## 58.22 Finite étale covers of punctured spectra, II

In this section we prove some variants of the material discussed in Section 58.19. Suppose we have a Noetherian local ring $(A, \mathfrak m)$ and $f \in \mathfrak m$. We set $X = \mathop{\mathrm{Spec}}(A)$ and $X_0 = \mathop{\mathrm{Spec}}(A/fA)$ and we let $U = X \setminus \{ \mathfrak m\}$ and $U_0 = X_0 \setminus \{ \mathfrak m\}$ be the punctured spectrum of $A$ and $A/fA$. All of this is exactly as in Situation 58.19.1. The difference is that we will consider the restriction functor

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

In other words, we will not try to lift finite étale coverings of $U_0$ to all of $U$, but just to some open neighbourhood $U'$ of $U_0$ in $U$.

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 $1$. $\square$

Lemma 58.22.2. In Situation 58.19.1 assume

1. $A$ has a dualizing complex and is $f$-adically complete,

2. every irreducible component of $X$ not contained in $X_0$ has dimension $\geq 3$.

Then the restriction functor

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

is fully faithful.

Proof. To prove this we may replace $A$ by its reduction by the topological invariance of the fundamental group, see Lemma 58.8.3. Then the result follows from Lemma 58.17.3 and Algebraic and Formal Geometry, Lemma 52.15.8. $\square$

Lemma 58.22.3. In Situation 58.19.1 assume

1. $A$ is $f$-adically complete,

2. $f$ is a nonzerodivisor.

3. $H^1_\mathfrak m(A/fA)$ is a finite $A$-module.

Then the restriction functor

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

is fully faithful.

Proof. Follows from Lemma 58.17.3 and Algebraic and Formal Geometry, Lemma 52.15.9. $\square$

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).