Galois group of residue field extension Let $E$ and $K$ be number fields, so that $E/K$ is normal, $G:= \textrm{Gal}(E/K)$, $\mathcal{O}_E$ and $\mathcal{O}_K$ their respective rings of integers.
The theory of Artin $L$-series relies on the fact that for any prime ideal $\mathfrak{p} \subset \mathcal{O}_K$, if $\mathfrak{P} \subset \mathcal{O}_E$ is a prime above $\mathfrak{p}$, and $G_{\mathfrak{P}}$ and $I_{\mathfrak{P}}$ are the corresponding decomposition and inertia subgroups, then we have the following isomorphism:
$$ G_{\mathfrak{P}}/I_{\mathfrak{P}} \cong \textrm{Gal}\left( (\mathcal{O}_E/\mathfrak{P})/(\mathcal{O}_K/\mathfrak{p}) \right) $$
But is this Galois group even well-defined for all prime ideals?
In Proposition 9.4, ch. I, §9, of $\textit{Algebraic Number Theory}$, Neukirch proves that $(\mathcal{O}_E/\mathfrak{P})/(\mathcal{O}_K/\mathfrak{p})$ is normal for any prime $\mathfrak{p} \subset \mathcal{O}_K$ and any choice of $\mathfrak{P} \subset \mathcal{O}_E$ above $\mathfrak{p}$. He then goes on to show that there is a surjective homomorphism
$$G_{\mathfrak{P}} \to \textrm{Gal}\left( (\mathcal{O}_E/\mathfrak{P})/(\mathcal{O}_K/\mathfrak{p}) \right)$$
and defines the inertia group to be the kernel of said homomorphism, thereby establishing the aforementioned isomorphism.
Now we know that if $\mathfrak{p} \subset \mathcal{O}_K$ is unramified over $\mathcal{O}_E$, then $(\mathcal{O}_E/\mathfrak{P})/(\mathcal{O}_K/\mathfrak{p})$ must be separable. But if $\mathfrak{p}$ ramifies over $\mathcal{O}_E$, how can we know that the corresponding residue field extension is separable?
Thank you for your attention.
 A: We know that the quotient rings are finite fields, and the field extension you are looking at has finite degree. But any finite field is a perfect field: any algebraic extension of a finite field is separable ,or if you prefer, any algebraic element over a finite field is separable. This is a well-known fact from Galois theory, but here is a quick proof.
We need to prove that any  monic irreducible polynomial $P$ over $\mathbb{F}_q$ is separable, which amounts to show that $P'\neq 0$ (since $P$ is irreducible, the gcd of $P$ and $P'$ is either $1$ or $P$. But in the later case,$P\mid P'$, and $P'=0$ for degree reasons).
Assume that $P'=0$. If $q=p^m, m\geq 1$, then $p$ is the characteristic, and one may see easily that $P=Q(X^p)$ for some $Q$. Now, elevation to the $p$-th power is surjective in $\mathbb{F}_q$, since $x=x^q=(x^{p^{m-1}})^p$. Hence, one may write $Q=X^k+a_{k-1}^p X^{k-1}+\cdots+ a_0^p$. But we then have $P=Q(X^p)=(X^k+a_{k-1}X^{k-1}+\cdots+a_0)^p$ since we are in characteristic $p$, contradicting the irreducibility of $P$.
Hence $P'\neq 0$, and $P$ is separable.
