# Every regular ideal contained in a maximal regular ideal?

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?

## 1 Answer

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.