Canonical Number Systems on Polynomial Quotients: A Finite-Generators Pipeline

1. Introduction

Conjecture 1.Let R be a ring with$\left|R\right|\le {\aleph }_0$that is finitely generated. Then R is isomorphic to an n-dimensional-base canonical number system (CNS).

The main results of this paper are the following theorems.

Definition 1 

(Canonical number system; following laus Scheicher and Jörg M. Thuswaldner [1], and Christiaan E. van de Woestijne [2]). Let R be a commutative ring and $\phi :R\to R$ a homomorphism with finite cokernel (i.e. $[R:\phi (R\left)\right]<\infty$). A finite set $D\subset R$ is adigit setif it is a complete set of representatives of $R/\phi \left(R\right)$ (contains exactly one element from each coset). We call $(R,\phi ,D)$ acanonical number system (CNS)if every $r\in R$ admits afinite expansion

$$r=\sum _{k=0}^{\ell }{\phi }^k\left(d_k\right)\mathrm{with}{d}_k\in D,$$

and the expansion isuniquewhenever D is irredundant (a true transversal).

Polynomial/base specialization. Let $P\left(x\right)\in \mathbb{Z}\left[x\right]$ be monic with $\left|P\right(0\left)\right|\ge 2$, put $R=\mathbb{Z}\left[x\right]/\left(P\right)$, and let $\overline{x}$ be the class of x in R. With the digit set $\mathcal{N}=\{0,1,\dots ,|P\left(0\right)|-1\}$ (a transversal of $R/\overline{x}R$), we say $(P,\mathcal{N})$ (or equivalently $(R,\overline{x},\mathcal{N})$) is a CNS if every $\gamma \in R$ has a unique finite expansion

$$\gamma =\sum _{k=0}^{\ell }{n}_k{\overline{x}}^k,n_k\in \mathcal{N}.$$

If P is irreducible and α is a root, this is the same as unique finite α-expansions in $R\cong \mathbb{Z}\left[\alpha \right]$; in this case α is called aCNS base.

Definition 2 

(n–dimensional CNS). Let E be a commutative ring and let L be an E–module.

Digits. Fix finite digit alphabets

$$D_1,D_2,\dots ,D_n\subset E,\left|D_i\right|<\infty \mathrm{for}\mathrm{each}i,$$

and set the product digit set

$$D:=D_1\times D_2\times \dots \times D_n.$$

Base. Fix a finite base frame

$$\mathbf{b}=(b_1,\dots ,b_n)\in L^n\left(\left|\mathbf{b}\right|=n<\infty \right).$$

Places. Fix a countable index set $I={\mathbb{N}}_0$ and an E–linear “shift” endomorphism $T:L\to L$. The place set is

$${\mathcal{B}}_{\infty }:=\{T^k\left(b_i\right):i=1,\dots ,n,k\in I\}\subset L.$$

Expansion. For $\mathbf{d}=(d_1,\dots ,d_n)\in D$ write $\langle \mathbf{d},\mathbf{b}\rangle :={\sum }_{i=1}^n{d}_i{b}_i\in L$. We say $(L,T,\mathbf{b},D)$ is ann–dimensional canonical number systemif every $r\in L$ admits a unique finite expansion

$$\begin{array}{|c|}\hline r={\displaystyle \sum _{k\in I}^{\mathrm{finite}}}〈{\mathbf{d}}_k,T^k\mathbf{b}〉,{\mathbf{d}}_k\in D.\\ \hline\end{array}$$

Equivalently (by E–linearity of T),

$$r={\displaystyle \sum _{k\in I}^{\mathrm{finite}}}{T}^k\left(\langle {\mathbf{d}}_k,\mathbf{b}\rangle \right).$$

Lemma 1  

(Finite–generators presentation). Let R be a commutative ring with 1 and suppose $R=\mathbb{Z}[a_1,\dots ,a_n]$ for some $a_1,\dots ,a_n\in R$. Then there exists a surjective ring homomorphism

$$\varphi :\mathbb{Z}[x_1,\dots ,x_n]\twoheadrightarrow R,\varphi \left(x_i\right)=a_i(1\le i\le n),$$

and hence an isomorphism of rings

$$R\cong \mathbb{Z}[x_1,\dots ,x_n]/ker\left(\varphi \right).$$

Proof. 

By the universal property of the polynomial algebra on a finite set of indeterminates (Stacks Project, Tag 00S0 [3]), there exists a unique ring morphism

$$\varphi :\mathbb{Z}[x_1,\dots ,x_n]{⟶R\mathrm{with}\varphi |}_{\mathbb{Z}}={\mathrm{id}}_{\mathbb{Z}},\varphi \left(x_i\right)=a_i.$$

Then

$$Im\left(\varphi \right)\supseteq \mathbb{Z}[\varphi \left(x_1\right),\dots ,\varphi \left(x_n\right)]=\mathbb{Z}[a_1,\dots ,a_n]=R,$$

hence $Im\left(\varphi \right)=R$ and $\varphi$ is surjective. The first isomorphism theorem yields

$$R\cong \mathbb{Z}[x_1,\dots ,x_n]/ker\left(\varphi \right).$$

□

$$\begin{array}{c}\mathrm{Fix}j\in \{1,\dots ,n\},\begin{array}{|c|}\hline \begin{array}{cc}\hfill E& :=\mathbb{Z}[x_1,\dots ,\widehat{x_j},\dots ,x_n]\hfill \\ & =\{f\in \mathbb{Z}[x_1,\dots ,x_n]:{deg}_{x_j}f=0\}\hfill \\ & =\mathbb{Z}[x_1,\dots ,x_{j-1},x_{j+1},\dots ,x_n]\hfill \end{array}\\ \hline\end{array}\\ L:=\mathbb{Z}[x_1,\dots ,x_n],T:L\to L,T\left(h\right)=x_{j}h,\mathbf{b}=\left(1\right)\in L,\langle g,\mathbf{b}\rangle =g\in E.\end{array}$$

$$\begin{array}{|c|}\hline \forall f\in L\exists !g_0,\dots ,g_d\in E:f={\displaystyle \sum _{k=0}^d}{T}^k\left(\langle g_k,\mathbf{b}\rangle \right)={\displaystyle \sum _{k=0}^d}{g}_k{x}_j^k.\\ \hline\end{array}$$

Lemma 2  

(Digit folding for an n–dimensional CNS). Let E be a commutative ring, L an E–module, and let $T,U\in {\mathrm{End}}_E\left(L\right)$ commute ($TU=UT$). Fix a finite base frame ${\mathbf{b}}^{\circ }=(b_1,\dots ,b_m)\in L^m$ and an integer $M\ge 1$. Form the expanded base

$$\tilde{\mathbf{b}}:=(U^0{b}_1,\dots ,U^{M-1}{b}_1;\dots ;U^0{b}_m,\dots ,U^{M-1}{b}_m)\in L^{mM}.$$

Let the digit alphabets be finite $D_1,\dots ,D_m\subset E$ and set $D:=D_1\times \dots \times D_m$, $\tilde{D}:=D^M$ (blockwise Cartesian product).

Assume $(L,T,\tilde{\mathbf{b}},\tilde{D})$ is an $mM$–dimensional CNS, i.e. every $r\in L$ has a unique finite expansion

$$r={\displaystyle \sum _{k\ge 0}^{\mathrm{finite}}}〈{\tilde{\mathbf{d}}}_k,T^k\tilde{\mathbf{b}}〉,{\tilde{\mathbf{d}}}_k={\left(d_{k,i,j}\right)}_{1\le i\le m,0\le j<M}\in \tilde{D}.$$

Define thefolded digit set

$$D_i^{\circ }:=\left\{{\displaystyle \sum _{j=0}^{M-1}}{d}_j{u}^j|d_j\in D_i\right\}\subset E\left[u\right]/\left(u^M\right)(1\le i\le m),D^{\circ }:=D_1^{\circ }\times \dots \times D_m^{\circ },$$

and let $E\left[u\right]$ act on L via $u·v:=U\left(v\right)$.

Then every $r\in L$ admits a finitefoldedCNS expansion with base ${\mathbf{b}}^{\circ }$:

$$\begin{array}{|c|}\hline r={\displaystyle \sum _{k\ge 0}^{\mathrm{finite}}}〈{\mathbf{e}}_k,T^k{\mathbf{b}}^{\circ }〉,{\mathbf{e}}_k=(e_{k,1},\dots ,e_{k,m})\in D^{\circ },e_{k,i}={\displaystyle \sum _{j=0}^{M-1}}{d}_{k,i,j}{u}^j.\\ \hline\end{array}$$

If the original expansion with $(\tilde{\mathbf{b}},\tilde{D})$ is unique and T is injective, then the folded expansion with $({\mathbf{b}}^{\circ },D^{\circ })$ is also unique.

2. Proofs of Main Results

In this section, we will prove the Conjecture Section 1

From the Lemma 1 and Conjecture Section 1 we need to know the $ker\left(\varphi \right)$

Setup 1 [Finite–generators presentation] Let R be a commutative ring with 1, generated (as a ring) by $a_1,\dots ,a_n\in R$. Choose a presentation

$$R\cong \mathbb{Z}[y_1,\dots ,y_m]/J$$

for some ideal $J\subset \mathbb{Z}[y_1,\dots ,y_m]$, and polynomials $p_i\left(y\right)\in \mathbb{Z}[y_1,\dots ,y_m]$ such that the image of $p_i\left(y\right)$ in R equals $a_i$. Define

$$\varphi :\mathbb{Z}[x_1,\dots ,x_n]⟶R,\varphi \left(x_i\right)=a_i.$$

Setup 2 [Big ring and graph ideal] Set the big ring

$$S:=\mathbb{Z}[x_1,\dots ,x_n,y_1,\dots ,y_m].$$

Define the graph ideal in S by

$$I:=J+\left(x_1-p_1\left(y\right),\dots ,x_n-p_n\left(y\right)\right)\subset S.$$

Proposition 1  

(Kernel as an elimination ideal). With I as above,

$$ker\left(\varphi \right)=I\cap \mathbb{Z}[x_1,\dots ,x_n].$$

Because

Proposition 2.  

Let $\varphi :\mathbb{Z}[x_1,\dots ,x_n]\to R$ be a ring homomorphism and put $K:=ker\left(\varphi \right)◃\mathbb{Z}[x_1,\dots ,x_n]$. If $K=(f_1,\dots ,f_r)$ for some $f_i\in \mathbb{Z}[x_1,\dots ,x_n]$, then

$$\mathbb{Z}[x_1,\dots ,x_n]/K\cong \mathbb{Z}[x_1,\dots ,x_n]/(f_1,\dots ,f_r).$$

Proof. 

Set $J:=(f_1,\dots ,f_r)$. Since $J=K$, define

$$\Psi :\mathbb{Z}[x_1,\dots ,x_n]/J⟶\mathbb{Z}[x_1,\dots ,x_n]/K,\Psi (a+J):=a+K.$$

This is a well-defined ring homomorphism because $a+J=b+J\iff a-b\in J=K\iff a+K=b+K$. Define $\Theta :\mathbb{Z}[x_1,\dots ,x_n]/K\to \mathbb{Z}[x_1,\dots ,x_n]/J$ by $\Theta (a+K):=a+J$. Then

$$(\Theta \circ \Psi )(a+J)=a+J,(\Psi \circ \Theta )(a+K)=a+K,$$

so $\Psi$ is an isomorphism with inverse $\Theta$. □

Thus we have

Lemma 3  

(Third isomorphism for quotients). Let A be a ring and let $J\subseteq K$ be ideals. The canonical map

$$\pi :A/J⟶A/K,\overline{a}⟼\overline{a}$$

is surjective with kernel $K/J$. Hence

$$(A/J)/(K/J)\cong A/K.$$

In particular, if $K=(f_1,\dots ,f_r)$ then $A/K\cong A/(f_1,\dots ,f_r)$.

Proof. 

By the correspondence theorem for ideals, the ideals of $A/J$ are exactly the quotients $L/J$ with L an ideal of A containing J; in particular $K/J$ is an ideal of $A/J$. The map $\pi$ is clearly surjective, and $ker\pi =\{\overline{a}\in A/J:a\in K\}=K/J$. Apply the first isomorphism theorem. See, e.g., ([4], Prop. 1.1 and Cor. 1.3) or ([3], Isomorphism theorems for rings). □

Lemma 4  

(Finite standard monomials). Assume there exist integers $M_1,\dots ,M_n\ge 1$ such that in R every monomial $x^{\alpha }=x_1^{{\alpha }_1}\dots x_n^{{\alpha }_n}$ is congruent to a $\mathbb{Z}$–linear combination of monomials with ${\alpha }_i<M_i$ for each i. Equivalently, the set

$$\mathcal{B}:=\left\{\overline{x^{\alpha }}\in R:0\le {\alpha }_i<M_i(1\le i\le n)\right\}$$

is a finite $\mathbb{Z}$–basis of R. Let $m:=\left|\mathcal{B}\right|$ and list $\mathbf{b}=(b_1,\dots ,b_m)$ the elements of $\mathcal{B}$.

Lemma 5  

(Digits and place map). Fix a prime p which is not a zero divisor in R and define

$$T:R\to R,T\left(r\right):=pr,D:=\left\{{\displaystyle \sum _{j=1}^m}{d}_j{b}_j|d_j\in \{0,1,\dots ,p-1\}\right\}\subset R.$$

Theorem 1  

(Folded CNS). Assume the pre–folding CNS of Theorem 2 is realized in the module form by commuting endomorphisms $T,U\in {End}_E\left(L\right)$ and data

$$\tilde{\mathbf{b}}=\left(U^0{b}_1,\dots ,U^{M-1}{b}_1;\dots ;U^0{b}_m,\dots ,U^{M-1}{b}_m\right),\tilde{D}=D^M,$$

for some finite frame ${\mathbf{b}}^{\circ }=(b_1,\dots ,b_m)$ and $M\ge 1$. Then there exists afoldeddigit set $D^{\circ }\subset E\left[u\right]/\left(u^M\right)$ and the same base frame ${\mathbf{b}}^{\circ }$ such that every $r\in L$ has a finite folded expansion

$$r={\displaystyle \sum _{k\ge 0}^{\mathrm{finite}}}〈{\mathbf{e}}_k,T^k{\mathbf{b}}^{\circ }〉,{\mathbf{e}}_k\in D^{\circ }.$$

If the pre–folding expansion is unique and T is injective, then the folded expansion is also unique. Consequently, $(L,T,{\mathbf{b}}^{\circ },D^{\circ })$ is an m–dimensional CNS.

Proof. 

Apply Lemma 2 with the given $(L,T,U,{\mathbf{b}}^{\circ },M)$. The lemma constructs $D^{\circ }$ and yields the folded expansion; its final clause gives uniqueness under the stated hypothesis on T. □

Thus from Lemma 1 and Lemma Section 2 we have

$$\mathrm{Let}\varphi :\mathbb{Z}[x_1,\dots ,x_n]\twoheadrightarrow R,K:=ker\left(\varphi \right).$$

$$R\cong \mathbb{Z}[x_1,\dots ,x_n]/K\cong \mathbb{Z}[x_1,\dots ,x_n]/(f_1,\dots ,f_r)$$

Thus from Lemma 1

$$\mathbb{Z}[x_1,\dots ,x_n]/(f_1,\dots ,f_r)\mathrm{is}\mathrm{an}n-\mathrm{dimensional}\mathrm{CNS}$$

$$⟹R\cong \mathrm{an}n-\mathrm{dimensional}\mathrm{CNS}.$$

3. Example

Example 1 

(Pipeline on). $\mathbb{Z}\left[i\right]\cong \mathbb{Z}\left[x\right]/(x^2+1)$Step 1: Presentation and kernel via graph ideal.Let $R=\mathbb{Z}\left[t\right]/(t^2+1)$ and set $a_1=\overline{t}\in R$. Define $\varphi :\mathbb{Z}\left[x\right]\to R$, $\varphi \left(x\right)=a_1$. Build the big ring $S=\mathbb{Z}[x,y]$ and the graph ideal

$$I:=(y^2+1,x-y)\subset S.$$

Then $K:=ker\left(\varphi \right)=I\cap \mathbb{Z}\left[x\right]=(x^2+1)$ (eliminating y gives $x^2+1$). Hence

$$R\cong \mathbb{Z}\left[x\right]/K\cong \mathbb{Z}\left[x\right]/(x^2+1).$$

(This mirrors the “big ring & graph ideal” step and kernel elimination in the paper.) [3]

Step 2: Finite standard monomials and pre–folding CNS. In $\mathbb{Z}\left[x\right]/(x^2+1)$ every monomial reduces to a $\mathbb{Z}$–combination of $\{1,x\}$, so $\mathcal{B}=\{\overline{1},\overline{x}\}$ is a finite $\mathbb{Z}$–basis. Fix a prime $p=5$ (a non–zero–divisor in R), put

$$T:R\to R,T\left(r\right)=5r,D:=\{d_0·\overline{1}+d_1·\overline{x}:d_0,d_1\in \{0,1,2,3,4\}\}.$$

Then every $r=a+b\overline{x}\in R$ has a (unique)pre–foldingbase–5 expansion

$$r={\displaystyle \sum _{k=0}^{\ell }}{5}^k\left(d_{k,0}·\overline{1}+d_{k,1}·\overline{x}\right),a=\sum _k{d}_{k,0}{5}^k,b=\sum _k{d}_{k,1}{5}^k,d_{k,j}\in \{0,\dots ,4\}.$$

Concrete digits. Take $r=37+11\overline{x}$. Write $37=2+2·5+1·5^2$ and $11=1+2·5$. Thus

$$r=\underset{k=0}{\underbrace{(2+1\overline{x})}}+5\underset{k=1}{\underbrace{(2+2\overline{x})}}+5^2\underset{k=2}{\underbrace{(1+0\overline{x})}}.$$

Step 3: Fold along a finite–power coordinate (digit folding). Let U be multiplication by $\overline{x}$; since ${\overline{x}}^2=-1$, only the two powers ${\overline{x}}^0,{\overline{x}}^1$ survive in normal form ($M=2$). Apply the folding lemma with base frame $b^{\circ }=\left(\overline{1}\right)$, so the expanded frame is $(\overline{1},\overline{x})$. Pack each pair of coefficients at level k into a single folded digit $e_k=d_{k,0}+d_{k,1}u\in \mathbb{Z}\left[u\right]/\left(u^2\right)$, where u acts on R by $u·v:=U\left(v\right)=\overline{x}v$. For our r:

$$e_0=2+1u,e_1=2+2u,e_2=1+0u.$$

The folded CNS expansion is

$$\begin{array}{|c|}\hline r={\displaystyle \sum _{k=0}^2}{5}^k\langle e_k,T^k{b}^{\circ }\rangle =(2+1u)+5(2+2u)+5^2(1+0u),\\ \hline\end{array}$$

which is unique because the pre–folding expansion was unique and T is injective.

Conclusion. We exhibited the full pipeline:

$$\mathbb{Z}\left[i\right]\cong \mathbb{Z}\left[x\right]/(x^2+1)⟹\mathrm{pre}\text{–}\mathrm{folding}\mathrm{base}\text{–}5\mathrm{CNS}\mathrm{on}\{\overline{1},\overline{x}\}⟹\mathrm{folded}1\text{–}\dim \mathrm{CNS}\mathrm{on}\left(\overline{1}\right).$$

This exemplifies the definitions and the folding mechanism used in the paper.

Example 2. 

${\mathbb{F}}_3$ (and ${\mathbb{F}}_3^n$) as a CNS Step 1 (Presentation via a kernel). Let $\varphi :\mathbb{Z}\to {\mathbb{F}}_3$ be the reduction mod 3 map, $\varphi \left(1\right)=\overline{1}$. Then $ker\left(\varphi \right)=3\mathbb{Z}$ and

$${\mathbb{F}}_3\cong \mathbb{Z}/ker\left(\varphi \right)=\mathbb{Z}/\left(3\right).$$

Equivalently, for any $n\ge 1$ one may take

$$\Phi :\mathbb{Z}[x_1,\dots ,x_n]⟶{\mathbb{F}}_3{,\Phi |}_{\mathbb{Z}}=\mathrm{mod}3,\Phi \left(x_i\right)=0,$$

with $ker(\Phi )=(3,x_1,\dots ,x_n)$, so ${\mathbb{F}}_3\cong \mathbb{Z}[x_1,\dots ,x_n]/(3,x_1,\dots ,x_n)$.

Step 2 (CNS on ${\mathbb{F}}_3$).Work with the 1–dimensional module form of Definition 2:

$$E=\mathbb{Z},L=\mathbb{Z},T:L\to L,T\left(m\right)=3m,\mathbf{b}=\left(1\right),D=\{0,1,2\}\subset L.$$

Passing to the quotient $L/3L\cong {\mathbb{F}}_3$, every $\overline{a}\in {\mathbb{F}}_3$ has the uniquefiniteexpansion (indeed, length 0)

$$\begin{array}{|c|}\hline \overline{a}=〈d_0,T^0\mathbf{b}〉=d_0·\overline{1},d_0\in \{0,1,2\}.\\ \hline\end{array}$$

Here the “place” exponents are just 0 (so the ‘exponential vector’ is trivially $1^1$), and the digit set is exactly $\{0,1,2\}$.

Step 3 (n–dimensional version with identity frame).Let $L={\mathbb{Z}}^n$ with the standard basis $\mathbf{b}=(e_1,\dots ,e_n)$ (theidentity vector frame), keep $T=3{\mathrm{id}}_L$, and take the product digit set

$$D={\{0,1,2\}}^n=\{(d_1,\dots ,d_n):d_i\in \{0,1,2\}\}.$$

Modulo 3, we obtain $L/3L\cong {\mathbb{F}}_3^n$, and every $\overline{v}\in {\mathbb{F}}_3^n$ has a unique expansion of length 0:

$$\begin{array}{|c|}\hline \overline{v}=〈{\mathbf{d}}_0,T^0\mathbf{b}〉=d_1{\overline{e}}_1+\dots +d_n{\overline{e}}_n,{\mathbf{d}}_0=(d_1,\dots ,d_n)\in {\{0,1,2\}}^n.\\ \hline\end{array}$$

Thus ${\mathbb{F}}_3$ (and, with the identity frame, ${\mathbb{F}}_3^n$) is a canonical n–dimensional number system whose digit vectors are drawn from ${\{0,1,2\}}^n$ and whose “place exponents” are the all–ones multiindex $1^n$ (i.e. only the $k=0$ place occurs after mod 3).