$\newcommand\spec{\operatorname{Spec}}$Hartshorne Exercise II.3.15 a) states the following:
Let $X$ be a scheme of finite type over a field $k$ then the following are equivalent (in which case we call $X$ geometrically irreducible):
- $X_{\bar k}$ is irreducible, where $\bar k$ is the algebraic closure of $k$
- $X_{k_s}$ is irreducible, where $k_s$ is the separable closure of $k$
- $X_K$ is irreducible for any field extension $K/k$.
where we have the typical notation abuse that we also see with algebraic varieties in classical algebraic geometry $X_K := X \times_{\spec k} \spec K$ for any field extension $K/k$.
I'm not really curious about how to answer this question (but maybe someone can still give a hint). What I am really curious about is why is it assumed that $X$ is finite type over $k$ (So $X$ is finitely covered by spectra of finitely generated $k$-algebras). I'm not sure why for instance locally of finite type (or even a less strict condition) is not sufficient or is it that the proof is easier if we assume $X$ is of finite type over $k$?