The Stacks project

Proof. This follows immediately from the second description of $\int _\pi w$ given above. To prove it from the definition, you use that if $E/F$ is a finite extension of fields and $F'/F$ is another field extension, then writing $(E \otimes _ F F')_{red} = \prod E'_ i$ as a product of fields finite over $F'$, we have

To prove this equality pick an algebraically closed field extension $\Omega /F'$ and observe that

where we have used Fields, Lemma 9.14.8. $\square$

Proof. Consider a diagram

as in Definition 37.75.2 for the morphism $f'$. For any $v' \in V'$ we have to show that $\int _{\pi '} (w' \circ h')$ is constant in an open neighbourhood of $v'$. By Lemma 37.75.1 (and the fact that étale morphisms are open) we may replace $V'$ by any étale neighbourhood of $v'$. After replacing $V'$ by an étale neighbourhood of $v'$ we may assume that $U' = U'_1 \amalg \ldots \amalg U'_ n$ where each $U'_ i$ has a unique point $u'_ i$ lying over $v'$ such that $\kappa (u'_ i)/\kappa (v')$ is purely inseparable, see Lemma 37.41.5. Clearly, it suffices to prove that $\int _{U'_ i \to V'} w'|_{U'_ i}$ is constant in a neighbourhood of $v'$. This reduces us to the case discussed in the next paragraph.

We have $v' \in V'$ and there is a unique point $u'$ of $U'$ lying over $v'$ with $\kappa (u')/\kappa (v')$ purely inseparable. Denote $x \in X$ and $y \in Y$ the image of $u'$ and $v'$. We can find an étale neighbourhood $(V, v) \to (Y, y)$ and an open $U \subset X_ V$ such that $\pi : U \to V$ is finite and such that there is a unique point $u \in U$ lying over $v$ which maps to $x \in X$ via the projection $h : U \to X$ such that moreover $\kappa (u)/\kappa (v)$ is purely inseparable. This is possible by the lemma used above. Consider the morphism

Since $u$ and $u'$ both map to $x \in X$ there is a point $u'' \in U''$ mapping to $(u, u')$. Denote $v'' \in V''$ the image of $u''$. After replacing $V', v'$ by $V'', v''$ we may assume that the composition $V' \to Y' \to Y$ factors through a map of étale neighbourhoods $(V', v') \to (V, v)$ such that the induced morphism $X'_{V'} = X_{V'} \to X_ V$ sends $u'$ to $u$. Inside the base change $X'_{V'} = X_{V'}$ we have two open subschemes, namely $U'$ and the inverse image $U_{V'}$ of $U \subset X_ V$. By construction both contain a unique point lying over $v'$, namely $u'$ for both of them. Thus after shrinking $V'$ we may assume these open subsets are the same; namely, $U' \setminus (U' \cap U_{V'})$ and $U_{V'} \setminus (U' \cap U_{V'})$ have a closed image in $V'$ and these images do not contain $v'$. Thus $U' = U_{V'}$ and we find a cartesian diagram as in Lemma 37.75.1. Since $\int _\pi (w \circ h)$ is locally constant by assumption we conclude. $\square$

Proof. Immediate from the definition. $\square$

Proof. Let us set $w_{g \circ f}(x) = w_ f(x) w_ g(f(x))$ for $x \in X$. Consider a diagram

where $W \to Z$ is étale, $U \subset X_ W$ is open, and $U \to W$ finite. We have to show that $\int _\pi w_{g \circ f}|_ U$ is locally constant. Choose a point $w \in W$. By Lemma 37.75.1 (and the fact that étale morphisms are open) it suffices to show that $\int _\pi w_{g \circ f}|_ U$ is constant after replacing $(W, w)$ by an étale neighbourhood. After replacing $(W, w)$ by an étale neighbourhood we may assume $U = U_1 \amalg \ldots \amalg U_ n$ where each $U_ i$ has a unique point $u_ i$ lying over $w$ such that $\kappa (u_ i)/\kappa (w)$ is purely inseparable, see Lemma 37.41.5. Clearly, it suffices to show that $\int _{U_ i \to W} w_{g \circ f}|_{U_ i}$ is constant in an étale neighbourhood of $w$. This reduces us to the case discussed in the next paragraph.

We have $w \in W$ and there is a unique point $u \in U$ lying over $w$ with $\kappa (u)/\kappa (w)$ purely inseparable. Consider the point $v = f(u) \in Y$. After replacing $(W, w)$ by an elementary étale neighbourhood we may assume there is an open neighbourhood $V \subset Y_ W$ of $v$ such that $V \to W$ is finite, see Lemma 37.41.1. Then $f_ W^{-1}(V) \cap U$ is an open neighbourhood of $u$ where $f_ W : X_ W \to Y_ W$ is the base change of $f$ to $W$. Hence after Zariski shrinking $W$, we may assume $f_ W(U) \subset V$. Thus we obtain morphisms

and $U \to V$ is finite as $V \to W$ is separated (because finite). Since $w_ f$ and $w_ g$ are weightings of $f$ and $g$ we see that $\int _ a w_ f|_ U$ is locally constant on $V$ and $\int _ b w_ g|_ V$ is locally constant on $W$. Thus after shrinking $W$ one more time we may assume these functions are constant say with values $n$ and $m$. It follows immediately that $\int _\pi w_{g \circ f}|_ U = \int _{b \circ a} w_{g \circ f}|_ U$ is constant with value $nm$ as desired. $\square$

Proof. Since the property is preserved by base change, see Lemma 37.75.3, it suffices to prove that $f$ is open. Since we may also replace $X$ by any open of $X$, it suffices to prove that $f(X)$ is open. Let $y \in f(X)$. Choose $x \in X$ with $f(x) = y$. It suffices to prove that $f(X)$ contains an open neighbourhood of $y$ and it suffices to do so after replacing $Y$ by an étale neighbourhood of $y$. By étale localization of quasi-finite morphisms, see Section 37.41, we may assume there is an open neighbourhood $U \subset X$ of $x$ such that $\pi = f|_ U : U \to Y$ is finite. Then $\int _\pi w|_ U$ is locally constant and has positive value at $y$. Hence $\pi (U)$ contains an open neighbourhood of $y$ and the proof is complete. $\square$

Proof. Consider a diagram as in Definition 37.75.2. Let $u \in U$ with images $x, y, v$ in $X, Y, V$. Then we claim that

and

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 10.52.13. 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 étale 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 29.48.2) 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

Last equality by Algebra, Lemma 10.52.12. The final number is the rank of $f_*\mathcal{O}_ X$ at $y$ as desired. $\square$

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. $\square$