Extending finite morphisms of curves to finite morphisms of arithmetic surfaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:37:57Z http://mathoverflow.net/feeds/question/58843 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58843/extending-finite-morphisms-of-curves-to-finite-morphisms-of-arithmetic-surfaces Extending finite morphisms of curves to finite morphisms of arithmetic surfaces Ariyan Javanpeykar 2011-03-18T14:27:04Z 2011-03-24T13:44:11Z <p>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 $K$. </p> <p>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 $Y$.</p> <p><strong>Question 1.</strong> Does the normalization of $\mathcal{X}$ in the function field of $Y$ provide me with such an extension?</p> <p>I would like to know the control we have on the branch locus in this case.</p> <p><strong>Question 2.</strong> 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 $Y$.)</p> <p>That is, we know the horizontal components of $D$. But can we say something about the vertical components? </p> <p><strong>Example.</strong> Take a finite etale cover of the projective line over $K$. </p> <p><strong>Example.</strong> Take a Belyi morphism $\pi:Y\longrightarrow \mathbf{P}^1_{\mathbf{Q}}$, i.e., the branch locus of $\pi$ is contained in <code>$\{0,1,\infty\}$</code>. 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 $p$?</p> http://mathoverflow.net/questions/58843/extending-finite-morphisms-of-curves-to-finite-morphisms-of-arithmetic-surfaces/58857#58857 Answer by Emerton for Extending finite morphisms of curves to finite morphisms of arithmetic surfaces Emerton 2011-03-18T17:03:01Z 2011-03-18T17:03:01Z <p>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.</p> http://mathoverflow.net/questions/58843/extending-finite-morphisms-of-curves-to-finite-morphisms-of-arithmetic-surfaces/59431#59431 Answer by Qing Liu for Extending finite morphisms of curves to finite morphisms of arithmetic surfaces Qing Liu 2011-03-24T13:44:11Z 2011-03-24T13:44:11Z <p>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. </p> <p>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.</p>