Lemma 70.12.2. Let $X$ be an algebraic space over the field $k$. Let $k \subset k'$ be a field extension. Then $X$ is geometrically connected over $k$ if and only if $X_{k'}$ is geometrically connected over $k'$.

**Proof.**
If $X$ is geometrically connected over $k$, then it is clear that $X_{k'}$ is geometrically connected over $k'$. For the converse, note that for any field extension $k \subset k''$ there exists a common field extension $k' \subset k'''$ and $k'' \subset k'''$. As the morphism $X_{k'''} \to X_{k''}$ is surjective (as a base change of a surjective morphism between spectra of fields) we see that the connectedness of $X_{k'''}$ implies the connectedness of $X_{k''}$. Thus if $X_{k'}$ is geometrically connected over $k'$ then $X$ is geometrically connected over $k$.
$\square$

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