Lemma 58.31.8. Let $S$ be an integral normal Noetherian scheme with generic point $\eta$. Let $f : X \to S$ be a smooth morphism with geometrically connected fibres. Let $\sigma : S \to X$ be a section of $f$. Let $Z \to X_\eta$ be a finite étale Galois cover (Section 58.7) with group $G$ of order invertible on $S$ such that $Z$ has a $\kappa (\eta )$-rational point mapping to $\sigma (\eta )$. Then there exists a finite étale Galois cover $Y \to X$ with group $G$ whose restriction to $X_\eta$ is $Z$.

Proof. First assume $S = \mathop{\mathrm{Spec}}(R)$ is the spectrum of a discrete valuation ring $R$ with closed point $s \in S$. Then $X_ s$ is an effective Cartier divisor in $X$ and $X_ s$ is regular as a scheme smooth over a field. Moreover the generic fibre $X_\eta$ is the open subscheme $X \setminus X_ s$. It follows from More on Algebra, Lemma 15.112.2 and the assumption on $G$ that $Z$ is tamely ramified over $X$ in codimension $1$. Let $Z' \to X$ be as in Lemma 58.31.6. Observe that the action of $G$ on $Z$ extends to an action of $G$ on $Z'$. By Lemma 58.31.7 we see that $Z' \to X$ is finite étale over an open neighbourhood of $\sigma (y)$. Since $X_ s$ is irreducible, this implies $Z \to X_\eta$ is unramified over $X$ in codimension $1$. Then we get a finite étale morphism $Y \to X$ whose restriction to $X_\eta$ is $Z$ by Lemma 58.31.3. Of course $Y \cong Z'$ (details omitted; hint: compute étale locally) and hence $Y$ is a Galois cover with group $G$.

General case. Let $U \subset S$ be a maximal open subscheme such that there exists a finite étale Galois cover $Y \to X \times _ S U$ with group $G$ whose restriction to $X_\eta$ is isomorphic to $Z$. Assume $U \not= S$ to get a contradiction. Let $s \in S \setminus U$ be a generic point of an irreducible component of $S \setminus U$. Then the inverse image $U_ s$ of $U$ in $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ is the punctured spectrum of $\mathcal{O}_{S, s}$. We claim $Y \times _ S U_ s \to X \times _ S U_ s$ is the restriction of a finite étale Galois cover $Y'_ s \to X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ with group $G$.

Let us first prove the claim produces the desired contradiction. By Limits, Lemma 32.20.3 we find an open subscheme $U \subset U' \subset S$ containing $s$ and a morphism $Y'' \to U'$ of finite presentation whose restriction to $U$ recovers $Y' \to U$ and whose restriction to $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ recovers $Y'_ s$. Moreover, by the equivalence of categories given in the lemma, we may assume after shrinking $U'$ there is a morphism $Y'' \to U' \times _ S X$ and there is an action of $G$ on $Y''$ over $U' \times _ S X$ compatible with the given morphisms and actions after base change to $U$ and $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$. After shrinking $U'$ further if necessary, we may assume $Y'' \to U \times _ S X$ is finite étale, see Limits, Lemma 32.20.4. This means we have found a strictly larger open of $S$ over which $Y$ extends to a finite étale Galois cover with group $G$ which gives the contradiction we were looking for.

Proof of the claim. We may and do replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$. Then $S = \mathop{\mathrm{Spec}}(A)$ where $(A, \mathfrak m)$ is a local normal domain. Also $U \subset S$ is the punctured spectrum and we have a finite étale Galois cover $Y \to X \times _ S U$ with group $G$. If $\dim (A) = 1$, then we can construct the extension of $Y$ to a Galois covering of $X$ by the first paragraph of the proof. Thus we may assume $\dim (A) \geq 2$ and hence $\text{depth}(A) \geq 2$ as $S$ is normal, see Algebra, Lemma 10.157.4. Since $X \to S$ is flat, we conclude that $\text{depth}(\mathcal{O}_{X, x}) \geq 2$ for every point $x \in X$ mapping to $s$, see Algebra, Lemma 10.163.2. Let

$Y' \longrightarrow X$

be the finite morphism constructed in Lemma 58.21.5 using $Y \to X \times _ S U$. Observe that we obtain a canonical $G$-action on $Y$. Thus all that remains is to show that $Y'$ is étale over $X$. In fact, by Lemma 58.26.3 (for example) it even suffices to show that $Y' \to X$ is étale over the (unique) generic point of the fibre $X_ s$. This we do by a local calculation in a (formal) neighbourhood of $\sigma (s)$.

Choose an affine open $\mathop{\mathrm{Spec}}(B) \subset X$ containing $\sigma (s)$. Then $A \to B$ is a smooth ring map which has a section $\sigma : B \to A$. Denote $I = \mathop{\mathrm{Ker}}(\sigma )$ and denote $B^\wedge$ the $I$-adic completion of $B$. Then $B^\wedge \cong A[[x_1, \ldots , x_ d]]$ for some $d \geq 0$, see Algebra, Lemma 10.139.4. Of course $B \to B^\wedge$ is flat (Algebra, Lemma 10.97.2) and the image of $\mathop{\mathrm{Spec}}(B^\wedge ) \to X$ contains the generic point of $X_ s$. Let $V \subset \mathop{\mathrm{Spec}}(B^\wedge )$ be the inverse image of $U$. Consider the finite étale morphism

$W = Y \times _{(X \times _ S U)} V \longrightarrow V$

By the compatibility of the construction of $Y'$ with flat base change in Lemma 58.21.5 we find that the base chang $Y' \times _ X \mathop{\mathrm{Spec}}(B^\wedge ) \to \mathop{\mathrm{Spec}}(B^\wedge )$ is constructed from $W \to V$ over $\mathop{\mathrm{Spec}}(B^\wedge )$ by the procedure in Lemma 58.21.5. Set $V_0 = V \cap V(x_1, \ldots , x_ d) \subset V$ and $W_0 = W \times _ V V_0$. This is a normal integral scheme which maps into $\sigma (S)$ by the morphism $\mathop{\mathrm{Spec}}(B^\wedge ) \to X$ and in fact is identified with $\sigma (U)$. Hence we know that $W_0 \to V_0 = U$ completely decomposes as this is true for its generic fibre by our assumption on $Z \to X_\eta$ having a $\kappa (\eta )$-rational point lying over $\sigma (\eta )$ (and of course the $G$-action then implies the whole fibre $Z_{\sigma (\eta )}$ is a disjoint union of copies of the scheme $\eta = \mathop{\mathrm{Spec}}(\kappa (\eta ))$). Finally, by Lemma 58.26.1 we have

$W_0 \times _ U V \cong W$

This shows that $W$ is a disjoint union of copies of $V$ and hence $Y' \times _ X \mathop{\mathrm{Spec}}(B^\wedge )$ is a disjoint union of copies of $\mathop{\mathrm{Spec}}(B^\wedge )$ and the proof is complete. $\square$

Comment #3594 by shanbei on

Second to the last line in the statement of this Lemma:

"such that $Y$ has" --> "such that $Z$ has"

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