Lemma 33.7.6. Let $k'/k$ be an extension of fields. Let $X$ be a scheme over $k$. Assume $k$ separably algebraically closed. Then the morphism $X_{k'} \to X$ induces a bijection of connected components. In particular, $X$ is geometrically connected over $k$ if and only if $X$ is connected.
Proof. Since $k$ is separably algebraically closed we see that $k'$ is geometrically connected over $k$, see Algebra, Lemma 10.48.4. Hence $Z = \mathop{\mathrm{Spec}}(k')$ is geometrically connected over $k$ by Lemma 33.7.5 above. Since $X_{k'} = Z \times _ k X$ the result is a special case of Lemma 33.7.4. $\square$
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