Let $X$ be a scheme with $p$ a closed point of $X$. If the stalk of $p$ is a field, then is $p$ an isolated point?(this means $\{p\}$ is open in $X$)
The question is local, so we assume $X=\operatorname{Spec}A$, then $p$ is closed means $\mathfrak{p}$ is maximal in $A$, $\mathcal{O}_{X,p}$ is a field means that for any $t\in\mathfrak{p}$, there exists an element $s\in A-\mathfrak{p}$ such that $ts=0$.
But how to find an ideal $I$ of $A$ such that $V(I)=\operatorname{Spec}A-\{p\}$?
Edit: If the above condition is not sufficient to prove $\{p\}$ is open, I can add the condition that $X$ is locally of finite presentation over a field $k$.
You have a topological problem so you can assume that $A$ is reduced. The fact that $O_{X,p}$ is a field means that $P$ is a minimal prime ideal. If you assume that $A$ is noetherian then it has finitely many minimal prime ideals. Take an element from $P_i-P$ for each minimal prime ideal and take their product $a$ now $a$ is contained in every minimal prime ideal but $P$. The fact that $P$ is also maximal means that $a$ is contained in every prime ideal but $P$ which is what you want. (I think I didn't use the reduced condition!)
I think this is not true for non-noetherian ring for example any ring whose underlying topological space of its spec is $\mathbb{Z}_p$.