Let $R$ be a commutative ring with identity and $A$ and $B$ be two proper ideals of $R$ such that $A+B=R$ and for each $r^2=r\in R$ we have either $r\not\in A$ or $r-1\not\in B$. How can we prove the existance of two maximal ideals $m\_1$ and $m\_2$ of $R$ such that $A\subseteq m\_1$, $B\subseteq m\_2$ and $\{r\in m\_1\mid r^2=r\}=\{r\in m\_2\mid r^2=r\}$?



The existence of two maximal ideals with the same set of idempotents
Sketch: First, if $e$ is an idempotent in $A$, show that $B$ can be replaced with $B+Re$, and the hypotheses still holds. Use this to reduce to the case that $A$ and $B$ contain the same idempotents.


Second, if $e$ is an idempotent of $R$ with $e,1-e\notin A$ (and hence also not in $B$) show that we can replace $A$ and $B$ with the new pair $A+Re$ and $B+Re$, still satisfying the same conditions. Use this to reduce to the case that $A$ and $B$ contain the same idempotents, and either an idempotent or its complement belongs to $A$.


Now, extend to any maximal ideals containing $A$ and $B$, and note that a maximal ideal cannot contain an idempotent and its complement at the same time.


