adding an n-th root to Q_p 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. 
 A: 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.
A: If $n$ is prime to $p$, then ${\mathbb Q}\_p(a^{1/n})$ is unramified if $n | v_p(a)$,
and is tamely ramified otherwise.
To see this, we note that we may first of all divide $a$ by powers of $p^n$, and so assume
that $0 \leq v_p(a) < n.$  
If in fact $v_p(a)=0$, i.e. $a$ is a unit, then the extension is unramified, and the ring
of integers is equal to ${\mathbb Z}\_p[a^{1/n}]$ (by Hensel's lemma, since $x^n - a$
is then a separable equation mod $p$).
Otherwise, if $0 < v_p(a) < n,$ we get a tamely ramified extension (essentially by the
definition of tamely ramified).
If $p | n$ then the situation is a little more complicated.  For example, if $n = p$
and $0 < v_p(a) < p,$ then the extension is wildy ramified.
If $a$ is a unit, then we may write $a = \zeta u,$ where $\zeta$ is a $(p-1)$st root of 1
and $u \equiv 1 \bmod p,$ and since $\zeta^p = \zeta,$ we see that
${\mathbb Q}\_p(a^{1/p}) = {\mathbb Q}\_p(u^{1/p}).$
Now (supposing that $p$ is odd, for simplicity) if $u \equiv 1 \bmod p^2,$ then $u$ is
in fact a $p$th power in ${\mathbb Q}\_p,$ and so the extension is trivial. On
the other hand, if $u \equiv 1 \bmod p,$ but not mod $p^2$, then the extension is
wildy ramified of degree $p$, with ring of integers equal to ${\mathbb Z}\_p[u^{1/p}].$
To see this last claim, note that if $X^p - u = 0,$ and we write $Y = X - 1$,
then $(Y  + 1)^p - u = 0,$ i.e. $Y^p + pY^{p-1} + \cdots + p Y + (u-1) = 0,$
and so $Y$ satisfies an Eisenstein polynomial of degree $p$.  This implies that the extension is wildly ramified of degree $p$, that $Y$ is a uniformizer in the extension,
and that the ring of integers is equal to ${\mathbb Z}\_p[Y] = {\mathbb Z}\_p[u^{1/p}].$
Added in response to Keith Conrad's comments below: As Keith points out, the extension
${\mathbb Q}_p(a^{1/n})$ is not really well-defined unless ${\mathbb Q}_p$ contains the
$n$th roots of $1$, or equivalently, if $n$ divides $p-1$ (or 2 if $p = 2$).
But note e.g. if $p$ does not divide $n$, then adding the $n$th roots of unity gives
an unramified extension of ${\mathbb Q}_p(a^{1/n})$, and so the ramification behaviour
is independent of the choice of $n$th root, while in the case when $n = p$ also treated above,
adjoining the $p$th roots of unity is a tamely ramified extension of ${\mathbb Q}_p$,
so the claims regarding wild ramification are independent of the choice of $p$th root.
