\section{Introduction}
A ring $R$ is called \textit{left Archimedean}, if for each nonunit element $r\in R$ we have $\bigcap_{n\in \mathbb{N}}r^nR=\{0\}$. The right Archimedean rings are defined in a similar way. In \cite{Sheldon}, it is shown that if $R$ is a domain and $\bigcap_{n\in \mathbb{N}} a^nR=0$, then the quotient field $Q(R[[x/a]])$ has infinite transcendent degree over the quotient field $Q(R[[x]])$. A very important result in this direction is  that if $R$ is a domain (or is completely integrally closed) and satisfies \emph{ACCP}, then $R$ is Archimedean, see \cite{andersonanti,Sheldon}. However, each \emph{ACCP}-domain is not necessarily Archimedean, see for example, \cite[p. 1127]{Dumitrescu}. There are similar results  in \cite{ArnoldII,condo}.   


Ribenboim in \cite{Ribenboimsemisimple} defined the ring of
generalized power series $R[[S]]$ consisting of all maps from $S$ to
$R$ whose support is Artinian and narrow  with  the pointwise
addition and the convolution multiplication. This construction
provided interesting examples of rings, for example see \cite{Ribenboim (1995a),Ribenboim (1995b)}. In \cite{Ribenboim}, Ribenboim gave a sufficient condition for the ring $R[[S]]$ (not necessarily commutative) to be Noetherian.  However, Frohn  gave an example to show that \emph{ACCP} does not rise to the power series ring in general (see \cite{counter}). 

In \cite{zim}, R. Mazurek and M. Ziembowski, introduced a ``twisted" version of the Ribenboim
construction and studied when it produces a von Neumann regular ring.   
They also proved  that if $R$ is a domain and $\omega$ an endomorphism of  $R$, then  $R$ satisfies \emph{ACCPL}, $S$ is an \emph{ACCPL}-monoid and $\omega_s$  is injective  for each $s\in S$ if and only if $R[[S,\omega]]$ is an \emph{ACCPL}-domain. 

As a similar result, for a strictly ordered monoid $S$ and a monoid homomorphism $\omega : S \rightarrow \End (R)$, in \cite{paykan} it is proved that if the attached skew generalized power series ring $R[[S,\omega]]$ is a right Archimedean domain then $R$ is a right Archimedean domain, $S$ is a right Artinian and narrow monoid and $\omega_s$ is an injective which preserves nonunits of $R$ for any $s \in S$.
Next step in this direction, 
we show that when $R$ is right (resp. left) Archimedean and satisfies the \emph{ACC} on  annihilators, then $R[[x;\alpha]]$ is a right (resp. left) Archimedean ring. 
In section \ref{archh}, we give an example to justify that the condition given in Theorem \ref{rigidar} is necessary on the ring $R$, to be able to pass the property of being Archimedean between $R$ and $R[[x;\alpha]]$. 








A ring $R$ is \textit{reduced}, if the condition $a^2=0$ results in $a=0$. According to Krempa \cite{krem}, an endomorphism $\alpha$ of a ring $R$ is said to be \textit{rigid} if $a\alpha(a)=0$ implies $a=0$, for $a \in R$. A ring $R$ is said to be $\alpha$-\textit{rigid} if there exists a rigid endomorphism $\alpha$ of $R$. Clearly, every domain $D$ with a monomorphism $\alpha$ is rigid. It is clear that, $\alpha$-rigid rings are reduced, (see e.g., \cite{ore}).
The following Lemma  by Hong in \cite{ore} is very useful  in the following sections.

\begin{lemma}\label{jav}
Let $R$ be a $\alpha$-rigid ring and $a,b \in R$. Then 
\begin{itemize}
\item[(1)] $R$ is reduced.
\item[(2)] If $ab=0$, then  we have
$a\alpha^{n}(b)=\alpha^{n}(a)b=0$  for any positive integer $n$.

\end{itemize}
\end{lemma}
\begin{proof}
(1) Let $a^2=0$. Then $a\alpha(a)\alpha(a)\alpha^2(a)=0$. So $a\alpha(a)\alpha(a\alpha(a))=0$. Because $R$ is rigid $a\alpha(a)=0$. Another use of it gives us $a=0$.

(2) Let $ab=0$. So $ba=0$. This implies that $a\alpha^n(ba)\alpha^{2n}(b)=0$. So $a\alpha^n(b)\alpha^n(a\alpha^n(b))=0$. Since $R$ is $\alpha-$rigid, then $a\alpha^n(b)=0$. Similarly $\alpha^n(a)b=0$.
\end{proof}

\medskip
Throughout this paper, all monoids and rings are with identity element that is inherited by submonoids and subrings and preserved under homomorphisms, but neither monoids nor rings are assumed to be - necessarily -  commutative.\par


\section{Archimedean skew power series rings}\label{archh}
In this section, we give the conditions on the ring $R$ such that the property of being Archimedean can be transferred from $R$ to the skew power series ring $ R[[x;\alpha]]$ and vice versa. The following results in \cite[Theorems 4.1]{paykan} describe such a transfer with the assumption that $ R[[S,\omega]]$ is a domain. 

Let  $\omega: S\longrightarrow \End(R)$ be a monoid homomorphism, where $R$ is a ring and $(S,\le)$ is a strictly ordered monoid. If $R[[S,\omega]]$ is a left (resp. right) Archimedean domain, then one can see that $R$ is a left (resp. right) Archimedean domain. It can be proved straightforward from the definition and the fact that $\bigcap _{n\in \mathbb{N}}(c_a)^nR[[S,\omega]]= 0$ for nonunit $a$. Also, $S$ is a left (resp. right) Artinian and  narrow monoid because of the fact that $\bigcap _{n\in \mathbb{N}}(\textbf{e}_s)^nR[[S,\omega]]=0$ for nonunit $s\in S$. Moreover, $\omega _s$ is injective (and preserve nonunits of $R$) for any $s\in S$. 

As mentioned in previous section, if $(S,.,\leq)$ is a (in particular positive) strictly totally ordered monoid and $0\neq f \in R[[S,\omega]]$, then $\supp( f )$ is a nonempty well-ordered subset of $S$. The smallest element of $\supp(f)$ is denoted by $\pi( f )$ and we set $\theta(f):=f(\pi(f))$. In particular, for a power series $f(x)=\sum f_nx^n$, with the first nonzero coefficient is $f_m$, we have $\pi(f)=m$ and $\theta(f)=f_m$.
 Let $U(R)$ be the set of unit elements of $R$. Then it is shown  in \cite[Proposition 3.2]{zim} that if $\pi(f) \in U(S)$ and $\theta (f) \in U(R)$, then $f \in U(R[[S,\omega]])$, where $f\in R[[S,\omega]] \setminus \{0\}$. 
\begin{theorem}\cite[Theorems 4.1]{paykan}\label{paykan}
Let $R$ be a ring, $(S,\le)$ a strictly ordered monoid and $\omega: S\longrightarrow \End(R)$ a monoid homomorphism. Then we have the following:
\begin{itemize}
\item[{\em (i)}] The ring $R[[S,\omega]]$ is a right Archimedean domain if and only if $R$ is a
right Archimedean domain, $\omega _s$ is injective and preserves nonunits of $R$ for any $s\in S$. 

\item[{\em (ii)}] The ring $R[[S,\omega]]$ is a left Archimedean domain if and only if $R$ is a left Archimedean domain and $\omega _s$ is injective for any $s\in S$. 
\end{itemize}
\end{theorem}



 Before stating the main result of the paper, we first recall a few definitions. An ideal $P$ of $R$ is \textit{completely prime} if $ab\in P$ implies $a\in P$ or $b\in P$ for $a,b\in R$.  Let $\alpha$ an endomorphism of $R$. An ideal $I$ of $R$ is called an \textit{$\alpha$-ideal} if $\alpha (I)\subseteq I$ and $I$ is called \textit{$\alpha$-invariant} if $\alpha ^{-1}(I)=I$. Note that if $I$ is an $(\alpha)$-ideal, then $\bar{\alpha}:R/I\longrightarrow R/I$ defined by $\bar{\alpha}(a + I)=\alpha (a)+I$ for $a\in R$ is an endomorphism of the factor ring $R/I$. 

We prove a similar result in the realm of reduced rings. Reduced rings are natural generalization of domains, for example direct product of domains is reduced. Therefore, Archimedean rings can also keep certain properties similar to the Archimedean domains.


 





\begin{theorem}\label{rigidar}
Let  $R$ be a $\alpha$-rigid ring satisfying the ACC on annihilators and 
$\alpha$  be a surjective endomorphism.
Then:
\begin{enumerate}
\item[\em{(1)}] $R$ is right Archimedean and $\alpha$ preserves nonunits of $R$ if and only if the reduced ring $R[[x;\alpha]]$ is right Archimedean .
\item[\em{(2)}] $R$ is left Archimedean if and only if the reduced ring $R[[x;\alpha]]$ is left Archimedean.
\end{enumerate} 

\end{theorem}



\begin{proof}
We will give the proof for the right-sided version. The proof for the left-sided version follows on the same line if the order of the multiplication is changed from right to left throughout the proof. We recall that for a power series $g(x)=g_0+g_1x+\cdots $ we have $\pi(g):=m$ where $g_m\neq 0$  and $g_i=0$ for $0\leq i<m$. Also $\theta(g)=g_m$ is the leading term of the power series. We set $\pi(0)=-\infty$. 



\medskip
 First, assume that $R$ is right Archimedean.  According to \cite{moussavi}, if $R$ is rigid, then $R[[x;\alpha]]$ is a reduced ring.  Let $A=R[[x;\alpha]]$ and 
 \[
 \Gamma := \{g \in A\; | \;  \text{ $g$ is a nonunit in $A$ and}\; \bigcap_{n\in \mathbb{N}}Ag^n\neq \{0\}\}.
 \]
We need to show that $\Gamma=\emptyset$. Assume that $\Gamma \neq \emptyset$.
We define  the set of leading terms $I_f:=\{\theta (g)|g\in fA\}\cup \{0\}$. Then, clearly $I_f$ is a right ideal of $R$.
Let
\[
T:=\{\Ann\big(\bigcup_{i\in \mathbb{N}}\;I_{c_ig}\big)|
 \;  c_i \in R,\; g\in \Gamma \; \text{and}\;  \; c_ig\neq 0,\;c_ig^i=c_jg^j \; \forall \; i,j\in \mathbb{N}\}.
\]
 Then $T \neq \emptyset$ because we assumed that $\Gamma \neq \emptyset$. Note that $\Ann\big(I_{c_ig}\big)=\Ann\big(I_{c_ig^i}\big)$ because $R$ is rigid. Considering that $R$ satisfies \emph{ACC} on annihilators, and then applying Zorn's Lemma results in the fact that $T$ has a maximal element. Let $V:=\Ann\big(\bigcup_{i\in \mathbb{N}}I_{a_if}\big)$ be the maximal element of $T$ for some $f\in
\Gamma$ such that
\begin{align}\label{sharto}

 a_i \in R, \; a_if\neq 0,\; \text{and}\; a_if^i=a_jf^j \; \forall \; i,j\in \mathbb{N}.
\end{align}

\textbf{Step 1.} \textit{We claim that $V$ is a two-sided completely prime ideal of $A$. }

From Lemma \ref{jav}, we see that $R$ is reduced because it is rigid. It implies that annihilators in $R$ are two-sided ideals. Let $ab\in V$ for some $a,b\in R$ and $b\notin V$. We need to show that $a\in V$.
First, we show that $a\in \Ann(I_{ba_if})$ for all $i \in \mbox{$\mathbb{N}$}$. Observe that if $ba_ifh=0$ for all $h\in A$ and $i \in \mbox{$\mathbb{N}$}$, then $b\in V$ which contradicts our assumption that $b\notin V$.  Therefore, $ba_jfA \neq \{0\}$ for some $j \in \mbox{$\mathbb{N}$}$. Take $ba_jfh \in ba_jfA $ such that $ba_jfh \neq 0$. This means that $\bigcup_i I_{ba_if}\neq 0$. Also, for all $i\in \mathbb{N}$ and for all $h\in A$ we have  $ab\theta (a_ifhb)=0$ because $ab \in V$ and $a_ifhb\in a_ifA$. Assume that $\theta (a_ifhb):=tb$, then $\theta (ba_ifh)=bt$ because $R$ is reduced. We conclude that $ab\theta (ba_ifh)=0$, because $R$ is reduced  - since in every reduced ring $wyz=0$ implies $wzy=0$. 

Take $w=a$ and $y=bt$ and $z=b$. 
So $abbt=ab^2t=0$. Again by using that $R$ is reduced, we get $abt=0$. So $a \theta (ba_i fh)=0$ for all $i \in \mbox{$\mathbb{N}$}$ and $h\in A$; and we conclude that $a\in \Ann (\bigcup_{i\in \mathbb{N}} I_{ba_if})$. 

Let $z\in V$. So $z\theta(a_ifl)=0$ for all $i>0$ and $l\in A$. In particular, $z\theta(a_ifhb)=0$ for all $h\in A$. Considering that $R$ is reduced, we get $z\theta(ba_ifh)=0$. This implies that $z\in \Ann(\bigcup_{i\in \mathbb{N}} I_{ba_if})$, which means that 
\begin{align}\label{shar}
V\subseteq \Ann(\bigcup_{i\in \mathbb{N}} I_{ba_if}).
\end{align}
Also, since  $ba_ifA \neq 0$, for some $i \in \mbox{$\mathbb{N}$}$ and $ba_if^i=ba_jf^j \; \forall \; i,j\in \mathbb{N}$, we see that $\Ann(\bigcup_{i\in \mathbb{N}} I_{ba_if})\in T$. The equation
(\ref{shar}) and the maximal property of $V$ in $T$ yields that
$V=\Ann(\bigcup_{i\in \mathbb{N}} I_{ba_if})$. This implies that
$a\in V$ and that $V$ is a two sided completely prime ideal.

\textbf{Step 2.}
\textit{We show that $V$ is $\alpha$-invariant.} 

Let $r\in \alpha^{-1}(V)$. Then there exists $v\in V$ such that
$\alpha^{-1}(v)=r$ and we have $\alpha(r)=v$. By definition
of $V$, we have $\alpha(r)u=0$ for all $u\in \bigcup_{i\in
\mathbb{N}}(I_{a_if})$. Then, from Lemma~\ref{jav}, we have  $ru=0$ and it shows that
$r\in V$ and $\alpha^{-1} (V)\subseteq V$.

To show the reverse inclusion, take $v\in V$. Then for all $u\in \bigcup_{i\in \mathbb{N}}(I_{a_if})$ we have $vu=0$. Again, by Lemma~\ref{jav}, we have $\alpha(v)u=0$ . It shows that $\alpha(v)\in V$. Since $\alpha$ is surjective and rigid, hence an
automorphism, we see that $v \in \alpha^{-1}(V)$ and therefore, $V\subseteq \alpha^{-1}(V)$. From this we conclude that $V$ is an $\alpha$-invariant.


\textbf{Step 3.} \textit{ We show that $W:=(\frac{R}{V})[[x;\overline{\alpha}]]$ is a well-defined Archimedean domain.}

We already know that $V$ is a two-sided completely prime ideal, so the
factor ring $U:=\frac{R}{V}$ is a well-defined domain. Then it is enough to show that $U$ is an Archimedean domain because then by using
Theorem \ref{paykan} we can conclude that $U[[x;\overline{\alpha}]]$ is an Archimedean
domain, where $\overline{\alpha}: R/V \longrightarrow R/V$ defined by $\overline{\alpha}(a+V)=\alpha(a)+V$ for all $a\in R$ is an endomorphism of factor ring $R/V$.


To show that $U$ is an Archimedean
domain, we take  $\overline{z}\in \bigcap _{i\in
\mathbb{N}}U\bar{y}^i $. Then $\overline{z}=\bar{z_1} \bar{y}=\bar{z_2} \bar{y}^2=\cdots$ for some $\overline{z_i}\in U$. It gives, \begin{align}\label{farzado}
z+l_0=z_1y+l_1=z_2y^2+l_2=\cdots.
\end{align}
for some $l_i \in V$. We choose $h\in A$ and $i\in
\mathbb{N}$. By multiplying  \ref{farzado}  with $\theta (a_ifh)$ and using that $l_i \in V$ for all $i$, we obtain
$\theta(a_ifh)z=\theta(a_ifh)z_1y=\cdots$. Then $\theta(a_ifh)z=0$, because $R$
is Archimedean and $y$ is not unit in $R$, otherwise $\overline{y}$ becomes a unit in $U$. As $h$ and $i$ are
chosen arbitrarily, we obtain that $z\in \Ann( I_{a_if})$ for all $i\in \mathbb{N}$ and therefore, $z\in V$. This shows, $\overline{z}=0$, as required. Therefore, $U$ is Archimedean.

\textbf{Step 4.} \textit{We claim that $\overline{f}$ is nonunit in $W=R/V[[x;\overline{\alpha}]]$.} 

Assume that there exists $g\in A$
such that $\overline{f}\overline{g}-\overline{1}=\overline{0}$. It means that all coefficients of $fg-1$ belong to $V$. In particular, $\theta (fg-1) \in V$ and we have  $\theta (fg-1)\theta (a_if)=0$ for all $i\in\mathbb{N}$. Assume that $f(x)=\sum f_lx^l$ and $g(x)=\sum g_l x^l$. We show that $f_0g_0$ is not a unit.

\textbf{Case 1.}  Assume that $\pi(fg-1)>0$. 
Then we have
 $f_0g_0=\theta(fg)=1$. 
Thus, 
$f_0$ is a unit.
This implies that $f$ is a unit, which contradicts to our assumption that $f$ is non-unit. Therefore, we conclude that $\pi(fg-1)> 0$ is not possible.

\textbf{Case 2.} Now, let $\pi(fg-1)=0$. Note that $(fg)_0-1\in V$ and  $(fg)_0-1\neq 0,-1$, because otherwise it shows that  $V=A$, which is impossible. Since $(fg)_0\neq 0$, then $(fg)_0$ is a nonunit, otherwise $fg$ is unit, \cite[Proposition 3.2]{zim}, which is not possible.
 Therefore, $f_0g_0$ is a nonunit. 

We assumed that $\overline{f}$ is a unit, which implies that $\overline{f}^i$ is a unit for all $i \in \mbox{$\mathbb{N}$}$. For each $i \in \mbox{$\mathbb{N}$}$, let $\overline{b_i}$ be the inverse of $\overline{f}^i$. Then we have  $\overline{b_i}\overline{f}^i=\overline{b_j}\overline{f}^j=1$, for all $i,j \in \mbox{$\mathbb{N}$}$. 
It means that for all $i,j \in \mbox{$\mathbb{N}$}$ we have $\overline{b_i}\overline{f}^i-\overline{b_j}\overline{f}^j=0$ and  all the coefficients of $b_if^i-b_jf^j$ belong to $V$. It leads us to have 
$\theta(a_1f)(b_if^i-b_jf^j)=0$  - because $R$ is reduced. 

Therefore, we obtain
\begin{align}\label{equ}
\theta (a_1f)=\theta (a_1f)b_1f=\theta (a_1f)b_2f^2=\cdots.
\end{align}

We know from equation (\ref{sharto}) that  $\theta (a_1f)\neq 0$, which gives $\theta (a_1f)(b_if^i)_0\neq
0$.
 Let $b_i(x)=\sum b_{i,j}x^j$ and  $t_i:=\theta (a_1f)b_{i,0}$ and $r:=f_0$. 
One can see that $(b_if^i)_0
=b_{i,0}f^i_0=b_{i,0}r^i$.
So $0\neq \theta (a_1f)(b_if^i)_0=t_ir^i$.
Thus, $t_{i+1}r^{i+1}=t_ir^i\neq 0$ for all $i$. Note that $r\neq 0$ is a non-unit because $f$ is non-unit. It sums up to the result that $R$  is non-Archimedean which is contradiction to our hypothesis. Therefore, $\overline{f}$ must be a non-unit.

\textbf{Step 5.} \textit{Finally, we are ready to show that $A$ is an Archimedean reduced ring.}

From equation (\ref{sharto}), we have
$\overline{a_1}\overline{f}=\overline{a_2}\overline{f^2}=\cdots$.
In Step 3 and 4, we showed that $W$ is Archimedean domain and $\overline{f}$ is a non-unit in $W$, so we must have $\overline{a_1}\overline{f}=\overline{0}$ and consequently, $\overline{a_1}=\overline{0}$ or
$\overline{f}=\overline{0}$. If $\overline{a_1}=\overline{0}$, then all coefficients of $a_1=\sum a_{1,j}x^j$ belong to $V$ and $a_{1,j}\theta(a_1f)=0$ for all $j\in \mathbb{N}$. 
Again, by using Lemma \ref{jav} we obtain
\begin{align*}
\big( \theta (a_1f)\big)^2 =\bigg(\theta(a_1f)\bigg)\bigg(\theta(a_1f)\bigg)
&=\theta(a_1f)\left(\sum_{x+y=\pi(a_1f)}a_{1,x}\alpha^x(f_y)\right)\nonumber\\
&=\sum_{x+y=\pi(a_1f)}\left(\theta(a_1f)a_{1,x}\alpha^x(f_y)\right)
\end{align*}
Because $\theta (a_1f)a_{1,x}=0$ for all  $x\in S$, we see that
$\theta (a_1f)^2=0$. It gives that $\theta (a_1f)=0$ because $R$ is reduced. This gives us a contradiction to our assumption that $a_if \neq 0$ for all $i \in \mbox{$\mathbb{N}$}$. Therefore, $\overline{a_1} \neq 0$. 
Then we must have $\overline{f}=\overline{0}$, that is, all coefficient of $f$ must belong to $V$ and $f_j\theta (a_1f)=0$ for all $j\in \mathbb{N}$. Then, from Lemma~\ref{jav}, we see that $\theta(a_1f)\alpha^l\left(f_j\right)=0$. This gives,  
\begin{align}
\bigg(\theta(a_1f)\bigg)^2=\sum_{x+y=\pi(a_1f)}a_{1,x}\alpha^x(f_y)\theta(a_1f)=0
\end{align}
which again implies that $\theta((a_1f))^2=0$ and contradicts our assumption that $a_if \neq 0$ for all $i \in \mbox{$\mathbb{N}$}$. Therefore, our primary assumption $\Gamma \neq \emptyset$ is false. It shows that $A$ is an Archimedean ring, which completes the proof of the "only if" part.

Conversely,  assume that $\bigcap _{n=1}^{\infty}Ra^n \neq 0$ for some $a\in R$. Then $0\neq \bigcap _{n=1}^{\infty}Ra^n \subseteq \bigcap _{n=1}^{\infty} Aa^n$. But $A$ is a right Archemidean ring, so $a$ should be a unit in $A$. Thus, $fa=1$ for some $f\in A$. This means that $(fa)_0=1$. 
So we get $f_0a=1$. Hence $a$ is a unit in $R$, which leads us to the conclusion that $R$ is a right Archemidean ring.

We show that $\alpha$ preserves nonunits. If $b\in R$ is a nonunit element and $\alpha(b)$ is a unit, then $c\alpha(b)=1$ for some $c\in R$. Assume that $f(x)b=1$, then $f_0b=1$ which is not possible. So $b$ is not a unit in $A$. Also, $c^nxb^{n}=c^n\alpha(b^n)x=x=c^mxb^m$. So $x\in \bigcap_{n} Ab^n$, while $b$ is not a unit in $A$. This contradiction shows that $\alpha(b)$ should not be a unit in $R$.
\end{proof}

In the case $\alpha=id_R$ - but $R$ is still not necessarily commutative - we conclude the next corollary immediately.
 
\begin{corollary}\label{archcor}
Let $R$ be a right (resp. left) Archimedean  reduced ring which satisfies the ACC on annihilators.
Then the power series ring $R[[x]]$ is a right (resp. left)
Archimedean  reduced ring.
\end{corollary}










In Theorem~\ref{rigidar}, we proved that under some conditions on the ring $R$, it is possible to transfer the property of being Archimedean from $R$ to $R[[x;\alpha]]$. 
It is natural to ask that what are the minimum conditions required on $R$ to allow such a transfer from $R$ to $R[[S,\omega]]$. With the help of the next example, we justify that the condition rigid required in Theorem~\ref{rigidar} cannot be changed to reduced.





\begin{example}\label{reducekhubu}
Let $\mathbb{F}$ be any field and put $S:=(\mathbb{N}\cup \{0\})\times (\mathbb{Z})$
with usual addition and usual order.

 Consider $\mathbb{F}[[S]]= \mathbb{F}[[(\mathbb{N}\cup \{0\})\times (\mathbb{Z})]]:=T$. If $\mathbb{F}[[\mathbb{Z}]]:=R$, then one can see that $T=R[[\mathbb{N}\cup \{0\}]]:=R[[S]]$. Also, we can see that $R$ is isomorphic to the topological closure of $\mathbb{F}((x))$ where the metric is induced from the p-adic valuation $\deg(\cdot)$ and $x:=e_{(1,0)}\in T$ - we use the standard notation $e_s$ for $s\in S$ in the  skew generalized  power series rings. 

Define $x:=e_{(1,0)}\in T$ and $y:=e_{(0,1)}\in T$.
One can see that $T$ is an skew generalized power series rings of $R$ where $S$ is a positive stirctly totally ordered monoid. 
Then, $T$ is a Noetherian integral domain (i.e. a reduced ring). Also, $T$ satisfies both \emph{ACCP} and \emph{ACC} on annihilators.
Take $\alpha :T\rightarrow T$ defined by 
\begin{align}
\alpha (1)=1, \quad \alpha (x)=y^{-1}, \quad \alpha (y)=y. 
\end{align} 
Note that $\alpha$ is not a monomorphism, and therefore, not rigid because $\alpha (xy^2-y)=0$ and $xy^2-y \neq 0$.  So $T$ is not an $\alpha$-rigid ring.
 
Let $U:=T[[z;\alpha]]$ and $f_k(z)=x\sum _{n=k}^{\infty}y^n z \in T$. Then 
\begin{align}
f_{k+1}(z).x=\big (\sum _{n=k+1}^{\infty} xy^nz \big )x=\sum _{n=k+1}^{\infty}x\alpha (x)y^n z=\sum _{n=k+1}^{\infty}xy^{n-1}z=\sum _{n=k}^{\infty}xy^nz=f_k(z).
\end{align}
It shows that $ f_k U \subseteq  f_{k+1} U$ and we obtain an ascending chain of principal ideals in $T$. In order to show that $f_k U \subsetneq  f_{k+1} U$, it is enough to show that $x$ is not unit in $U$. Suppose that $h(z)x=1$ for some $h(z) \in U$. Let $h(z)=\sum _{i=0}^{\infty}g_i(x,y,y^{-1})z^i$, then 
$h(z)x=g_0(x,y,y^{-1})x+(\sum _{i=1}^{\infty}g_i(x,y,y^{-1})z^i)x=1$.
It implies, $g_0(x,y,y^{-1})x=1$ showing that $x$ is a unit in $T$, which is not true. Therefore, we obtain a non-stablizing chain in $U$ given by $f_1 T \subsetneq  f_{2} U\subsetneq \ldots$. Hence, $U$ does not satisfy \emph{ACCPR}.
\end{example}

In \cite{lantz}, the authors give another example which also shows that the $\alpha$-rigid condition in Theorem~\ref{rigidar} is not superfluous. A ring is $\alpha$-compatible if $a\alpha(b)=0$ results in $ab=0$ - see more information about $\alpha$-compatible rings in \cite{moussavi}. It is known that the set of $\alpha$-rigid rings are the intersection of the set of $\alpha$-compatible rings and reduced rings. We proved that Archimedean property can be shifted from the $\alpha$-rigid rings and propose a counterexample for the reduced rings. The remaining natural question which seeks an answer is:
\begin{question}
Let  $R$ be an $\alpha$-compatible ring satisfying the \emph{ACC} on annihilators. Furthermore, let  $\alpha$ be a surjective endomorphism in $R$. Then $R$ is right (resp. left) Archimedean if and only if the reduced ring $R[[x;\alpha]]$ is right (resp. left) Archimedean.
\end{question} 

