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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)