The Stacks project

Lemma 58.31.7. 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 58.31.5. Observe that the action of $G$ on $Z$ extends to an action of $G$ on $Z'$. By Lemma 58.31.6 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 58.31.2. 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 32.19.3 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 32.19.4. 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. $\square$


Comments (1)

Comment #3594 by shanbei on

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

"such that has" --> "such that has"

There are also:

  • 4 comment(s) on Section 58.31: Tame ramification

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EYJ. Beware of the difference between the letter 'O' and the digit '0'.