(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let [itex]L_1/K,\,L_2/K[/itex] be extensions of p-adic fields, at least one of which is Galois, with ramification indices [itex]e_1,\,e_2 [/itex]. Suppose that [itex](e_1,\,e_2) = 1 [/itex]. Show that [itex]L_1 L_2/K [/itex] has ramification index [itex]e_1 e_2[/itex].

2. Relevant equations

I have most of the proof done: I'm trying to show [itex]e_1 e_2 | e [/itex] where [itex] e[/itex] is the ramification index of [itex]L_1 L_2/K [/itex], and then show [itex]e | e_1 e_2 [/itex]. It's easy enough to show that [itex]e_1 e_2 | e [/itex]: if we define [itex]e_1 ' [/itex] to be the ramification index between [itex]L_2, \, L_1 L_2 [/itex] and likewise [itex]e_2 ' [/itex] between [itex]L_1, \, L_1 L_2 [/itex], then [itex]e = e_1 e_2' = e_2 e_1' [/itex] by multiplicative property of ramification indices. So both [itex]e_1,\,e_2 [/itex] divide [itex]e [/itex] and are coprime so [itex]e_1 e_2 | e [/itex].

All I need now is to show that [itex]e_1' | e_1 [/itex], then [itex]e_1' e_2 = e | e_1 e_2[/itex] (or alternatively to show [itex]e_2' | e_2 [/itex]). However, I can't seem to manage this: indeed quite the opposite, I keep deducing [itex]e_1 | e_1' [/itex] and [itex]e_2 | e_2' [/itex] by using the 2 forms of [itex]e [/itex] and coprimality. I also, as far as I know, haven't used the fact one of the extensions is Galois (or that they are of p-adic fields) yet, unless I have forgotten a precondition for one of the results I've used. Could anyone help complete my proof? Thanks! -S

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# Homework Help: Quick help finishing off a proof: extension of p-adic fields

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