Let $f:X\rightarrow Y$ be a ramified cover of Riemann surfaces or algebraic curves over $\mathbb{C}$. My question is can one in terms of the ramification data of $f$, determine whether the cover is Galois or not. I am Specially interested in the following question: if $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ are both ramified Galois covers of curves, when $g\circ f:X\rightarrow Z$ is (or isn't) a Galois cover?
3 Answers
I assume that the surfaces are connected. Composition of two regular covering maps need not be regular even if the coverings are unbranched: A (usual) covering map is regular iff it is defined by a normal subgroup of the fundamental group. If you have a composition $$ X \stackrel{f}{\to} Y \stackrel{g}{\to} Z $$ of regular covers then $\pi_1(X)\triangleleft \pi_1(Y)$ and $\pi_1(Y)\triangleleft \pi_1(Z)$. Hence, $\pi_1(X)$ is a subnormal subgroup of $\pi_1(Z)$. Subnormal subgroups need not be normal. However, if, say, $\pi_1(X)< \pi_1(Y)$ is a characteristic subgroup then it will be normal in $\pi_1(Z)$. This is the simplest condition I know to ensure that composition of regular covering maps is again regular. The same applies to ramified covers once you remove branch points and their preimages.
The ramification data doesn't generally suffice to determine whether the cover is Galois. If $f:X\to Y$ and $g:Y\to Z$ are ramified Galois covers, then usually $g\circ f$ won't be Galois, for instance because there will usually be a point $z$ of $Z$ which has two preimages under $g\circ f$ having different ramification indices.

$\begingroup$ Thanks. Your remark about $g\circ f$ sounds interesting. Why having two preimages with different ramification data prevents the cover to be Galois? is it s wellknown result that the preimages of points in Galois covers should have the same ramification indices? $\endgroup$ Jul 31, 2013 at 22:47
There are many automorphisms of $Y$ over $Z$. One can pull back the cover $X$ along any of these automorphisms of $Y$, producing a new cover of $Y$. If the composition is Galois, then these new covers are in fact isomorphic to the original cover  this is because the map $Gal(XZ) \to Gal(YZ)$ is surjective. Taking a lift of an element of $Gal(YZ)$ produces an isomorphism between the original cover and the pullback.
The converse is also true. If such an isomorphism exist, they give us enough elements of the automorphism group of $X$ over $Z$ that, combined with the automorphisms of $X$ over $Y$, the cover is Galois!
So this explains why the two preimages with different ramification data will prevent the cover from being Galois. Because then the pullback of the cover along an automorphism which permutes those two preimages will not be isomorphic to the original  it will have different ramification data.

2$\begingroup$ For a finite extension of number fields, the story is similar: ramification indices of all prime ideals over the same prime must be the same. Also residue field degrees of all prime ideals over the same prime must be the same. Moreover, there is a converse result in this case: if, for each prime downstairs, all prime ideals above it have equal ramification indices and equal residue field degrees then the extension is Galois, and the hypothesis on ramification is actually not needed to get the conclusion. Over C residue field degrees are all 1, so no information is stored there. $\endgroup$– KConradAug 1, 2013 at 6:42

$\begingroup$ Thanks. Was a beautiful argument, although I did not understand the converse argument that how the existence of such an isomorphism leads to Galoisness property. possibly by enough elemnts of $X$ over $Z$, you means enough elements of $Y$ over $Z$ that combined with automorphisms of $X$ over $Y$ makes the composite Galois. But even in this case I don't understand why? I should think more on this. $\endgroup$ Aug 1, 2013 at 8:17

1$\begingroup$ Consider the group of automorphisms of the trio $X => Y=> Z$ fixing $Z$. As a subgroup, it has all the automorphisms of $X$ over $Y$. The quotient group maps to the group of automorphisms of $Y$ over $Z$, and if this condition is satisfied, that's surjective. So the order of the group is at least $[X:Y][Y:Z]=[X:Z]$, so the cover is Galois. $\endgroup$ Aug 1, 2013 at 19:31