$f(\mathrm{Nil}_{*}R)\subseteq\mathrm{Nil}_{*}S$ and there exists an epimorphism from $R/\mathrm{Nil}_{*}R$ to $S/\mathrm{Nil}_{*}S$ 
Let $f:R\rightarrow S$ be a ring epimorphism. Prove that $f(\mathrm{Nil}_{*}R)\subseteq\mathrm{Nil}_{*}S$ and there exists an epimorphism from $R/\mathrm{Nil}_{*}R$ to $S/\mathrm{Nil}_{*}S$

Here $\mathrm{Nil}_{*}$ is the prime radical. More precisely, if $R$ is a ring then $\mathrm{Nil}_{*}R=\sqrt{0}$.
Thanks a lot. 

If the rings in question are commutative, then $\mathrm{Nil}_*$ is the ideal of nilpotents. If $x^n=0$, then $f(x)^n = f(x^n) = 0$, so $f(\mathrm{Nil}_*R) \subseteq\mathrm{Nil}_*S$. The "first homomorphism theorem" then gives you an induced map from $R/\mathrm{Nil}_*R$ to $S/\mathrm{Nil}_*S$, which is surjective since $f$ is.
