Let $\pi:Y\longrightarrow X$ be a finite morphism of smooth projective geometrically connected curves over a number field $K$. Let $h:\mathcal{X}\longrightarrow \textrm{Spec} \ O_K$ be a regular integral flat projective $O_K$-scheme with generic fibre $X$. Here $O_K$ is the ring of integers of $K$.

I would like to extend $\pi$ to a finite morphism $p:\mathcal{Y} \longrightarrow \mathcal{X}$, where $\mathcal{Y}$ is a normal integral flat projective $O_K$-scheme with generic fibre $Y$.

Question 1. Does the normalization of $\mathcal{X}$ in the function field of $Y$ provide me with such an extension?

I would like to know the control we have on the branch locus in this case.

Question 2. Let $S$ in $X$ be the branch locus of $\pi$. Can I describe the branch locus $D$ of $p:\mathcal{Y}\longrightarrow \mathcal{X}$ in terms of the data $h:\mathcal{X}\longrightarrow \textrm{Spec} \ O_K$ and $S$? (Here $p$ is the normalization of $\mathcal{X}$ in the function field of $Y$.)

That is, we know the horizontal components of $D$. But can we say something about the vertical components?

Example. Take a finite etale cover of the projective line over $K$.

Example. Take a Belyi morphism $\pi:Y\longrightarrow \mathbf{P}^1_{\mathbf{Q}}$, i.e., the branch locus of $\pi$ is contained in $\{0,1,\infty\}$. Take the normalization $p:\mathcal{Y}\longrightarrow \mathbf{P}^1_{\mathbf{Z}}$ of $\mathbf{P}^1_{\mathbf{Z}}$ in the function field of $Y$, where $Y$ is a smooth projective connected curve over $\mathbf{Q}$. Now, what can we say about the vertical components of the branch locus of $p$?

  • $\begingroup$ In your first example, you really want to consider an étale cover $P^1$ ? $\endgroup$ – Qing Liu Mar 24 '11 at 13:28
  • $\begingroup$ Yes. I admit it being a silly example. In fact, the vertical components of the normalization of P^1_{O_K} in P^1_L is just $P^1_{O_L}\longrightarrow P^1_{O_K}$ for which the branch locus depends only on the extension $O_K \subset O_L$. Either I'm wrong or this is indeed very silly. $\endgroup$ – Ariyan Javanpeykar Mar 24 '11 at 19:34

The vertical ramification can't be seen from the branch locus $D$. For example, consider $$ Y : y^2 = f(x), \quad f(x)\in O_K[x]$$ (say with $f(x)$ monic and separable in all residue fields of $O_K$) and $$ Y' : y^2 = tf(x), \quad t\in O_K.$$ Then $Y\to \mathbb P^1_K$ extends in an obvious way to $\mathcal Y\to \mathbb P^1_S$ and there is no vertical ramification. On the other hand, $\mathcal Y'\to \mathbb P^1_S$ is ramified at the places where $t$ is not a square up to unit. Both covers have the same branch locus.

To be more positive, you can sometimes kill the vertical ramification after finite extension of $K$. This is the case when you have a tamely ramified cover $Y\to X$ (Abhyankar's lemma, see SGA 1). This have nices applications. For example, suppose $\mathcal X$ is smooth and $Y\to X$ is a Galois cover of degree invertible in $S$. Suppose further that the horizontal branch locus $\overline{D}$ is étale over $S$, then after finite extension $S'/S$ (killing the vertical ramification), $\mathcal Y$ becomes smooth (so the original $Y$ has potentially good reduction). This is a result of Grothendieck on the specialization of tame fundamental groups. If I remember well it can also be fund in Fulton's paper "Hurwitz Schemes and Irreducibility of Moduli of Algebraic Curves" with a direct proof.


Yes, the normalization $\mathcal Y$ of $\mathcal X$ in $K(Y)$ will give you such an extension. Since $\mathcal X$ is regular and $\mathcal Y$ is normal, purity of the branch locus shows that the branch locus is a divisor on $\mathcal Y$ (as you probably know). However, my feeling is that it is difficult to say much more than this in general.

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    $\begingroup$ Thank you for your answer. I guess it seemed like too much to hope for. $\endgroup$ – Ariyan Javanpeykar Mar 18 '11 at 17:12
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    $\begingroup$ Just to be precise, the flatness of $\mathcal Y\to\mathcal X$ crucially uses the fact that $\mathcal X$ is $2$-dimensional. $\endgroup$ – Torsten Ekedahl Mar 18 '11 at 18:03
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    $\begingroup$ The condition on the dimension of $\mathcal X$ is crucial to have the flatness (hence purity) of $\mathcal Y\to \mathcal X$, but in higher dimension, the purity still holds (Zariski-Nagata-Zariski purity, SGA1, exp. X. 3.1). $\endgroup$ – Qing Liu Mar 24 '11 at 13:27

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