Lemma 33.8.11. Let $k$ be a field. Let $X$ be a scheme over $k$. If $X$ is irreducible and has a dense set of $k$-rational points, then $X$ is geometrically irreducible.

Proof. Let $k'/k$ be a finite extension of fields and let $Z, Z' \subset X_{k'}$ be irreducible components. It suffices to show $Z = Z'$, see Lemma 33.8.8. By Lemma 33.8.10 we have $p(Z) = p(Z') = X$ where $p : X_{k'} \to X$ is the projection. If $Z \not= Z'$ then $Z \cap Z'$ is nowhere dense in $X_{k'}$ and hence $p(Z \cap Z')$ is not dense by Morphisms, Lemma 29.48.7; here we also use that $p$ is a finite morphism as the base change of the finite morphism $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$, see Morphisms, Lemma 29.44.6. Thus we can pick a $k$-rational point $x \in X$ with $x \not\in p(Z \cap Z')$. Since the residue field of $x$ is $k$ we see that $p^{-1}(\{ x\} ) = \{ x'\}$ where $x' \in X_{k'}$ is a point whose residue field is $k'$. Since $x \in p(Z) = p(Z')$ we conclude that $x' \in Z \cap Z'$ which is the contradiction we were looking for. $\square$

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