Lemma 38.15.5. In Situation 38.15.1. If there exists an impurity $(g : T \to S, t' \leadsto t, \xi )$ of $\mathcal{F}$ above $s$ with $g$ quasi-finite at $t$, then there exists an impurity $(g : T \to S, t' \leadsto t, \xi )$ such that $(T, t) \to (S, s)$ is an elementary étale neighbourhood.

**Proof.**
Let $(g : T \to S, t' \leadsto t, \xi )$ be an impurity of $\mathcal{F}$ above $s$ such that $g$ is quasi-finite at $t$. After shrinking $T$ we may assume that $g$ is locally of finite type. Apply More on Morphisms, Lemma 37.41.1 to $T \to S$ and $t \mapsto s$. This gives us a diagram

where $(U, u) \to (S, s)$ is an elementary étale neighbourhood and $V \subset T \times _ S U$ is an open neighbourhood of $v = (t, u)$ such that $V \to U$ is finite and such that $v$ is the unique point of $V$ lying over $u$. Since the morphism $V \to T$ is étale hence flat we see that there exists a specialization $v' \leadsto v$ such that $v' \mapsto t'$. Note that $\kappa (t') \subset \kappa (v')$ is finite separable. Pick any point $\zeta \in X_{v'}$ mapping to $\xi \in X_{t'}$. By Divisors, Lemma 31.7.3 we see that $\zeta \in \text{Ass}_{X_ V/V}(\mathcal{F}_ V)$. Moreover, the closure $\overline{\{ \zeta \} }$ does not meet the fibre $X_ v$ as by assumption the closure $\overline{\{ \xi \} }$ does not meet $X_ t$. In other words $(V \to S, v' \leadsto v, \zeta )$ is an impurity of $\mathcal{F}$ above $S$.

Next, let $u' \in U'$ be the image of $v'$ and let $\theta \in X_ U$ be the image of $\zeta $. Then $\theta \mapsto u'$ and $u' \leadsto u$. By Divisors, Lemma 31.7.3 we see that $\theta \in \text{Ass}_{X_ U/U}(\mathcal{F})$. Moreover, as $\pi : X_ V \to X_ U$ is finite we see that $\pi \big (\overline{\{ \zeta \} }\big ) = \overline{\{ \pi (\zeta )\} }$. Since $v$ is the unique point of $V$ lying over $u$ we see that $X_ u \cap \overline{\{ \pi (\zeta )\} } = \emptyset $ because $X_ v \cap \overline{\{ \zeta \} } = \emptyset $. In this way we conclude that $(U \to S, u' \leadsto u, \theta )$ is an impurity of $\mathcal{F}$ above $s$ and we win. $\square$

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