Lemma 58.22.1 (0BLN)—The Stacks project

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Part 3: Topics in Scheme Theory
Chapter 58: Fundamental Groups of Schemes
Section 58.22: Finite étale covers of punctured spectra, II
Lemma 58.22.1 (cite)

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Lemma 58.22.1. In Situation 58.19.1. Let $U' \subset U$ be open and contain $U_0$. Assume for $\mathfrak p \subset A$ minimal with $\mathfrak p \in U'$, $\mathfrak p \not\in U_0$ we have $\dim (A/\mathfrak p) \geq 2$. Then

\[ \textit{FÉt}_{U'} \longrightarrow \textit{FÉt}_{U_0},\quad V' \longmapsto V_0 = V' \times _{U'} U_0 \]

is a faithful functor. Moreover, there exists a $U'$ satisfying the assumption and any smaller open $U'' \subset U'$ containing $U_0$ also satisfies this assumption. In particular, the restriction functor

\[ \mathop{\mathrm{colim}}\nolimits _{U_0 \subset U' \subset U\text{ open}} \textit{FÉt}_{U'} \longrightarrow \textit{FÉt}_{U_0} \]

is faithful.

Proof. By Algebra, Lemma 10.60.13 we see that $V(\mathfrak p)$ meets $U_0$ for every prime $\mathfrak p$ of $A$ with $\dim (A/\mathfrak p) \geq 2$. Thus the displayed functor is faithful for a $U$ as in the statement by Lemma 58.17.5. To see the existence of such a $U'$ note that for $\mathfrak p \subset A$ with $\mathfrak p \in U$, $\mathfrak p \not\in U_0$ with $\dim (A/\mathfrak p) = 1$ then $\mathfrak p$ corresponds to a closed point of $U$ and hence $V(\mathfrak p) \cap U_0 = \emptyset $. Thus we can take $U'$ to be the complement of the irreducible components of $X$ which do not meet $U_0$ and have dimension $1$. $\square$

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