**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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