The goal of this exercise is to prove a special case of Dedekind's prime ideal factorization theorem (details below).
I've managed to work through parts (a) to (e), but I can't find a nice way to prove (f) - the only arguments I've come up with are quite long, and don't use the hint or any previous part of the exercise. My suspicion is that I'm missing something very obvious that allows a nice solution. I'd be very grateful if someone could either confirm my suspicion and point me in the right direction, or deny it so I can at least get some closure!
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 $f$.
The exercise: (Edited to shorten)
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 $\frak{p}$.
(a) Show that $f_i({\alpha}) \in \frak{P}$ for some $i$. We can assume that $f_1({\alpha}) \in \frak{P}$.
(b) Prove that $f \geq deg(f_1(x))$.
(c) Use that $f(x)$ is separable mod $\frak{p}$ to show that $deg(f_1(x)) \geq |D_{\frak{P}}| = ef$.
(d) Conclude that $e = 1$ and $f = deg(f_1(x))$. Thus $\frak{p}$ is unramified in $L$.
(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.
(f) Show that $\mathfrak{P}_i$ is generated by $\mathfrak{p}$ and $f_i(\alpha)$. Hint: let $I_i = \mathfrak{p} \mathcal{O}_L + f_i(\alpha) \mathcal{O}_L$. Use $I_i \subseteq \mathfrak{P}_i$ and $I_1 ... I_g \subseteq \mathfrak{p} \mathcal{O}_L$ to show $I_i = \mathfrak{P}_i$.
Many thanks,
Rob
As the hint suggests, clearly the ideal $I_i\subset\mathcal{O}_L$ generated by $\mathfrak{p}$ and $f_i(\alpha)$ is contained in $\mathfrak{P}_i$, and hence $$I_1\cdots I_g\subset\mathfrak{P}_1\cdots\mathfrak{P}_g=\mathfrak{p}\mathcal{O}_L.$$ The ideal $I_i$ is prime because $f_i$ is irreducible over $\mathcal{O}_K/\mathfrak{p}$, so it follows that $I_i=\mathfrak{P}_j$ for some $j$. Because $I_i\subset\mathfrak{P}_i$ we must have $i=j$.