$f(\mathrm{Nil}_{*}R)\subseteq\mathrm{Nil}_{*}S$ 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. 
 A: I assume $\mathrm{Nil}_\ast$ is the intersection of all prime ideals.
Let $x\in \mathrm{Nil}_\ast R$. If there is a prime $\mathfrak p\subset S$ such that $y=f(x)\notin \mathfrak p$, then $x$ does not lie in $f^{-1}(\mathfrak p)$, but this is impossible as $f^{-1}(\mathfrak p)\subset R$ is prime.
By the isomorphism theorems, the inclusion $f(\mathrm{Nil}_\ast R)\subseteq \mathrm{Nil}_\ast S$ is exactly what you need in order to complete the dotted arrow below: 

Just to be clear: you would like to define the map downstairs by 
$$
r+\mathrm{Nil}_\ast R\mapsto f(r)+\mathrm{Nil}_\ast S.
$$
But this is well-defined if and only if whenever $r-r'\in\mathrm{Nil}_\ast R$ we have that $f(r)-f(r')\in\mathrm{Nil}_\ast S$, which means exactly "$f(\mathrm{Nil}_\ast R)\subseteq\mathrm{Nil}_\ast S$".
The diagram above will be commutative, so the surjectivity of the map downstairs is clear (all maps are surjective).
A: 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.
