I am confused by an application of Abhyankhar's lemma in the proof of Theorem 3.4 of Deligne-Rapoport.


Here is the question with only the relevant parts of the text:


Let $X$ and $Y$ be two curves over $\mathbb{Z}[1/n]$ (ie relative dimension 1). Let $U\subseteq X$ be an open set and let $C$ be its complement. Assume $X$ and $Y$ are normal and that $X$ and $C$ are both smooth over $\mathbb{Z}[1/n]$. Finally, there is a map $f:Y\to X$ identifying $Y$ as the normalization of $X$ in the function field of $Y$. Let $U'$ be the open $f^{-1}(U)\subseteq Y$ and let $C'$ be its complement. Assume lastly that $f|\_{U'}:U'\to U$ is finite etale. How can I conclude that the maps $Y\to \operatorname{Spec}\mathbb{Z}[1/n]$ and $C'\to \operatorname{Spec}\mathbb{Z}[1/n]$ are smooth by an application of Abhyrankhar's lemma? 


Let $P$ be a component of $C$. Let us replace $X$ by a Zariski open subset $X$ so that $P$ is cut out by one equation, $f$. Let $n$ be the degree of ramification of $P$ in $Y$, and let $X'=X[T]/(T^n-f)$. Abhyankhar's lemma suggests that $U'\to U$ when pulled back to $X'$ extends to an etale cover even over the locus $\{f=0\}$. Thus we get a curve $Z$ which fits into the following commutative diagram:
$$\require{AMScd} \begin{CD} Z @>>> Y\\ @VVV @VVV \\ X' @>>> X\end{CD}$$
where we know the left most arrow is etale.


Now as $X'$ can be computed to be smooth over $\mathbb{Z}[1/n]$, to conclude that $Y$ is smooth it suffices to show the top map is etale. If $Y$ were known to be regular, then Zariski-Nagata purity would facilitate this check, however I cannot seem to figure out why $Y$ should be regular (which it definitely is if it is to be smooth over $\mathbb{Z}[1/n]$!)


Thanks for any help.



Application of Abhyankhar's lemma
Let's work locally near a point $x \in P$. Consider the composed cover $Z \to X' \to X$. Use the fact that etale-locally, each cover splits into irreducible components, where each irreducible component contains at most one point of the fiber over $x$, and those that contain one point are finite. (See [here](https://stacks.math.columbia.edu/tag/04GE), (13) to reduce to the finite case and then (9)). So we may assume that whichever irreducible component of $Z$ contains a point over $x$ contains only one such point, and is finite, which because it is etale over $X'$, implies that it is a finite etale cover of $X'$ of degree $1$ and thus is $X'$. So locally on $X$ we get a map $X' \to Y$. Because $n$ is the ramification index of $Y$, over $U$ this is a surjective map of finite etale covers of the same degree, hence an isomorphism, and because $X'$ is normal it is an isomorphism everywhere. Thus because $X'$ is smooth, $Y$ is smooth.


The point of this is that we can also view Abyankhar's lemma as an etale-local classification result for etale covers with tame ramification around a smooth normal crossings divisor. This is how I generally think of it and it might be how Deligne and/or Rapoport think of it as well.


