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. 



adding an n-th root to Q_p
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.


