Exercise 5.6(e) of Cox's Primes of the Form $x^2+ny^2$

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I am having a lot of trouble proving part (e) on the exercise 5.6 in David Cox's book "Primes of the Form $x^2+ny^2$". The result, proposition 5.11, is a special case of Dedekind's prime ideal factorization theorem, which according to the book is much easier and simple to prove than the general case. The proof is done through exercise 5.6, but I can't find a way to prove (e) in an easy way without going back to the proof of the general case. I leave the result and the full exercise statement below, credits to the user Rob Smith for the transcription.


The result to be proved:

Let $K\subset L$ be a Galois extension of number fields, where $L=K(\alpha)$ for some $\alpha\in\mathcal{O}_L$. Let $f(x)$ be the monic minimal polynomial of $\alpha$ over $K$, so that $f(x)\in \mathcal{O}_K[x]$. If $\frak{p}$ is prime in $\mathcal{O}_K$ and $f(x)$ is separable modulo $\frak{p}$, then

If $f(x)\equiv f_1(x)\cdots f_g(x)\mod \frak{p}$, where the $f_i(x)$ are distinct and irreducible modulo $\frak{p}$, then ${\frak{P}}_i={\frak{p}}\mathcal{O}_L+f_i(\alpha)\mathcal{O}_L$ is a prime ideal of $\mathcal{O}_L$, ${\frak{P}}_i\neq{\frak{P}}_j$ for $i\neq j$, and $${\frak{p}}\mathcal{O}_L= {\frak{P}}_1\cdots {\frak{P}}_g.$$ Furthermore, all of the $f_i(x)$ have the same degree, which is the inertial degree $f$.


So the exercise in question states: let $\frak{P}$ be a prime of $\mathcal{O}_L$ containing $\frak{p}$, and let $D_{\frak{P}} = \{\sigma \in Gal(L/K) : \sigma(\frak{P}) = \frak{P}\}$ be the decomposition group. Recall that $|D_{\frak{P}}| = ef$, where $e$ and $f$ are the ramification index and inertial degree, respectively, of $\frak{P}$ over $\frak{p}$.

I've managed to prove the following (questions (a) to (d) in the exercise): $f_i({\alpha}) \in \frak{P}$ for some $i$, so we can assume that $f_1({\alpha}) \in \frak{P}$; $f \geq deg(f_1(x))$; $deg(f_1(x)) \geq |D_{\frak{P}}| = ef$ because $f(x)$ is separable mod $\frak{p}$; and $e = 1$ and $f = deg(f_1(x))$. The issue I have is with part (e):

(e) Show that $\mathfrak{p} \mathcal{O}_L = \mathfrak{P}_1 ... \mathfrak{P}_g$ where $\mathfrak{P}_i$ is prime in $\mathcal{O}_L$ and $f_i(\alpha) \in \mathfrak{P}_i$. This shows the $f_i(x)$ all have the same degree.


Thanks in advance!