Nilpotent or just nil idempotent ideal?

Let $R$ be a ring with zero Krull dimension and $I$ be an idempotent ideal contained in the Jacobson radical $J(R)$ of $R$. Could one infer just with these hypotheses that $I$ is a nilpotent ideal?

I know that Krull dimension being zero implies that $J(R)$ is equal to the set of nilpotent elements of $R$ (whence a nil ideal), and also $R/J(R)$ is von-Neumann regular. So, I only deduce that $I$ is a nil ideal.

Thanks for any hint or suggestion!

This example works also here. The only thing to show is $\dim R=0$, but this is obvious.
• @user26857 Why $\dim R=0$? I think we must show that $R$ has only one prime ideal. How? – karparvar Mar 22 '16 at 17:49
• If $X_1\in P$ and $X_2^2-X_1\in P$ then $X_2^2\in P$, so $X_2\in P$, and so on. – user26857 Mar 22 '16 at 17:51
• @user26857 And if $X_1$ does not belong to $P$.... – karparvar Mar 23 '16 at 6:38
• @karparvar As far as I know, the prime ideals of a quotient ring $R/I$ have the form $P/I$ with $P\subset R$ prime containing $I$. – user26857 Mar 23 '16 at 9:00