← Notes on Rieg Theory
Older Versions__OldVer__main2023_05_10.tex
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\begin{document}
\title{Notes on Rieg Theory}
\author{Ryuya Hora}
\date{\today}
\maketitle
\definecolor{shadecolor}{gray}{0.9}
\maketitle
\begin{abstract}
In ring theory classes, we are told that \dq{natural numbers do not have the structure of a ring, but only a semiring (or a \emph{rig}).} In fact, it does not have the structure of a ring like $2-3=-1 \notin \N$, but it does have a rich algebraic structure in another direction, exponentials $2^3=8$.
In combinatorics, when considering the action of a finite group on a finite set, one sometimes constructs a Burnside rig of the isomorphism classes. However, even in this case, if we do not extend it to a ring, it has an exponential structure.
In this article, we investigate the properties of such semirings with exponentials, which we call \emph{riegs}, and observe their connections with bicartesian closed categories and toposes as categorifications.
Let us spoil some of the wonders of rieg theory. In contrast to field theory, there are only a finite number of possible \dq{positive characteristics} of a rieg! They are $1,2,6,42$, and $1806$.
\end{abstract}
% \input{preceeding_studies}
\newpage
\tableofcontents
\newpage
\newpage
\section{Introduction}\label{SectionIntroduction}
\input{introduction}
\newpage
\section{Basic theory}
\subsection{Definitions}
In this note, we refer to a semiring as a \emph{rig}, because it is a ring without \textbf{n}egatives.
Since what we will consider is a ring with \textbf{e}xponentials instead of \textbf{n}egatives, we call it a \emph{rieg}.
We will give two
% seemingly different (but, of course, equivalent)
definitions of riegs. The first one is more conceptual and easy to remember, and the latter equational one is theoretically convenient.
First, recall that, for a commutative monoid $(A, e, \ast)$, the set of endomorphisms $\End{(A, e, \ast)}$ has a rig structure. The addition is elementwise addition, and the multiplication is the composition.
\begin{definition}[Rieg, conceptual definition]
A \emph{rieg} is a commutative rig $(A, 0, 1, +, \ti )$ equipped with a rig homomorphism
\[A \to \End{(A, 1,\ti)}.\]
\end{definition}
As an equational theory, we can redefine it with numerous axioms:
\begin{definition}[Rieg, equational theory]
The equational theory of riegs consists of
\begin{itemize}
\item two $0$-ary operations, $0,1$ and
\item three $2$-ary operations, $+, \ti, \ex$
\end{itemize}
and following equations:\footnote{some of them are derivable from others}
\begin{itemize}
\item $0+x=x$
\item $x+0=x$
\item $(x+y)+z=x+(y+z)$
\item $x+y=y+x$
\item $1\ti x =x$
\item $x\ti 1 =x$
\item $(x\ti y)\ti z=x\ti (y\ti z)$
\item $x\ti y=y\ti x$
\item $x\ti 0=0$
\item $x\ti (y+z)=(x\ti y) + (x\ti z)$
\item $0\ti x=0$
\item $(x+y)\ti z=(x\ti z)+(y\ti z)$
\item $1 \ex x = 1$
\item $(x\ti y)\ex z = (x\ex z)\ti (y\ex z)$
\item $x \ex 0 = 1$
\item $x\ex (y+z) =(x\ex y) \ti (x\ex z)$
\item $x\ex 1 = x$
\item $x\ex (y\ti z) = (x\ex y)\ex z$.
\end{itemize}
\end{definition}
\memo{This is a little bit different from Tarski's high school algebra problem. How about the rieg case of the problem? I believe this is the \dq{right} question.}
From now, $x^y$ denotes $x\ex y$ and $xy$ denotes $x\ti y$, as usual.
For example, with this notation and elimination of obvious redundant axioms, the above definitions look a little simpler like this:
\begin{itemize}
\item $0+x=x$
\item $(x+y)+z=x+(y+z)$
\item $x+y=y+x$
\item $1 x =x$
\item $(x y) z=x (y z)$
\item $x y=y x$
\item $x 0 =0$
\item $x (y+z)=x y + x z$
\item $1 ^ x = 1$
\item $(x y)^ z = x^ z y^ z$
\item $x ^ 0 = 1$
\item $x^ {y+z} =x^ y x^ z$
\item $x^ 1 = x$
\item $x^{y z} = (x^ y)^ z$.
\end{itemize}
% The prototypical example is the rieg of natural numbers.
% \begin{example}[Natural numbers]
% $(\N,0,1,+,\ti,\ex)$.
% \end{example}
We will see numerous examples of riegs in this note!
\memo{semiring with tetrations}
From this definition, we obtain the notion of rieg homomorphisms, as usual.
\begin{definition}
A rieg homomorphism is a function that preserves all (five) operations $0,1,+,\ti,\ex$.
\end{definition}
The category of riegs is referred to as $\Rieg$.
\subsection{Basic properties}
The first thing to point out is the following equality which follows immediately from the axiom $x^0=1$.
\begin{proposition}
$0^0=1$.
\end{proposition}
\memo{If $0^x=1$, then is $x$ zero? No, Heyting algebras}
This is a natural algebraic requirement, and we will later observe that it is also consistent with a category-theoretic construction. For example, the number of functions from an empty set to an empty set is one.
Both riegs and rings are variants of rigs, but with different atmospheres. In fact, when those two structures coexist, an anomaly of $0^{-1}$ immediately occurs!
\begin{proposition}[Riegs and Rings]
If a rieg $R$ is a ring, then it is a zero ring. \footnote{As a rieg, late we will call it the \emph{trivial rieg}.}
\end{proposition}
\begin{proof}
Since $0=1-1$ in a ring, we have
$1=0^0=0^{1-1}=0^{1}\ti 0^{-1}=0 \ti 0^{-1}=0$.
Therefore, for any $x\in R$, we have
$x=1x=0x=0$.
\end{proof}
This is a simple but typical argument that transitions between addition and multiplication. The key feature of rieg theory
% , which is not found in mere semicyclic theory,
is that addition and multiplication are closely interconnected, transformed into each other by exponential $\ex$.
\memo{Is $0$ the only element that has additive inverse?}
\begin{remark}[Relation to Dioids]\label{RemarkDioidsandRiegs}
Dioids are another variant of rigs that is also almost disjoint to rings. For the definition, see Remark \ref{RemarkDefinitionOfDioids} or the introductory book \cite{gondran2008graphs}. Until now, the author could not find a direct connection between dioids and riegs. At least, there is no inclusion relation. The parity rieg (Example \ref{ExampleParityRieg}) is a rieg, but not dioid. Conversely, the dioid of extended non-negative real numbers $[0, \infty]$ cannot have a rieg structure (Remark \ref{RemarkExtendedReals}).
\end{remark}
Several category-theoretic properties follow immediately from the mere existence of an equation-theoretic definition. Some of them are noted below. However, if you are not familiar with category theory, you can ignore them, since most of the content of this paper is understandable without them.
\begin{proposition}[Limits and colimits in $\Rieg$]
The category of riegs $\Rieg$ has all small limits and colimits.
Furthermore, the forgetful functor $U\colon \Rieg \to \Set$ (preserves and) creates all limits.
\end{proposition}
This means we can construct arbitrary small limits using limits in $\Set$. For example, the terminal object is a singleton with obvious operations, which we call it the \emph{trivial rieg} (see Example \ref{ExampleTrivialRieg}). For a given family of riegs $\{R_{\lambda}\}_{\lambda\in \Lambda}$, its categorical product is given by the product set $\prod_{\lambda\in \lambda} R_{\lambda}$ with the index-wise operations.
\begin{proposition}[Free forgetful adjunction]
The forgetful functor $U\colon \Rieg \to \Set$ has a left adjoint $F\colon \Set \to \Rieg$
\end{proposition}
Furthermore, we have a variation of the homomorphism theorem:
\begin{proposition}[Surj-inj factorization system]
A rieg homomorphism is uniquely (up to the canonical isomorphism) decomposed into the composition of a surjective homomorphism followed by an injective homomorphism.
\end{proposition}
\memo{cite context and write monadic aspects}
% \begin{example}[Initial rieg: natural numbers]
% The rieg of natural numbers $\N$ is the initial object of $\Rieg$.
% \end{example}
% \begin{example}[Terminal rieg: trivial rieg]\label{ExampleTrivialRieg}
% As for every equational theory, the terminal rieg consists of only one element $0=1$. We call it the \emph{trivial rieg}.
% \end{example}
% For categorical (small) colimits, the existence follows from categorical universal algebra. However, concrete calculations tend to be exhausting.
An element of the free rieg $F(\{x\})$ looks like
\[(3^{x}+x^{x^{x}})^{x^{2^{x}}+5^{x+0^{x}}+1}+0^{x} x.\]
\begin{question}
Is there a combinatorial expression of elements of free algebras?
\end{question}
\begin{question}
There is the unique rieg homomorphism for the free rieg for $\{x\}$ to $\N$ that sends $x$ to $0$. For example,
\[x\ex \ex n = \underbrace{x \ex x \ex \dots \ex x}_{n}\] is sent to $0$ or $1$, depending on the parity of $n$. What is that function? \memo{combinatorial game theory}
% whether $n$ is odd or even.
\end{question}
% \begin{question}
% What functions can be \dq{programmed} as an element of the free algebra? \memo{Using overflow riegs, there is a strong limitation. But for finite sequences?}
% \end{question}
% \section{Examples}
\newpage
\section{Examples}
\subsection{Riegs of Numbers}
\subsubsection{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 $\{0, \text{positive finite}, \infty\}$}
\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}
\subsubsection{Cardinal numbers}
\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 $\beth_{\omega_1}$).}
\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 $$
\subsubsection{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.
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}.
\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}
\subsubsection{Lists of numbers}
\subsubsection{Dual numbers}
\memo{Constructed by 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]
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.
\subsubsection{Polynomials}
Is it possible to have more than one rieg structures 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 $\End(\N[x],\ti,1).$
\newpage
\subsection{Riegs of Modular arithmetic}\label{SectionCharacteristics}
\subsubsection{Two motivations}
% \subsection{Modular arithmetic with exponentials}
\subsubsection*{Elmentary motivation}
In this section, we treat modular arithmetic with exponentials. As an introduction, let us consider the following typical elementary number theory problem, which can be solved by modular arithmetic:
\[2^n +5 =m^2.\]
Consider the remainder by $8$. If $n$ is greater than or equal to $3$, the left-hand side is congruent with $5$, and the right-hand side is one of square remainders of $8$, which are $0,1,4$, and it's absurd. Therefore, it suffices to check all the cases where $n=0,1,2$, and we find that $(n,m)=(2,3)$ is the only solution.
In this example, when considering exponentiation modulo $8$, we should divide \dq{first exceptions}, namely three numbers $n=0,1,2.$
The theory of modulo riegs is involved here. The modular arithmetic of modulo $b$ requires first exceptions and that the number of them, $a$ must (not surprisingly) be greater than or equal to any number that appears as an exponent in the prime factorization of $b$. The rieg of modulo $b$ with first exceptoins less than $a$ is denoted by $\Mr{a}{b}$. For exmple, in the above equation, we utilized the modulo rieg $\Mr{3}{8}$. In this section we classify what $a,b$ is possible.
\subsubsection*{Theoretical motivation}
There is another more theoretical motivation to classify all modulo riegs.
For rings, the quotient rings of the initial ring $\Z$ are easily classified and described as $\Z/n\Z$ using natural numbers $n$. This classification is important even if you are not interested in integer ring $\Z$ itself. It is because it can be used to define the \emph{characteristic} of a ring $R$ (or usually, fields), considering the unique ring homomorphism
\[\Z \to R \text{ in }\Ring\]
and its surjection-injection factorization.
We can establish similar characteristics theory for riegs, utilizing quotient algebras of the initial rieg $\N$ (i.e. modulo riegs) and
% We call them \emph{modulo riegs}.
% As in the ring case, we can discuss the theory of rieg's \dq{characteristic},
considering the unique rieg homomorphism
\[\N \to R\text{ in }\Rieg\]
and its surjection-injection factorization.
Just as examining the prime ideals of an integer ring $\Z$ reveals that the characteristics of any fields are either zero or prime, examining the modulo rieg (of $\N$) leads to properties about arbitrary riegs.
For example, see Corollary \ref{CorollaryClassificationOfCharacteristics}.
\subsubsection{Classification of modulo riegs}
Before we get into the theoretical considerations, let us list a few examples.
\begin{example}[$\N$ itself]
$\N$ itself is a trivial modulo rieg.
\end{example}
\begin{example}[Trivial rieg]
A singleton $\{\ast\}$ has a unique rieg structure. We call it \emph{trivial rieg}. It is the terminal rieg and can be regarded as a modulo rieg by the unique homomorphism $\N \to \{\ast\}$.
\end{example}
\begin{example}[Truth value rieg]
As we will see later, the set of truth values $\2 = \{\bot, \top\}$ has a natural rieg structure, as $(\2, \bot, \top, \lor, \land, \leftarrow)$.
% \begin{itemize}
% \item $0$ is the $\bot$
% \item $1$ is $\top$
% \item $x+y$ is $x \lor y$
% \item $xy$ is $x\land y$
% \item $x^y$ is $y \to x$
% \end{itemize}
The proposition
\[\text{IsPositive}\colon \N \twoheadrightarrow \2\]
is a surjection and defines a modulo rieg.
% The set $\{\text{zero},\text{positive}\}$ and the natural surjection $\N \to \{\text{zero},\text{positive}\}$ induces a rieg structure on $\{\text{zero},\text{positive}\}$.
% For several reasons, we call this rieg $\2$.
\end{example}
\begin{example}[Parity rieg]
As a bit more non-trivial example, the natural surjection
\[\N \twoheadrightarrow \{0, \text{positive odd}, \text{positive even} \}\]
induces a modulo rieg. We call it the \emph{parity rieg}.
\end{example}
The starting point for modulo rieg classification is the classification of quotients of the additive monoid $\N$.
% By the classification of quotients of the additive monoid $(\N,0,+)$,
It is known that all non-trivial congruence $\sim$ on $(\N,+ 0)$ is of the form
\[n \sim m \iff (n=m)\lor ((n,m\geq a) \land (n\equiv m \mod{b})) \]
% \[n \sim m \iff (n\equiv m \mod{b})\land ((n=m)\lor (n,m\geq a)) \]
with $a\geq0$ and $b>0$. Let $\Mr{a}{b}$ denote the quotient additive monoid. It is quicker to look at the picture of $\Mr{3}{4}$ than to look at the logical symbols (Figure \ref{PictureModuleRieg}).
\begin{figure}
\begin{shaded}
\centering
\begin{tikzpicture} [scale = 2]
\fill[black] (0,2) circle (0.06) node[above]{};
\fill[black] (1,2) circle (0.06) node[above]{};
\fill[black] (2,2) circle (0.06) node[above]{};
\fill[black] (3,2) circle (0.06) node[above]{};
\fill[black] (4-0.36,1+0.36) circle (0.06) node[above]{};
\fill[black] (3,0+0.36+0.36) circle (0.06) node[above]{};
\fill[black] (2+0.36,1+0.36) circle (0.06) node[above]{};
\draw[black, thick](3,1+0.36) circle (0.64);
\draw[black, thick] (0,2) -- (3,2);
\end{tikzpicture}
\caption{Picture of $\Mr{3}{4}$}
\label{PictureModuleRieg}
\end{shaded}
\end{figure}
For example, (the underlying additive monoid of) the trivial rieg is $\Mr{0}{1}$, the truth value rieg $\2$ is $\Mr{1}{1}$, and the parity rieg is $\Mr{1}{2}$.
Before stating the classification theorem, we introduce two terminologies.
\begin{definition}[Tame number]
A positive integer $n$ is said to be \emph{tame} if, for any prime factor $p$ of $n$, $p-1$ also divides $n$.
\end{definition}
The list of tame numbers begins as:
\[1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 32, 36, 40, 42, 48, 54, 60, 64, 72, 80, 84, 96, 100, 108, \dots.\]
Notice that there are infinitely many tame numbers since $2^k$ is tame.
The reason for considering tame numbers is the following algebraic paraphrase.
Simply put, \dq{$n$ is tame if and only if exponentials are also well-defined by $\mod{n}$.}
\begin{lemma}[Algebraic characterization of tame numbers]\label{LemmaAlgebraicCharacterizationOfTameNumbers}
For a positive integer $n$, the following conditions are equivalent:
\begin{enumerate}
\item $n$ is tame.
\item for any $x \in (\Z/n\Z)^{\times}$, $x^n \equiv 1 \mod{n}.$
\end{enumerate}
\end{lemma}
\begin{proof}
First, we will prove (2), assuming (1).
Let $n={p_1}^{k_1}\dots {p_l}^{k_l}$ be the prime factorization of $n$.
Then we have a group isomorphism
\[
(\Z/n\Z)^{\times} \cong (\Z/ {p_1}^{k_1} \Z)^{\times}\times \dots \times (\Z/ {p_l}^{k_l} \Z)^{\times}.
\]
Therefore, it is enough to prove that, for any $1\leq i \leq l$ and $x \in (\Z/ {p_i}^{k_i} \Z)^{\times}$, $x^n \equiv 1 \mod{{p_i}^{k_i}}$. But this immediately follows from tameness of $n$, since $\# (\Z/ {p_i}^{k_i} \Z)^{\times} = (p-1)p^{k_i -1}$, which divide $n$.
Conversely, we will prove (1) from (2). Let $p$ be an odd prime factor of $n$. (The case where $p=2$ is obvious.) The multiplicative group $(\Z/n\Z)^{\times}$ has a element $x$ whose order is $p-1$, because $(\Z/ {p}^{k} \Z)^{\times}$ is cyclic. \footnote{This is not obvious by definition, but well-known.} By the assumption $x^n \equiv 1 \mod n$, we have $p-1 \mid n$.
% \[(\Z/ p^{k_l} \Z)^{\times} \simeq \Z/(p-1)p^{}\]
\end{proof}
\begin{remark}[Carmichael numbers]
This is reminiscent of Carmichael numbers, which are defined as a positive integer $n$ that satisfies
\begin{itemize}
\item for any $x \in (\Z/n\Z)^{\times}$, $x^{n-1} \equiv 1 \mod{n}.$
\end{itemize}
Although the definitions are very similar, the only tame Carmichael numbers are obviously only trivial ones, namely $1$ and $2$. In light of these facts, rather than the boring name \dq{tame numbers,} perhaps we should call them anti-Carmichael numbers (?)
\end{remark}
\begin{definition}[Dimention]
A \emph{dimension} of a positive integer $n$ is the maximum value of the exponent that appears in the prime factorization of $n$. It is denoted by $\di(n)$.
\end{definition}
For example, the dimentions of small numbers are $\di(1)=0, \di(2)=1, \di(3)=1, \di(4)=2, \di(5) = 1, \di(6)=1, \di(7)=1, \di(8)=3$.
\begin{remark}[Radical]\label{RemarkRadical}
A dimension is inextricably linked to the notion of \emph{radicals} of positive integers.
\memo{ABC conjecture} Recall that the radical of a positive integer $n$, often denoted by $\rad(n)$, is a positive integer obtained by replacing all exponents in the prime factorization of $n$ by $1$. For example, $\rad(5400)=\rad(2^3\cdot 3^3\cdot 5^2)=2\cdot 3\cdot 5 = 30$.
The dimension $\dim(n)$ is the minimum $d$ such that $n$ divide $\rad(n)^d$, and conversely $\rad(n)$ is the maximum $1$-dimensional divisor of $n$.
\end{remark}
\begin{theorem}[Classification of modulo riegs \cite{burris1993tarski}]
For $a\geq 0$ and $b>0$, $\Mr{a}{b}$ has a rieg structure if and only if $b$ is a tame number whose dimension is at most $a$.
% \begin{enumerate}
% \item $a\geq \di(b)$
% \item $b$ is tame.
% \end{enumerate}
\end{theorem}
\begin{proof}
Assuming $\Mr{a}{b}$ has a rieg structure, we will prove $b$ is a tame number whose dimension is at most $a$. Let $[x] \in \Mr{a}{b}$ denote the equivalence class that contains $x\in \N$. For an arbitrary natural number $x$ that is coprime with $b$, we have
\[
[x^{a+b}]=[x]^{[a+b]}=[x]^{[a]}=[x^{a}],
\]
hence \[x^{a+b} \equiv x^{a} \mod{b},\] and \[x^b \equiv 1 \mod{b}.\] By Lemma \ref{LemmaAlgebraicCharacterizationOfTameNumbers}, $b$ is tame. Next, we move on to the dimension of $b$. For any non-negative integer $m\geq 0$, we have
\[[\rad(b) ^a]=[\rad(b)]^{[a]}=[\rad(b)]^{[a+mb]}=[\rad(b)^{a+mb}],\]
and \[\rad(b) ^a\equiv \rad(b) ^{a+mb} \mod{b}.\]
By taking a sufficiently large $m$ such that $a+mb\geq \dim(b)$, we obtain \[\rad(b) ^a\equiv 0 \mod{b},\] equivalently, $a\geq \dim(b)$.
Conversely, assuming that $b$ is a tame number whose dimension is at most $a$, we will prove the rig $\Mr{a}{b}$ has a rieg structure. It is enough to prove that $\ex$ is well-defined.\footnote{Here, we implicitly use the fact of universal algebra: if an equivalence relation is compatible with all operations, then its quotient set has the unique (same type of) algebraic structure such that the canonical surjection is homomorphic.} If $a=0$, then $b=1$, and this case is trivial. In the rest of the proof, we assume $a>0$.
We begin with the easier case: assuming $[x_0]=[x_1]$, we will prove $[{x_0} ^{y}]=[{x_1}^{y}]$. If $x_0 = x_1$ or $y=0$, this is trivial. Suppose $x_0 < x_1$ and $y>0$. Since $x_0 \neq x_1$, we have $a\leq x_0,x_1$ and ${x_0} \equiv {x_1} \mod{b}$. Since $y>0$, we have $a\leq x_0,\leq {x_0}^y\leq {x_1}^{y}$. As a result, we obtian $a\leq {x_0}^y, {x_1}^{y}$ and ${x_0}^y \equiv {x_1}^{y}\mod{b}$, which mean $[{x_0} ^{y}]=[{x_1}^{y}]$.
Lastly, assuming $[y_0]=[y_1]$, we will prove $[x ^{y_0}]=[x^{y_1}]$. If $y_0=y_1$, this is trivial. Therefore, we can assume $0<a\leq y_0, y_1$ and $y_0\equiv y_1 \mod{b}$. Since the cases where $x=0,1$ are trivial, \footnote{For $x=0$ case, we used $0< y_0,y_1$.} we can also assume $x\geq 2$. Under those assumptions, because we have $a\leq y_0 \leq x^{y_0}$ and $a\leq y_1 \leq x^{y_1}$, it is enough to prove
\[x^{y_0}\equiv x^{y_1} \mod{b}.\]
Take an arbitrary prime factor $p$ of $b$ and let $k$ be the number of times that $p$ divides $b$. We prove
\[x^{y_0}\equiv x^{y_1} \mod{p^k}.\]
If $p$ divides $x$, both sides of the equation are equal to zero, because \[k\leq \dim(b) \leq a \leq y_0,y_1.\]
If $x$ is coprime with $p$, it follows from tameness of $b$ and $\# (\Z/p^k \Z)^{\times} = (p-1)p^{k-1}$.
The proof is complete.
\end{proof}
As a result, we obtain the complete (infinite) list of modulo riegs (Figure \ref{PictureListOfModuloRiegs}).
\begin{figure}
\centering
\begin{shaded}
\begin{align*}
a=0 &:\ \Mr{0}{1}\\
a=1 &:\ \Mr{1}{1},\Mr{1}{2},\Mr{1}{6},\Mr{1}{42},\Mr{1}{1806}\\
a=2 &:\ \Mr{2}{1},\Mr{2}{2},\Mr{2}{4},\Mr{2}{6},\Mr{2}{12},\Mr{2}{18},\Mr{2}{20},\Mr{2}{36},\Mr{2}{42},\Mr{2}{60},\Mr{2}{84},\dots \\
a=3 &:\ \Mr{3}{1},\Mr{3}{2},\Mr{3}{4},\Mr{3}{6},\Mr{3}{8},\Mr{3}{12},\Mr{3}{18},\Mr{3}{20},\Mr{3}{24},\Mr{3}{36},\Mr{3}{40}, \dots \\
a=4 &:\ \Mr{4}{1},\Mr{4}{2},\Mr{4}{4},\Mr{4}{6},\Mr{4}{8},\Mr{4}{12},\Mr{4}{16},\Mr{4}{18},\Mr{4}{20},\Mr{4}{24},\Mr{4}{36}, \dots \\
\vdots&\\
\dq{a=\infty}&:\ \N
\end{align*}
\caption{The complete list of modulo riegs}
\label{PictureListOfModuloRiegs}
\end{shaded}
\end{figure}
As promised, this classification theorem implies non-trivial properties about general riegs. The following corollary is particularly impressive.
% As a corollary of the classification theorem, we obtain the following proposition:
\begin{corollary}[Classification of characteristics, case $a=1$ \cite{burris1993tarski}]\label{CorollaryClassificationOfCharacteristics}
For a rieg $R$, if there exists the minimum positive integer $n$ such that
\[1=1+\underbrace{1+\dots +1}_{n},\]
then there are only five possible values of $n$, $n=1,2,6,42,1806$.
\end{corollary}\memo{I thought this might be a new result.}
\begin{proof}
This is equivalent to saying that $1,2,6,42,1806$ are the only tame numbers of dimension at most $1$. If $n$ is a $1$-dimensional tame number and $p$ is the maximum prime factor, then $n/p$ is also a ($0$ or $1$-dimensional) tame number and $p-1$ is a divisor of $n/p$. Therefore, we can inductively generate all $1$-dimensional tame numbers. By concrete calculations, we can check that the inductive procedure stops when $n=1806$.
The concrete calculation is conducted as follows:
Start with $n=1$. Adding $1$ to the divisor of $n=1$ and checking if it is prime, we find prime $1+1=2$ and newly obtain $n=1*2=2$.
Again, adding $1$ to the divisors of (newly obtained) $n=2$ and checking if they are prime, we find two prime numbers $2$ and $3$. Thus we obtain a new number $n=2*3=6$.
We repeat this procedure.
Adding $1$ to the divisors of $n=6$ and checking if they are prime, we find prime numbers $2, 3$, and $7$ and obtain $n=6*7=42$.
Again, adding $1$ to the divisors of $n=42$ and checking if they are prime, we find primes $2, 3, 7$, and $43$ and newly obtain $n=42*43=1806$. Adding $1$ to the divisors of $n=1806$ and checking if they are prime, we find only the already found primes $2,3,7,43$. (For example, $86+1=3\cdot 29$, $301+1=2\cdot 151$, and $1806+1 = 13\cdot 139$.) Thus this procedure is over, and all possible $n$ are $1,2,6,42,1806$.
\end{proof}
\memo{Sylvester's sequence}
\begin{conjecture}
There are infinitely many 2-dimensional tame numbers.
\end{conjecture}
\cite{OEISsyl}\cite{burris1993tarski}
By a computer calculation, we know there is a $2$-dimensional tame number bigger than $10^{10000}$.
\memo{In the case of $1$-dimensional tame numbers, there are only $4$ different prime number appear, $p=2,3,7,43$. By computer calculation, we observed that at least $1000$ prime numbers appear in $2$-dimensional tame numbers.}
% However, not all $\Mr{a}{b}$ has a rieg structure. The obvious restriction is
% \begin{itemize}
% \item $a=0 \implies b=1$,
% \end{itemize} and for something a bit more non-trivial,
% \begin{itemize}
% % \item the order of elements of $(\Z/b\Z)^{\times}$ must be divisors of $b$.
% \item if a prime number $p$ divides $b$, then $p-1$ must divide $b$.
% \end{itemize}
% As we will see later, for $a=1$, there are exactly five possible $b$, namely $1,2,6,42,1806$.
\subsubsection{Characteristics of riegs}
\begin{definition}
The \emph{characteristic} of a rieg is \memo{write}
\end{definition}
\memo{Solve some Diophantus equation utilizing those riegs}
% \section{Relation to other algebraic structures}
\subsubsection{Application: riegs of three elements}
$n=1,2$: Unique
\memo{Utilize characteristics theory}
\newpage
\subsection{Riegs of Propositions}
\subsubsection{The rieg of truth values}
\begin{example}[Rieg of truth values]\label{ExampleRiegOfTruthValues}
The set of truth values $\2 = \{\bot, \top\}$ has a natural rieg structure, as $(\2, \bot, \top, \lor, \land, \leftarrow)$. This is isomorphic to the overflow rieg with threshold $2$.
\end{example}
\begin{example}[Powerset]
For any set $X$, its powerset $\Po{X}$ has a rieg structure with $(\Po{X}, \emptyset, X, \cup,\cap, \leftarrow)$, where $A\leftarrow B$ is defined to be $A\cup B^{\co}$. More generally, every Heyting algebra\footnote{or CCC} is a rieg. \memo{Then, there is a free construction of \dq{associated Heyting algebras}}
When $X=\emptyset$, its powerset rieg is the trivial rieg. When $X$ is a singleton, its power rieg is the rieg of truth values.
\end{example}
% \subsection*{Relation to Heyting algebras}
\subsubsection{Heyting algebras are riegs}
A Heyting algebra is not just related but is a rieg satisfying additional equations.
\begin{proposition}
Every Heyting algebra $(H,0,1,\lor,\land, \leftarrow)$ is a rieg.
\end{proposition}
\begin{proposition}
A rieg $A$ is a Heyting algebra, if and only if it satisfies the following additional equations:
\begin{itemize}
% \item $x\ti x=x$
% \item $1+1=1$ %$x+x = x$
\item $x^x =1$
\item $x(y+1)=x$ %$$xy+y=y$
\item $x^{y+1}=x$ %$y y^x = y$
% \item $(x+y)y=y$
\item $x y^x = xy$
\end{itemize}
\end{proposition}
\begin{corollary}
The category of Heyting algebras $\HeyAlg$ is a reflective full subcategory of the category of riegs $\Rieg$.
\[
\begin{tikzcd}
\HeyAlg \ar[r, shift right, hookrightarrow]&\Rieg \ar[l, shift right]
\end{tikzcd}
\]
\end{corollary}
The construction of the left adjoint is the usual one: quotient by new equations. For example, the associated Heyting algebra of $\N$ is the Heyting algebra of truth values $\{\bot, \top\}.$\footnote{This particular example is a trivial case, since a left adjoint preserves colimits.}
The relation between Heyting algebras and riegs is not an addition of structures, but an addition of properties. In this respect, it is closer to the relationship between abelian groups and groups rather than to the relationship between rings and groups.
\memo{Full sub. Birkhoff's theorem}
\memo{Is it Malcev?}
\newpage
\subsubsection{Associated Heyting algebra of a rieg}
\memo{Sheaf topos over a top.sp.}
\memo{Or, more generally, how about a topos?}
\subsection{Riegs of Structures}
\subsubsection{Bicartesian closed category forms a rieg}
We have seen that every Heyting algebra is a rieg. More generally, in this section, we will observe that every \emph{bicartesian closed category} gives an example of riegs.
\begin{definition}[Bicartesian closed category]
A category $\C$ is \emph{bicartesian closed},
if $\C$ is
\begin{itemize}
\item bicartesian (i.e. has all finite products and finite coproducts), and
\item cartesian closed.
\end{itemize}
\end{definition}
Examples include the category of sets, finite sets, posets, categories, groupoids, directed graphs, and group actions. Every topos is bicartesian closed.
We will see those examples in the following subsections.
\begin{proposition}
For a bicartesian closed category $\C$, the set of all isomorphism classes $\K{\C}$ equipped with the five operations induced from bicartesian closed structure, is a rieg.
\end{proposition}
\begin{remark}[Size matter]
\end{remark}
\memo{Explain, why not monoidal, but cartesian}
\memo{For bicartesian closed bicategories? finite groupoids, finite groups}
\memo{Rieg of species? Differential operator?}
\subsubsection{Riegs of sets}
\subsubsection{Riegs of functions: Liouville's divisor theorem}
\begin{example}[Multisets of natural numbers]
Let $\mN$ be a set of all multiset of natural numbers. Each multiset will be denoted just like lists of natural numbers. For example, \[(), (3,1,4), (1,2,3,1,1,0,0) \in \mN.\] Note here that the list may be rearranged as desired, \[(1,2,3,1,1,0,0) = (0,0,1,1,1,2,3)\in \mN\]
This set of multisets has a standard rieg structure.
The addition is concatenations of lists. The product is a component-wise product.
\begin{align*}
(3,1,4)+ (1,2,7,0)&=(3,1,4,1,2,7,0)\\
(3,1,4)\ti (1,2,7,0)&= (3,6,21,0,1,2,7,0,4,8,28,0)
% (3\ti 1,3\ti 2,3\ti 7, 3\ti 0, 1\ti 1, 1\ti 2, 1\ti 7, 1\ti 0, 4\ti 1, 4\ti 2, 4\ti 7, 4\ti 0)
\end{align*}
This rig structure is isomorphic to the rig of (formal) Dirichlet polynomials $\N[\frac{1}{n^x}\mid n=0,1,2,\dots]$ \cite{spivak2020dirichlet}. However, exponentials are different:
\[(n_1, \dots,n_k)^{(m)}=({n_1}^m , \dots, {n_k}^m ).\]
There is a unique extension of this exponential to all exponentials. For example,
\begin{align*}
(3,1,4)^{ (2,0)} &= (3,1,4)^{ (2)} (3,1,4)^{ (0)} \\
&=(9,1,16)(1,1,1)\\
&=(9,9,9,1,1,1,16,16,16).
\end{align*}
\end{example}
\memo{Write the generalization and $\mult{1}$}
\subsubsection{Riegs of graphs}
\subsubsection{Riegs of group actions: Burnside riegs}
\memo{rieg homomorphism to $\N$, which induced by atomic geometric morphism}
\paragraph{Rieg of involutions}
$A^x = n + \frac{n^2 -n}{2} x$, where $n= \#A$.
\subsubsection{Riegs of loops: Counting repeating dicimals}
\subsection{Riegs of Higher structures}
% \subsubsection{Bicartesian closed bicategory}
\subsubsection{Riegs of categories}
\subsubsection{Riegs of finite groups}
Since the category of finite groupoids $\Groupoidfin$ is bicartesian closed, we have the riegs of finite groupoids, $\K{\Groupoidfin}$.
Furthermore, since all five operations are compatible with equivalences, we have its quotient $\biK{\Groupoidfin}$.
\memo{Using the rieg structure, compute the number of subgroups of order $2$, faster!}
\memo{compute $n$-dimensional representation of a product group}
\subsection{Functorial construction of riegs}
\subsubsection{Functor from a locally bicartesian closed category}
\subsubsection{Examples of induced rieg homomorphisms}
\subsubsection{Categories and discrete fibrations}
\memo{categorification of the multiset construction}
\subsubsection{Topological spaces and \'etale maps}
\subsubsection{Quasitoposes}
\memo{Heyting algebra}
\cite{nlab:quasitopos}
\begin{definition}
\end{definition}
\begin{example}
The category of (finite) categories and (finite) discrete fibrations.
\end{example}
\begin{example}
The category of topological spaces and local homeomorphisms.
\end{example}
% \section{Rieg of multisets, Liouville's divisor theorem}
% \section{Riegs from toposes, repeating decimal}
\memo{CC and CCC functor, \'etendue, logical morphism, directed graph, and group actions}
\newpage
% \input
\section{Structures of riegs}
\subsection{Canonical preorder of a rieg}
\begin{definition}[Canonical preorder of a monoid]
The \emph{canonical preorder} of a monoid $(M, \ast, e)$ is defined by \[x\leq y \iff \exists z\in M ( x \ast z = y).\]
\end{definition}
\begin{definition}[Canonical preorder of a rig]
The \emph{canonical preorder} of a rig or rieg is the canonical preorder with respect to the addition.
\end{definition}
\begin{example}
The canonical preorder of the natural number rieg $\N$ is the usual total order.
\end{example}
This preorder is used in several contexts.
\begin{remark}[Dioids]\label{RemarkDefinitionOfDioids}
A rig $R$ is called \emph{dioid} if the canonical preorder is antisymmetric (i.e., partial order). This structure is used for algorithm theory \cite{gondran2008graphs}. As mentioned in Remark \ref{RemarkDioidsandRiegs}, until now, the author does not know the direct connection between dioids and riegs.
\end{remark}
\begin{definition}[Minimal element]
An element $x\in R$ of a rieg $R$ is said to be minimal if it satisfies the following two conditions:
\begin{enumerate}
\item $\lnot(x\leq 0)$
\item For any $0\leq y \leq x$, $y=0$ or $y=x$.
% \item If $x=y+y'$, then $y=0$ or $y'=0$.
\end{enumerate}
\end{definition}
The second condition is equivalent to saying that ``
If $x=y+y'$, then $y=0$ or $y'=0$." This condition is some sense of indecomposability.
\begin{question}
Can we replace the first condition with a more simply looking condition $x \neq 0$? In other words,
Is $0$ the only minimum element in a rieg? Is $0$ the only element that has the additive inverse?
No! by the dual number rieg, see subsubsection \ref{SubsubsectionDualNumbers}.
\end{question}
\subsection{Connected Element}
\begin{definition}
An element $x\in R$ of a rieg $R$ is said to be \emph{connected} if it satisfies the following two conditions:
\begin{enumerate}
% \item $(-)^{x}\colon R \to R$ is a rig homomorphism.
\item $0^x= 0$
\item $(a+b)^x=a^x + b^x$ .
\end{enumerate}
\end{definition}
In other words, $x$ is connected if $(-)^{x}\colon R \to R$ is a rig homomorphism.
\begin{proposition}[Internally connected objects and connected elements]
\end{proposition}
\subsection{Connectedly based rieg}
\begin{proposition}
For a connectedly based rieg,
\[0^x = \]
\end{proposition}
\memo{Heyting, locally connected, strict initial}
% \input{FutureWorks}
% \appendix
% \input{Programs}
\printbibliography
\appendix
\section{Rigs}
\subsection{definition}
\subsection{augmentated rigs}
For our purpose of applying rieg theory to enumerative combinatorics, it is natural to consider homomorphism from a ri(e)g to $\N$. We call such a structure, augmentation. \memo{It's conventional in ring theory}
\begin{definition}
For a rig $R$, an ($\N$-)augmentation of $R$ is a rig homomorphism $R\to \N$.
\end{definition}
\end{document}