Frobenius templates in certain $\mathbf{2\times 2}$ matrix rings^∗

Let $\mathbb{N}$ be the set of nonnegative integers, $\mathbb{Z}$ the set of integers, $\mathbb{Z}^{+}\vcentcolon=\mathbb{N}\setminus\left\{0\right\}$, and, for $\alpha_{1},\dots,\alpha_{n}\in\mathbb{N}$, $MN(\alpha_{1},\dots,\alpha_{n})\vcentcolon=\{\sum_{i=1}^{n}\lambda_{i}\alpha_{i}:\lambda_{1},\dots,\lambda_{n}\in\mathbb{N}\}$. A list in a set $S$ is a finite sequence of elements from $S$. If $n\in\mathbb{Z}^{+}$ and $(\alpha_{1},\dots,\alpha_{n})$ is a list in $\mathbb{N}$, we will say $\alpha_{1},\dots,\alpha_{n}$ are coprime to mean $\gcd(\alpha_{1},\dots,\alpha_{n})=1$. If we assume $\gcd(\alpha_{1},\dots,\alpha_{n})\neq 1$ when $\alpha_{1}=\dots=\alpha_{n}=0$, then a list in $\mathbb{N}$ is coprime if and only if the list contains positive integers and the positive integers in the list are coprime. For an additive group $G$, if $S\subseteq G$ and $g\in G$, then let $g+S\vcentcolon=\left\{g+s:s\in S\right\}$.

Here is a 19th century theorem, due to Sylvester and Frobenius and others, restated to suit our purpose:

The classical Frobenius problem [3], sometimes called the “coin problem,” is to evaluate $\chi(\alpha_{1},\dots,\alpha_{n})$ at coprime lists $(\alpha_{1},\dots,\alpha_{n})$ in $\mathbb{N}$. Frobenius used to discuss this problem in his lectures [1], so this area of number theory happens to be named after him. For $n=2$, there is a formula: if $\alpha,\beta\in\mathbb{N}$ and $\gcd(\alpha,\beta)=1$, then $\chi(\alpha,\beta)=(\alpha-1)(\beta-1)$. For $n>2$ there are no known general formulas, although there are formulas for special cases and there are algorithms; see [8]. The classical Frobenius problem inspired the results in [4], which in turn inspired Nicole Looper [6] to generalize the classical Frobenius problem by introducing the notion of a Frobenius template (defined below), allowing one to pose similar problems in rings other than $\mathbb{Z}$.

All of our rings will contain a multiplicative identity, 1, and most will be commutative. An additive monoid in a ring $R$, is a subset of $R$ that is closed under addition and contains the identity $0$; monoids, unlike groups, do not have to contain inverses. A Frobenius template in a ring $R$ is a triple $(A,C,U)$ such that:

1. (i)

   $A$ is a nonempty subset of $R$,

2. (ii)

   $C$ and $U$ are additive monoids in $R$, and

3. (iii)

   for all lists $(\alpha_{1},\dots,\alpha_{n})$ in $A$,
   $MN(\alpha_{1},\dots,\alpha_{n})=\{\sum_{i=1}^{n}\lambda_{i}\alpha_{i}:\lambda_{1},\dots,\lambda_{n}\in C\}$ is a subset of $U$.

For any valid template, the properties above guarantee that $MN(\alpha_{1},\dots,\alpha_{n})$ will also be an additive monoid in $R$. The Frobenius set of a list $\alpha_{1},\dots,\alpha_{n}\in A$ is $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)\vcentcolon=\left\{w\in R:w+U\subseteq MN(\alpha_{1},\dots,\alpha_{n})\right\}$. Notice that $w\in\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$ implies that $w\in MN(\alpha_{1},\dots,\alpha_{n})$ since $0\in U$, so $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)\subseteq MN(\alpha_{1},\dots,\alpha_{n})\subseteq U$. Notice also that the assumption that $U$ is closed under addition guarantees that if $w\in\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$, then $w+U\subseteq\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$.

For a given template $(A,C,U)$, the corresponding Frobenius problem is the following pair of tasks:

1. 1.

   Determine for which lists $\alpha_{1},\dots,\alpha_{n}\in A$ it is true that $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)\neq\emptyset$.

2. 2.

   For lists $\alpha_{1},\dots,\alpha_{n}$ such that $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)\neq\emptyset$, describe the set $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$.

Note that in all the Frobenius templates that have been studied so far, it has always been the case that $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$, when nonempty, is a finite union of sets of the form $w+U$ for some $w\in R$.

The classical Frobenius problem revolves around the template $(\mathbb{N},\mathbb{N},\mathbb{N})$ in the ring $\mathbb{Z}$. As discussed earlier, if $n=2$ and $\alpha_{1},\alpha_{2}\in\mathbb{N}$, then $\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ is nonempty if (and only if) $\alpha_{1},\alpha_{2}$ are coprime, and $\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)=\chi(\alpha_{1},\alpha_{2})+\mathbb{N}=(\alpha_{1}-1)(\alpha_{2}-1)+\mathbb{N}$ in such cases. So the classical Frobenius problem has been completely solved when $n=2$.

Our definition above of a Frobenius template is slightly less general than Looper’s definition in [6]. The difference is that in Looper’s templates, $U=U(\alpha_{1},\dots,\alpha_{n})$ is allowed to vary with the list $\alpha_{1},\dots,\alpha_{n}$. This is absolutely necessary in order to get interesting and nontrivial results when the ring involved is a subring of $\mathbb{C}$, the complex numbers, that is not contained in $\mathbb{R}$, the real numbers. The ring of Gaussian integers, $\mathbb{Z}[i]=\{a+bi:a,b\in\mathbb{Z}\}$, is such a ring, a particularly famous member of the family of rings $\{\mathbb{Z}[\sqrt{m}i]:m\in\mathbb{Z}^{+}\}$.

To see why we must let the element $U$ of a Frobenius template in such a ring depend on the list $\alpha_{1},\dots,\alpha_{n}\in A$, here is an example in $\mathbb{Z}[\sqrt{3}i]$. Suppose $C=\{a+b\sqrt{3}i:a,b\in\mathbb{N}\}=\left\{re^{i\theta}\in\mathbb{Z}[\sqrt{3}i]:0\leq r\text{ and }0\leq\theta\leq\frac{\pi}{2}\right\}$ and $A=C\setminus\{0\}$. Then consider $\alpha=2e^{i\frac{\pi}{3}}=1+\sqrt{3}i\in A$ and $\beta=1\in A$. What should $U$ be for this list? We set $U(\alpha,\beta)=\{re^{i\theta}\in\mathbb{Z}[\sqrt{3}i]:0\leq r\text{ and }0\leq\theta\leq\frac{5\pi}{6}\}$, an angular sector in $\mathbb{Z}[\sqrt{3}i]$. We so choose $U(\alpha,\beta)$ because $MN(\alpha,\beta)=\left\{\lambda_{1}(2e^{i\frac{\pi}{3}})+\lambda_{2}:\lambda_{1},\lambda_{2}\in C\right\}\subseteq U(\alpha,\beta)$, and no angular sector in $\mathbb{Z}[\sqrt{3}i]$ properly contained in $U(\alpha,\beta)$ contains $MN(\alpha,\beta)$. Further, if $\hat{U}$ is an additive monoid in $\mathbb{Z}[\sqrt{3}i]$ which properly contains $U(\alpha,\beta)$, then no translate $w+\hat{U}$, $w\in\mathbb{C}$, is contained in $MN(\alpha,\beta)$. Thus $U(\alpha,\beta)$ is the only possible additive monoid in $\mathbb{Z}[\sqrt{3}i]$ containing $MN(\alpha,\beta)$ such that $\textnormal{Frob}\left(\alpha,\beta\right)=\{w:w+U(\alpha,\beta)\subseteq MN(\alpha,\beta)\}$ might possibly be nonempty.

We leave to the reader the verification of the claims in the previous paragraph. We hope that the main point is clear: letting $U$ vary with the list $\alpha_{1},\dots,\alpha_{n}\in A\subseteq\mathbb{Z}[\sqrt{m}i]$ in the formulation of Frobenius problems in $\mathbb{Z}[\sqrt{m}i]$ is necessitated by the nature of multiplication in the complex numbers. For an appreciation of Frobenius problems in such settings, see [4] and [5].

We do not take on similar difficulties here. In the next section we give some more or less obvious results in various Frobenius templates, then review previous results concerning templates in the rings $\mathbb{Z}[\sqrt{m}]$ with $m\in\mathbb{Z}^{+}\setminus\{n^{2}:n\in\mathbb{Z}^{+}\}$. In the third section, we classify the Frobenius set in a pleasing modification of the classical Frobenius template. In the fourth section, we solve Frobenius problems in rings of $2\times 2$ (upper) triangular matrices with constant diagonal and entries from a ring $Q$, where different choices of $Q$ allow different templates. In the last section, we further generalize the idea of a Frobenius template and explore an example of this generalization.

Throughout this section, $R$ will be a ring with multiplicative identity 1. A list $(\alpha_{1},\dots,\alpha_{n})$ in $R$ spans unity in $R$ if and only if $1=\lambda_{1}\alpha_{1}+...+\lambda_{n}\alpha_{n}$ for some $\lambda_{1},\dots,\lambda_{n}\in R$.

As a warm-up, here are some valid Frobenius templates $(A,C,U)$ with, in most cases, trivial solutions to the corresponding Frobenius problems. In each example, $(\alpha_{1},\dots,\alpha_{n})$ denotes a list in $A$. Check that $A$ is nonempty, $C$ and $U$ are additive monoids in $R$, and $U$ always contains $MN(\alpha_{1},\dots,\alpha_{n})$.

1. 1.

   In the template $(R,\left\{0\right\},\left\{0\right\})$, we have $MN(\alpha_{1},\dots,\alpha_{n})=\left\{0\right\}$ and
   $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)=\left\{0\right\}$. This result extends to any template $(A,C,U)$ with $C=U=\left\{0\right\}$.

2. 2.

   Similar to example 1, templates of the form $(\left\{0\right\},C,\left\{0\right\})$ always have $MN(0)=\left\{0\right\}$ and $\textnormal{Frob}\left(0\right)=\left\{0\right\}$.

3. 3.

   In the template $(R,R,R)$, Proposition 2.1 shows that $(\alpha_{1},\dots,\alpha_{n})$ spans unity in $R$ if $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)\neq\emptyset$. Conversely, if $(\alpha_{1},\dots,\alpha_{n})$ spans unity, then $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)=R$.

4. 4.

   Suppose that $I$ is an ideal of $R$ and the template is $(\{1\},I,I)$. Clearly $MN(1)=I=U$, so $U\subseteq\textnormal{Frob}\left(1\right)$. Recall that $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$ is a subset of $MN(\alpha_{1},\dots,\alpha_{n})$; thus $\textnormal{Frob}\left(1\right)=U$.

5. 5.

   Again, let $I$ be an ideal of $R$, and now consider the template $(I,R,I)$. When $I=R$, this is example 3. In contrast, let $I$ be a proper ideal. Then no list $(\alpha_{1},\dots,\alpha_{n})$ spans unity in $R$ — but this is precisely because $1\notin I$, so no facile conclusion based on Proposition 2.1 presents itself. Frobenius problems for this class of templates could be interesting. For instance, if the template is $(2\mathbb{Z},\mathbb{Z},2\mathbb{Z})$ in $\mathbb{Z}$, where $2\mathbb{Z}$ is the ideal of even integers, it is straightforward to prove that $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)\neq\emptyset$ if and only if the integers $\absolutevalue{\frac{\alpha_{1}}{2}},\dots,\absolutevalue{\frac{\alpha_{n}}{2}}$ are coprime. In this case, $MN(\alpha_{1},\dots,\alpha_{n})=2\mathbb{Z}$ and $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)=2\text{Frob}^{\prime}\left(\frac{\alpha_{1}}{2},\dots,\frac{\alpha_{n}}{2}\right)$, where $\text{Frob}^{\prime}$ denotes Frobenius sets with respect to the classical template $(\mathbb{N},\mathbb{N},\mathbb{N})$.

6. 6.

   If the template is $((0,\infty),[0,\infty),[0,\infty))$ in $\mathbb{R}$, then $MN(\alpha_{1},\dots,\alpha_{n})=[0,\infty)=\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$ for every list $(\alpha_{1},\dots,\alpha_{n})$. On the other hand, if we restrict the coefficients to the set of rational numbers $\mathbb{Q}$ with the template $((0,\infty),\mathbb{Q}\cap[0,\infty),[0,\infty))$ in $\mathbb{R}$, then $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$ is always empty, as $MN(\alpha_{1},\dots,\alpha_{n})$ is countable and any translate of $[0,\infty)$ is uncountable.

7. 7.

   Suppose that $R_{1}$ and $R_{2}$ are rings with associated Frobenius templates $(A_{1},C_{1},U_{1})$ and $(A_{2},C_{2},U_{2})$, respectively. Let $A_{1}\times A_{2}$ be the Cartesian product of $A_{1}$ and $A_{2}$. The same goes for $C_{1}\times C_{2}$ and $U_{1}\times U_{2}$, which are additive monoids (under componentwise addition) in the product ring $R_{1}\times R_{2}$. It is simple to see that $(A_{1}\times A_{2},C_{1}\times C_{2},U_{1}\times U_{2})$ is a valid Frobenius template in $R_{1}\times R_{2}$. Let $(\beta_{1},\dots,\beta_{n})$ be a list in $A_{1}\times A_{2}$ with each $\beta_{i}=(\gamma_{i},\mu_{i})$ for $i=1,\dots,n$. If $\textnormal{Frob}\left(\beta_{1},\dots,\beta_{n}\right)\neq\emptyset$ in $(A_{1}\times A_{2},C_{1}\times C_{2},U_{1}\times U_{2})$, then
   $\textnormal{Frob}\left(\beta_{1},\dots,\beta_{n}\right)=\textnormal{Frob}_{1}(\gamma_{1},\dots,\gamma_{n})\times\textnormal{Frob}_{2}(\mu,\dots,\mu_{2})$, where $\textnormal{Frob}_{1}$ and $\textnormal{Frob}_{2}$ are Frobenius sets in $(A_{1},C_{1},U_{1})$ and $(A_{2},C_{2},U_{2})$, respectively. For $m\in\mathbb{Z}^{+}$, this result extends to the product ring $R_{1}\times\dots\times R_{m}$.

Besides the Gaussian integers [4, 5], prior work on generalized Frobenius problems has concentrated on templates in the real subring $\mathbb{Z}[\sqrt{m}]\vcentcolon=\{a+b\sqrt{m}:a,b\in\mathbb{Z}\}$, where $m\in\mathbb{Z}^{+}$ is not a perfect square. Below, we showcase some highlights from this work. If $m$ is a positive integer with irrational square root, then $\mathbb{Z}[\sqrt{m}]$ is a subring of $\mathbb{R}$. Let $\mathbb{N}[\sqrt{m}]\vcentcolon=\{a+b\sqrt{m}:a,b\in\mathbb{N}\}$ and $\mathbb{Z}[\sqrt{m}]^{+}\vcentcolon=\mathbb{Z}[\sqrt{m}]\cap[0,\infty)$, which are both additive monoids in $\mathbb{Z}[\sqrt{m}]$ that contain $1$.

1. 1.

   In the template $(\mathbb{N}[\sqrt{m}],\mathbb{N}[\sqrt{m}],\mathbb{N}[\sqrt{m}])$, the result below is proven in [6]:

   Theorem 2.1.

   If $\alpha_{1},\dots,\alpha_{n}\in\mathbb{N}[\sqrt{m}]$, $\alpha_{i}=a_{i}+b_{i}\sqrt{m}$, $a_{i},b_{i}\in\mathbb{N}$, $i=1,\dots,n$, then $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)\neq\emptyset$ if and only if $(\alpha_{1},\dots,\alpha_{n})$ spans unity in $\mathbb{Z}[\sqrt{m}]$ and at least one of $a_{1},\dots,a_{n}$, $b_{1},\dots,b_{n}$ is zero.

   The result stated in [6], Theorem 3, is slightly weaker than the formulation above, but needlessly so, since the proof in [6] proves the statement given here.

2. 2.

   In the template $(\mathbb{Z}[\sqrt{m}]^{+},\mathbb{Z}[\sqrt{m}]^{+},\mathbb{Z}[\sqrt{m}]^{+})$, the following is nearly proven in [2]:

   Theorem 2.2.

   If $\alpha_{1},\dots,\alpha_{n}\in\mathbb{Z}[\sqrt{m}]^{+}$, then the following are equivalent:

   1. (i)

      $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)\neq\emptyset$;

   2. (ii)

      $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)=\mathbb{Z}[\sqrt{m}]^{+}$;

   3. (iii)

      $(\alpha_{1},\dots,\alpha_{n})$ spans unity in $\mathbb{Z}[\sqrt{m}]$.

   Since Theorem 2 of [2] proves $(iii)\implies(ii)$, we simply appeal to Proposition 2.1 to prove the statement above, which is somewhat stronger than Theorem 2 in [2].

   Also in [2], with reference to the template $(\mathbb{N}[\sqrt{m}],\mathbb{N}[\sqrt{m}],\mathbb{N}[\sqrt{m}])$,
   $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$ is determined in a very special class of cases. Recall that for coprime positive integers $c_{1},\dots,c_{n}$, $\chi\left(c_{1},\dots,c_{n}\right)$ is the smallest $w\in\mathbb{Z}^{+}$ such that $w+\mathbb{N}\subseteq\left\{\sum_{i=1}^{n}\lambda_{i}c_{i}:\lambda_{1},\dots,\lambda_{n}\in\mathbb{N}\right\}$.

   Theorem 2.3.

   For $a_{1},\dots,a_{r},b_{1},\dots,b_{s}\in\mathbb{N}$,
   $(a_{1},\dots,a_{r},b_{1}\sqrt{m},\dots,b_{s}\sqrt{m})$ is a list in $\mathbb{N}[\sqrt{m}]$. We have

   $$\begin{split}&\textnormal{Frob}\left(a_{1},\dots,a_{r},b_{1}\sqrt{m},\dots,b_{s}\sqrt{m}\right)\neq\emptyset\\ &\iff a_{1},\dots,a_{r},b_{1}m,\dots,b_{s}m\text{ are coprime,}\end{split}$$

   and, in such cases,

   $$\begin{split}&\textnormal{Frob}\left(a_{1},\dots,a_{r},b_{1}\sqrt{m},\dots,b_{s}\sqrt{m}\right)\\ &=\chi(a_{1},\dots,a_{r},b_{1}m,\dots,b_{s}m)+\chi(a_{1},\dots,a_{r},b_{1},\dots,b_{s})\sqrt{m}+\mathbb{N}[\sqrt{m}]\text{.}\end{split}$$

   Note that the case $s=0$ is allowed in this result. Also, if the integers $a_{1},\dots,a_{r},b_{1}m,\dots,b_{s}m$ are coprime, then so are $a_{1},\dots,a_{r},b_{1},\dots,b_{s}$, so the second part of Theorem 2.3 uses well-defined expressions.

3. 3.

   With $\chi(a_{1},\dots,a_{n})$ as above, recall that in the $n=2$ case of the classical template $(\mathbb{N},\mathbb{N},\mathbb{N})$ in $\mathbb{Z}$, $\chi(a_{1},a_{2})=(a_{1}-1)(a_{2}-1)$ for coprime $a_{1},a_{2}$. In a tour de force in [7], Kim proves a similar formula for the template $(\mathbb{N}[\sqrt{m}],\mathbb{N}[\sqrt{m}],\mathbb{N}[\sqrt{m}])$ in $\mathbb{Z}[\sqrt{m}]$:

   Theorem 2.4.

   Suppose $\alpha=a+b\sqrt{m},\beta=c+d\sqrt{m}$, $a,b,c,d\in\mathbb{N}$, $a+b,c+d>0$, $abcd=0$ (see Theorem 2.1, above), and suppose that $\alpha,\beta$ span unity in $\mathbb{Z}[\sqrt{m}]$. Then $\textnormal{Frob}\left(\alpha,\beta\right)=(\alpha-1)(\beta-1)(1+\sqrt{m})+\mathbb{N}[\sqrt{m}]$.

What’s next? We propose these categories of rings where one might discover results of interest of the Frobenius type.

1. 1.

   Subrings of algebraic extensions of $\mathbb{Q}$.

   Prior work on the Gaussian integers and the rings $\mathbb{Z}[\sqrt{m}]$ has put us into the foothills of a mountain range in this area. Besides the extensions of $\mathbb{Q}$ of finite degree, what about the ring of algebraic integers? Or, a better choice to start with, the ring of real algebraic integers?

2. 2.

   Polynomial rings.

   Somebody we know is working on this, but we haven’t heard from him for a year or so.

3. 3.

   Rings of square matrices.

   Did we say our rings have to be commutative? No, we did not. Keep in mind that in noncommutative rings, the precise definition of
   $MN(\alpha_{1},\dots,\alpha_{n})$ becomes important, since for $\alpha_{1},\dots,\alpha_{n}$ from $A$ and $\lambda_{1},\dots,\lambda_{n}$ from $C$, the linear combinations $\sum_{i=1}^{n}\lambda_{i}\alpha_{i}$ and $\sum_{i=1}^{n}\alpha_{i}\lambda_{i}$ are not necessarily equal.

The classical template $(\mathbb{N},\mathbb{N},\mathbb{N})$ uses nonnegative integer coefficients. What happens when we modify the available coefficients? Consider the template $(\mathbb{N},(n+\mathbb{N})\cup\left\{0\right\},\mathbb{N})$ for some $n\in\mathbb{Z}^{+}$. When $n=1$, this is the classical template.

Remember that when $a_{1},\dots,a_{k}\in\mathbb{N}$ are coprime, $\chi\left(a_{1},\dots,a_{k}\right)$ is the unique integer such that $\textnormal{Frob}\left(a_{1}\dots,a_{k}\right)=\chi\left(a_{1},\dots,a_{k}\right)+\mathbb{N}$ with respect to the classical template. In the results that follow, $\chi\left(a_{1},\dots,a_{k}\right)$ retains this meaning, whereas Frob and $MN$ reference the modified template $(\mathbb{N},(n+\mathbb{N})\cup\left\{0\right\},\mathbb{N})$.

While reading the next proof, remember that $\chi\left(a,b\right)=(a-1)(b-1)$ for coprime $a,b\in\mathbb{N}$, so $\chi\left(a,b\right)-1=ab-a-b$.

For a ring $Q$ with multiplicative identity $1$, let $R$ be the set $Q^{2}=Q\times Q$ under coordinatewise addition, with multiplication in $R$ defined by $(a,b)\cdot(c,d)=(ac,ad+bc)$. $R$ is a ring with multiplicative identity $(1,0)$, isomorphic to the ring of upper triangular $2\times 2$ matrices over $Q$ with constant diagonal: $(a,b)\sim\begin{bmatrix}a&b\\ 0&a\end{bmatrix}$. If $Q$ is commutative, then $R$ is commutative.

We will concentrate on the cases when $Q$ is a subring of the real field $\mathbb{R}$ containing $1$, and the Frobenius template is

Moving on, we now know a sufficient condition for $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)=\emptyset$ in a large swath of templates. The natural question is whether $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$ is ever nonempty, and the answer is yes.

Proposition 4.1 can be generalized, mutatis mutandis, to $Q$ being any linearly ordered commutative ring with unity. The same holds for Proposition 4.2, with the additional stipulation that $a_{1}$ is a unit. In both cases, the ordering is compatible with addition and multiplication.

In other words, setting $Q$ as a subfield of $\mathbb{R}$ yields a boring Frobenius template, since we can shrink elements of $\mathbb{R}$ using coefficients in $c_{i},d_{i}\in Q\cap(0,1)$. If we prohibit such shrinking, the template becomes much more interesting. Notice that this modification is similar to Section 3’s modification of the classical template.