← Notes on Rieg Theory
Older Versions__2023_07_23Ver__RiegOfNumbers.tex
\section{Riegs of Numbers}
\subsection{Non-negative numbers}
It's time to play with concrete examples! In this section, we will see \dq{numerical} examples. The prototypical example is the rieg of natural numbers.
\begin{example}[Natural numbers]
The set of non-negative integers (or natural numbers) with usual operations $(\N,0,1,+,\ti,\ex)$ is a rieg. Note that $0^0=1$, as mentioned earlier. We call it the rieg of natural numbers. This is the initial object of the category $\Rieg$.
\end{example}
% \memo{$\{0,1,2,\dots \infty\}$ is a rieg. (Quotient rieg of cardinal arithmetic) and its quotient $\{0, \text{positive finite}, \infty\}$}
\memo{$\{0,1,2,\dots \infty\}$ is a rieg. (Quotient rieg of cardinal arithmetic). But it is not the case for its quotient rig $\{0, \text{positive finite}, \infty\}$, since $1^{\infty} \neq 2^{\infty}$}\memo{How about $\{0,1, \mathrm{plural}, \infty\}$? $\ch = (2,1)$}
% \memo{$\{0,1,2,\dots \infty\}$ is not a rieg at lest in the naive way. (Quotient rieg of cardinal arithmetic), since $1^\infty \noteq $ and its quotient $\{0, \text{positive finite}, \infty\}$}
\memo{We can construct \dq{$p$-adic} numbers or profinite constructions. $\lim \Mr{a}{b}$m $\lim \Mr{n}{1}$ and $\lim\Mr{n}{p^n}NOOO$}
\begin{example}[Non-negative real numbers]
The set of non-negative real numbers $[0,\infty)$ has the natural rieg structure, defining
\[
0^x=
\left\{
\begin{array}{ll}
1 & (x=0) \\
0 & (x>0).
\end{array}
\right.
\]
The rieg of natural numbers is a subrieg of this rieg. However, the subset of all non-negative rational numbers is not subrieg, since it is not closed under exponentials.
\memo{The rieg $[0,\infty)$ is not a $\Top$-internal rieg, since $\ex$ is not continuous. Then, the rig of non-negative continuous functions is not closed under element-wise exponentials. How about non-negative $L^1$ functions?}
\memo{(J. Koizumi) exp-preserving function $(0,\infty)\to (0,\infty)$ is either the constant function to $1$ or the identity function.}
\end{example}
\begin{remark}[The rig ${[0,\infty]}$ is not a rieg]\label{RemarkExtendedReals}
It is natural to ask whether this rieg structure can be extended to the rig $[0,\infty]$. The answer is no. The rig ${[0,\infty]}$ cannot have a rieg structure. Suppose that there is an extension to a rieg structure. First, since $2^{\infty}=2^{\infty+1}= 2^{\infty} \ti 2$, we have $2^{\infty}\in \{0,\infty\}$. And, since $2^{\infty}(0.5)^{\infty} = 1^{\infty} =1$, we have $2^{\infty}\in (0,\infty)$, contradiction.
\end{remark}
\begin{question}
What's the subrieg of $[0,\infty)$ generated by non-negative rational numbers? Are there any non-trivial equations? Is it isomorphic to \dq{free rieg} of a rig $\mathbb{Q}_{\geq}?$
\end{question}
\subsection{Finite system of numbers}
\begin{example}[Overflow riegs]
Consider finite memory versions of the natural number rieg.
% \emph{overflow riegs}.
For any natural number $n$, we define \emph{the overflow rieg with threshold $n$}. The underlying set is the set with $n+1$-elements
\[\{0,1,2, \dots n-1, \TB\}.\]
The operations are basically the same as the riegs of natural numbers, but values greater than or equal to $n$ overflow and are changed into $\TB$. Intuitively, the symbol $\TB$ means \dq{too big!}. For example, in the overflow rieg with threshold $5$, we have $0+1=1$, $2^3=\TB$, and
\begin{align*}
0^\TB \cdot 2^3+1^{4+2} +2^{1^{3+2}}
&=0\cdot \TB+1^{\TB} +2^{1^{\TB}} \\
&=0+1+2^{1} \\
&=3.
\end{align*}
\end{example}
\begin{example}[Trivial rieg]\label{ExampleTrivialRieg}
The \emph{trivial rieg} is a singleton $\{\TB\}$ equipped with the unique rieg structure. In other words, it is the overflow rieg with threshold $0$ and the terminal object of the category $\Rieg$.
\end{example}
The overflow rieg with threshold $1$ will appear soon (Example \ref{ExampleRiegOfTruthValues}) as the rieg of truth values.
\memo{threshold $2$ is also theoretically important: $\{0,1, \mathrm{plural}\}$. $2^{\infty}\neq 1^{\infty}$}
Is it possible to perform modular arithmetic in rieg theory? In fact, this question is of great importance in the development of general rieg theory, especially the theory of characteristics, and will be discussed in detail in Section \ref{SectionCharacteristics}.
Let's briefly touch upon it here. First, simply considering $\Z/n\Z$ does not work well, because it is a ring not rieg (except $n=0$). However, by separating out $0$ or, more generally, small natural numbers, it is sometimes possible to develop modular arithmetic.
\begin{example}[Parity rieg]\label{ExampleParityRieg}
The set $\{0, \text{positive odd}, \text{positive even} \}$ equipped with the intended operations, which are well-defined is a rieg. We call it the \emph{parity rieg}.
\end{example}
Actually, for any $n$, it is possible to construct a finite rieg $M$ and decompose the unique rig homomorphism
\[\N \twoheadrightarrow \Z/n\Z\]
into a rieg homomorphism followed by a rig homomorphism
\[\N \twoheadrightarrow M \twoheadrightarrow \Z/n\Z.\]
% with a unique surjective rig homomorphism $M \twoheadrightarrow \Z/n\Z$,
In this sense, one can do modular arithmetic, including exponentiation! See Section \ref{SectionCharacteristics}.
\begin{example}[$\Mr{a}{b}$]
\end{example}
\memo{For any $n\in \N$, we can construct a finite rieg that has surjective rig homomorphism to $\Z/n\Z$. In this sense, we can treat exponentials in modular arithmetic. If you want to calculate exponential $\mod{5}$, you should use $\Mr{2}{20}$ Is it \dq{free rieg} of the rig?}
\memo{Rieg of functions with first-order differential coefficients}
\memo{What's called the exponential ring of complex numbers}
\memo{Cardinal arithmetic}
\memo{Some example from Heyting, e.g. max-min of [0,1]}
\memo{max-plus algebra}
\subsection{Numbers with Infinites}
There are some riegs, wth \dq{infinites}. This cannot happen for a (non-trivial) ring, because $0+\infty = 1+ \infty$ implies $0=1$.
\begin{example}[Cardinal numbers]
The cardinal arithmetic also has the structure of a rieg. However, one must be careful about the size problem, since the class of all cardinal numbers is a proper class, not a set.
The familiar solution is to assume the existence of a Grothendieck universe, but there is no need to go to such great lengths. We can simply take a strong limit cardinal $\kappa$ (for example, $\beth_{\omega}$) and consider the set of all cardinal numbers less than $\kappa$. Then this set has a rieg structure by cardinal arithmetic. When $\kappa = \aleph_0$, this is the rieg of natural numbers.
\memo{The regularity of $\kappa$ is not needed here, but in some situations, it may be useful to take a strong limit cardinal with cofinality greater than $\omega$, such as $\beth_{\omega_1}$).}
\memo{Relation to topos}
\end{example}
\memo{Under the GCH (generalized continuum hypothesis), the rieg induced from $\beth_{\omega}$($=\aleph_{\omega}$ by the GCH) is worth mentioning.}
\memo{And, forcing can access this rieg structure.}
% The next example cannot be realized with substitution operations. In this sense, it is $$
\begin{example}[Complemented natural numbers]
$\N\cup\{\infty\}$ has a canonical rieg structure, which is obtained as the limit of the overflow riegs
\[
\begin{tikzcd} \dots\ar[r]&\mathrm{OverFlow}_{2}\ar[r]&\mathrm{OverFlow}_{1}\ar[r]&\mathrm{OverFlow}_{0}.
\end{tikzcd}
\]
In this sense, this rieg is the overflow rieg with threshold $\infty$.
\memo{$[0,\infty]$ does not have a rieg structure}
Notice that this construction is similar to the construction of $p$-adic integers. In subsection \ref{SubsectionProfinite}, we will see the profinite completion of $\N$, which is a refinement of $\N\cup\{\infty\}$
\end{example}
\subsection{Lists of numbers}
\subsection{Dual numbers}
\memo{Constructed by Yuhi Kamio}
Recall that a dual number is a number plus an \dq{infinitesimal number}
\[a+b\ep \ (a,b \in \R)\]
The ring of infinitesimal numbers is denoted by $\R[\ep]$ and algebraically defined as $\R[\ep]=\R[x]/(x^2).$
There is a similar rieg, which we call the \emph{dual number rieg.}
\begin{definition}[Dual numbers, concrete definition]\label{SubsubsectionDualNumbers}
The dual number rieg $\N[\ep]$ has $\N \times \Z$ as the underlying set and its element $(n,m)$ is denoted by $n+m\epsilon$. Three operations are defined as follows:
\begin{enumerate}
\item $(n+m\ep)+ (n'+m'\ep)=(n+n')+(m+m')\ep$
\item $(n+m\ep)\ti (n'+m'\ep)=(nn')+(nm'+n'm)\ep$
\item $(n+m\ep)\ex (n'+m'\ep)=\underbrace{(n+m\ep)\ti \dots \ti (n+m\ep)}_{n' \text{times}} = (n^{n'}, n^{n'-1}n'm\ep).$
\end{enumerate}
\end{definition}
This definition has a more conceptual paraphrase, with which we can easily prove that it's actually a rieg. First notice that the projection
\[\N[\ep]\to \N\colon n+m\ep \mapsto n\]
is a rig homomorphism to the initial rig $\N$. Thus we have the composite rig homomorphism
\[\N[\ep]\to \N\to \End(\N[\ep],\ti,1),\]
and this gives the exponential structure of the dual number rieg.
More generally, we have the following proposition.
\begin{proposition}[Rieg structure by a natural number evaluation]\label{PropositionNNevaluation}
For a rig $R$ and a rig homomorphism (or say, \dq{natural number evaluation})
\[v\colon R \to \N,\]
$R$ admits an induced rieg structure
\[R \to \N \to \End(R,\ti,1).\]
\end{proposition}
In other words, the exponential is given by
\[x^y = \underbrace{x\ti\dots \ti x}_{v(y)\text{ times}}.\]
For example, we can replace the additive group $\Z\ep$ in the definition of the dual number rieg with an arbitrary commutative monoid.
\begin{example}\label{ExamplePolynomialexp}
In proposition \ref{PropositionNNevaluation}, letting $v\colon R \to \N$ be the evaluation function
\[\mathrm{ev}_0 \colon \N[x]\to \N \colon x\mapsto 0,\]
we have a rieg structure on $\N[x]$ with
\[f(x)^{g(x)} = f(x)^{g(0)}= \underbrace{f(x)\ti\dots \ti f(x)}_{g(0)\text{ times}}.\]
\end{example}
\subsection{Polynomials}
Is it possible to have more than one rieg structure on a single rig? The answer is (not at all surprisingly) yes. A typical example is the rig of polynomials $\N[x]$. Since $\N[x]$ is the free rig with a generator $x$, rieg structures on $\N[x]$
\[\N[x] \to \End(\N[x],\ti,1)\]
bijectively correspond to the elements of $\End(\N[x],\ti,1).$
\begin{example}[Shift operator]
\end{example}
\newpage