Tameness for the Galois closure of a map of curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:45:47Z http://mathoverflow.net/feeds/question/60743 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60743/tameness-for-the-galois-closure-of-a-map-of-curves Tameness for the Galois closure of a map of curves Randy Reddick 2011-04-05T21:05:05Z 2011-04-07T17:41:22Z <p>Say we are working over some $K=\overline{K}$, of characteristic $p>0$. Let $\phi: Y\rightarrow X$ be a nonconstant map of smooth projective curves. To this map we can associate a map $\psi: Z\rightarrow X$, where on the level of fields this is the Galois closure of $k(X)\subseteq k(Y)$. I would like to know about the tameness of this map.</p> <p>Let $e_P$ denote the ramification indices (with the maps understood to be either $\psi$ or $\phi$ depending on where $P$ lives). Now obviously if $p|e_P$ and if $Q$ lies above $P$, $p|e_Q$ as well, so $\psi$ has wild ramification at $Q$. I am wondering when we can ensure this map is (everywhere) tamely ramified. For instance if $d=deg(\phi) &lt; p$, then the degree of the Galois closure of $k(Y)$ over $k(X)$ has degree dividing $d!$, and hence $\psi$ remains tame.</p> <p>My question is this: Suppose we can show for each $P\in Y$ such that $e_P \geq p$ that each point above $P$ is tamely ramified. Can we conclude that $\psi$ is (everywhere) tamely ramified? It seems to me that this isn't true but I cannot produce a counterexample. It would be fortuitous if it were true, however. Any help is greatly appreciated.</p> http://mathoverflow.net/questions/60743/tameness-for-the-galois-closure-of-a-map-of-curves/60807#60807 Answer by Holger Partsch for Tameness for the Galois closure of a map of curves Holger Partsch 2011-04-06T11:59:01Z 2011-04-07T17:41:22Z <p>Look at Lemma 2.1.3 i.v) from Grothendieck and Murre: "The Tame Fundamental Group of a Formal Neighbourhood of a divsors with Normal Crossings on a Scheme".</p> <p>It says when given a tame field extension $L \supset K$, then its Galois closure will again be tame.</p> <p>Here, tameness is just defined with respect to one valution of $K$. But the proof should apply in your situation as well.</p> http://mathoverflow.net/questions/60743/tameness-for-the-galois-closure-of-a-map-of-curves/60853#60853 Answer by Alexey Zaytsev for Tameness for the Galois closure of a map of curves Alexey Zaytsev 2011-04-06T21:13:08Z 2011-04-06T21:23:38Z <p>For the first glance it should follow form the Abhaynkar's lemma (see "Algebraic Function Fields" by Stichtenoth, Theorem 3.9.1) and the fact the Galois closure is the composite of all the different embeddings of L over K into fixed algebraic closure of K (so each of them has the same properties of tame ramifications). Then we just apply the lemma and get the result that $p=char(K)$ does not divide $e_P$ for any place P in K. </p>