The Stacks project

Proof. Let $(g : T \to S, t' \leadsto t, \xi )$ be any impurity of $\mathcal{F}$ above $s$. We apply Limits, Lemma 32.14.1 to $t \in T$ and $Z = \overline{\{ \xi \} }$ to obtain an open neighbourhood $V \subset T$ of $t$, a commutative diagram

and a closed subscheme $Z' \subset X_{T'}$ such that

1. the morphism $b : T' \to S$ is locally of finite presentation,

2. we have $Z' \cap X_{a(t)} = \emptyset $, and

3. $Z \cap X_ V$ maps into $Z'$ via the morphism $X_ V \to X_{T'}$.

As $t'$ specializes to $t$ we may replace $T$ by the open neighbourhood $V$ of $t$. Thus we have a commutative diagram

where $b \circ a = g$. Let $\xi ' \in X_{T'}$ denote the image of $\xi $. By Divisors, Lemma 31.7.3 we see that $\xi ' \in \text{Ass}_{X_{T'}/T'}(\mathcal{F}_{T'})$. Moreover, by construction the closure of $\overline{\{ \xi '\} }$ is contained in the closed subset $Z'$ which avoids the fibre $X_{a(t)}$. In this way we see that $(T' \to S, a(t') \leadsto a(t), \xi ')$ is an impurity of $\mathcal{F}$ above $s$.

Thus we may assume that $g : T \to S$ is locally of finite presentation. Let $Z = \overline{\{ \xi \} }$. By assumption $Z_ t = \emptyset $. By More on Morphisms, Lemma 37.24.1 this means that $Z_{t''} = \emptyset $ for $t''$ in an open subset of $\overline{\{ t\} }$. Since the fibre of $T \to S$ over $s$ is a Jacobson scheme, see Morphisms, Lemma 29.16.10 we find that there exist a closed point $t'' \in \overline{\{ t\} }$ such that $Z_{t''} = \emptyset $. Then $(g : T \to S, t' \leadsto t'', \xi )$ is the desired impurity. $\square$