Lemma 37.75.3. Let $f : X \to Y$ be a locally quasi-finite morphism. Let $w : X \to \mathbf{Z}$ be a weighting. Let $f' : X' \to Y'$ be the base change of $f$ by a morphism $Y' \to Y$. Then the composition $w' : X' \to \mathbf{Z}$ of $w$ and the projection $X' \to X$ is a weighting of $f'$.
Proof. Consider a diagram
\[ \xymatrix{ X' \ar[d]_{f'} & U' \ar[l]^{h'} \ar[d]^{\pi '} \\ Y' & V' \ar[l]_{g'} } \]
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
\[ U'' = U \times _ X U' \longrightarrow V \times _ Y V' = V'' \]
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. $\square$
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