The Stacks project

Proof. It follows from Algebra, Lemma 10.156.3 that $\mathcal{O}_{Y, y}^{sh} \to \mathcal{O}_{X, x}^{sh}$ is finite. Since $Y$ is geometrically unibranch there is a unique minimal prime $\mathfrak p$ in $\mathcal{O}_{Y, y}^{sh}$, see More on Algebra, Lemma 15.106.5. Write

as a finite product of fields. We set $w(x) = \sum [K_ i : \kappa (\mathfrak p)]_ s$.

Since this definition is clearly insensitive to étale localization, in order to show that $w$ is a weighting we reduce to showing that if $f$ is a finite morphism, then $\int _ f w$ is locally constant. Observe that the value of $\int _ f w$ in a generic point $\eta $ of $Y$ is just the number of points of the geometric fibre $X_{\overline{\eta }}$ of $X \to Y$ over $\eta $. Moreover, since $Y$ is unibranch a point $y$ of $Y$ is the specialization of a unique generic point $\eta $. Hence it suffices to show that $(\int _ f w)(y)$ is equal to the number of points of $X_{\overline{\eta }}$. After passing to an affine neighbourhood of $y$ we may assume $X \to Y$ is given by a finite ring map $A \to B$. Suppose $\mathcal{O}_{Y, y}^{sh}$ is constructed using a map $\kappa (y) \to k$ into an algebraically closed field $k$. Then

by Algebra, Lemma 10.153.4 and the lemma used above. Observe that the minimal prime $\mathfrak p$ of $\mathcal{O}_{Y, y}^{sh}$ maps to the prime of $A$ corresponding to $\eta $. Hence we see that the desired equality holds because the number of points of a geometric fibre is unchanged by a field extension. $\square$