# Does each finite morphism of curves have a model whose minimal resolution is semi-stable

Let $\pi:Y\to X$ be a finite morphism of smooth projective geometrically connected curves over a number field $K$.

Question. Does there exist a finite field extension $L/K$ and a regular model $\mathcal{X}/O_L$ for $X_L/L$ such that the minimal resolution of singularities of the normalization of $\mathcal{X}$ in $Y_L$ is semi-stable over $O_L$?

The answer to this question is yes if $X=\mathbf{P}^1_K$. (Take $\mathcal{X} = \mathbf{P}^1_{O_L}$, where $L/K$ can be chosen using Corollary 2.8 in Liu's Stable reduction of finite covers of curves. ) More generally, if $X$ has potentially good reduction over $K$, the same argument works.

Unfortunately, I can't seem to make this work for higher genus curves. What am I missing?

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Why not just compose with a morphism $g:X\to \mathbb{P}^1_K$? By what you say, there is a field extension $L/K$ such that the minimal resolution of a proper model of $X_L$ is semistable, and there also exists such a field extension $F/K$ for $Y_F$. It seems that on the compositum, the base changes of both these models are semistable. –  Jason Starr Nov 30 '11 at 1:07
How does the morphism $\pi$ (after base changing to the compositum) extend to a morphism of the above kind? In fact, in general what you get is an extension $\mathcal{Y} \longrightarrow \mathcal{X}$ of $\pi_L$ (for some suitable $L/K$) with $\mathcal{X}$ normal, $\mathcal{Y}$ normal and the min resolution of $\mathcal{Y}$ semi-stable over the base. What I'm looking for is a way to get $\mathcal{X}$ to be regular. Let me explain a bit more carefully that your strategy leads to the statement above. In fact, let $M$ be the compositum of $L$ and $F$ in your comment, and..................... –  Ariyan Javanpeykar Nov 30 '11 at 1:34
... normalize $\mathbf{P}^1_{O_M}$ in $Y$. This gives a sequence of finite morphisms $\mathcal{Y}\to \mathcal{X} \to \mathbf{P}^1_{O_M}$ with $\mathcal{Y}$ and $\mathcal{X}$ normal, the min resolution of $\mathcal{Y}$ semi-stable and the min res. of $\mathcal{X}$ semi-stable. We would be done if we could factor $\mathcal{Y}\to \mathcal{X}$ through the min res of $\mathcal{X}$, but this should not be possible in general. –  Ariyan Javanpeykar Nov 30 '11 at 1:34

If $X$ has no (potentially) good reduction, then the answer to your question is no. More precisely, there always exists a finite cover $Y\to X$ such that for any finite extension $L/K$, no regular semi-stable model of $Y_L$ dominates a regular semi-stable model of $X_L$.
Suppose we are given a finite morphism of semi-stable models $\mathcal Y\to\mathcal X$ with $\mathcal Y$ regular but $\mathcal X$ is singular. I claim that in this case, no semi-stable regular model of $Y$ can dominate a semi-stable regular model of $X$ dominating $\mathcal X$.
For simplicity, I will work over a strictly henselian DVR. Let $x_0$ be a singular point of $\mathcal X$ and let $y_0$ be a point of $\mathcal Y$ lying over $x_0$. Then $x_0$ is a double point in its fiber $\mathcal X_s$ and similarly for $y_0$. Let $\mathcal Y'$ be a regular semi-stable model dominating $\mathcal Y$ and dominating a desingularization of $\mathcal X$. As $\mathcal X'\to \mathcal X$ is not an isomorphism above $x_0$, $\mathcal Y'\to\mathcal Y$ is not an isomorphism above $y_0$. So some irreducible component $\Gamma$ of $\mathcal Y'_s$ must be contracted to $y_0$ in $\mathcal Y$. A smooth point $\mathcal Y'$ contained in $\Gamma$ lifts to a section of $\mathcal Y'$. This section is mapped to a section of $\mathcal Y$ passing through $y_0$. But as $\mathcal Y$ is regular, its sections are contained in its smooth locus. Contradiction because $y_0$ is not a smooth point.
With some extra work, one can show that the same property holds over any finite extension $L/K$ (one has to desingularize first the singular points of $\mathcal Y_{\mathcal O_L}$.).
Finally, for any $X$ with potentially bad reduction at a prime $\mathfrak p$ and for any integer $n\ge 2$ prime to $\mathfrak p$, there exists (after enlarging $K$) a cyclic étale cover $Y\to X$ of degree $n$ with the required property on the double points. See §6.3 and especially Prop. 6.6 in this paper.