The Stacks project

Proof. The only interesting fact in this lemma is the commutative algebra result given in More on Algebra, Lemma 15.114.9. $\square$

Proof. Let $\xi \in X \setminus U$ be a generic point of an irreducible component of $X \setminus U$ of codimension $1$. Then $\mathcal{O}_{X, \xi }$ is a discrete valuation ring. As in the discussion above, write $Y \times _ U \mathop{\mathrm{Spec}}(K_\xi ) = \mathop{\mathrm{Spec}}(L_\xi )$. Denote $B_\xi$ the integral closure of $\mathcal{O}_{X, \xi }$ in $L_\xi$. Our assumption that $Y$ is unramified over $X$ in codimension $1$ signifies that $\mathcal{O}_{X, \xi } \to B_\xi$ is finite étale. Thus we get $Y_\xi \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi })$ finite étale and an isomorphism

over $\mathop{\mathrm{Spec}}(K_\xi )$. By Limits, Lemma 32.20.3 we find an open subscheme $U \subset U' \subset X$ containing $\xi$ and a morphism $Y' \to U'$ of finite presentation whose restriction to $U$ recovers $Y$ and whose restriction to $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi })$ recovers $Y_\xi$. Finally, the morphism $Y' \to U'$ is finite étale after possible shrinking $U'$ to a smaller open by Limits, Lemma 32.20.4. Repeating the argument with the other generic points of $X \setminus U$ of codimension $1$ we may assume that we have a finite étale morphism $Y' \to U'$ extending $Y \to U$ to an open subscheme containing $U' \subset X$ containing $U$ and all codimension $1$ points of $X \setminus U$. We finish by applying Lemma 58.21.6 to $Y' \to U'$. Namely, all local rings $\mathcal{O}_{X, x}$ for $x \in X \setminus U'$ are regular and have $\dim (\mathcal{O}_{X, x}) \geq 2$. Hence we have purity for $\mathcal{O}_{X, x}$ by Lemma 58.21.3. $\square$

Proof. This is a special case of Lemma 58.31.2. First, $D$ is nowhere dense in $X$ (see discussion in Divisors, Section 31.13) and hence $X \setminus D$ is dense in $X$. Second, the ring $\mathcal{O}_{X, x}$ is a regular local ring for all $x \in D$ by Algebra, Lemma 10.106.7 and our assumption that $\mathcal{O}_{D, x}$ is regular. $\square$

Proof. Before we start we note that $\mathcal{O}_{X, x}$ is a regular local ring for all $x \in D$. This follows from Algebra, Lemma 10.106.7 and our assumption that $\mathcal{O}_{D, x}$ is regular. Below we will also use that regular rings are normal, see Algebra, Lemma 10.157.5.

To prove the lemma we may work locally on $X$. Thus we may assume $X = \mathop{\mathrm{Spec}}(A)$ and $D \subset X$ is given by a nonzerodivisor $f \in A$. Then $Y = \mathop{\mathrm{Spec}}(B)$ as a finite étale scheme over $A_ f$. Let $\mathfrak p_1, \ldots , \mathfrak p_ r$ be the minimal primes of $A$ over $f$. Then $A_ i = A_{\mathfrak p_ i}$ is a discrete valuation ring; denote its fraction field $K_ i$. By assumption

is a finite product of fields each tamely ramified with respect to $A_ i$. Choose $e \geq 1$ sufficiently divisible (namely, divisible by all ramification indices for $L_{ij}$ over $A_ i$ as in More on Algebra, Remark 15.111.6). Warning: at this point we do not know that $e$ is invertible on $A$.

Consider the finite free $A$-algebra

Observe that $f' = x$ is a nonzerodivisor in $A'$ and that $A'/f'A' \cong A/fA$ is a regular ring. Set $B' = B \otimes _ A A' = B \otimes _{A_ f} A'_{f'}$. By Abhyankar's lemma (More on Algebra, Lemma 15.114.4) we see that $\mathop{\mathrm{Spec}}(B')$ is unramified over $\mathop{\mathrm{Spec}}(A')$ in codimension $1$. Namely, by Lemma 58.31.1 we see that $\mathop{\mathrm{Spec}}(B')$ is still at least tamely ramified over $\mathop{\mathrm{Spec}}(A')$ in codimension $1$. But Abhyankar's lemma tells us that the ramification indices have all become equal to $1$. By Lemma 58.31.3 we conclude that $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A'_{f'})$ extends to a finite étale morphism $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A')$.

For a point $x \in D$ corresponding to $\mathfrak p \in V(f)$ denote $A^{sh}$ a strict henselization of $A_\mathfrak p = \mathcal{O}_{X, x}$. Observe that $A^{sh}$ and $A^{sh}/fA^{sh} = (A/fA)^{sh}$ (Algebra, Lemma 10.156.4) are regular local rings, see More on Algebra, Lemma 15.45.10. Observe that $A'$ has a unique prime $\mathfrak p'$ lying over $\mathfrak p$ with identical residue field. Thus

is a strictly henselian local ring finite over $A^{sh}$ (Algebra, Lemma 10.156.3). Since $f'$ is a nonzerodivisor in $(A')^{sh}$ and since $(A')^{sh}/f'(A')^{sh} = A^{sh}/fA^{sh}$ is regular, we conclude that $(A')^{sh}$ is a regular local ring (see above). Observe that the induced extension

of fraction fields has degree $e$ (and not less). Since $A' \to C$ is finite étale we see that $A^{sh} \otimes _ A C$ is a finite product of copies of $(A')^{sh}$ (Algebra, Lemma 10.153.6). We have the inclusions

and each of these rings is Noetherian and normal; this follows from Algebra, Lemma 10.163.9 for the ring in the middle. Taking total quotient rings, using the product decomposition of $A^{sh} \otimes _ A C$ and using Fields, Lemma 9.24.3 we conclude that there is an isomorphism

of $Q(A^{sh})$-algebras for some finite set $I$ and integers $e_ i | e$. Since $A^{sh} \otimes _ A B$ is a normal ring, it must be the integral closure of $A^{sh}$ in its total quotient ring. We conclude that we have an isomorphism

over $A^{sh}_ f$ because the algebras $A^{sh}[x]/(x^{e_ i} - f)$ are regular and hence normal. The discriminant of $A^{sh}[x]/(x^{e_ i} - f)$ over $A^{sh}$ is $e_ i^{e_ i}f^{e_ i - 1}$ (up to sign; calculation omitted). Since $A_ f \to B$ is finite étale we see that $e_ i$ must be invertible in $A^{sh}_ f$. On the other hand, since $A_ f \to B$ is tamely ramified over $\mathop{\mathrm{Spec}}(A)$ in codimension $1$, by Lemma 58.31.1 the ring map $A^{sh}_ f \to A^{sh} \otimes _ A B$ is tamely ramified over $\mathop{\mathrm{Spec}}(A^{sh})$ in codimension $1$. This implies $e_ i$ is nonzero in $A^{sh}/fA^{sh}$ (as it must map to an invertible element of the fraction field of this domain by definition of tamely ramified extensions). We conclude that $V(e_ i) \subset \mathop{\mathrm{Spec}}(A^{sh})$ has codimension $\geq 2$ which is absurd unless it is empty. In other words, $e_ i$ is an invertible element of $A^{sh}$. We conclude that the pullback of $Y$ to $\mathop{\mathrm{Spec}}(A^{sh})$ is indeed a finite disjoint union of standard tamely ramified morphisms.

To finish the proof, we write $A^{sh} = \mathop{\mathrm{colim}}\nolimits A_\lambda$ as a filtered colimit of étale $A$-algebras $A_\lambda$. The isomorphism

descends to an isomorphism

for suitably large $\lambda$. After increasing $\lambda$ a bit more we may assume $e_ i$ is invertible in $A_\lambda$. Then $\mathop{\mathrm{Spec}}(A_\lambda ) \to \mathop{\mathrm{Spec}}(A)$ is the desired étale neighbourhood of $x$ and the proof is complete. $\square$

Proof. This is just an addendum to Lemma 58.31.5 and in fact the truth of this lemma follows almost immediately if you've read the proof of that lemma. But we can also deduce the lemma from the result of Lemma 58.31.5. Namely, taking the normalization of $X$ in $Y$ commutes with étale base change, see More on Morphisms, Lemma 37.19.2. Hence we see that we may prove the statements on the local structure of $Y' \to X$ étale locally on $X$. Thus, by Lemma 58.31.5 we may assume that $X = \mathop{\mathrm{Spec}}(A)$ where $A$ is a Noetherian ring, that we have a nonzerodivisor $f\in A$ such that $A/fA$ is regular, and that $Y$ is a finite disjoint union of spectra of rings $A_ f[x]/(x^ e - f)$ where $e$ is invertible in $A$. We omit the verification that the integral closure of $A$ in $A_ f[x]/(x^ e - f)$ is equal to $A' = A[x]/(x^ e - f)$. (To see this argue that the localizations of $A'$ at primes lying over $(f)$ are regular.) We omit the details. $\square$

Proof. By the valuative criterion of properness applied to the finite morphism $Y' \to X$ we see that $\mathop{\mathrm{Spec}}(K)$-valued points of $Y$ matching $t|_{\mathop{\mathrm{Spec}}(K)}$ as maps into $X$ lift uniquely to morphisms $t' : \mathop{\mathrm{Spec}}(R) \to Y'$. Thus statement (1) make sense.

Choose an étale neighbourhood $(U, u) \to (X, t(\mathfrak m_ R))$ such that $U = \mathop{\mathrm{Spec}}(A)$ and such that $Y' \times _ X U \to U$ has a description as in Lemma 58.31.6 for some $f \in A$. Then $\mathop{\mathrm{Spec}}(R) \times _ X U \to \mathop{\mathrm{Spec}}(R)$ is étale and surjective. If $R'$ denotes the local ring of $\mathop{\mathrm{Spec}}(R) \times _ X U$ lying over the closed point of $\mathop{\mathrm{Spec}}(R)$, then $R'$ is a discrete valuation ring and $R \subset R'$ is an unramified extension of discrete valuation rings (More on Algebra, Lemma 15.44.4). The assumption on $t$ signifies that the map $A \to R'$ corresponding to

maps $f$ to a uniformizer $\pi \in R'$. Now suppose that

for some $e_ i \geq 1$. Then we see that

The rings $R'[x]/(x^{e_ i} - f)$ are discrete valuation rings (More on Algebra, Lemma 15.114.2) and hence have no map into the fraction field of $R'$ unless $e_ i = 1$.

Proof of (1). In this case the map $t' : \mathop{\mathrm{Spec}}(R) \to Y'$ base changes to determine a corresponding map $t'' : \mathop{\mathrm{Spec}}(R') \to Y' \times _ X U$ which must map into a summand corresponding to $i \in I$ with $e_ i = 1$ by the discussion above. Thus clearly we see that $Y' \times _ X U \to U$ is étale along the image of $t''$. Since being étale is a property one can check after étale base chamge, this proves (1).

Proof of (2). In this case the assumption implies that $e_ i = 1$ for all $i \in I$. Thus $Y' \times _ X U \to U$ is finite étale and we conclude as before. $\square$

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

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

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

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. $\square$

Proof. If $S$ is Noetherian, then this is the result of Lemma 58.31.8. The general case follows from this by a standard limit argument. We strongly urge the reader to skip the proof.

We can write $S = \mathop{\mathrm{lim}}\nolimits S_ i$ as a directed limit of a system of schemes with affine transition morphisms and with $S_ i$ of finite type over $\mathbf{Z}$, see Limits, Proposition 32.5.4. For each $i$ let $S \to S'_ i \to S_ i$ be the normalization of $S_ i$ in $S$, see Morphisms, Section 29.53. Combining Algebra, Proposition 10.162.16 Morphisms, Lemmas 29.53.15 and 29.53.13 we conclude that $S'_ i$ is of finite type over $\mathbf{Z}$, finite over $S_ i$, and that $S'_ i$ is an integral normal scheme such that $S \to S'_ i$ is dominant. By Morphisms, Lemma 29.53.5 we obtain transition morphisms $S'_{i'} \to S'_ i$ compatible with the transition morphisms $S_{i'} \to S_ i$ and with the morphisms with source $S$. We claim that $S = \mathop{\mathrm{lim}}\nolimits S'_ i$. Proof of claim omitted (hint: look on affine opens over a chosen affine open in $S_ i$ for some $i$ to translate this into a straightforward algebra problem). We conclude that we may write $S = \mathop{\mathrm{lim}}\nolimits S_ i$ as a directed limit of a system of normal integral schemes $S_ i$ with affine transition morphisms and with $S_ i$ of finite type over $\mathbf{Z}$.

For some $i$ we can find a smooth morphism $X_ i \to S_ i$ of finite presentation whose base change to $S$ is $X \to S$. See Limits, Lemmas 32.10.1 and 32.8.9. After increasing $i$ we may assume the section $\sigma$ lifts to a section $\sigma _ i : S_ i \to X_ i$ (by the equivalence of categories in Limits, Lemma 32.10.1). We may replace $X_ i$ by the open subscheme $X_ i^0$ of it studied in More on Morphisms, Section 37.29 since the image of $X \to X_ i$ clearly maps into it (openness by More on Morphisms, Lemma 37.29.6). Thus we may assume the fibres of $X_ i \to S_ i$ are geometrically connected. After increasing $i$ we may assume $|G|$ is invertible on $S_ i$. Let $\eta _ i \in S_ i$ be the generic point. Since $X_\eta$ is the limit of the schemes $X_{i, \eta _ i}$ we can use the exact same arguments to descent $Z \to X_\eta$ to some finite étale Galois cover $Z_ i \to X_{i, \eta _ i}$ after possibly increasing $i$. See Lemma 58.14.1. After possibly increasing $i$ once more we may assume $Z_ i$ has a $\kappa (\eta _ i)$-rational point mapping to $\sigma _ i(\eta _ i)$. Then we apply the lemma in the Noetherian case and we pullback to $X$ to conclude. $\square$