Lemma 33.8.15. Let $k$ be a field. Let $X \to \mathop{\mathrm{Spec}}(k)$ be locally of finite type. Assume $X$ has finitely many irreducible components. Then there exists a finite separable extension $k'/k$ such that every irreducible component of $X_{k'}$ is geometrically irreducible over $k'$.

**Proof.**
Let $\overline{k}$ be a separable algebraic closure of $k$. The assumption that $X$ has finitely many irreducible components combined with Lemma 33.8.14 (3) shows that $X_{\overline{k}}$ has finitely many irreducible components $\overline{T}_1, \ldots , \overline{T}_ n$. By Lemma 33.8.14 (2) there exists a finite extension $\overline{k}/k'/k$ and irreducible components $T_ i \subset X_{k'}$ such that $\overline{T}_ i = T_{i, \overline{k}}$ and we win.
$\square$

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