What are some more examples that are elementary of GCD domains that are not unique factorization domains?
For my personal notes, I'm looking for examples of rings that belong to one rung but not the next in Wikipedia's hierarchy:
rngs ⊃ rings ⊃ commutative rings ⊃ integral domains ⊃
integrally closed domains ⊃ GCD domains ⊃ unique factorization domains ⊃
principal ideal domains ⊃ Euclidean domains ⊃
fields ⊃ algebraically closed fields
I'm trying to fill out each level with examples that are as elementary as possible.
I'm looking for examples of a non-UFD GCD domains besides the algebraic integers and the ring of entire functions on $\mathbb{C}$ or the monoid ring $(k, \mathbb{Q}_{\ge 0})$.
One example that I thought of earlier today are the multiplicative functions from $\mathbb{N} \to \mathbb{Z}$, let's call this ring $R$. Addition, subtraction, and multiplication are pointwise. Multiplicativeness is a little bit of a red herring here, since I'm effectively just treating these functions as sequences of integers.
These are the functions such that $f$ is constantly zero or all of the conditions hold:
- $f(0)$ is $0$.
- $f(1)$ is $1$.
- $f(n)$ for positive $n$ is $f(p_1)^{a_1}f(p_2)^{a_2}\cdots$ where $p_1^{a_1}p_2^{a_2}\cdots$ is the prime factorization of $n$.
The GCD of two functions in $R$ is defined componentwise on $(f(0), g(0)), (f(1), g(1)), \{ (f(p), g(p)): \;\text{$p$ prime} \}$. The GCD is only defined up to a unit, but that's okay because any multiplicative function with image $\{\pm 1\}$ is a unit here.
However, we can show that this function does not have unique factorization by considering the identity function $\text{id}$.
$\text{id}$ is an identity function under composition (temporarily disregarding the fact that the domain and codomain are not identical). However, under multiplication, which is pointwise, $\text{id}$ can be expressed as the following infinite product (temporarily disregarding the fact that $R$ does not have enough structure to allow us to take limits):
$$ \text{id} = \prod_{p \;\text{prime}} \Delta[p] $$
In the above $\Delta[p]$ is the function that sends $0$ to $0$, $1$ to $1$, and every prime $q$ to $q$ if $p=q$ and $1$ otherwise.
The existence of the above infinite product means that any purported factorization of $\text{id}$ (which must be finite) can be further broken down. Thus $\text{id}$ doesn't have a unique factorization up to units, thus $R$ is not a UFD.
