The Stacks project

Proof. For every $x \in X$ we are going to find an étale neighourhood $(U, u) \to (X, x)$ with $U = \mathop{\mathrm{Spec}}(A)$ such that the base change $Y \times _ X U \to U$ is a finite disjoint union of standard tamely ramified morphisms as in Example 58.31.4. We will assume $x \in D$; the case $x \not\in D$ follows from Étale Morphisms, Lemma 41.18.3 and taking $e = 1$ and $f = 1$ in Example 58.31.4.

In this paragraph we reduce to the case where $X$ is the spectrum of a strictly henselian local ring and $x$ is the closed point. Namely, shrinking $X$ we may assume $X = \mathop{\mathrm{Spec}}(A)$ and $D \subset X$ is given by a nonzerodivisor $f \in A$. Then $Y$ is affine as a finite étale covering of $\mathop{\mathrm{Spec}}(A_ f)$. Write $Y = \mathop{\mathrm{Spec}}(B)$. Let $A^{sh}$ be the strict henselization of $\mathcal{O}_{X, x}$. Since $A \to A^{sh}$ is flat and $x \in D$, we see that $f$ maps to a nonzerodivisor of the maximal ideal of $A^{sh}$. Observe that $A^{sh}/fA^{sh} = (A/fA)^{sh}$ (Algebra, Lemma 10.156.4) is a regular local ring as the strict henselization of $\mathcal{O}_{D, x}$, see More on Algebra, Lemma 15.45.10. By Lemma 58.31.1 the base change of $Y$ to $\mathop{\mathrm{Spec}}(A^{sh})$ is tamely ramified in codimension $1$. Assume we've proven the assertion for the base change of $Y$ to $\mathop{\mathrm{Spec}}(A^{sh})$. Since $A^{sh}$ is strictly henselian any étale neighbourhood has a section and we conclude that we have an isomorphism

where $I$ is a finite set and each $e_ i$ is an integer invertible in $A^{sh}$. Write $A^{sh} = \mathop{\mathrm{colim}}\nolimits A_\lambda $ as a filtered colimit of étale $A$-algebras $A_\lambda $. The displayed isomorphism descends to an isomorphism

for suitably large $\lambda $, see for example Algebra, Lemma 10.127.8. 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 $x$.

Assume $X = \mathop{\mathrm{Spec}}(A)$ where $A$ is a strictly henselian local ring and $x \in X$ corresponds to the maximal ideal $\mathfrak m$ of $A$. Let $f \in \mathfrak m$ be the nonzerodivisor cutting out $D$. Then $A/fA$ is regular and by Algebra, Lemma 10.106.7 we find that $A$ is regular too. We will use some properties of regular local rings, for example that they are normal domains, see Algebra, Lemmas 10.106.2 and 10.157.5. In particular $A$ is a domain. As above we see that $Y = \mathop{\mathrm{Spec}}(B)$ is affine and $A_ f \to B$ is finite étale. Hence $B$ is normal too (Algebra, Lemma 10.163.9) and we conclude $Y$ is a finite disjoint union of spectra of normal domains by Algebra, Lemma 10.37.16. Thus we may assume $B$ is a domain too. Since $A$ and $A/fA$ are domains, we conclude that $f$ is a prime element of $A$ generating a height $1$ prime $\mathfrak p = fA$. Observe that $A_\mathfrak p$ is a discrete valuation ring with uniformizer $f$.

Let $K$ be the fraction field of $A$ and let $L$ be the fraction field of $B$. The tame ramification assumption means that $L$ is tamely ramified with respect to $A_\mathfrak p$. Let $e \geq 1$ be the product of the ramification indices for $L$ over $A_\mathfrak p$ as in More on Algebra, Remark 15.112.6. Then $e$ is invertible in $\kappa (\mathfrak p)$ but at this point we do not know that $e$ is invertible in $A/fA$ or in $A$.

Consider the finite free $A$-algebra

This is a local finite extension of a strictly henselian local ring and hence strictly henselian. Observe that $f' = x$ is a nonzerodivisor in $A'$ and that $A'/f'A' \cong A/fA$ is a regular ring. So as before $A'$ is regular and a fortiori a domain. Set $B' = B \otimes _ A A' = B \otimes _{A_ f} A'_{f'}$. By Abhyankar's lemma (More on Algebra, Lemma 15.115.4) 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 $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A')$. However, since $A'$ is strictly henselian, we conclude that $\mathop{\mathrm{Spec}}(C)$ is a finite disjoint union of copies of $\mathop{\mathrm{Spec}}(A')$. Conclusion: there exists at least one morphism $\mathop{\mathrm{Spec}}(A'_{f'}) \to \mathop{\mathrm{Spec}}(B')$. We conclude that there exists an inclusion $B \subset A'_{f'}$ of $A_ f$-algebras.

It follows that we have inclusions $K \subset L \subset K[f^{1/e}]$. By More on Algebra, Lemma 15.115.2 we conclude that $L$ is equal to $K[f^{1/n}]$ for some divisor $n$ of $e$. The normality of $B$ then implies that $B \cong A_ f[y]/(y^ n - f)$; details omitted. However, the ramification locus of the ring map $A_ f \to A_ f[y]/(y^ n - f)$ is cut out by $ny^{n - 1}$ which must imply that $n$ is a unit in $A_ f$. Since $f$ is a prime element, this means that $n = u f^ s$ for some unit $u$ of $A$ and $s \in \mathbf{Z}$. Since $n$ maps to a unit in $\kappa (\mathfrak p)$ we find $s = 0$ and the proof is complete. $\square$