Membership and elasticity in certain affine monoids

TL;DR

This paper investigates the membership problem and elasticity of elements in specific affine monoids of dimension 2 with low embedding dimensions, providing insights into their algebraic structure.

Contribution

It offers new criteria for membership and elasticity calculations in affine monoids of dimension 2 with embedding dimensions 2 and 3.

Findings

Characterization of membership conditions in these monoids

Formulas or methods for computing elasticity

Enhanced understanding of the structure of low-dimensional affine monoids

Abstract

For affine monoids of dimension 2 with embedding dimension 2 and 3, we study the problem of determining when a vector is an element of the monoid, and the problem of determining the elasticity of a monoid element.

Equations

$${\pi }_{\mathcal{A}}:{\mathbb{N}}_0^p⟶S;\mathbf{u}=(u_1,\dots ,u_p)⟼\sum _{i=1}^p{u}_i{\mathbf{a}}_i$$

$$\varphi (\left[\begin{array}{c}a\\ b\end{array}\right])<\varphi (\left[\begin{array}{c}a+c\\ b+d\end{array}\right])<\varphi (\left[\begin{array}{c}c\\ d\end{array}\right]).$$

$$bc-ad=1$$

$$s=(\delta -t)u+(\alpha +ct)v+(\beta -at)w$$

$$t=\{\begin{array}{ll}\lfloor \frac{\beta }{a}\rfloor & \frac{x}{y}\le \frac{a}{b}\\ \delta & \frac{x}{y}\ge \frac{a}{b}\end{array}.$$

$$\{\begin{array}{ll}\delta +\alpha +\beta +\lfloor \frac{\beta }{a}\rfloor (c-a-1)& \frac{x}{y}\le \frac{a}{b}\\ \delta +\alpha +\beta +\delta (c-a-1)& \frac{x}{y}\ge \frac{a}{b}\end{array},$$

$$\text{sign}(z)=\{\begin{array}{ll}1& z>0\\ 0& z=0\\ -1& z<0\end{array}.$$

$$\rho (ks)={\left(\frac{c}{a}\frac{ya-x(b-1)}{yc-x(d-1)}\right)}^{\tau }.$$

$$\rho (ks)={\left(c\frac{y(c-a)-x(d-b)}{yc-x(d-1)}\right)}^{\tau }.$$

$$\underset{k\to \infty }{\lim }\rho (ks)=\{\begin{array}{ll}{\left(\frac{c}{a}\frac{ya-x(b-1)}{yc-x(d-1)}\right)}^{\tau }& \frac{x}{y}\le \frac{a}{b}\\ {\left(c\frac{y(c-a)-x(d-b)}{yc-x(d-1)}\right)}^{\tau }& \frac{x}{y}\ge \frac{a}{b}\end{array}.$$

Full text

1 Introduction

Let $\mathbb{N}$ denote the set of positive integers, $\mathbb{N}_{0}$ denote the set of nonnegative integers, and $\mathbb{Q}^{\star}$ denote the set of nonnegative rational numbers adjoined with $+\infty$. An affine monoid, $S$, is a finitely generated submonoid of $\mathbb{N}_{0}^{r}$, with operation $+$, for some positive integer $r$. They are of substantial interest (see, e.g., [4, 8, 11]). In the remainder, we restrict to the case $r=2$. Any affine monoid is cancellative ($\mathbf{a}+\mathbf{b}=\mathbf{a}+\mathbf{c}$ implies $\mathbf{b}=\mathbf{c}$), reduced (its only unit is [math], the identity element), and torsion free ($k\mathbf{a}=k\mathbf{b}$ for $k\in\mathbb{N}$ implies $\mathbf{a}=\mathbf{b}$). Let $S$ be an affine monoid minimally generated by $\mathcal{A}:=\{\mathbf{a}_{1},\ldots,\mathbf{a}_{p}\}\subset\mathbb{N}_{0}^{r}$, that is to say $S=\langle\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{p}\rangle:=\mathbb{N}_{0}\mathbf{a}_{1}+\cdots+\mathbb{N}_{0}\mathbf{a}_{p}$ and no proper subset of $\mathcal{A}$ generates $S$. We say the embedding dimension of $S$ is $p$. For a general introduction to monoids and their invariants, see [5].

The monoid map

[TABLE]

is sometimes known as the factorization homomorphism associated to $\mathcal{A}$, and if $\pi_{\mathcal{A}}(\mathbf{u})=s$, $\mathbf{u}$ is called a factorization of $s$. For every $s\in S$, the set $\mathsf{Z}(s):=\pi_{\mathcal{A}}^{-1}(s)$ is called the set of factorizations of $s$. Given $s\in S$, for $\mathbf{u}=(u_{1},\ldots,u_{p})\in\mathsf{Z}(s)$, define the length of the factorization $\mathbf{u}$, to be $|\mathbf{u}|=u_{1}+\cdots+u_{p}$, and define the set of lengths of $s$ as $\mathsf{L}(s)=\{|\mathbf{u}|:\mathbf{u}\in\mathsf{Z}(\mathbf{a})\}$. Define the elasticity of $s\in S$ as $\rho(s)=\frac{\max(\mathsf{L}(s))}{\min(\mathsf{L}(s))}$, and the elasticity of $S$ to be $\rho(S)=\sup\{\rho(s):s\in S\setminus\{0\}\}$. The elasticity is a very important monoid invariant (see, e.g., [2, 3, 6, 7]).

The monoid elasticity $\rho(S)$ for affine monoids is known (see, e.g., [9]). In this note, our main tool will be the function $\phi:\mathbb{Z}^{2}\to\mathbb{Q}^{\star}$ given by $\phi:\left[\begin{smallmatrix}a\\ b\end{smallmatrix}\right]\mapsto\frac{a}{b}$, with $\frac{a}{0}$ conventionally taken to be $+\infty$. Our main focus will be $S\subseteq\mathbb{N}_{0}^{2}$, with embedding dimension $2$ and $3$.

We will compute the elasticity of individual monoid elements. We also provide membership tests for arbitrary elements of $\mathbb{N}_{0}^{2}$. We will show that for a given $s\in\mathbb{N}_{0}^{2}$, membership in $S$ and $\rho(s)$ are largely determined by $\phi(s)$.

2 Preliminaries

We begin with the observation that $\mathbb{Q}^{\star}$ is ordered, and the semigroup operation (commonly known as the mediant) preserves this order. This property is well-known; its proof is included for completeness.

Lemma 1**.**

Let $a,b,c,d\in\mathbb{N}_{0}$ with $\phi(\left[\begin{smallmatrix}a\\ b\end{smallmatrix}\right])<\phi(\left[\begin{smallmatrix}c\\ d\end{smallmatrix}\right])$. Then

[TABLE]

Proof: We prove only the nontrivial case $bd\neq 0$. Then $ad<bc$ by hypothesis. If we add $ab$ to both sides and divide by $b(b+d)$, we conclude $\frac{a}{b}<\frac{a+c}{b+d}$ which gives the first inequality. If we instead add $cd$ to both sides and divide by $d(b+d)$, we get the second inequality.   QED

Corollary 2**.**

Let $u,v\in\mathbb{N}_{0}^{2}$ with $\phi(u)<\phi(v)$. Let $s\in\langle u,v\rangle$. Then $\phi(u)\leq\phi(s)\leq\phi(v)$.

Proof: Strict inequality is lost if $s=u+u$ or similar.  QED

Let $GL(2)$ denote the set of $2\times 2$ unimodular matrices (i.e. with determinant $\pm 1$), with entries from $\mathbb{Z}$. Let $\left[\begin{smallmatrix}u&v\end{smallmatrix}\right]$ denote the $2\times 2$ matrix whose first column is $u$, and whose second column is $v$. Let $\left[\begin{smallmatrix}\mathcal{A}\end{smallmatrix}\right]$ denote a similar matrix whose columns are the monoid generators.

Corollary 3**.**

Let $u,v\in\mathbb{N}_{0}^{2}$ with $\phi(u)<\phi(v)$. Let $s\in\langle u,v\rangle$. Let $A\in GL(2)$. Suppose that $Au,Av\in\mathbb{N}_{0}^{2}$. Then $As\in\langle Au,Av\rangle$, and either $\phi(Au)\leq\phi(As)\leq\phi(Av)$ or $\phi(Av)\leq\phi(As)\leq\phi(Au)$.

Proof: Since $s\in\langle u,v\rangle$, there is some vector $w$ with $\left[\begin{smallmatrix}u&v\end{smallmatrix}\right]\left[\begin{smallmatrix}w\end{smallmatrix}\right]=\left[\begin{smallmatrix}s\end{smallmatrix}\right]$. Then $A\left[\begin{smallmatrix}u&v\end{smallmatrix}\right]\left[\begin{smallmatrix}w\end{smallmatrix}\right]=A\left[\begin{smallmatrix}s\end{smallmatrix}\right]$, hence $\left[\begin{smallmatrix}Au&Av\end{smallmatrix}\right]\left[\begin{smallmatrix}w\end{smallmatrix}\right]=\left[\begin{smallmatrix}As\end{smallmatrix}\right]$. Hence $As\in\langle Au,Av\rangle$. We apply Corollary 2 in one of two ways, depending on whether $\phi(Au)\leq\phi(Av)$ or $\phi(Au)\geq\phi(Av)$.   QED

Given some $u=\left[\begin{smallmatrix}a\\ b\end{smallmatrix}\right]\in\mathbb{N}_{0}^{2}$, we say that it is $\phi$-minimal if $\gcd(a,b)=1$; otherwise we could take a smaller $u^{\prime}=\left[\begin{smallmatrix}a/\gcd(a,b)\\ b/\gcd(a,b)\end{smallmatrix}\right]$ with $\phi(u)=\phi(u^{\prime})$. Henceforth we assume that all of our monoid generators are $\phi$-minimal.

Lemma 4**.**

Let $u=\left[\begin{smallmatrix}a\\ b\end{smallmatrix}\right],v=\left[\begin{smallmatrix}c\\ d\end{smallmatrix}\right]\in\mathbb{N}_{0}^{2}$. Suppose that both are $\phi$-minimal and $\phi(u)=\phi(v)$. Then $u=v$.

Proof: If $bd=0$, then $b=d=0$ and $a=c=1$; hence, $u=v$. Otherwise $ad=bc$. Since $\gcd(a,b)=1$, $a|c$. Since $\gcd(c,d)=1$, $c|a$. Since $a,c\in\mathbb{N}_{0}$, $a=c$. Similarly, $b=d$.  QED

Since all monoid generators are distinct, by Lemma 4, they must also have distinct $\phi$-values. Henceforth, we may assume, without loss of generality, that our monoid generators are given in strictly increasing $\phi$ order.

We now recall Hermite Normal Form, an analog of row echelon form for matrices over non-fields like $\mathbb{Z}$. For every rectangular matrix $M$ with integer entries, there is an associated square unimodular matrix $U$ such that $UM$ is (a) upper triangular; and (b) the pivot in each nonzero row is strictly to the right of the previous row; and (c) all entries of $M$ are nonnegative integers. For an introduction to these and other properties of HNF, see [1].

Now, for $M=\left[\begin{smallmatrix}u&v\end{smallmatrix}\right]$, applying HNF we have the first column of $UM$ as $\left[\begin{smallmatrix}g\\ 0\end{smallmatrix}\right]$, where $g$ is the $\gcd$ of the entries of $u$. Since $u$ is $\phi$-minimal, $g=1$. Hence, we have $UM=\left[\begin{smallmatrix}1&b\\ 0&a\end{smallmatrix}\right]$, with $a,b\in\mathbb{N}_{0}$. We now consider a row-swapped HNF, defined as $U^{\prime}=\left[\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix}\right]U$, so $U^{\prime}M=\left[\begin{smallmatrix}0&a\\ 1&b\end{smallmatrix}\right]$. Note that $U^{\prime}u,U^{\prime}v\in\mathbb{N}_{0}^{2}$, so by Corollary 3, if $s\in\langle u,v\rangle$ then $\phi(U^{\prime}u)\leq\phi(U^{\prime}s)\leq\phi(U^{\prime}v)$. Further, note that $0=\phi(U^{\prime}u)$ and $\phi(U^{\prime}v)>0$. Henceforth we will assume without loss of generality that our first generator is $\left[\begin{smallmatrix}0\\ 1\end{smallmatrix}\right]$.

We now recall Smith Normal Form, a non-field analog of the linear algebra theorem giving invertible $U,V$ with $UMV=\left[\begin{smallmatrix}I&0\\ 0&0\end{smallmatrix}\right]$, a block matrix. For any rectangular matrix $M$ with integer entries, there are associated square unimodular matrices $U,V$ such that $UMV=\left[\begin{smallmatrix}D&0\\ 0&0\end{smallmatrix}\right]$, where $D=diag(d_{1},d_{1}d_{2},\ldots,d_{1}d_{2}\cdots d_{k})$. Of particular interest to us are the $d_{i}$, the so-called determinantal divisors of $M$, which satisfy that $d_{i}$ is the gcd of all the $i\times i$ minors of $M$. For example, $d_{1}(M)$ is the gcd of all the entries of $M$.

The determinantal divisors of $M$ are not disturbed upon multiplication (on either side) by any unimodular matrix. Further, they are not disturbed by appending a column that is a $\mathbb{Z}$-linear combination of the other columns. For an introduction to these and other properties of SNF, see [10] or [1].

Given a single generator $u$, because we have assumed it is $\phi$-minimal, the determinantal divisor $d_{1}(\left[\begin{smallmatrix}u\end{smallmatrix}\right])=1$. Consequently, for any invertible $U^{\prime}$, we must have $d_{1}(\left[\begin{smallmatrix}U^{\prime}u\end{smallmatrix}\right])=1$. In particular, applying our row-swapped HNF preserves $\phi$-minimality.

We provide our first membership test for our affine monoid, of arbitrary embedding dimension.

Lemma 5**.**

Let $S=\langle\mathcal{A}\rangle$, and let $v\in\mathbb{N}_{0}^{2}$. Set $M=\left[\begin{smallmatrix}\mathcal{A}\end{smallmatrix}\right]$ and $M^{\prime}=\left[\begin{smallmatrix}\mathcal{A}&v\end{smallmatrix}\right]$. If $d_{2}(M)\neq d_{2}(M^{\prime})$, then $v\notin S$.

Proof: If $v\in S$, then removing the last column of $M^{\prime}$ (which gives $M$) will not change the determinantal divisors.  QED

3 Embedding Dimension 2

In this section, we fix the case of $S=\langle u,v\rangle$, with $u=\left[\begin{smallmatrix}0\\ 1\end{smallmatrix}\right],v=\left[\begin{smallmatrix}a\\ b\end{smallmatrix}\right]$, and $\gcd(a,b)=1$. Note that $d_{2}(\left[\begin{smallmatrix}u&v\end{smallmatrix}\right])=a$. Consider some $s=\left[\begin{smallmatrix}x\\ y\end{smallmatrix}\right]\in\mathbb{N}_{0}^{2}$. We have proved that if $s\in S$, then $0\leq\phi(s)\leq\frac{a}{b}$, and that $d_{2}(\left[\begin{smallmatrix}u&v&s\end{smallmatrix}\right])=d_{2}(\left[\begin{smallmatrix}u&v\end{smallmatrix}\right])=a$. It turns out that these two necessary conditions for membership are sufficient.

Theorem 6**.**

With notation as above, $s\in S$ if and only if both of the following hold:

$0\leq\frac{x}{y}\leq\frac{a}{b}$; and 2. 2.

$a|x$.

Further, if $s\in S$, then $\rho(s)=1$.

Proof: Suppose first that $s\in\langle u,v\rangle$. By Corollary 2, $\phi(u)\leq\phi(s)\leq\phi(v)$. Note that $d_{2}(\left[\begin{smallmatrix}0&a\\ 1&b\end{smallmatrix}\right])=a$, as the $2\times 2$ minor is $-a$. Note also that one of the $2\times 2$ minors of $\left[\begin{smallmatrix}\mathcal{A}&s\end{smallmatrix}\right]$ has determinant $-x$, so we must have $a|x$.

Suppose now that the two conditions hold, i.e. there is some $k\in\mathbb{N}_{0}$ with $x=ka$. If $k=0$, then $s=y\left[\begin{smallmatrix}0\\ 1\end{smallmatrix}\right]$. No other factorization is possible, as even one copy of $v$ will disturb the [math].

Otherwise, since $\frac{x}{y}\leq\frac{ka}{kb}=\frac{x}{kb}$, we must have $y\geq kb$. Hence we may write $\left[\begin{smallmatrix}x\\ y\end{smallmatrix}\right]=k\left[\begin{smallmatrix}a\\ b\end{smallmatrix}\right]+(y-kb)\left[\begin{smallmatrix}0\\ 1\end{smallmatrix}\right]$, which proves $s\in\langle u,v\rangle$. No other factorization is possible, by a back-substitution-type argument: $u$ does not affect the first coordinate, so we must have $k$ copies of $v$ and hence $y-kb$ copies of $u$.   QED

This provides an alternate proof of the well-known fact that in embedding dimension 2, $\rho(S)=1$.

4 Embedding Dimension 3

We turn now to the case of embedding dimension $3$. Henceforth, we fix the case of $S=\langle u,v,w\rangle$, with $u=\left[\begin{smallmatrix}0\\ 1\end{smallmatrix}\right],v=\left[\begin{smallmatrix}a\\ b\end{smallmatrix}\right],w=\left[\begin{smallmatrix}c\\ d\end{smallmatrix}\right]$, $\phi(u)<\phi(v)<\phi(w)$, and $\gcd(a,b)=1=\gcd(c,d)$. Set $M=\left[\begin{smallmatrix}0&a&c\\ 1&b&d\end{smallmatrix}\right]$. We will also fix $s=\left[\begin{smallmatrix}x\\ y\end{smallmatrix}\right]\in\mathbb{N}_{0}^{2}$.

We first offer a simple way to compute the determinantal divisor $d_{2}$ below.

Lemma 7**.**

With notation as above, $d_{2}(M)=\gcd(a,c)$.

Proof: Since $\gcd(a,c)$ divides each entry of the first row of each $2\times 2$ submatrix, it divides each minor. Hence $\gcd(a,c)|d_{2}(M)$. Considering the submatrices $\left[\begin{smallmatrix}0&a\\ 1&b\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}0&c\\ 1&d\end{smallmatrix}\right]$, we find that $d_{2}(M)$ divides each of $a,c$. Hence $d_{2}(M)|\gcd(a,c)$.   QED

Similarly to the embedding dimension 2 case, if $s\in S$, we must have $0\leq\phi(s)\leq\frac{c}{d}$, and $d_{2}(\left[\begin{smallmatrix}M&s\end{smallmatrix}\right])=d_{2}(\left[\begin{smallmatrix}M\end{smallmatrix}\right])=\gcd(a,c)$. Further, we must have $x\in\langle a,c\rangle$, since only $v,w$ have nonzero first coordinates to contribute to $x$. Unfortunately, in general these necessary conditions are not sufficient, as the following example demonstrates.

Example 8**.**

Consider $u=\left[\begin{smallmatrix}0\\ 1\end{smallmatrix}\right],v=\left[\begin{smallmatrix}11\\ 10\end{smallmatrix}\right],w=\left[\begin{smallmatrix}10\\ 3\end{smallmatrix}\right],s=\left[\begin{smallmatrix}199\\ 119\end{smallmatrix}\right]$. Note that $\phi(s)<2<\phi(w)$, and that $d_{2}(\left[\begin{smallmatrix}M&s\end{smallmatrix}\right])=d_{2}(\left[\begin{smallmatrix}M\end{smallmatrix}\right])=1$. $199$ can be factored (uniquely) in $\langle 11,10\rangle$ as $199=9\cdot 11+10\cdot 10$. However, $9v+10w=\left[\begin{smallmatrix}199\\ 120\end{smallmatrix}\right]$. Including $u$’s will not help, so $s\notin S$.

If $x\in\langle a,c\rangle$, then we can impose a restriction on its representation, as follows.

Proposition 9**.**

Let $a,c\in\mathbb{N}$ with $\gcd(a,c)=1$. If $x\in\langle a,c\rangle$, then there are $\alpha,\beta\in\mathbb{N}_{0}$ with $x=\alpha a+\beta c$ and $0\leq\alpha<c$.

Proof: Since $x\in\langle a,c\rangle$, there are some $\alpha^{\prime},\beta^{\prime}\in\mathbb{N}_{0}$ with $x=\alpha^{\prime}a+\beta^{\prime}b$. But also $x=(\alpha^{\prime}-tc)a+(\beta^{\prime}+ta)c$ for all integer $t$. Choose $t\geq 0$ maximal with $\alpha^{\prime}-tc\geq 0$, set $\alpha=\alpha^{\prime}-tc,\beta=\beta^{\prime}+ta$, and observe that $0\leq\alpha<c$.  QED

We will frequently use the canonical factorization of $x$ in $\langle a,c\rangle$ from Proposition 9, which we call $\alpha(x),\beta(x)$.

Despite the setback of Example 8, with an additional restriction, we can solve the membership problem. Henceforth, we add the following standing hypothesis.

[TABLE]

Note that $(\star)$ implies that $1=\gcd(a,b)=\gcd(a,c)=\gcd(b,d)=\gcd(c,d)=1$. Hence, condition $(\star)$ alone implies $\phi$-minimality on $v,w$, and also $d_{2}(M)=1$.

Theorem 10**.**

With notation as above, $s\in S$ if and only if both

$0\leq\frac{x}{y}\leq\frac{c}{d}$; and 2. 2.

$x\in\langle a,c\rangle$.

Proof: If $s\in S$, both conditions are easily seen to hold.

Suppose now that the two conditions hold. Take $\alpha,\beta$ as in Proposition 9. We now prove that $y\geq\alpha b+\beta d$. Supposing otherwise, we have $y\leq\alpha b+\beta d-1$. Since $\alpha<c$, $-\alpha>-c$, and hence $(ad-bc)\alpha>-c$. Adding $\beta cd$ to both sides, with a bit of algebra we get $\alpha ad+\beta cd>\alpha bc+\beta cd-c$, or $\frac{\alpha a+\beta c}{\alpha b+\beta d-1}>\frac{c}{d}$. But then $\frac{x}{y}>\frac{c}{d}$, which contradicts hypothesis. Hence $y\geq\alpha b+\beta d$. Then we write $s=(y-\alpha b-\beta d)\left[\begin{smallmatrix}0\\ 1\end{smallmatrix}\right]+\alpha\left[\begin{smallmatrix}a\\ b\end{smallmatrix}\right]+\beta\left[\begin{smallmatrix}c\\ d\end{smallmatrix}\right]$, and hence $s\in S$.  QED

We turn now to the elasticity problem. The different factorizations of $s$ in $S$ all come from different factorizations of $x$ in $\langle a,c\rangle$, by the following.

Lemma 11**.**

With notation as above, given $\alpha^{\prime},\beta^{\prime}\in\mathbb{N}_{0}$ with $x=\alpha^{\prime}a+\beta^{\prime}c$, there is exactly one $\delta=\delta(\alpha^{\prime},\beta^{\prime})\in\mathbb{Z}$ with $s=\delta u+\alpha^{\prime}v+\beta^{\prime}w$.

Proof: If $s=\delta u+\alpha^{\prime}v+\beta^{\prime}w$, then $y=\delta+\alpha^{\prime}b+\beta^{\prime}d$. We solve for $\delta$ uniquely. If $\delta\geq 0$, then $s=\delta u+\alpha^{\prime}v+\beta^{\prime}w$ is a factorization of $s$ in $S$.   QED

Henceforth, we define function $\delta(\alpha,\beta)$, applying Lemma 11 to the factorization from Proposition 9.

We call a factorization of $s$ extreme if it is either of minimal or maximal length. The extreme factorizations are given in the following theorem; there are two cases based on whether $\frac{x}{y}$ is in $(0,\frac{a}{b}]$ or $[\frac{a}{b},\frac{c}{d})$. Recall that $\lfloor z\rfloor$ denotes the greatest integer that is less than or equal to $z$.

Theorem 12**.**

With notation as above, the extreme factorizations of $s$ are

[TABLE]

for $t=0$ and for

[TABLE]

These extreme factorizations have lengths $\delta+\alpha+\beta$ and

[TABLE]

respectively.

Proof: Note that, since $\gcd(a,c)=1$, all factorizations of $x$ in $\langle a,c\rangle$ are given by $x=(\alpha+ct)a+(\beta-at)c$, for various integer $t$. Note that $\alpha+ct\geq 0$ precisely when $t\geq 0$, by our choice of $\alpha$.

By Lemma 11, for each choice of $t$ there is a unique $\delta_{t}=\delta(\alpha+ct,\beta-at)$ with $s=\delta_{t}u+(\alpha+ct)v+(\beta-at)w$. Hence $y=\delta_{t}+(\alpha+ct)b+(\beta-at)d=\delta_{t}+\alpha b+\beta d+t$, so $\delta_{t}=y-\alpha b-\beta d-t$. The factorization length (of $s$ in $S$) is $(\alpha+ct)+(\beta-at)+(y-\alpha b-\beta d-t)=(\alpha+\beta+y-\alpha b-\beta d)+t(c-a-1)$. In particular, the length varies linearly with $t$; one extreme is when $t=0$, and the other is when $t$ is maximal.

There are two upper bounds on $t$, both of which must hold. One is that $\beta-at\geq 0$ (else the coefficient of $w$ would not be in $\mathbb{N}_{0}$), while the other is that $0\leq\delta_{t}=y-\alpha b-\beta d-t=\delta-t$. Now we compare the two bounds of $\frac{\beta}{a}$ and $\delta$. We have $\frac{\beta}{a}\leq\delta$ exactly when $\alpha ab+\beta cb\leq\alpha ab+\beta ad+\delta a$, which holds exactly when $xb\leq ya$ or $\frac{x}{y}\leq\frac{a}{b}$. In this case, we use the $\frac{\beta}{\alpha}$ bound and get the other for free; in the other case it is the reverse.

Substituting $t=0$ and $t=\lfloor\frac{\beta}{a}\rfloor$ (or $t=\delta$), we find the lengths as above.   QED

Note that the sign of $c-a-1$ determines which of the two extreme factorizations is minimal and which is maximal. In particular, we have the following.

Corollary 13**.**

With notation as above, if $c=a+1$, then $\rho(S)=1$.

Proof: By Theorem 12, each $s\in S$ has $|\mathsf{L}(\mathbf{s})|=1$.  QED

Corollary 14**.**

With notation as above, we fix $a,b,c,d,x,\alpha,\beta$ and suppose that $\beta(x)<a$. Then, for every $y\geq\frac{bx}{a}$, $\rho(\left[\begin{smallmatrix}x\\ y\end{smallmatrix}\right])=1$.

Proof: Our hypotheses force $\frac{x}{y}\leq\frac{a}{b}$ and $\lfloor\frac{\beta}{a}\rfloor=0$. Although $\delta$ will vary based on $y$, all factorizations of $\left[\begin{smallmatrix}x\\ y\end{smallmatrix}\right]$ have the same length.   QED

5 Multiples of $s\in S$

We now fix $s\in S$, and consider factorizations of $ks=\left[\begin{smallmatrix}kx\\ ky\end{smallmatrix}\right]\in S$ for various $k\in\mathbb{N}$. For any individual $k$, we can of course compute $\rho(ks)$ using Theorem 12, but we seek $\rho(ks)$, or estimates thereto, for all the various choices of $k$. We offer three such results, two specific and one general. For convenience, we recall the sign function given by

[TABLE]

Our special results determine $\rho(ks)$ exactly, independently of $k$, but are for periodic values of $k$ only. There are two, based on whether or not $\frac{x}{y}\leq\frac{a}{b}$.

Theorem 15**.**

With notation as above, set $\tau=\textrm{sign}(c-a-1)$. Suppose thet $ac|k$ and $\frac{x}{y}\leq\frac{a}{b}$. Then

[TABLE]

Proof: Let $k^{\prime}\in\mathbb{N}$ with $k=k^{\prime}ac$. We have $\alpha(kx)=0$ and $\beta=\beta(kx)=k^{\prime}ax$. We calculate $\delta=\delta(0,\beta)=ky-\beta d=ak^{\prime}(cy-dx)$. One of the extreme factorization lengths will be $\delta+\beta=ak^{\prime}(cy-dx)+ak^{\prime}x=ak^{\prime}(cy-(d-1)x)$. The other will be $\delta+\beta+\lfloor\frac{\beta}{a}\rfloor(c-a-1)=ak^{\prime}(cy-(d-1)x)+k^{\prime}x(c-a-1)$.   QED

We now give our second special result, for the case of $k$ a multiple of $c$ and $\frac{x}{y}\geq\frac{a}{b}$. Note that again the elasticity is independent of $k$.

Theorem 16**.**

With notation as above, set $\tau=\textrm{sign}(c-a-1)$. Suppose thet $c|k$ and $\frac{x}{y}\geq\frac{a}{b}$. Then

[TABLE]

Proof: Let $k^{\prime}\in\mathbb{N}$ with $k=k^{\prime}c$. We have $\alpha(kx)=0$ and $\beta=\beta(kx)=k^{\prime}x$. We calculate $\delta=\delta(0,\beta)=ky-\beta d=k^{\prime}(cy-dx)$. One of the extreme factorization lengths will be $\delta+\beta=k^{\prime}(cy-dx)+k^{\prime}x=k^{\prime}(cy-(d-1)x)$. The other will be $\delta+\beta+\delta(c-a-1)=k^{\prime}(cy-(d-1)x)+k^{\prime}(cy-dx)(c-a-1)$.   QED

The following is a general result for all $k$. In particular, it implies that $\rho(ks)$ is largely predicted by $\phi(s)$, with this prediction becoming more accurate as $k\to\infty$. Note also that the limiting values agree, as expected, with the values in Theorems 15, 16.

Theorem 17**.**

With notation as above, set $\tau=\textrm{sign}(c-a-1)$. Then

[TABLE]

Proof: We set $\alpha=\alpha(kx),\beta=\beta(kx)$, with $kx=\alpha a+\beta c$, and $0\leq\alpha<c$. Note that $\beta=\frac{kx-\alpha a}{c}$. We calculate $\delta=ky-\alpha b-\beta d=ky-\alpha b-(kx-\alpha a)\frac{d}{c}=k(y-x\frac{d}{c})-\alpha(b-\frac{ad}{c})=k(y-x\frac{d}{c})-\frac{\alpha}{c}$.

Rather than taking $\rho(ks)$ as the ratio of $\max\mathsf{L}(ks)$ to $\min\mathsf{L}(ks)$, we will instead take $\rho$ as the ratio of $\frac{1}{k}\max\mathsf{L}(ks)$ to $\frac{1}{k}\min\mathsf{L}(ks)$. One of these will be $\frac{1}{k}(\delta+\alpha+\beta)=\frac{1}{k}\left(k(y-x\frac{d}{c})-\frac{\alpha}{c}+\alpha+\frac{kx-\alpha a}{c}\right)=y-x\frac{d-1}{c}+\frac{\alpha}{k}\frac{c-a-1}{c}$. In the limit, the last term vanishes, leaving $y-x\frac{d-1}{c}$.

We consider the case of $\frac{x}{y}\leq\frac{a}{b}$. The other term we will have in our ratio limit will be $\frac{1}{k}\left(\delta+\alpha+\beta+\lfloor\frac{\beta}{a}\rfloor(c-a-1)\right)=y-x\frac{d-1}{c}+\frac{\alpha}{k}\frac{c-a-1}{c}+\frac{1}{k}\lfloor\frac{\beta}{a}\rfloor(c-a-1)$ Now, $\frac{\beta}{a}=k\frac{x}{ac}-\frac{\alpha}{c}$. In the limit we will get $y-x\frac{d-1}{c}+\frac{x}{ac}(c-a-1)$. We simplify to $y-x\frac{b-1}{a}$. This gives the first formula.

Finally, we turn to the case of $\frac{x}{y}\geq\frac{a}{b}$. The other term we will have in our ratio limit will be $\frac{1}{k}\left(\delta+\alpha+\beta+\delta(c-a-1)\right)=y-x\frac{d-1}{c}+\frac{\alpha}{k}\frac{c-a-1}{c}+\frac{c-a-1}{k}\left(k(y-x\frac{d}{c})-\frac{\alpha}{c}\right)$. In the limit we will get $y-x\frac{d-1}{c}+(c-a-1)(y-x\frac{d}{c})=(c-a)y-(d-b)x$. This gives the second formula.   QED

We close by noting that the functions appearing in Theorems 15, 16, and 17 are quite simple, being linear fractional transformations in the variable $\frac{x}{y}=\phi(s)$.