The Stacks project

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. $\square$