For a non-empty set $S$ and associative binary operation $\ast$, the association $M=(S,\ast)$ is called a monoid if $S$ is closed under the operation $\ast$, and there exists an identity element $e\in S$ such that for all $a\in S$, $a\ast e=e=e\ast a$.
A monoid is called commutative/abelian if its operation is commutative.
A submonoid is a subset of the elements of a monoid that in themselves form a monoid under the same operation.
A monoid $(M,\ast)$ is cancellative if for all $a,b,c\in M$ we have:
For a commutative cancellative monoid $M$, if for some $x,y\in M$ we have $x\mid y$ ($x$ divides $y$), then there must exist a $z\in M$ such that $xz=y$.
This describes divisibility within the set of the monoid $M$. It may be important to denote divisibility within some specific set or monoid $S$, and so we write $x\mid_S y$. For example, if $x\mid y$ and $x,y\in\N$ then we could also write $x\mid_\N y$.
A unit of a monoid $M$, similar to a unit of a ring, is an element $x\in M$ for which there exists an $x'\in M$ such that $x\ast x'=e$.
Going forward, we'll assume $M$ to be a commutative cancellative monoid, and let $\unitsof M$ be the set of units of $M$.
A non-unit $x\in M$ (that is $x\in\nonunitsof M$) is irreducible/an atom if whenever $x=yz$ then either $y\in\unitsof M$ or $z\in\unitsof M$ (one is a unit, but not both). We write the $\atomset(M)$ for the set of atoms of $M$.
A non-unit $x\in M$ is prime if whenever $x\mid yz$ for some $y,z\in M$, either $x\mid y$ or $x\mid z$ (divides one, but not the other). An element being prime implies it is an atom.
A monoid $M$ is called atomic if every non-unit $x$ can be written as a product of atoms of $M$.
The unique factorization monoid $(\N,\times)$ is atomic in that every integer greater than 1 can be written as a product of powers of atoms (or more specifically here primes) that is unique up to the order of the atoms, as in the Fundamental Theorem of Arithmetic.
Since $M_{a,b}$ is a submonoid of $(\N,\times)$, it is also atomic.
Going forward, all monoids will be assumed to be atomic.
A monoid $M$ is factorial if every non-unit $x\in M$ has a unique factorization into atoms of $M$. $M$ is factorial if and only if all of its atoms are prime.
For a non-unit $x\in\nonunitsof M$, we define which we call the length set of $x$. These sets collected together is The length set of $M$.
The ratio $\rho(x)=\sup\lengthset(x)/\inf\lengthset(x)$ is called the elasticity of $x$, and the elasticity of the monoid is defined by
If $\rho(M)=1$ then $M$ is called half-factorial. For example, $M_{1,4}$ is half-factorial in that any reducible $x$ has factorization(s) of length 2 ($\rho(x)=2/2=1$) and any atom $u$ has $\rho(u)=1$, thus $\rho(M)=1$.
A monoid $M$ is said to be fully elastic if for every rational $q$ with $1\le q<\rho(M)$ there exists an $x\in\nonunitsof M$ such that $\rho(x)=q$; if there exists an $x\in\nonunitsof M$ such that $\rho(M)=\rho(x)$ then the elasticity of $M$ is said to be accepted.
Given $x\in\nonunitsof M$ we write its length set in increasing order where $n_i<n_{i+1}$ for $i\le i\le k-1$. The delta set of $x$ is defined by and the delta set of $M$ by
Let $M$ and $N$ be commutative, cancellative, atomic monoids, and $\sigma:M\to N$ a monoid homomorphism. The map $\sigma$ is a transfer homomorphism if:
For a set $P$, the free commutative monoid $\fcm(P)$ TODO: DEFINE
Let $M$ be a monoid. A divisor theory for $M$ is a free commutative monoid $\fcm(P)$ and a monoid homomorphism $\sigma:M\to\fcm(P)$ satisfying the following properties:
A monoid $M$ which has a divisor theory is said to be a Krull monoid. The generators $P$ are said to be the prime divisors of $M$, and the quotient monoid $\fcm(P)/\sigma(M)$ is known as the class group of $M$.
Given $a,b\in\N$ with $0<a\le b$ and $a^2\equiv a\bmod b$, define the arithmetic congruence monoid of $a$ and $b$:
Let $x,y,z\in\N$ such that $x=yz$. If $x,y\in M_{1,b}$ then $z\in M_{1,b}$.
Said another way, if for $x,y\in M_{1,b}$ we have that $x\mid_\N y$, then $x\mid_{M_{1,b}}y$. From this we say that $M_{1,b}$ is saturated in $\N$.
In contrast, $M_{b,b}$ is not since, for example, $2\mid_\N 6$ but $2\nmid_{M_{2,2}} 6$ since $6\notin M_{2,2}$.
Let $P=\{ p\in\N: p\text{ is prime and }\gcd(p,b)=1 \}$, and $\fcm(P)$ the free commutative monoid generated by $P$.
(For example, let $b=3$. Although $2$ it is prime, since $4\in M_{1,3}$ and $\gcd(2,4)=2\ne 1$, therefore $2\notin P$. I suppose it might be useful in the future to denote $P_b$.)
Let $P=\{ p\in\N: p\text{ is prime and }\gcd(p,b)=1 \}$. The free commutative monoid $\fcm(P)\le(\N,\times)$ and the homomorphism $\iota:M_{1,b}\hookrightarrow\fcm(P)$ form a divisor theory for $M_{1,b}$. Thus $M_{1,b}$ is Krull. (The hooked arrow $\hookrightarrow$ is to denote that $\iota$ is an inclusion map, symbolizing that $M_{1,b}$ is a subset of $\fcm(P)$.)
The ACM class $M_{1,b}$ being Krull is an instance of a general theorem that all saturated submonoids of factorial monoids are Krull, and all other ACMs besides $\N$ are not Krull. For ACMs that are multiples (as in $M_{b,b}$, $M_{b,2b}$, etc.), this is due to the correspondence with numerical monoids and translates of $(\N^n,+)$, both of which are not Krull.