Lemma 33.8.7. Let $k'/k$ be an extension of fields. Let $X$ be a scheme over $k$. Set $X' = X_{k'}$. Assume $k$ separably algebraically closed. Then the morphism $X' \to X$ induces a bijection of irreducible components.

Proof. Since $k$ is separably algebraically closed we see that $k'$ is geometrically irreducible over $k$, see Algebra, Lemma 10.47.5. Hence $Z = \mathop{\mathrm{Spec}}(k')$ is geometrically irreducible over $k$. by Lemma 33.8.5 above. Since $X' = Z \times _ k X$ the result is a special case of Lemma 33.8.4. $\square$

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