I have been working on the exercises of the book "Algebraic number theory" published by Frölich and Cassels. In one exercise, the following statement is given as a fact:
Let $K$ be a number field and $m\geq 2$. Let $v$ be a non-archimedean prime of $K$ which does not divide $m$, and let $a\in K^*$. Then the extension $K(a^{\frac{1}{m}})/K$ is unramified at $v$ if and only if $v(a)$ is divisible by $m$.
I have trouble writing down a proof for this statement, even after reviewing the chapter on Kummer theory inside the same book. In this chapter, it is proven that the discriminant of the extension $K(a^{\frac{1}{m}})/K$ divides $m^ma^{m-1}$, and hence a prime $v$ is unramified in particular if it does not divide $ma$ (ie in our case if $v(a)=0$, given that $v(m)$ is already $0$), which makes sense to me. However I fail to see how to prove/justify the more general statement above. Could someone give me a hint or a reference for this ?
Suppose $v$ is normalized, i.e., its image is exactly $\mathbb{Z}$. $v$ can be extended (though in general not unique) from $K$ to $K(a^{1/m})$, denote this extension also by $v$.
If $m\nmid v(a)$, then from $mv(a^{1/m}) = v(a)$ we see that $v(a^{1/m})$ is not an integer, so $v$ is ramified.
If $m\mid v(a)$, let $\pi$ be a uniformizer of $K$ for $v$, write $a = \pi^{mn} b$ for $v(b) = 0$, then $K(a^{1/m}) = K(b^{1/m})$. The discriminant now divides $m^m b^{m-1}$, which has valuation $0$, therefore $v$ is unramified.