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.


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 $d=deg(\phi) < p$, then the degree of the Galois closure of $k(Y)$ over $k(X)$ has degree dividing $d!$, and hence $\psi$ remains tame.


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.



Tameness for the Galois closure of a map of curves
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".


It says when given a tame field extension $L \supset K$, then its Galois closure will again be tame.


Here, tameness is just defined with respect to one valution of $K$. But the proof should apply in your situation as well.


