A maximal regular ideal of commutative ring is prime?

Question

Let $A$ be a commutative ring. An ideal is called regular if it is generated by idempotents. If we consider a maximal (between regular ideals) regular ideal, is true that it is prime?

Answer 1

It seems that $F\times\mathbb Z_4$ is a counterexample for any field $F$. Namely, $I=F\times\{0\}$ is a maximal regular ideal generated by the idempotent $(1,0)$, but $(0,2)^2=(0,0)\in I$ and $(0,2)\notin I$, so it is not prime.