One is $\mathbb{C}[x]/(x(x-1))$.
–
YoungsuMay 29 '13 at 18:25

2

Given $P$, it's quite unclear whether you're asking that $a$ be idempotent for every $a$ such that $P=Ann(a)$ or that $a$ be idempotent for some $a$ such that $P=Ann(a)$. If the latter, take the direct sum of two domains.
–
Steven LandsburgMay 29 '13 at 18:25

@Steven Landsburg:The latter case is what I mean:$P=Ann_R(a)$ for some idempotent $a$.
–
QEDMay 30 '13 at 10:13

every$a$ such that $P=Ann(a)$ or that $a$ be idempotent forsome$a$ such that $P=Ann(a)$. If the latter, take the direct sum of two domains. – Steven Landsburg May 29 '13 at 18:25