This is about Exercise 7 from here.
Let $K$ be a number field of degree $n$ with Galois group $S_n$ whose discriminant$D$ is squarefree. Prove that the Galois closure of $K$ is unramified over all finite places of $\Bbb Q(\sqrt D)$.
Let $L$ be the Galois closure. We have to show that for all rational primes $p\mid D$, the ramification index of $p$ in $L$ is $2$. There is a hint to the exercise suggesting to first prove that there is exactly one ramified prime above $p$ in $K$ and then deduce that the inertia group of $p$ in $L$ has order $2$.
The first part is not difficult, it follows from the usual bounds on the valuation of the different, so we have $(p)={\frak p_1^2 p_2\cdots p}_r$ for distinct primes ${\frak p_1,\dots,p}_r$ of $K$.
Here is my argument for the second part. Let $K=\Bbb Q(\alpha)$ for some algebraic integer $\alpha$ and let $f$ be its minimal polynomial. Let $\frak P$ be a prime of $L$ lying above $p$. Consider the completion $L_{\frak P}$. It is the splitting field of $f$ over $\Bbb Q_p$. Note also that over $\Bbb Q_p$ $f$ factors as $f=f_1\cdots f_r$ corresponding to the primes ${\frak p_1,\dots\frak p}_r$. For $2\leq i\leq r$, ${\frak p}_i$ is unramified over $p$, so $f_i$ generates an unramified extension of $\Bbb Q_p$. Also $f_1$ must generate a degree $2$ totally ramified extension. Now $L_{\frak P}$ is the composite of all the splitting fields of the $f_i$ and from this it is clear that $e_{L_{\frak P}/\Bbb Q_p}=2$.
Question: Is this correct? I didn't use the hypothesis that $\operatorname{ Gal}(L/\Bbb Q)=S_n$. Also I am wondering if there is a way that avoids the completion and proves this directly, e.g. using the inertia group as the hint suggested. I was thinking that perhaps one could show that because the way $p$ splits in $K$, at most two of the roots of $f$ become equal mod $\frak P$ (or at least one could choose $f$ like that). Then it would be clear that the only non-trivial element of the inertia subgroup $I_{\frak P}$ could be the transposition swapping said roots (like in this answer).