Lemma 33.7.14. Let $k$ be a field. Let $X$ be a scheme over $k$. Assume $X$ is connected and has a point $x$ such that $k$ is algebraically closed in $\kappa (x)$. Then $X$ is geometrically connected. In particular, if $X$ has a $k$-rational point and $X$ is connected, then $X$ is geometrically connected.

**Proof.**
Set $T = \mathop{\mathrm{Spec}}(\kappa (x))$. Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. The assumption on $k \subset \kappa (x)$ implies that $T_{\overline{k}}$ is irreducible, see Algebra, Lemma 10.47.8. Hence by Lemma 33.7.13 we see that $X_{\overline{k}}$ is connected. By Lemma 33.7.7 we conclude that $X$ is geometrically connected.
$\square$

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