The Stacks project

Proof. By Lemma 53.12.2 and the discussion above (which used Varieties, Lemma 33.20.2 and Algebra, Lemma 10.52.12) it suffices to prove the results on the multiplicity $d_ x$ of $x$ in $R$. Part (1) was proved in the discussion above. In the discussion above we proved (2) and (3) only in the case where $\kappa (x)$ is separable over $k$. In the rest of the proof we give a uniform treatment of (2) and (3) using material on differents of quasi-finite Gorenstein morphisms.

First, observe that $f$ is a quasi-finite Gorenstein morphism. This is true for example because $f$ is a flat quasi-finite morphism and $X$ is Gorenstein (see Duality for Schemes, Lemma 48.25.7) or because it was shown in the proof of Discriminants, Lemma 49.12.6 (which we used above). Thus $\omega _{X/Y}$ is invertible by Discriminants, Lemma 49.16.1 and the same remains true after replacing $X$ by opens and after performing a base change by some $Y' \to Y$. We will use this below without further mention.

Choose affine opens $U \subset X$ and $V \subset Y$ such that $x \in U$, $y \in V$, $f(U) \subset V$, and $x$ is the only point of $U$ lying over $y$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(B)$. Then $R \cap U$ is the different of $f|_ U : U \to V$. By Discriminants, Lemma 49.9.4 formation of the different commutes with arbitrary base change in our case. By our choice of $U$ and $V$ we have

where $e_ x$ is the ramification index as in the statement of the lemma. Let $C = \mathcal{O}_{X, x}/(s^{e_ x})$ viewed as a finite algebra over $\kappa (y)$. Let $\mathfrak {D}_{C/\kappa (y)}$ be the different of $C$ over $\kappa (y)$ in the sense of Discriminants, Definition 49.9.1. It suffices to show: $\mathfrak {D}_{C/\kappa (y)}$ is nonzero if and only if the extension $\mathcal{O}_{Y, y} \subset \mathcal{O}_{X, x}$ is tamely ramified and in the tamely ramified case $\mathfrak {D}_{C/\kappa (y)}$ is equal to the ideal generated by $s^{e_ x - 1}$ in $C$. Recall that tame ramification means exactly that $\kappa (x)/\kappa (y)$ is separable and that the characteristic of $\kappa (y)$ does not divide $e_ x$. On the other hand, the different of $C/\kappa (y)$ is nonzero if and only if $\tau _{C/\kappa (y)} \in \omega _{C/\kappa (y)}$ is nonzero. Namely, since $\omega _{C/\kappa (y)}$ is an invertible $C$-module (as the base change of $\omega _{A/B}$) it is free of rank $1$, say with generator $\lambda $. Write $\tau _{C/\kappa (y)} = h\lambda $ for some $h \in C$. Then $\mathfrak {D}_{C/\kappa (y)} = (h) \subset C$ whence the claim. By Discriminants, Lemma 49.4.8 we have $\tau _{C/\kappa (y)} \not= 0$ if and only if $\kappa (x)/\kappa (y)$ is separable and $e_ x$ is prime to the characteristic. Finally, even if $\tau _{C/\kappa (y)}$ is nonzero, then it is still the case that $s \tau _{C/\kappa (y)} = 0$ because $s\tau _{C/\kappa (y)} : C \to \kappa (y)$ sends $c$ to the trace of the nilpotent operator $sc$ which is zero. Hence $sh = 0$, hence $h \in (s^{e_ x - 1})$ which proves that $\mathfrak {D}_{C/\kappa (y)} \subset (s^{e_ x - 1})$ always. Since $(s^{e_ x - 1}) \subset C$ is the smallest nonzero ideal, we have proved the final assertion. $\square$