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 $K_\infty$?
