Lemma 53.13.10. Let $X$ be a smooth curve over a field $k$. Let $\overline{x} \in X_{\overline{k}}$ be a closed point with image $x \in X$. The ramification index of $\mathcal{O}_{X, x} \subset \mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is the inseparable degree of $\kappa (x)/k$.

**Proof.**
After shrinking $X$ we may assume there is an étale morphism $\pi : X \to \mathbf{A}^1_ k$, see Morphisms, Lemma 29.36.20. Then we can consider the diagram of local rings

The horizontal arrows have ramification index $1$ as they correspond to étale morphisms. Moreover, the extension $\kappa (x)/\kappa (\pi (x))$ is separable hence $\kappa (x)$ and $\kappa (\pi (x))$ have the same inseparable degree over $k$. By multiplicativity of ramification indices it suffices to prove the result when $x$ is a point of the affine line.

Assume $X = \mathbf{A}^1_ k$. In this case, the local ring of $X$ at $x$ looks like

where $P$ is an irreducible monic polynomial over $k$. Then $P(t) = Q(t^ q)$ for some separable polynomial $Q \in k[t]$, see Fields, Lemma 9.12.1. Observe that $\kappa (x) = k[t]/(P)$ has inseparable degree $q$ over $k$. On the other hand, over $\overline{k}$ we can factor $Q(t) = \prod (t - \alpha _ i)$ with $\alpha _ i$ pairwise distinct. Write $\alpha _ i = \beta _ i^ q$ for some unique $\beta _ i \in \overline{k}$. Then our point $\overline{x}$ corresponds to one of the $\beta _ i$ and we conclude because the ramification index of

is indeed equal to $q$ as the uniformizer $P$ maps to $(t - \beta _ i)^ q$ times a unit. $\square$

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