## Tag `0BLE`

## 52.16. Finite étale covers of punctured spectra, I

We first prove some results á la Lefschetz.

Situation 52.16.1. Let $(A, \mathfrak m)$ be a Noetherian local ring 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 $A/fA$.

Recall that for a scheme $X$ the category of schemes finite étale over $X$ is denoted $\textit{FÉt}_X$, see Section 52.5. In Situation 52.16.1 we will study the base change functors $$ \xymatrix{ \textit{FÉt}_X \ar[d] \ar[r] & \textit{FÉt}_U \ar[d] \\ \textit{FÉt}_{X_0} \ar[r] & \textit{FÉt}_{U_0} } $$ In many case the right vertical arrow is faithful.

Lemma 52.16.2. In Situation 52.16.1. Assume one of the following holds

- $\dim(A/\mathfrak p) \geq 2$ for every minimal prime $\mathfrak p \subset A$ with $f \not \in \mathfrak p$, or
- every connected component of $U$ meets $U_0$.
Then $$ \textit{FÉt}_U \longrightarrow \textit{FÉt}_{U_0},\quad V \longmapsto V_0 = V \times_U U_0 $$ is a faithful functor.

Proof.Let $a, b : V \to W$ be two morphisms of schemes finite étale over $U$ whose restriction to $U_0$ are the same. Assumption (1) means that every irreducible component of $U$ meets $U_0$, see Algebra, Lemma 10.59.12. The image of any irreducible component of $V$ is an irreducible component of $U$ and hence meets $U_0$. Hence $V_0$ meets every connected component of $V$ and we conclude that $a = b$ by Étale Morphisms, Proposition 40.6.3. In case (2) the argument is the same using that the image of a connected component of $V$ is a connected component of $U$. $\square$Before we prove something more interesting, we need a couple of lemmas.

Lemma 52.16.3. In Situation 52.16.1. Let $V \to U$ be a finite morphism. Let $A^\wedge$ be the $\mathfrak m$-adic completion of $A$, let $X' = \mathop{\mathrm{Spec}}(A^\wedge)$ and let $U'$ and $V'$ be the base changes of $U$ and $V$ to $X'$. If $Y' \to X'$ is a finite morphism such that $V' = Y' \times_{X'} U'$, then there exists a finite morphism $Y \to X$ such that $V = Y \times_X U$ and $Y' = Y \times_X X'$.

Proof.This is a straightforward application of More on Algebra, Proposition 15.78.15. Namely, choose generators $f_1, \ldots, f_t$ of $\mathfrak m$. For each $i$ write $V \times_U D(f_i) = \mathop{\mathrm{Spec}}(B_i)$. For $1 \leq i, j \leq n$ we obtain an isomorphism $\alpha_{ij} : (B_i)_{f_j} \to (B_j)_{f_i}$ of $A_{f_if_j}$-algebras because the spectrum of both represent $V \times_U D(f_if_j)$. Write $Y' = \mathop{\mathrm{Spec}}(B')$. Since $V \times_U U' = Y \times_{X'} U'$ we get isomorphisms $\alpha_i : B'_{f_i} \to B_i \otimes_A A^\wedge$. A straightforward argument shows that $(B', B_i, \alpha_i, \alpha_{ij})$ is an object of $\text{Glue}(A \to A^\wedge, f_1, \ldots, f_t)$, see More on Algebra, Remark 15.78.10. Applying the proposition cited above (and using More on Algebra, Remark 15.78.19 to obtain the algebra structure) we find an $A$-algebra $B$ such that $\text{Can}(B)$ is isomorphic to $(B', B_i, \alpha_i, \alpha_{ij})$. Setting $Y = \mathop{\mathrm{Spec}}(B)$ we see that $Y \to X$ is a morphism which comes equipped with compatible isomorphisms $V \cong Y \times_X U$ and $Y' = Y \times_X X'$ as desired. $\square$Lemma 52.16.4. In Situation 52.16.1 assume $A$ is henselian or more generally that $(A, (f))$ is a henselian pair. Let $A^\wedge$ be the $\mathfrak m$-adic completion of $A$, let $X' = \mathop{\mathrm{Spec}}(A^\wedge)$ and let $U'$ and $U'_0$ be the base changes of $U$ and $U_0$ to $X'$. If $\textit{FÉt}_{U'} \to \textit{FÉt}_{U'_0}$ is fully faithful, then $\textit{FÉt}_U \to \textit{FÉt}_{U_0}$ is fully faithful.

Proof.Assume $\textit{FÉt}_{U'} \longrightarrow \textit{FÉt}_{U'_0}$ is a fully faithful. Since $X' \to X$ is faithfully flat, it is immediate that the functor $V \to V_0 = V \times_U U_0$ is faithful. Since the category of finite étale coverings has an internal hom (Lemma 52.5.4) it suffices to prove the following: Given $V$ finite étale over $U$ we have $$ \mathop{Mor}\nolimits_U(U, V) = \mathop{Mor}\nolimits_{U_0}(U_0, V_0) $$ The we assume we have a morphism $s_0 : U_0 \to V_0$ over $U_0$ and we will produce a morphism $s : U \to V$ over $U$.By our assumption there does exist a morphism $s' : U' \to V'$ whose restriction to $V'_0$ is the base change $s'_0$ of $s_0$. Since $V' \to U'$ is finite étale this means that $V' = s'(U') \amalg W'$ for some $W' \to U'$ finite and étale. Choose a finite morphism $Z' \to X'$ such that $W' = Z' \times_{X'} U'$. This is possible by Zariski's main theorem in the form stated in More on Morphisms, Lemma 36.38.3 (small detail omitted). Then $$ V' = s'(U') \amalg W' \longrightarrow X' \amalg Z' = Y' $$ is an open immersion such that $V' = Y' \times_{X'} U'$. By Lemma 52.16.3 we can find $Y \to X$ finite such that $V = Y \times_X U$ and $Y' = Y \times_X X'$. Write $Y = \mathop{\mathrm{Spec}}(B)$ so that $Y' = \mathop{\mathrm{Spec}}(B \otimes_A A^\wedge)$. Then $B \otimes_A A^\wedge$ has an idempotent $e'$ corresponding to the open and closed subscheme $X'$ of $Y' = X' \amalg Z'$.

The case $A$ is henselian (slightly easier). The image $\overline{e}$ of $e'$ in $B \otimes_A \kappa(\mathfrak m) = B/\mathfrak mB$ lifts to an idempotent $e$ of $B$ as $A$ is henselian (because $B$ is a product of local rings by Algebra, Lemma 10.148.3). Then we see that $e$ maps to $e'$ by uniqueness of lifts of idempotents (using that $B \otimes_A A^\wedge$ is a product of local rings). Let $Y_1 \subset Y$ be the open and closed subscheme corresponding to $e$. Then $Y_1 \times_X X' = s'(X')$ which implies that $Y_1 \to X$ is an isomorphism (by faithfully flat descent) and gives the desired section.

The case where $(A, (f))$ is a henselian pair. Here we use that $s'$ is a lift of $s'_0$. Namely, let $Y_{0, 1} \subset Y_0 = Y \times_X X_0$ be the closure of $s_0(U_0) \subset V_0 = Y_0 \times_{X_0} U_0$. As $X' \to X$ is flat, the base change $Y'_{0, 1} \subset Y'_0$ is the closure of $s'_0(U'_0)$ which is equal to $X'_0 \subset Y'_0$ (see Morphisms, Lemma 28.24.15). Since $Y'_0 \to Y_0$ is submersive (Morphisms, Lemma 28.24.11) we conclude that $Y_{0, 1}$ is open and closed in $Y_0$. Let $e_0 \in B/fB$ be the corresponding idempotent. By More on Algebra, Lemma 15.10.8 we can lift $e_0$ to an idempotent $e \in B$. Then we conclude as before. $\square$

The following lemma will be superseded by Lemma 52.16.6 below.

Lemma 52.16.5. In Situation 52.16.1. Assume $f$ is a nonzerodivisor, that $A$ has depth $\geq 3$, and that $A$ is henselian or more generally $(A, (f))$ is a henselian pair. Then $$ \textit{FÉt}_U \longrightarrow \textit{FÉt}_{U_0},\quad V \longmapsto V_0 = V \times_U U_0 $$ is a fully faithful functor.

Proof.By Lemma 52.16.4 we may assume $A$ is a complete local Noetherian ring. The functor is faithful by Lemma 52.16.2 (to see the assumption of that lemma holds, apply Algebra, Lemma 10.71.9). Since the category of finite étale coverings has an internal hom (Lemma 52.5.4) it suffices to prove the following: Given $V$ finite étale over $U$ we have $$ \mathop{Mor}\nolimits_U(U, V) = \mathop{Mor}\nolimits_{U_0}(U_0, V_0) $$ If we have a morphism $U_0 \to V_0$ over $U_0$, then we obtain an decomposition $V_0 = U_0 \amalg V'_0$ into open and closed subschemes. We will show that this implies the same thing for $V$ thereby finishing the proof.For $n \geq 1$ let $U_n$ be the punctured spectrum of $A/f^{n + 1}A$ and let $V_n \to U_n$ be the base change of $V \to U$. By Étale Morphisms, Theorem 40.15.2 we conclude that there is a unique decomposition $V_n = U_n \amalg V'_n$ into open and closed subschemes whose base change to $U_0$ recovers the given decomposition.

Since $A$ has depth $\geq 3$ and $f$ is a nonzerodivisor, we see that $A/fA$ has depth $\geq 2$ (Algebra, Lemma 10.71.7). This implies the vanishing of $H^0_\mathfrak m(A/fA)$ and $H^1_\mathfrak m(A/fA)$, see Dualizing Complexes, Lemma 45.11.1. This in turn tells us that $A/fA \to \Gamma(U_0, \mathcal{O}_{U_0})$ is an isomorphism, see Local Cohomology, Lemma 48.6.2. As $f$ is a nonzerodivisor we obtain short exact sequences $$ 0 \to A/fA \xrightarrow{f^n} A/f^{n + 1}A \to A/f^n A \to 0 $$ Induction on $n$ shows that $H^0_\mathfrak m(A/f^{n + 1}A) = H^1_\mathfrak m(A/f^{n + 1}A) = 0$ for all $n$. Hence the same reasoning shows that $A/f^{n + 1}A \to \Gamma(U_n, \mathcal{O}_{U_n})$ is an isomorphism. Combined with the decompositions above this determines a map $$ \Gamma(V, \mathcal{O}_V) \to \mathop{\mathrm{lim}}\nolimits \Gamma(V_n, \mathcal{O}_{V_n}) \to \mathop{\mathrm{lim}}\nolimits \Gamma(U_n, \mathcal{O}_{U_n}) = A $$ Since $V \to U$ is affine, this $A$-algebra map corresponds to a section $U \to V$ as desired. $\square$

In the following lemma we prove fully faithfulness under very weak assumptions. Note that the assumptions do not imply that $U$ is a connected scheme, but the conclusion guarantees that $U$ and $U_0$ have the same number of connected components.

Lemma 52.16.6. In Situation 52.16.1. Assume

- $f$ is a nonzerodivisor,
- $H^1_\mathfrak m(A)$ is finite,
- $H^2_\mathfrak m(A)$ is annihilated by a power of $f$, and
- $A$ is henselian or more generally $(A, (f))$ is a henselian pair.
Then $$ \textit{FÉt}_U \longrightarrow \textit{FÉt}_{U_0},\quad V \longmapsto V_0 = V \times_U U_0 $$ is a fully faithful functor.

Proof.By Lemma 52.16.4 we may assume that $A$ is a Noetherian complete local ring. (The assumptions carry over; use Dualizing Complexes, Lemma 45.9.3.)Assume $A$ is complete in addition to the other conditions. We will show that given $\pi : V \to U$ finite étale, the set of connected components of $V$ agrees with the set of connected components of $V_0$. This will prove the lemma because the category of finite étale covers has internal hom (Lemma 52.5.4) and images of sections are connected components (Étale Morphisms, Proposition 40.6.1). Some details omitted.

Set $\mathcal{B} = \pi_*\mathcal{O}_V$. This is a finite locally free $\mathcal{O}_U$-algebra. Thus $\text{Ass}(\mathcal{B}) = \text{Ass}(\mathcal{O}_U)$. Assumption (2) means that $H^0(U, \mathcal{O}_U)$ is a finite $A$-module and equivalently that $j_*\mathcal{O}_U$ is coherent (Local Cohomology, Lemma 48.6.2). By Local Cohomology, Proposition 48.6.7 and the agreement of $\text{Ass}$ we see that the same holds for $\mathcal{B}$ and we conclude that $B = \Gamma(U, \mathcal{B}) = \Gamma(V, \mathcal{O}_V)$ is a finite $A$-algebra.

Next, using that $H^2_\mathfrak m(A) = H^1(U, \mathcal{O}_U)$ is annihilated by $f^n$ for some $n$ we see that $H^1(U, \mathcal{B}) = H^1(V, \mathcal{O}_V)$ is annihilated by $f^m$ for some $m$, see Local Cohomology, Lemma 48.11.3.

At this point we apply Local Cohomology, Lemma 48.13.4 to the scheme $V$ over $\mathop{\mathrm{Spec}}(A)$ and the sheaf $\mathcal{O}_V$ with $p = 0$. Since $f$ is a nonzerodivisor in $A$ the $f$-power torsion subsheaf of $\mathcal{O}_V$ is zero. The first short exact sequence of the lemma collapses to become $$ H^0 = \mathop{\mathrm{lim}}\nolimits H^0(V, \mathcal{O}_V/f^n\mathcal{O}_V) = \mathop{\mathrm{lim}}\nolimits H^0(V_n, \mathcal{O}_{V_n}) $$ where $V_n \subset V$ is the closed subscheme cut out by $f^{n + 1}$. Since $H^1(V, \mathcal{O}_V)$ is annihilated by a power of $f$ we see that the Tate module $T_f(H^1(V, \mathcal{O}_V))$ is zero. On the other hand, since $A$ is complete and $B = H^0(V, \mathcal{O}_V)$ is a finite $A$-module it is complete (Algebra, Lemma 10.96.1) hence derived complete (More on Algebra, Proposition 15.80.5) and hence equal to its derived $f$-adic completion. Thus we see that $H^0 = B$. Since $$ V_0 \subset V_1 \subset V_2 \subset \ldots $$ are nilpotent thickenings the connected components of these schemes agree. Correspondingly the maps $$ \ldots \to H^0(V_2, \mathcal{O}_{V_2}) \to H^0(V_1, \mathcal{O}_{V_1}) \to H^0(V_0, \mathcal{O}_{V_0}) $$ induce bijections between idempotents. Hence the map $B \to H^0(V_0, \mathcal{O}_{V_0})$ induces a bijection between idempotents and we conclude. $\square$

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```
\section{Finite \'etale covers of punctured spectra, I}
\label{section-pi1-punctured-spec}
\noindent
We first prove some results \'a la Lefschetz.
\begin{situation}
\label{situation-local-lefschetz}
Let $(A, \mathfrak m)$ be a Noetherian local ring and $f \in \mathfrak m$.
We set $X = \Spec(A)$ and $X_0 = \Spec(A/fA)$ and we
let $U = X \setminus \{\mathfrak m\}$ and
$U_0 = X_0 \setminus \{\mathfrak m\}$ be the punctured spectrum of
$A$ and $A/fA$.
\end{situation}
\noindent
Recall that for a scheme $X$ the category of schemes finite
\'etale over $X$ is denoted $\textit{F\'Et}_X$, see
Section \ref{section-finite-etale}.
In Situation \ref{situation-local-lefschetz}
we will study the base change functors
$$
\xymatrix{
\textit{F\'Et}_X \ar[d] \ar[r] & \textit{F\'Et}_U \ar[d] \\
\textit{F\'Et}_{X_0} \ar[r] & \textit{F\'Et}_{U_0}
}
$$
In many case the right vertical arrow is faithful.
\begin{lemma}
\label{lemma-faithful}
In Situation \ref{situation-local-lefschetz}.
Assume one of the following holds
\begin{enumerate}
\item $\dim(A/\mathfrak p) \geq 2$ for every minimal prime
$\mathfrak p \subset A$ with $f \not \in \mathfrak p$, or
\item every connected component of $U$ meets $U_0$.
\end{enumerate}
Then
$$
\textit{F\'Et}_U \longrightarrow \textit{F\'Et}_{U_0},\quad
V \longmapsto V_0 = V \times_U U_0
$$
is a faithful functor.
\end{lemma}
\begin{proof}
Let $a, b : V \to W$ be two morphisms of schemes finite \'etale over $U$
whose restriction to $U_0$ are the same. Assumption (1)
means that every irreducible component of $U$ meets $U_0$, see
Algebra, Lemma \ref{algebra-lemma-one-equation}.
The image of any irreducible component of $V$ is an
irreducible component of $U$ and hence meets $U_0$.
Hence $V_0$ meets every connected component of $V$ and
we conclude that $a = b$ by \'Etale Morphisms, Proposition
\ref{etale-proposition-equality}.
In case (2) the argument is the same using that the image
of a connected component of $V$ is a connected component of $U$.
\end{proof}
\noindent
Before we prove something more interesting, we need a couple of lemmas.
\begin{lemma}
\label{lemma-fill-in-missing}
In Situation \ref{situation-local-lefschetz}. Let $V \to U$ be a finite
morphism. Let $A^\wedge$ be the $\mathfrak m$-adic completion of $A$,
let $X' = \Spec(A^\wedge)$ and let $U'$ and $V'$ be the base changes of
$U$ and $V$ to $X'$. If $Y' \to X'$ is a finite morphism such that
$V' = Y' \times_{X'} U'$, then there exists a finite morphism $Y \to X$
such that $V = Y \times_X U$ and $Y' = Y \times_X X'$.
\end{lemma}
\begin{proof}
This is a straightforward application of
More on Algebra, Proposition \ref{more-algebra-proposition-equivalence}.
Namely, choose generators $f_1, \ldots, f_t$ of $\mathfrak m$.
For each $i$ write $V \times_U D(f_i) = \Spec(B_i)$.
For $1 \leq i, j \leq n$ we obtain an isomorphism
$\alpha_{ij} : (B_i)_{f_j} \to (B_j)_{f_i}$ of $A_{f_if_j}$-algebras
because the spectrum of both represent $V \times_U D(f_if_j)$.
Write $Y' = \Spec(B')$. Since $V \times_U U' = Y \times_{X'} U'$
we get isomorphisms $\alpha_i : B'_{f_i} \to B_i \otimes_A A^\wedge$.
A straightforward argument shows that $(B', B_i, \alpha_i, \alpha_{ij})$
is an object of $\text{Glue}(A \to A^\wedge, f_1, \ldots, f_t)$, see
More on Algebra, Remark \ref{more-algebra-remark-glueing-data}.
Applying the proposition cited above (and using
More on Algebra, Remark \ref{more-algebra-remark-formal-glueing-algebras}
to obtain the algebra structure) we find an $A$-algebra $B$ such that
$\text{Can}(B)$ is isomorphic to $(B', B_i, \alpha_i, \alpha_{ij})$.
Setting $Y = \Spec(B)$ we see that $Y \to X$ is a morphism
which comes equipped with compatible isomorphisms
$V \cong Y \times_X U$ and $Y' = Y \times_X X'$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-fully-faithful-henselian-completion}
In Situation \ref{situation-local-lefschetz} assume $A$ is henselian
or more generally that $(A, (f))$ is a henselian pair.
Let $A^\wedge$ be the $\mathfrak m$-adic completion of $A$,
let $X' = \Spec(A^\wedge)$ and let $U'$ and $U'_0$ be the base changes of
$U$ and $U_0$ to $X'$. If $\textit{F\'Et}_{U'} \to \textit{F\'Et}_{U'_0}$
is fully faithful, then $\textit{F\'Et}_U \to \textit{F\'Et}_{U_0}$
is fully faithful.
\end{lemma}
\begin{proof}
Assume $\textit{F\'Et}_{U'} \longrightarrow \textit{F\'Et}_{U'_0}$
is a fully faithful. Since $X' \to X$ is faithfully flat, it is
immediate that the functor $V \to V_0 = V \times_U U_0$ is faithful.
Since the category of finite \'etale coverings has an internal hom
(Lemma \ref{lemma-internal-hom-finite-etale})
it suffices to prove the following: Given $V$ finite \'etale over $U$
we have
$$
\Mor_U(U, V) = \Mor_{U_0}(U_0, V_0)
$$
The we assume we have a morphism $s_0 : U_0 \to V_0$ over $U_0$ and we will
produce a morphism $s : U \to V$ over $U$.
\medskip\noindent
By our assumption there does exist a morphism $s' : U' \to V'$
whose restriction to $V'_0$ is the base change $s'_0$ of $s_0$.
Since $V' \to U'$ is finite \'etale this means that $V' = s'(U') \amalg W'$
for some $W' \to U'$ finite and \'etale.
Choose a finite morphism $Z' \to X'$ such that $W' = Z' \times_{X'} U'$.
This is possible by Zariski's main theorem in the form stated in
More on Morphisms, Lemma
\ref{more-morphisms-lemma-quasi-finite-separated-pass-through-finite}
(small detail omitted).
Then
$$
V' = s'(U') \amalg W' \longrightarrow X' \amalg Z' = Y'
$$
is an open immersion such that $V' = Y' \times_{X'} U'$.
By Lemma \ref{lemma-fill-in-missing} we can find $Y \to X$ finite
such that $V = Y \times_X U$ and $Y' = Y \times_X X'$.
Write $Y = \Spec(B)$ so that $Y' = \Spec(B \otimes_A A^\wedge)$.
Then $B \otimes_A A^\wedge$ has an idempotent $e'$
corresponding to the open and closed subscheme $X'$ of $Y' = X' \amalg Z'$.
\medskip\noindent
The case $A$ is henselian (slightly easier). The image $\overline{e}$
of $e'$ in $B \otimes_A \kappa(\mathfrak m) = B/\mathfrak mB$ lifts to an
idempotent $e$ of $B$ as $A$ is henselian (because $B$ is a product of
local rings by Algebra, Lemma \ref{algebra-lemma-characterize-henselian}).
Then we see that $e$ maps to $e'$ by uniqueness of lifts of idempotents
(using that $B \otimes_A A^\wedge$ is a product of local rings).
Let $Y_1 \subset Y$ be the open and closed subscheme corresponding to $e$.
Then $Y_1 \times_X X' = s'(X')$ which implies that $Y_1 \to X$ is
an isomorphism (by faithfully flat descent) and gives the desired section.
\medskip\noindent
The case where $(A, (f))$ is a henselian pair. Here we use that $s'$ is
a lift of $s'_0$. Namely, let $Y_{0, 1} \subset Y_0 = Y \times_X X_0$
be the closure of $s_0(U_0) \subset V_0 = Y_0 \times_{X_0} U_0$.
As $X' \to X$ is flat, the base change $Y'_{0, 1} \subset Y'_0$
is the closure of $s'_0(U'_0)$ which is equal to $X'_0 \subset Y'_0$
(see Morphisms, Lemma
\ref{morphisms-lemma-flat-base-change-scheme-theoretic-image}).
Since $Y'_0 \to Y_0$ is submersive
(Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology})
we conclude that $Y_{0, 1}$ is open and closed in $Y_0$.
Let $e_0 \in B/fB$ be the corresponding idempotent.
By More on Algebra, Lemma
\ref{more-algebra-lemma-characterize-henselian-pair}
we can lift $e_0$ to an idempotent $e \in B$.
Then we conclude as before.
\end{proof}
\noindent
The following lemma will be superseded by
Lemma \ref{lemma-fully-faithful-minimal} below.
\begin{lemma}
\label{lemma-fully-faithful}
In Situation \ref{situation-local-lefschetz}.
Assume $f$ is a nonzerodivisor, that $A$ has depth $\geq 3$, and that
$A$ is henselian or more generally $(A, (f))$ is a henselian pair. Then
$$
\textit{F\'Et}_U \longrightarrow \textit{F\'Et}_{U_0},\quad
V \longmapsto V_0 = V \times_U U_0
$$
is a fully faithful functor.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-fully-faithful-henselian-completion} we may assume $A$
is a complete local Noetherian ring. The functor is faithful by
Lemma \ref{lemma-faithful} (to see the assumption of that lemma holds, apply
Algebra, Lemma \ref{algebra-lemma-depth-dim-associated-primes}).
Since the category of finite \'etale coverings has an internal hom
(Lemma \ref{lemma-internal-hom-finite-etale})
it suffices to prove the following: Given $V$ finite \'etale over $U$ we have
$$
\Mor_U(U, V) = \Mor_{U_0}(U_0, V_0)
$$
If we have a morphism $U_0 \to V_0$ over $U_0$, then we obtain an
decomposition $V_0 = U_0 \amalg V'_0$ into open and closed subschemes.
We will show that this implies the same thing for $V$ thereby
finishing the proof.
\medskip\noindent
For $n \geq 1$ let $U_n$ be the punctured spectrum of $A/f^{n + 1}A$
and let $V_n \to U_n$ be the base change of $V \to U$. By
\'Etale Morphisms, Theorem \ref{etale-theorem-remarkable-equivalence}
we conclude that there is a unique decomposition
$V_n = U_n \amalg V'_n$
into open and closed subschemes whose base change to $U_0$ recovers
the given decomposition.
\medskip\noindent
Since $A$ has depth $\geq 3$ and $f$ is a nonzerodivisor, we see
that $A/fA$ has depth $\geq 2$
(Algebra, Lemma \ref{algebra-lemma-depth-drops-by-one}).
This implies the
vanishing of $H^0_\mathfrak m(A/fA)$ and $H^1_\mathfrak m(A/fA)$, see
Dualizing Complexes, Lemma \ref{dualizing-lemma-depth}.
This in turn
tells us that $A/fA \to \Gamma(U_0, \mathcal{O}_{U_0})$ is an isomorphism, see
Local Cohomology, Lemma
\ref{local-cohomology-lemma-finiteness-pushforwards-and-H1-local}.
As $f$ is a nonzerodivisor we obtain short exact sequences
$$
0 \to A/fA \xrightarrow{f^n} A/f^{n + 1}A \to A/f^n A \to 0
$$
Induction on $n$ shows that
$H^0_\mathfrak m(A/f^{n + 1}A) = H^1_\mathfrak m(A/f^{n + 1}A) = 0$
for all $n$. Hence the same reasoning shows that
$A/f^{n + 1}A \to \Gamma(U_n, \mathcal{O}_{U_n})$
is an isomorphism.
Combined with the decompositions above this determines a map
$$
\Gamma(V, \mathcal{O}_V) \to
\lim \Gamma(V_n, \mathcal{O}_{V_n}) \to
\lim \Gamma(U_n, \mathcal{O}_{U_n}) = A
$$
Since $V \to U$ is affine, this $A$-algebra map corresponds to
a section $U \to V$ as desired.
\end{proof}
\noindent
In the following lemma we prove fully faithfulness under very weak assumptions.
Note that the assumptions do not imply that $U$ is a connected scheme, but
the conclusion guarantees that $U$ and $U_0$ have the same number of
connected components.
\begin{lemma}
\label{lemma-fully-faithful-minimal}
\begin{reference}
\cite[Corollary 1.11]{Bhatt-local}
\end{reference}
In Situation \ref{situation-local-lefschetz}. Assume
\begin{enumerate}
\item $f$ is a nonzerodivisor,
\item $H^1_\mathfrak m(A)$ is finite,
\item $H^2_\mathfrak m(A)$ is annihilated by a power of $f$, and
\item $A$ is henselian or more generally $(A, (f))$ is a henselian pair.
\end{enumerate}
Then
$$
\textit{F\'Et}_U \longrightarrow \textit{F\'Et}_{U_0},\quad
V \longmapsto V_0 = V \times_U U_0
$$
is a fully faithful functor.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-fully-faithful-henselian-completion}
we may assume that $A$ is a Noetherian complete local ring.
(The assumptions carry over; use
Dualizing Complexes, Lemma \ref{dualizing-lemma-torsion-change-rings}.)
\medskip\noindent
Assume $A$ is complete in addition to the other conditions.
We will show that given $\pi : V \to U$ finite \'etale, the set
of connected components of $V$ agrees with the set of connected
components of $V_0$. This will prove the lemma because the
category of finite \'etale covers has internal hom
(Lemma \ref{lemma-internal-hom-finite-etale})
and images of sections are connected components
(\'Etale Morphisms, Proposition \ref{etale-proposition-properties-sections}).
Some details omitted.
\medskip\noindent
Set $\mathcal{B} = \pi_*\mathcal{O}_V$. This is a finite locally free
$\mathcal{O}_U$-algebra. Thus
$\text{Ass}(\mathcal{B}) = \text{Ass}(\mathcal{O}_U)$.
Assumption (2) means that $H^0(U, \mathcal{O}_U)$ is a finite
$A$-module and equivalently that $j_*\mathcal{O}_U$ is coherent
(Local Cohomology, Lemma
\ref{local-cohomology-lemma-finiteness-pushforwards-and-H1-local}).
By Local Cohomology, Proposition \ref{local-cohomology-proposition-kollar}
and the agreement of $\text{Ass}$
we see that the same holds for $\mathcal{B}$ and we conclude
that $B = \Gamma(U, \mathcal{B}) = \Gamma(V, \mathcal{O}_V)$
is a finite $A$-algebra.
\medskip\noindent
Next, using that $H^2_\mathfrak m(A) = H^1(U, \mathcal{O}_U)$
is annihilated by $f^n$ for some $n$ we see that
$H^1(U, \mathcal{B}) = H^1(V, \mathcal{O}_V)$
is annihilated by $f^m$ for some $m$, see
Local Cohomology, Lemma \ref{local-cohomology-lemma-annihilate-Hp}.
\medskip\noindent
At this point we apply Local Cohomology, Lemma
\ref{local-cohomology-lemma-formal-functions-principal} to
the scheme $V$ over $\Spec(A)$ and the sheaf $\mathcal{O}_V$
with $p = 0$. Since $f$ is a nonzerodivisor in $A$ the $f$-power torsion
subsheaf of $\mathcal{O}_V$ is zero. The first short exact sequence
of the lemma collapses to become
$$
H^0 = \lim H^0(V, \mathcal{O}_V/f^n\mathcal{O}_V) =
\lim H^0(V_n, \mathcal{O}_{V_n})
$$
where $V_n \subset V$ is the closed subscheme cut out by $f^{n + 1}$.
Since $H^1(V, \mathcal{O}_V)$ is annihilated by a power
of $f$ we see that the Tate module $T_f(H^1(V, \mathcal{O}_V))$ is zero.
On the other hand, since $A$ is complete and
$B = H^0(V, \mathcal{O}_V)$ is a finite $A$-module
it is complete (Algebra, Lemma \ref{algebra-lemma-completion-tensor})
hence derived complete
(More on Algebra,
Proposition \ref{more-algebra-proposition-derived-complete-modules})
and hence equal to its derived $f$-adic completion.
Thus we see that $H^0 = B$.
Since
$$
V_0 \subset V_1 \subset V_2 \subset \ldots
$$
are nilpotent thickenings the connected components of these schemes
agree. Correspondingly the maps
$$
\ldots \to
H^0(V_2, \mathcal{O}_{V_2}) \to
H^0(V_1, \mathcal{O}_{V_1}) \to
H^0(V_0, \mathcal{O}_{V_0})
$$
induce bijections between idempotents. Hence the map
$B \to H^0(V_0, \mathcal{O}_{V_0})$ induces a bijection between
idempotents and we conclude.
\end{proof}
```

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