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I have the following problem: suppose $F/K$ is an abelian (Galois) extension of number fields, Galois group G, and $\mathfrak{p}$ is a prime of K, $\mathfrak{P}$ a prime of F dividing $\mathfrak{p}$; then let M be the subfield of F fixed by the decomposition group $G_{\mathfrak{P}} \subset G$. Then I want to show that $\mathfrak{p}$ splits completely in M; all ramification indices $e_i$ and residue class degrees $f_i$ are equal to 1 in M/K.

So, I know that the degree of an extension is the product of e, f and the number of primes (with all $e_i$ are equal, say to e, all $f_i$ equal to f). I know $|G_{\mathfrak{P}}|=ef$ is the product of the ramification index and residue class degree. However, I'm struggling to show that $\mathfrak{p}$ splits completely in M. I feel like I have a lot of options and I'm not sure what's best: I could try to show directly that each $e_i$ and $f_i=1$ , or try to play around with the tower law and the degree $[M:K]$, or try to make use of the properties of the decomposition group or something similar. Nothing I've done so far has got me anywhere however - could anyone help please? Thanks in advance! -Tom

Edit: I was an idiot and accidentally cleared my cookies without thinking, sorry. Have made an account to stop this happening again - could you possibly link this to this account so I can add comments? Apologies!

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@Tom: I've flagged for moderator attention, so your request will soon be dealt with. – t.b. Nov 13 '11 at 3:41
up vote 1 down vote accepted

Here are some ideas.

Let $\mathfrak{P}_M = \mathfrak{P} \cap M$. You can use the multiplicativity of ramification degrees to show that $e(\mathfrak{P}/\mathfrak{P}_M) \leq e$, and similarly that $f(\mathfrak{P}/\mathfrak{P}_M) \leq f$. But $[F : M] = ef$ and $\mathfrak{P}$ is the only prime of $F$ lying above $\mathfrak{P}_M$. You can combine these last two facts to show that the above inequalities have to be equalities.

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(1) If $\mathfrak{P}_1,...,\mathfrak{P}_n$ are the primes of $\mathcal{O}_F$ above $\mathfrak{p}$ and $G_{\mathfrak{p}} = \langle G_{\mathfrak{P}_1},...,G_{\mathfrak{P}_k}\rangle.$ Then $\mathfrak{p}$ splits completely in $F^{G_{\mathfrak{p}}}.$

(2) In an Abelian extension, $F/K,$ the group $G_{\mathfrak{p}} = \langle G_{\beta_1},...,G_{\beta_k}\rangle = G_{\mathfrak{P}}$ for any $\mathfrak{P}|\mathfrak{p}.$

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