Every closed point is open in a scheme locally of finite type over a field

Question

Let $X$ be a scheme locally of finite type over a field $k$, then we know the closed points in $X$ is very dense.

If we assume that every closed point is also open in $X$, then is $X$ a discrete scheme?

Thanks in advance.

Answer 1

Here's the relevant topological fact: Let $X$ be a topological space such that (1) every closed point of $X$ is open, and (2) every nonempty closed subset of $X$ contains a closed point. Then $X$ is discrete. Indeed, for any $x \in X$, the closure $\overline{\{x\}}$ contains a closed point $y$, which is also an open point, so $\overline{\{x\}} \setminus \{y\}$ is closed, which means $x = y$ by the definition of closure. Thus every point of $X$ is closed and open, so $X$ is discrete.

The only algebraic input needed here is that locally Noetherian schemes (which includes schemes locally of finite type over a field) have the property that every nonempty closed subset contains a closed point. (For reference, this is Lemma 28.5.9 (tag 02IL) in the Stacks Project.)