Let $R$ be a commutative ring, $I$ be a minimal ideal of $R$. Prove that for all $y$ belong to $Rad(R)$, $yI=0$. ($Rad(R)$ denotes the Jacobson radical of $R$)
$Rad(R)$ equals the intersection of all maximal right ideals of the ring. It is also true that $Rad(R)$ equals the intersection of all maximal left ideals within the ring. In that case, $Rad(R)$ equals the intersection of all maximal ideals of the ring. So take $y$ belongs to $Rad(R)$, $y$ must belong to some maximal ideal $M$ of $R$. Clearly, $yI$ belongs to $MI$. I want $MI=0$. Is that a trivial fact?
Thank you very much.
Since $I$ is simple, $R/M \cong I$ For some maximal ideal M of R.
Clearly the annihilator of $R/M$ is M, therefore... (Can you take it from here?)