In ch. I, §9, p. 56 of Algebraic Number Theory, Neukirch gives a proof of the statement (Proposition 9.4) that for a normal extension $L/K$ of algebraic number fields, any prime $\mathfrak{p} \subset \mathcal{O}_K$ and any prime $\mathfrak{P} \subset \mathcal{O}_L$ above $\mathfrak{p}$, the corresponding residue field extension $(\mathcal{O}_L/\mathfrak{P})/(\mathcal{O}_K/\mathfrak{p})$ is normal.
The idea of the proof is the following: You let $\theta \in \mathcal{O}_L$ be a representative of an element $\overline{\theta} \in \mathcal{O}_L/\mathfrak{P}$, and let $f(x) \in K[x]$, $\overline{g}(X) \in (\mathcal{O}_K/\mathfrak{p})[X]$ be the minimal polynomials of $\theta$ and $\overline{\theta}$ respectively over $K$ and $\mathcal{O}_K/\mathfrak{p}$ respectively.
Then comes the first step that puzzles me: He says that $\overline{\theta} = \theta\ (\textrm{mod}\ \mathfrak{P})$ is a root of the polynomial $\overline{f}(x) = f(x)\ (\textrm{mod}\ \mathfrak{p})$, and thence concludes that $\overline{g}(x)$ divides $\overline{f}(x)$ (presumably in $(\mathcal{O}_K/\mathfrak{p})[X]$?).
My question is: What does this mean? If we want to show that $\overline{g}(x)$ divides $\overline{f}(x)$, then $\overline{f}(x)$ should be living in $(\mathcal{O}_K/\mathfrak{p})[X]$, and hence be a function $\overline{f}: \mathcal{O}_K/\mathfrak{p} \to \mathcal{O}_K/\mathfrak{p}$, right? So the expression should be given more explicitly as: $$\overline{f}(x\ (\textrm{mod}\ \mathfrak{p})) = f(x)\ (\textrm{mod}\ \mathfrak{p}).$$ But this $\overline{f}$ is evidently only well-defined if $f$ fixes $\mathfrak{p}$, which is not necessarily the case.
How is this to be understood?
All input would be highly appreciated.