\section{Introduction} \label{sec:intro}

The study of group rings and algebras dates back to the mid-nineteen century. Most of the initial research in this area focused mainly on the structure of the groups of units of group rings and the isomorphic problem for groups over a given coefficient ring; see Highman~\cite{gH40} and Sehgal~\cite{sS69}, respectively. However, most of the early initial ring-theoretical study of group rings and algebras was mainly carried out on a non-commutative setting. 

It was not until the seventies that the study of a commutative version of group rings started earning substantial attention. In a series of papers, Gilmer and Parker began a systematic treatment of semigroup rings (see \cite{rG74,GP74,GP75}). They particularly focused on the following problem: given a commutative ring $R$ and a commutative semigroup $S$, establish conditions under which the semigroup ring $R[X;S]$ of $S$ over $R$ satisfies certain algebraic property. Clearly, the determination of such conditions often requires a fair understanding of the algebraic properties of $S$ and $R$. As the structure of commutative semigroups most of the time cannot be derived from that one of abelian groups, a new research direction in commutative algebra had emerged.

Since then semigroup rings and algebras have received a significant amount of consideration. Semigroup algebras, in particular, permeate through many fields under active research, including algebraic combinatorics~\cite{BCMP98}, discrete geometry~\cite{BG02}, and functional analysis~\cite{mA04}. Gilmer in \cite{rG84} offers a comprehensive exposition on the advances of commutative semigroup ring theory until mid 1980.

Even though additive semigroups of rationals have an interesting atomic structure~\cite{fG17} and a complex arithmetic of factorizations~\cite{fG18a}, the semigroup rings they determine have not been explored in much detail so far. Examples of such semigroup rings, however, have sporadically shown up in the literature; nontrivial examples have been given in \cite[Section~1]{aG16} and \cite[Example~2.1]{AAZ90}. In this paper, we study semigroup rings and algebras of additive submonoids of $\mathbb{Q}_{\ge 0}$, which we simply call Puiseux rings and algebras, respectively. In particular, we will construct three families of Puiseux algebras exhibiting different factorization properties.

\noindent {\bf Outline.} In Section~\ref{sec:background} and Section~\ref{sec:semigroup algebras of PM}, as an excuse to settle down notation, we introduce most of the main general concepts we will be using later. In Section~\ref{sec:Irreducibility Criteria}, we extend the Gauss's Lemma and the Eisenstein's Criterion from polynomial rings to Puiseux rings. Then, in Section~\ref{sec:atomic PA}, we show that the Puiseux algebras of prime reciprocal Puiseux monoids are atomic. In the first part of Section~\ref{sec:antimatter PA}, we exhibit an infinite class of antimatter Puiseux algebras over algebraically closed fields, while in the second part of the same section, we exhibit another infinite class of antimatter Puiseux algebras over any given field of finite characteristic. Finally, we characterize the irreducible elements of Puiseux algebras whose exponent semigroups are root-closed positive submonoids of $(\mathbb{Q},+)$, and we prove that elements in such algebras have at most one factorization.
\medskip

\section{Notation and Background} \label{sec:background}

Throughout this sequel, we set $\mathbb{N} := \{1,2, \dots\}$ and $\mathbb{N}_0 := \mathbb{N} \cup \{0\}$. We also let $\mathbb{P}$ denote the set of prime numbers. In addition, for $X \subseteq \mathbb{R}$ and $r \in \mathbb{R}$, we define $X_{\ge r} := \{x \in X \mid x \ge r\}$; with a similar spirit we use the notations $X_{< r}$ and $X_{> r}$. If $q \in \mathbb{Q}_{> 0}$, then we call the unique $a,b \in \mathbb{N}$ such that $q = a/b$ and $\gcd(a,b)=1$ the \emph{numerator} and \emph{denominator} of $q$ and denote them by $\mathsf{n}(q)$ and $\mathsf{d}(q)$, respectively. Finally, for $S \subseteq \mathbb{Q}_{> 0}$, we set
\[
	\mathsf{n}(S) := \{\mathsf{n}(s) \mid s \in S\} \quad \text{ and } \quad \mathsf{d}(S) := \{\mathsf{d}(s) \mid s \in S\}.
\]

Withing the scope of our exposition, each monoid is tacitly assumed to be commutative and cancellative. Unless we specify otherwise, monoids are written additively. For a monoid $M$, we let $U(M)$ denote the set of units (i.e., invertible elements) of $M$. When $U(M) = \{0\}$, we say that $M$ is \emph{reduced}. Also, we set $M^\bullet := M \setminus \{0\}$. The difference group of $M$ is denoted by $\mathsf{gp}(M)$, i.e.,
\[
	\mathsf{gp}(M) := \{x - y \mid x,y \in M\}.
\]
For $S \subseteq M$, we write $\langle S \rangle$ for the smallest (under inclusion) submonoid of $M$ containing $S$. If $M = \langle S \rangle$, then we call $S$ a \emph{generating set} of $M$. In addition, $M$ is called \emph{finitely generated} if it can be generated by a finite set, while $M$ is called \emph{cyclic} if it can be generated by a singleton. Further basic definitions and concepts on commutative cancellative monoids can be found in \cite[Chapter~2]{pG01}.

If $y,z \in M$, then $y$ \emph{divides} $z$ \emph{in} $M$, in symbols, $y \mid_M z$, if there exists $x \in M$ such that $z = x+y$. An element $p \in M \setminus U(M)$ is called \emph{prime} provided that for all $x,y \in M$ with $p \mid_M x + y$, either $p \mid_M x$ or $p \mid_M y$. In addition, an element $a \in M \setminus U(M)$ is an \emph{atom} if for any $x,y \in M$ such that $a = x+y$ either $x \in U(M)$ or $y \in U(M)$. The set of all atoms of $M$ is denoted by $\mathcal{A}(M)$, and $M$ is \emph{atomic} if $M = \langle \mathcal{A}(M) \rangle$. On the other hand, we call $M$ \emph{antimatter} if $M$ does not contain any atoms, i.e., $\mathcal{A}(M)$ is empty. A subset $I$ of $M$ is an \emph{ideal} of $M$ if $I + M \subseteq I$. The ideal $I$ is \emph{principal} if $I = x + M$ for some $x \in M$. The monoid $M$ satisfies the \emph{ascending chain condition on principal ideals} (or \emph{ACCP}) if each increasing sequence of principal ideals of $M$ eventually stabilizes; in this case, $M$ is called an \emph{ACCP} \emph{monoid}. It is well known that each ACCP monoid is atomic \cite[Proposition~1.1.4]{GH06b}.

A multiplicative monoid $F$ is \emph{free abelian} on a subset $P$ of $F$ provided that every element $a \in F$ can be written uniquely in the form
\[
	a = \prod_{p \in P} p^{\pval(a)},
\]
where $\pval(a) \in \mathbb{N}_0$ and $\pval(a) > 0$ only for finitely many $p \in P$. For each set $P$, there exists a unique (up to canonical isomorphisms) monoid $F$ that is free abelian on $P$. The fundamental theorem of arithmetic states that the multiplicative monoid $\mathbb{N}$ is free abelian on $\mathbb{P}$. In this case, we can extend $\pval$ to $\mathbb{Q}_{\ge 0}$ for each $p \in \mathbb{P}$ by setting $\pval(r) := \pval(\mathsf{n}(r)) - \pval(\mathsf{d}(r))$ for each $r \in \mathbb{Q}_{> 0}$ and $\pval(0) = \infty$. The map $\pval$ is called the $p$-\emph{adic valuation} on $\mathbb{Q}_{\ge 0}$. The next two properties of the p-adic valuation, used throughout the paper without explicit mention, are easy to verify:
\begin{eqnarray*}
	&& \pval(rs) = \pval(r) + \pval(s) \ \ \text{for all} \ \ r,s \in \mathbb{Q}_{\ge 0}; \vspace{10pt} \\
	&& \pval(r + s) \ge \min \{ \pval(r), \pval(s)\} \ \ \text{for all} \ \ r,s \in \mathbb{Q}_{\ge 0}. \vspace{3pt}
\end{eqnarray*}

Suppose now that $M$ is a reduced monoid. The free abelian monoid on $\mathcal{A}(M)$, denoted by $\mathsf{Z}(M)$, is called the \emph{factorization monoid} of $M$, and elements in $\mathsf{Z}(M)$ are called \emph{factorizations}. If $z = a_1 \dots a_n \in \mathsf{Z}(M)$ for some $n \in \mathbb{N}_0$ and $a_1, \dots, a_n \in \mathcal{A}(M)$, then $n$ is the \emph{length} of $z$ and is denoted by $|z|$. The \emph{factorization homomorphism} of $M$ is the unique homomorphism $\phi \colon \mathsf{Z}(M) \to M$ satisfying $\phi(a) = a$ for all $a \in \mathcal{A}(M)$. For each $x \in M$, the set
\[
	\mathsf{Z}(x) := \mathsf{Z}_M(x) := \phi^{-1}(x) \subseteq \mathsf{Z}(M)
\]
is the \emph{set of factorizations} of $x$. Clearly, the monoid $M$ is atomic if and only if $\mathsf{Z}(x)$ is nonempty for all $x \in M$. Finally, for each $x \in M$, the \emph{set of lengths} of $x$ is defined by
\[
	\mathsf{L}(x) := \mathsf{L}_M(x) := \{|z| \ : \ z \in \mathsf{Z}(x)\}.
\]


\section{Semigroup Rings of Puiseux Monoids} \label{sec:semigroup algebras of PM}

A \emph{Puiseux monoid} is an additive submonoid of $\mathbb{Q}_{\ge 0}$. Every additive submonoid of $\mathbb{Q}$ which is not a group is isomorphic to a Puiseux monoid \cite[Theorem~2.9]{rG84}. The following result will be referenced later.

\begin{prop} \cite[Theorem~3.10]{fG17} \label{prop:non-dense PM are atomic}
	A Puiseux monoid is atomic provided that it does not have $0$ as a limit point.
\end{prop}

For a stronger version of Proposition~\ref{prop:non-dense PM are atomic}, see \cite[Proposition~4.5]{fG16}. The atomic structure and algebraic properties of Puiseux monoids has been recently studied (see~\cite{fG18} and references therein).

We say that a Puiseux monoid $M$ is \emph{prime reciprocal} if there exist an infinite subset $P$ of $\mathbb{P}$ and a subset $S$ of $\mathbb{Q}_{> 0}$ with pairwise distinct denominators such that $M = \langle S \rangle$ and $\mathsf{d}(S) = P$. Prime reciprocal Puiseux monoids are atomic. Indeed, we have the following stronger result.

\begin{theorem} \cite[Theorem~5.5]{GG17} \label{thm:primary PM are hereditarily atomic}
	Submonoids of a prime reciprocal Puiseux monoid are atomic. \\
\end{theorem}

For a ring $R$ and a semigroup $S$, consider the set $R[X;S]$ comprising all functions $f \colon S \to R$ satisfying that $\{s \in S \mid f(s) \neq 0 \}$ is finite. We shall conveniently represent an element $f \in R[X;S]$ by
\[

	f = \sum_{s \in S} f(s)X^s = \sum_{i=1}^n f(s_i)X^{s_i},
\]
where $s_1, \dots, s_n$ are precisely those $s \in S$ satisfying that $f(s) \neq 0$. Addition and multiplication on $R[X;S]$ are defined as for polynomials. Under these operations, $R[X;S]$ is a ring, which is called the \emph{semigroup ring of} $S$ \emph{over} $R$. Following Gilmer~\cite{rG84}, we will write $R[S]$ instead of $R[X;S]$. We call the semigroup (resp., monoid) $S$ the \emph{semigroup} (resp., \emph{monoid}) \emph{of exponents} of $R[S]$. When $R$ is a field, $R[S]$ naturally becomes an algebra, which is called the \emph{semigroup algebra of} $S$ \emph{over} $R$. As in this paper we are mainly concerned with semigroup rings/algebras over Puiseux monoids, the next definition will save us from repetitive long descriptions of our main objects.

\begin{definition}
	Let $R$ be a ring, and let $M$ be a Puiseux monoid. Then we say that $R[M]$ is a \emph{Puiseux ring}. When $R$ is a field, we say that $R[M]$ is a Puiseux algebra.
\end{definition}

Let $R[M]$ be a Puiseux ring. We say that $f = \alpha_1X^{q_1} + \dots + \alpha_k X^{q_k} \in F[M] \setminus \{0\}$ is represented in \emph{canonical form} if $\alpha_i \neq 0$ for $i = 1, \dots, k$ and $q_1 > \dots > q_k$. In this case, the set $\{q_1, \dots, q_k\}$ is called the \emph{support} of $f$ and is denoted by $\mathsf{Supp}(f)$. Observe that any element of $R[M]$ has a unique representation in canonical form. In this case, $\deg(f) := q_1$ is called the \emph{degree} of $f$. Clearly, when $R$ is an integral domain,
\[
	\deg(fg) = \deg(f) + \deg(g)
\]
for all nonzero elements $f$ and $g$ in $R[M]$. As it is customary for polynomials, we say that $f$ is a \emph{monomial} if $k = 1$.

If $R$ is an integral domain and $M$ is a Puiseux monoid, then $R[M]$ is also an integral domain \cite[Theorem~8.1]{rG84} and the set of units of $R[M]$ is $R^\times$ \cite[Theorem~11.1]{rG84}. Additionally, $R[M]$ is Noetherian if and only if $R$ is Noetherian and $M$ is finitely generated \cite[Theorem~7.7]{rG84}. The following proposition is an immediate implication of~\cite[Theorem~7.2(2)]{rG84}

\begin{prop} \label{prop:induced homomorphism of semigroup algebras}
	Suppose that $\phi \colon M \to M'$ be a monoid homomorphism, and let $\phi^* \colon R[M] \to R[M']$ be the ring homomorphism determined by $X^s \mapsto X^{\phi(s)}$. If $\phi$ is injective (resp., surjective), then $\phi^*$ is injective (resp., surjective).
\end{prop}

It is not hard to see that, unless $M \cong (\mathbb{N}_0,+)$, no monomial of $R[M]$ can be a prime element; this is a consequence of the trivial fact that non-cyclic Puiseux monoids do not contain prime elements. Furthermore, if $F$ is a field, then a monomial in $F[M]$ is irreducible if and only if its degree is in $\mathcal{A}(M)$ \cite[Lemma~3.1]{CM11}. Thus, the next result holds.

\begin{prop} \label{prop:F[M] antimatter implies M antimatter}
	If $F[M]$ contains no irreducible elements, then $M$ is antimatter.
\end{prop}
\medskip

\section{Irreducibility Criteria} \label{sec:Irreducibility Criteria}

In this section we establish extended versions of the Gauss's Lemma and Eisenstein's Criterion for Puiseux rings over GCD-domains. Besides extending these important irreducibility criteria, the purpose of this section is to illustrate the effectiveness of composing an element of a Puiseux ring $R[M]$ with an appropriate integral power of $X$ to obtain a polynomial of $R[X]$.

To begin with, recall that an integral domain $R$ is a GCD-domain if any finite subset $S = \{r_1, \dots, r_n\}$ of $R^\bullet$ has a greatest common divisor in $R$; the set of all greatest common divisors of $S$ is denoted by GCD$(S)$ or GCD$(r_1, \dots, r_n)$.


\begin{definition}
	Let $M$ be a Puiseux monoid, and let $R$ be a GCD-domain. Suppose that $f = \alpha_1 X^{q_1} + \dots + \alpha_k X^{q_k} \in R[M]$ is written in canonical form. Then GCD$(\alpha_1, \dots, \alpha_k)$ is called the \emph{content} of $f$ and is denoted by $\mathsf{c}(f)$. If $\mathsf{c}(f) \subseteq R^\times$, then $f$ is called \emph{primitive}.
\end{definition}

Let $R$ be a GCD-domain. As it is the case for polynomials in $R[X]$, the content of an element of a Puiseux ring $R[M]$ is a set of associate elements of $R$. For a GCD-domain $R$, the primitive version of the Gauss's Lemma states that $\mathsf{c}(fg) = \mathsf{c}(f) \mathsf{c}(g)$ for all $f,g \in R[X]^\bullet$. Let us extend this result to the class of Puiseux rings.

\begin{prop}[Extended Gauss's Lemma I]
	Let $M$ be a Puiseux monoid, and let $R$ be a GCD-domain. Then $\mathsf{c}(f g) = \mathsf{c}(f) \mathsf{c}(g)$ for all $f,g \in R[M]^\bullet$.
\end{prop}

\begin{proof}
	Take $m \in \mathbb{N}$ to be a common multiple of the set $\mathsf{d}(\mathsf{Supp}(f)) \cup \mathsf{d}(\mathsf{Supp}(g))$. Then we have that both $f(X^m)$ and $g(X^m)$ are polynomials in $R[X]$. Now it follows by the primitive version of the Gauss's Lemma that $\mathsf{c}(f(X^m) g(X^m)) = \mathsf{c}(f(X^m)) \mathsf{c}(g(X^m))$. On the other hand, $\mathsf{c}(f) = \mathsf{c}(f(X^m))$ and $\mathsf{c}(g) = \mathsf{c}(g(X^m))$, from which the proposition follows.
\end{proof}

On the other hand, if $F$ denotes the field of fractions of the GCD-domain $R$, the irreducibility version of the Gauss's Lemma states that a non-constant polynomial $f$ with coefficients in $R$ is irreducible in $R[X]$ if and only if it is irreducible in $F[X]$ and $\mathsf{c}(f) \subseteq R^\times$. We also extend this result to the context of Puiseux rings.

\begin{prop}[Extended Gauss's Lemma II]
	Let $M$ be a Puiseux monoid, $R$ a GCD-domain, and $F$ the field of fractions of $R$. Then $f \in R[M] \setminus R$ is irreducible in $R[M]$ if and only if $f$ is primitive in $R[M]$ and irreducible in $F[M]$.
\end{prop}

\begin{proof}
	For the direct implication, suppose that $f \in R[M] \setminus R$ is irreducible in $R[M]$. Since $f \notin R$, the fact that it is irreducible implies that $\mathsf{c}(f) \subseteq R^\times$. Thus, $f$ is primitive. Now suppose that $f = g \, h$ for some $g,h \in F[M]$. Taking $m$ to be a common multiple of the set $\mathsf{d}(\mathsf{Supp}(g)) \cup \mathsf{d}(\mathsf{Supp}(h))$ as we did in the previous proof, we obtain that $f(X^m), g(X^m)$, and $h(X^m)$ are polynomials in $R[X]$. Since we have $f(X^m) = g(X^m) h(X^m)$, Gauss's Lemma guarantees the existence of $\alpha, \beta \in F$ such that $\alpha \, \beta = 1$ and $\alpha \, g(X^m), \beta \, h(X^m) \in R[X]$. This implies that both $\alpha \, g$ and $\beta \, h$ belongs to $R[M]$. Since $f$ is irreducible in $R[M]$, either $\alpha g(X^m) \in F$ or $\beta h(X^m) \in F$, which means that either $g \in F$ or $h \in F$. Hence $f$ is irreducible in $F[M]$.
	
	To argue the reverse implication, suppose, by way of contradiction, that $f$ reduces in $R[M]$, that is, there exist non-units elements $g,h \in R[M]^\bullet$ such that $f = g \, h$. Since $f$ is primitive, $\mathsf{c}(f) \subseteq R^\times$. Observe now that $g \notin R$; otherwise $g$ would divide any representative of $\mathsf{c}(f) \subseteq R^\times$ in $R$, and so $g$ itself would be a unit of $R$. Similarly, we can see that $h \notin R$. As a result, both $g$ and $h$ are nonzero and non-units in $F[M]$, contradicting the fact that $f$ is an irreducible element of $F[M]$.
\end{proof}

As mentioned before, the technique used in the proofs of both versions of the Gauss's Lemma is very useful to reduce problems from a Puiseux ring $R[M]$ to the ring of polynomials $R[X]$. Indeed, can also use it to established an extended version of the Eisenstein's Criterion for Puiseux rings.

\begin{prop}[Extended Eisenstein's Criterion] \label{prop:Eisenstein Criterion for PA}
	Let $M$ be a Puiseux monoid, and let $R$ be an integral domain with field of fractions $F$. Suppose that the element
	\[
		f = \alpha_n X^{q_n} + \dots + \alpha_1 X^{q_1} + \alpha_0 \in R[M]
	\]
	is written in canonical form. If there exists a prime ideal $P$ of $R$ satisfying that
	\begin{enumerate}
		\item $\alpha_n \notin P$,
		\vspace{3pt}
		\item $\alpha_j \in P$ if $0 \le j < n$, and
		\vspace{3pt}
		\item $\alpha_0 \notin P^2$,
	\end{enumerate}
	then $f$ cannot be a product of two elements of $R[M] \setminus R$.
\end{prop}

\begin{proof}
	Suppose that $f = g \, h$ for some elements $g,h \in R[M]$. Taking $m$ to be a common multiple of the set $\mathsf{d}(\mathsf{Supp}(g)) \cup \mathsf{d}(\mathsf{Supp}(h))$, we obtain that $f(X^m), g(X^m) \in R[X]$, which yields the following factorization in $R[X]$:
	\begin{align} \label{eq:factorized polynomial}
		f(X^m) = \alpha_n X^{mq_n} + \dots + \alpha_1 X^{mq_1} + \alpha_0 = g(X^m) h(X^m).
	\end{align}
	Since the coefficients of $f(X^m)$ still are $\alpha_0, \dots, \alpha_n$, which satisfy the hypotheses of the standard Eisenstein's Criterion, $f(X^m)$ cannot be written as a product of two non-constant polynomials of $R[X]$. Therefore either $g(X^m) \in R$ or $h(X^m) \in R$. This means that either $g \in R$ or $h \in R$, which completes the proof.
\end{proof}

\begin{cor}
	Let $M$ be a nontrivial Puiseux monoid. Then for each $q \in M^\bullet$, there exists an irreducible element of $\mathbb{Q}[M]$ of degree $q$.
\end{cor}

\begin{proof}
	It suffices to show that, for any $q \in M^\bullet$, the element $f = X^q + 2 \in \mathbb{Q}[M]$ is irreducible. Using the Extended Eisenstein's Criterion with the prime ideal $(2)$, one obtains that $f$ cannot be expressed as a product of two elements in $\mathbb{Z}[M] \setminus \mathbb{Z}$. Since $f$ is primitive, it is indeed irreducible in $\mathbb{Z}[M]$. Now the Extended Gauss's Lemma II ensures that $f$ is irreducible in $\mathbb{Q}[M]$.
\end{proof}

\section{Atomic Puiseux Algebras} \label{sec:atomic PA}

Let us start by recalling that an integral domain $R$ is atomic if each nonzero non-unit element of $R$ can be expressed as a product of irreducibles. Our purpose in this section is to construct an infinite class of non-finitely generated atomic Puiseux algebras.

Given a Puiseux algebra $F[M]$, it is not hard to verify that the atomicity of $M$ is a necessary condition for $F[M]$ to be atomic. Conversely, it is natural to ask for which atomic Puiseux monoids $M$ the algebra $F[M]$ is also atomic. It was proved in \cite[Section~2]{AAZ90} that, for any field $F$ and $M = \langle 1/p \mid p \in \mathbb{P} \rangle$, the algebra $F[M]$ satisfies the ascending chain condition on principal ideals (ACCP). Motivated by this example, we will prove that, for each prime reciprocal Puiseux monoid $M$, the algebra $F[M]$ (over any field $F$) satisfies the ACCP and, by Proposition~\ref{prop:ACCP implies atomic}, is atomic. The following result is standard and a direct consequence of \cite[Theorem~1.3(a)]{AAZ90}.

\begin{prop} \label{prop:ACCP implies atomic}
	If $R$ is an integral domain satisfying the ACCP, then $R$ is atomic.
\end{prop}




Prime reciprocal Puiseux monoids are crucial in our construction of a class of atomic Puiseux algebra. This is because prime reciprocal Puiseux monoids satisfy the ACCP property, which can be easily transferred to their corresponding Puiseux algebras.

\begin{theorem} \label{thm:each PPM satisfies the ACCP}
	Each prime reciprocal Puiseux monoid satisfies the ACCP.
\end{theorem}

\begin{proof}
	Let $\{p_n\}$ be a strictly increasing sequence of prime numbers, and let $\{a_n\}$ be a sequence of positive integers with $p_n \nmid a_n$. Now set $M = \langle a_n/p_n \mid n \in \mathbb{N} \rangle$. It is not hard to verify that for all $x \in M$ there exist $k,n \in \mathbb{N}_0$ and $0 \le \alpha_i < p_i$ such that
	\begin{align} \label{eq:decomposition existence}
		x = n + \sum_{i=1}^k \alpha_i \frac{a_i}{p_i}.
	\end{align}
	Let us proceed to check that the sum decomposition in (\ref{eq:decomposition existence}) is unique. To do this, suppose that $k', m \in \mathbb{N}_0$ and $0 \le \beta_i < p_i$ satisfy that
	\begin{align} \label{eq:decomposition uniqueness}
		n + \sum_{i=1}^k \alpha_i \frac{a_i}{p_i} = m + \sum_{i=1}^{k'} \beta_i \frac{a_i}{p_i}.
	\end{align}

	Suppose, without loss of generality, that $k = k'$. After isolating $(\alpha_i - \beta_i)a_i/p_i$ in (\ref{eq:decomposition uniqueness}) and applying the $p_i$-adic valuation, we obtain that $p_i \mid \alpha_i - \beta_i$, which implies that $\alpha_i = \beta_i$ for each $i \in \{1, \dots, k\}$. As a consequence, $n = m$ and we can conclude that the uniqueness of the decomposition in~(\ref{eq:decomposition existence}) holds.
	

	
	Let us assume, by way of contradiction, that $M$ does not satisfy the ACCP. Then there exists a strictly decreasing sequence $\{q_n\}$ of elements in $M$ such that
	\begin{align} \label{eq:chain of ideals}
		q_1 + M \subsetneq q_2 + M \subsetneq \cdots.
	\end{align}
	Now suppose that in the unique sum decomposition of $q_1$ as in (\ref{eq:decomposition existence}), $N$ is the integer and $p_{n_1}, \dots, p_{n_k}$ are the distinct prime denominators of the atoms with nonzero coefficients. By (\ref{eq:chain of ideals}), for each $n \in \mathbb{N}$ there exists $r_{n+1} \in M$ such that $q_n = q_{n+1} + r_{n+1}$ and, therefore,
	\[
		\sum_{i=1}^n r_{i+1} \le q_{n+1} + \sum_{i=1}^n r_{i+1} = q_1
	\]
	for each $n \in \mathbb{N}$. Thus, $\lim_{n \to \infty} r_n = 0$. Then, for any finite subset $A$ of $\mathcal{A}(M)$ there exist a large enough $\ell \in \mathbb{N}$ and $a \in \mathcal{A}(M)$ such that $a \mid_M r_\ell$ and $a \notin A$. Therefore we can find $t \in \mathbb{N}$ and (possibly repeated) prime numbers $p_{m_1}, \dots, p_{m_t}$ satisfying that $|\{p_{m_1}, \dots, p_{m_t}\}| > N + k$ and $a_{m_i}/p_{m_i} \in \mathcal{A}(r_{i+1})$ for each $i = 1, \dots, t$. Take $z_i \in \mathsf{Z}(r_i)$ containing the atom $a_{m_i}/p_{m_i}$, and take $z_{t+1} \in \mathsf{Z}(q_{t+1})$. As $q_1 = q_{t+1} + \sum_{i=1}^t r_{i+1}$, we have $z = z_{t+1} + \sum_{i=1}^t z_i \in \mathsf{Z}(q_1)$. By the uniqueness of the sum decomposition in (\ref{eq:decomposition existence}), the factorization $z$ contains at least $\mathsf{d}(a)$ copies of each atom $a$ for which $\mathsf{d}(a) \in P := \{p_{m_1}, \dots, p_{m_t}\} \setminus \{p_{n_1}, \dots, p_{n_k}\}$. Since $|\{p_{m_1}, \dots, p_{m_t}\}| > N + k$, it follows that $|P| > N$. Thus,
	\[
		N \ge \sum_{a \in \mathcal{A}(M) \, \mid \, \mathsf{d}(a) \in P} \! \! \! \mathsf{d}(a) a \ge |P| > N,
	\]
	which is a contradiction. Hence $M$ satisfies the ACCP. The fact that $M$ was taken to be an arbitrary Puiseux monoid completes the proof.
\end{proof}
\medskip

The next standard result is used to prove Theorem~\ref{thm:infinite family of atomic Puiseux algebras}. We include its proof here for the sake of completeness.

\begin{prop} \label{prop:PF of an ACCP PM is again ACCP}
	Let $F$ be a field, and let $M$ be a reduced monoid. If $M$ satisfies the ACCP, then the semigroup algebra $F[M]$ also satisfies the ACCP.
\end{prop}

\begin{proof}
	Let $(f_1) \subseteq (f_2) \subseteq \cdots$ be an ascending chain of principal ideals of $F[M]$. Then the sequence $\{\deg(f_n) + M\}$ of subsets of $M$ is an ascending chain of principal ideals, which stabilizes because $M$ satisfies the ACCP. As $M$ is reduced, there exists $N \in \mathbb{N}$ such that $\deg(f_n) = \deg(f_N)$ for all $n \ge N$. This, along with the fact that $f_n$ divides $f_N$ in $F[M]$ for each $n \ge N$ implies that $(f_n) = (f_N)$ for all $n \in N$.
\end{proof}
\medskip

We are now in a good position to exhibit a large class of atomic Puiseux algebras.

\begin{theorem} \label{thm:infinite family of atomic Puiseux algebras}
	For each field $F$, there is an infinite class of non-finitely generated atomic Puiseux algebras over $F$
\end{theorem}



\begin{proof}
	Let $\{P_n\}$ be a sequence of pairwise disjoint infinite subsets of $\mathbb{P}$. For each $n \in \mathbb{N}$, consider the prime reciprocal Puiseux monoid
	\[
		M_n := \bigg \langle \frac{1}{p} \ \bigg{|} \ p \in P_n \bigg \rangle.
	\]
	We claim that the members of the class $\mathcal{C} := \{F[M_n] \mid n \in \mathbb{N}\}$ of Puiseux algebras are atomic and pairwise non-isomorphic. The atomicity follows by combining Proposition~\ref{prop:each PPM satisfies the ACCP}, Proposition~\ref{prop:PF of an ACCP PM is again ACCP}, and Proposition~\ref{prop:ACCP implies atomic}.
	
	To show that the algebras in $\mathcal{C}$ are pairwise non-isomorphic, suppose, by way of contradiction, that $F[M_j] \cong F[M_k]$ for distinct $j,k \in \mathbb{N}$. Let $\phi \colon F[M_j] \to F[M_k]$ be an algebra isomorphism. Since $\phi$ fixes $F$, it follows that $\phi(X^q) \notin F$ for any $q \in M_j^\bullet$. This implies that $\deg(\phi(X)) \in M_k^\bullet$. As $P_j$ is unbounded there exists $p \in P_j$ such that $p > \mathsf{n}( \deg( \phi(X) ) )$. Observe that
	\begin{align} \label{eq: degrees 1}
		\deg \big( \phi(X) \big) = \deg \big( \phi \big( X^{\frac{1}{p}} \big)^p \big) = p \deg \big( \phi \big( X^{\frac{1}{p}} \big) \big).
	\end{align}
	Because $p \notin P_k$, it does not divide any element in $\mathsf{d}(M_k^\bullet)$. Thus, it follows from (\ref{eq: degrees 1}) that $p \mid \mathsf{n}( \deg \phi (X) )$, which contradicts that $p > \mathsf{n}( \deg( \phi(X) ) )$. Hence the Puiseux algebras in $\mathcal{C}$ are atomic and pairwise non-isomorphic, as desired.
\end{proof}

We have seen how to construct an infinite family of non-finitely generated atomic Puiseux algebras using atomic Puiseux monoids. In contrast, Puiseux monoids cannot be used to construct an infinite family of non-isomorphic half-factorial Puiseux algebras. We conclude this section illustrating this observation.

Puiseux algebras satisfying certain algebraic properties have been characterized in terms of their exponent Puiseux monoids by Gilmer.

\begin{theorem} \cite[Theorem~8.4]{GP74}
	For a nontrivial Puiseux monoid $M$, the following conditions are equivalent:
	\begin{enumerate}
		\item $M \cong (\mathbb{N}_0,+)$;
		\vspace{3pt}
	
		\item $F[M]$ is a PID;
		\vspace{3pt}
		\item $F[M]$ is a Euclidean domain;
		\vspace{3pt}
		\item $F[M]$ is a Dedekind domain.
	\end{enumerate}
\end{theorem}

\begin{definition}
	An atomic monoid $M$ is called \emph{half-factorial} provided that for all $x \in M^\bullet$, the fact that $z, z' \in \mathsf{Z}(x)$ implies that $|z| = |z'|$. An integral domain is call \emph{half-factorial} if its multiplicative monoid is half-factorial.
\end{definition}

Let us proceed to characterize the Puiseux algebras that are half-factorial in terms of its exponent monoids.

\begin{prop}
	Let $F$ be a field, and let $M$ be a Puiseux monoid. Then $F[M]$ is half-factorial if and only if $M \cong (\mathbb{N}_0,+)$.
\end{prop}

\begin{proof}
	The reverse implication follows immediately from Proposition~\ref{prop:induced homomorphism of semigroup algebras} and the fact that $F[\mathbb{N}_0] \cong F[X]$. For the direct implication, suppose that $F[M]$ is half-factorial. Since $F[M]$ is, in particular, atomic, $M$ is also atomic. In addition, it is not hard to verify that $\mathsf{L}_M(q) = \mathsf{L}_{F[M]}(X^q)$ for all $q \in M^\bullet$. This implies that $M$ is a half-factorial monoid. Now take $a, a' \in \mathcal{A}(M)$ and set $x = \mathsf{n}(a) \mathsf{n}(a')$. Since we have $\mathsf{d}(a) \mathsf{n}(a') a = x = \mathsf{d}(a') \mathsf{n}(a) a'$, it follows that $\{ \mathsf{d}(a) \mathsf{n}(a'), \mathsf{d}(a') \mathsf{n}(a) \} \subseteq \mathsf{L}_{M}(x)$. As $M$ is half-factorial, $a = a'$. Hence $|\mathcal{A}(M)| = 1$ and so $M \cong (\mathbb{N}_0,+)$.
\end{proof}

\begin{cor}
	There is only one half-factorial Puiseux algebra up to isomorphism.
\end{cor}
\medskip

\section{Antimatter Puiseux Algebras} \label{sec:antimatter PA}

Let $M$ be a monoid, and recall that $\mathsf{gp}(M)$ denotes the difference group of~$M$. An element $x \in \mathsf{gp}(M)$ is called a \emph{root element} of $M$ if $nx \in M$ for some $n \in \mathbb{N}$. The subset $\widetilde{M}$ of $\mathsf{gp}(M)$ consisting of all root elements of $M$ is called the \emph{root closure} of $M$. If $\widetilde{M} = M$, then $M$ is called \emph{root-closed}.

\begin{theorem} \cite[Theorem~13.5]{rG84} \label{thm:root closed characterization}
	Let $M$ be a Puiseux monoid containing $1$. The following conditions are equivalent:
	\begin{enumerate}
		\item $M$ is root-closed; \vspace{3pt}
		\item $\mathsf{gp}(M) \cap \mathbb{Q}_{\ge 0} = M$; \vspace{3pt}
		\item $M$ is the union of an ascending sequence of cyclic additive submonoids of $\mathbb{Q}$. 
	\end{enumerate}
\end{theorem}

\begin{example}
	Since the Puiseux monoid $\mathbb{Q}_{\ge 0}$ satisfies condition~(2) of Theorem~\ref{thm:root closed characterization}, it is root-closed. To find an ascending sequence of cyclic submonoids of $\mathbb{Q}$ whose union is $\mathbb{Q}_{\ge 0}$, take $\{p_n\}$ to be an enumeration of $\mathbb{P}$ and notice that
	\[
		\mathbb{Q}_{\ge 0} = \bigcup_{n \in \mathbb{N}} P_n, \ \text{ where } \ P_n := \bigg \langle \frac{1}{p_1^n \dots p_n^n} \bigg \rangle.
	\]
\end{example}

Recall that an integral domain is called a Bezout domain if the addition of two principal ideals is again a principal ideal and is called a Pr\"ufer domain provided that each nonzero finitely generated ideal is invertible. Root-closed Puiseux monoids are important because the Puiseux algebras they determine are both Bezout domains and Pr\"ufer domains.

\begin{theorem} \cite[Theorem~13.5]{rG84}
	Let $F$ be a field, and let $M$ be a Puiseux monoid containing 1. The following conditions are equivalent:
	\begin{enumerate}
		\item $M$ is root-closed; \vspace{3pt}
		\item $F[M]$ is a Bezout domain; \vspace{3pt}
		\item $F[M]$ is a Pr\"ufer domain.
	\end{enumerate} 
\end{theorem}

Root-closed Puiseux monoids exhibit extreme atomic structure, i.e., they are either atomic or antimatter, as the next proposition illustrates.

\begin{prop} \label{prop:atomicity of root closed PM}
	For a root-closed nontrivial Puiseux monoid $M$, the following conditions hold.
	\begin{enumerate}
		\item $M$ is atomic if and only if $M$ is cyclic. \vspace{3pt}
		\item $M$ is antimatter if and only if $0$ is a limit point of $M$. \vspace{3pt}
		\item $M$ is either atomic or antimatter.
	\end{enumerate}
\end{prop}

\begin{proof}
	Let us argue condition~(1). Suppose first that $M$ is atomic. Since $M$ is root-closed, Theorem~\ref{thm:root closed characterization} ensures the existence of a sequence $\{q_n\}$ of rational numbers such that $\langle q_1 \rangle \subseteq  \langle q_2 \rangle \subseteq \cdots$ and $M = \cup_{n \in \mathbb{N}} \, \langle q_n \rangle$. As $M$ is nontrivial, it must contain an atom $a$. Clearly, $a = q_j$ for some $j \in \mathbb{N}$. Because $a$ is an atom, the fact that $a \in \langle q_n \rangle$ for each $n \ge j$ forces $q_n = a$ for each $n \ge j$. Hence $M = \langle a \rangle$. The backward implication follows trivially.
	
	Now we verify condition~(2). Suppose that $M$ is antimatter. Because $M$ is nontrivial, it is not a group. Thus, it cannot be atomic and antimatter simultaneously. Then $M$ is not atomic. By Proposition~\ref{prop:non-dense PM are atomic}, one has that $0$ is a limit point of $M$, and the direct implication follows. Conversely, suppose that $0$ is a limit point of $M$, and let the sequence of rationals $\{q_n\}$ be defined as in the previous paragraph. Because we have that $\mathcal{A}(M) \subseteq \{q_n \mid n \in \mathbb{N}\}$, it suffices to check that $q_n \notin \mathcal{A}(M)$ for any $n \in \mathbb{N}$. To see this, observe that because $\lim q_n = 0$, for each $n \in \mathbb{N}$ there exists $m > n$ such that $q_m < q_n$ and so $q_m$ strictly divides $q_n$ in $M$. Hence $M$ is antimatter.
	
	Condition~(3) follows immediately from conditions~(1) and~(2), along with Proposition~\ref{prop:non-dense PM are atomic}.
\end{proof}

The following example, on the other hand, exhibits an antimatter Puiseux monoid that fails to be root-closed.

\begin{example} \label{ex:antimatter PM that is NOT root closed}
	The Puiseux monoid $M =: \big\langle 1/2^n, 1/3^n \mid n \in \mathbb{N} \big\rangle$ is clearly antimatter. Suppose, by way of contradiction, that $M$ is root-closed. By Theorem~\ref{thm:root closed characterization}, there exists an ascending sequence $\{\langle r_n \rangle \}$ of cyclic submonoids of $M$ such that
	\[
		M = \bigcup_{n \in \mathbb{N}} \langle r_n \rangle.
	\]
	Take $k \in \mathbb{N}$ such that $1/2 \in \langle r_k \rangle$. It is easy to see that $\mathsf{n}(r_k) = 1$, which implies that $\mathsf{n}(r_n) = 1$ for each $n \ge k$. Let $\mathsf{d}(r_k) = 2^a 3^b$ for some $a,b \in \mathbb{N}_0$. Since $r = 1/2^{a+1} + 1/3^{b+1}$ satisfies that $\mathsf{d}(r) = 2^{a+1} 3^{b+1}$, it follows that $r \in M \setminus \langle r_k \rangle$. Now take $\ell \in \mathbb{N}$ such that $r \in \langle r_\ell \rangle$. Then $\ell > k$ and, therefore, $\mathsf{n}(r_\ell) = 1$. Because $r$ is an integer multiple of $r_\ell$, we have that $\mathsf{d}(r_{\ell}) = 2^c 3^d$, where $c \ge a+1$ and $d \ge b+1$. Hence
	\[
		\frac 16 = 2^{c-1} 3^{d-1} \frac 1{2^c 3^d} \in \langle r_\ell \rangle \subseteq M.
	\]
	So there exist $s,t \in \mathbb{N}$ and nonnegative coefficients $\alpha_1, \dots \alpha_t$ and $\beta_1, \dots, \beta_s$ with either $\alpha_t \neq 0$ or $\beta_s \neq 0$ satisfying that
	\begin{align} \label{eq:antimatter not ascending cyclic}
		\frac 16 = \sum_{i=1}^t \frac{\alpha_i}{2^i} + \sum_{i=1}^s \frac{\beta_i}{3^i}.
	\end{align}
	After simplifying if necessary, we can assume that $2 \nmid \alpha_i$ and $3 \nmid \beta_j$ for $i = 1,\dots,t$ and $j=1,\dots,s$. If $t \neq 0$, then multiplying (\ref{eq:antimatter not ascending cyclic}) by $2^t$ one obtains that $t = 1$. Similarly, if $s \neq 0$, then $s = 1$. The fact that $s,t \in \{0,1\}$ forces $1/6 \in \{\alpha_1/2, \beta_1/3, \alpha_1/2 + \beta_1/3\}$, which is a contradiction.
\end{example}

\begin{remark}
	The fact that $1/6 \in M$ in Example~\ref{ex:antimatter PM that is NOT root closed} could have also been achieved by verifying first that
	\[
		\mathsf{gp}(M) = \bigg\{\frac{n}{2^r3^s} \ \bigg{|} \ n \in \mathbb{Z} \ \text{and} \ r,s \in \mathbb{N}_0 \bigg\},
	\]
	and then using the root-closedness of $M$, along with the fact that $5/6 \in M$, to conclude that $1/6 \in M$. 
\end{remark}

We are ready now to exhibit a class of Puiseux algebras without irreducible elements.

\begin{definition}
	An integral domain $R$ is called \emph{antimatter} provided that $R$ has no irreducible elements. 
\end{definition}

Antimatter domains have been studied by Coykendall, Dobbs, and Mullins in \cite{CDM99}. As the next proposition reveals, root-closed Puiseux monoids can be used to construct antimatter Puiseux algebras over algebraically closed fields.

\begin{prop} \label{prop:antimatter PA induced by root closed PM}
	Let $F$ be an algebraically closed field, and let $M$ be a nontrivial root-closed Puiseux monoid. Then the following conditions hold.
	\begin{enumerate}
		\item If $M$ is not antimatter, then $F[M] \cong F[X]$. \vspace{3pt}
		\item If $M$ is antimatter, then $F[M]$ is an antimatter Bezout domain.
	\end{enumerate}
\end{prop}

\begin{proof}
	If $M$ is not antimatter, then Proposition~\ref{prop:atomicity of root closed PM} ensures that $M$ is isomorphic to $(\mathbb{N}_0,+)$. Then Proposition~\ref{prop:induced homomorphism of semigroup algebras} guarantees that $F[M] \cong F[X]$, and condition~(1) follows. On the other hand, condition~(2) is a restatement of \cite[Theorem~1]{ACHZ07}.
\end{proof}

\begin{example}
	The root-closedness of $M$ in Proposition~\ref{prop:antimatter PA induced by root closed PM} is necessary for condition~(1) to hold. To see this, consider the Puiseux monoid $M = \langle 1/3, 1/2^n \mid n \in \mathbb{N} \rangle$. It is not antimatter as $1/3$ is an atom. However, $(X^{1/2}) \subsetneq (X^{1/2^2}) \subsetneq (X^{1/2^3}) \subsetneq \cdots$ is an ascending chain of ideals of $F[M]$ that does not stabilize, while $F[X]$ happens to be a Noetherian domain. 
\end{example}

On the other hand, we wonder whether the root-closedness of $M$ is necessary for condition~(2) in Proposition~\ref{prop:antimatter PA induced by root closed PM} to hold.

\begin{question}
	Is there a non-root-closed and antimatter Puiseux monoid such that the algebra $F[M]$ is antimatter over any (or some) algebraically closed field $F$?
\end{question}

We proceed to exhibit a second class of antimatter Puiseux algebras.

\begin{theorem} \label{thm:infinitely many PA over algebraically closed fields}
	Let $F$ be an algebraically closed field. There are infinitely many non-isomorphic antimatter Puiseux algebras over $F$.
\end{theorem}

\begin{proof}
	Consider the set of primes $P := \mathbb{P} \setminus \{\text{char}(F)\}$ and, for each $p \in P$, define $M_p = \langle 1/p^n \mid n \in \mathbb{N} \rangle$. Now consider the class of Puiseux algebras
	\[
		\mathcal{C} := \{F[M_p] \mid p \in P\}.
	\]
	Clearly, each $M_p$ is an antimatter monoid. In addition, because $1/p^{n+1}$ divides $1/p^n$ in $M_p$, it follows that $M_p$ is the union of an ascending sequence of cyclic submonoids of $\mathbb{Q}$, namely, $M_p = \cup_{n \in \mathbb{N}} \langle 1/p^n \rangle$. Thus, each $M_p$ is root-closed by Theorem~\ref{thm:root closed characterization}. Since $F$ is algebraically closed, Proposition~\ref{prop:antimatter PA induced by root closed PM}(2) guarantees that each algebra $F[M_p]$ is an antimatter Bezout domain.
	
	Finally, let us verify that the algebras in $\mathcal{C}$ are pairwise non-isomorphic. Suppose, by way of contradiction, that there exist $p,q \in P$ with $p \neq q$ such that $F[M_p] \cong F[M_q]$. Let $\phi \colon F[M_p] \to F[M_q]$ be an algebra isomorphism. As in the proof of Theorem~\ref{thm:infinite family of atomic Puiseux algebras}, we have that $\deg(\phi(X^{1/p^j})) \neq 0$ for every $j \in \mathbb{N}$. Take $n \in \mathbb{N}$ such that $n > \pval(\deg(\phi(X)))$. Since $p \neq \text{char}(F)$,
	\begin{align} \label{eq:degrees}
		\pval \big( \deg \big( \phi(X) \big) \big) = \pval \big( p^n \deg \big( \phi \big( X^{\frac{1}{p^n}} \big) \big) \big) = n + \pval \big( \deg \big( \phi \big( X^{\frac{1}{p^n}} \big) \big) \big).
	\end{align}
	Since $\mathsf{d}(M_q)$ consists only of powers of $q$ and $\deg \big( \phi \big( X^{1/p^n} \big) \big) \in M_q^\bullet$, it follows that $\pval \big( \deg \big( \phi \big( X^{1/p^n} \big) \big) \big) \ge 0$. Therefore $\pval \big( \deg \big( \phi(X) \big) \big) \ge n$ as a consequence of (\ref{eq:degrees}), which rises a contradiction. Hence $\mathcal{C}$ is a countable collection of pairwise non-isomorphic antimatter Puiseux algebras, which completes the proof.
\end{proof}
\medskip

Finally, we construct a family of antimatter Puiseux algebra over perfect fields of finite characteristic, now using certain divisible Puiseux monoids. Let $M$ be a Puiseux monoid. For $n \in \mathbb{N}$ and $x \in M$, we say that $x$ is $n$-\emph{divisible} if $x/n \in M$, and we say that $M$ is $n$-\emph{divisible} if $x$ is $n$-divisible for each $x \in M$. Clearly, every $n$-divisible Puiseux monoid is antimatter provided that $n \ge 2$.

\begin{theorem} \label{thm:characteristic p and p-divisible implies that F[M] is antimatter}
	For any perfect field $F$ of finite characteristic, there are infinitely many non-isomorphic antimatter Puiseux algebras over $F$.
\end{theorem}

\begin{proof}
	Let $p$ be the characteristic of $F$. First, we introduce the following class of Puiseux monoids parameterized by the prime numbers in $\mathbb{P} \setminus \{p\}$:
	\[
		M_q := \bigg \langle \frac 1{p^iq^j} \ \bigg{|} \ i,j \in \mathbb{N} \bigg \rangle.
	\]
	Clearly, $M_q$ is antimatter and $p$-divisible for each $q \in \mathbb{P} \setminus \{p\}$. Now consider the class of Puiseux algebra $\mathcal{F}$ over the field $F$ determined by the Puiseux monoids $M_q$, namely,
	\[
		\mathcal{C} = \big\{ F[M_q] \mid q \in \mathbb{P} \setminus \{p\} \big\}.
	\]
	Once again, we can use an argument similar to that one given in the proof of Theorem~\ref{thm:infinite family of atomic Puiseux algebras} to show that the algebras in $\mathcal{C}$ are pairwise non-isomorphic: let $\phi \colon M_q \to M_{q'}$ be an algebra isomorphism for distinct $q, q' \in \mathbb{P} \setminus \{p\}$, then take $n > \mathsf{v}_q(\deg(\phi(X^{1/p})))$, and finally notice that
	\[
		n > \mathsf{v}_q \big( \deg \big( \phi(X^{\frac{1}{p}} \big) \big) \big) = \mathsf{v}_q \big( q^n \deg \big( \phi \big( X^{\frac{1}{p q^n}} \big) \big) \big) = n + \mathsf{v}_q \big(\deg \big( \phi \big( X^{\frac{1}{p q^n}} \big) \big) \big) \ge n,
	\]
	which is a contradiction. As a result, the Puiseux algebras in the collection $\mathcal{C}$ are pairwise non-isomorphic, as desired.
	
	It only remains to verify that each algebra in $\mathcal{C}$ is an antimatter domain. To do so, take $M \in \mathcal{C}$ and consider the element
	\[
		f = \alpha_1 X^{q_1} + \dots + \alpha_n X^{q_n} \in F[M] \setminus F.
	\]
	As $F$ is a perfect field of characteristic $p$, the Frobenius homomorphism $x \mapsto x^p$ is surjective and, therefore, for each $i = 1,\dots,n$, there exists $\beta_i \in F$ with $\alpha_i = \beta_i^p$ for some $\beta_i \in F$. Also, the $p$-divisibility of $M$ ensures that $q_i/p \in M$ for $1 \le i \le n$. As a consequence, we have that $f = \big( \beta_1 X^{q_1/p} + \dots + \beta_n X^{q_n/p} \big)^p$ is not irreducible in $F[M]$. Hence $F[M]$ is an antimatter Puiseux algebra.
\end{proof}
\vspace{6pt}

\begin{remark} 
	Theorem~\ref{thm:characteristic p and p-divisible implies that F[M] is antimatter} is a version of \cite[Theorem~5(2)]{ACHZ07}, which states that if $R$ is an antimatter GCD-domain whose quotient field is perfect of finite characteristic, then $R[\mathbb{Q}_{\ge 0}]$ is also an antimatter GCD-domain.
\end{remark}
\medskip

Note that in order to construct the family of antimatter Puiseux algebras in Theorem~\ref{thm:characteristic p and p-divisible implies that F[M] is antimatter}, it has been convenient to use root-closed Puiseux monoids. In contrast to Theorem~\ref{thm:characteristic p and p-divisible implies that F[M] is antimatter}, the next proposition implies that a Puiseux algebra of a root-closed monoid over a field of characteristic zero is never antimatter. First, let us introduce a simple lemma we need to use.

\begin{lemma} \label{lem:denominaotr division in a root-closed monoid}
	Let $M$ be a root-closed Puiseux monoid containing $1$. Then $1/d \in M$ for each $d \in \mathsf{d}(M^\bullet)$.
\end{lemma}

\begin{proof}
	Take $r \in M^\bullet$ satisfying that $\mathsf{d}(r) = d$. In addition, take $a,b \in \mathbb{N}_0$ such that $a \mathsf{n}(r) - b \, \mathsf{d}(r) = 1$. Since $1 \in M$, it follows that $\mathbb{N}_0 \subseteq \mathsf{gp}(M)$. Then we have that
	\[
		\frac{1}{d} = \frac{a \mathsf{n}(r) - b \, \mathsf{d}(r)}{d} = a r - b \in \mathsf{gp}(M).
	\]
	Since $M$ is root-closed, Theorem~\ref{thm:root closed characterization} guarantees that $1/d \in \mathsf{gp}(M) \cap \mathbb{Q}_{\ge 0} = M$, which completes the proof.
\end{proof}

\begin{prop} \label{prop:irreducibles in a PA of a root closed PM}
	Let $F$ be a field, and let $M$ be a root-closed Puiseux monoid. Then an element $f \in F[M] \setminus F$ is irreducible if and only if $f(X^m)$ is irreducible in $F[X]$ for every positive common multiple $m$ of $\mathsf{d}(\mathsf{Supp}(f))$.
\end{prop}

\begin{proof}
	Suppose first that $f$ is an irreducible element of $F[M]$. Let $m$ be a common multiple of $\mathsf{d}\big(\mathsf{Supp}(f) \big)$. Then $f(X^m)$ is an element of $F[X]$. Take $g, h \in F[X]$ such that $f(X^m) = g(X) \, h(X)$. Since $M$ is root-closed, Lemma~\ref{lem:denominaotr division in a root-closed monoid} ensures that $g(X^{1/m}), h(X^{1/m}) \in F[M]$. Therefore $f(X) = a(X^{1/m}) b(X^{1/m})$ in $F[M]$. Since $f$ is irreducible in $F[M]$ either $g(X^{1/m}) \in F$ or $h(X^{1/m}) \in F$, which implies that either $g \in F$ or $h \in F$. Hence $f(X^m)$ is irreducible in $F[X]$.
	
	Conversely, suppose that $f \in F[M]$ satisfies that $f(X^m)$ is an irreducible polynomial in $F[X]$ for every common multiple $m$ of the set $\mathsf{d}(\mathsf{Supp}(f))$. To argue that $f$ is irreducible in $F[M]$ suppose that $f = g \, h$ for some $g, h \in F[M]$. Let $m_0$ be a common multiple of the set $\mathsf{d}(\mathsf{Supp}(g)) \cup \mathsf{d}(\mathsf{Supp}(h))$. Then we have the equality $f(X^{m_0}) = g(X^{m_0})h(X^{m_0})$ in the ring of polynomials $F[X]$. Since $f(X^{m_0})$ is irreducible in $F[X]$, either $g(X^{m_0}) \in F$ or $h(X^{m_0}) \in F$ and, therefore, either $g \in F$ or $h \in F$. This implies that $f$ is irreducible in $F[M]$.
\end{proof}
\medskip

\begin{definition}
	An integral domain $R$ is called an \emph{unrestricted unique factorization domain} (or simply a \emph{U-UFD}) if every element that can be factored into irreducibles has a unique factorization, that is, if $\alpha_1 \dots \alpha_n = \beta_1 \dots \beta_m$ with each  $\alpha_i$ and $\beta_j$ irreducible in $R$, then $m=n$ and there exists a permutation $\sigma \in S_n$ such that $\alpha_i$ is an associate of $\beta_{\sigma(i)}$ for each $i = 1,\dots,n$.
\end{definition}

The concept of U-UFD was introduced in~\cite{CZ04} and studied in connection with integral domains where irreducible elements are prime. We conclude this paper proving that Puiseux algebras of root-closed Puiseux monoids, are \emph{unrestricted UFD}.


\begin{theorem} \label{thm:U-UFD Puiseux algebras}
	If $M$ is a root-closed Puiseux monoid, then $F[M]$ is a U-UFD.
\end{theorem}

\begin{proof}
	Suppose that $f \in F[M]$ can be written as a product of irreducible elements of $F[M]$. Now, for $k,\ell \in \mathbb{N}$ let $g_1, \dots, g_k$ and $h_1, \dots, h_\ell$ be irreducible elements in $F[M]$ satisfying that
	\begin{equation} \label{eq:two products of irreducibles}
		g_1 \cdots g_k = f = h_1 \cdots h_\ell.
	\end{equation}
	Take $m$ to be a common multiple of the set
	\[
		\mathsf{d}\big(\mathsf{Supp}(f)\big) \bigcup \bigg(\bigcup_{i=1}^k\mathsf{d}\big(\mathsf{Supp}(g_i)\big) \bigg) \bigcup  \bigg(\bigcup_{j=1}^\ell\mathsf{d}\big(\mathsf{Supp}(h_j)\big) \bigg).
	\]
	Note that $f(X^m)$, $g_i(X^m)$ and $h_j(X^m)$ are polynomials in $F[X]$ for every $i = 1, \dots, k$ and $j = 1, \dots, \ell$. In addition, it follows from Proposition~\ref{prop:irreducibles in a PA of a root closed PM} that the mentioned polynomials are irreducibles in $F[X]$. Substituting $X$ by $X^m$ in~(\ref{eq:two products of irreducibles}) and using the fact that $F[X]$ is a UFD, one finds that $\ell = k$ and $g_i(X^m) = h_{\sigma(i)}(X^m)$ for some permutation $\sigma \in S_k$ and every $i = 1, \dots, k$. This, in turns, implies that $g_i = h_{\sigma(i)}$ for $i = 1, \dots, k$. Hence $|\mathsf{Z}_{F[M]}(f)| \in \{0,1\}$ for any nonzero non-unit element $f$, that is, $F[M]$ is a U-UFD.
\end{proof}




\section*{Acknowledgements}
	During the preparation of this paper, the author was supported by the NSF-AGEP fellowship.
	
