The Stacks project

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). $\square$

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. $\square$

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. $\square$

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, Lemma 52.15.6. $\square$

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 52.15.5. $\square$

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. $\square$