Timeline for What prevents a cover to be Galois?
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| Aug 1, 2013 at 19:31 | comment | added | Will Sawin | 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. | |
| Aug 1, 2013 at 8:17 | comment | added | Darius Math | Thanks. Was a beautiful argument, although I did not understand the converse argument that how the existence of such an isomorphism leads to Galois-ness 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. | |
| Aug 1, 2013 at 6:42 | comment | added | KConrad | 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. | |
| Aug 1, 2013 at 2:56 | history | answered | Will Sawin | CC BY-SA 3.0 |