Extension of idempotent ideals Let $R$ be a Noetherian commutative ring with $1$. If $R[[x]]$ denotes the ring of formal power series over $R$ and $I$ is an idempotent ideal of $R$ I want to know whether the extension of $I$ in $R[[x]]$ is also idempotent.
Since $R$ is Noetherian, one has $I[[x]]=IR[[x]]$. So, if a series $a_0+a_1x+a_2x^2+...\in I[[x]]$ we could re-write it by substituting such finite summations: $a_t=\sum_i x_{ti}y_{ti}$ with $x_{ti},y_{ti}\in I$, $(t=0,1,...)$. Now, may it be true that the second series is in $(I[[x]])^2$?
Thanks for any cooperation!

Every finitely generated idempotent ideal in any commutative ring is generated by a single idempotent element.  Indeed, by Nakayama's lemma, $I^2=I$ implies there is $r\in R$ such that $1-r\in I$ and $rI=0$.  Then $r(1-r)=0$, so $r=r^2$ is an idempotent.  For any $i\in I$, $i=i-ri=(1-r)i$, so $I$ is generated by the idempotent $1-r$.
It is now trivial that $I[[x]]$ is also an idempotent ideal, since it is also generated by the idempotent element $1-r$.
