The following is part of an argument in Vakil's Rising Sea, p. 238.
Let $\pi : X \to \operatorname{Spec}{B}$ be a morphism of schemes with $X$ reduced and $g \in B$. Consider the morphism $$\pi^{\#}_{B_g} : B_g \to \Gamma(\operatorname{Spec}{B_g}, \pi_*\mathcal{O}_X).$$ If $\pi^{\#}_{B_g}(r/g^n) = 0$, then $\pi^{\#}_B(rg) = 0$.
Vakil sketches the following: Because $r$ vanishes on $\operatorname{Spec}{B_g}$ and $g$ vanishes on $V(g)$, we get that $rg$ vanishes on $\operatorname{Spec}{B} = \operatorname{Spec}{B_g} \cup V(g)$.
What does this mean formally? The set $V(g)$ is not open, so we cannot just apriori restrict the action of the sheaf morphism to $V(g)$. So possibly $\pi^{\#}(g_{\mathfrak{p}}) = 0$ for $\mathfrak{p} \in V(g)$ but I'm not sure why or if this is true.
Best Answer
In this case, Vakil is using "vanishing" to mean evaluating to $0$ in the residue field at every point of $V(g)$. Since $V(g)\subset\operatorname{Spec}B$ as a set consists of the prime ideals containing $g$, the evaluation map $B\to B_p/pB_p$ sends $g$ to zero.
(Let me point out that the reason this suffices for his purposes in this proof is that on a reduced scheme, any function which evaluates to zero in every residue field is actually zero.)