The Stacks project

Lemma 36.67.5. Let $f : X \to Y$ be a morphism of schemes. Assume $f$ is locally quasi-finite, locally of finite presentation, and flat. Then there is a positive weighting $w : X \to \mathbf{Z}_{> 0}$ of $f$ given by the rule that sends $x \in X$ lying over $y \in Y$ to

\[ w(x) = \text{length}_{\mathcal{O}_{X, x}} (\mathcal{O}_{X, x}/\mathfrak m_ y \mathcal{O}_{X, x}) [\kappa (x) : \kappa (y)]_ i \]

where $[\kappa ' : \kappa ]_ i$ is the inseparable degree (Fields, Definition 9.14.7).

Proof. Consider a diagram as in Definition 36.67.2. Let $u \in U$ with images $x, y, v$ in $X, Y, V$. Then we claim that

\[ \text{length}_{\mathcal{O}_{X, x}} (\mathcal{O}_{X, x}/\mathfrak m_ y \mathcal{O}_{X, x}) = \text{length}_{\mathcal{O}_{U, u}} (\mathcal{O}_{U, u}/\mathfrak m_ v \mathcal{O}_{U, u}) \]

and

\[ [\kappa (x) : \kappa (y)]_ i = [\kappa (u) : \kappa (v)]_ i \]

The first equality follows as $\mathcal{O}_{X, x} \to \mathcal{O}_{U, u}$ is a flat local homomorphism such that $\mathfrak m_ y \mathcal{O}_{U, u} = \mathfrak m_ v \mathcal{O}_{U, u}$ and $\mathfrak m_ x \mathcal{O}_{U, u} = \mathfrak m_ u$ (because $\mathcal{O}_{Y, y} \to \mathcal{O}_{V, v}$ and $\mathcal{O}_{X, x} \to \mathcal{O}_{U, u}$ are unramified) and hence the equality by Algebra, Lemma 10.51.13. The second equality follows because $\kappa (v)/\kappa (y)$ is a finite separable extension and $\kappa (u)$ is a factor of $\kappa (x) \otimes _{\kappa (y)} \kappa (v)$ and hence the inseparable degree is unchanged. Having said this, we see that formation of the function in the lemma commutes with ├ętale base change. This reduces the problem to the discussion of the next paragraph.

Assume that $f$ is a finite, flat morphism of finite presentation. We have to show that $\int _ f w$ is locally constant on $Y$. In fact, $f$ is finite locally free (Morphisms, Lemma 28.46.2) and we will show that $\int _ f w$ is equal to the degree of $f$ (which is a locally constant function on $Y$). Namely, for $y \in Y$ we see that

\begin{align*} (\textstyle {\int }_ f w)(y) & = \sum \nolimits _{f(x) = y} \text{length}_{\mathcal{O}_{X, x}} (\mathcal{O}_{X, x}/\mathfrak m_ y \mathcal{O}_{X, x}) [\kappa (x) : \kappa (y)]_ i [\kappa (x) : \kappa (y)]_ s \\ & = \sum \nolimits _{f(x) = y} \text{length}_{\mathcal{O}_{X, x}} (\mathcal{O}_{X, x}/\mathfrak m_ y \mathcal{O}_{X, x}) [\kappa (x) : \kappa (y)] \\ & = \text{length}_{\mathcal{O}_{Y, y}}((f_*\mathcal{O}_ X)_ y/ \mathfrak m_ y (f_*\mathcal{O}_ X)_ y) \end{align*}

Last equality by Algebra, Lemma 10.51.12. The final number is the rank of $f_*\mathcal{O}_ X$ at $y$ as desired. $\square$


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