In "On definable susbsets of p-adic fields", Macintyre 1976, he states the following fact that, let $x\in Q\_p^\*$ such that $v(x) = 0$ (where $v$ is the p-adic valuation) and $k\in N$ then there exists $n\in Z$ such that $x/n\in (Q\_p)^k$ (ie the $k$-th powers).


This property is evident if $k$ is prime to $p$ by Hensel's lemma (the form I use is that let $f\in Z\_p[X]$ and $a\in Z\_p$ such that res$(f(a)) = 0$ and res$(f'(a))\neq 0$ then there exists $b\in Z\_p$ such that $f(b) = 0$ and res$(a) =$ res$(b)$). But if not, then I don't know how to prove it.


If anyone has any ideas on the subject I'd be very grateful.



k-th powers in the field of p-adics
The basic idea is that it is possible to extract $k^\mathrm{th}$-roots of elements of $\mathbb{Z}\_p$ that are close enough to 1. This is because you can compute explicitely the radius of convergence of the series $(1+X)^{1/k}$ (because you know the $p$-adic valuation of the factorials, hence of the binomial coefficients). Let us say that the radius of convergence is greater that $p^{-r}$.


Now you can write $x= n + yp^r$, with $y\in \mathbb{Z}\_p$ and $n$ invertible in $\mathbb{Z}\_p$, since $v(x)=0$. Then $x/n = 1+ n^{-1}yp^r$ has a $k^\mathrm{th}$-root.


