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Let $S$ be a finite set of primes and $K/\mathbb{Q}$ a finite Galois extension unramified outside $S$. For each primes $p\notin S$, let $n_p$ be the number of primes of $K$ above $p$ and $f_p$ the degree of the residue field extension. An elementary result in algebraic number theory says,

$n_pf_p=[K:\mathbb{Q}]$.

With the setting, my question is: for any pair of integer $(n,f)$ such that $nf=[K:\mathbb{Q}]$, does there exist $p\notin S$ such that $n_p=n$ and $f_p=f$?

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The answer is no. Consider $ K = \mathbf Q(\sqrt{2}, \sqrt{3}) $, and let $ p $ be any prime unramified in this extension. We will show that $ p $ cannot be inert. Assume the contrary, and let $ G $ be the decomposition group of $ p $, which is necessarily the whole Galois group. There is then a surjection from $ G $ to the Galois group of the residue class field extension. This extension is the quartic extension of $ \mathbb F_p $, so its Galois group over $ \mathbb F_p $ is $ C_4 $. But then, we must have a surjection $ G \cong C_2 \times C_2 \to C_4 $. Such a surjection would be an isomorphism, which is absurd. Therefore, no prime of $ \mathbf Z $ is inert in $ K $, thus the equation $ n_p f_p = [K : \mathbf Q] $ has no solution for $ f_p = [K : \mathbf Q] $ and $ n_p = 1 $. Similar counterexamples are easily constructed using the same idea.

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  • $\begingroup$ I don't follow at "the Galois group of the residue class field extension". You mean for a prime ideal $q$ above $p$ in $O_K$, we look at $L = O_K / q$, and $[L : \mathbb{F}_p] = f_p = 4$, so $L \simeq \mathbb{F}_{p^4}$ whose Galois group is $Aut(\mathbb{F}_{p^4}/\mathbb{F}_{p}) \simeq C_4$, that's it ? $\endgroup$ – reuns Sep 21 '16 at 13:24
  • $\begingroup$ Since $ p $ is inert in $ \mathcal O_K $, the prime ideal above $ p $ is just $ p $ itself. But yes, that is correct: we look at the extension $ (\mathcal O_K/(p)) / \mathbf Z/p \mathbf Z $. $\endgroup$ – Starfall Sep 21 '16 at 13:26

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