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Let $A$ be a (not necessarily unital)commutative ring. We define a regular ideal $I$ if exists element $e \in A$, such that $e$ is the unit in $A/I$.
Can we deduce, that every proper regular ideal is contained in a maximal regular ideal?
It seems that this is another Zorn's lemma argument: let $m_i$ be a chain of proper, regular ideals then $\bigcup m_i$ is a proper ideal, but is it regular?
The set of ideals of $A/I$ is in bijection with the set of ideals of $A$ which contain $I$. Since $A/I$ is unital, there exists a maximal ideal in $A/I$ and this corresponds to a maximal ideal $J$ of $A$ containing $I$. Note that if $e \in A$ is such that $e + I \in A/I$ is the unit, then $e + J \in A/J \cong (A/I)/(J/I)$ is the unit, too. In fact, we have $ea - a, ae - a \in I \subseteq J$ for all $a \in A$. Note that this shows that every ideal containing a regular ideal is already regular.
