The Stacks project

Proof. Consider the diagram

where $\overline{K}$ is the separable algebraic closure of $K$, and where $\overline{k}$ is the separable algebraic closure of $k$. By Lemma 33.8.7 the morphism $X_{\overline{K}} \to X_{\overline{k}}$ induces a bijection between irreducible components. Hence it suffices to show the lemma for the morphisms $X_{\overline{k}} \to X$ and $X_{\overline{K}} \to X_ K$. In other words we may assume that $K = \overline{k}$.

The morphism $p : X_{\overline{k}} \to X$ is integral, flat and surjective. Flatness implies that generalizations lift along $p$, see Morphisms, Lemma 29.25.9. Hence generic points of irreducible components of $X_{\overline{k}}$ map to generic points of irreducible components of $X$. Integrality implies that $p$ is universally closed, see Morphisms, Lemma 29.44.7. Hence we conclude that the image $p(\overline{T})$ of an irreducible component is a closed irreducible subset which contains a generic point of an irreducible component of $X$, hence $p(\overline{T})$ is an irreducible component of $X$. This proves the first assertion. If $T \subset X$ is an irreducible component, then $p^{-1}(T) =T_ K$ is a nonempty union of irreducible components, see Lemma 33.8.9. Each of these necessarily maps onto $T$ by the first part. Hence the map is surjective. $\square$