Extending finite morphisms of curves to finite morphisms of arithmetic surfaces - MathOverflow

Extending finite morphisms of curves to finite morphisms of arithmetic surfaces

2011-03-18T14:27:04Z

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 $p$?

Answer by Emerton for Extending finite morphisms of curves to finite morphisms of arithmetic surfaces

2011-03-18T17:03:01Z

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.

Answer by Qing Liu for Extending finite morphisms of curves to finite morphisms of arithmetic surfaces

2011-03-24T13:44:11Z

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.