TITLE: adding an n-th root to Q_p
QUESTION [9 upvotes]: What can be said about extensions à la $\mathbb{Q}_p(\sqrt[n]{a})/\mathbb{Q}_p$? Ramification behaviour, valuation ring, ...? 
I find it hard to say anything general - for example, as a function of the $p$-adic valuation of $n$ and/or $a$. Of course some special cases are rather easy to handle, and I understand what happens when $v_p(a) = 0$. This might be a hard question, or a question for which there is a standard reference - I didn't find one - or something rather easy, in which case I'm just missing something.

REPLY [5 votes]: It's possible to delve more deeply into the ramification structure in the more complicated wild case referred to in Emerton's answer. If $n=p^m$, then Coleman calculated the conductor of the Kummer extension 
${\mathbb Q}_p(\zeta_n,\sqrt[n]{a})/{\mathbb Q}_p(\zeta_n)$ 
using his reciprocity law and one should be able to use this to determine the ramification filtration of 
${\mathbb Q}_p(\sqrt[n]{a})/\mathbb{Q}_p$.
Romyar Sharifi does just that 
here
to determine the ramification groups of the maximal (nonabelian) Kummer extension ${\mathbb Q}_p\left(\sqrt[p^{\infty}]{{\mathbb Q}_p^\times}\right)/\mathbb{Q}_p$.
In particular, one finds that the (upper) ramification jumps are of the form $i$, $i+\frac{1}{p-1}$ and $i+\frac{1}{p(p-1)}$ for integers $i$.
In a similar vein and also by computing conductors, Viviani 
finds the ramification groups of the extension 
${\mathbb Q}_p(\zeta_{p^m},\sqrt[p^m]{a})/{\mathbb Q}_p$.
as long as $p^2$ doesn't divide $a$.
One trick he uses is to notice that for instance if $p$ exactly divides $a$ then
$$\frac{(1-\zeta_p)}{\sqrt[p]{a}\;\sqrt[p^2]{a}\cdots\sqrt[p^m]{a}}$$
is a uniformizer of 
${\mathbb Q}_p(\zeta_p,\sqrt[p^m]{a})$ 
and observes that the proofs could be simplified and generalized if one was able to write down uniformizers in further extensions.
So does anyone know how to do this ? Can one explicitly write down a uniformizer for the field 
${\mathbb Q}_p(\zeta_{p^m},\sqrt[p^m]{a})$ in a similar way ?
At a stretch, this might have some application in integral $p$-adic Hodge theory. The arithmetically profinite extension $K(\zeta_{p^\infty},\sqrt[p^\infty,]{\pi})/K$ where $\pi$ is a prime in $K$ makes an appearance in Tong Liu's extension of Breuil and Kisin's work and the more one knows about this extension and it's field of norms the better for calculations/proofs.