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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 $a=ub$.

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    $\begingroup$ A ring without a $1$ should work (for instance, $C_0(\mathbb{R})$), because it has no units :) $\endgroup$ – Prahlad Vaidyanathan Dec 2 '13 at 9:59
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    $\begingroup$ I tried with $\dfrac{1}{x}$ and $e^{-x}$ but it didn't work. Will you please help? $\endgroup$ – Anupam Dec 2 '13 at 12:04
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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 $$ (y+h(1-yz),x) = \mathbb{Q}[x,y,z] $$ But, by setting $x=0, y=z$, one gets $$ (z+h(1-z^2)) = \mathbb{Q}[z] $$ Check that this is not possible.

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