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

###### Abstract

The classical Frobenius problem is to find the largest integer that cannot be
written as a linear combination of a given set of positive, coprime integers
using nonnegative integer coefficients. Prior work has generalized the
classical Frobenius problem from integers to Frobenius problems in other
rings. This paper explores Frobenius problems in various rings of (upper)
triangular $2\times 2$ matrices with constant diagonal.

Timothy Eller

Georgia Southern University<EMAIL_ADDRESS>

Jakub Kraus

University of Michigan<EMAIL_ADDRESS>
Yuki Takahashi

Grinnell College<EMAIL_ADDRESS>
Zhichun (Joy) Zhang

Swarthmore College<EMAIL_ADDRESS>
Key words and phrases: Frobenius template, Frobenius problem, coin problem,
monoid, ring.

AMS Subject Classification: 11D07, 11B05.

##  1 Introduction

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:

######  Theorem 1.1.

If $\alpha_{1},\dots,\alpha_{n}\in\mathbb{N}$ are coprime, then for some
$w\in\mathbb{N}$, $w+\mathbb{N}\subseteq MN(\alpha_{1},\dots,\alpha_{n})$.

Note that because $0\in\mathbb{N}$, if $w+\mathbb{N}\subseteq
MN(\alpha_{1},\dots,\alpha_{n})$ then
$w\in MN(\alpha_{1},\dots,\alpha_{n})$. Note also that because $\mathbb{N}$ is
closed under addition, if $w+\mathbb{N}\subseteq
MN(\alpha_{1},\dots,\alpha_{n})$ and $b\in\mathbb{N}$, then
$w+b+\mathbb{N}\subseteq w+\mathbb{N}\subseteq
MN(\alpha_{1},\dots,\alpha_{n})$. Thus, if we define, for
$\alpha_{1},\dots,\alpha_{n}\in\mathbb{Z}^{+}$,
$\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)\vcentcolon=\\{w\in\mathbb{N}:w+\mathbb{N}\subseteq
MN(\alpha_{1},\dots,\alpha_{n})\\}$, we have that either
$\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)=\emptyset$ or
$\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)=\chi(\alpha_{1},\dots,\alpha_{n})+\mathbb{N}$,
where $\chi(\alpha_{1},\dots,\alpha_{n})$ is the least element of
$\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$.

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.

##  2 Some fundamentals and known results

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
$\lambda_{1},\dots,\lambda_{n}\in R$.

######  Proposition 2.1.

Let $(A,C,U)$ be a Frobenius template in $R$ such that $1\in U$. If
$(\alpha_{1},\dots,\alpha_{n})$ is a list in $A$ such that
$\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)\neq\emptyset$, then
$(\alpha_{1},\dots,\alpha_{n})$ spans unity in $R$.

###### Proof.

Let $w\in\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$, i.e.
$w\in R$ and $w+U\subseteq MN(\alpha_{1},\dots,\alpha_{n})$. Since $0\in U$
and $1\in U$, it follows that $w,w+1\in MN(\alpha_{1},\dots,\alpha_{n})$.
Therefore, for some $\lambda_{1},\dots,\lambda,\gamma_{1},\dots,\gamma_{n}\in
C$, $w=\sum_{i=1}^{n}\lambda_{i}\alpha_{i}$ and
$w+1=\sum_{i=1}^{n}\gamma_{i}\alpha_{i}$, so
$1=\sum_{i=1}^{n}(\gamma_{i}-\lambda_{i})\alpha_{i}$. ∎

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.

##  3 Modifying the classical template

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
$(\mathbb{N},(n+\mathbb{N})\cup\left\\{0\right\\},\mathbb{N})$.

######  Proposition 3.1.

If $a_{1},\dots,a_{k}\in\mathbb{N}$ are coprime, then

$(a_{1}+\dots+a_{k})n+\chi\left(a_{1},\dots,a_{k}\right)+\mathbb{N}\subseteq\textnormal{Frob}\left(a_{1},\dots,a_{k}\right)\text{.}$

###### Proof.

Suppose $\omega\in\mathbb{N}$ and
$\omega\geq(a_{1}+\dots+a_{k})n+\chi\left(a_{1},\dots,a_{k}\right)$. We will
show $\omega\in\textnormal{Frob}\left(a_{1},\dots,a_{k}\right)$, i.e.

$\omega+\mathbb{N}\subseteq
MN(a_{1},\dots,a_{k})=\left\\{\sum_{i=1}^{k}\lambda_{i}a_{i}:\lambda_{1},\dots,\lambda_{k}\in(n+\mathbb{N})\cup\left\\{0\right\\}\right\\}\text{.}$

Suppose $f\in\mathbb{N}$, $f\geq\omega$. It suffices to show that $f\in
MN(a_{1},\dots,a_{k})$. Since $f\geq\omega$, we have that
$f-(a_{1}+\dots+a_{k})n\geq\chi\left(a_{1},\dots,a_{k}\right)$, so there exist
coefficients $\gamma_{1},\dots,\gamma_{k}\in\mathbb{N}$ such that
$f=(a_{1}+\dots+a_{k})n+\sum_{i=1}^{k}\gamma_{i}a_{i}=\sum_{i=1}^{k}(\gamma_{i}+n)a_{i}$.
Taking $\lambda_{i}=\gamma_{i}+n\geq n$ for $i=1,\dots,k$ yields the desired
result. ∎

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

######  Proposition 3.2.

Let $a,b\in\mathbb{N}$ and $n\in\mathbb{Z}^{+}$. If $a,b$ are coprime and
$n-1$ is not divisible by $a$ or $b$, then
$\textnormal{Frob}\left(a,b\right)=(a+b)n+\chi\left(a,b\right)+\mathbb{N}$.

###### Proof.

By the preceding proposition, it suffices to show that
$\textnormal{Frob}\left(a,b\right)\subseteq(a+b)n+\chi\left(a,b\right)+\mathbb{N}$.
So let
$t\in\mathbb{N}\setminus\left((a+b)n+\chi\left(a,b\right)+\mathbb{N}\right)$,
and suppose that $t\in\textnormal{Frob}\left(a,b\right)$. We will find a
contradiction, which will complete the proof.

Since $t\in\textnormal{Frob}\left(a,b\right)$, we have $t+\mathbb{N}\subseteq
MN(a,b)$, so $an+bn+\chi\left(a,b\right)-1\geq t$ is an element of $MN(a,b)$,
so there exist
$\lambda_{1},\lambda_{2}\in(n+\mathbb{N})\cup\left\\{0\right\\}$ such that

$\chi\left(a,b\right)-1=\lambda_{1}a+\lambda_{2}b-an-
bn=(\lambda_{1}-n)a+(\lambda_{2}-n)b\text{.}$

Now, if both $\lambda_{1}$ and $\lambda_{2}$ are $\geq n$, then
$\chi\left(a,b\right)-1$ equals a linear combination of $a$ and $b$ with
nonnegative integer coefficients, which is impossible. We also cannot have
$\lambda_{1}=0=\lambda_{2}$ since $\chi\left(a,b\right)+an+bn>1$. So consider
the case that $\lambda_{1}=0$ and $\lambda_{2}\geq n$. In that case,
$(\lambda_{2}-n)b=\chi\left(a,b\right)-1+an=ab-b+a(n-1)$, so
$(\lambda_{2}-n-a+1)b=a(n-1)$, so $b$ divides $a(n-1)$. But $a$ and $b$ are
coprime, so this implies that $b$ divides $n-1$, which we have assumed to be
false. Similarly $\lambda_{1}\geq n$ and $\lambda_{2}=0$ would imply that
$(\lambda_{1}-n)a=\chi\left(a,b\right)-1+bn=ab-a+b(n-1)$, yielding the
contradiction that $a$ divides $n-1$. Thus, all cases yield a contradiction,
so our supposition that $t\in\textnormal{Frob}\left(a,b\right)$ must be false.
∎

##  4 $\mathbf{2\times 2}$ triangular matrices with constant diagonal

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

$\begin{split}&\left(A(Q),C(Q),U(Q)\right)\\\
&=\left(\left(Q\cap(0,\infty)\right)\times\left(Q\cap[0,\infty)\right),\left(Q\cap[0,\infty)\right)^{2},\left(Q\cap[0,\infty)\right)^{2}\right).\end{split}$

######  Proposition 4.1.

Let $Q$ be a subring of $\mathbb{R}$ containing $1$. For any positive integer
$n$, let $(\alpha_{1},\dots,\alpha_{n})$ be a list in
$A(Q)\setminus\left(Q\times\left\\{0\right\\}\right)$. Then
$\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)=\emptyset$.

###### Proof.

Let $\alpha_{i}=(a_{i},b_{i})$, $i=1,\dots,n$. Elements of
$MN(\alpha_{1},\dots,\alpha_{n})$ look like

$\begin{split}(t,u)&=\sum_{i=1}^{n}(a_{i},b_{i})\cdot(c_{i},d_{i})\\\
&=\sum_{i=1}^{n}(a_{i}c_{i},a_{i}d_{i}+b_{i}c_{i})\\\
&=\left(\sum_{i=1}^{n}a_{i}c_{i},\sum_{i=1}^{n}\left(a_{i}d_{i}+b_{i}c_{i}\right)\right),\end{split}$

where $c_{i}\geq 0$ and $d_{i}\geq 0$ for all $i=1,\dots,n$.

Suppose that $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$ is
nonempty, so there is a tuple $(t,u)\in Q^{2}$ such that
$(t,u)+(Q\cap[0,\infty))^{2}\subseteq MN(\alpha_{1},\dots,\alpha_{n})$.
Therefore, for all $q\geq 0$, we have $(t,u)+(q,0)=(t+q,u)\in
MN(\alpha_{1},\dots,\alpha_{n})$. So for larger and larger $q$,
$t+q=\sum_{i=1}^{n}a_{i}c_{i}(q)$ and
$u=\sum_{i=1}^{n}(a_{i}d_{i}(q)+b_{i}c_{i}(q))$ for some nonnegative integers
$c_{i}(q)$ and $d_{i}(q)$ ($i=1,\dots,n$) that vary with $q$. Clearly
$\sum_{i=1}^{n}a_{i}c_{i}(q)=t+q\to\infty$ as $q\to\infty$. Because each fixed
$a_{i},b_{i}>0$, and each varying $c_{i}(q),d_{i}(q)\geq 0$, it follows that
$\max_{1\leq i\leq n}c_{i}(q)\to\infty$ as $q\to\infty$, and consequently
$u=\sum_{i=1}^{n}(a_{i}d_{i}(q)+b_{i}c_{i}(q))\to\infty$ as $q\to\infty$. But
$u$ is a given constant; it does not vary with $q$. Therefore, our supposition
must be false, so
$\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)=\emptyset$. ∎

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 4.2.

Let $Q$ be a subfield of $\mathbb{R}$. For any positive integer $n$, let
$(\alpha_{1},\dots,\alpha_{n})$ be a sequence of $2$-tuples
$\alpha_{i}=(a_{i},b_{i})\in A(Q)$ (for $i=1,\dots,n$) satisfying $b_{1}=0$.
Then
$\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)=MN(\alpha_{1},\dots,\alpha_{n})=U(Q)$.

###### Proof.

We have

$MN(\alpha_{1},\dots,\alpha_{n})=\left\\{\left(\sum_{i=1}^{n}a_{i}c_{i},\sum_{i=1}^{n}a_{i}d_{i}+\sum_{i=2}^{n}b_{i}c_{i}\right):c_{i},d_{i}\in
Q\cap[0,\infty)\right\\}\text{.}$

It will suffice to show that
$(0,0)\in\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$. Let
$f,g\geq 0$, $f,g\in Q$, so that $f,g\in(0,0)+U(Q)$. Then use the coefficients
$c_{1}=\frac{f}{a_{1}}$, $d_{1}=\frac{g}{a_{1}}$, and $c_{j}=d_{j}=0$ for
$j=2,\dots,n$ to show that $(f,g)\in MN(\alpha_{1},\dots,\alpha_{n})$, so
$(0,0)\in\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$. ∎

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.

To make the following result fit on the page, we will temporarily adopt the
convention that for a tuple $(a,b)$, the notation $(a,b)_{+}$ stands for
$(a,b)+[0,\infty)^{2}$.

######  Proposition 4.3.

Let $Q=\mathbb{R}$, and change $C(\mathbb{R})$ to be
$C=\left\\{(0,0)\right\\}\cup[1,\infty)^{2}$. Let
$\alpha_{1}=(a_{1},0),\alpha_{2}=(a_{2},b_{2})$ form a list
$(\alpha_{1},\alpha_{2})$ of tuples in $A$. If $b_{2}=0$, then
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)=(\min(a_{1},a_{2}),\min(a_{1},a_{2}))_{+}$.
If $b_{2}>0$, then

$\begin{split}&\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)=\\\
&\begin{cases}(a_{1},a_{1})_{+}&a_{1}\leq a_{2}\text{ and }a_{1}\leq b_{2}\\\
\left(a_{1},a_{1}\right)_{+}\cup\left(\frac{a_{1}a_{2}}{b_{2}}+a_{1},b_{2}\right)_{+}&b_{2}<a_{1}\leq
a_{2}\\\
\left(a_{1},a_{2}\right)_{+}\cup\left(a_{2},\frac{a_{1}b_{2}}{a_{2}}+a_{2}\right)_{+}&a_{1}>a_{2}\text{
and }a_{2}\leq b_{2}\\\
\left(a_{1},a_{2}\right)_{+}\cup\left(a_{2},\frac{a_{1}b_{2}}{a_{2}}+a_{2}\right)_{+}\cup\left(\frac{(a_{2})^{2}}{b_{2}}+a_{1},b_{2}\right)_{+}&b_{2}<a_{2}<a_{1}\end{cases}\end{split}$

Simple modifications to the proof of Proposition 4.1 will prove that
Proposition 4.1 holds in this template. Since
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\subseteq\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$,
Proposition 4.3 combines with Proposition 4.1 to prove that if
$\alpha_{i}=(a_{i},b_{i})\in(0,\infty)\times[0,\infty)$ for $i=1,\dots,n$ and
$n>1$, then $\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$ is
nonempty if and only if some $b_{i}=0$.

###### Proof.

Case 0: Suppose $b_{2}=0$, so elements of $MN(\alpha_{1},\alpha_{2})$ look
like $(a_{1}c_{1}+a_{2}c_{2},a_{1}d_{1}+a_{2}d_{2})$ for
$c_{1},c_{2},d_{1},d_{2}\in[0,\infty)\setminus(0,1)$. Notice that
$f\in(0,\min(a_{1},$
$a_{2}))$ or $g\in(0,\min(a_{1},a_{2}))$ implies $(f,g)\notin
MN(\alpha_{1},\alpha_{2})$. Therefore, if
$(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$, then
$t,u\geq\min(a_{1},a_{2})$. The converse is trivial, so this case is done.

The set $MN(\alpha_{1},\alpha_{2})$ is

$\left\\{\left(a_{1}c_{1}+a_{2}c_{2},a_{1}d_{1}+a_{2}d_{2}+b_{2}c_{2}\right):c_{1},c_{2},d_{1},d_{2}\in[0,\infty)\setminus(0,1)\right\\}\text{.}$

The trickiness of the remaining cases is that both coordinates share the
coefficient $c_{2}$. Similar to the last paragraph, notice that
$f\in(0,\min(a_{1},a_{2}))$ or $g\in(0,\min(a_{1},a_{2},b_{2}))$ implies
$(f,g)\notin MN(\alpha_{1},\alpha_{2})$; therefore,
$(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ implies
$t\geq\min(a_{1},a_{2})$ and $u\geq\min(a_{1},a_{2},b_{2})$.

Case 1: $a_{1}\leq a_{2}$ and $a_{1}\leq b_{2}$. By remarks above, in this
case
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\subseteq(a_{1},a_{1})\cup[0,\infty)^{2}$.
On the other hand, if $f,g\geq a_{1}$, then
$(f,g)=(\frac{f}{a_{1}},\frac{g}{a_{1}})\alpha_{1}+(0,0)\alpha_{2}$, so
$(f,g)\in MN(\alpha_{1},\alpha_{2})$. Therefore,
$(a_{1},a_{1})+[0,\infty)^{2}\subseteq\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$,
so
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)=(a_{1},a_{1})+[0,\infty)^{2}$.

Case 2: $0<b_{2}<a_{1}\leq a_{2}$. Then
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\subseteq(a_{1},b_{2})+[0,\infty)^{2}$
and, by the proof in Case 1,

$(a_{1},a_{1})+[0,\infty)^{2}=[a_{1},\infty)^{2}\subseteq\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\text{.}$

Therefore, to determine $\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$
in this case, it is sufficient to determine which
$(t,u)\in[a_{1},\infty)\times[b_{2},a_{1})$ are in
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$.

If $(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\subseteq
MN(\alpha_{1},\alpha_{2})$, then for some
$c_{1},c_{2},d_{1},d_{2}\in\left\\{0\right\\}\cup[1,\infty)$,
$t=a_{1}c_{1}+a_{2}c_{2}$ and $u=a_{1}d_{1}+a_{2}d_{2}+c_{2}b_{2}$. Then
$b_{2}\leq u<a_{1}\leq a_{2}$ implies that $d_{1}=d_{2}=0$ and
$c_{2}=\frac{u}{b_{2}}$. Then we have that
$t=a_{1}c_{1}+a_{2}\frac{u}{b_{2}}$. If $t<a_{1}+a_{2}\frac{u}{b_{2}}$, then
$c_{1}=0$ and $t=a_{2}\frac{u}{b_{2}}$. But this would imply that for any
$t^{\prime}$ such that $t<t^{\prime}<a_{1}+a_{2}\frac{u}{b_{2}}$,
$(t^{\prime},u)\notin MN(\alpha_{1},\alpha_{2})$, and this would contradict
the assumption that
$(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$.

Therefore, if $(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$
and $b_{2}\leq u<a_{1}$, then $t\geq a_{1}+a_{2}\frac{u}{b_{2}}$. But
$(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ implies that
$(t,u^{\prime})\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ for
every $u^{\prime}\geq u$. Therefore, $t\geq
a_{1}+a_{2}\frac{u^{\prime}}{b_{2}}$ for every $u^{\prime}$ satisfying $u\leq
u^{\prime}<a_{1}$. Therefore, $t\geq a_{1}+\frac{a_{1}a_{2}}{b_{2}}$.

Thus

$\displaystyle(a_{1},a_{1})+[0,\infty)^{2}$
$\displaystyle\subseteq\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$
$\displaystyle\subseteq\left((a_{1},a_{1})+[0,\infty)^{2}\right)\cup\left(\bigg{[}\frac{a_{1}a_{2}}{b_{2}}+a_{1},\infty\bigg{)}\times[b_{2},a_{1})\right)$
$\displaystyle\subseteq\left((a_{1},a_{1})+[0,\infty)^{2}\right)\cup\left(\left(\frac{a_{1}a_{2}}{b_{2}}+a_{1},b_{2}\right)+[0,\infty)^{2}\right)\text{,}$

so the proof in this case will be over if we show that
$\left(\frac{a_{1}a_{2}}{b_{2}}+a_{1},b_{2}\right)$ is in
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$. Suppose that
$f\geq\frac{a_{1}a_{2}}{b_{2}}+a_{1}$ and $g\geq b_{2}$. We will see that
$(f,g)\in MN(\alpha_{1},\alpha_{2})$. We may as well assume that $g<a_{1}$.
Then
$(f,g)=\left(\frac{1}{a_{1}}\left(f-\frac{ga_{2}}{b_{2}}\right),0\right)\alpha_{1}+\left(\frac{g}{b_{2}},0\right)\alpha_{2}$,
so $(f,g)\in MN(\alpha_{1},\alpha_{2})$ because $\frac{g}{b_{2}}\geq 1$ and
$\frac{1}{a_{1}}\left(f-\frac{ga_{2}}{b_{2}}\right)\geq\frac{1}{a_{1}}\left(\frac{a_{1}a_{2}}{b_{2}}+a_{1}-\frac{a_{1}a_{2}}{b_{2}}\right)=1$.

Case 3: $a_{2}<a_{1}$ and $a_{2}\leq b_{2}$. By arguments on display above,
$(a_{1},a_{1})+[0,\infty)^{2}\subseteq\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\subseteq(a_{2},a_{2})+[0,\infty)^{2}$.
But it is also easy to see that
$(a_{1},a_{2})\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$
($\implies(a_{1},a_{2})+[0,\infty)^{2}\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$):
if $f\geq a_{1}$ and $g\geq a_{2}$, then
$(f,g)=\left(\frac{f}{a_{1}},0\right)\alpha_{1}+\left(0,\frac{g}{a_{2}}\right)\alpha_{2}\in
MN(\alpha_{1},\alpha_{2})$.

It remains to determine
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\cap\left([a_{2},a_{1})\times[a_{2},\infty)\right)$.
Suppose that $(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ and
$a_{2}\leq t<a_{1}$. Let
$c_{1},c_{2},d_{1},d_{2}\in\left\\{0\right\\}\cup[1,\infty)$ satisfy
$t=a_{1}c_{1}+a_{2}c_{2}$, $u=a_{1}d_{1}+a_{2}d_{2}+c_{2}b_{2}$. Then
$a_{2}\leq t<a_{1}$ implies that $c_{1}=0$ and $c_{2}=\frac{t}{a_{2}}$. If
$u<a_{2}+\frac{tb_{2}}{a_{2}}$ then $d_{1}=d_{2}=0$ and
$u=\frac{tb_{2}}{a_{2}}$. But then if
$u<u^{\prime}<a_{2}+\frac{tb_{2}}{a_{2}}$, $(t,u^{\prime})\notin
MN(\alpha_{1},\alpha_{2})$, contradicting the assumption that
$(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$.

Therefore, $u\geq a_{2}+\frac{tb_{2}}{a_{2}}$. But
$(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ implies that
$(t^{\prime},u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ for all
$t^{\prime}$ such that $t<t^{\prime}<a_{1}$. Therefore, $u\geq
a_{2}+\frac{t^{\prime}b_{2}}{a_{2}}$ for all such $t^{\prime}$. Therefore,
$u\geq a_{2}+\frac{a_{1}b_{2}}{a_{2}}$.

To finish the proof in this case, it will suffice to show that
$[a_{2},a_{1})\times[a_{2}+\frac{a_{1}b_{2}}{a_{2}},\infty)\subseteq\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$,
and for that it will suffice to show that if $a_{2}\leq f<a_{1}$ and
$a_{2}+\frac{a_{1}b_{2}}{a_{2}}\leq g$ then $(f,g)\in
MN(\alpha_{1},\alpha_{2})$. For such $(f,g)$,
$(f,g)=\left(\frac{f}{a_{2}},\frac{1}{a_{2}}\left(g-\frac{fb_{2}}{a_{2}}\right)\right)\alpha_{2}$,
so $(f,g)\in MN(\alpha_{1},\alpha_{2})$ since $\frac{f}{a_{2}}\geq 1$ and
$\frac{1}{a_{2}}\left(g-\frac{fb_{2}}{a_{2}}\right)\geq\frac{1}{a_{2}}\left(a_{2}+\frac{a_{1}b_{2}}{a_{2}}-\frac{a_{1}b_{2}}{a_{2}}\right)=1$.

Case 4: $0<b_{2}<a_{2}<a_{1}$. Clearly
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\subseteq(a_{2},b_{2})+[0,\infty)^{2}$.
If $a_{1}\leq f$ and $a_{2}\leq g$ then
$(f,g)=\left(\frac{f}{a_{1}},0\right)\alpha_{1}+\left(0,\frac{g}{a_{2}}\right)\alpha_{2}$,
so $(f,g)\in MN(\alpha_{1},\alpha_{2})$; therefore,
$(a_{1},a_{2})+[0,\infty)^{2}\subseteq\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$.

Next, we shall show that

$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\cap\left([a_{2},a_{1})\times[b_{2},\infty)\right)=[a_{2},a_{1})\times\bigg{[}a_{2}+\frac{a_{1}b_{2}}{a_{2}},\infty\bigg{)}\text{,}$

by an argument that will be familiar to anyone who has read the proofs in
cases 2 and 3.

Suppose that $(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ and
$a_{2}\leq t<a_{1}$. Then for some
$c_{2},d_{1},d_{2}\in\left\\{0\right\\}\cup[1,\infty)$, $t=c_{2}a_{2}$ and
$u=a_{1}d_{1}+a_{2}d_{2}+\frac{t}{a_{2}}b_{2}$. If
$u<a_{2}+\frac{tb_{2}}{a_{2}}$, then $d_{1}=d_{2}=0$ and
$u=\frac{tb_{2}}{a_{2}}$. But then for all
$u^{\prime}\in\left(u,\frac{tb_{2}}{a_{2}}\right)$, $(t,u^{\prime})\notin
MN(\alpha_{1},\alpha_{2})$, which contradicts the assumption that
$(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$.

Therefore, $u\geq a_{2}+\frac{tb_{2}}{a_{2}}$. But then the fact that
$(t^{\prime},u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ for all
$t^{\prime}$ satisfying $t<t^{\prime}<a_{1}$ implies that $u\geq
a_{2}+\frac{t^{\prime}b_{2}}{a_{2}}$ for all such $t^{\prime}$. Therefore
$u\geq a_{2}+\frac{a_{1}b_{2}}{a_{2}}$, which shows that

$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\cap\left([a_{2},a_{1})\times[b_{2},\infty)\right)\subseteq[a_{2},a_{1})\times\bigg{[}a_{2}+\frac{a_{1}b_{2}}{a_{2}},\infty\bigg{)}\text{.}$

On the other hand, if $a_{2}\leq f<a_{1}$ and $g\geq
a_{2}+\frac{a_{1}b_{2}}{a_{2}}$, then
$(f,g)=\left(\frac{f}{a_{2}},\frac{1}{a_{2}}\left(g-\frac{fb_{2}}{a_{2}}\right)\right)\alpha_{2}$.
From $\frac{f}{a_{2}}\geq 0$ and

$\frac{1}{a_{2}}\left(g-\frac{fb_{2}}{a_{2}}\right)\geq\frac{1}{a_{2}}\left(a_{2}+\frac{a_{1}b_{2}}{a_{2}}-\frac{a_{1}b_{2}}{a_{2}}\right)=1,$

we conclude that $(f,g)\in MN(\alpha_{1},\alpha_{2})$. Thus

$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\cap\left([a_{2},a_{1})\times[b_{2},\infty)\right)=[a_{2},a_{1})\times\bigg{[}a_{2}+\frac{a_{1}b_{2}}{a_{2}},\infty\bigg{)}.$

To finish Case 4, we will determine
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\cap\left([a_{1},\infty)\times[b_{2},a_{2})\right)$.
Suppose that $(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$,
with $t\geq a_{1}$ and $b_{2}\leq u<a_{2}$. Then for some
$c_{1},c_{2}\in\left\\{0\right\\}\cup[1,\infty)$, $t=a_{1}c_{1}+a_{2}c_{2}$
and $u=b_{2}c_{2}$. Then $t=a_{1}c_{1}+\frac{a_{2}u}{b_{2}}$. If
$t<a_{1}+\frac{ua_{2}}{b_{2}}$, then $c_{1}=0$ and $t=\frac{ua_{2}}{b_{2}}$.

Then for all $t^{\prime}$ such that $t<t^{\prime}<a_{1}+\frac{ua_{2}}{b_{2}}$,
$(t^{\prime},u)\notin MN(\alpha_{1},\alpha_{2})$, which contradicts the
assumption that $(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$.
Therefore, $t\geq a_{1}+\frac{ua_{2}}{b_{2}}$. But then
$(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ implies that
$(t,u^{\prime})\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$ for all
$u^{\prime}$ satisfying $u<u^{\prime}<a_{2}$. Therefore, $t\geq
a_{1}+\frac{u^{\prime}a_{2}}{b_{2}}$ for each such $u^{\prime}$. Therefore,
$t\geq a_{1}+\frac{a_{2}^{2}}{b_{2}}$. Thus
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\cap\left([a_{1},\infty)\times[b_{2},a_{2})\right)\subseteq[a_{1}+\frac{a_{2}^{2}}{b_{2}},\infty)\times[b_{2},a_{2})$.
On the other hand, suppose that $a_{1}+\frac{a_{2}^{2}}{b_{2}}\leq f$ and
$b_{2}\leq g<a_{2}$. Then
$(f,g)=\left(\frac{1}{a_{1}}\left(f-\frac{ga_{2}}{b_{2}}\right),0\right)\alpha_{1}+\left(\frac{g}{b_{2}},0\right)\alpha_{2}$.
Thus, since $\frac{g}{b_{2}}\geq 1$ and
$\frac{1}{a_{1}}\left(f-\frac{ga_{2}}{b_{2}}\right)\geq\frac{1}{a_{1}}\left(a_{1}+\frac{a_{2}^{2}}{b_{2}}-\frac{a_{2}^{2}}{b_{2}}\right)=1$,
$(f,g)\in MN(\alpha_{1},\alpha_{2})$. Thus
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)\cap[a_{1},\infty)\times[b_{2},a_{2})=[a_{1}+\frac{a_{2}^{2}}{b_{2}},\infty)\times[b_{2},a_{2})$.
This, together with previous results, proves the claim in Case 4. ∎

We now shift focus to $Q=\mathbb{Z}$. Recall that
$\chi\left(a_{1},\dots,a_{n}\right)$ is well-defined for coprime positive
integers $a_{1},\dots,a_{n}$, with respect to the classical Frobenius template
$(\mathbb{N},\mathbb{N},\mathbb{N})$.

######  Proposition 4.4.

For $n\geq 2$, let $(\alpha_{1},\dots,\alpha_{n})$ be a list of $2$-tuples
$\alpha_{i}=(a_{i},b_{i})\in\mathbb{Z}^{+}\times\mathbb{N}$ (for
$i=1,\dots,n$) satisfying $\gcd(a_{1},\dots,a_{n})=1$ and $b_{1}=0$. Then

$\left(\chi\left(a_{1},\dots,a_{n}\right),\chi\left(a_{1},\dots,a_{n}\right)+(a_{1}-1)\sum_{i=2}^{n}b_{i}\right)+\mathbb{N}^{2}\subseteq\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right).$

###### Proof.

Let $t=\chi\left(a_{1},\dots,a_{n}\right)$ and
$u=\chi\left(a_{1},\dots,a_{n}\right)+(a_{1}-1)(\sum_{j=2}^{n}b_{j})$. It’s
sufficient to show that
$(t,u)\in\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$, i.e.
$(t,u)+\mathbb{N}^{2}\subseteq MN(\alpha_{1},\dots,\alpha_{n})$. Pick an
arbitrary $(f,g)\in(t,u)+\mathbb{N}^{2}$, i.e. choose integers
$f\geq\chi\left(a_{1},\dots,a_{n}\right)$ and
$g\geq\chi\left(a_{1},\dots,a_{n}\right)+\sum_{j=2}^{n}b_{j}(a_{1}-1)$. To
complete the proof, we will show that

$(f,g)\in
MN(\alpha_{1},\dots,\alpha_{n})=\left\\{\left(\sum_{i=1}^{n}a_{i}c_{i},\sum_{i=1}^{n}a_{i}d_{i}+\sum_{i=2}^{n}b_{i}c_{i}\right):c_{i},d_{i}\in\mathbb{N}\right\\}.$

Since $f\geq\chi\left(a_{1},\dots,a_{n}\right)$, we can write
$f=\sum_{i=1}^{n}a_{i}c_{i}$ for some nonnegative integers
$c_{1},\dots,c_{n}$. For each $i\in\left\\{2,\dots,n\right\\}$, use Euclidean
division to write $c_{i}=a_{1}q_{i}+\tilde{c}_{i}$ for some new nonnegative
integers $q_{i}$ and $\tilde{c}_{i}\leq(a_{1}-1)$, then set
$\tilde{c}_{1}=c_{1}+\sum_{i=2}^{n}a_{i}q_{i}$ so that
$\sum_{i=1}^{n}a_{i}c_{i}=f=\sum_{i=1}^{n}a_{i}\tilde{c}_{i}$. Can we find
nonnegative $d_{1},\dots,d_{n}$ such that
$g=\sum_{i=1}^{n}a_{i}d_{i}+\sum_{i=2}^{n}b_{i}\tilde{c}_{i}$? Well,
$g-\sum_{i=2}^{n}b_{i}\tilde{c}_{i}\geq
g-\sum_{i=2}^{n}b_{i}(a_{1}-1)\geq\chi\left(a_{1},\dots,a_{n}\right)$, so yes.
Therefore, $(f,g)\in MN(\alpha_{1},\dots,\alpha_{n})$. ∎

######  Corollary 4.4.1.

For $n\geq 2$, let $(\alpha_{1},\dots,\alpha_{n})$ be a list of $2$-tuples
$\alpha_{i}=(a_{i},b_{i})\in\mathbb{Z}^{+}\times\mathbb{N}$ (for
$i=1,\dots,n$) satisfying $\gcd(a_{1},\dots,a_{n})=1$.
$\textnormal{Frob}\left(\alpha_{1},\dots,\alpha_{n}\right)$ is nonempty if and
only if at least one $b_{i}=0$.

###### Proof.

Combine propositions 4.1 and 4.4. ∎

We have seen results like this before, such as Theorem 2.1 of Section 2 and
the paragraph preceding the proof of Proposition 4.3. Combining this corollary
with the next result completely solves the Frobenius problem in the case $n=2$
and $Q=\mathbb{Z}$.

######  Proposition 4.5.

Suppose that $a_{1},a_{2}\in\mathbb{Z}^{+},b\in\mathbb{N}$, $a_{1}$ and
$a_{2}$ are coprime, and $\alpha_{1}=(a_{1},0),\alpha_{2}=(a_{2},b)$. Then

$\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)=\left(\chi\left(a_{1},a_{2}\right),\chi\left(a_{1},a_{2}\right)+b(a_{1}-1)\right)+\mathbb{N}^{2}.$

###### Proof.

Let $(t,u)\in\mathbb{N}\times\mathbb{N}$. By Proposition 4.4,
$t\geq\chi\left(a_{1},a_{2}\right)$ and
$u\geq\chi\left(a_{1},a_{2}\right)+b(a_{1}-1)$ implies that
$(t,u)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$. For the
converse, we will prove the contrapositive. Suppose that
$t<\chi\left(a_{1},a_{2}\right)$. Then $\chi\left(a_{1},a_{2}\right)-1\geq t$
is not in $\left\\{a_{1}c_{1}+a_{2}c_{2}:c_{1},c_{2}\in\mathbb{N}\right\\}$,
so $(\chi\left(a_{1},a_{2}\right)-1,u)$ is not in
$\\{(a_{1}c_{1}+a_{2}c_{2},a_{1}d_{1}+a_{2}d_{2}+bc_{2}):c_{1},c_{2},d_{1},d_{2}\in\mathbb{N}\\}=MN(\alpha_{1},\alpha_{2})$,
yet $(\chi\left(a_{1},a_{2}\right)-1,u)\in(t,u)+\mathbb{N}^{2}$. Hence
$(t,u)\notin\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$.

Now assume that $u<\chi\left(a_{1},a_{2}\right)+b(a_{1}-1)$. Set
$f=a_{1}c_{1}+a_{2}(a_{1}-1)$ for some nonnegative $c_{1}$ large enough such
that $f\geq t$. Consider any alternative expression of $f$ as
$f=a_{1}\tilde{c}_{1}+a_{2}\tilde{c}_{2}$ using nonnegative coefficients
$\tilde{c}_{1},\tilde{c}_{2}$. Recall that $f=a_{1}c_{1}+a_{2}(a_{1}-1)$, so
$a_{1}(c_{1}-\tilde{c}_{1})=a_{2}(\tilde{c}_{2}-(a_{1}-1))$. Then
$a_{1}|(\tilde{c}_{2}-(a_{1}-1))$ since $a_{1}$ and $a_{2}$ are coprime, so
$\tilde{c}_{2}\equiv a_{1}-1\pmod{a_{1}}$. And $\tilde{c}_{2}$ is nonnegative,
so $\tilde{c}_{2}\geq(a_{1}-1)$, so $\tilde{c}_{2}=(a_{1}-1)+a_{1}k$ for some
nonnegative integer $k$. Set $g=\chi\left(a_{1},a_{2}\right)+b(a_{1}-1)-1$ so
that $g\geq u$; hence $(f,g)\in(t,u)+\mathbb{N}^{2}$. Suppose that $(f,g)\in
MN(\alpha_{1},\alpha_{2})$; there are nonnegative coefficients $d_{1}$ and
$d_{2}$ such that
$g=a_{1}d_{1}+a_{2}d_{2}+b\tilde{c}_{2}=a_{1}d_{1}+a_{2}d_{2}+b((a_{1}-1)+a_{1}k)$.
Consequently, $g-b(a_{1}-1)=a_{1}(d_{1}+bk)+a_{2}d_{2}$ for nonnegative
integers $(d_{1}+bk)$ and $d_{2}$, so $g-b(a_{1}-1)\in MN(a_{1},a_{2})$. But
this is impossible, because $g-b(a_{1}-1)=\chi\left(a_{1},a_{2}\right)-1\notin
MN(a_{1},a_{2})$, so our supposition must be false; $(f,g)\notin
MN(\alpha_{1},\alpha_{2})$. Therefore,
$(t,u)\notin\textnormal{Frob}\left(\alpha_{1},\alpha_{2}\right)$. ∎

Proposition 4.5 shows that when $n=2$, the set inclusion in the conclusion of
Proposition 4.4 is an equality. However, for lists of length $>2$, there are
counterexamples to the reverse set inclusion of Proposition 4.4, so
Proposition 4.5 does not generalize to longer lists of tuples. The following
is a counterexample: Let $\alpha_{1}=(3,0)$, $\alpha_{2}=(5,2)$, and
$\alpha_{3}=(7,4)$. Then it can be shown that
$(5,16)\in\textnormal{Frob}\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)+\mathbb{N}^{2}$,
but
$(5,16)\notin\\{\chi\left(a_{1},\dots,a_{n}\right)+(a_{1}-1)\left(\sum_{i=2}^{n}b_{i}\right)\\}+\mathbb{N}^{2}$.
In fact, it can be shown that
$\textnormal{Frob}\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)=(5,9)+\mathbb{N}^{2}$.
Furthermore, for $\beta_{1}=(3,0)$, $\beta_{2}=(5,1)$, and $\beta_{3}=(7,4)$
we have

$\textnormal{Frob}\left(\beta_{1},\beta_{2},\beta_{3}\right)=\left((5,9)+\mathbb{N}^{2}\right)\cup\left((8,7)+\mathbb{N}^{2}\right)\text{,}$

so the Frobenius set might even be a union of two sets, each of the form
$(a,b)+\mathbb{N}^{2}$, neither contained in the other. This situation
resembles that of the classical template, in the sense that the Frobenius
problem is completely solved for lists of length $2$, but not for lists of
length $>2$.

##  5 A more general template

Here we broaden our horizons and pass from the templates $(A,C,U)$ where $A$
and $C$ are thought of as subsets of the same overlying ring, to templates
$(A^{\prime},C^{\prime},U^{\prime})$ where $A^{\prime}$ is a monoid and
$C^{\prime}$ is a set of functions acting on $A^{\prime}$. The first kind of
template, i.e. the only kind hitherto discussed in this paper, can be
considered a special case of the second; furthermore, templates of the second
kind cannot in general be interpreted as examples of the first kind. The
different entries in the new kind of template have the same roles as the
corresponding entries in the original kind of template.

For the sake of presentation, we will showcase a certain example of this new
kind of template and leave the precise definitions to the reader. Let
$A=U=\mathbb{N}\times\mathbb{N}\times\mathbb{N},$ and let $C$ be the set of
upper triangular matrices in $M_{3}(\mathbb{N})$. In the ring $\mathbb{Z}^{3}$
with coordinate addition and multiplication, $A=U=\mathbb{N}^{3}$ is a monoid.
However, $C$ is not contained in $\mathbb{Z}^{3},$ so this template is
different from those considered previously. Note that $C$ contains an
isomorphic copy of $\mathbb{N}^{3}$ via the semiring embedding given by

$\begin{bmatrix}x\\\ y\\\ z\\\ \end{bmatrix}\mapsto\begin{bmatrix}x&0&0\\\
0&y&0\\\ 0&0&z\\\ \end{bmatrix}.$

Consider a general pair of tuples $(a,b,c)^{T}$ and $(d,e,f)^{T}\in A$. A
general member of

$MN\left(\begin{bmatrix}a\\\ b\\\ c\\\ \end{bmatrix},\begin{bmatrix}d\\\ e\\\
f\\\ \end{bmatrix}\right)$

has the form

$\begin{bmatrix}u&v&w\\\ 0&x&y\\\ 0&0&z\\\ \end{bmatrix}\begin{bmatrix}a\\\
b\\\ c\\\ \end{bmatrix}+\begin{bmatrix}u^{\prime}&v^{\prime}&w^{\prime}\\\
0&x^{\prime}&y^{\prime}\\\ 0&0&z^{\prime}\\\ \end{bmatrix}\begin{bmatrix}d\\\
e\\\ f\\\ \end{bmatrix}\text{,}$

with all entries coming from $\mathbb{N}$. As expected,
$\textnormal{Frob}\left((a,b,c)^{T},(d,e,f)^{T}\right)$ is defined to be

$\left\\{\mathbf{w}\in MN\left(\begin{bmatrix}a\\\ b\\\ c\\\
\end{bmatrix},\begin{bmatrix}d\\\ e\\\ f\\\
\end{bmatrix}\right):\mathbf{w}+\mathbb{N}^{3}\subseteq
MN\left(\begin{bmatrix}a\\\ b\\\ c\\\ \end{bmatrix},\begin{bmatrix}d\\\ e\\\
f\\\ \end{bmatrix}\right)\right\\}\text{.}$

Hence, by our understanding of the classical Frobenius problem, we see that

$\textnormal{Frob}\left(\begin{bmatrix}a\\\ b\\\ c\\\
\end{bmatrix},\begin{bmatrix}d\\\ e\\\ f\\\
\end{bmatrix}\right)=\begin{bmatrix}\chi(a,b,c,d,e,f)\\\ \chi(b,c,e,f)\\\
\chi(c,f)\\\ \end{bmatrix}+\mathbb{N}^{3},$

when nonempty, which is true if and only if $\gcd(c,f)=1$.

From the case $k=2$ it is straightforward to see what
$\textnormal{Frob}\left(\alpha_{1},...,\alpha_{k}\right)$ is for arbitrary
$k\in\mathbb{Z}^{+}$ and $\alpha_{1},...,\alpha_{k}\in A$. It is not difficult
to generalize these results to $m\times 1$ column vectors and $m\times m$
matrices for $m>3$. Therefore these cases are no longer interesting, except
for the connection between them and the classical Frobenius problem. We can
get more challenging problems by restricting the matrices in $C$. For
instance, we could require the matrices to be symmetric, or upper triangular
with constant diagonal.

These examples point to a generalized Frobenius template
$(A^{\prime},C^{\prime},U^{\prime})$ in which $A^{\prime}$ (or perhaps
$A^{\prime}\cup\left\\{0\right\\}$) and $U^{\prime}$ are monoids in a ring
$R$, and $C^{\prime}$ is a monoid in the ring of endomorphisms of $(R,+)$ such
that each $\varphi\in C$ maps $A$ into $U.$

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