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!
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$\begingroup$ What are your thoughts on the problem? What have you tried and where did you get stuck? $\endgroup$– ServaesCommented Nov 27, 2018 at 22:55
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1$\begingroup$ @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. $\endgroup$– AlgebrusCommented Nov 27, 2018 at 22:59
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1$\begingroup$ @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 $u$? $\endgroup$– AlgebrusCommented Nov 27, 2018 at 23:32
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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 $\Bbb Q_3$.
Now that I have your attention, I would like to encourage you to become friends with the stronger version of Hensel, which is not about finding roots, but about lifting a factorization over the residue field to a factorization over the original complete field. I’m away from home right now, so I can’t give a precise reference, but it’s in the excellent book of Gouvêa.
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$\begingroup$ Thank you for your answer! I am aware that there exists a factorization-version of HLL, but it has the disadvantage that it does not generalize (to my knowledge) to multivariate polynomials. $\endgroup$– AlgebrusCommented Nov 29, 2018 at 0:29
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1$\begingroup$ Maybe not. But the stronger version does all sorts of things that the weak does not, and it’s really useful. $\endgroup$– LubinCommented Nov 29, 2018 at 6:10