TITLE: Associated elements in a ring
QUESTION [8 upvotes]: Please help me to find elements $a,b$ in a ring $R$ such that $a\mid b$ and $b\mid a$, but there does not exist any unit $u$ in $R$ such that $a=ub$.

REPLY [9 votes]: This is a non-trivial example taken from some notes I had lying around (I was being flippant earlier simply because it does not make sense to define a unit in a non-unital ring) :
Consider $R = \mathbb{Q}[x,y,z]/(x-xyz)$, and denote by $\overline{f}$ the image of $f\in \mathbb{Q}[x,y,z]$ in $R$. Now note that
$$
\overline{x} = \overline{xy}\overline{z}
$$
and hence
$$
\overline{x} \mid \overline{xy} \text{ and } \overline{xy} \mid \overline{x} \text{ in } R
$$
I claim that there does not exist $\overline{f} \in R^{\ast}$ such that
$$
\overline{f}\overline{x} = \overline{xy}
$$
Suppose such an $f \in \mathbb{Q}[x,y,z]$ existed, then $fx - xy \in (x-xyz)$, whence $f-y \in (1-yz)$. So there exists $h \in \mathbb{Q}[x,y,z]$ such that
$$
f = y + h(1-yz)
$$
Now suppose $\overline{f}$ is a unit, then it must follow that
$$
(y+h(1-yz),x-xyz) = \mathbb{Q}[x,y,z]
$$
But, by setting $x=0, y=z$, one gets
$$
(z+h(1-z^2)) = \mathbb{Q}[z]
$$
Check that this is not possible.