Let $\pi:Y\longrightarrow \mathbf{P}^1_{\mathbf{Z}}$ be a finite surjective flat morphism of schemes, where $Y$ is a normal integral flat projective 2-dimensional $\mathbf{Z}$-scheme, with branch locus $D$. Let us suppose that $\pi$ is tamely ramified.

Question 1. Does this mean that for every prime number $p$ such that some vertical component of $D$ maps to $p$ in $\textrm{Spec} \mathbf{Z}$ does not divide the degree of $\pi$?

By Abhyankar's Lemma, there exists a number field $K$ with ring of integers $O_K$ such that the branch locus $D\subset (B_{O_K})_{\textrm{hor}}$ of the morphism $\pi_{O_K}:Y_{O_K}\longrightarrow \mathbf{P}^1_{O_K}$ is horizontal.

By embedded resolution, there exists a projective birational morphism $f:X\longrightarrow \mathbf{P}^1_{O_K}$ such that $f^\ast D$ is a divisor with normal crossings.

Question 2. Since $D$ is horizontal, does it follow that $f^\ast D$ is horizontal. I thought so but wasn't completely sure.

Question 3. Does the (edit: normalization of the) fibre product $Y^\prime$ of $Y_{O_K}$ and $X$ over $\mathbf{P}^1_{O_K}$ give a morphism $Y^\prime\longrightarrow X$ with branch locus a horizontal normal crossings divisor? I would be surprised if the branch locus picked up vertical components...

  • $\begingroup$ Do you mean "combining" in the title? $\endgroup$ Commented Apr 8, 2011 at 14:33
  • $\begingroup$ Yeah. I changed it. Thnx. $\endgroup$ Commented Apr 8, 2011 at 14:47

1 Answer 1


Question 1: it is not exactly the meaning of tameness (ramification index prime to $p$ and separable residue extension). But if $p>\deg \pi$, then $\pi$ is tame at $p$.

Question 2: The strict transform of $D$ is horizontal because it is finite birational to $D$, but the preimage of $D$ by $f$ is not horizontal in general.

Question 3: I don't think so. Consider the following example. Let $x$ be a coordinate of $P^1$ and consider the cover induced by the equation $y^3=x^3+p^2$. Then $D$ is the divisor $V(x^3+p^2)$. To solve the singularity of $D$, you have to blowup $P^1$ along the point $(x,p)$. Then you get a vertical ramification at $p$. (In general you need to normalize the fiber product of $X$ with $Y$).

  • $\begingroup$ Q1. Is it elementary to see that the residue extension is separable if $p>\deg \pi$? Q3. I apologize for my sloppiness. In the question (edited now) I should have put the "normalization of the fibre product". In any case, your example shows that one cannot expect this to hold. $\endgroup$ Commented Jul 21, 2011 at 15:08
  • $\begingroup$ Q1: yes, because the degrees of the residue extensions are less or equal to $\deg\pi$. $\endgroup$
    – Qing Liu
    Commented Jul 21, 2011 at 15:33

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