Does each finite morphism of curves have a model whose minimal resolution is semi-stable - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T00:41:47Z http://mathoverflow.net/feeds/question/82232 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82232/does-each-finite-morphism-of-curves-have-a-model-whose-minimal-resolution-is-semi Does each finite morphism of curves have a model whose minimal resolution is semi-stable Ariyan Javanpeykar 2011-11-29T23:43:26Z 2011-11-30T23:29:30Z <p>Let $\pi:Y\to X$ be a finite morphism of smooth projective geometrically connected curves over a number field $K$. </p> <p><strong>Question.</strong> 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 $O_L$?</p> <p>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 <em>Stable reduction of finite covers of curves</em>. ) More generally, if $X$ has potentially good reduction over $K$, the same argument works.</p> <p>Unfortunately, I can't seem to make this work for higher genus curves. What am I missing?</p> http://mathoverflow.net/questions/82232/does-each-finite-morphism-of-curves-have-a-model-whose-minimal-resolution-is-semi/82328#82328 Answer by Qing Liu for Does each finite morphism of curves have a model whose minimal resolution is semi-stable Qing Liu 2011-11-30T23:29:30Z 2011-11-30T23:29:30Z <p>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 $X_L$. </p> <p>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 $\mathcal X$. </p> <p>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. </p> <p>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 $\mathcal Y_{\mathcal O_L}$.). </p> <p>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 <a href="http://www.math.u-bordeaux.fr/~liu/articles/modcove.pdf" rel="nofollow">this paper</a>. </p>