This is a problem in Atiyah Macdonald Commutative Algebra.(Problem 5.22) $S$ is a subring of an inregral domain $R$. $R$ is finitely generated $S$ algebra. If Jacobson radical of $S$ is 0, then Jacobson radical of $R$ is 0.
I followed the hint provided the book as following.
The purpose is to show for each element of $R$, there is a maximal ideal of $R$ avoiding it. Pick a $0\neq v\in R$. Since we have injective maps $S\to R$ and $R\to R_v$, we embed $S$ in $R_v$ as a subring and this $R_v$ is finitely generated as well over $S$.
It is clear that there is $0\neq s\in A$ such that we can extend ring homomorphism $\phi:S\to\Omega$ to $\tilde{\phi}:R_v\to\Omega$ where $\Omega$ is some algebraically closed field. Here $\Omega$ is chosen to be the algebraically closed field of $S/m$ where $s\not\in m$ as jacobson radical of $S$ is 0. It is clear that $v$ under the map $R_v\to\Omega$ is not zero or we will have $1$ being sent to $0$.
However, I am stucked at showing $Ker(R_v\to\Omega)\subset R_v$ is maximal ideal since this will lead to $Ker(R_v\to\Omega)\cap B$ is maximal. It is not even clear that this will be maximal. However, I can show if $R$'s jacobson radical is trivial, $R[x]$'s is trivial. I also noticed that quotient and localization do not commute with infinite intersection. What should I do at this point?