By Lemma \ref{lemma-rh} and the discussion above (which used Varieties, Lemma \ref{varieties-lemma-algebraic-scheme-dim-0} and Algebra, Lemma \ref{algebra-lemma-pushdown-module}) 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. \medskip\noindent 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 \ref{duality-lemma-flat-morphism-from-gorenstein-scheme}) or because it was shown in the proof of Discriminants, Lemma \ref{discriminant-lemma-discriminant-quasi-finite-morphism-smooth} (which we used above). Thus $\omega_{X/Y}$ is invertible by Discriminants, Lemma \ref{discriminant-lemma-gorenstein-quasi-finite} 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. \medskip\noindent 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 $U = \Spec(A)$ and $V = \Spec(B)$. Then $R \cap U$ is the different of $f|_U : U \to V$. By Discriminants, Lemma \ref{discriminant-lemma-base-change-different} formation of the different commutes with arbitrary base change in our case. By our choice of $U$ and $V$ we have $$ A \otimes_B \kappa(y) = \mathcal{O}_{X, x} \otimes_{\mathcal{O}_{Y, y}} \kappa(y) = \mathcal{O}_{X, x}/(s^{e_x}) $$ 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 \ref{discriminant-definition-different}. 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 \ref{discriminant-lemma-tau-nonzero} 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.