# p-th powers in p-adic field

Denote by $$K$$ the completion of $$\bigcup_{n \geq 1} \mathbb{Q}_p (\zeta_{p^n})$$, where $$\zeta_{p^n}$$ is a $$p^n$$-th root of unity. Is it true that any element in $$K$$ is a $$p$$-th power of some element in $$K$$?

I am grateful for any help!

• What are your thoughts on the problem? What have you tried and where did you get stuck? Commented Nov 27, 2018 at 22:55
• @Servaes: Well, first it suffices to prove that every $x \in K$ of absolute value $1$ has a $p$-th root. Now the obvious idea would be Hensel's Lifting Lemma, but the derivative has the factor $p$, which is the main problem. Commented Nov 27, 2018 at 22:59
• @Servaes: Write $f(x) = x^p-u$ for some $u \in K$ with $|u|=1$. If $f(\alpha) = 0$, then this means that $|\alpha|=1$ as well. But then $|f'(\alpha)|=|p|=p^{-1}$, so we need to construct $\alpha \in K$ such that $|\alpha|=1$ and $|f(\alpha)|<|f'(\alpha)|^2 = p^{-2}$. But why does such an $\alpha$ exist for any given $u$? Commented Nov 27, 2018 at 23:32

Not true, except conceivably for $$p=2$$.
Your extension is (the completion of) an abelian field over $$\Bbb Q_p$$, but even in the case $$p=3$$, $$\sqrt[3]3$$ isn’t in it, ’cause $$\Bbb Q_3(\zeta_3,\sqrt[3]3\,)$$ is nonabelian over $$\Bbb Q_3$$.