April 13, 2026
The study of semirings is usually traced to Vandiver’s 1934 paper, where the algebraic consequences of dropping additive cancellation (and, more generally, additive inverses) were investigated [1]. Their ideal theory was developed early on by Bourne, who introduced and studied analogues of radicals in semirings [2]. By now, semiring theory has matured into a broad subject with several standard references and applications, including the monographs of Golan and of Hebisch-Weinert [3], [4].
We study the polynomial semiring \(\mathbb{N}_0[X]\). Fixing \(\alpha\in\mathbb{R}_{>0}\), evaluation at \(\alpha\) defines an additive submonoid \[\mathbb{N}_0[\alpha]=\{f(\alpha): f(X)\in \mathbb{N}_0[X]\}\subseteq \mathbb{R}_{\ge 0}.\] Recent work of Correa-Morris and Gotti shows that such monoids arise naturally in the study of algebraic valuations of polynomial semirings and develops a detailed understanding of their atomic structure and factorization phenomena [5]. Recent progress can be found in [6]–[9]. This places \(\mathbb{N}_0[\alpha]\) within the general framework of factorization theory in commutative cancellative monoids, a subject whose modern development grew out of the failure of unique factorization in algebraic number theory and is now treated in the monograph of Geroldinger and Halter-Koch. [10].
In [11], the authors study the admissibility of pairs \((\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right|,\left|\mathcal{S}(\mathbb{N}_0[\alpha])\right|)\) where the latter denotes the number of strong atoms. In the same paper, the authors give a detailed classification of rank \(2\) semirings of the form \(\mathbb{N}_0[\alpha]\). We note that Theorem 5.4 of [11] yields a partial answer to the conjecture stated in [5] (Conjecture 5.14): For every \(n\in\mathbb{N}\) there is an algebraic number \(\alpha\in\mathbb{R}_{>0}\) such that \(\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right|- \deg \mathfrak{m}_\alpha(X) = n\), for even integers \(n\).
In this paper, we aim to study the cardinality of the set of atoms \(\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right|\). Our first main result is a complete characterization of polynomials of the form \(\mathfrak{m}_\alpha(X)=p(X)-c\), where \(p(X)\in \mathbb{N}_0[X]\) (see Theorem 10). Our next main result links the weak Perron numbers to infinitely generated monoids; this is Theorem 11.
Next, we apply our results to obtain a partial classification of rank-\(3\) monoids of the form \(\mathbb{N}_0[\alpha]\) according to their generation pattern; this is done in Section 5. Most cases follow directly from the theorems in Section 4. For the form \(\mathfrak{m}_\alpha(X) = X^3 + aX^2 - bX - c\), we give necessary conditions for finite generation of \(\mathbb{N}_0[\alpha]\) and provide a family of monoids showing that these conditions are not sufficient. We also prove that it is never an \(LFM\), and we give a family of monoids with \(\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right| = 5\). For the form \(\mathfrak{m}_\alpha(X) = X^3 - aX^2 + bX - c\), we provide a necessary and sufficient condition for infinite generation and necessary and sufficient conditions for \(\mathbb{N}_0[\alpha]\) to be a proper \(LFM\).
We write \(\mathbb{N}_0:=\mathbb{N}\cup\{0\}\).
For integers \(m\le n\) we use \(\llbracket m,n\rrbracket := \{m,m+1,\dots,n\}\).
If \((M,+,0)\) is a commutative monoid and \(S\subseteq M\), then \[\langle S\rangle \;:=\;\left\{\sum_{i=1}^k s_i \;:\;k\in\mathbb{N},\;s_i\in S\right\}\cup\{0\}\] denotes the submonoid generated by \(S\).
Definition 1. A commutative semiring* is a set \(R\) with operations \((+, \cdot)\) such that \((R,+,0)\) is a commutative
monoid, \((R,\cdot,1)\) is a commutative monoid, multiplication distributes over addition, and \(0\cdot r=r\cdot 0=0\) for all \(r\in R\).
The semiring \(\mathbb{N}_0[X]\) is the set of polynomials with coefficients in \(\mathbb{N}_0\) with the usual addition and multiplication of polynomials.*
Definition 2. Fix \(\alpha\in\mathbb{R}_{>0}\). The evaluation map \[\operatorname{ev}_\alpha:\mathbb{N}_0[X]\to \mathbb{R}_{\ge 0},\qquad f(X)\mapsto f(\alpha),\] has image \[\mathbb{N}_0[\alpha]\;:=\;\operatorname{ev}_\alpha(\mathbb{N}_0[X]) \;\subseteq\;(\mathbb{R}_{\ge0},+),\] which we view as a commutative cancellative monoid under addition.
Definition 3. Let \(m_\alpha(X)\in\mathbb{Q}[X]\) be the minimal polynomial* of \(\alpha\) over \(\mathbb{Q}\). Let \(\mathfrak{m}_\alpha(X)\in\mathbb{Z}[X]\) be the unique primitive integer polynomial with positive leading coefficient whose image in \(\mathbb{Q}[X]\) is \(m_\alpha(X)\) (so \(\mathfrak{m}_\alpha(\alpha)=0\) and \(\gcd\) of its coefficients is \(1\)).*
For any \(f(X)=\sum_i a_iX^i\in\mathbb{Z}[X]\), define its positive and negative parts \[f^+(X):=\sum_{a_i>0} a_iX^i\in\mathbb{N}_0[X],\qquad f^-(X):=\sum_{a_i<0} (-a_i)X^i\in\mathbb{N}_0[X],\] so that \(f=f^+-f^-\) and \(\operatorname{supp}(f^+)\cap\operatorname{supp}(f^-)=\varnothing\). We call \((f^+,f^-)\) the (canonical) minimal pair* of \(f\). In particular, we write \(\mathfrak{m}_\alpha = p_\alpha - q_\alpha\) with \((p_\alpha,q_\alpha)=(\mathfrak{m}_\alpha^+,\mathfrak{m}_\alpha^-)\).*
Definition 4. Let \((M,+,0)\) be a commutative cancellative monoid. An element \(u\in M\) is a unit* if it has an additive inverse in \(M\). A nonunit \(a\in M\setminus\{0\}\) is an atom (or irreducible) if \(a=b+c\) with \(b,c\in M\) implies that \(b\) or \(c\) is a unit. We write \(\mathcal{A}(M)\) for the set of atoms of \(M\). The monoid \(M\) is atomic if every nonzero nonunit is a finite sum of atoms, and antimatter if \(\mathcal{A}(M)=\varnothing\).*
Definition 5. Assume \(M\) is atomic. A factorization* of \(x\in M\setminus\{0\}\) is an expression \(x=a_1+\cdots+a_k\) with \(k\in\mathbb{N}\) and \(a_i\in\mathcal{A}(M)\). Two factorizations are identified if they differ only by permuting the atoms. Let \(\mathsf Z(x)\) be the set of factorizations of \(x\), and for \(z\in\mathsf Z(x)\) let \(|z|:=k\) be its length. The set of lengths is \(\mathsf L(x):=\{|z|:z\in\mathsf Z(x)\}\).*
\(M\) is a UFM* (unique factorization monoid) if \(|\mathsf Z(x)|=1\) for every \(x\ne 0\).*
\(M\) is an HFM* (half-factorial monoid) if \(|\mathsf L(x)|=1\) for every \(x\ne 0\).*
\(M\) is an LFM* (length-factorial monoid) if the map \(\mathsf Z(x)\to\mathbb{N}\), \(z\mapsto |z|\) is injective for every \(x\ne 0\). We say \(M\) is a proper LFM if it is an LFM but not a UFM.*
\(M\) is an FFM* (finite factorization monoid) if \(\mathsf Z(x)\) is finite for every \(x\ne 0\).*
Definition 6. A commutative monoid \(M\) is finitely generated (FGM) if \(M=\langle S\rangle\) for some finite \(S\subseteq M\).
Definition 7. [12] Let \(\alpha\) be a real algebraic number and let \(\mathrm{Conj}(\alpha)\) be the set of Galois conjugates of \(\alpha\) in \(\mathbb{C}\). We say \(\alpha\) is a weak Perron number* if \(\alpha>0\) and \(|\beta|\le \alpha\) for every \(\beta\in\mathrm{Conj}(\alpha)\). We say \(\alpha\) is a Perron number if \(\alpha>0\) and \(|\beta|< \alpha\) for every \(\beta\in\mathrm{Conj}(\alpha)\) such that \(\beta \neq \alpha\).*
Definition 8. A monic polynomial \(H(X)\in \mathbb{Z}[X]\) is called a negative-tail polynomial* if \[H(X)=X^n-\sum_{i=0}^{n-1}a_iX^i\] for some \(n\in\mathbb{N}\) and coefficients \(a_0,\ldots,a_{n-1}\in\mathbb{N}_0\).*
Throughout this paper, we consider only commutative cancellative monoids.
Let \(\alpha\) be a positive real algebraic number over \(\mathbb{Q}\). Let \(\mathfrak{m}_{\alpha}\) be the primitive polynomial of minimal degree over
\(\mathbb{Z}\) with \(\mathfrak{m}_{\alpha}(\alpha)=0\). Let \(p_{\alpha}(X),\;q_\alpha(X)\in\mathbb{N}_0[X]\) such that \(\mathfrak{m}_{\alpha}(X)=p_{\alpha}(X)-q_{\alpha}(X)\).
Proposition 1 ([5], Theorem 4.2). For each algebraic \(\alpha \in \mathbb{R}_{>0}\), the monoid \(\mathbb{N}_0[\alpha]\) is atomic if and only if \(1 \in \mathcal{A}(\mathbb{N}_0[\alpha])\), and \(\mathbb{N}_0[\alpha]\) is antimatter otherwise.
Also, if \(\mathbb{N}_0[\alpha]\) is atomic, then there exists \(\sigma \in \mathbb{N} \cup \{\infty\}\) such that \[\label{eq:atoms} \mathcal{A}(\mathbb{N}_0[\alpha]) = \{ \alpha^n : n \in [0,\sigma) \cap \mathbb{N}_0 \}.\qquad{(1)}\]
If \(\mathbb{N}_0[\alpha]\) is finitely generated (and so atomic), then \[\sigma = \min \left\{ n \in \mathbb{N} : \alpha^n \in \langle \alpha^j : j \in \llbracket 0, n-1 \rrbracket \rangle \right\}.\]
We deduce that, whenever \(\mathbb{N}_0[\alpha]\) is an FGM, we have \(\sigma\geq \deg \mathfrak{m}_\alpha\).
Note that \(\mathbb{N}_0[\alpha]\) is finitely generated if and only if \(\mathbb{N}_0[\alpha]\) is atomic and \(\mathfrak{m}_\alpha\) divides a negative
tail polynomial.
By [5] Theorem 5.4, we have \(\sigma=\deg \mathfrak{m}_\alpha\) if and only if \(\mathbb{N}_0[\alpha]\) is a UFM if and only if \(\mathfrak{m}_\alpha\) is a negative tail polynomial. We also recall the following results that we will directly use in this paper:
Proposition 2 ([5], Proposition 4.5). Let \(\alpha \in \mathbb{R}_{>0}\) be an algebraic number with minimal polynomial \(\mathfrak{m}_{\alpha}(X)\). Then the following statements hold.
If \(\alpha \notin \mathbb{Q}\) and \(|\mathfrak{m}_{\alpha}(0)| \neq 1\), then \(\mathbb{N}_0[\alpha]\) is atomic.
If \(\mathfrak{m}_{\alpha}(X)\) has more than one positive root, then \(\mathbb{N}_0[\alpha]\) is atomic.
Proposition 3 ([5], Proposition 5.6). If \(\mathbb{N}_0[\alpha]\) is a finitely generated monoid (FGM) for some algebraic \(\alpha \in \mathbb{R}_{>0}\), then \(m_\alpha(X) \in \mathbb{Z}[X]\) and its only positive root is \(\alpha\) (counting multiplicity).
Proposition 4 ([5], Theorem 5.4). Let \(\alpha \in \mathbb{R}_{>0}\) be an algebraic number. The following statements hold:
(1) If \(\mathbb{N}_0[\alpha]\) is a UFM, then it is finitely generated.
(2) Suppose that \(\alpha\) has algebraic degree \(d\), minimal polynomial \(m_{\alpha}(X)\), and minimal pair \((p(X), q(X))\). Then the following conditions are equivalent:
(a) *$\mathbb{N}_0[\alpha]$ is a UFM;*
(b) *$\mathbb{N}_0[\alpha]$ is an HFM;*
(c) *$\deg m_{\alpha}(X) = |\mathcal{A}(\mathbb{N}_0[\alpha])|$;*
(d) *$p(X) = X^{d}$ for some $d \in \mathbb{N}$.*Theorem 5 ([5], Theorem 5.9). Let \(\alpha \in \mathbb{R}_{>0}\) be an algebraic number. The following conditions are equivalent:
(a) \(\mathbb{N}_0[\alpha]\) is a proper LFM;
(b) \(\mathcal{A}(\mathbb{N}_0[\alpha]) = \{\alpha^{j} : j \in \llbracket 0, \deg m_{\alpha}(X) \rrbracket \}\).
We also recall the following theorem:
Theorem 6 (Descartes’ Rule of Signs). Let \[f(X) = a_n X^n + \cdots + a_1 X + a_0 \in \mathbb{R}[X],\] with \(a_n \neq 0\). The number of positive roots of \(f\), counted with multiplicity, is at most the number of sign variations in the coefficient sequence \((a_n, \ldots, a_0)\), and differs from it by an even integer.
We can directly deduce from this rule the following lemma:
Lemma 1. Suppose that \(\mathfrak{m}_\alpha(X)\) has exactly two sign variations. If \(\mathfrak{m}_\alpha\) has a positive root and \(\mathbb{N}_0[\alpha]\) is atomic, then \(\mathbb{N}_0[\alpha]\) is infinitely generated.
Proof. By the theorem above, the number of positive roots of \(\mathfrak{m}_\alpha\), counted with multiplicity, is even. Since \(\mathfrak{m}_{\alpha}\) has a positive root, it
must have at least two positive roots, counted with multiplicity.
Because \(\mathfrak{m}_\alpha\) is irreducible over \(\mathbb{Q}\) and the ground field has characteristic \(0\), the polynomial is separable. Thus all roots
are simple, so it has at least two distinct positive roots, then by proposition 2 and proposition 3 \(\mathbb{N}_0[\alpha]\) is infinitely generated. ◻
Theorem 7 ([13]). Suppose that the irreducible polynomial \(f(x)\in \mathbb{Z}[x]\) has \(m\) roots, at least one of which is real, on the circle \(|z|=c\) Then \(f(x)=g(x^{m})\), where \(g(x)\) has no more than one real root on any circle in \(\mathbb{C}\).
Proposition 8 ([14], Proposition 4.4). Let \(\alpha \in \mathbb{A}\cap(0,1)\). If \(\alpha^{-1}\) is a Perron number with no positive conjugate aside from itself, then there exists a polynomial \(h(x)\in \mathbb{Z}[x]\) such that \(h(x)m_\alpha(x)\in x\mathbb{N}_0[x]-1\) and \(h(x)m_\alpha(x)\) is simple.
Proposition 9 ([11], Proposition 5.1). Let \(\alpha\) be an algebraic number with minimal polynomial \(m(x)\in \mathbb{Q}[x]\) such that \(m(x^k)\) is irreducible in \(\mathbb{Q}[x]\) for some \(k\in \mathbb{N}_{\ge 2}\). If \(\beta\) is a root of \(m(x^k)\), then \(\left|\mathcal{A}(\mathbb{N}_0[\beta])\right| = k\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right|\).
Theorem 10. Write \(\mathfrak{m}_\alpha(X)=p_\alpha(X)-q_\alpha(X)\) with \(q_\alpha(X)=c\in\mathbb{N}\).
If \(p_\alpha(X)=X^m\), then \(\mathbb{N}_0[\alpha]\) is a UFM.
Otherwise, we have two cases:
If \(c=1\) then \(\mathbb{N}_0[\alpha]\) is antimatter.
If \(c>1\) then \(\mathbb{N}_0[\alpha]\) is atomic and \(\mathbb{N}_0[\alpha]\) is infinitely generated.
Proof.
by Proposition 4.
If \(c=1\), we deduce from \(\mathfrak{m}_\alpha(\alpha)=0\) that \(1=p_\alpha(\alpha)\), thus by Proposition 1, \(\mathbb{N}_0[\alpha]\) is antimatter.
If \(c>1\), by Proposition 2, \(\mathbb{N}_0[\alpha]\) is atomic.
Let \(p_\alpha(X)=X^m+p_{m-1}X^{m-1}+\dots+p_1X\) where \((p_1,\dots,p_{m-1})\) is a non-zero vector in \(\mathbb{N}_0^{m-1}\). Suppose by contradiction that \(\mathbb{N}_0[\alpha]\) is finitely generated, then there exists a negative tail polynomial \(H(X)\) such that \(H(X)=\mathfrak{m}_\alpha(X) \cdot Q(X)\) with \(Q(X)\in\mathbb{Z}[X]\). Since \(H\) and \(\mathfrak{m}_{\alpha}\) are both monic in \(X\) then by Gauss’s lemma \(Q(X)\) is also monic. Let \(m= \deg \mathfrak{m}_\alpha\), and \(n=\deg H\). Write \[Q(X)=X^{n-m}+q_{n-m-1}X^{n-m-1}+\dots+q_0,\;q_0,\dots,q_{n-m}\in \mathbb{Z}.\]
We will first prove that all the coefficients \(q_i\) are nonnegative by comparing the coefficients of \(H(X)\) with \(m_\alpha(X).Q(X)\) in an increasing degree. The degree zero coefficient, implies that \(-cq_0\leq 0\), which implies that \(q_0\geq 0\), the degree \(1\) implies that \(-cq_1+p_1q_0\leq 0\), which implies that \(q_1\geq \frac{p_1q_0}{c}\geq 0\), continuing in the same manner, we find that \(\sum\limits_{i+j=n-m-1}p_iq_j-cq_{n-m-1}\leq 0\) which implies that \(q_{n-m-1}\geq \frac{\sum\limits_{i+j=n-m-1}p_iq_j}{c}\geq 0\). Which proves the claim.
Now starting the comparison at degree \(n-1\), we have \(p_{m-1}+q_{n-m-1}\leq 0\), and since both terms on the left-hand side are non-negative, they must be both zero. Taking the degree \(n-2\), we get \(p_{m-2}+q_{n-m-2}\leq 0\), which again implies that both terms must be zero. Continuing in the same manner, if \(n-m\geq m\), we get at degree \(n-m+1\), \(p_1+q_{n-2m+1}\leq 0\), which again implies that both terms must be zero, and this will contradict the fact that not all the \(p_i\)s are zero. Otherwise, if \(n-m<m\), then at degree \(m\), we get \(p_{m-(n-m)}+q_0\leq 0\), which implies that all the \(q_i\)s are zero, which is again impossible since this will imply that \(\mathfrak{m}_\alpha=X^m-c\).
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Theorem 11. If \(\mathbb{N}_0[\alpha]\) is a finitely generated monoid then \(\alpha\) is a weak Perron number. Conversely, suppose that \(\alpha\) is an algebraic integer, that \(\alpha\) is weak Perron, and that \(\alpha\) is the unique positive conjugate of its minimal polynomial. Then \(\mathbb{N}_0[\alpha]\) is finitely generated.
Proof. We begin with the forward direction. Since \(\mathbb{N}_0[\alpha]\) is an FGM, then there exists \(n \in \mathbb{N}\) such that \(\alpha^n =
\sum\limits_{i=0}^{n-1}c_i \alpha^i\) where \(c_i \in \mathbb{N}_0\). Define \(Q(t) = t^n - \sum\limits_{i=0}^{n-1}c_i t^i\). Suppose \(\alpha\) is
not a weak Perron number, then there exists a conjugate \(\beta\) of \(\alpha\) such that \(|\beta| > \alpha\). Since \(Q(\alpha)=0\) we must also have \(Q(\beta)=0\). Therefore, \(\beta^n=\sum\limits_{i=0}^{n-1}c_i \beta^i\) and by the triangle inequality we get \(|\beta|^n \leq \sum\limits_{i=0}^{n-1}c_i |\beta|^i\). \(Q(X)\) is ultimately positive and it has only one sign change, thus, by Theorem 6, it has at most one positive root, which must be \(\alpha\). Since \(|\beta| > \alpha\) then \(Q(|\beta|)>0\) and
therefore we get \(|\beta|^n > \sum\limits_{i=0}^{n-1}c_i |\beta|^i\) which is a contradiction.
For the reverse direction, assume that \(\alpha\) is an algebraic integer, that \(\alpha\) is weak Perron, and that \(\alpha\) is the unique positive
conjugate of its minimal polynomial. For the case \(\alpha=1\), \(\mathbb{N}_0[\alpha]= \mathbb{N}_0\) which is finitely generated. We assume \(\alpha \neq
1\).
We show that \(\alpha>1\). Suppose, by contradiction, that \(0<\alpha<1\). Let \(\alpha=\alpha_1,\alpha_2,\dots,\alpha_d\) be the conjugates of \(\alpha\). Since \(\alpha\) is weak Perron, we have \(|\alpha_i|\leq \alpha<1\) for every \(i\in\{1,\dots,d\}\). Hence \(|m_\alpha(0)|=\prod_{i=1}^d |\alpha_i| <1\) which is a contradiction since \(\alpha\) is an algebraic integer. It follows that \(\alpha>1\).
Let \(k\) be the number of roots of \(m_\alpha(X)\) on the circle \(|z|=\alpha\). Since \(\alpha\) itself is such a root, we have \(k\geq 1\). By Theorem 7, there exists a polynomial \(g(X)\in \mathbb{Z}[X]\) such that \(m_\alpha(X)=g(X^k)\). Moreover, since \(m_\alpha(X)\) is irreducible over \(\mathbb{Q}\), the polynomial \(g\) is also irreducible over \(\mathbb{Q}\). Set \(\lambda=\alpha^k\). Then \(g(\lambda)=0\), and since \(\alpha\) is an algebraic integer, so is \(\lambda\).
We claim that \(\lambda\) is a Perron number. Let \(\delta\) be a conjugate of \(\lambda\), equivalently a root of \(g\). If \(\eta^k=\delta\), then \(m_\alpha(\eta)=g(\eta^k)=g(\delta)=0\). Thus \(\eta\) is a conjugate of \(\alpha\). Since \(\alpha\) is weak Perron, we have \(|\eta|\leq \alpha\). Therefore \(|\delta|=|\eta|^k\leq \alpha^k=\lambda\). Hence \(\lambda\) is weak Perron.
It remains to show that no conjugate of \(\lambda\) distinct from \(\lambda\) has modulus \(\lambda\). Suppose that \(\delta\) is a root of \(g\) with \(|\delta|=\lambda\). Let \(\omega_0=1,\omega_1,\dots,\omega_{k-1}\) be the \(k\)-th distinct roots of unity. We have \(m_\alpha(\omega_i\alpha)=g(\alpha^k)=g(\lambda)=0\). Therefore, the \(k\)-th roots of \(m_\alpha\) that lie on the circle \(|z|=\alpha\) are \(\alpha \omega_0, \alpha \omega_1 \dots,\alpha \omega_{k-1}\). Let \(\mu \in \mathbb{C}\) with \(\mu ^ k = \delta\). We have \(g(\delta)=0\) implies that \(g(\mu^k)=0\) implies that \(m_\alpha(\mu)=0\) with \(|\mu|=\sqrt[k]{|\delta|} = \sqrt[k]{\lambda}=\alpha\). Therefore \(\mu = \omega_i \alpha\) for a certain \(i\). This implies \(\delta=\mu^k=\lambda\). So \(\lambda\) is a Perron number.
We now show that \(\lambda\) is the unique positive conjugate of its minimal polynomial. Let \(\delta>0\) be a positive root of \(g\). Then \(\delta^{1/k}>0\) is a positive root of \(f\), because \(f(\delta^{1/k})=g(\delta)=0\). Since \(\alpha\) is the unique positive root of \(f\), we get \(\delta^{1/k}=\alpha\), and therefore \(\delta=\alpha^k=\lambda\). Hence \(\lambda\) has no positive conjugate aside from itself.
If \(k \neq 1\), set \(\gamma=\lambda^{-1}\). Then \(\gamma\in(0,1)\), and \(\gamma^{-1}=\lambda\) is a Perron number with no positive conjugate aside from itself. By Proposition 8, there exists a polynomial \(h(X)\in\mathbb{Z}[X]\) such that \(h(X)m_\gamma(X)\in X\mathbb{N}_0[X]-1\). Therefore there exists a polynomial \(F(X)\in X\mathbb{N}_0[X]\) such that \(h(X)m_\gamma(X)=F(X)-1\). Evaluating at \(\gamma\) gives \(F(\gamma)=1\). Write \(F(X)=\sum_{i=1}^{N}c_iX^i\) where \(c_1,\dots,c_N\in\mathbb{N}_0\). Then \(1=\sum_{i=1}^{N}c_i\gamma^i =\sum_{i=1}^{N}c_i\lambda^{-i}\). Multiplying by \(\lambda^N\) gives \(\lambda^N=\sum_{i=1}^{N}c_i\lambda^{N-i}\). Which implies that \(\mathbb{N}_0[\lambda]\) is finitely generated by Proposition 1.
By Proposition 9, applied to \(\lambda\) and to the root \(\alpha\) of \(g(X^k)\), gives \(\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right| = k\left|\mathcal{A}(\mathbb{N}_0[\lambda])\right|\). Thus \(\mathcal{A}(\mathbb{N}_0[\alpha])\) is finite and nonempty. By Proposition 1, \(\mathbb{N}_0[\alpha]\) is therefore atomic and finitely generated.
If \(k=1\), set \(\gamma=\alpha^{-1}\) and the above argument follows. ◻
Remark 1. The hypotheses in the previous reverse direction cannot be weakened in an obvious way. The condition that \(\alpha\) be an algebraic integer cannot be omitted. Indeed, take \(\alpha=\frac{3}{2}\). The condition that \(\alpha\) be the unique positive conjugate cannot be omitted as well. Take \(\alpha=2+\sqrt{2}\).
As before, let \(\alpha\) be an algebraic positive real number over \(\mathbb{Q}\). Let \(\mathfrak{m}_{\alpha}\) be the primitive polynomials of minimal degree over \(\mathbb{Z}\) with \(\mathfrak{m}_{\alpha}(\alpha)=0\). In this section, we will restrict our attention to the case \(\deg \mathfrak{m}_\alpha = 3\). Let \(p_{\alpha}(X),\;q_\alpha(X)\in\mathbb{N}_0[X]\) such that \(m_{\alpha}(X)=p_{\alpha}(X)-q_{\alpha}(X)\).
In the case when \(\mathfrak{m}_\alpha\) is not monic or when \(\alpha\) is not the unique positive root, by Proposition 3 \(\mathbb{N}_0[\alpha]\) is infinitely generated or antimatter. For the rest of this section, we suppose that \(\mathfrak{m}_\alpha\) is monic with \(\alpha\) its unique positive root.
Also note that if \(\alpha < 1\) then for any \(n \in \mathbb{N}\), \(\alpha^n\) cannot be written as an \(\mathbb{N}_0\)-linear combination of \(\{1, \alpha, \dots, \alpha ^ {n-1}\}\). In view of Theorem 1, \(\mathbb{N}_0[\alpha]\) must be infinitely generated or antimatter. For the rest of this section, we will assume that \(\alpha > 1\).
Write \(\mathfrak{m}_\alpha(X) = X^3 \pm aX^2 \pm bX \pm c\) where \(a,b,c \in \mathbb{N}_0\). Note that \(c \neq 0\) since \(\mathfrak{m}_\alpha\) is irreducible. Since \(\mathfrak{m}_\alpha(X)\) has a unique positive root and it is eventually positive, then \(\mathfrak{m}_\alpha(0) < 0\) which forces \(\mathfrak{m}_\alpha\) to have the form \(\mathfrak{m}_\alpha(x) = X^3 \pm aX^2 \pm bX - c\). We assume that \(a,b,c \neq 0\) and we treat the cases where they are zero separately. At the end of this section, there is a summary of the classification.
For \(\mathfrak{m}_\alpha(X) = X^3+aX^2-bX-c\), if \(\alpha\) is not a weak Perron number, then \(\mathbb{N}_0[\alpha]\) is infinitely generated or antimatter by Theorem 11. If \(\mathbb{N}_0[\alpha]\) is an FGM, we have the following necessary conditions on the coefficients of \(\mathfrak{m}_\alpha(X)\):
Proposition 12. Suppose \(\mathbb{N}_0[\alpha]\) is an FGM and \(\mathfrak{m}_\alpha(X) = X^3+aX^2-bX-c\) then \(b\geq a^2\) and \(b^3 \geq a^3c\)
Proof. Let \(\beta\) and \(\gamma\) be the conjugates of \(\alpha\) over \(\mathbb{Q}\). Since \(\alpha\) is a weak Perron number, we have \(|\beta| \leq \alpha\) and \(|\gamma| \leq \alpha\).
By Viète’s formulas we have \(\alpha\beta\gamma=c\) which implies \(c\leq \alpha^3\). We also have \(\alpha^3+a\alpha^2-b\alpha = c \leq \alpha^3\) which
implies \(\frac{b}{a} \geq \alpha\). Combining both inequalities we get \(b^3 \geq a^3c\). For the other inequality, we have by Viète’s equations \(\beta + \gamma +
\alpha = -a\) which implies \(a + \alpha \leq -\beta - \gamma \leq 2\alpha\). Combining with \(\alpha \leq \frac{b}{a}\), the inequality follows.
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The converse of the above proposition is not true in general, even if \(\mathbb{N}_0[\alpha]\) is atomic. In the next example, we will prove that the above conditions do not imply weak perron.
Example 1. Let \(\mathfrak{m}_\alpha(X)=X^3+X^2-bX-2\) where \(b\geq 6\) is an integer. By checking the divisors of \(2\), we can see that \(\mathfrak{m}_\alpha(X)\) has no rational root, and hence is irreducible. We have \(\mathfrak{m}_\alpha(\sqrt{b}).\mathfrak{m}_\alpha(0)<0\), thus \(\alpha\) is between \(0\) and \(\sqrt{b}\). Note that \(\alpha\) is the unique positive root for \(\mathfrak{m}_\alpha\) since by Theorem 6 it has \(1\). We also have \(\mathfrak{m}_\alpha(-\sqrt{b})>0\) with \(\mathfrak{m}_\alpha\) of odd degree, so it must have a root \(\beta<-\sqrt{b}\). Thus we have \(|\beta|>\sqrt{b}>\alpha\), and therefore \(\alpha\) is not a weak Perron number. Since \(|\mathfrak{m}_\alpha(0)|\neq 1\), then \(\mathbb{N}_0[\alpha]\) is atomic by proposition 2. In view of theorem 11, \(\mathbb{N}_0[\alpha]\) is infinitely generated.
Proposition 13. If \(\mathbb{N}_0[\alpha]\) is atomic and \(\mathfrak{m}_\alpha(X) = X^3+aX^2-bX-c\) then \(\mathbb{N}_0[\alpha]\) is not an LFM.
Proof. By Proposition 4, \(\mathbb{N}_0[\alpha]\) is not a UFM. Suppose it is a proper LFM, by Proposition 5 with Proposition 1 there exists a polynomial \(H(X) = X^4 - \sum \limits_{i=0}^{3} c_i X^i\) where \(c_i\) are all nonnegative having \(\alpha\) as a root. As such, it can be written as \(H(X) = \mathfrak{m}_\alpha(X)(X+d)\) with \(d\) being a nonzero integer. After developing, considering the coefficient degree \(0\), we get \(d > 0\). The coefficient of degree \(3\) implies that \(d < -a\) which is a contradiction. ◻
The next example illustrates the case when \(\mathfrak{m}_\alpha\) satisfies the necessary conditions of proposition 12, and the size of the set of atoms is \(5\).
Example 2. Let \(\mathfrak{m}_\alpha(X)=X^3+X^2-pX-2p\) where \(p\) is a prime number strictly greater than \(3\). By the rational root test, the only possible rational roots are \(\pm1,\pm2,\pm p,\pm2p\), and none of them is a root, and hence is irreducible. We have \(\mathfrak{m}_\alpha(1)<0\), and \(\mathfrak{m}_\alpha\) is ultimately positive, therefore it has a positive root \(\alpha>1\). Note that \(\alpha\) is the unique positive root for \(\mathfrak{m}_\alpha\) since by Theorem 6 it can have at most \(1\). Since \(|\mathfrak{m}_\alpha(0)|\neq 1\), then \(\mathbb{N}_0[\alpha]\) is atomic by Proposition 2. In view of Proposition 13, \(\mathbb{N}_0[\alpha]\) is not an \(LFM\), thus \(1,\;\alpha,\;\alpha^2,\;\alpha^3,\;\alpha^4\) are atoms. We will prove that \(\alpha^5\) is not an atom. Indeed, \(\alpha\) is a root of \(X^5+(1-p)X^3+(2-p)X^2-4p\).
Suppose \(\mathfrak{m}_\alpha(X) = X^3-aX^2+bX-c\), with a positive root \(\alpha\) such that \(\mathbb{N}_0[\alpha]\) is atomic. Since \(\mathfrak{m}_\alpha(-x)\) has no sign changes, by theorem 6, \(\mathfrak{m}_\alpha\) has no negative roots. If it has
three positive roots, then by Proposition 3 it cannot be finitely generated. For the rest of this subsection, we suppose that \(\alpha\) is the unique positive root of \(\mathfrak{m}_\alpha\). Denote \(\beta\) and \(\overline{\beta}\) the conjugates of
\(\alpha\).
We have the following lemma:
Lemma 2. The root \(\alpha\) is a weak perron if and only if \(b^3\leq a^3c\).
Proof. By Viète’s equations, we have \(c=\alpha\beta\overline{\beta}\). We deduce the following: \[\begin{align} \alpha\;\text{is a weak perron} &\Leftrightarrow c\leq \alpha^3\\ &\Leftrightarrow \sqrt[3]{c}\leq \alpha\\ &\Leftrightarrow \mathfrak{m}_\alpha(\sqrt[3]{c})\leq 0\;\text{by continuity of }\mathfrak{m}_\alpha\\ &\Leftrightarrow b^3\leq a^3c. \end{align}\] ◻
Corollary 1. Assume that \(\mathfrak{m}_\alpha(X)=X^3-aX^2+bX-c\) and that \(\alpha\) is the unique positive root of \(\mathfrak{m}_\alpha\). Then \(\mathbb{N}_0[\alpha]\) is infinitely generated if and only if \(b^3>a^3c\).
Proof. First suppose that \(b^3>a^3c\). Then \(\alpha\) is not weak Perron, by Lemma 2. By Theorem 11, \(\mathbb{N}_0[\alpha]\) is infinitely generated.
Conversely, suppose that \(b^3\leq a^3c\). Then Lemma 2 implies that \(\alpha\) is weak Perron. Since \(\alpha\) is an algebraic integer and is the unique positive conjugate of its minimal polynomial, then the converse implication of Theorem 11 applies, and so \(\mathbb{N}_0[\alpha]\) is finitely generated. ◻
We deduce a family of infinitely generated monoids \(\mathbb{N}_0[\alpha_p]\) of rank \(3\), where \(\alpha_p\) is a positive root of \(X^3-X^2+pX-2p^2\), where \(p>2\) is a prime.
Proposition 14. \(\mathbb{N}_0[\alpha]\) is a proper LFM if and only if \(b\leq a^2\) and \(\frac{b}{a}\leq \lfloor\frac{c}{b}\rfloor\).
Proof. First note that \(\mathbb{N}_0[\alpha]\) is not a \(UFM\) by proposition 4. Let \(Q(X)=X+d\) where \(d\in\mathbb{Z}\). For \(\mathfrak{m}_\alpha Q\) to be a negative tail polynomial it is necessary and sufficient that the following system admits an integer solution for \(d\): \[\begin{align} -cd\leq 0,\;bd-c\leq 0,\;-ad+b\leq 0,\;-a+d\leq 0, \end{align}\] this system is solvable if and only if \(b\leq a^2\) and \(\frac{b}{a}\leq \lfloor\frac{c}{b}\rfloor\). ◻
We note that in the case when \(b>a^2\), one can prove that \(\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right|\geq 6\). The next two examples illustrate cases where \(\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right|= 5\) and \(\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right|= 7\).
Example 3. In this example \(\alpha\) is a root of \(\mathfrak{m}_\alpha(X)=X^3-3X^2+5X-8\) and \(\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right|= 5\).
Example 4. In this example \(\alpha\) is a root of \(\mathfrak{m}_\alpha(X)=X^3-2X^2+5X-20\) and \(\left|\mathcal{A}(\mathbb{N}_0[\alpha])\right|= 7\).
| Form of \(\mathfrak{m}_\alpha(X)\) | Conclusion for \(\mathbb{N}_0[\alpha]\) | Reference | ||||
|---|---|---|---|---|---|---|
| \(X^3-c\) | UFM whenever \(X^3-c\) is irreducible. | Proposition 4 | ||||
| \(X^3+bX-c\) | Antimatter if \(c=1\); infinitely generated if \(c>1\). | Theorem 10 | ||||
| \(X^3-bX-c\) | UFM. | Proposition 4 | ||||
| \(X^3+aX^2-c\) | Antimatter if \(c=1\); infinitely generated if \(c>1\). | Theorem 10 | ||||
| \(X^3-aX^2-c\) | UFM. | Proposition 4 | ||||
| \(X^3+aX^2+bX-c\) | Antimatter if \(c=1\); infinitely generated if \(c>1\). | Theorem 10 | ||||
| \(X^3-aX^2-bX-c\) | UFM. | Proposition 4 | ||||
| \(X^3+aX^2-bX-c\) | If \(\mathbb{N}_0[\alpha]\) is an FGM, then \(b\ge a^2\) and \(b^3\ge a^3c\). | Proposition 12 | ||||
| If \(\mathbb{N}_0[\alpha]\) is atomic, then \(\mathbb{N}_0[\alpha]\) is never an LFM. | Proposition 13 | |||||
| \(X^3-aX^2+bX-c\) | \(\mathbb{N}_0[\alpha]\) is a proper LFM if and only if \(b\le a^2\) and \(\frac{b}{a}\le \left\lfloor \frac{c}{b}\right\rfloor\). | Proposition 14 | ||||
| \(b^3 \leq a^3c\) if and only if \(\mathbb{N}_0[\alpha]\) is finitely generated. | Corollary 1 |
The authors are grateful to the anonymous referee for a careful reading of the manuscript and for several valuable comments. In particular, the referee encouraged us to reconsider the converse direction relating weak Perron numbers to finite generation.