Let $R$ be a ring with identity with $J(R)^2=0$, where $J(R)$ is the Jacobson radical of $R$. Also, assume $R/J(R)$ is Boolean. I want to show that each one-sided (say, right) ideal $I$ containing $J(R) $ is two-sided.
My try:
When $R/J(R)$ is Boolean, it is inferred that $J(R)$ is a semiprime ideal, i.e., it is equal to its radical. So, to solve the problem it suffices to show that for all $r\in R$ and $i\in I$, $ri\in J(R)$. But, $ri-(ri)^2\in J(R)$, and its square is zero.... .
Any suggestion would be thanked!
Isn't it actually very elementary?
The right/left/two-sided ideals of $R$ containing $J(R)$ correspond to those of $R/J(R)$. Since $R/J(R)$ is boolean, hence commutative, you just have two-sided ideals.