Let $R$ be an integral domain. Define a "reduced class" in $R$ to be any subset $X$ of $R$ that is a multiplicative submonoid and such that every element of $R$ has exactly one associate (unit multiple) in $X$. In particular, $X$ is a skeleton of $R$ viewed as a category corresponding to the divisibility preorder |.
Examples of "reduced classes" include the nonnegative integers in $\mathbb{Z}$, the set $\{0,1\}$ in any field, and the monic polynomials together with the zero polynomial in $K[x]$, where $K$ is any field.
Question:
Does a "reduced class" always exist in any integral domain, assuming the axiom of choice?
If the answer is no, then please give a specific example of an integral domain that does not contain a "reduced class".
I think this will work as a counterexample . . .
Let $R=A/I$, where $A=F_3[x,y]$, and $I$ is the ideal of $A$ given by $$I=(x^2+y^2)$$
Then $I$ is a prime ideal of $A$, so $R$ is an integral domain.
Suppose $S\subset R$ is a reduced class.
For $a\in A$, let $\bar{a}$ denote the corresponding element of $R$.
Since the only units of $R$ are $\pm\bar{1}$, it follows for all $a\in A$, either $\bar{a}\in S$ or $-\bar{a}\in S$, hence $\bar{a}^2\in S$.
In particular, we have $\bar{x}^2,\bar{y}^2\in S$.
Then from $$x^2\equiv -y^2\;(\text{mod}\;I)$$ we get $\bar{x}^2=-\bar{y}^2$, hence from $\bar{x}^2\in S$, we get $-\bar{y}^2\in S$.
Since $\bar{y}\ne 0$, it follows that $-\bar{y}^2\ne\bar{y}^2$.
But then since $\bar{y}^2\in S$ and $-\bar{y}^2\in S$, we have a contradiction, since $-\bar{y}^2$ and $\bar{y}^2$ are distinct associates.