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.
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.