Proof.
It is clear that (1) implies (2). Assume (2). Let k'/k be a finite purely inseparable field extension (for example k = k'). 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 normal ring. Hence we deduce (3) by Algebra, Lemma 10.165.1.
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 normal by assumption and (1) is proved.
\square
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