In a book I am reading (Cox, Primes), one of the exercises asks to prove the following version of Kummer's Theorem regarding factorization of primes in an extension:
Suppose $L/K$ is Galois, $L=K(\alpha)$, $\alpha\in\mathcal{O}_L$, $f(x)$ the monic minimal polynomial of $\alpha$ over $K$ (thus $f(x)\in\mathcal{O}_K[x]$). Suppose $\mathfrak{p}$ is prime in $\mathcal{O}_K$ and that $f(x)$ is separable modulo $\mathfrak{p}$, say $$f(x)\equiv f_1(x)\cdots f_g(x)\mod{\mathfrak{p}}$$ where the $f_i$ are distinct and irreducible. Then all of the $f_i$ have the same degree, $\mathfrak{p}$ is unramified in $L$, and the inertial degree of each factor is the degree of the $f_i$.
(There is actually more detail, but this is all I need for my question.)
Following the order in the exercise, one first shows that if $\mathfrak{P}$ is an ideal over $\mathfrak{p}$, then $f_i(\alpha)\in\mathfrak{P}$ for some $i$ (assume $i=1$). This is pretty straightforward.
The next step is: If $\sigma\in D_{\mathfrak{P}}$, the decomposition group of $\mathfrak{P}$, then $f_1(\sigma(\alpha)) = \sigma(f_1(\alpha))\in\mathfrak{P}$. Separability then implies that $\text{deg}(f_1(x))\ge |D_{\mathfrak{P}}|$.
I don't see how to prove the degree statement. Presumably the idea is to show that the $\sigma(\alpha)$ are incongruent modulo $\mathfrak{P}$, thus providing at least $|D_{\mathfrak{P}}|$ roots of $f_1$ in $\mathcal{O}_L/\mathfrak{P}$. But as the answer below points out, this is in general not true (and is in fact equivalent to what we are eventually trying to prove, that $\mathfrak{p}$ is unramified in $L$).
No two $\sigma(\alpha)$ are equal does NOT imply that no two are congruent modulo $\mathfrak{P}$ (in fact this discrepancy is precisely measured by the inertia group).