Let $K$ be a number field and $a\in K^*$ of infinite order in $K^*$. How do I show that $$[K(\sqrt[n]{a},\zeta_n):K]\geq C\cdot n\cdot\varphi(n)$$ holds for all positive integers $n$, with a positive constant $C$ which only depends on $K$ and $a$?

Similar estimates often occur in connection with variations of Artin's primitive root conjecture. Hooley 1967 proved this for $K=\mathbb{Q}$ and Lenstra 1977 proved it for all $n$ which are coprime to some integer.

Usually one would approach the problem by Kummer theory and show that the set of $m$ such that $\sqrt[m]{a}$ is contained in some $K(\zeta_n)$ is finite. Thus, it suffices to show that for any prime $p$ the radical extension $K(\sqrt[p^e]{a})/K$ cannot be abelian for infinitely many $e$. This seems obvious since $a$ is not a root of unity, but I couldn't work out a proof yet.

I am thankful for any help.


The paper "The order of reductions of an algebraic integer" by Antonella Perruca (on arXiv here) contains a complete determination of the degree of $[K(\sqrt[n]{a}, \zeta_{n}) : K]$ when $a \in K^{\times}$ is not a root of unity and $n$ is the power of a prime. It follows from the results of this paper that the bound you seek is true when $n$ is a prime power. (Incidentally, this paper uses a result of Schinzel from 1977 that classifies when $K(\sqrt[p^{e}]{a})/K$ is abelian.)

For the case of a general positive integer $n$ (which you may understand), it suffices to understand the intersections between $K_{n} = K(\sqrt[n]{a}, \zeta_{n})$ and $K_{\ell^{k}} =K(\sqrt[\ell^{k}]{a}, \zeta_{\ell^{k}})$ where $n$ is a product of primes $< \ell$. Degree considerations show that this intersection must be contained in $K(\zeta_{\ell})$ and as a consequence $(K_{n} \cap K_{\ell^{k}})/K$ must be ramified only at $\ell$. Since $K_{n}$ is ramified only at the primes that divide $n$ and the numerator and denominator of $N_{K/\mathbb{Q}}(a)$, a non-trivial intersection can occur only for finitely many $\ell$, and the degree of such an intersection is bounded by an absolute constant.


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