A special ideal in a local ring Let $R $ be a local commutative ring with identity (may or may not be Noetherian), and let $0\ne a\in R $ be such that for every ideal $I $ of $R $  either $a\in I $ or there exists $r\in R $ such that $0\ne rI\subseteq \langle a\rangle $. How can we deduce that $\langle a\rangle $ is comparable with every ideal, that is, for every ideal $J $ we have either $\langle a\rangle \subseteq J$
or $J\subseteq\langle a\rangle $ ?

You can't deduce it : take $R$ to be any local domain and $a$ such that $\langle a \rangle$ is not comparable to every ideal. Then it satisfies the condition because for any nonzero ideal $I$, $0\neq aI\subset \langle a \rangle$. 
For instance, take $\mathbb Z [X]$ localized at $(2,X)$. In that ring take $a=2$ and $J=(X)$. Then $a\notin J$ (if it were, there would be $2 P = QX$ with $P\notin (2,X)$, looking at the minimal degree coefficient yields a contradiction) and $J$ is not included in $\langle a \rangle$ (similar proof)
