Let $F_\infty=\bigcup_{n\geq1}\operatorname{Q}(2^{1/2^n})$ , $K_\infty=\bigcup_{n\geq1}\operatorname{Q}(\zeta_{2^n})$, what is the intersection?

Question

Let $F_\infty=\bigcup_{n\geq1}\operatorname{Q}(2^{1/2^n})$ and $K_\infty=\bigcup_{n\geq1}\operatorname{Q}(\zeta_{2^n})$. What is the intersection $F_\infty\cap K_\infty$? (Here $\zeta_{2^n}$ is a primitive $2^n$th root of unity.)

Jyrki Lahtonen used the method of algebraic number theory to deduce that $$\operatorname{Q}(2^{1/p^n})\cap\operatorname{Q}(\zeta_{p^n})=\operatorname{Q},\quad p\neq2\ \text{an odd prime number}.$$ See here. I guess that $$F_\infty\cap K_\infty=\operatorname{Q}(\sqrt{2}).$$ We know that $F_\infty$ is a real field and the maximal real subfield $K^+_\infty$ of $K_\infty$ is an extension of $\operatorname{Z}_2$ ($2$-adic integers).

My question is

1. For every $2^n$, does $F_\infty$ contain just one field of degree $2^n$?

2. What is the maximal sub-abelian extension of $F_\infty$? (I mean the intersection of $F_\infty$ and $\operatorname{Q}^{\operatorname{ab}}$.)

If we replace $2$ with any odd prime $p$, we can also define $F_\infty=\bigcup_{n\geq1}\operatorname{Q}(p^{1/p^n})$ and $K_\infty=\bigcup_{n\geq1}\operatorname{Q}(\zeta_{p^n})$ and can also ask the same question: what is the intersection of $F_\infty$ and $K_\infty$?

Answer 1

1. The answer to your first question is yes, $F_{\infty}$ has only one subfield of degree $2^n$. Indeed, writing $F_n = \mathbb{Q}(2^{\frac{1}{2^n}})$, you have $F_n \subset F_m$ whenever $n\leq m$. Every finite subextension of $F_{\infty}|\mathbb{Q}$ comes from a subextension of some $F_m|\mathbb{Q}$, so you may want to establish the subextension of $F_m|\mathbb{Q}$. Such question are well answered by Galois theory.

Let $L|K$ is a finite Galois extension of group $G$, there is a decreasing bijection between subextensions of $L|K$ and subgroups of $G$, given by $F|K \mapsto \textrm{Gal}(L|F)$. If $F$ is a subextension of $L|K$, there is a decreasing bijection between subgroups of $G$ containing $\textrm{Gal}(L|F)$, and subextensions of $F|K$.

So let $F_m^{\textrm{gal}}$ be the Galois extension of $F_m|\mathbb{Q}$. You have $F_m^{\textrm{gal}}=F_m(\zeta_{2^m}) = \mathbb{Q}(2^{\frac{1}{2^m}},\zeta_{2^m})$, and let $G=\operatorname{Gal}(F_m^{\textrm{gal}} / \Bbb Q)$ be its Galois group, which is generated by $\tau : 2^{\frac{1}{2^n}} \mapsto 2^{\frac{1}{2^n}} \zeta_{2^n}, \ \zeta_{2^n} \mapsto \zeta_{2^n}$ and $\textrm{Gal}(F_n^{\textrm{gal}}|F_n)$, itself a subgroup of $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ thus isomorphic to some $(\mathbb{Z}/2^i\mathbb{Z})^{\times}$, which we write an element as $\sigma_a : \zeta_{2^n} \mapsto \zeta_{2^n}^a$. We have $\sigma_a \tau \sigma_a^{-1}(2^{\frac{1}{2^n}}) = 2^{\frac{1}{2^n}} \zeta_{2^n}^a = \tau^a(2^{\frac{1}{2^n}})$, and $\sigma_a \tau \sigma_a^{-1} (\zeta_{2^n}) = \sigma_a \tau(\zeta_{2^n}^{a^{-1}}) = \zeta_{2^n} = \tau^a(\zeta_{2^n})$. Thus $\sigma_a \tau \sigma_a^{-1} = \tau^a$, and we have an isomorphism $G \cong \mathbb{Z}/2^n\mathbb{Z} \rtimes (\mathbb{Z}/2^i\mathbb{Z})^{\times}$.
Subfields of $F_n$ are in bijection with subgroups of $\mathbb{Z}/2^n\mathbb{Z} \rtimes (\mathbb{Z}/2^i\mathbb{Z})^{\times}$ that contains $(\mathbb{Z}/2^i\mathbb{Z})^{\times}$. Such a subgroup $H$ have to contains some power of $\tau$, and by the rule $\sigma_a \tau \sigma_a^{-1} = \tau^a$, the powers of $\tau$ occuring in $H$ is a subgroup of $\mathbb{Z}$, thus of the form $e\mathbb{Z}$, where $e$ divides $2^n$ the order of $\tau$. Thus $H$ is generated by $\tau^e$ and $(\mathbb{Z}/2^i\mathbb{Z})^{\times}$ with $e|2^n$, and vice versa. We thus count exactly $n+1$ subfields of $F_n$, which are the one we know $\mathbb{Q}(2^{\frac{1}{2^d}})$.
So $F_n$ has exactly one subextension of degree $2^d$ which is $F_d$, and thus $F_{\infty}$ has for only subextension of degree $2^n$ the extension $F_n$.

2. Thus $F_m \cap \mathbb{Q}^{\textrm{ab}} = F_2$. Indeed its fintie subextensions are exactly the $F_n$ by the first question, and $2^{\frac{1}{2^n}}$ is in $\mathbb{Q}^{\textrm{ab}}$ if and only if $n\leq 1$. Indeed $2^{\frac{1}{4}}$ is already not in $\mathbb{Q}^{\textrm{ab}}$ because $\textrm{Gal}(F_2^{\textrm{gal}}|\mathbb{Q}) = \mathbb{Z}/4\mathbb{Z} \rtimes (\mathbb{Z}/4\mathbb{Z})^{\times}$ which is not abelian.

Thus, your initial question may found an answer : $F_{\infty} \cap K_{\infty}$ is a subfield of $F_{\infty} \cap \mathbb{Q}^{\textrm{ab}} = F_2$.

ps : As the maximal real subfield of $\mathbb{Q}(\zeta_n)$ is $\mathbb{Q}(\textrm{cos}(\frac{2 \pi}{n}))$, and $\textrm{cos}(\frac{\pi}{2^{n+1}}) = \frac{1}{2}\sqrt{2+\sqrt{2+...+\sqrt{2}}}$ with $n$ roots, you may try to prove that indeed $\textrm{cos}(\frac{2\pi}{2^4})$ is not in the $F_n$, that is $\sqrt{2+\sqrt{2}}$ is not in the $F_n$, and you may check this by direct computation.