Let $X$ be a locally noetherian scheme and let $x\in X$ a point such that $\{x\}$ is locally closed in $X$. Show that $\dim\overline{\{x\}}\le1$.
This is proposition 5.1.10 in EGA IV.
Let $Z$ be the subscheme with underlying space $\overline{\{x\}}$. We assume that $Z$ is reduced, and since $\dim Z=\sup_{z\in Z} \dim \mathcal O_{Z,z}$ it suffices to show that $\dim \mathcal O_{Z,z}\le1$ for all $z\in Z$. In this case we may assume that the ring $A=\mathcal O_{Z,z}$ is local noetherian integral. By the following theorem, if there exists an $f$ such that $A_f$ is a field, then $\dim A\le1$.
Artin-Tate's theorem: Let $A$ be a noetherian integral domain. Then $A$ is semi-local of dimension $\le1$ if and only if there exists an $f\in A$ such that $A_f$ is a field.
I don't know how to find such a $f$. Could anybody give a proof?
Given $z\in Z$, let $\DeclareMathOperator\spec{spec}\spec A$ be an affine open around $z$. Since $\{x\}$ is locally closed in $X$, it is an open in $Z$, and so an open in $\spec A$. Since basic affine opens form a base for the topology, we see that $\{x\}\supset\spec A_f$ for some $f\in A$, and hence must have $\{x\}=\spec A_f$, i.e. $A_f=\kappa(x)$ is a field. Since $\mathscr O_{Z,z}$ is a localization of $A$, we conclude that $\mathscr O_{Z,z}[1/f]=A_f$ is a field as well.