There is a question on the site that discusses why divisibility on an integrity domain $A$ is not an order. Is there a characterization of the case when divisibility forms an order? I think I could have:
$|$ forms an order on $A \iff U(A) = \{1\}$
So the condition is that the set of units is trivial. To help me to prove it I have that:
Given $a,b \in A - \{0\}$ then $a \sim b \iff a|b \land b |a$.
$\sim$ is the associates relation ($a \sim b \iff \exists u \in U(A). b = ua$). I need to prove antisymmetry:
$\forall a,b \in A. a|b \land b|a \implies a = b$.
My attempt
$\Rightarrow)$ Let's assume that I pick two non-zero elements $a,b \in A$ and assume that $a|b \land b|a$ then we know that $a \sim b$. If $a = b$ we are done. Otherwise there exist some unit $u \in U(A) - \{1\}$ such that $b = ua$.
I don't know how to get a contradiction.
$\Leftarrow)$ If $U(A) = \{1\}$ and I take two elements $a,b \in A - \{0\}$ then if $a|b \land b|a \implies a \sim b$ and the only unit is 1 so they must be equal. If one of the elements is zero and still $a|b \land b|a$ clearly the other must be zero as well.
What I need
I need help with the right implication. However, if does not hold then I would ask you for an alternative characterization.
I had to paraphrase what you have written so far:
If $u\neq 1$, then $ua\neq a$, and hence $b\neq a$ and $|$ is not an order. Hence the assumption that $u\neq 1$ is incorrect, and the only possibility is $u=1$.
I think your characterization is spot on. Divisibility will only form an order when $1$ is the only unit. But it always forms a pre-order. That is why the concept of associate elements is so useful: it allows us to substitute the divisibility pre-order on elements with the divisibility (partial-)order on associate classes.