Situation 58.19.1. 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 $A/fA$.

## 58.19 Finite étale covers of punctured spectra, I

We first prove some results á la Lefschetz.

Recall that for a scheme $X$ the category of schemes finite étale over $X$ is denoted $\textit{FÉt}_ X$, see Section 58.5. In Situation 58.19.1 we will study the base change functors

In many case the right vertical arrow is faithful.

Lemma 58.19.2. In Situation 58.19.1. 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

is a faithful functor.

**Proof.**
Case (2) is immediate from Lemma 58.17.5. Assumption (1) implies every irreducible component of $U$ meets $U_0$, see Algebra, Lemma 10.60.13. Hence (1) follows from (2).
$\square$

Before we prove something more interesting, we need a couple of lemmas.

Lemma 58.19.3. In Situation 58.19.1. 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 15.89.15. 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 15.89.10. Applying the proposition cited above (and using More on Algebra, Remark 15.89.19 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 58.19.4. In Situation 58.19.1 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 58.5.4) it suffices to prove the following: Given $V$ finite étale over $U$ we have

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 37.43.3 (small detail omitted). Then

is an open immersion such that $V' = Y' \times _{X'} U'$. By Lemma 58.19.3 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 10.153.3). 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 29.25.16). Since $Y'_0 \to Y_0$ is submersive (Morphisms, Lemma 29.25.12) 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 15.11.6 we can lift $e_0$ to an idempotent $e \in B$. Then we conclude as before. $\square$

In Situation 58.19.1 fully faithfulness of the restriction functor $\textit{FÉt}_ U \longrightarrow \textit{FÉt}_{U_0}$ holds under fairly mild assumptions. In particular, the assumptions often do not imply $U$ is a connected scheme, but the conclusion guarantees that $U$ and $U_0$ have the same number of connected components.

Lemma 58.19.5. In Situation 58.19.1. Assume

$A$ has a dualizing complex,

the pair $(A, (f))$ is henselian,

one of the following is true

$A_ f$ is $(S_2)$ and every irreducible component of $X$ not contained in $X_0$ has dimension $\geq 3$, or

for every prime $\mathfrak p \subset A$, $f \not\in \mathfrak p$ we have $\text{depth}(A_\mathfrak p) + \dim (A/\mathfrak p) > 2$.

Then the restriction functor $\textit{FÉt}_ U \longrightarrow \textit{FÉt}_{U_0}$ is fully faithful.

**Proof.**
Let $A'$ be the $\mathfrak m$-adic completion of $A$. We will show that the hypotheses remain true for $A'$. This is clear for conditions (a) and (b). Condition (c)(ii) is preserved by Local Cohomology, Lemma 51.11.3. Next, assume (c)(i) holds. Since $A$ is universally catenary (Dualizing Complexes, Lemma 47.17.4) we see that every irreducible component of $\mathop{\mathrm{Spec}}(A')$ not contained in $V(f)$ has dimension $\geq 3$, see More on Algebra, Proposition 15.109.5. Since $A \to A'$ is flat with Gorenstein fibres, the condition that $A_ f$ is $(S_2)$ implies that $A'_ f$ is $(S_2)$. References used: Dualizing Complexes, Section 47.23, More on Algebra, Section 15.51, and Algebra, Lemma 10.163.4. Thus by Lemma 58.19.4 we may assume that $A$ is a Noetherian complete local ring.

Assume $A$ is a complete local ring in addition to the other assumptions. By Lemma 58.17.1 the result follows from Algebraic and Formal Geometry, Lemmas 52.15.5 and 52.15.7. $\square$

Lemma 58.19.6. In Situation 58.19.1. Assume

$H^1_\mathfrak m(A)$ and $H^2_\mathfrak m(A)$ are annihilated by a power of $f$, and

$A$ is henselian or more generally $(A, (f))$ is a henselian pair.

Then the restriction functor $\textit{FÉt}_ U \longrightarrow \textit{FÉt}_{U_0}$ is fully faithful.

**Proof.**
By Lemma 58.19.4 we may assume that $A$ is a Noetherian complete local ring. (The assumptions carry over; use Dualizing Complexes, Lemma 47.9.3.) By Lemma 58.17.1 the result follows from Algebraic and Formal Geometry, Lemma 52.15.6.
$\square$

Lemma 58.19.7. In Situation 58.19.1 assume $A$ has depth $\geq 3$ and $A$ is henselian or more generally $(A, (f))$ is a henselian pair. Then the restriction functor $\textit{FÉt}_ U \to \textit{FÉt}_{U_0}$ is fully faithful.

**Proof.**
The assumption of depth forces $H^1_\mathfrak m(A) = H^2_\mathfrak m(A) = 0$, see Dualizing Complexes, Lemma 47.11.1. Hence Lemma 58.19.6 applies.
$\square$

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