Solubility of the quintic? Over the p-adics, every Galois group is solvable.  Does this imply that the quintic (and higher-order polynomials for that matter) can be solved by radicals over $\mathbb{Q}_p$?
EDIT: The original place I learned that the p-adic galois groups were solvable was in Milne's Algebraic Number Theory text (Chapter 7, Cor 7.59).  
As was pointed out the comments, I should clarify that I meant to ask 2 questions.  Namely, whether the general quintic can be solved by radicals in this context (still no) and whether any given one can be (which I now believe is yes).  
 A: The fact is true for every field $K$ of characteristic 0: every finite algebraic extension of $K$ with solvable Galois group is inside an extension obtained by adding radicals (i.e. solutions of equations $x^n=a$). The field is separable since its characteristic is 0. Hilbert's Theorem 90 (see Google) holds over every field of characteristic 0. Therefore if a  finite extension $K'$ of $K$ contains primitive roots of 1 of degree $n$, then any extension $E$ of $K'$ with cyclic Galois group of order $n$ is obtained by adding a root of the equation $x^n=a$ for some $a\in K'$. Now you can apply the fundamental theorem of Galois theory. 
A: Dave Dummit's paper "Solving Solvable Quintics" http://www.ams.org/journals/mcom/1991-57-195/S0025-5718-1991-1079014-X/S0025-5718-1991-1079014-X.pdf constructs a sextic out of the coefficients of the quintic which has a rational root if and only if the quintic has a solvable Galois Group.  Although this is stated over $\mathbb{Q}$ I believe that it applies over any field of characteristic 0 or > 5.
A: Try Lazard (Daniel),
Solving quintics by radicals, in The legacy of Niels Henrik Abel, 207–225, Springer, Berlin, 2004.
MR2077574 (2005g:12002) says : 
Let $F$ be a field of characteristic different from 2 and 5. Let $f$ be a univariate irreducible polynomial of degree 5 over $F$. The polynomial $f$ is said to be solvable by radicals if the Galois group over $F$ of the field generated by all the roots of $f$ is solvable. In the paper the author gives a formula for solving by radicals any polynomial $f$ of degree 5 which is solvable by radicals. The field extension which is generated by the radicals which appear in the result is always minimal, when only one root is produced, as well as when all roots are given. This formula has been implemented in Maple.
Reviewed by Jerzy Urbanowicz.
A: Even though every extension of $\mathbb{Q}_p$ is solvable, I don't think one can write down a formula for the solution to a general quintic in terms of radicals; if one could, then the $S_5$-extension $\mathbb{Q}_p(r_1,...,r_5)/\mathbb{Q}_p(e_1,...,e_5)$ would be solvable, where the $e_i$ are the elementary symmetric polynomials in the $r_i$.
A: If you want to solve the quintic over the $p$-adics, two cases naturally arise : $p\neq5$ and $p=5$.
Suppose first that $p\neq5$ and let $f\in\mathbf{Q}_p[T]$ be an irreducible polynomial of degree $5$.  Then the extension $K$ obtained by adjoining a root $\alpha$ of $f$ to $\mathbf{Q}_p$ is always contained in $F(\root5\of{F^\times})$, where $F=\mathbf{Q}_p(\root5\of1)$.  So you see immediately that $\alpha$ can be expressed by radicals.  
In fact, $f$ can always be taken to be of the form $f=T^5-x$ in the generic case when $K$ is (totally) ramified over $\mathbf{Q}_p$, so if you wish I can claim to have solved the quintic by radicals by just saying that $\alpha=\root5\of x$.
Now let $f\in\mathbf{Q}_5[T]$ be an irreducible polynomial of degree $5$.  Then the extension $K$ obtained by adjoining a root $\alpha$ of $f$ to $\mathbf{Q}_5$ is always contained in $F(\root5\of{F^\times})$, where $F=\mathbf{Q}_5(\root4\of{\mathbf{Q}_5^\times})$.  Here again you see that $\alpha$ can be expressed by radicals.  
There is nothing special about the prime $5$ or the base field $\mathbf{Q}_p$.  You can replace $5$ by any prime $l$, and $\mathbf{Q}_p$ by any finite extension thereof, and you will get similar results depending on whether $l\neq p$ or $l=p$.
You can even replace $\mathbf{Q}_p$ by a finite extension of $\mathbf{F}_p((\pi))$, provided you replace the "radical" $\root p\of{x}$ (which denotes a root of the binomial $T^p-x$) by its characteristic-$p$ cousin $\wp^{-1}(x)$ (which denotes a root of the trinomial $T^p-T-x$). 
