Ramified primes in radical extension of number fields Let $ K $ be a number field, $ n\ge2$ be a positive integer and  $a \in K^*$.
How does one show 
in the simplest possible way that   a prime ideal $\mathfrak {p}$ of $ K $ not dividing $ n$ is unramified in $ K (\sqrt [n]{a})/K $ if and only if the exact exponent of  $\mathfrak {p}$ in the decomposition in primes ideals of $ a\mathcal{O}_K $ is a multiple of n?
Thanks a lot in advance.
 A: I think that the « simplest » proof  consists in passing to the corresponding  extension of local fields, i.e. of the completions $L_w /K_v$ at the (additive) valuations given by the prime ideal $P$ under study. 
Choose a $n$-th root $\alpha$ of $a$ .
 If $L_w/K_v$ is unramified,  $n.w(\alpha)= w(a)= v(a)$, so $v(a)$ is a multiple of $n$. 
Conversely, suppose that $v(a)$ is a multiple of $n$. Multiplying $a$ by a $n$-th power if necessary, we may assume that $0 \le v(a) \le n – 1$ , so the hypothesis becomes $v(a) = 0$, i.e. $a$ is a $v$-unit. It is classically known that $L_w /K_v$ is unramified iff the relative discriminant $\delta$ of the extension is a $v$-unit (see e.g. Serre’s « Local Fields », chap. III, §5, coroll. 1). Moreover, the discriminant  $\delta$ is the norm of the different $\Delta$, and the latter is the principal ideal generated by $f’(\alpha)$, where $f$ is the minimal polynomial of $\alpha$  (op. cit., §7, coroll. 2). Since $f$ divides $X^n – a$, it follows that $f’(\alpha)$ divides $n\alpha^{n-1}$, hence $\delta$ divides the norm of $n\alpha^{n-1}$. Assuming further that $n$ is a $v$-unit, we get that $\delta$ is a $v$-unit, and the extension is unramified.
Note that the assumption that $P$ does not divide $n$ is necessary, because e.g. if $n = p$ := the rational prime under $P$ and $ K_v$ contains a primitive $p$-th root $\zeta$ of 1, it is known that $L_w /K_v$ is unramified iff $v(a – 1) \ge pv(\zeta - 1)  $ (see e.g. exercise 2.12 of Cassels-Fröhlich’s « Algebraic Number Theory »). 
