Proof.
Looking over the proof of Lemma 55.11.9 we see that in order to get equality, the inequalities (55.11.9.1), (55.11.9.2), and (55.11.9.3) have to be equalities.
Let C_ i be the irreducible component containing x. Equality in (55.11.9.1) shows via Algebraic Curves, Lemma 53.18.4 that C_{i, \overline{k}}^\nu \to C_{i, \overline{k}} is an isomorphism. Hence C_{i, \overline{k}} is smooth and part (1) holds.
Next, let C_ i \subset X_ k be another irreducible component. Then we may assume we have D = D' + C_ i as in the induction step in the proof of Lemma 55.11.9. Equality in (55.11.9.2) immediately implies that \kappa _ i/k is finite separable. Equality in (55.11.9.3) implies either a_{ij} = 1 for some j or that there is a unique C_ j \subset D' meeting C_ i and a_{ij} = w_ i. In both cases we find that C_ i has a \kappa _ i-rational point c and c = C_ i \cap C_ j scheme theoretically. Since \mathcal{O}_{X, c} is a regular local ring, this implies that the local equations of C_ i and C_ j form a regular system of parameters in the local ring \mathcal{O}_{X, c}. Then \mathcal{O}_{C_ i, c} is regular by (Algebra, Lemma 10.106.3). We conclude that C_ i \to \mathop{\mathrm{Spec}}(\kappa _ i) is smooth at c (Algebra, Lemma 10.140.5). It follows that C_ i is geometrically integral over \kappa _ i (Varieties, Lemma 33.25.10). To finish we have to show that C_ i is smooth over \kappa _ i. Observe that
C_{i, \overline{k}} = C_ i \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(\overline{k}) = \coprod \nolimits _{\kappa _ i \to \overline{k}} C_ i \times _{\mathop{\mathrm{Spec}}(\kappa _ i)} \mathop{\mathrm{Spec}}(\overline{k})
where there are [\kappa _ i : k]-summands. Thus if C_ i is not smooth over \kappa _ i, then each of these curves is not smooth, then these curves are not normal and the normalization morphism drops the genus (Algebraic Curves, Lemma 53.18.4) which is disallowed because it would drop the geometric genus of C_ i/k contradicting [\kappa _ i : k] g_ i = g_{geom}(C_ i/k).
\square
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