Associated elements in a ring [duplicate]

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 $a=ub$.

• A ring without a $1$ should work (for instance, $C_0(\mathbb{R})$), because it has no units :) Dec 2, 2013 at 9:59
• I tried with $\dfrac{1}{x}$ and $e^{-x}$ but it didn't work. Will you please help? Dec 2, 2013 at 12:04

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.