Let $R$ be a ring with identity. If $I$ is a finitely generated ideal of R such that $I^2=I$, must $I$ be generated by an idempotent?
I've known that it holds for $R$ commutative. So I'm interesting with the case $R$ noncommutative. And by saying "$I$ be generated by an idempotent" I mean that the idempotent commutes with each element of $I$.
I've found a rather simple counterexample. Let $C_2$ denote the cyclic group of order $2$; consider the abelian group $A:=C_2\times C_2$ and write $a:=(\overline 1,\overline 0),\ b:=(\overline 0,\overline 1).$ Define a bilinear 2-operation on $A$ by: $a^2=a,\ b^2=0,\ ab=b,\ ba=0.$ It's easy to check that this operation is associative, so it gives a ring structure on $A$. Moreover, under this ring structure we have no unit but $A^2=A$ still holds. In fact, this ring is generated by an idempotent, namely $a$, but $a$ does not commute with each other element. It seems that I seriously underestimated the complexity of noncommutative rings.