When a maximal ideal is not contained the union of the others Let $R$ be a commutative ring with 1 and $a\in R$ such that $a\not=a^2$ and assume that there is an ideal $I$ of $R$ such that $a\not\in I$ and if $J$ is an ideal of $R$, then $(a\not\in J)\Rightarrow (J\subseteq I)$. Set $m:=\{r\in R\mid ra\in I\}$ and assume that $m $ is a maximal ideal.

Can we show that $m\not\subseteq \bigcup_{n\in\operatorname{Max}(R)\setminus \{m\}}n$, where $\operatorname{Max}(R)$ is the set of all maximal ideals of $R$? If that is not true under what conditions on $a$ it is true?

I am studying zero-dimensional rings in which every prime ideal has the form $m$ as above for some element $a$.  In the case that $a=a^2$, then the result is true since then $a$ is a local idempotent and $m$ is the only maximal ideal that contains $1-a$.  I'm hoping it is still true in general that there is a local idempotent $e$ such that $1-e\in m$, which would imply the result.
 A: Here is a counterexample.  Let $k$ be a field and let $S$ be the ring of eventually constant sequences with values in $k$; we write $s(\infty)$ for the eventually constant value of an element $s\in S$.  Let $I\subset S$ be the ideal of sequences that are eventually $0$, and let $R=S[a]/(a^2, Ia)$.  Every element of $R$ can be written uniquely in the form $s+ca$ for $s\in S$ and $c\in k$.  Such an element divides $a$ iff either $s(\infty)=0$ and $c\neq 0$ (multiply by $t\in S$ that vanishes on the support of $s$ and such that $t(\infty)=1/c$) or $s(\infty)\neq 0$ (multiply by $a/s(\infty)$).  So, the elements that do not divide $a$ are just the ideal $I$, which thus is the largest ideal that does not contain $a$.  Your ideal $m$ is then $I+(a)$, which is a maximal ideal.  But every element of $m$ is in some other maximal ideal: for each $n\in\mathbb{N}$ there is a maximal ideal consisting of those elements $s+ca$ such that $s(n)=0$, and every element of $I+(a)$ has the form $s+ca$ for $s\in I$ and thus $s$ vanishes at some (in fact, all but finitely many) $n\in\mathbb{N}$.
On the other hand, it is true if you assume the annihilator of $a$ is a principal ideal.  Indeed, let $J\subseteq R$ be the annihilator of $a$.  Note that $m$ is exactly the set of elements of $R$ that are not units mod $J$, since $r\not\in m$ iff $ra\not\in I$ iff $ra\mid a$ iff $a(rs-1)=0$ for some $s\in R$ and the last condition says exactly that $s$ is an inverse of $r$ mod $J$.  So, $m$ is the unique maximal ideal containing $J$.  If $J$ is principal, this means $m$ cannot be covered by other maximal ideals, since $m$ is the only maximal ideal that contains a generator of $J$.
