The Frobenius problem over number fields with a real embedding

Abstract

Given a number field K with at least one real embedding, we generalize the notion of the classical Frobenius problem to the ring of integers ${\mathfrak {O}}_K$ of K by describing certain Frobenius semigroups, $\textrm{Frob}(\alpha _1,\ldots ,\alpha _n)$, for appropriate elements $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K$. We construct a partial ordering on $\textrm{Frob}(\alpha _1,\ldots ,\alpha _n)$, and show that this set is completely described by the maximal elements with respect to this ordering. We also show that $\textrm{Frob}(\alpha _1,\ldots ,\alpha _n)$ will always have finitely many such maximal elements, but in general, the number of maximal elements can grow without bound as n is fixed and $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K$ vary. Explicit examples of the Frobenius semigroups are also calculated for certain cases in real quadratic number fields.

1 Introduction and summary of results

It is well known that if $\alpha _1,\ldots ,\alpha _n\in {\mathbb {N}}:=\{0,1,2,\ldots \}$ are nonzero and coprime, then there is some smallest positive integer $\chi (\alpha _1,\ldots ,\alpha _n)$ with the property that for any integer $N\geqslant \chi (\alpha _1,\ldots ,\alpha _n)$, there are natural numbers $x_1,\ldots ,x_n\in {\mathbb {N}}$ for which

$$\begin{aligned} x_1\alpha _1+\cdots +x_n\alpha _n=N. \end{aligned}$$

The classical Frobenius problem concerns explicitly finding the number $\chi (\alpha _1,\ldots ,\alpha _n)$. When $n=2$, it is known (see [1]) that

$$\begin{aligned} \chi (\alpha _1,\alpha _2)=(\alpha _1-1)(\alpha _2-1), \end{aligned}$$

and more complicated formulas are known for $\chi$ when $n=3$ (see [13]). We can restate the classical Frobenius problem as follows: Define a submonoid ${{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n)$ of ${\mathbb {N}}$ by

$$\begin{aligned} {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n)=\left\{ \left. \sum _{i=1}^{n}x_i\alpha _i \, \right| x_1,\ldots ,x_n\in {\mathbb {N}}\right\} . \end{aligned}$$

Then the classical Frobenius problem is to determine the semigroup

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)= & {} \{w\in {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n) \mid w+{\mathbb {N}}\subseteq {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n)\}\\= & {} \chi (\alpha _1,\ldots ,\alpha _n)+{\mathbb {N}}, \end{aligned}$$

and the above shows that in the case $n=2$, we have

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\alpha _1,\alpha _2)=(\alpha _1-1)(\alpha _2-1)+{\mathbb {N}}. \end{aligned}$$

Using this new statement of the classical Frobenius problem, we can generalize to certain commutative rings with unity in the spirit of Johnson and Looper’s paper [8]:

Definition 1

Let R be a commutative ring with unity that is finitely generated as a ${\mathbb {Z}}$-module. Then we define a Frobenius template (or simply template) for R to be a triple ${\mathcal {T}}=(A,C,U)$ consisting of

• (1) a subset $A\subseteq R$ containing 1;

• (2) a nonzero additive monoid $C\subseteq R$; and

• (3) a function U that assigns to each collection of nonzero distinct elements $\alpha _1,\ldots ,\alpha _n\in A$ that generate R as a ${\mathbb {Z}}$-module, an additive submonoid $U(\alpha _1,\ldots ,\alpha _n)$ of R for which

  $$\begin{aligned} U(\alpha _1,\ldots ,\alpha _n)\supseteq {{\,\textrm{SG}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n):=\left\{ \left. \sum _{i=1}^{n}x_i\alpha _i \, \right| x_1,\ldots ,x_n\in C\right\} . \end{aligned}$$

We call the Frobenius template ${\mathcal {T}}=(A,C,U)$ Frobenius if for all nonzero $\alpha _1,\ldots ,\alpha _n\in A$ that generate R as a ${\mathbb {Z}}$-module, there is some $w\in {{\,\textrm{SG}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)$ for which

$$\begin{aligned} w+U(\alpha _1,\ldots ,\alpha _n)\subseteq {{\,\textrm{SG}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n). \end{aligned}$$

This notion of a Frobenius template is slightly different from what originally appeared in [8], since we have replaced the notion of $\alpha _1,\ldots ,\alpha _n\in A$ being coprime (i.e., having no common non-unit divisors) with the much stronger notion of $\alpha _1,\ldots ,\alpha _n\in A$ generating R as a ${\mathbb {Z}}$-module. Of course, an arbitrary ring R may not be finitely generated as a ${\mathbb {Z}}$-module, so this assumption is added into the above definition in order to avoid having a useless concept in cases where this fails. While the rest of this paper deals with cases where R has characteristic zero and is a finitely generated free ${\mathbb {Z}}$-module, it could be interesting to consider Frobenius templates for rings in which one of these conditions fails, in which case if R is not finitely generated as a ${\mathbb {Z}}$-module, then the definition of the Frobenius template would have to be altered in order to allow for an infinite indexed family of elements $\{\alpha _i\}_{i\in I}\subseteq A$. Another key property that all rings considered in Frobenius templates in this paper will have is that they can be naturally embedded into the real numbers, so they inherit the standard total ordering that ${\mathbb {R}}$ has. It could also be interesting to consider rings without this property.

Based on the above results about rephrasing the classical Frobenius problem in terms of determining certain semigroups, we can generalize the Frobenius problem to some rings:

Definition 2

If ${\mathcal {T}}=(A,C,U)$ is a Frobenius template for a ring R and $\alpha _1,\ldots ,\alpha _n\in A$ generate R as a ${\mathbb {Z}}$-module, then the Frobenius problem for ${\mathcal {T}}$ associated to the elements $\alpha _1,\ldots ,\alpha _n\in A$ is to determine the Frobenius semigroup

$$\begin{aligned} {{\,\textrm{Frob}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n):= & {} \{w\in {{\,\textrm{SG}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n) \ | \ w\\{} & {} +U(\alpha _1,\ldots ,\alpha _n)\subseteq {{\,\textrm{SG}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)\}. \end{aligned}$$

If the template ${\mathcal {T}}$ is Frobenius, then the Frobenius problem associated to ${\mathcal {T}}$ is to determine the semigroup ${{\,\textrm{Frob}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)$ for all $\alpha _1,\ldots ,\alpha _n\in A$ that generate R as a ${\mathbb {Z}}$-module. We will often drop the subscript ${\mathcal {T}}$ if the template in use is clear from context.

Given a Frobenius template ${\mathcal {T}}=(A,C,U)$ over a ring R and nonzero elements $\alpha _1,\ldots ,\alpha _n\in A$ that generate R as a ${\mathbb {Z}}$-module, the requirement that $C\subseteq R$ is a monoid shows that ${{\,\textrm{SG}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)$ is also a monoid. It is then an immediate consequence of Definition 2 that the following are equivalent:

$$\begin{aligned}{} & {} 0\in {{\,\textrm{Frob}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n) \iff {{\,\textrm{SG}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)=U(\alpha _1,\ldots ,\alpha _n) \\{} & {} \quad \iff {{\,\textrm{Frob}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)={{\,\textrm{SG}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n). \end{aligned}$$

Example 1

If we slightly relax the condition for the Frobenius template that $\alpha _1,\ldots ,\alpha _n$ generate R as a ${\mathbb {Z}}$-module, then we can relate this notion to generalized numerical semigroups. Consider the Frobenius template on $R={\mathbb {Z}}^d$ given by ${\mathcal {T}}=({\mathbb {N}}^d,\Delta ,{\mathbb {N}}^d)$, where $\Delta =\{(x,\ldots ,x) \mid x\in {\mathbb {N}}\}\subseteq {\mathbb {Z}}^d$. If $\alpha _1,\ldots ,\alpha _n\in {\mathbb {N}}^d$, then we have that

$$\begin{aligned} {{\,\textrm{SG}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)={\mathbb {N}}\alpha _1+\cdots +{\mathbb {N}}\alpha _n. \end{aligned}$$

In particular, if $S\subseteq {\mathbb {N}}^d$ is a generalized numerical semigroup (i.e., a submonoid of ${\mathbb {N}}^d$ for which ${\mathcal {H}}(S):={\mathbb {N}}^d\setminus S$ is finite) and $\alpha _1,\ldots ,\alpha _n\in S$ is a generating set for S, then ${{\,\textrm{SG}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)=S$. Thus for such $\alpha _1,\ldots ,\alpha _n\in S$, the Frobenius problem associated to ${\mathcal {T}}$ for $\alpha _1,\ldots ,\alpha _n$ is to determine the Frobenius semigroup

$$\begin{aligned} {{\,\textrm{Frob}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)=\{w\in S \mid w+{\mathbb {N}}^d\subseteq S\}. \end{aligned}$$

If we define ${{\,\textrm{FA}\,}}(S)$ to be the collection of maximal elements of ${\mathcal {H}}(S)$ with respect to the pointwise partial order $\leqslant _\textrm{p}$ for ${\mathbb {Z}}^d$ (i.e., the Frobenius allowable gaps, see [12]), then we see that if $w\in {{\,\textrm{FA}\,}}(S)$ then

$$\begin{aligned} w+e_i+{\mathbb {N}}^d\subseteq S \end{aligned}$$

for all $i=1,\ldots ,d$, where $e_i=(0,\ldots ,1,\ldots ,0)\in {\mathbb {N}}^d$ is the ith standard basis vector for ${\mathbb {Z}}^d$. Thus $w+e_i\in {{\,\textrm{Frob}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)$ for all $i=1,\ldots d$, and moreover the elements $w\in {\mathcal {H}}(S)$ for which $w+e_i\in {{\,\textrm{Frob}\,}}_{\mathcal {T}}(\alpha _1,\ldots ,\alpha _n)$ for all $i=1,\ldots ,d$ are precisely the elements of ${{\,\textrm{FA}\,}}(S)$. This means that solving the Frobenius problem associated to ${\mathcal {T}}$ for certain sets of elements in ${\mathbb {N}}^d$ is closely related to the Frobenius allowable gaps for numerical semigroups in ${\mathbb {N}}^d$.

From Definition 2, we see that the classical Frobenius problem is about the ring $R={\mathbb {Z}}$, and the Frobenius template in question is ${\mathcal {T}}=({\mathbb {N}},{\mathbb {N}},{\mathbb {N}})$ (where the third ${\mathbb {N}}$ is the constant function that assigns to any tuple of natural numbers the submonoid ${\mathbb {N}}$ of ${\mathbb {Z}}$), since nonzero integers $\alpha _1,\ldots ,\alpha _n\in {\mathbb {N}}$ generate ${\mathbb {Z}}$ as a ${\mathbb {Z}}$-module if and only if they are coprime. Work has been done on finding and studying certain interesting Frobenius templates (with the requirement of elements generating the ring as a ${\mathbb {Z}}$-module replaced with other requirements) for the ring $R={\mathbb {Z}}\left[ \sqrt{m}\right]$, where $m\in {\mathbb {Z}}$ is not a square, in [3, 5, 6, 8, 10], and [9].

In this paper, we look into creating and studying an interesting Frobenius template for the ring of integers of a number field that has at least one real embedding (henceforth, if K is such a number field then we fix $\sigma _1:K\hookrightarrow {\mathbb {C}}$ to be one of the embeddings of K into ${\mathbb {C}}$ for which $\sigma _1(K)\subseteq {\mathbb {R}}$). Some results are based on similar results in [7], but we weaken the restriction there of the number field being totally real to that of the number field having a real embedding.

Definition 3

Let K be a number field with at least one real embedding and with ring of integers ${\mathfrak {O}}_K$, and define

$$\begin{aligned} {\mathfrak {O}}_K^+=\{\alpha \in {\mathfrak {O}}_K \mid \sigma _1(\alpha )\geqslant 0\}. \end{aligned}$$

For any $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module, let the positive rational cone generated by $\alpha _1,\ldots ,\alpha _n$ be the set

$$\begin{aligned} C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)=\left\{ \left. \sum _{i=1}^{n}x_i\alpha _i \, \right| x_1,\ldots ,x_n\in {\mathbb {Q}}_{\geqslant 0}\right\} \subseteq \{\alpha \in K \mid \sigma _1(\alpha )\geqslant 0\}. \end{aligned}$$

Let $C_{\mathbb {Q}}\cap {\mathfrak {O}}_K$ denote the function that assigns to each such collection $\alpha _1,\ldots ,\alpha _n$ the submonoid $C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K$ of ${\mathfrak {O}}_K^+$.

With the above notation, we have that $C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K\supseteq {\mathbb {N}}\alpha _1+\cdots +{\mathbb {N}}\alpha _n$, so we can form the Frobenius template

$$\begin{aligned} {\mathcal {T}}=({\mathfrak {O}}_K^+,{\mathbb {N}},C_{\mathbb {Q}}\cap {\mathfrak {O}}_K). \end{aligned}$$

Note that if we take $K={\mathbb {Q}}$ then ${\mathfrak {O}}_K={\mathbb {Z}}$, ${\mathfrak {O}}_K^+={\mathbb {N}}$, and $C_{\mathbb {Q}}={\mathbb {N}}$, so $C_{\mathbb {Q}}\cap {\mathfrak {O}}_K={\mathbb {N}}$ (i.e., it assigns to any such collection $\alpha _1,\dots ,\alpha _n$ the submonoid ${\mathbb {N}}\subseteq {\mathbb {Z}}$). Hence this Frobenius template reduces down to the classical Frobenius template when $K={\mathbb {Q}}$. We now arrive at the first main theorem of this paper.

Theorem 1

Let K be a number field with a real embedding. Then the template ${\mathcal {T}}=({\mathfrak {O}}_K^+,{\mathbb {N}},C_{\mathbb {Q}}\cap {\mathfrak {O}}_K)$ is Frobenius.

After proving this, we begin to look at the structure of the Frobenius semigroup ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ in the template $({\mathfrak {O}}_K^+,{\mathbb {N}},C_{\mathbb {Q}}\cap {\mathfrak {O}}_K)$. In particular, we define a partial ordering on ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ and show that ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ contains maximal elements with respect to this ordering. Furthermore, we show that the set ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ of all such maximal elements satisfies the following:

Theorem 2

Let K be a number field with a real embedding and let $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ be nonzero elements that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Then ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ is a finite set, and ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ is equal to the finite union

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)=\bigcup _{\mu \in {\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)} (\mu +C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K). \end{aligned}$$

After establishing these properties of the Frobenius semigroups, we give an explicit calculation of certain Frobenius semigroups for real quadratic number fields. Lastly, we show the remarkable result that in general, the size of ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ can be unbounded, even if n is fixed.

2 Some preliminary results

For completeness, we give a summary of results from [2] that are used in this paper, and also some general results about matrices that will be used later on. Let $A\in {\mathbb {Z}}^{d\times n}$, $1\leqslant d< n$, be a matrix satisfying

• (I1) $\gcd \{\det (A') \mid A' \text { is a } d\times d \text { minor of } A\}=1$;

• (I2) $\left\{ \left. x\in {\mathbb {R}}_{\geqslant 0}^n \, \right| Ax=0\right\} =0.$

Let ${\mathcal {F}}(A)\subseteq {\mathbb {Z}}^d$ denote the set

$$\begin{aligned} {\mathcal {F}}(A)=\{Ax \mid x\in {\mathbb {N}}^n\}, \end{aligned}$$

let

$$\begin{aligned} C_{\mathbb {R}}(A)=\left\{ Ax \left| \, x\in {\mathbb {R}}^n_{\geqslant 0}\!\right. \right\} \end{aligned}$$

be the positive cone generated by the columns of A, and similarly let

$$\begin{aligned} C_{\mathbb {Q}}(A)=\left\{ Ax \left| \, x\in {\mathbb {Q}}^n_{\geqslant 0}\!\right. \right\} \end{aligned}$$

be the positive rational cone generated by the columns of A. Then the following is a slightly weaker version of Lemma 1.1 in [2].

Lemma 3

(Lemma 1.1, [2]) Let $A\in {\mathbb {Z}}^{d\times n}$, $1\leqslant d< n$, be an integral matrix satisfying conditions (I1) and (I2). Then for any integer vector $w\in {{\,\textrm{int}\,}}(C_{\mathbb {R}}(A))\cap {\mathbb {Z}}^d$ (where ${{\,\textrm{int}\,}}(C_{\mathbb {R}}(A))=\{Ax \mid x\in {\mathbb {R}}^n_{>0}\}$), there is some positive number $N\geqslant 0$ so that if $t\geqslant N$, then

$$\begin{aligned} (tw+C_{\mathbb {R}}(A))\cap {\mathbb {Z}}^d\subseteq {\mathcal {F}}(A). \end{aligned}$$

Note that if we take $t\in {\mathbb {N}}$ sufficiently large in the above lemma, then $tw\in {\mathbb {Z}}^d$, so Lemma 3 shows that we will have

$$\begin{aligned} tw+C_{\mathbb {R}}(A)\cap {\mathbb {Z}}^d\subseteq {\mathcal {F}}(A) \end{aligned}$$

for all large enough positive integers t.

The following lemma, whose proof is based on [4], will be important in the proof of Theorem 1.

Lemma 4

Let R be a commutative ring with unity and $A\in R^{d\times n}$ be a matrix for which the corresponding R-module homomorphism $R^n\rightarrow R^d$ is surjective. If $A_1,\ldots ,A_N$ denote the $d\times d$ minors of A (i.e., the matrices resulting from selecting d distinct columns of A), then $\det (A_1),\ldots ,\det (A_N)$ generate the unit ideal in R. In particular, if $R={\mathbb {Z}}$ then $\gcd (\det (A_1),\ldots ,\det (A_N))=1$.

Proof

Because the R-module homomorphism $R^n\rightarrow R^d$ corresponding to A is surjective, and because free R-modules are projective, we know that there is an R-module homomorphism $R^d\rightarrow R^n$ for which

commutes, where the map $R^d\rightarrow R^d$ is the identity. Hence there is a matrix $B\in R^{n\times d}$ corresponding to the map $R^d\rightarrow R^n$ for which $AB=I$ is the $d\times d$ identity matrix. If $B_1,\ldots ,B_N$ denote the $d\times d$ minors of B (i.e., the matrices resulting from selecting d distinct rows of B), then the Cauchy-Binet formula shows that

$$\begin{aligned} 1=\det (I)=\det (AB)=\sum _{i=1}^{N}\det (A_i)\det (B_i), \end{aligned}$$

so $\det (A_1)R+\cdots +\det (A_N)R=R$.

If $R={\mathbb {Z}}$ then we know that $\det (A_1){\mathbb {Z}}+\cdots +\det (A_N){\mathbb {Z}}={\mathbb {Z}}$, so it follows that $\gcd (\det (A_1),\ldots ,\det (A_N))=1$. $\square$

3 The Frobenius problem

Using the results of Sect. 2, we can prove the first main theorem of this paper. This proof is based on ideas present in the proof of Theorem 1.1 in [7].

Theorem 1

Let K be a number field with a real embedding. Then the template ${\mathcal {T}}=({\mathfrak {O}}_K^+,{\mathbb {N}},C_{\mathbb {Q}}\cap {\mathfrak {O}}_K)$ is Frobenius.

Proof

Fix nonzero elements $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Then we know that $n\geqslant d:=[K:{\mathbb {Q}}]$, which is the rank of ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. We first deal with the case where $n=d$, meaning $\alpha _1,\ldots ,\alpha _d$ span the free ${\mathbb {Z}}$-module ${\mathfrak {O}}_K$ of rank d, and are thus a basis for ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. In this case, if $\beta \in C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _d)\cap {\mathfrak {O}}_K$, we have

$$\begin{aligned} \beta =\sum _{i=1}^{d}x_i\alpha _i=\sum _{i=1}^{d}y_i\alpha _i, \end{aligned}$$

where $x_1,\ldots ,x_d\in {\mathbb {Q}}_{\geqslant 0}$ and $y_1,\ldots ,y_d\in {\mathbb {Z}}$. Then the fact that K is the field of fractions of ${\mathfrak {O}}_K$, and that ${\mathbb {Q}}$ is a flat ${\mathbb {Z}}$-module, shows that $\alpha _1,\ldots ,\alpha _d$ is also a basis for K as a ${\mathbb {Q}}$-vector space, and in particular, they are ${\mathbb {Q}}$-linearly independent. Hence each $x_i=y_i$, so each $x_i\in {\mathbb {N}}$, which means that $\beta \in {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _d)$, and thus

$$\begin{aligned} C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _d)\cap {\mathfrak {O}}_K={{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _d). \end{aligned}$$

Then ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _d)={{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _d)$, so it is nonempty.

Now suppose that $n>d$, and let $\beta _1,\ldots ,\beta _d\in {\mathfrak {O}}_K$ be a basis for ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Let $a_{ij}\in {\mathbb {Z}}, i=1,\ldots ,n,j=1,\ldots ,d$, be the unique integers for which

$$\begin{aligned} \alpha _i=\sum _{j=1}^{d}a_{ij}\beta _j. \end{aligned}$$

Let $\varphi :{\mathfrak {O}}_K\rightarrow {\mathbb {Z}}^d$ be the isomorphism associated to the ${\mathbb {Z}}$-basis $\beta _1,\ldots ,\beta _d$, so $\varphi (\alpha _i)=(a_{i1},\ldots ,a_{id})$, and let $A=(a_{ij})^\textrm{T}\in {\mathbb {Z}}^{d\times n}$ be the matrix whose columns are the $\varphi (\alpha _i)$. Note that $\varphi :{\mathfrak {O}}_K\rightarrow {\mathbb {Z}}^d$ is the restriction of the vector space isomorphism $K\rightarrow {\mathbb {Q}}^d$ associated to the same basis $\beta _1,\ldots ,\beta _d$ for K as a ${\mathbb {Q}}$-vector space. Let $\sigma _1,\ldots ,\sigma _d:K\hookrightarrow {\mathbb {C}}$ be the d distinct embeddings of K into ${\mathbb {C}}$ (where $\sigma _1$ is still the embedding we fixed earlier with $\sigma _1(K)\subseteq {\mathbb {R}}$), and let $\sigma :K\rightarrow {\mathbb {C}}^d$ be the Minkowski embedding of K into ${\mathbb {C}}^d$, given by

$$\begin{aligned} \sigma (x)=(\sigma _1(x),\ldots ,\sigma _d(x)) \end{aligned}$$

for $x\in K$. Note that $\sigma$ is a ${\mathbb {Q}}$-linear map because each $\sigma _i$ must fix ${\mathbb {Q}}$. Let $B=(\sigma _i(\beta _j))\in {\mathbb {C}}^{d\times d}$ and $C=(\sigma _i(\alpha _j))\in {\mathbb {C}}^{d\times n}$ be the matrices whose columns are the vectors $\sigma (\beta _1),\ldots ,\sigma (\beta _d)\in {\mathbb {C}}^d$ and $\sigma (\alpha _1),\ldots ,\sigma (\alpha _n)\in {\mathbb {C}}^d$, respectively. Then

$$\begin{aligned} (BA)_{ij}=\sum _{k=1}^{d}B_{ik}A_{kj}=\sum _{k=1}^{d}\sigma _i(\beta _k)a_{jk} =\sigma _i\!\left( \sum _{k=1}^{d}a_{jk}\beta _k\right) =\sigma _i(\alpha _j)=C_{ij}, \end{aligned}$$

so $BA=C$. Note that the facts that $\sigma _1(K)\subseteq {\mathbb {R}}$ and that $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ are nonzero show that the first row of C will have all strictly positive entries.

We now claim that the matrix $A\in {\mathbb {Z}}^{d\times n}$ satisfies conditions (I1) and (I2). Because $\alpha _1,\ldots ,\alpha _n$ generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module, we know that ${\mathbb {Z}}\alpha _1+\cdots +{\mathbb {Z}}\alpha _n={\mathfrak {O}}_K$, so

$$\begin{aligned} \varphi ({\mathfrak {O}}_K)={\mathbb {Z}}\varphi (\alpha _1)+\cdots +{\mathbb {Z}}\varphi (\alpha _n)=A{\mathbb {Z}}^n, \end{aligned}$$

where we used the fact that the columns of $A\in {\mathbb {Z}}^{d\times n}$ are the $\varphi (\alpha _1),\ldots ,\varphi (\alpha _n)$. But $\varphi :{\mathfrak {O}}_K\rightarrow {\mathbb {Z}}^d$ is an isomorphism, so $\varphi ({\mathfrak {O}}_K)={\mathbb {Z}}^d$, and thus $A{\mathbb {Z}}^n={\mathbb {Z}}^d$, so the linear transformation ${\mathbb {Z}}^n\rightarrow {\mathbb {Z}}^d$ associated to the matrix A is surjective. Lemma 4 then shows that

$$\begin{aligned} \gcd \{\det (A') \ | \ A' \text { is a } d\times d \text { minor of } A\}=1, \end{aligned}$$

so condition (I1) is satisfied. Now suppose that $x=(x_1,x_2,\ldots ,x_n)\in {\mathbb {R}}_{\geqslant 0}^n$ is such that $Ax=0$. Then

$$\begin{aligned} BAx=Cx= & {} \begin{pmatrix} \sigma _1(\alpha _1) &{} \cdots &{} \sigma _1(\alpha _n)\\ \sigma _2(\alpha _1) &{} \cdots &{} \sigma _2(\alpha _n)\\ \vdots &{} \ddots &{} \vdots \\ \sigma _d(\alpha _1) &{} \cdots &{} \sigma _d(\alpha _n) \end{pmatrix}\begin{pmatrix} x_1\\ x_2\\ \vdots \\ x_n \end{pmatrix}\\= & {} \begin{pmatrix} x_1\sigma _1(\alpha _1)+\cdots +x_n\sigma _1(\alpha _n)\\ x_1\sigma _2(\alpha _1)+\cdots +x_n\sigma _2(\alpha _n)\\ \vdots \\ x_1\sigma _d(\alpha _1)+\cdots +x_n\sigma _d(\alpha _n) \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ \vdots \\ 0 \end{pmatrix}. \end{aligned}$$

Because $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ are nonzero, we know that each $\sigma _1(\alpha _i)>0$, so the fact that the $x_i\geqslant 0$ shows that in order for $x_1\sigma _1(\alpha _1)+\dots +x_n\sigma _1(\alpha _n)=0$, we must have that each $x_i=0$. Hence

$$\begin{aligned} \left\{ \left. x\in {\mathbb {R}}_{\geqslant 0}^{n} \,\right| Ax=0\right\} =0, \end{aligned}$$

so the matrix A also satisfies condition (I2).

We now apply Lemma 3 to show that if $\gamma \in {{\,\textrm{int}\,}}(C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n))\cap {\mathfrak {O}}_K$ (where ${{\,\textrm{int}\,}}(C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n))$ denotes strictly positive rational linear combinations of $\alpha _1,\ldots ,\alpha _n$), then there is some nonzero $t\in {\mathbb {N}}$ for which

$$\begin{aligned} t\gamma +C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K\subseteq {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n). \end{aligned}$$

If this is true, then we can take $\gamma =\sum \limits _{i=1}^{n}x_i\alpha _i$, where $x_1,\ldots ,x_n\in {\mathbb {N}}$ are nonzero, so $\gamma \in {{\,\textrm{int}\,}}(C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n))\cap {\mathfrak {O}}_K$ and $t\gamma \in {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n)$, and thus $t\gamma \in {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$, so ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ is nonempty. Then it would follow that for any $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module, the set ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ is nonempty, so we would know that the template ${\mathcal {T}}=\left( {\mathfrak {O}}_K^+,{\mathbb {N}},C_{\mathbb {Q}}\cap {\mathfrak {O}}_K\right)$ is Frobenius.

In order to apply Lemma 3 in this manner, we must first transition from the situation we have in ${\mathfrak {O}}_K$ to the situation given in Lemma 3. Because $\varphi (\alpha _1),\ldots ,\varphi (\alpha _n)\in {\mathbb {Z}}^d$ are the columns of the matrix A, we know that

$$\begin{aligned} \varphi (C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n))=C_{\mathbb {Q}}(A),\qquad \varphi ({{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n))={\mathcal {F}}(A), \end{aligned}$$

and the fact that $\varphi :{\mathfrak {O}}_K\rightarrow {\mathbb {Z}}^d$ is a ${\mathbb {Z}}$-module isomorphism shows that $\varphi ({\mathfrak {O}}_K)={\mathbb {Z}}^d$. It follows that if $t\in {\mathbb {N}}$ and $\gamma \in {{\,\textrm{int}\,}}(C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n))\cap {\mathfrak {O}}_K$, then

$$\begin{aligned} \varphi (t\gamma +C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K)&=t\varphi (\gamma ) +\varphi (C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n))\cap \varphi ({\mathfrak {O}}_K)\\&=t\varphi (\gamma )+C_{\mathbb {Q}}(A)\cap {\mathbb {Z}}^d. \end{aligned}$$

Hence

$$\begin{aligned} t\gamma +C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K\subseteq {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n) \end{aligned}$$

if and only if

$$\begin{aligned} t\varphi (\gamma )+C_{\mathbb {Q}}(A)\cap {\mathbb {Z}}^d\subseteq {\mathcal {F}}(A). \end{aligned}$$

But the fact that $\gamma \in {{\,\textrm{int}\,}}(C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n))\cap {\mathfrak {O}}_K$ shows that $\varphi (\gamma )\in {{\,\textrm{int}\,}}(C_{\mathbb {Q}}(A))\cap {\mathbb {Z}}^d\subseteq {{\,\textrm{int}\,}}(C_{\mathbb {R}}(A))\cap {\mathbb {Z}}^d$, so Lemma 3 shows that for all sufficiently large $t\in {\mathbb {N}}$,

$$\begin{aligned} t\varphi (\gamma )+C_{\mathbb {Q}}(A)\cap {\mathbb {Z}}^d\subseteq t\varphi (\gamma ) +C_{\mathbb {R}}(A)\cap {\mathbb {Z}}^d\subseteq {\mathcal {F}}(A), \end{aligned}$$

and thus

$$\begin{aligned} t\gamma +C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K \subseteq {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n). \end{aligned}$$

$\square$

Now that we have shown whenever $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module, the Frobenius semigroup

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)= & {} \{w\in {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n) \mid w\\{} & {} +C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K\subseteq {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n)\} \end{aligned}$$

is nonempty, we can begin to describe it in certain cases. The simplest case is when $\alpha _1,\ldots ,\alpha _n$ are linearly independent, in which case they form an integral basis for ${\mathfrak {O}}_K$. In this case, the beginning of the proof of Theorem 1 shows that the following is true.

Lemma 5

Let K be a degree-d number field with at least one real embedding, and suppose that $\beta _1,\ldots ,\beta _d\in {\mathfrak {O}}_K^+$ is a basis for ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Then

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d)={{\,\textrm{SG}\,}}(\beta _1,\ldots ,\beta _d). \end{aligned}$$

The following lemma is an immediate consequence of the definitions.

Lemma 6

Let K be a number field with a real embedding, and suppose that $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Then if $\alpha \in {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n)$,

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n,\alpha )={{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n). \end{aligned}$$

Proof

If $\alpha \in {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n)$ then we know that ${{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n,\alpha )={{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n)$ and $C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n,\alpha )=C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)$. Hence $C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n,\alpha )\cap {\mathfrak {O}}_K=C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K$, so the definition of the Frobenius semigroup shows that

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n,\alpha )={{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n). \end{aligned}$$

$\square$

The converse to Lemma 6 will be true provided that the elements $\alpha _1,\ldots ,\alpha _n$ form a basis for ${\mathfrak {O}}_K$.

Corollary 7

Let K be a number field with a real embedding and $\beta _1,\ldots ,\beta _d\in {\mathfrak {O}}_K^+$ be a basis for ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. If $\alpha \in {\mathfrak {O}}_K^+$, then

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d,\alpha )={{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d) \end{aligned}$$

if and only if $\alpha \in {{\,\textrm{SG}\,}}(\beta _1,\ldots ,\beta _d)$.

Proof

The ‘if’ part is handled by Lemma 6, so suppose that ${{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d,\alpha )={{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d)$. Then by Lemma 5, we know that ${{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d,\alpha )={{\,\textrm{SG}\,}}(\beta _1,\ldots ,\beta _d)$, so in particular, $0\in {{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d,\alpha )$, and thus ${{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d,\alpha )={{\,\textrm{SG}\,}}(\beta _1,\ldots ,\beta _d,\alpha )$. It follows that ${{\,\textrm{SG}\,}}(\beta _1,\ldots ,\beta _d,\alpha )={{\,\textrm{SG}\,}}(\beta _1,\ldots ,\beta _d)$, so we have that $\alpha \in {{\,\textrm{SG}\,}}(\beta _1,\ldots ,\beta _d)$. $\square$

It is not always true that

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)\subseteq {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n,\alpha ) \end{aligned}$$

for $\alpha _1,\ldots ,\alpha _n,\alpha \in {\mathfrak {O}}_K^+$. Suppose that $\beta _1,\ldots ,\beta _d\in {\mathfrak {O}}_K^+$ is a basis for ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module, and $\alpha \in {\mathfrak {O}}_K^+$ is such that $0\notin {{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d,\alpha )$. Then $0\in {{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d)$, so ${{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d)\not \subseteq {{\,\textrm{Frob}\,}}(\beta _1,\ldots ,\beta _d,\alpha )$. See Sect. 5 for an explicit example of a number field K with a real embedding and such an $\alpha \in {\mathfrak {O}}_K^+$, as well as an example of elements $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module, are not a basis for ${\mathfrak {O}}_K$, and for which ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)={{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n)$.

4 The structure of the Frobenius set

Let K be a number field with a real embedding $\sigma _1:K\hookrightarrow {\mathbb {C}}$ and fix $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Define a partial ordering $\preccurlyeq$ on ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ by $w\preccurlyeq v$ for $w,v\in {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ if and only if

$$\begin{aligned} w+C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K\subseteq v+C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K. \end{aligned}$$

Note that if $w\preccurlyeq v$ then $w=v+\gamma$ for some $\gamma \in C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K$, so the fact that $\sigma _1(\gamma )\geqslant 0$ implies the inequality $\sigma _1(w)\geqslant \sigma _1(v)$. The relation $\preccurlyeq$ is clearly reflexive and transitive, and the previous observation implies that $\preccurlyeq$ is also antisymmetric. This relation is also compatible with addition in ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$, in the sense that if $u,v,w\in {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ and $u\preccurlyeq v$, then $u+w\preccurlyeq v+w$. Furthermore, note that if $v\in {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ and $w\in v+C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K$, then $w=v+\gamma$ for some $\gamma \in C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K$, and thus

$$\begin{aligned} w+C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K&=v+\gamma +C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K\\&\subseteq v+C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K\subseteq {{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n), \end{aligned}$$

which shows that both $w\in {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ and $w\preccurlyeq v$. Thence ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ will not contain any minimal elements with respect to $\preccurlyeq$, since if $v\in {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ then we can choose any

$$\begin{aligned} w\in v+C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K \subseteq {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n) \end{aligned}$$

that is distinct from v, and then we will have $w\preccurlyeq v$ but $w\ne v$. However, we claim that ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ has maximal elements with respect to the partial ordering $\preccurlyeq$, and furthermore, there are only finitely many such maximal elements.

Lemma 8

Let K be a number field with a real embedding and fix nonzero $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Then ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ contains an element that is maximal with respect to $\preccurlyeq$. Furthermore, if $w\in {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$, then $w\preccurlyeq \mu$ for some maximal element $\mu \in {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$.

Proof

For brevity, set $C=C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K, S={{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n),$ and $F={{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$. We first claim that for any $w\in S$, there are finitely many $v\in S$ for which $\sigma _1(w)\geqslant \sigma _1(v)$. Suppose that this is not the case, so there are an infinite number of distinct elements $v_1,v_2,\ldots \in S$ for which $\sigma _1(w)\geqslant \sigma _1(v_i)$ for every $i=1,2,\dots$. Let $v_{ij}\in {\mathbb {N}}$ be (not necessarily unique) natural numbers for which

$$\begin{aligned} v_i=\sum _{j=1}^{n}v_{ij}\alpha _j. \end{aligned}$$

If there is some integer $N>0$ such that $v_{ij}\leqslant N$ for all $i=1,2,\dots$ and $j=1,\ldots ,n$, then there could be at most $(N+1)^n$ choices for the $v_i$, since the coefficient of each $\alpha _j$ in some representation of $v_i$ as a sum of the $\alpha _j$ must be one of the natural numbers $0,1,\ldots ,N$. Hence the coefficients $v_{ij}$ must grow without bound as $i\rightarrow \infty$ and j ranges from 1 to n. The fact that all of the $\sigma _1(\alpha _j)$ are positive then shows that at least one $v_i$ must be such that $\sigma _1(v_i)>\sigma _1(w)$, contradicting the fact that $\sigma _1(w)\geqslant \sigma _1(v_i)$ for all i. Hence there are finitely many $v\in S$ for which $\sigma _1(w)\geqslant \sigma _1(v)$. If $w,v\in F\subseteq S$ and $w\preccurlyeq v$, then $\sigma _1(w)\geqslant \sigma _1(v)$, so it also follows that for any $w\in F$, there are only finitely many $v\in F$ for which $w\preccurlyeq v$.

We now apply Zorn’s lemma to show that F has a maximal element with respect to $\preccurlyeq$. If $\Sigma \subseteq F$ is a finite chain in F, then $\Sigma$ clearly has an upper bound in F with respect to $\preccurlyeq$, so suppose that $\Sigma \subseteq F$ is an infinite chain in F. Furthermore, assume that $\Sigma$ does not have an upper bound in F. Then for any $w\in \Sigma$, we can find some $v_1\in \Sigma \setminus \{w\}$ for which $w\preccurlyeq v_1$; else $v_1\preccurlyeq w$ for all $v_1\in \Sigma \setminus \{w\}$ because $\Sigma$ is totally ordered, and thus w will be an upper bound of $\Sigma$. Similarly, we can find some $v_2\in \Sigma \setminus \{w,v_1\}$ for which $w\preccurlyeq v_1\preccurlyeq v_2$; else $v_2\preccurlyeq v_1$ for all $v_2\in \Sigma \setminus \{w,v_1\}$ because $\Sigma$ is totally ordered, and thus $v_1$ will be an upper bound of $\Sigma$. Continuing in this manner, we can find an infinite number of distinct elements $v_1,v_2,\ldots \in \Sigma \subseteq F$ for which $w\preccurlyeq v_1\preccurlyeq v_2\preccurlyeq \cdots$, which contradicts the fact that there are only finitely many $v\in F$ for which $w\preccurlyeq v$. Hence $\Sigma$ must have an upper bound, which means that we can apply Zorn’s lemma to conclude that F has a maximal element with respect $\preccurlyeq$.

Now suppose that $w\in F$, let $F_w\subseteq F$ be the set

$$\begin{aligned} F_w=\{v\in F \ | \ w\preccurlyeq v\}, \end{aligned}$$

and give $F_w$ the induced ordering from $\preccurlyeq$ on F. Note that by the observations in the first paragraph of this proof, $\#F_w<\infty$, and because $w\in F_w$. Furthermore, if $\mu$ is a maximal element of $F_w$ then $\mu$ is a maximal element of F, for if $v\in F{\setminus }\{\mu \}$ and $\mu \preccurlyeq v$ then $w\preccurlyeq \mu \preccurlyeq v$, and thus $v\in F_w$, so $\mu \preccurlyeq v$ is not possible by the maximality of $\mu$. The fact that $\#F_w<\infty$ implies that that $F_w$ has a maximal element $\mu$, and then $\mu$ is a maximal element of F for which $w\preccurlyeq \mu$. $\square$

Now define ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)\subseteq {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ to be the set of all maximal elements of ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ with respect to $\preccurlyeq$, and note that by Lemma 8. If we combine all the statements of Lemma 8, then we arrive at the following nice characterization of ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$.

Lemma 9

Let K be a number field with a real embedding and $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ be nonzero elements that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Then

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)=\bigcup _{\mu \in {\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)} (\mu +C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K). \end{aligned}$$

Proof

Let

$$\begin{aligned}{} & {} C=C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K,\quad S={{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n),\\{} & {} F={{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n),\quad {\mathfrak {M}}={\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n). \end{aligned}$$

The inclusion

$$\begin{aligned} F\supseteq \bigcup _{\mu \in {\mathfrak {M}}}(\mu +C) \end{aligned}$$

is obvious, since if $\mu \in F$ then $\mu +C\subseteq F$. Now, Lemma 8 shows that if $w\in F$ then there is some $\mu \in {\mathfrak {M}}$ for which $w\preccurlyeq \mu$, and thus $w\in \mu +C$, so we immediately get the inclusion in the other direction. $\square$

The above lemma provides the first main ingredient in the proof of Theorem 2. If $n\geqslant 1$, recall that the pointwise partial order $\leqslant _{\textrm{p}}$ on ${\mathbb {Z}}^n$ is defined by $(x_1,\ldots ,x_n)\leqslant _\textrm{p} (y_1,\ldots ,y_n)$ if and only if $x_i\leqslant y_i$ for all $i=1,\ldots ,n$. Then Dickson’s lemma says that the following is true:

Lemma 10

(Dickson’s Lemma) Let $n\geqslant 1$. Then any nonempty set $S\subseteq {\mathbb {N}}^n$ has a finite and nonzero number of minimal elements with respect to $\leqslant _\textrm{p}$.

Proof

Let k be a field and consider the polynomial ring $k[x_1,\ldots ,x_n]$ and the ideal ${\mathfrak {a}}=\langle x^s \mid s\in S\rangle \subseteq k[x_1,\ldots ,x_n]$, where we are using the notation $x^\alpha =x_1^{\alpha _1}\cdots x_n^{\alpha _n}$ when $\alpha =(\alpha _1,\ldots ,\alpha _n)\in {\mathbb {N}}^n$. If S is finite then the claim of the lemma is obvious, so suppose that S is infinite. Then because ${\mathbb {N}}^n$ is countable, we can write $S=\{s_1,s_2,\dots \}$. By Hilbert’s Basis Theorem, $k[x_1,\ldots ,x_n]$ is a Noetherian ring, so there is some smallest nonzero $m\in {\mathbb {N}}$ for which

$$\begin{aligned} \langle x^{s_1},\ldots ,x^{s_m}\rangle =\langle x^{s_1},\ldots ,x^{s_{m+1}}\rangle =\cdots , \end{aligned}$$

and thus ${\mathfrak {a}}=\langle x^{s_1},\ldots ,x^{s_m}\rangle$. Then if $s\in S$ is not one of the $s_1,\ldots ,s_m$, we have $x^s\in {\mathfrak {a}}$, so there are $f_1,\ldots ,f_m\in k[x_1,\ldots ,x_n]$ for which

$$\begin{aligned} x^s=\sum _{i=1}^{m}f_ix^{s_i}. \end{aligned}$$

But this means that at least one monomial term in one of the $f_i$ must be some nonzero element of k times $x^{s-s_i}$, so $0\leqslant _\textrm{p}s-s_i$, and thus $s_i\leqslant _\textrm{p}s$. Hence s cannot be a minimal element of S, so $s_1,\ldots ,s_m$ are all the possible minimal elements of S. It follows that S can have only finitely many minimal elements, and choosing the minimal elements (with respect to $\leqslant _\textrm{p}$) out of the $s_1,\ldots ,s_m$ shows that S has at least one minimal element. $\square$

Using Dickson’s lemma, we can show that the set ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ will always have finitely many elements, meaning ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ will always have finitely many maximal elements with respect to $\preccurlyeq$. This allows us to prove Theorem 2.

Theorem 2

Let K be a number field with a real embedding and let $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ be nonzero elements that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Then ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ is a finite set, and ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ is equal to the finite union

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)=\bigcup _{\mu \in {\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)} (\mu +C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K). \end{aligned}$$

Proof

As before, set

$$\begin{aligned}{} & {} C=C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K,\quad S={{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n),\\{} & {} F={{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n),\quad {\mathfrak {M}}={\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n). \end{aligned}$$

Because $\alpha _1,\ldots ,\alpha _n$ span ${\mathfrak {O}}_K$ by assumption, we have a surjective ${\mathbb {Z}}$-module homomorphism $f:{\mathbb {Z}}^n\rightarrow {\mathfrak {O}}_K$ given by

$$\begin{aligned} f(x_1,\ldots ,x_n)=\sum _{i=1}^{n}x_i\alpha _i. \end{aligned}$$

Define a set $W=f^{-1}(F)\cap {\mathbb {N}}^n\subseteq {\mathbb {N}}^n$, and note that for any $w\in F$, there is some $w^*\in W$ for which $f(w^*)=w$, since every $w\in F$ must also be in $S=f({\mathbb {N}}^n)$. Theorem 1 then shows that W is nonempty because F is. We now claim that if $\mu \in {\mathfrak {M}}$ and $\mu ^*\in f^{-1}(\mu )\cap {\mathbb {N}}^n$, then $\mu ^*$ is a minimal element of W with respect to the pointwise partial ordering $\leqslant _\textrm{p}$. Suppose that $w^*\in W$ is such that $f(w^*)=w\in F$ and $w^*\leqslant _\textrm{p} \mu ^*$. Let $w^*=(w_1,\ldots ,w_n)$ and $\mu ^*=(\mu _1,\ldots ,\mu _n)$, so the fact that $w^*\leqslant _\textrm{p} \mu ^*$ implies that each $w_i\leqslant \mu _i$, and thus there are $y_1,\ldots ,y_n\in {\mathbb {N}}$ for which $\mu _i=w_i+y_i$. Hence

$$\begin{aligned} w+\sum _{i=1}^{n}y_i\alpha _i=f((w_1,\ldots ,w_n)+(y_1,\ldots ,y_n))=f(\mu _1,\ldots ,\mu _n)=\mu , \end{aligned}$$

so

$$\begin{aligned} \mu +C=w+\sum _{i=1}^{n}y_i\alpha _i+C\subseteq w+C, \end{aligned}$$

where we used the fact that each $y_i\in {\mathbb {N}}$, so each $y_i\alpha _i\in S\subseteq C$. But $w\in F$, so the maximality of $\mu$ implies that $\mu =w$, and thus each $y_i=0$ because each $\sigma _1(\alpha _i)>0$, so the only way for $\sum \limits _{i=1}^{n}y_i\alpha _i=0$ when $y_1,\ldots ,y_n\in {\mathbb {N}}$ is for all of $y_1,\ldots ,y_n$ to be zero. Then $\mu ^*=w^*$, which means that $\mu ^*$ must be a minimal element of W with respect to the pointwise partial ordering $\leqslant _\textrm{p}$. By Dickson’s lemma, there are finitely many minimal elements of $W\subseteq {\mathbb {N}}^n$, so it follows that there must also be finitely many maximal elements of F. Hence ${\mathfrak {M}}$ is a finite set, and Lemma 9 shows that F may be written as the finite union

$$\begin{aligned} F=\bigcup _{\mu \in {\mathfrak {M}}}(\mu +C). \end{aligned}$$

$\square$

The union in Theorem 2 is interesting since it is the smallest possible way of writing ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ as a union of translates of the set $C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K$. This is because if W is some collection of elements of ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ for which

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)=\bigcup _{w\in W}(w+C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K), \end{aligned}$$

(1)

then each $\mu \in {\mathfrak {M}}$ must be contained in some $w+C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha )\cap {\mathfrak {O}}_K$, so $\mu =w$ for some $w\in W$ and thus ${\mathfrak {M}}\subseteq W$. It turns out that if we add a few more conditions to the set W, then we can get a valuable characterization of when a given set of elements in ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ will be the set of maximal elements. Even more so, this characterization allows us to avoid having to actually check the maximality of certain elements in many explicit calculations of the Frobenius semigroup.

Lemma 11

Let K be a number field with a real embedding and let $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ be distinct nonzero elements that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Suppose that $\mu _1,\ldots ,\mu _m\in {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ satisfy

• (1) for all $w\in {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$, there is some $i=1,\ldots ,m$ for which $w\preccurlyeq \mu _i$;

• (2) for all distinct $i,j\in \{1,\ldots ,m\}$, .

Then ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)=\{\mu _1,\ldots ,\mu _m\}$.

Proof

As before, let

$$\begin{aligned}{} & {} C=C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K,\quad S={{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n),\\{} & {} F={{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n),\quad {\mathfrak {M}}={\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n). \end{aligned}$$

Then condition (1) shows that if $\mu \in {\mathfrak {M}}$ then $\mu \preccurlyeq \mu _i$ for some i. Hence $\mu =\mu _i$ by the maximality of $\mu$, so ${\mathfrak {M}}\subseteq \{\mu _1,\ldots ,\mu _m\}$. Now fix $i=1,\ldots ,m$. Then Lemma 8 shows that $\mu _i\preccurlyeq \mu$ for some maximal $\mu \in {\mathfrak {M}}$, and condition (1) shows that $\mu \preccurlyeq \mu _j$ for some $j=1,\ldots ,m$, and thus $\mu =\mu _j$. Hence $\mu _i\preccurlyeq \mu =\mu _j$, so condition (2) shows that $j=i$ and $\mu =\mu _i$. Hence $\{\mu _1,\ldots ,\mu _m\}\subseteq {\mathfrak {M}}$, so ${\mathfrak {M}}=\{\mu _1,\ldots ,\mu _m\}.$ $\square$

An argument identical to the one given in Lemma 11 also shows that if $W\subseteq {{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ is a set satisfying Eq. (1), and no two elements of W precede one another, then W must be the set of maximal elements ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$. In particular, this means that W must be finite. It is thus not possible to write ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ as an infinite union of translates of the set $C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K$, where no translate is a subset of another one. It is also interesting to consider how the distinct maximal elements of ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ are related to each other. The next lemma begins to answer this question.

Lemma 12

Let K be a number field with a real embedding and let $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Then any two distinct elements of ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ are ${\mathbb {Z}}$-linearly independent.

Proof

Let

$$\begin{aligned}{} & {} C=C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)\cap {\mathfrak {O}}_K,\quad S={{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n),\\{} & {} F={{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n),\quad {\mathfrak {M}}={\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n), \end{aligned}$$

and suppose that this is not true. Then there are distinct $\mu _1,\mu _2\in {\mathfrak {M}}$ and nonzero $y_1,y_2\in {\mathbb {Z}}$ for which

$$\begin{aligned} y_1\mu _1+y_2\mu _2=0. \end{aligned}$$

Because $\sigma _1(\mu _1),\sigma _1(\mu _2)>0$, precisely one of the $y_i$ must be less than zero, so we may assume that there are nonzero $x_1,x_2\in {\mathbb {N}}$ such that $x_1<x_2$ (if $x_1=x_2$ then it trivially follows that $\mu _1=\mu _2$) and for which we have an equation in the form

$$\begin{aligned} x_1\mu _1-x_2\mu _2=0, \end{aligned}$$

so $\mu _2=(x_1/x_2)\mu _1$. Then $x_1/x_2<1$, so $1-x_1/x_2\in {\mathbb {Q}}_{\geqslant 0}$, and thus $\mu _1-\mu _2=(1-x_1/x_2)\mu _1\in C_{\mathbb {Q}}(\alpha _1,\ldots ,\alpha _n)$ because $\mu _1\in S$ by definition. But $\mu _1-\mu _2\in {\mathfrak {O}}_K$, so it follows that $\mu _1-\mu _2\in C$, and thus

$$\begin{aligned} \mu _1+C=\mu _2+(\mu _1-\mu _2)+C\subseteq \mu _2+C. \end{aligned}$$

The maximality of $\mu _1$ then implies that $\mu _1=\mu _2$, which contradicts the original assumption that $\mu _1,\mu _2\in {\mathfrak {M}}$ were distinct. $\square$

While it may be tempting to try to generalize this to show that any number of maximal elements of ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)$ are linearly independent, results in Sect. 6 show that this fails.

5 An example of a simple Frobenius semigroup for quadratic extensions

We now offer an explicit calculation of the Frobenius semigroup for certain collections of three elements in real quadratic number fields. Note that if K is a real quadratic number field then all embeddings of K into ${\mathbb {C}}$ are real, so we let $\sigma _1:K\hookrightarrow {\mathbb {C}}$ be the inclusion of K into ${\mathbb {C}}$. We first prove some basic properties about the $C_{\mathbb {Q}}\cap {\mathfrak {O}}_K$ and ${{\,\textrm{SG}\,}}$ sets in this case.

Lemma 13

Let K be a real quadratic number field with positive integral basis $\beta _1,\beta _2\in {\mathfrak {O}}_K^+$, and suppose that $\alpha =a_2\beta _2-a_1\beta _1\in {\mathfrak {O}}_K^+$, where $a_1,a_2\in {\mathbb {N}}$ are nonzero. Then

$$\begin{aligned} C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K=\left\{ y_1\beta _1+y_2\beta _2 \left| \, y_1,y_2\in {\mathbb {Z}}, \ y_2\geqslant 0 \,\text { and } y_2\geqslant -\frac{a_2}{a_1}y_1\right. \right\} .\nonumber \\ \end{aligned}$$

(2)

Proof

Let C denote the set on the right hand side of Eq. (2), and first suppose that $y_1\beta _1+y_2\beta _2\in C$, with $y_1\in {\mathbb {Z}}, y_2\in {\mathbb {N}}$, and $y_2\geqslant -(a_2/a_1)y_1$. Then

$$\begin{aligned} \frac{y_2}{a_2}\alpha =y_2\beta _2-\frac{y_2a_1}{a_2}\beta _1. \end{aligned}$$

By definition, we know that

$$\begin{aligned} \frac{y_2a_1}{a_2}+y_1=\frac{y_2a_1+y_1a_2}{a_2}\geqslant 0, \end{aligned}$$

so it follows that

$$\begin{aligned} y_2\beta _2+y_1\beta _1=\frac{y_2}{a_2}\alpha +\left( \frac{y_2a_1}{a_2}+y_1\right) \!\beta _1\in C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K, \end{aligned}$$

and thus $C\subseteq C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K$. Now suppose that $z,z_1,z_2\in {\mathbb {Q}}_{\geqslant 0}$ and $z\alpha +z_1\beta _1+z_2\beta _2\in {\mathfrak {O}}_K$. Then

$$\begin{aligned} z\alpha +z_1\beta _1+z_2\beta _2=(z_1-za_1)\beta _1+(z_2+za_2)\beta _2, \end{aligned}$$

so $z_1-za_1,z_2+za_2\in {\mathbb {Z}}$ because $\beta _1,\beta _2$ is a basis for ${\mathfrak {O}}_K$. Furthermore,

$$\begin{aligned} -\frac{a_2}{a_1}(z_1-za_1)=-\frac{a_2z_1}{a_1}+za_2\leqslant z_2+za_2, \end{aligned}$$

since $a_1,a_2,z_1,z_2\geqslant 0$. Hence $z\alpha +z_1\beta _1+z_2\beta _2\in C$, so it follows that $C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K=C$. $\square$

In this special case of a quadratic extension, the elements of ${{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$ will also satisfy certain nice properties.

Lemma 14

Let K be a real quadratic number field with positive integral basis $\beta _1,\beta _2\in {\mathfrak {O}}_K^+$, and suppose that $\alpha =a_2\beta _2-a_1\beta _1\in {\mathfrak {O}}_K^+$, where $a_1,a_2\in {\mathbb {N}}$ are nonzero. Furthermore, assume that $q,r\in {\mathbb {N}}$ are such that $0\leqslant r\leqslant a_1$ and $y_1=a_1q+r$. If $y_2\in {\mathbb {Z}}$, then

• (a) if $0<r<a_1$, we have $y_2\beta _2-y_1\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$ if and only if $y_2\geqslant (q+1)a_2$; and

• (b) if $r=0$, meaning $y_1=qa_1$, we have $y_2\beta _2-y_1\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$ if and only if $y_2\geqslant qa_2$.

Proof

(a): First suppose that $y_2\geqslant (q+1)a_2$. Then

$$\begin{aligned} y_2\beta _2-y_1\beta _1= & {} (q+1)\alpha +((q+1)a_1-y_1)\beta _1\\{} & {} +(y_2-(q+1)a_2)\beta _2\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ) \end{aligned}$$

since both $(q+1)a_1-y_1=a_1-r\geqslant 0$ and $y_2-(q+1)a_2\geqslant 0$. Conversely, suppose that $y_2\beta _2-y_1\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$, so there are $z,z_1,z_2\in {\mathbb {N}}$ for which

$$\begin{aligned} y_2\beta _2-y_1\beta _1=z\alpha +z_1\beta _1+z_2\beta _2=(z_1-za_1)\beta _1+(z_2+za_2)\beta _2. \end{aligned}$$

Hence $za_1-z_1=y_1$ and $z_2+za_2=y_2$, so $za_1-z_1=a_1q+r$, and the fact that $r>0$ shows that

$$\begin{aligned} za_1=a_1q+r+z_1>a_1q, \end{aligned}$$

so $z\geqslant q+1$, and thus

$$\begin{aligned} y_2=z_2+za_2\geqslant z_2+(q+1)a_2\geqslant (q+1)a_2. \end{aligned}$$

(b): First suppose that $y_2\geqslant qa_2$. Then

$$\begin{aligned} y_2\beta _2-qa_1\beta _1=q\alpha +(y_2-qa_2)\beta _2\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ) \end{aligned}$$

because $y_2-qa_2\geqslant 0$. Conversely, suppose that $y_2\beta _2-qa_1\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$, so there are are $z,z_1,z_2\in {\mathbb {N}}$ for which

$$\begin{aligned} y_2\beta _2-qa_1\beta _1=z\alpha +z_1\beta _1+z_2\beta _2=(z_1-za_1)\beta _1+(z_2+za_2)\beta _2. \end{aligned}$$

Hence $za_1-z_1=qa_1$ and $z_2+za_2=y_2$, so $za_1=qa_1+z_1\geqslant qa_1$, and thus $z\geqslant q$, so

$$\begin{aligned} y_2=z_2+za_2\geqslant z_2+qa_2\geqslant qa_2. \end{aligned}$$

$\square$

See Figs. 1 and 2 for a geometric interpretation of Lemmas 13 and 14. Using Lemmas 13 and 14, we can explicitly calculate what the Frobenius semigroup looks like for certain elements of ${\mathfrak {O}}_K^+$. Specifically, let $\beta _1,\beta _2\in {\mathfrak {O}}_K^+$ be a positive integral basis for ${\mathfrak {O}}_K$, and suppose that $\alpha \in {\mathfrak {O}}_K^+$ is in the form

$$\begin{aligned} \alpha =ab\beta _2-a\beta _1, \end{aligned}$$

where $a,b\in {\mathbb {N}}$ are nonzero natural numbers. Then we claim that

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )=(a-1)\beta _1+C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K. \end{aligned}$$

To do this, we first show the following, which is a stronger version of Lemma 13.

Lemma 15

Let K be a real quadratic number field and let $\beta _1,\beta _2\in {\mathfrak {O}}_K^+$ be an integral basis for ${\mathfrak {O}}_K$. If $a,b\in {\mathbb {N}}$ are nonzero and $\alpha =ab\beta _2-a\beta _1\in {\mathfrak {O}}_K^+$, then

$$\begin{aligned} C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K= & {} \left\{ \left. \frac{n}{a}\alpha +n_1\beta _1 +n_2\beta _2 \right| n,n_1,n_2\in {\mathbb {N}}\right\} \nonumber \\ {}= & {} {{\,\textrm{SG}\,}}\!\left( \beta _1,\beta _2, \frac{\alpha }{a}\right) . \end{aligned}$$

(3)

Proof

The inclusion

$$\begin{aligned} C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\supseteq {{\,\textrm{SG}\,}}\!\left( \beta _1,\beta _2,\frac{\alpha }{a}\right) \end{aligned}$$

is obvious because a divides the coefficients of both $\beta _1$ and $\beta _2$ in the expansion of $\alpha$ as a linear combination of them. Now suppose that $x,x_1,x_2\in {\mathbb {Q}}_{\geqslant 0}$ are such that

$$\begin{aligned} x\alpha +x_1\beta _1+x_2\beta _2=(x_1-xa)\beta _1+(x_2+xab)\beta _2\in C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K. \end{aligned}$$

If $x=0$ then there is nothing to show, so suppose that $x\ne 0$. Then $x_2+xab\in {\mathbb {N}}$ because it must be an integer and all of x, a, b are nonzero, and $x_1-xa\in {\mathbb {Z}}$, so

$$\begin{aligned} (x_2+xab)+b(x_1-xa)=x_2+bx_1\in {\mathbb {N}}. \end{aligned}$$

By the division algorithm, there exist $n,n_2\in {\mathbb {N}}$ for which $x_2+xab=nb+n_2$ and $0\leqslant n_2<b$. Define $n_1\in {\mathbb {Z}}$ so that $x_1-xa=n_1-n$. If $x_1=x_2=0$ then x will clearly be in the form we want it to be, so suppose that at least one of $x_1$ or $x_2$ is nonzero, and thus strictly positive. Then we have that

$$\begin{aligned} 0<x_2+bx_1=x_2+xab+b(x_1-xa)=nb+n_2+n_1b-nb=n_2+n_1b. \end{aligned}$$

Because $0\leqslant n_2<b$, the above shows that

$$\begin{aligned} n_1>-\frac{n_2}{b}>-1, \end{aligned}$$

and thus $n_1\geqslant 0$. Hence $n_1\in {\mathbb {N}}$, so we have $x_2+xab=nb+n_2$ and $x_1-xa=n_1-n$, where $n,n_1,n_2\in {\mathbb {N}}$. It follows that

$$\begin{aligned} x\alpha +x_1\beta _1+x_2\beta _2&=(x_1-xa)\beta _1+(x_2+xab)\beta _2\\&=(n_1-n)\beta _1+(nb+n_2)\beta _2\\&=\frac{n}{a}(ab\beta _2-a\beta _1)+n_1\beta _1+n_2\beta _2\\&=\frac{n}{a}\alpha +n_1\beta _1+n_2\beta _2, \end{aligned}$$

so we have the inclusion in the other direction in Eq. (3), and consequently Eq. (3) is true. $\square$

Using the results of Lemmas 13, 14, and 15, we can calculate the Frobenius semigroup ${{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ in certain cases.

Proposition 16

Let K be a real quadratic number field and $\beta _1,\beta _2\in {\mathfrak {O}}_K^+$ be an integral basis. If $a,b\in {\mathbb {N}}$ are nonzero and $\alpha =ab\beta _2-a\beta _1\in {\mathfrak {O}}_K^+$, then

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )=(a-1)\beta _1+C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K. \end{aligned}$$

Proof

By Lemma 11, in order to show that ${\mathfrak {M}}(\beta _1,\beta _2,\alpha )=\{(a-1)\beta _1\}$, it will suffice to show that $(a-1)\beta _1\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$, and if $w\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$, then $w\preccurlyeq (a-1)\beta _1$. Note that $n_1\beta _1+n_2\beta _2\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$ whenever $n_1,n_2\in {\mathbb {N}}$, so, in view of Lemma 15, we need only show that

$$\begin{aligned} (a-1)\beta _1+\frac{n}{a}\alpha \in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ) \end{aligned}$$

for all $n\in {\mathbb {N}}$ in order to show that $(a-1)\beta _1+C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K\subseteq {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$. If $0\leqslant n<a$, then

$$\begin{aligned} (a-1)\beta _1+\frac{n}{a}\alpha= & {} (a-1)\beta _1+nb\beta _2-n\beta _1\\= & {} nb\beta _2 +(a-1-n)\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2)\subseteq {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ) \end{aligned}$$

because $n\leqslant a-1$. If $n\geqslant a$, then we can perform the division algorithm to get $q,r\in {\mathbb {N}}$ for which $0\leqslant r<a$ and $n=aq+r$, so

$$\begin{aligned} (a-1)\beta _1+\frac{n}{a}\alpha =(a-1)\beta _1+\frac{r}{a}\alpha +q\alpha \in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ) \end{aligned}$$

by the above. Hence $(a-1)\beta _1\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$.

We now claim that if $w\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$, then $w\preccurlyeq (a-1)\beta _1$. In order to do this, we make use of the isomorphism $\varphi :{\mathfrak {O}}_K\rightarrow {\mathbb {Z}}^2$ associated to the basis $\beta _1,\beta _2$ for ${\mathfrak {O}}_K$.

Fig. 1

An illustration of the sets $\varphi ({{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ))$ and $\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$ in ${\mathbb {Z}}^2$. Black points represent elements of $\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$, and the shaded region represents points of $\varphi ({{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ))$. The thick black line represents the set $\varphi (C_{\mathbb {Q}}(\alpha )\cap {\mathfrak {O}}_K)$, which along with the positive $\beta _1$ axis, marks the boundary of the set $\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$

Note that the rational multiples of $\varphi (\alpha )$ in $\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$ all lie along the line $y=-bx$ in ${\mathbb {Z}}^2$. The region obtained by shifting the cone $\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$ to the right by $(a-1)\varphi (\beta _1)$ will be bounded by the lines $y=0$ and $y=-bx+b(a-1)$. It follows that if $x_1\in {\mathbb {Z}}$, $x_2\in {\mathbb {N}}$, and $x_1\beta _1+x_2\beta _2\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$, then $x_1\beta _1+x_2\beta _2\preccurlyeq (a-1)\beta _1$ if and only if $x_2\geqslant -bx_1+b(a-1)$.

In order to show that every element in ${{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ precedes $(a-1)\beta _1$, we leverage Lemma 14 and use the observations at the end of the previous paragraph. Suppose that $x_1,x_2\in {\mathbb {N}}$ and $w=x_1\beta _1+x_2\beta _2\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$, so we claim that $w\preccurlyeq (a-1)\beta _1$. Then

$$\begin{aligned} \frac{x_1+1}{a}\alpha +x_1\beta _1+x_2\beta _2=((x_1+1)b+x_2)\beta _2-\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ), \end{aligned}$$

so Lemma 14 shows that we must have

$$\begin{aligned} (x_1+1)b+x_2\geqslant ab. \end{aligned}$$

This is equivalent to

$$\begin{aligned} x_2\geqslant -bx_1+b(a-1), \end{aligned}$$

meaning $x_1\beta _1+x_2\beta _2\preccurlyeq (a-1)\beta _1$.

Now suppose that $x_1,x_2\in {\mathbb {N}}$ are nonzero and $w=x_2\beta _2-x_1\beta _1\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ (we can ignore the case where $x_2<0$ because no point in that form will be in ${{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$). By the division algorithm, there are $q,r\in {\mathbb {N}}$ for which $x_1=qa+r$ and $0\leqslant r<a$. Then since $(a-r+1)\alpha /a\in C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K$ by Lemma 15, we know that

$$\begin{aligned}&\frac{a-r+1}{a}\alpha +x_2\beta _2-x_1\beta _1\\&\quad =(a-r+1)b\beta _2-(a-r+1)\beta _1+x_2\beta _2-x_1\beta _1\\&\quad =((a-r+1)b+x_2)\beta _2-((q+1)a+1)\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ), \end{aligned}$$

and Lemma 14 shows that in order for this to be true, we must have

$$\begin{aligned} (a-r+1)b+x_2\geqslant (q+2)ab, \end{aligned}$$

which implies that

$$\begin{aligned} x_2\geqslant qab+2ab-ab+rb-b=bx_1+b(a-1). \end{aligned}$$

Hence $x_2\beta _2-x_1\beta _1\preccurlyeq (a-1)\beta _1$, so Lemma 11 shows that $(a-1)\beta _1$ is the only maximal element of ${{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$, and thus

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )=(a-1)\beta _1+C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K. \end{aligned}$$

$\square$

If we take $a=1$, then Lemma 15 and Proposition 16 show that

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,b\beta _2-\beta _1)=C_{\mathbb {Q}}(\beta _1,\beta _2,b\beta _2-\beta _1) \cap {\mathfrak {O}}_K={{\,\textrm{SG}\,}}(\beta _1,\beta _2,b\beta _2-\beta _1). \end{aligned}$$

This gives us an example of a number field K with a real embedding and a collection of elements $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ that is not a basis for ${\mathfrak {O}}_K$ for which ${{\,\textrm{Frob}\,}}(\alpha _1,\ldots ,\alpha _n)={{\,\textrm{SG}\,}}(\alpha _1,\ldots ,\alpha _n)$. Furthermore, if we take $a>1$ then we get a collection of elements in the form $\alpha =ab\beta _2-a\beta _1\in {\mathfrak {O}}_K^+$ for which ${{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ never contains 0, and thus

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\beta _1,\beta _2)\not \subseteq {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha ). \end{aligned}$$

The existence of such elements $\alpha \in {\mathfrak {O}}_K^+$ was promised at the end of Sect. 3.

6 The number of maximal elements

From the results of Sects. 4 and 5, it may be tempting to conclude that for a number field K with a real embedding and a fixed value of n, the number $\#{\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ will be bounded above as the nonzero elements $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ range over spanning sets of ${\mathfrak {O}}_K$. However, we now show that this is not the case, and in fact, when $K/{\mathbb {Q}}$ is a real quadratic extension we can make $\#{\mathfrak {M}}(\alpha _1,\alpha _2,\alpha _3)$ arbitrarily large. In particular, we have the following:

Proposition 17

Let K be a real quadratic number field and fix a positive integral basis $\beta _1,\beta _2\in {\mathfrak {O}}_K^+$ for ${\mathfrak {O}}_K$ with $\beta _1<\beta _2$. If $m>1$ is an integer and $\alpha =(m+1)\beta _2-m\beta _1$, then

$$\begin{aligned} {\mathfrak {M}}(\beta _1,\beta _2,\alpha )=\{(m-i)\beta _1+(i-1)\beta _2 \mid i=1,\ldots ,m\}, \end{aligned}$$

and thus

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )=\bigcup _{i=1}^{m}\big ((m-i)\beta _1+(i-1)\beta _2 +C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K\big ). \end{aligned}$$

Proof

Note that $\alpha \in {\mathfrak {O}}_K^+$ because $\beta _1<\beta _2$. For each $i=1,\ldots ,m$, let

$$\begin{aligned} \mu _i=(m-i)\beta _1+(i-1)\beta _2, \end{aligned}$$

so we claim ${\mathfrak {M}}(\beta _1,\beta _2,\alpha )=\{\mu _1,\ldots ,\mu _m\}$.

To do this, we first show that each $\mu _i\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$, and then apply Lemma 11. Recall, by Lemma 13, that the set $C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K$ is given by

$$\begin{aligned} C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K=\left\{ x_1\beta _1+x_2\beta _2 \left| \, x_1,x_2\in {\mathbb {Z}}, \ x_2\geqslant 0 \,\text { and } x_2\geqslant -\frac{m+1}{m}x_1\right. \right\} . \end{aligned}$$

If $x_1,x_2\in {\mathbb {N}}$ then it is clear that $\mu _i+x_1\beta _1+x_2\beta _2\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$ because $\mu _i,x_1\beta _1+x_2\beta _2\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$, so suppose that $x_1,x_2\in {\mathbb {N}}$ and $x_2\beta _2-x_1\beta _1\in C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K$. First assume that $x_1\leqslant m$. Then

$$\begin{aligned} \mu _i+x_2\beta _2-x_1\beta _1=(m-i-x_1)\beta _1+(i-1+x_2)\beta _2, \end{aligned}$$

and if $x_1\leqslant m-i$ then the above is clearly in ${{\,\textrm{SG}\,}}(\beta _1,\beta _2)$, so suppose that $x_1>m-i$. Then the fact that $x_1\leqslant m$ implies that $x_1\leqslant 2m-i$, and we can thus write

$$\begin{aligned} \mu _i+x_2\beta _2-x_1\beta _1&=((m+1)\beta _2-m\beta _1)+((2m-i-x_1)\beta _1+(x_2+i-2-m)\beta _2)\\&=\alpha +(2m-i-x_1)\beta _1+(x_2+i-2-m)\beta _2. \end{aligned}$$

We know that $2m-i-x_1\geqslant 0$, and the fact that $x_1>m-i$ implies that $x_1\geqslant m-i+1$. The fact that $x_2\beta _2-x_1\beta _1\in C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K$, combined with Lemma 13, then shows that

$$\begin{aligned} x_2\geqslant \frac{m+1}{m}x_1\geqslant \frac{m+1}{m}(m-i+1) =m+2+\frac{1}{m}-i-\frac{i}{m}. \end{aligned}$$

Hence

$$\begin{aligned} x_2-m-2+i\geqslant \frac{1}{m}-\frac{i}{m}\geqslant \frac{1}{m}-1>-1, \end{aligned}$$

so $x_2-m-2+i\geqslant 0$, and thus

$$\begin{aligned} \mu _i+x_2\beta _2-x_1\beta _1=\alpha +(2m-i-x_1)\beta _1+(x_2+i-2-m) \beta _2\in {{\,\textrm{SG}\,}}(\alpha ,\beta _1,\beta _2).\nonumber \\ \end{aligned}$$

(4)

Now suppose that $x_1>m$, $x_2\beta _2-x_1\beta _1\in C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K$, and write $x_1=qm+r_1$ for $q,r_1\in {\mathbb {N}}$ and $0\leqslant r_1<m$. If $x_2\geqslant (q+1)(m+1)$ and $r_1>0$ then part (a) of Lemma 14 shows that $x_2\beta _2-x_2\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$, and if $x_2\geqslant (q+1)(m+1)$ and $r_1=0$ then part (b) of Lemma 14 shows that $x_2\beta _2-x_1\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$, since $x_2\geqslant (q+1)(m+1)>q(m+1)$. We thus suppose that $x_2<(q+1)(m+1)=q(m+1)+m+1$. The condition that $x_2\beta _2-x_1\beta _1\in C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K$, combined with Lemma 13, then gives

$$\begin{aligned} x_2\geqslant \frac{m+1}{m}x_1=\frac{m+1}{m}(qm+r_1)=q(m+1)+r_1\frac{m+1}{m}. \end{aligned}$$

Hence $q(m+1)\leqslant x_2<(q+1)(m+1)$, so we can write $x_2=q(m+1)+r_2$ for some $0\leqslant r_2<m+1$, meaning

$$\begin{aligned} x_2\beta _2-x_1\beta _1=q(m+1)\beta _2+r_2\beta _2-qm\beta _1-r_1\beta _1=q\alpha +r_2\beta _2-r_1\beta _1. \end{aligned}$$

The fact that $0\leqslant r_1<m$, combined with what we showed in Eq. (4), then gives that

$$\begin{aligned} \mu _i+x_2\beta _2-x_1\beta _1=q\alpha +(\mu _i+r_2\beta _2-r_1\beta _1)\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ). \end{aligned}$$

Thus $\mu _1,\ldots ,\mu _m\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$.

We now apply Lemma 11 in order to show that ${\mathfrak {M}}(\beta _1,\beta _2,\alpha )=\{\mu _1,\ldots ,\mu _m\}$, so we must first show that for any distinct $i,j\in \{1,\ldots ,m\}$. Let $\varphi :{\mathfrak {O}}_K\rightarrow {\mathbb {Z}}^2$ be the ${\mathbb {Z}}$-module isomorphism associated to the basis $\beta _1,\beta _2$ for ${\mathfrak {O}}_K$, and let $\pi _i:{\mathfrak {O}}_K\rightarrow {\mathbb {Z}}$, $i=1,2$, be the projection of ${\mathfrak {O}}_K={\mathbb {Z}}\beta _1\oplus {\mathbb {Z}}\beta _2$ onto ${\mathbb {Z}}\beta _i$. Note that if $\gamma _1,\gamma _2\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ and $\pi _2(\gamma _1)<\pi _2(\gamma _2)$, then because $\pi _2(C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)\subseteq {\mathbb {N}}$ by Lemma 13. Hence if $i<j$ then because

$$\begin{aligned} \pi _2(\mu _i)=i-1<j-1=\pi _2(\mu _j). \end{aligned}$$

Now, note that the slope of the line in ${\mathbb {Z}}^2$ connecting $\varphi (\mu _i)$ to $\varphi (\mu _j)$ is $-1$, while by Lemma 13, the slope of the line extending out of $\varphi (\mu _i)$ that determines one edge of the boundary of $\varphi (\mu _i)+\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$ is

$$\begin{aligned} -\frac{m+1}{m}<-1. \end{aligned}$$

Hence $\varphi (\mu _j)$ cannot be contained in $\varphi (\mu _i)+\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$, so $\mu _j$ cannot be contained in $\mu _i+C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K$, and thus it is not possible that $\mu _j\preccurlyeq \mu _i$.

Fig. 2

An illustration of the sets $\varphi ({{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )),\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$, and $\varphi (\mu _i)+\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$ in ${\mathbb {Z}}^2$ for the case $m=4$. Black points represent elements of $\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$, the shaded region represents elements of $\varphi ({{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ))$, and the thick lines coming out of each $\mu _i$ represent part of the boundary of the set $\varphi (\mu _i)+\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$

We now show that if $w\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$, then $w\preccurlyeq \mu _i$ for some $i=1,\ldots ,m$. We first deal with the elements of ${{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ with a non-negative $\beta _1$ coefficient, meaning we claim that if $x_1,x_2\in {\mathbb {N}}$ and $w=x_1\beta _1+x_2\beta _2\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$, then $w\preccurlyeq \mu _i$ for some $i=1,\ldots ,m$. Note that in ${\mathbb {Z}}^2$, the points $\varphi (\mu _1),\ldots ,\varphi (\mu _m)$ all lie along the line $y=m-1-x$, and furthermore, the points $\varphi (\mu _1),\ldots ,\varphi (\mu _m)$ consist of all integral points along this line with non-negative x and y coordinates. Thus if $x_1,x_2\in {\mathbb {N}}$ and $w=x_1\beta _1+x_2\beta _2$, then $w\preccurlyeq \mu _i$ for some i if and only if $x_2\geqslant m-1-x_1$. If $x_1\geqslant m-1$ then $m-1-x_1\leqslant 0\leqslant x_2$ because $x_2\in {\mathbb {N}}$, so we know that $w\preccurlyeq \mu _i$ for some i. Now suppose that $w=x_1\beta _1+x_2\beta _2\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ and $x_1<m-1$. We claim that $w=x_1\beta _1+x_2\beta _2\preccurlyeq \mu _i$ for some $i=1,\ldots ,m$. By Lemma 13, we know that $(2+x_1)\beta _2-(1+x_1)\beta _1\in C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K$ because

$$\begin{aligned} \frac{m+1}{m}(1+x_1)=x_1+1+\frac{x_1+1}{m}<x_1+2. \end{aligned}$$

Now,

$$\begin{aligned} w+(2+x_1)\beta _2-(1+x_1)\beta _1=(2+x_1+x_2)\beta _2-\beta _1, \end{aligned}$$

so the fact that $w\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ shows that $(2+x_1+x_2)\beta _2-\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$, and Lemma 14 then shows that

$$\begin{aligned} 2+x_1+x_2\geqslant m+1. \end{aligned}$$

Then

$$\begin{aligned} x_2\geqslant m-1-x_1, \end{aligned}$$

so $w\preccurlyeq \mu _i$ for some i.

We now deal with the elements of ${{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ with a negative $\beta _1$ coefficient, meaning we claim that if $x_1,x_2\in {\mathbb {N}}$ and $w=x_2\beta _2-x_1\beta _1\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$, then $w\preccurlyeq \mu _m$. As before, note that the line in ${\mathbb {Z}}^2$ determining part of the boundary of the cone $\varphi (\mu _m)+\varphi (C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K)$ is $y=-\frac{m+1}{m}x+m-1$, so we have $w\preccurlyeq \mu _m$ if and only if $x_2\geqslant \frac{m+1}{m}x_1+m-1$. Let $x_1=qm+r$, where $q,r\in {\mathbb {N}}$ and $0\leqslant r< m$. Then if $r\geqslant 1$,

$$\begin{aligned} (m-r+2)\beta _2-(m-r+1)\beta _1\in C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K \end{aligned}$$

by Lemma 13, since

$$\begin{aligned} \frac{m+1}{m}(m-r+1)=m+1+\frac{m+1}{m}(1-r)\leqslant m+1+1-r=m-r+2. \end{aligned}$$

Adding w to this point yields

$$\begin{aligned} w+(m-r+2)\beta _2-(m-r+1)\beta _1&=(m-r+2+x_2)\beta _2-(m-r+1+x_1)\beta _1\\&=(m-r+2+x_2)\beta _2-((q+1)m+1)\beta _1, \end{aligned}$$

and the fact that $w\in {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ then shows that this element is in ${{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha )$. Lemma 14 then implies that we must have

$$\begin{aligned} m-r+2+x_2\geqslant (q+2)(m+1). \end{aligned}$$

It follows that

$$\begin{aligned} x_2&\geqslant q(m+1)+m+r=\frac{x_1-r}{m}(m+1)+m+r\\&=\frac{m+1}{m}x_1+m -\frac{r}{m}> \frac{m+1}{m}x_1+m-1, \end{aligned}$$

so $w=x_2\beta _2-x_1\beta _1\preccurlyeq \mu _m$.

If $r=0$ then $x_1=qm$, the point $2\beta _2-\beta _1\in C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K$ because $2\geqslant \frac{m+1}{m}$, and we have

$$\begin{aligned} w+2\beta _2-\beta _1= & {} x_2\beta _2-qm\beta _1+2\beta _2-\beta _1\\= & {} (2+x_2) \beta _2-(qm+1)\beta _1\in {{\,\textrm{SG}\,}}(\beta _1,\beta _2,\alpha ). \end{aligned}$$

By Lemma 14, this means that $2+x_2\geqslant (q+1)(m+1),$ so

$$\begin{aligned} x_2\geqslant q(m+1)+m-1=\frac{m+1}{m}x_1+m-1, \end{aligned}$$

and thus $w\preccurlyeq \mu _m$. It follows that conditions (1) and (2) of Lemma 11 are satisfied, so ${\mathfrak {M}}(\beta _1,\beta _2,\alpha )=\{\mu _1,\ldots ,\mu _m\}$, and thence the set ${{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ is given by

$$\begin{aligned} {{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )=\bigcup _{i=1}^{m}\big ((m-i)\beta _1+(i-1)\beta _2 +C_{\mathbb {Q}}(\beta _1,\beta _2,\alpha )\cap {\mathfrak {O}}_K\big ). \end{aligned}$$

$\square$

Note that in the above proof, the elements $\mu _2,\ldots ,\mu _{m-1}$ are sort of “fringe” maximal elements. That is, with the exception of $\mu _2,\ldots ,\mu _{m-1}$, every element of ${{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ will either precede $\mu _1$ or $\mu _m$.

While the above proposition shows that in the case of quadratic extensions, we can make $\#{\mathfrak {M}}(\alpha _1,\alpha _2,\alpha _3)$ arbitrarily large, we strongly believe that something similar will be the case for arbitrary number fields admitting at least one real embedding.

Conjecture 1

Let K be a number field with at least one real embedding and let $\beta _1,\ldots ,\beta _d\in {\mathfrak {O}}_K^+$ be a basis for ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Then for any positive integer $m\geqslant 1$, there is some $\alpha \in {\mathfrak {O}}_K^+$ for which $\#{\mathfrak {M}}(\beta _1,\ldots ,\beta _d,\alpha )\geqslant m$.

Given a number field K with a real embedding and $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module, another interesting question to consider is how the elements of ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ are related to each other. Lemma 12 shows that the elements of ${\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ are all pairwise linearly independent, but the calculations done in Sects. 5 and 6 suggest that there is likely much more to be seen. For example, all of the maximal elements in the set ${{\,\textrm{Frob}\,}}(\beta _1,\beta _2,\alpha )$ in Proposition 17 are collinear when considered as elements in ${\mathbb {Z}}^2$ under the isomorphism ${\mathfrak {O}}_K\rightarrow {\mathbb {Z}}^2$ associated to the basis $\beta _1,\beta _2$. Suppose that K is an arbitrary degree-d number field with at least one real embedding and an integral basis $\beta _1,\ldots ,\beta _d\in {\mathfrak {O}}_K^+$, and we have elements $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$. A natural question to ask is then whether there is some nice geometric description of the elements of ${\mathfrak {M}}(\beta _1,\ldots ,\beta _d,\alpha _1,\ldots ,\alpha _n)$ when they are considered as elements in ${\mathbb {Z}}^d$ under the isomorphism ${\mathfrak {O}}_K\rightarrow {\mathbb {Z}}^d$ associated to the basis $\beta _1,\ldots ,\beta _d$ for ${\mathfrak {O}}_K$?

It is also important to note that all of the explicit calculations of Frobenius semigroups that we have so far only consider cases where the collection of elements $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ contain a basis for ${\mathfrak {O}}_K$. If $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ span ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module, then there will certainly be some collection $\alpha _{i_1},\ldots ,\alpha _{i_d}$, where $d=[K:{\mathbb {Q}}]$, that is linearly independent. However, in this case the elements $\alpha _{i_1},\ldots ,\alpha _{i_d}$ need not span ${\mathfrak {O}}_K$. Focusing on the quadratic case, let us assume that $\beta _1,\beta _2\in {\mathfrak {O}}_K^+$ are linearly independent, and $a_1,a_2\in {\mathbb {Q}}_{\geqslant 0}$ are such that $\beta _1,\beta _2,a_2\beta _2-a_1\beta _1\in {\mathfrak {O}}_K^+$ span ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module. Then the nice conclusions of Lemmas 13 and 14 may not hold, since their proofs rely on the assumption that $\beta _1,\beta _2$ is an integral basis for ${\mathfrak {O}}_K$. This shows that calculations of the Frobenius semigroup in cases where the elements do not contain a basis could potentially be much more complicated.

Another thing to consider is given some degree-d number field K with a real embedding and $\alpha _1,\ldots ,\alpha _n\in {\mathfrak {O}}_K^+$ that generate ${\mathfrak {O}}_K$ as a ${\mathbb {Z}}$-module, is there some nice bound on $\#{\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ in terms of n, d, and $\alpha _1,\ldots ,\alpha _n$? Proposition 17 shows that this bound cannot depend on only n and d, but it does not rule out the possibility of the bound also depending on $\alpha _1,\ldots ,\alpha _n$. Specifically, there may be some bounds on $\#{\mathfrak {M}}(\alpha _1,\ldots ,\alpha _n)$ that depend on more number theoretic properties of ${\mathfrak {O}}_K$, perhaps similar in nature to some of the bounds appearing in [1, 2], or [7].