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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.

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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.

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  • $\begingroup$ That's certainly true, but the question was unclearly enough phrased that I thought that he was asking the converse: if the Galois closure of $L\supset K$ is tame over $L$, is then $L$ tame over $K$? I'll wait for OP's comment. $\endgroup$
    – Lubin
    Commented Apr 6, 2011 at 20:12
  • $\begingroup$ I was asking what Holger has posted. The converse seems clear to me. Just a silly error though--you meant iv) rather than v), but yes this is exactly the sort of result I was hoping for. $\endgroup$
    – Randall
    Commented Apr 6, 2011 at 20:26
  • $\begingroup$ No, the converse, as I stated it, is not true. $\endgroup$
    – Lubin
    Commented Apr 6, 2011 at 22:04
  • $\begingroup$ the converse is true, since if Q lies over S and S over P, where Q is place of a Galois closure of L/K S is place of L and P of K then e(Q|P)=e(Q|S)e(S|P) and hence if p does not divide e(Q|P) then p does not divide e(S|P) (p=char(K). $\endgroup$
    – Alexey Zaytsev
    Commented Apr 6, 2011 at 22:16
  • $\begingroup$ Sorry, it seems that we understand the converse in different sense. if the property being tame only over L, then you are right. $\endgroup$
    – Alexey Zaytsev
    Commented Apr 6, 2011 at 22:21
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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.

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