TITLE: Tameness for the Galois closure of a map of curves
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: 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.