Let $k$ be an algebraically closed field, and let $X$ be a scheme locally of finite type over $Spec(k)$. Consider a closed point $x\in X$ such that the local ring of $X$ at $x$ is isomorphic to $k$. Then is $x$ also open in $X$?
For example, when $X$ is affine, say $X=Spec(A)$, we let $\mathfrak{m}$ be the maximal ideal corresponding to $x$, then by assumption, the local ring $A_\mathfrak{m}$ is the base field $k$. I seem to be stuck here.