
> 
> Let $K$ be a number field and $v$ be it's one of $K$'s non-archimedian valuation.
> Then, I would like to prove $K\_v(a^{1/m}) /K\_v$ is unramified if
> only if $v(a)≡0 \pmod m$.
> 
> 
> 


This is from Silverman's 'the arithmetic of elliptic curves', p213.
I know unramified extension of local field is in bijection with extensions of the residue field. Thus, the unramified extension is generated by roots of unity of order prime to the character of residue field of local field.


But I don't have tactics to judge given extension is unramified or not.


Thank you in advance.



$K_v(a^{1/m}) /K_v$ is unramified if only if $v(a)≡0 \pmod m$
If $v(a)\not\equiv 0\pmod m$, ramification is easy: just consider the valuation of the element $a^{1/m}$.


The converse is a little subtler than you make it seem, and depending on how exactly you phrase it, it need not be true: for instance, if $m=p$ coincides with the residue characteristic of $v$, $a=1$ and we interpret $a^{1/m}$ as a primitive $p$-th root of unity, the extension *will* be ramified.


Silverman includes a number of assumptions which exclude this case: he takes $v\not\in S$, and a couple paragraphs above it is stated that we assume this imples $v(m)=0$. Assuming this is the case and that $v(a)\equiv 0\pmod m$, note that, by multiplying $a$ by some $m$-th power, we may assume $v(a)=0$. Now, the polynomial $X^m-a$ has discriminant $\pm m^ma^{m-1}$, which under all these assumptions has valuation zero. This implies the discriminant of the extension $K\_v(\sqrt[m]{a})/K\_v$ is a $v$-adic unit, which implies the extension is unramified.


