I saw this claim in a proof of Vakil's claim right after 6.5.B in FOAG that $\phi: A \to B$ a map of rings induces a dominant morphism of schemes if and only if the kernel is contained in the nilradical of $A$. I tried to show that directly, but I got stuck in the direction that starts by assuming kernel is contained in the nilradical. I think I am struggling with finishing the proof for the same reason I can't seem to show the corresponding direction under the claim in the subject line. I'd appreciate any hints or references. Thank you.
Update: just to show some work (which may be relevant for either problem), I have gone as far as to show that if $\ker \phi \subset \operatorname{nil}A$, we can show that $\phi^{-1}(\operatorname{nil}(B)) \subset \operatorname{nil}A$. This is easy to see by first noting that if $\phi(f)=0$, then $f^n = 0$ for some $n$. Moreover, if $\phi(g)^m = 0$, then $g^m \in \ker \phi$, so there is some $n$ s.t. $g^{mn} =0$.
Here's how one can approach the question in the subject line.
First recall or prove the following two facts.
(1) If $S\subset \operatorname{Spec}R$ is any subset of an affine scheme, then $\overline{S}=V(\bigcap_{\mathfrak{p}\in S}\mathfrak{p})$.
(2) $\sqrt{I}=\bigcap_{\mathfrak{p}\in V(I)}\mathfrak{p}$.
Now see if you can use these to simplify $\overline{f(V(I))}$:
$$\overline{f(V(I))}=V\big(\bigcap_{\mathfrak{p}\in f(V(I))}\mathfrak{p}\big)=\dots$$
Edit: Proof (sketch) of (1).
One inclusion is rather easy (the RHS is closed and contains $S$, hence contains the closure of $S$).
For the other inclusion, suppose $\mathfrak{q}$ is in the RHS, i.e. $\mathfrak{q}\supseteq \bigcap_{\mathfrak{p}\in S}\mathfrak{p}$. Then if $\mathfrak{q}\notin \overline{S}$ there is a basic open $D(f)$ containing $\mathfrak{q}$ such that $D(f)\cap S=\emptyset$. But this cannot hold since $\mathfrak{q}\in D(f)$ means $f\notin \mathfrak{q}$ while $D(f)\cap S = \emptyset$ means $f\in\mathfrak{p}$ for every $\mathfrak{p}\in S$, i.e. $f\in \bigcap_{\mathfrak{p}\in S}\mathfrak{p}\subseteq\mathfrak{q}$.