The following is a reproduction of Exercise 2 in ch. I. §9 of the English translation of J. Neukirch's $\textit{Algebraic Number Theory}$. There is only one edition.
$\textbf{Exercise 1.}$ If $L\mid K$ is a Galois extension of algebraic number fields, and $\mathfrak{P}$ a prime ideal which is unramified over $K$ (i.e., $\mathfrak{p} = \mathfrak{P} \cap K$ is unramified in $L$), then there is one and only one automorphism $\varphi_{\mathfrak{P}} \in G(L \mid K)$ such that
$$ \varphi_{\mathfrak{P}} a \equiv a^q\ (\textrm{mod}\ \mathfrak{P}) \quad \textrm{for all}\ a \in \mathcal{O},$$
where $q = [\kappa(\mathfrak{P}):\kappa(\mathfrak{p})]$. It is called the $\textbf{Frobenius automorphism}$. The decomposition group $G_{\mathfrak{P}}$ is cyclic and $\varphi_{\mathfrak{P}}$ is a generator of $G_{\mathfrak{P}}$.
I should explain Neukirch's somewhat idiosyncratic notation: $\mathcal{O}$ and $\mathcal{o}$ are the rings of integers of the number fields $L$ and $K$ respectively, $\kappa(\mathfrak{P}) := \mathcal{O}/\mathfrak{P}$ and $\kappa(\mathfrak{p}):=\mathcal{o}/\mathfrak{p}$, where $\mathfrak{P} \subset \mathcal{O}$ is a prime ideal lying above the prime ideal $\mathfrak{p} \subset \mathcal{o}$.
My question is the following: Generally, the Frobenius automorphism is defined in terms of the order of the base field, as opposed to the degree of the extension, so should it not be: $$q := \lvert \kappa(\mathfrak{p}) \rvert$$ as opposed to $$ q = [\kappa(\mathfrak{P}):\kappa(\mathfrak{p})]\ ?$$
This question is a follow-up to this post.