Unramified primes of splitting field I would like to show the following: 
Theorem: Let $K$ be a number field and and $L$ be the splitting field of a polynomial $f$ over $K$. If $f$ is separable modulo a prime $\lambda$ of $K$, then $L$ is unramified above $\lambda$. 
This should follow from the following theorem: 
Theorem: Let $L / K$ be a finite extension of number fields, and $B$ resp. $A$ the ring of integers of $L$ resp. $K$. Let $\mathfrak{p}$ be a prime of $K$ and $p$ the prime number lying under $\mathfrak{p}$. Let $\alpha \in B$. Let $f$ be the minimal polynomial of $\alpha$ over $K$, and let $\overline{f} = \overline{g_1}^{e_1} \cdots \overline{g_r}^{e_r}$ be the distinct irreducible factors of $f$ modulo $\mathfrak{p}$. If $p$ does not divide the order of $B / A[\alpha]$, then $\mathfrak{p}B = \mathfrak{P}_1^{e_1} \cdots \mathfrak{P_r}^{e_r}$. 
How can I do this?
Thanks a lot!
 A: I'll start by reformulating the first theorem:

Theorem: Let $F = K(\alpha)$, where $f$ is the minimal polynomial of $\alpha$ over $K$. Suppose that $\mathfrak p$ is a prime of $K$, and the $f(X)$ splits as a product of distinct irreducibles modulo $\mathfrak p$ (i.e. $f$ is separable mod $\mathfrak p$). Then $\mathfrak p$ is unramified in $F$.

Why is this the same thing? As the splitting field of $f$, $L$ is the compositum of the fields $K(\alpha_i)$ where the $\alpha_i$ are the roots of $f$. Each of these fields is isomorphic, so if $\mathfrak p$ is unramified in one of them, it is unramified in all of them, and hence it is unramfied in $L$.
Let $p$ be the rational prime lying under $\mathfrak p$.

Case 1:  $p\nmid [\mathcal O_F:\mathcal O_K[\alpha]]$.
Here, we are in the case of the second theorem. By assumption, $$\overline f=\overline g_1\cdots\overline g_n\pmod {\mathfrak p}$$
where the $g_i$ are distinct, so $\mathfrak p\mathcal O_F$ splits as a product of distinct primes. Hence it is unramified.

Case 2:  $p\mid [\mathcal O_F:\mathcal O_K[\alpha]]$.
In this case, the theorem does not apply directly as stated. However, the proof of the theorem shows that $\mathfrak p$ splits as a product of distinct primes in $\mathcal O_K[\alpha]$. If $\mathfrak P^2\mid \mathfrak p\mathcal O_F$ for some prime $\mathfrak P$ of $\mathcal O_L$, then, taking $\mathfrak q = \mathfrak P\cap\mathcal O_K[\alpha]$, we see that $\mathfrak q^2\mid \mathfrak p\mathcal O_K[\alpha]$, which does not happen.
