On the atomicity of monoid algebras

TL;DR

This paper demonstrates that even when both a monoid and a field are atomic, their monoid algebra may not be atomic, providing counterexamples across various ranks and characteristics.

Contribution

It provides the first negative answers to the open question about atomicity of monoid algebras over fields when the monoid is atomic, with explicit constructions.

Findings

Constructed atomic monoids of arbitrary infinite rank with non-atomic monoid algebras over any integral domain.

Exhibited atomic monoids of finite rank over fields of finite characteristic with non-atomic monoid algebras.

Built a rank 1 atomic monoid over the field with two elements where the monoid algebra is not atomic.

Abstract

Let $M$ be a commutative cancellative monoid, and let $R$ be an integral domain. The question of whether the monoid ring $R[x;M]$ is atomic provided that both $M$ and $R$ are atomic dates back to the 1980s. In 1993, Roitman gave a negative answer to the question for $M = \mathbb{N}_0$: he constructed an atomic integral domain $R$ such that the polynomial ring $R[x]$ is not atomic. However, the question of whether a monoid algebra $F[x;M]$ over a field $F$ is atomic provided that $M$ is atomic has been open since then. Here we offer a negative answer to this question. First, we find for any infinite cardinal $\kappa$ a torsion-free atomic monoid $M$ of rank $\kappa$ satisfying that the monoid domain $R[x;M]$ is not atomic for any integral domain $R$. Then for every $n \ge 2$ and for each field $F$ of finite characteristic we exhibit a torsion-free atomic monoid of rank $n$ such that $F[x;M]$ is not atomic. Finally, we construct a torsion-free atomic monoid $M$ of rank $1$ such that $\mathbb{Z}_2[x;M]$ is not atomic.

Equations

$$S_{\ge r}:=\{s\in S\mid s\ge r\}$$

$$[[x,y]]:=\{n\in \mathbb{Z}:x\le n\le y\}.$$

$$G=\bigg{\langle}\frac{1}{2^{n}\cdot p_{n}}\ \bigg{|}\ n\in\mathbb{N}\bigg{\rangle}$$

$$f=\sum _{s\in M}f(s)x^s=\sum _{i=1}^{n}f(s_i)x^{s_i},$$

$$\mathrm{deg}(fg)=\mathrm{deg}(f)+\mathrm{deg}(g)$$

$$(r,s)=\sum _{i=1}^k(a_i,0)+\sum _{j=1}^{\ell }(0,b_j)$$

$$\Omega :=\{a,b,c,s_n,t_n\mid n\in {\mathbb{N}}_0\}.$$

$$g=\sum _{\omega \in \Omega }{\mathsf{v}}_{\omega }(g)\omega ,$$

$$A:=\{c,s_n,t_n,a-nc-s_n,b-nc-t_n\mid n\in {\mathbb{N}}_0\}.$$

$$f=\sum _{j=1}^n{r}_j{x}^{v_j}\text{ and }g=\sum _{j=1}^m{h}_j{x}^{w_j}$$

$$f(x,y)=(1+x^{n})y^{n}+x^{n}=(1+x^{n})\bigg{(}y^{n}+\frac{x^{n}}{1+x^{n}}\bigg{)}.$$

$$x^n{h}_2(x{)}^q=c(1+x^n)h_1(x{)}^q$$

$$M_{p}:=\bigg{\langle}\frac{1}{p^{n}p_{n}}\ \bigg{|}\ p_{n}\neq p\bigg{\rangle}.$$

$$\big{(}X^{\frac{1}{p^{k}}}+Y^{\frac{1}{p^{k}}}+X^{\frac{1}{p^{k}}}Y^{\frac{1}{p^{k}}}\big{)}^{t}$$

$$g(X^{a{p}^k},Y^{a{p}^k})h(X^{a{p}^k},Y^{a{p}^k})=X^{a{p}^k}+Y^{a{p}^k}+X^{a{p}^k}{Y}^{a{p}^k}=(X^a+Y^a+X^a{Y}^a{)}^{p^k}.$$

$$g\big{(}X^{ap^{k}},Y^{ap^{k}}\big{)}=\big{(}X^{a}+Y^{a}+X^{a}Y^{a}\big{)}^{t}.$$

$$\displaystyle g=\big{(}X^{\frac{1}{p^{k+1}}}+Y^{\frac{1}{p^{k+1}}}+X^{\frac{1}{p^{k+1}}}Y^{\frac{1}{p^{k+1}}}\big{)}^{pt}.$$

$$3^{{\ell }_n-{\ell }_{n-1}}>2^{n+1}$$

$$a_{n}=\frac{1}{2^{n}}-\frac{1}{2^{2n}3^{\ell_{n}}}=\frac{1}{2^{n+1}}+\bigg{(}\frac{1}{2^{n+1}}-\frac{1}{2^{2n}3^{\ell_{n}}}\bigg{)}>\frac{1}{2^{n+1}}+\frac{1}{2^{2n+2}3^{\ell_{n+1}}}=b_{n+1}$$

$$a_n=\sum _{i=1}^N{\alpha }_i{a}_i+\sum _{i=1}^N{\beta }_i{b}_i$$

$$a_n=\sum _{i=n+1}^N{\alpha }_i{a}_i+\sum _{i=n+1}^N{\alpha }_i{b}_i=\sum _{i=n+1}^N{\alpha }_i\frac{1}{2^{i-1}},$$

$$a_n=({\alpha }_m-{\beta }_m)\frac{2^m{3}^{{\ell }_m}-1}{2^{2m}{3}^{{\ell }_m}}+\sum _{i=m}^N{\beta }_i\frac{1}{2^{i-1}}+\sum _{i=n+1}^{m-1}\frac{{\alpha }_i(2^i{3}^{{\ell }_i}-1)+{\beta }_i(2^i{3}^{{\ell }_i}+1)}{2^{2i}{3}^{{\ell }_i}}.$$

$$a_n\ge {\alpha }_m{a}_m\ge ({\alpha }_m-{\beta }_m)b_{m+1}\ge 3^{{\ell }_m-{\ell }_{m-1}}{b}_{m+1}>\frac{3^{{\ell }_m-{\ell }_{m-1}}}{2^{m+1}}>1,$$

$$b_n-{\alpha }_n{a}_n=\sum _{i=n+1}^N{\alpha }_i{a}_i+\sum _{i=n+1}^N{\beta }_i{b}_i$$

$$b_n-{\alpha }_n{a}_n=({\alpha }_m-{\beta }_m)\frac{2^m{3}^{{\ell }_m}-1}{2^{2m}{3}^{{\ell }_m}}+\sum _{i=m}^N{\beta }_i\frac{1}{2^{i-1}}+\sum _{i=n+1}^{m-1}\frac{{\alpha }_i(2^i{3}^{{\ell }_i}-1)+{\beta }_i(2^i{3}^{{\ell }_i}+1)}{2^{2i}{3}^{{\ell }_i}}.$$

$$x^{3^{n+1}}-1=\prod _{i=0}^{n+1}{Q}_{3^i}(x)$$

$$Q_{3^{n+1}}(x)=\frac{x^{3^{n+1}}-1}{{\prod }_{i=0}^n{Q}_{3^i}(x)}=\frac{x^{3^{n+1}}-1}{x^{3^n}-1}=f_n(x).$$

$$f\big{(}x^{6^{k}}\big{)}g\big{(}x^{6^{k}}\big{)}=\big{(}x^{6^{k}}\big{)}^{2}+x^{6^{k}}+1=\big{(}x^{2\cdot 3^{k}}+x^{3^{k}}+1\big{)}^{2^{k}}$$

$$\displaystyle f\big{(}x^{6^{k}}\big{)}=\big{(}x^{2\cdot 3^{k}}+x^{3^{k}}+1\big{)}^{t}=\big{(}\big{(}x^{6^{k}}\big{)}^{2\frac{1}{2^{k}}}+\big{(}x^{6^{k}}\big{)}^{\frac{1}{2^{k}}}+1\big{)}^{t}.$$

Full text

1. Introduction

In [3] R. Gilmer offers a very comprehensive summary of the theory of commutative semigroup rings developed until the 1980s. Many algebraic properties of a commutative ring $R$ and a monoid $M$ implying the corresponding property on the monoid ring $R[x;M]$ had been studied by that time, but still the following fundamental question was stated by Gilmer as an open problem.

Question 1.1**.**

[3, p. 189]** Let $M$ be a commutative cancellative monoid and let $R$ be an integral domain. Is the monoid domain $R[x;M]$ atomic provided that both $M$ and $R$ are atomic?

In 1990, D. Anderson et al. restated a special version of the above question in the context of polynomial rings, namely the case of $M=(\mathbb{N}_{0},+)$.

Question 1.2**.**

[1, Question 1]** If $R$ is an atomic integral domain, is the integral domain $R[x]$ also atomic?

Question 1.2 was answered negatively by M. Roitman in 1993. He constructed a class of atomic integral domains whose polynomial rings fail to be atomic [13]. In a similar direction, Roitman constructed examples of atomic integral domains whose corresponding power series rings fail to be atomic as well as examples of atomic power series rings over non-atomic integral domains [12]. Observe that Roitman’s negative answer to Question 1.2 gives a striking answer to Question 1.1, showing that $R[x;M]$ can fail to be atomic even if one takes $M$ to be the simplest nontrivial atomic monoid, namely $M=(\mathbb{N}_{0},+)$. This naturally suggests the question of whether the atomicity of $M$ implies the atomicity of $R[x;M]$ provided that $R$ is the taken to be one of the simplest nontrivial atomic integral domains, a field. Clearly, this is another special version of Question 1.1.

Question 1.3**.**

Let $F$ be a field. If $M$ is an atomic monoid, is the monoid algebra $F[x;M]$ also atomic?

There are many known classes of atomic monoids whose monoid algebras (over any field) happen to be atomic. For instance, if a monoid $M$ satisfies the ACCP (and, therefore, is atomic), then it is not hard to argue that for any field $F$ the monoid algebra $F[x;M]$ also satisfies the ACCP. Therefore $F[x;M]$ inherits the atomicity of $M$ as it is well known that integral domains satisfying the ACCP are atomic. In particular, every finitely generated monoid satisfies the ACCP and, thus, induces atomic monoid algebras. As BF-monoids also satisfy the ACCP [2, Corollary 1.3.3], one can use them to obtain many non-finitely generated atomic monoid algebras with rational (or even real) exponents; this is because submonoids of $(\mathbb{R}_{\geq 0},+)$ not having [math] as a limit point are BF-monoids [5, Proposition 4.5]. Furthermore, an infinite class of atomic submonoids of $(\mathbb{Q}_{\geq 0},+)$ (which are not BF-monoids) with atomic monoid algebras was exhibited in [4, Theorem 5.4].

The purpose of this paper is to provide a negative answer for Question 1.3. In Section 3 we find, for any infinite cardinal $\kappa$, a torsion-free atomic monoid $M$ of rank $\kappa$ satisfying that $R[x;M]$ is not atomic for any integral domain $R$. Then, in Section 4, for every $n\geq 2$ and for each field $F$ of finite characteristic we exhibit a torsion-free atomic monoid of rank $n$ such that $F[x;M]$ is not atomic. Finally, in Section 5, we construct a torsion-free atomic monoid $M$ of rank $1$ such that $\mathbb{Z}_{2}[x;M]$ is not atomic.

2. Background

2.1. General Notation

Throughout this paper, we set $\mathbb{N}:=\{1,2,\dots\}$ and we set $\mathbb{N}_{0}:=\mathbb{N}\cup\{0\}$. In addition, the symbols $\mathbb{Z}$, $\mathbb{Q}$, and $\mathbb{R}$ denote the sets of integers, rational numbers, and real numbers, respectively. For $S\subseteq\mathbb{R}$ and $r\in\mathbb{R}$, we set

[TABLE]

and, in a similar manner, we use the notation $S_{>r}$. For $x,y\in\mathbb{R}$ such that $x\leq y$, we let $\llbracket x,y\rrbracket$ denote the discrete interval between $x$ and $y$, i.e.,

[TABLE]

Information background and undefined terms related to atomic monoids and domains can be found in [3].

2.2. Atomic Monoids.

Every monoid here is tacitly assumed to be commutative and cancellative. Unless we specify otherwise, monoids are written additively. For a monoid $M$, we set $M^{\bullet}:=M\setminus\{0\}$, and we let $U(M)$ denote the set of invertible elements of $M$. The monoid $M$ is reduced provided that $U(M)=\{0\}$. On the other hand, $M$ is torsion-free provided that for all $x,y\in M$ the fact that $nx=ny$ for some $n\in\mathbb{N}$ implies that $x=y$. For $x,y\in M$, we say that $x$ divides $y$ in $M$ and write $x\mid_{M}y$ if $y=x+x^{\prime}$ for some $x^{\prime}\in M$. For $S\subseteq M$, we let $\langle S\rangle$ denote the smallest (under inclusion) submonoid of $M$ containing $S$. If $M=\langle S\rangle$, then we call $S$ a generating set of $M$. Further basic definitions and concepts on commutative cancellative monoids can be found in [7, Chapter 2].

An element $a\in M\setminus U(M)$ is an atom if for any $x,y\in M$ with $a=x+y$ 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 called atomic if each element of $M\setminus U(M)$ can be written as a sum of atoms. A subset $I$ of $M$ is an ideal of $M$ if $I+M\subseteq I$. An ideal $I$ is principal if $I=x+M$ for some $x\in M$, and $M$ satisfies the ascending chain condition on principal ideals (or ACCP) provided that every increasing sequence of principal ideals of $M$ eventually stabilizes. It is well known that every monoid satisfying the ACCP must be atomic [2, Proposition 1.1.4].

The Grothendieck group $\mathsf{gp}(M)$ of a monoid $M$ is the abelian group (unique up to isomorphism) satisfying that any abelian group containing a homomorphic image of $M$ will also contain a homomorphic image of $\mathsf{gp}(M)$. The rank of the monoid $M$ is defined to be the rank of the group $\mathsf{gp}(M)$, namely, the cardinality of a maximal integrally independent subset of $\mathsf{gp}(M)$. Equivalently, the rank of $M$ is the dimension of the $\mathbb{Q}$-vector space $\mathbb{Q}\otimes_{\mathbb{Z}}\mathsf{gp}(M)$. The monoid $M$ is called a totally ordered monoid with respect to a total order $\preceq$ on the underlying set of $M$ provided that $\preceq$ is compatible with the operation of $M$, i.e., for all $x,y,z\in M$ the inequality $x\preceq y$ implies that $x+z\preceq y+z$. If, in addition, $0\preceq x$ for all $x\in M$, then we say that $M$ is positive with respect to $\preceq$.

Some of the monoid algebras exhibited in this paper are constructed using submonoids of $(\mathbb{Q}_{\geq 0},+)$, which are called Puiseux monoids. Clearly, each Puiseux monoid is totally ordered with respect to the standard order of $\mathbb{Q}$. Although all Puiseux monoids we shall be using here are atomic, we should notice that there exists a huge variety of non-atomic Puiseux monoids. For instance, for any prime $p$, the Puiseux monoid $M=\langle 1/p^{n}\mid n\in\mathbb{N}_{0}\rangle$ is not atomic as $\mathcal{A}(M)$ is empty. On the other hand, there are atomic Puiseux monoids that do not satisfy the ACCP, and such monoids are crucial in this paper.

Example 2.1**.**

Let $p_{n}$ denote the $n^{\text{th}}$ odd prime. The Puiseux monoid

[TABLE]

was introduced by A. Grams in [6] to construct the first example of an atomic integral domain that does not satisfy the ACCP. It is not hard to check that $G$ is an atomic monoid. However, the increasing chain of principal ideals $\{1/2^{n}+G\}$ does not stabilize and, therefore, $G$ does not satisfy the ACCP. We call $G$ the Gram’s monoid.

2.3. Monoid Algebras.

As usual, for a commutative ring $R$ with identity, $R^{\times}$ denotes the set of units of $R$. An integral domain is atomic (resp., satisfies the ACCP) if its multiplicative monoid is atomic (resp., satisfies the ACCP).

For a monoid $M$ and an integral domain $R$, we let $R[x;M]$ denote the ring consisting of all functions $f\colon M\to R$ satisfying that $\{s\in M\mid f(s)\neq 0\}$ is finite. We shall conveniently write a generic element $f\in R[x;M]\setminus\{0\}$ in one of the following ways:

[TABLE]

where $s_{1},\dots,s_{n}$ are precisely those $s\in M$ with $f(s)\neq 0$. The ring $R[x;M]$ is, indeed, an integral domain [3, Theorem 8.1], and the set of units of $R[x;M]$ is $R^{\times}$ [3, Theorem 11.1]. As a result, $R[x;M]$ is called the monoid domain of $M$ over $R$ or, simply, a monoid domain. When $R$ is a field we call $R[x;M]$ a monoid algebra.

Let us assume now that $M$ is a totally ordered monoid with respect to a given order $\preceq$. In this case, we write the elements $f\in R[x;M]\setminus\{0\}$ in canonical form, that is $f=\alpha_{1}x^{q_{1}}+\dots+\alpha_{k}x^{q_{k}}$ with $\alpha_{i}\neq 0$ for $i=1,\dots,k$ and $q_{1}\succ\dots\succ q_{k}$. Clearly, there is only one way to write $f$ in canonical form. The element $q_{1}\in M$, denoted by $\deg(f)$, is called the degree of $f$. As for the case of rings of polynomials over integral domains, the identity

[TABLE]

holds for all $f,g\in R[x;M]\setminus\{0\}$. As it is customary for polynomials, we say that $f$ is a monomial provided that $k=1$.

The following two results, which we shall be using later, are well known and have rather straightforward proofs.

Theorem 2.2**.**

1. (1)

Let $M$ be a monoid satisfying the ACCP. Then for any field $F$ the monoid algebra $F[x;M]$ satisfies the ACCP. 2. (2)

Let $R$ be an integral domain. If $R$ satisfies the ACCP, then $R$ is atomic.

3. Non-Atomic Monoid Domains

Our goal in this section is to construct an atomic monoid $M$ with the property that for each integral domain $R$ the monoid domain $R[x;M]$ fails to be atomic.

Proposition 3.1**.**

For monoids $M$ and $N$ the following statements hold.

1. (1)

If $M$ and $N$ are atomic, then $M\times N$ is atomic. 2. (2)

If $M$ and $N$ satisfy the ACCP, then $M\times N$ satisfies the ACCP.

Proof.

To argue (1), suppose that $M$ and $N$ are atomic. Clearly, $(a,0)$ and $(0,b)$ belong to $\mathcal{A}(M\times N)$ when $a\in\mathcal{A}(M)$ and $b\in\mathcal{A}(N)$. Therefore, for any atomic decompositions $r=\sum_{i=1}^{k}a_{i}$ and $s=\sum_{j=1}^{\ell}b_{j}$ of $r\in M$ and $s\in N$,

[TABLE]

is an atomic decomposition of $(r,s)$ in $M\times N$. Hence $M\times N$ is atomic, and (1) follows. To argue (2), assume that $M$ and $N$ both satisfy the ACCP. Let $\{(a_{n},b_{n})+M\times N\}$ be an increasing chain of principal ideals in $M\times N$. Then $\{a_{n}+M\}$ and $\{b_{n}+N\}$ are increasing chains of principal ideals in $M$ and $N$, respectively. As $\{a_{n}+M\}$ and $\{b_{n}+N\}$ stabilize, $\{(a_{n},b_{n})+M\times N\}$ must also stabilize. As a result, $M\times N$ satisfies the ACCP, which completes the proof. ∎

Theorem 3.2**.**

There exists a reduced torsion-free atomic monoid $M$ with rank $\aleph_{0}$ such that for each integral domain $R$, the monoid domain $R[x;M]$ is not atomic.

Proof.

Consider the abelian group $G$ freely generated by the set

[TABLE]

For each element $g\in G$, one can write

[TABLE]

where $\mathsf{v}_{\omega}(g)\in\mathbb{Z}$ and $\mathsf{v}_{\omega}(g)=0$ for all but finitely many elements $\omega\in\Omega$. Observe that $\mathsf{v}_{\omega}(g+h)=\mathsf{v}_{\omega}(g)+\mathsf{v}_{\omega}(h)$ for all $g,h\in G$. Clearly, there exists a unique total order $\preceq$ on $G$ satisfying the following conditions:

1. (1)

the sequences $\{s_{n}\}$ and $\{t_{n}\}$ are both decreasing; 2. (2)

$a\succ b\succ c\succ s_{n}\succ t_{0}$ for every $n\in\mathbb{N}_{0}$; 3. (3)

$\preceq$ is lexicographic on $G$ with respect to the order already assigned to the elements of $\Omega$.

From now on we treat $G$ as a totally ordered group with respect to the order $\preceq$. Consider the submonoid $M$ of $G$ generated by the set

[TABLE]

Notice that $a,b\in M$ and $s\succ 0$ for all $s\in A$. As a result, $M$ is a positive monoid with respect to $\preceq$ and, therefore, $M$ is reduced.

To argue that $M$ is atomic it suffices to check that $A$ is a minimal set of generators of $M$. That $c\notin\langle A\setminus\{c\}\rangle$ follows from the fact that $c$ is the only element $s\in A$ with $\mathsf{v}_{c}(s)\succ 0$. Also, for each $n\in\mathbb{N}$, $s_{n}$ is the only element $s\in A$ with $\mathsf{v}_{s_{n}}(s)\succ 0$ and, therefore, $s_{n}\notin\langle A\setminus\{s_{n}\}\rangle$. A similar argument shows that $t_{n}\notin\langle A\setminus\{t_{n}\}\rangle$ for any $n\in\mathbb{N}_{0}$. In addition, for every $n\in\mathbb{N}_{0}$, the element $a-nc-s_{n}$ is the only $s\in A$ satisfying that $\mathsf{v}_{s_{n}}(s)\prec 0$, and so $a-nc-s_{n}\notin\langle A\setminus\{a-nc-s_{n}\}\rangle$. In a similar way one finds that $b-nc-t_{n}\notin\langle A\setminus\{b-nc-t_{n}\}\rangle$ for any $n\in\mathbb{N}_{0}$. Hence $A$ is a minimal set of generators of $M$, which implies that $M$ is an atomic monoid with $\mathcal{A}(M)=A$.

Let us check now that the rank of $M$ is $\aleph_{0}$. Since $a=nc+s_{n}+(a-nc-s_{n})$, we have that $a\in M$. In a similar way, one can see that $b\in M$. Thus, $\Omega\subseteq M$, which implies that the free commutative monoid $F(\Omega)$ on $\Omega$ is a submonoid of $M$. On the other hand, it is clear that $G$ is the Grothendieck group $F(\Omega)$. Then $G=\mathsf{gp}(F(\Omega))\subseteq\mathsf{gp}(M)\subseteq G$ and, therefore, $\mathsf{gp}(M)=G$. Hence the rank of $M$ is $\aleph_{0}$, as desired.

Now let $R$ be an integral domain. We proceed to show that $x^{a}+x^{b}\in R[x;M]$ cannot be expressed as a product of irreducibles. Suppose, by way of contradiction, that $x^{a}+x^{b}=fg$ for some $f,g\in R[x;M]$ such that $f$ is an irreducible in $R[x;M]$ with $\mathsf{v}_{a}(\deg(f))\succ 0$ and $g\notin R[x;M]^{\times}=R^{\times}$. Write

[TABLE]

in canonical forms. As $v_{1}+w_{1}=a\in A$, one finds that $\mathsf{v}_{a}(v_{1})=1$ and $\mathsf{v}_{a}(w_{1})=0$. In addition, for $j\in\llbracket 1,m\rrbracket$ we have that $w_{j}\preceq w_{1}$ and so $\mathsf{v}_{a}(w_{j})=0$. Now $v_{n}+w_{m}=b$ implies that $\mathsf{v}_{a}(v_{n})=0$ and, therefore, there exists a smallest index $i$ with $\mathsf{v}_{a}(v_{i})=0$. As $\mathsf{v}_{a}(v_{1})=1$ we have that $i>1$.

Claim 1. $v_{i^{\prime}}+w_{j}\neq v_{i}+w_{1}$ when $(i^{\prime},j)\neq(i,1)$.

Proof of Claim 1. First, note that if $i^{\prime}>i$, then $v_{i^{\prime}}\prec v_{i}$ and $w_{j}\preceq w_{1}$; this implies that $v_{i^{\prime}}+w_{j}\prec v_{i}+w_{1}$. On the other hand, if $i^{\prime}=i$, then $j>1$ and so $w_{j}\prec w_{1}$; this also implies that $v_{i^{\prime}}+w_{j}\prec v_{i}+w_{1}$. Lastly, if $i^{\prime}<i$, then $\mathsf{v}_{a}(v_{i^{\prime}})>0$ by the minimality of the index $i$. This implies that $\mathsf{v}_{a}(v_{i^{\prime}}+w_{j})>0=\mathsf{v}_{a}(v_{i}+w_{1})$, and our claim follows.

By Claim 1, we have that $x^{v_{i}+w_{1}}$ occurs in $x^{a}+x^{b}$. Because $\mathsf{v}_{a}(v_{i}+w_{1})=0$, the equality $x^{v_{i}+w_{1}}=x^{b}$ must hold and, as a consequence, $v_{i}+w_{1}=v_{n}+w_{m}$. Thus, $m=1$ and so it immediately follows that $g=x^{s}$ for some $s\in M$.

Claim 2. $s\in\langle c\rangle$.

Proof of Claim 2. It is clear that $s$ is a common divisor of $a$ and $b$ in $M$. Take $a^{\prime},b^{\prime}\in M$ such that $a=a^{\prime}+s$ and $b=b^{\prime}+s$. From $a=a^{\prime}+s$, one obtains that $0\leq\max\{\mathsf{v}_{b}(a^{\prime}),\mathsf{v}_{b}(s)\}\leq\mathsf{v}_{b}(a)=0$, and so $\mathsf{v}_{b}(a^{\prime})=\mathsf{v}_{b}(s)=0$. This, in turn, implies that $\mathsf{v}_{t_{n}}(a^{\prime})\geq 0$ and $\mathsf{v}_{t_{n}}(s)\geq 0$ for every $n\in\mathbb{N}_{0}$. Thus, $\mathsf{v}_{t_{n}}(s)=0$ for every $n\in\mathbb{N}_{0}$. In a similar manner one can verify that $\mathsf{v}_{a}(s)=\mathsf{v}_{s_{n}}(s)=0$ for every $n\in\mathbb{N}_{0}$. Thus, $s\in\langle c\rangle$, as claimed.

Finally, suppose that $s=kc$ for some $k\in\mathbb{N}$. Then $f=x^{a-kc}+x^{b-kc}$. For $c^{\prime}:=(a-(k+2)c-s_{k+2})+s_{k+2}\in M$, one sees that $2c+c^{\prime}=a-kc$. Thus, $2c\mid_{M}a-kc$. This implies that $x^{2c}\mid_{R[x;M]}x^{a-kc}$. Similarly, we have that $x^{2c}\mid_{R[x;M]}x^{b-kc}$. As a result, the non-unit non-irreducible element $x^{2c}$ divides $f$ in $R[x;M]$, which contradicts the fact that $f$ is irreducible in $R[x;M]$. Then $x^{a}+x^{b}$ cannot be expressed as a product of irreducibles in $R[x;M]$, which implies that $R[x;M]$ is not atomic. ∎

For any integral domain $R$ and monoids $M$ and $N$, there is a canonical ring isomorphism $F[x;M\times N]\cong F[y;M][z;N]$ induced by the assignment $x^{(a,b)}\mapsto y^{a}z^{b}$. To avoid having ordered pairs as exponents, we identify $R[x;M\times N]$ with $F[y;M][z;N]$ and write the elements of $F[x,M\times N]$ as polynomial expressions in two variables.

Corollary 3.3**.**

For any infinite cardinal $\kappa$, there exists a reduced torsion-free atomic monoid $M$ with rank $\kappa$ such that for any integral domain $R$ the monoid domain $R[x;M]$ is not atomic.

Proof.

By Theorem 3.2 there exists a reduced torsion-free atomic monoid $M^{\prime}$ with rank $\aleph_{0}$ such that for any integral domain $R$, the monoid domain $R[x;M^{\prime}]$ is not atomic. Let $M_{\kappa}$ be the free commutative monoid of rank $\kappa$, and set $M:=M^{\prime}\times M_{\kappa}$. Clearly, $M$ is reduced and torsion-free. It follows from Proposition 3.1(1) that $M$ is atomic, and the equality $\mathsf{gp}(M)=\mathsf{gp}(M^{\prime})\times\mathsf{gp}(M_{\kappa})$ ensures that the rank of $M$ is $\kappa$. Now let $R$ be an integral domain. Then $R[x;M]\cong R[y;M^{\prime}][z;M_{\kappa}]$. Since $M_{\kappa}$ is a torsion-free monoid and the monoid domain $R[y;M^{\prime}]$ is not atomic, [9, Proposition 1.4] guarantees that $R[x;M]$ is not atomic. ∎

In the next two sections, we shall construct reduced torsion-free atomic monoids with monoid algebras that are not atomic over finite-characteristic fields.

4. Non-atomic Monoid Algebras of Finite Characteristic

In this section, we find, for any given field $F$ of finite characteristic, a rank-$2$ totally ordered atomic monoid $M$ such that $F[x;M]$ is not atomic.

Let $r$ and $m$ be integers with $m>0$ and $\gcd(r,m)=1$. Recall that the order of $r$ modulo $m$ is the smallest $n\in\mathbb{N}$ for which $r^{n}\equiv 1\pmod{m}$, and that $r$ is a primitive root modulo $m$ if its order modulo $m$ equals $\phi(m)$, where $\phi$ is the Euler’s totient function. It is well known that for any odd prime $p$ and positive integer $k$, there exists a primitive root modulo $p^{k}$.

Lemma 4.1**.**

Let $F$ be a field of finite characteristic $p$ and $n\in\mathbb{N}$ be such that $p\nmid n$. Then the polynomial $x^{n}+y^{n}+x^{n}y^{n}$ is irreducible in $F[x,y]$.

Proof.

Set $f(x,y)=y^{n}(1+x^{n})+x^{n}$. Since $1+x^{n}$ and $x^{n}$ are relatively primes in $F[x]$, the polynomial $f(x,y)$ is primitive as a polynomial on $y$ over $F[x]$. By Gauss’s Lemma, arguing that $f(x,y)$ is irreducible in $F[x][y]$ amounts to proving that it is irreducible in $F(x)[y]$, where $F(x)$ is the field of fractions of $F[x]$. We can write now

[TABLE]

Set $a_{x}=\frac{x^{n}}{1+x^{n}}$. Then $f(x,y)$ is irreducible in $F[x,y]$ if and only if $y^{n}+a_{x}$ is irreducible in $F(x)[y]$. Using [8, Theorem 8.1.6], one can guarantee the irreducibility of $y^{n}+a_{x}$ by verifying that $a_{x}\notin 4F(x)^{4}$ when $4$ divides $n$ and that $-a_{x}\notin F(x)^{q}$ for any prime $q$ dividing $n$. To prove that these two conditions hold suppose, by way of contradiction, that $a_{x}\in cF(x)^{q}$, where $c\in\{-1,4\}$ and $q$ is either $4$ or a prime dividing $n$. Take $h_{1}(x),h_{2}(x)\in F[x]\setminus\{0\}$ such that $h_{1}(x)$ and $h_{2}(x)$ are relatively prime in $F[x]$ and a_{x}=c\big{(}h_{1}(x)/h_{2}(x)\big{)}^{q}. Then we have that

[TABLE]

From (4.1), one can deduce that $h_{2}(x)^{q}$ and $1+x^{n}$ are associates in $F[x]$, and so there exists $\alpha\in F^{\times}$ such that $h_{2}(x)^{q}=\alpha(1+x^{n})$. Taking derivatives in both sides of $h_{2}(x)^{q}=\alpha(1+x^{n})$ and using that $p\nmid n$, we obtain that $h_{2}(x)=x^{m}$ for some $m\in\mathbb{N}$, yielding that $c(1+x^{n})h_{1}(x)^{q}=x^{n+mq}$. However, this contradicts that $1+x^{n}$ does not divide $x^{n+mq}$ in $F[x]$. Hence $f(x,y)$ is irreducible in $F[x,y]$. ∎

Motivated by the Gram’s monoid, in the next example we exhibit a family of Puiseux monoids indexed by prime numbers whose members are atomic but do not satisfy the ACCP.

Example 4.2**.**

Let $\{p_{n}\}$ be a sequence consisting of all prime numbers ordered increasingly. For each prime $p$ consider the Puiseux monoid

[TABLE]

A very elementary argument of divisibility can be used to check that $M_{p}$ is atomic for each prime $p$. On the other hand, $M_{p}$ contains the strictly increasing sequence of principal ideals $\{1/p^{n}+M_{p}\}$ and, therefore, $M_{p}$ does not satisfy the ACCP. Notice that $M_{2}$ is precisely the Gram’s monoid.

Theorem 4.3**.**

For each field $F$ of finite characteristic $p$, there exists a torsion-free rank-$2$ atomic monoid $M$ such that the monoid algebra $F[x;M]$ is not atomic.

Proof.

Take $M:=M_{p}\times M_{p}$, where $M_{p}$ is the atomic monoid parametrized by $p$ exhibited in Example 4.2. It is clear that $M$ is torsion-free and has rank $2$. In addition, $M$ is atomic by Proposition 3.1(1). Now we claim that each non-unit factor of $f:=X+Y+XY$ in $F[x;M]$ has the form

[TABLE]

for some $k\in\mathbb{N}_{0}$ and $t\in\mathbb{N}$. To prove our claim, let $g\in F[x;M]$ be a non-unit factor of $f$, and take $h\in F[x;M]$ such that $f=g\,h$. Then there exist $k\in\mathbb{N}_{0}$ and $a\in\mathbb{N}$ with $p\nmid a$ such that $g(X^{ap^{k}},Y^{ap^{k}})$ and $h(X^{ap^{k}},Y^{ap^{k}})$ are both in the polynomial ring $F[X,Y]$. After changing variables, we obtain

[TABLE]

By Lemma 4.1, the polynomial $X^{a}+Y^{a}+X^{a}Y^{a}$ is irreducible in the polynomial ring $F[X,Y]$. Since $F[X,Y]$ is a UFD, there exists $t\in\mathbb{N}$ such that

[TABLE]

Going back to the original variables, we obtain g(X,Y)=\big{(}X^{\frac{1}{p^{k}}}+Y^{\frac{1}{p^{k}}}+X^{\frac{1}{p^{k}}}Y^{\frac{1}{p^{k}}}\big{)}^{t}, which establishes our initial claim.

Let us proceed to verify that $F[x;M]$ is not atomic. Notice that $f$ is not irreducible as f=\big{(}X^{\frac{1}{p}}+Y^{\frac{1}{p}}+X^{\frac{1}{p}}Y^{\frac{1}{p}}\big{)}^{p}. By the argument given in the previous paragraph, any factor $g$ of $f$ in a potential decomposition into irreducibles of $F[x;M]$ must be of the form \big{(}X^{\frac{1}{p^{k}}}+Y^{\frac{1}{p^{k}}}+X^{\frac{1}{p^{k}}}Y^{\frac{1}{p^{k}}}\big{)}^{t} and, therefore,

[TABLE]

Since $X^{\frac{1}{p^{k+1}}}+Y^{\frac{1}{p^{k+1}}}+X^{\frac{1}{p^{k+1}}}Y^{\frac{1}{p^{k+1}}}\in F[x;M]$, the equality (4.2) would contradict that $g$ is an irreducible element of $F[x;M]$. Thus, the algebra $F[x;M]$ is not atomic. ∎

Corollary 4.4**.**

For each field $F$ of finite characteristic $p$ and each $r\in\mathbb{N}_{\geq 2}$, there exists a totally ordered atomic monoid $M$ with rank $r$ such that $F[x;M]$ is not atomic.

Proof.

By Theorem 4.3, there exists a torsion-free rank-$2$ atomic monoid $M^{\prime}$ such that the monoid algebra $F[y;M^{\prime}]$ is not atomic. Now take $M:=M^{\prime}\times\mathbb{N}_{0}^{r-2}$. Clearly, $M$ has rank $r$. On the other hand, Proposition 3.1(1) ensures that $M$ is atomic. Also, $M$ is totally orderable because it is torsion-free and cancellative [3, Corollary 3.4]. Finally, since $F[x;M]\cong F[y;M^{\prime}][z;\mathbb{N}_{0}^{r-2}]$, it follows from [9, Proposition 1.4] and the fact that $F[y;M^{\prime}]$ is not atomic that $F[x;M]$ is not atomic. ∎

5. Non-atomic Monoid Algebras with Rational Exponents

The purpose of this section is to find an atomic Puiseux monoid $M$ such that the algebra $\mathbb{Z}_{2}[x;M]$ is not atomic. Since every Puiseux monoid is totally ordered and has rank $1$, this result will complement Corollary 4.4.

For $q\in\mathbb{Q}_{>0}$, let $a,b\in\mathbb{N}$ be the unique positive integers such that $q=a/b$ and $\gcd(a,b)=1$. Then we denote $b$ by $\mathsf{d}(q)$. In addition, if $S\subseteq\mathbb{Q}_{>0}$, then we denote the set $\{\mathsf{d}(s)\mid s\in S\}$ by $\mathsf{d}(S)$.

Proposition 5.1**.**

There exists an atomic Puiseux monoid $M$ satisfying the following two conditions:

1. (1)

M\subseteq\mathbb{Z}\big{[}\frac{1}{2},\frac{1}{3}\big{]}; 2. (2)

$\frac{1}{2^{n}}\in M$* for each $n\in\mathbb{N}_{0}$.*

Proof.

Let $\{\ell_{n}\}$ be a strictly increasing sequence of positive integers satisfying that

[TABLE]

for every $n\in\mathbb{N}$. Now set $A=\{a_{n},b_{n}\mid n\in\mathbb{N}\}$, where $a_{n}:=\frac{2^{n}3^{\ell_{n}}-1}{2^{2n}3^{\ell_{n}}}$ and $b_{n}:=\frac{2^{n}3^{\ell_{n}}+1}{2^{2n}3^{\ell_{n}}}$. It is clear that $1>b_{n}>a_{n}$ for every $n\in\mathbb{N}$. In addition,

[TABLE]

for every $n\in\mathbb{N}$. Therefore the sequence $b_{1},a_{1},b_{2},a_{2},\dots$ is strictly decreasing and bounded from above by $1$. Consider now the Puiseux monoid $M=\langle A\rangle$. Clearly, $M$ satisfies condition (1). On the other hand, $\frac{1}{2^{n}}=a_{n+1}+b_{n+1}\in M$ for every $n\in\mathbb{N}_{0}$. Thus, $M$ also satisfies condition (2).

All we need to prove is that $M$ is atomic. It suffices to verify that $A$ is a minimal generating set of $M$ [2, Proposition 1.1.7]. Suppose, by way of contradiction, that this is not the case. Then there exists $n\in\mathbb{N}$ such that $M=\langle A\setminus\{a_{n}\}\rangle$ or $M=\langle A\setminus\{b_{n}\}\rangle$.

CASE 1: $M=\langle A\setminus\{a_{n}\}\rangle$. In this case,

[TABLE]

for some $N\in\mathbb{N}_{\geq n}$ and nonnegative integer coefficients $\alpha_{i}$’s and $\beta_{i}$’s ($i\in\llbracket 1,N\rrbracket$) such that $\alpha_{n}=0$ and either $\alpha_{N}>0$ or $\beta_{N}>0$. Since the sequence $b_{1},a_{1},b_{2},a_{2},\dots$ is strictly decreasing, $\alpha_{i}=\beta_{i}=0$ for $i\in\llbracket 1,n\rrbracket$. Notice that $\alpha_{i}=\beta_{i}$ cannot hold for all $i\in\llbracket n+1,N\rrbracket$; otherwise,

[TABLE]

which is impossible because $3\mid\mathsf{d}(a_{n})$. Set m=\max\big{\{}i\in\llbracket n+1,N\rrbracket\mid\alpha_{i}\neq\beta_{i}\big{\}}. First assume that $\alpha_{m}>\beta_{m}$. Then we can rewrite (5.1) as follows:

[TABLE]

After multiplying both sides of the equality (5.2) by $2^{2N}3^{\ell_{m}}$, one can easily see that each summand involved in such an equality except perhaps $2^{2N-2m}(\alpha_{m}-\beta_{m})(2^{m}3^{\ell_{m}}-1)$ is divisible by $3^{\ell_{m}-\ell_{m-1}}$. Therefore $3^{\ell_{m}-\ell_{m-1}}$ must also divide $\alpha_{m}-\beta_{m}$. Now since $a_{m}>b_{m+1}>\frac{1}{2^{m+1}}$, we find that

[TABLE]

which is a contradiction. In a similar way we arrive at a contradiction if we assume that $\beta_{m}>\alpha_{m}$.

CASE 2: $M=\langle A\setminus\{b_{n}\}\rangle$. In this case, it is not hard to see that

[TABLE]

for some nonnegative coefficients $\alpha_{i}$’s ($i\in\llbracket n,N\rrbracket$) and $\beta_{j}$’s ($j\in\llbracket n+1,N\rrbracket$) such that either $\alpha_{N}>0$ or $\beta_{N}>0$. Observe that $\alpha_{n}\in\{0,1\}$ as it is obvious that $2a_{n}>b_{n}$. As before, there exists $m\in\llbracket n+1,N\rrbracket$ such that $\alpha_{m}\neq\beta_{m}$, and we can assume that such $m$ is as large as possible. If $\alpha_{m}>\beta_{m}$, then

[TABLE]

Since $\mathsf{d}(b_{n}-\alpha_{n}a_{n})\in\{2^{n-1}3^{\ell_{n}},2^{2n}3^{\ell_{n}}\}$, after multiplying the previous equation by $2^{2N}3^{\ell_{m}}$ we can see that $3^{\ell_{m}-\ell_{m-1}}$ divides $\alpha_{m}-\beta_{m}$. Now, an argument similar to that one given in CASE 1 can be used to obtain that $b_{n}>1$, which is a contradiction. We can proceed in a similar manner to obtain a contradiction if we assume that $\alpha_{m}<\beta_{m}$. Hence we have proved that $A$ is a minimal generating set of $M$, which means that $M$ is atomic with $\mathcal{A}(M)=A$. ∎

The following result will be used in the proof of Lemma 5.3.

Lemma 5.2**.**

[10, page 179]** If $p$ is an odd prime and $r$ is a primitive root modulo $p^{2}$, then $r$ is a primitive root modulo $p^{n}$ for every $n\geq 2$.

For a given field $F$, we let $Q_{n}(x)$ denote the $n^{\text{th}}$ cyclotomic polynomial over $F$. The next lemma is proposed as an exercise in [11, Chapter 3]. For the convenience of the reader, we provide a proof here.

Lemma 5.3**.**

For each $n\in\mathbb{N}$, the polynomial $x^{2\cdot 3^{n}}+x^{3^{n}}+1$ is irreducible in $\mathbb{Z}_{2}[x]$.

Proof.

One can verify that the order of $2$ modulo $9$ is $\phi(9)=6$. Then $2$ is a primitive root modulo $3^{2}$. It follows now by Lemma 5.2 that $2$ is also a primitive root modulo $3^{k}$ for every $k\geq 2$. Fix $n\in\mathbb{N}$ and set $f_{n}(x):=x^{2\cdot 3^{n}}+x^{3^{n}}+1$. From

[TABLE]

one obtains that

[TABLE]

Therefore $f_{n}(x)$ is the ${3^{n+1}}$-th cyclotomic polynomial over $\mathbb{Z}_{2}$ (see [11, Example 2.46]). Since $2$ is a primitive root module $3^{k}$ for any $k\geq 2$, the least positive integer $d$ satisfying that $2^{d}\equiv 1\pmod{3^{n+1}}$ is $\phi(3^{n+1})=2\cdot 3^{n}$. Hence [11, Theorem 2.47(ii)] guarantees that the polynomial $f_{n}(x)$ is irreducible. ∎

Theorem 5.4**.**

There exists an atomic Puiseux monoid $M$ such that $\mathbb{Z}_{2}[x;M]$ is not atomic.

Proof.

Let $M$ be an atomic Puiseux monoid satisfying conditions (1) and (2) of Proposition 5.1. First, we will argue that each factor of the element $x^{2}+x+1$ in $\mathbb{Z}_{2}[x;M]$ has the form \big{(}x^{2\frac{1}{2^{k}}}+x^{\frac{1}{2^{k}}}+1\big{)}^{t} for some $k\in\mathbb{N}_{0}$ and $t\in\mathbb{N}$. First, note that because $M$ contains $\langle 1/2^{k}\mid k\in\mathbb{N}_{0}\rangle$, it follows that $x^{2\frac{1}{2^{k}}}+x^{\frac{1}{2^{k}}}+1\in\mathbb{Z}_{2}[x;M]$ for all $k\in\mathbb{N}_{0}$. Now suppose that $f(x)$ is a factor of $x^{2}+x+1$ in $\mathbb{Z}_{2}[x;M]$, and take $g(x)\in\mathbb{Z}_{2}[x;M]$ such that $x^{2}+x+1=f(x)g(x)$. Then there exists $k\in\mathbb{N}_{0}$ such that

[TABLE]

in the polynomial ring $\mathbb{Z}_{2}[x]$. By Lemma 5.3, the polynomial $x^{2\cdot 3^{k}}+x^{3^{k}}+1$ is irreducible in $\mathbb{Z}_{2}[x]$. Since $\mathbb{Z}_{2}[x]$ is a UFD, there exists $t\in\mathbb{N}$ such that

[TABLE]

After changing variables in (5.5), one obtains that f(x)=\big{(}x^{2\frac{1}{2^{k}}}+x^{\frac{1}{2^{k}}}+1)^{t}. Thus, each factor of $x^{2}+x+1$ in $\mathbb{Z}_{2}[x;M]$ has the desired form.

Now suppose, by way of contradiction, that the monoid domain $\mathbb{Z}_{2}[x;M]$ is atomic. Then $x^{2}+x+1=\prod_{i=1}^{n}f_{i}(x)$ for some $n\in\mathbb{N}$ and irreducible elements $f_{1}(x),\dots,f_{n}(x)$ in $\mathbb{Z}_{2}[x;M]$. Since $f_{1}(x)$ is a factor of $x^{2}+x+1$, there exist $k\in\mathbb{N}_{0}$ and $t\in\mathbb{N}$ such that f_{1}(x)=\big{(}x^{2\frac{1}{2^{k}}}+x^{\frac{1}{2^{k}}}+1)^{t}. As $f_{1}(x)$ is irreducible, $t=1$. Now the equality f_{1}(x)=\big{(}x^{2\frac{1}{2^{k+1}}}+x^{\frac{1}{2^{k+1}}}+1\big{)}^{2} contradicts the fact that $f_{1}(x)$ is irreducible in $\mathbb{Z}_{2}[x;M]$. Hence $\mathbb{Z}_{2}[x;M]$ is not atomic. ∎

Acknowledgements

While working on this paper, the second author was partially supported by the UC Year Dissertation Fellowship. The authors would like to thank an anonymous referee for her/his careful review and very helpful suggestions.