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.)

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.
