BIG Intermediate fields in an infinite tower This question arose out of the answer by Brandon Carter to the following question:
Infinite dimensional intermediate subfields of an algebraic extension of an algebraic number field
Define an infinite tower of field extensions recursively, starting from an algebraic number field $K=K_1$ in this way:  That is [$K_{n+1}:K_n]=d_n, d_{n}>1$ with $K_{n+1}$ obtained from $K_n$  by adjoining a root of an irreducible polynomial of the kind $X^{d_n}-a$.    Define $K_\infty$ to be the union of all these $K_n$'s which is a field, and is an algebraic extension of the rationals. Does $K_\infty/K_1$ contain infinitely many intermediate fields of infinite degree over $K$? (I expect the answer to be negative following Brandon Carter's answer mentioned above.)
 A: Just like in the previous question, it is possible to have infinitely many such subfields or none at all.
For an example with infinitely many such subfields, let $p_n$ denote the $n$th prime. Then let $K_1 = \mathbf{Q}$ and define $K_n$ recursively by $K_n = K_{n-1}(\sqrt{p_{n-1}})$. Then $K_\infty = \mathbf{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \dots)$ has infinitely many subfields of infinite degree over $\mathbf{Q}$. For example, for any finite set $S$  of primes, let $K_\infty^{(S)}$ denote the extension of $\mathbf{Q}$ given by adjoining $\sqrt{p}$ for all $p \notin S$. Then $K_\infty/K_\infty^{(S)}$ is finite, and so $K_\infty^{(S)}/\mathbf{Q}$ is infinite. An analogous construction works for any base field.
On the other hand, let $K_1$ be any number field containing the $p$th roots of unity. Then Kummer theory implies that a $\mathbf{Z}_p$-extension of $K_1$ is constructed from the process you describe, and just as in the previous question, this has no subfields of infinite degree over $K_1$. In particular, a $\mathbf{Z}_2$-extension of any number field will work.
