**Proof.**
It is clear that (1) implies (2). Assume (2). This in particular implies that $\mathcal{O}_{X, x}$ is a regular local ring. Let $k'/k$ be a finite purely inseparable field extension. Consider the ring $\mathcal{O}_{X, x} \otimes _ k k'$. By Algebra, Lemma 10.46.7 its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$. Hence it is a local ring also (Algebra, Lemma 10.18.3). Therefore there is a unique point $x' \in X_{k'}$ lying over $x$ and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes _ k k'$. By assumption this is a regular ring. Hence we deduce (3) from the definition of a geometrically regular ring.

Assume (3). Let $k'/k$ be a field extension. Since $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ is surjective, also $X_{k'} \to X$ is surjective (Morphisms, Lemma 29.9.4). Let $x' \in X_{k'}$ be any point lying over $x$. The local ring $\mathcal{O}_{X_{k'}, x'}$ is a localization of the ring $\mathcal{O}_{X, x} \otimes _ k k'$. Hence it is regular by assumption and (1) is proved.
$\square$

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