# Residue field degrees for number fields

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 $f_p=f$?

## 1 Answer

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.

• 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 ? – reuns Sep 21 '16 at 13:24
• 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 $(\mathcal O_K/(p)) / \mathbf Z/p \mathbf Z$. – Starfall Sep 21 '16 at 13:26