I want to show in homomorphic image of Jacobson ring Nilradical is equal to the Jacobson radical. By Jacobson ring we know that every prime ideal is intersection of maximal ideals.
Where I stuck is the following. Given $f:A \to B$ surjective ring homomorphism and $A$ be a Jacobson ring. Consider $y \in Jac(B)$ then want to show that for any prime ideal $p$ of $B,$ $y \in p.$ Now if $f^{-1}(y) \cap f^{-1}(p)= \phi $ I cannot draw any contradiction. Please help me to prove this way.
Hint: We know that $B\cong A/I$ for some ideal $I\subset A$. Thus, what you want to show is that $\operatorname{nil}(A/I)=\mathcal{J}(A/I)$. Use the correspondence between prime ideals of $A$ and those of $A/I$.