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

[simba31415]

Homework Statement

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

Homework Equations

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

All I need now is to show that e_1' | e_1, then e_1' e_2 = e | e_1 e_2 (or alternatively to show e_2' | e_2). However, I can't seem to manage this: indeed quite the opposite, I keep deducing e_1 | e_1' and e_2 | e_2' by using the 2 forms of e 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

 

[morphism]

I would try to show that e\leq e_1e_2. If L_1/K is Galois, what kind of inequality does that impose on e_1?

 

[simba31415]

I would try to show that e\leq e_1e_2. If L_1/K is Galois, what kind of inequality does that impose on e_1?

Hi morphism, sorry it took me so long to reply - I didn't see anyone had responded to me. I've been having a think, but I can't think of any obvious inequalities which we can deduce for e_1: presumably we want e_1 \geq something for this. I know all the ramification indices for any prime splitting downstairs must be the same in this case, but I don't think that helps us here.

So it would suffice to show that e_1 \geq e_1', but while I could believe it I don't know if I could prove it. Do you think you could give me another hint perhaps? Sorry, it's probably something very obvious I'm overlooking!