array}{l}\displaystyle \varphi _{n}(x)=\varphi _{n-1}(x)-(\psi _{n}e_{t_{n}-1})(x)=0-\\\displaystyle -\sum _{(y,z)\in X_{x}(\psi _{n},e_{t_{n}-1})}\psi _{n}(y)\omega _{y}(e_{t_{n}-1}(z))=0{\mbox{.}}\end{array}}}$

So ${\textstyle \pi (\varphi _{n})=t_{n}>t_{n-1}}$. Thus

${\displaystyle t<t_{1}<\cdots <t_{n-1}<t_{n}{\mbox{.}}}$

Suppose that ${\textstyle t_{n}\in t^{+}\cup t_{1}^{+}\cup \cdots \cup t_{n-1}^{+}}$. Then there exists ${\textstyle i}$ such that

${\displaystyle t_{n}\not \in t_{n-1}^{+},\ldots ,t_{n}\not \in t_{i+1}^{+}{\mbox{,}}\quad {\mbox{but  }}t_{n}\in t_{i}^{+}{\mbox{.}}}$

Let ${\textstyle t_{n}=t_{i}+v}$ for some ${\textstyle v\in S}$. Then

${\displaystyle {\begin{array}{l}\displaystyle \varphi _{i}(t_{n})=(\psi _{i+1}e_{t_{i}}+\psi _{i+2}e_{t_{i}+1}+\cdots +\psi _{n}e_{t_{n}-1}+\varphi _{n})(t_{n})=\\\displaystyle =(\psi _{i+1}e_{t_{i}})(t_{n})+(\psi _{i+2}e_{t_{i}+1})(t_{n})+\cdots +\\\displaystyle +(\psi _{n}e_{t_{n}-1})(t_{n})+\varphi _{n}(t_{n}){\mbox{.}}\end{array}}}$

Note that, since ${\textstyle t_{n}\in t_{i}^{+}}$ and ${\textstyle t_{n}\not \in t_{i+1}^{+}}$, we see that

${\displaystyle {\begin{array}{l}\displaystyle (\psi _{i+1}e_{t_{i}})(t_{n})=\psi _{i+1}(v)\omega _{v}(e_{t_{i}}(t_{i}))=\\\displaystyle =\psi _{i+1}(v)\omega _{v}(1)=\psi _{i+1}(v){\mbox{,}}\end{array}}}$

and

${\displaystyle {\begin{array}{l}\displaystyle (\psi _{i+2}e_{t_{i}+1})(t_{n})=\\\displaystyle =\sum _{(y,z)\in X_{t_{n}}(\psi _{i+2},e_{t_{i}+1})}\psi _{i+2}(y)\omega _{y}(e_{t_{i}+1}(z))=0{\mbox{.}}\end{array}}}$

Thus

${\displaystyle {\begin{array}{l}\displaystyle \varphi _{i}(t_{n})=\psi _{i+1}(v)+\varphi _{n}(t_{n})=\varphi _{i}(v+t_{i})+\varphi _{n}(t_{n})=\varphi _{i}(t_{n})+\\\displaystyle +\varphi _{n}(t_{n}){\mbox{,}}\end{array}}}$

which implies that ${\textstyle \varphi _{n}(t_{n})=0}$, a contradiction. Hence ${\textstyle t_{n}\not \in t^{+}\cup t_{1}^{+}\cup \cdots \cup t_{n-1}^{+}}$. Now, we have the infinite strictly increasing sequence of ideals of ${\textstyle S}$

${\displaystyle t+S\subsetneqq (t+S)\cup (t_{1}+S)\subsetneqq (t+S)\cup (t_{1}+S)\cup (t_{2}+S)\subsetneqq \cdots }$

a contradiction with Lemma 1. Therefore ${\textstyle s\geq t}$. ■

Now, we are able to prove the main result of this paper.

Theorem 3.

Suppose that  ${\textstyle (S,\leq )}$is a strictly totally ordered monoid which is finitely generated and satisfies the condition that  ${\textstyle 0\leq s}$for every  ${\textstyle s\in S}$. Assume that  ${\textstyle \omega _{s}(1)=1}$for each  ${\textstyle s\in S}$. Then  ${\textstyle M_{R}}$is a Hopfian right  ${\textstyle R}$-module if and only if  ${\textstyle B_{A}}$is a Hopfian right  ${\textstyle A}$-module.

Proof.

Suppose that ${\textstyle M_{R}}$ is a Hopfian right ${\textstyle R}$-module. Let ${\textstyle \alpha :B_{A}\longrightarrow B_{A}}$ be any surjective ${\textstyle A}$-homomorphism. We want to prove that ${\textstyle \alpha }$ is injective to be an isomorphism. Define ${\textstyle f:B\longrightarrow M}$ via ${\textstyle f(\varphi )=\varphi (0)}$. Now, define ${\textstyle h:M_{R}\longrightarrow M_{R}}$ via ${\textstyle h(m)=f\alpha (d_{m}^{0})}$.

(1) ${\textstyle h}$ is an ${\textstyle R}$-homomorphism: For any ${\textstyle m\in M}$ and ${\textstyle r\in R}$, we have

${\displaystyle {\begin{array}{l}\displaystyle h(mr)=f\alpha (d_{mr}^{0})=f\alpha (d_{m}^{0}c_{r})=f(\alpha (d_{m}^{0}c_{r}))=\\\displaystyle =f(\alpha (d_{m}^{0})c_{r})=(\alpha (d_{m}^{0})c_{r})(0){\mbox{.}}\end{array}}}$

Since ${\textstyle 0\leq s}$ for every ${\textstyle s\in S}$, we get ${\textstyle X_{0}(\alpha (d_{m}^{0}),c_{r})=\lbrace (0,0)\rbrace }$ and so

${\displaystyle {\begin{array}{l}\displaystyle h(mr)=(\alpha (d_{m}^{0}))(0)\omega _{0}(c_{r}(0))=(\alpha (d_{m}^{0}))(0)\omega _{0}(r)=\\\displaystyle =(\alpha (d_{m}^{0}))(0)r=f\alpha (d_{m}^{0})r=h(m)r{\mbox{.}}\end{array}}}$

(2) ${\textstyle h}$ is a surjective map: For any ${\textstyle m\in M}$, there exists ${\textstyle \beta \in B}$ such that ${\textstyle \alpha (\beta )=d_{m}^{0}}$ since ${\textstyle \alpha }$ is surjective. Let ${\textstyle \psi =\beta -d_{\beta (0)}^{0}}$. Then

${\displaystyle \psi (0)=(\beta -d_{\beta (0)}^{0})(0)=\beta (0)-d_{\beta (0)}^{0}(0)=\beta (0)-\beta (0)=0{\mbox{.}}}$

So ${\textstyle \pi (\psi )>0}$ and using Lemma 2, ${\textstyle \pi (\alpha (\psi ))>0}$. Hence

${\displaystyle {\begin{array}{l}\displaystyle m=d_{m}^{0}(0)=\alpha (\beta )(0)=\alpha (\psi +d_{\beta (0)}^{0})(0)=(\alpha (\psi )+\\\displaystyle +\alpha (d_{\beta (0)}^{0}))(0)=\alpha (\psi )(0)+\alpha (d_{\beta (0)}^{0})(0)=0+\\\displaystyle +\alpha (d_{\beta (0)}^{0})(0)=\alpha (d_{\beta (0)}^{0})(0)=f\alpha (d_{\beta (0)}^{0})=\\\displaystyle =h(\beta (0)){\mbox{.}}\end{array}}}$

Hence ${\textstyle h}$ is a surjective ${\textstyle R}$-homomorphism which must be an isomorphism, since ${\textstyle M_{R}}$ is a Hopfian right ${\textstyle R}$-module. To prove that ${\textstyle \alpha }$ is injective, let ${\textstyle \varphi \in B_{A}}$ be such that ${\textstyle \alpha (\varphi )=0}$. It follows that ${\textstyle h(\varphi (0))=\alpha (\varphi )(0)=0}$. Then ${\textstyle \varphi (0)=0}$, since ${\textstyle h}$ is an ${\textstyle R}$-isomorphism. Suppose that ${\textstyle u\in S}$ and for any ${\textstyle v\in S}$ with ${\textstyle v<u,\varphi (v)=0}$. We show that ${\textstyle \varphi (u)=0}$. Define ${\textstyle \psi \in B_{A}}$ via

${\displaystyle \psi (s)=\lbrace {\begin{array}{cc}\varphi (u){\mbox{,}}&{\mbox{if}}s=u{\mbox{,}}\\0{\mbox{,}}&{\mbox{if  }}s\not =u{\mbox{.}}\end{array}}}$

Thus ${\textstyle (\varphi -\psi )(s)=0}$, for any ${\textstyle s\leq u}$, and it follows that ${\textstyle \pi (\varphi -\psi )>u}$. Using Lemma 2, we have ${\textstyle \pi (\alpha (\varphi -\psi ))>u}$. Hence

${\displaystyle {\begin{array}{l}\displaystyle \alpha (\psi )(u)=\alpha (\psi )(u)+0=\alpha (\psi )(u)+\alpha (\varphi -\psi )(u)=\alpha (\psi +\varphi -\\\displaystyle -\psi )(u)=\alpha (\varphi )(u)=0{\mbox{.}}\end{array}}}$

Consider the following computation, for any ${\textstyle s\in S}$

${\displaystyle (d_{\varphi (u)}^{0}e_{u})(s)=\sum _{(x,y)\in X_{s}(d_{\varphi (u)}^{0},e_{u})}{\underset {\varphi (u)}{\overset {0}{d}}}(x)\omega _{x}(e_{u}(y))=d_{\varphi (u)}^{0}(0)\omega _{0}(e_{u}(s))=\varphi (u)e_{u}(s)=\lbrace {\begin{array}{cc}\varphi (u){\mbox{,}}&{\mbox{if  }}s=u{\mbox{,}}\\0{\mbox{,}}&{\mbox{if  }}s\not =u{\mbox{.}}\end{array}}=\psi (s){\mbox{.}}}$

It follows that ${\textstyle \psi =d_{\varphi (u)}^{0}e_{u}}$. Thus

${\displaystyle {\begin{array}{l}\displaystyle h(\varphi (u))=f\alpha (d_{\varphi (u)}^{0})=\alpha (d_{\varphi (u)}^{0})(0)=\\\displaystyle =(\alpha (d_{\varphi (u)}^{0})e_{u})(u)=\alpha (d_{\varphi (u)}^{0}e_{u})(u)=\alpha (\psi )(u)=0{\mbox{,}}\end{array}}}$

which implies that ${\textstyle \varphi (u)=0}$, since ${\textstyle h}$ is an ${\textstyle R}$-isomorphism. Hence ${\textstyle \varphi (s)=0}$ for any ${\textstyle s\in S}$, and so ${\textstyle \varphi =0}$. Therefore ${\textstyle \alpha }$ is an ${\textstyle A}$-isomorphism and ${\textstyle B_{A}}$ is a Hopfian right ${\textstyle A}$-module.

Conversely, suppose that ${\textstyle B_{A}}$ is a Hopfian right ${\textstyle A}$-module. Let ${\textstyle h:M_{R}\longrightarrow M_{R}}$ be any surjective ${\textstyle R}$-homomorphism. We want to prove that ${\textstyle h}$ is injective.

Define ${\textstyle \alpha :B_{A}\longrightarrow B_{A}}$ via ${\textstyle \alpha (\varphi )(s)=h(\varphi (s))}$ for any ${\textstyle \varphi \in B_{A}}$ and ${\textstyle s\in S}$. We show that ${\textstyle \alpha }$ is an ${\textstyle A}$-isomorphism.

(1) ${\textstyle \alpha }$ is an ${\textstyle A}$-homomorphism: For any ${\textstyle \varphi \in B_{A},f\in A}$ and ${\textstyle s\in S}$, we set

${\displaystyle X_{1}=\lbrace (u,v)\in X_{s}(\varphi ,f)\vert h(\varphi (u))=\alpha (\varphi )(u)=0\rbrace \quad {\mbox{and}}}$

Then clearly ${\textstyle X_{2}=X_{s}(\alpha (\varphi ),f)}$. Consider the following computation, for any ${\textstyle s\in S}$

${\displaystyle {\begin{array}{l}\displaystyle \alpha (\varphi f)(s)=h((\varphi f)(s))=\\\displaystyle =h(\sum _{(u,v)\in X_{s}(\varphi ,f)}\varphi (u)\omega _{u}(f(v)))=\\\displaystyle =\sum _{(u,v)\in X_{s}(\varphi ,f)}h(\varphi (u))\omega _{u}(f(v))=\\\displaystyle =\sum _{(u,v)\in X_{1}}h(\varphi (u))\omega _{u}(f(v))+\\\displaystyle +\sum _{(u,v)\in X_{2}}h(\varphi (u))\omega _{u}(f(v))=0+\\\displaystyle +\sum _{(u,v)\in X_{2}}h(\varphi (u))\omega _{u}(f(v))=\\\displaystyle =\sum _{(u,v)\in X_{s}(\alpha (\varphi ),f)}\alpha (\varphi )(u)\omega _{u}(f(v))=(\alpha (\varphi )f)(s){\mbox{.}}\end{array}}}$

Thus ${\textstyle \alpha (\varphi f)=\alpha (\varphi )f}$.

(2) ${\textstyle \alpha }$ is a surjective map: For any ${\textstyle \psi \in B_{A}}$ and any ${\textstyle s\in supp(\psi )}$, there exists an element ${\textstyle m_{s}\in M}$ such that ${\textstyle h(m_{s})=\psi (s)}$, since ${\textstyle h}$ is a surjective map.

Define ${\textstyle \beta :S\longrightarrow M}$ via

${\displaystyle \beta (s)=\lbrace {\begin{array}{cc}m_{s}{\mbox{,}}&{\mbox{if  }}s\in supp(\psi ){\mbox{,}}\\0{\mbox{,}}&{\mbox{if  }}s\not \in supp(\psi ){\mbox{.}}\end{array}}}$

Clearly, ${\textstyle supp(\beta )\subseteq supp(\psi )}$, which implies that ${\textstyle supp(\beta )}$ is an artinian and narrow subset of ${\textstyle S}$ and thus ${\textstyle \beta \in B_{A}}$.

If ${\textstyle s\in supp(\psi )}$, then

${\displaystyle \alpha (\beta )(s)=h(\beta (s))=h(m_{s})=\psi (s){\mbox{.}}}$

If ${\textstyle s\not \in supp(\psi )}$, then

${\displaystyle \alpha (\beta )(s)=h(\beta (s))=h(0)=0{\mbox{.}}}$

Thus ${\textstyle \alpha (\beta )=\psi }$. Hence ${\textstyle \alpha }$ is a surjective ${\textstyle A}$-homomorphism which must be an isomorphism, since ${\textstyle B_{A}}$ is a Hopfian right ${\textstyle A}$-module. To prove that ${\textstyle h}$ is injective, let ${\textstyle m\in M}$ be such that ${\textstyle h(m)=0}$. Then for any ${\textstyle s\in S}$,

${\displaystyle \alpha (d_{m}^{0})(s)=h(d_{m}^{0}(s))=\lbrace {\begin{array}{cc}h(d_{m}^{0}(0))=h(m)=0{\mbox{,}}&{\mbox{if  }}s=0{\mbox{,}}\\h(0)=0{\mbox{,}}&{\mbox{if  }}s\not =0{\mbox{.}}\end{array}}}$

Thus ${\textstyle \alpha (d_{m}^{0})=0}$ and so ${\textstyle m=0}$, since ${\textstyle \alpha }$ is an ${\textstyle A}$-isomorphism. Therefore ${\textstyle h}$ is an ${\textstyle R}$-isomorphism and ${\textstyle M_{R}}$ is a Hopfian right ${\textstyle R}$-module. ■

If we set ${\textstyle \omega (s)=Id_{R}}$, for every ${\textstyle s\in S}$, we get the following result as a corollary.

Corollary 4 [9].

Suppose that  ${\textstyle (S,\leq )}$is a strictly totally ordered monoid which is finitely generated and satisfies the condition that  ${\textstyle 0\leq s}$for every  ${\textstyle s\in S}$. Then  ${\textstyle M_{R}}$is a Hopfian right  ${\textstyle R}$-module if and only if  ${\textstyle [[M^{S,\leq }]]_{[[R^{S,\leq }]]}}$is a Hopfian right  ${\textstyle [[R^{S,\leq }]]}$-module.

References

1. [1] G. Baumslag; Hopficity and abelian groups; J. Irwin, E.A. Walker (Eds.), Topics in Abelian Groups, Scott Foresmann and Company (1963), pp. 331–335
2. [2] R. Gilmer; Commutative Semigroup Rings; University of Chicago press, Chicago (1984)
3. [3] V. Hiremath; Hopfian rings and Hopfian modules; Indian J. Pure Appl. Math., 17 (1986), pp. 895–900
4. [4] P. Ribenboim; Semisimple rings and Von Neumann regular rings of generalized power series; J. Algebra, 198 (1997), pp. 327–338
5. [5] K. Varadarajan; Hopfian and Co-Hopfian objects; Publ. Mat., 36 (1992), pp. 293–317
6. [6] K. Varadarajan; A note on the Hopficity of ${\textstyle M[x]}$ or ${\textstyle M[[x]][J]}$; Nat. Acad. Sci. Lett., 15 (1992), pp. 53–56
7. [7] K. Varadarajan; Generalized power series modules; Comm. Algebra, 29 (3) (2001), pp. 1281–1294
8. [8] R. Zhao, Y. Jiao; Principal quasi-Baerness of modules of generalized power series; Taiwanese J. Math., 15 (2) (2011), pp. 711–722
9. [9] L. Zhongkui; A note on Hopfian modules; Comm. Algebra, 28 (6) (2000), pp. 3031–3040
10. [10] L. Zhongkui; Triangular matrix representations of rings of generalized power series; Acta Math. Sin. (Engl. Ser.), 22 (2006), pp. 989–998