Converse to a proposition on algebraically closed fields

Question

This is a follow up to a previous question. Let us call a field $F$ root-closed if every element $x$ of $F$ has at least one $n$-th root for every positive integer $n$. It is very easy to show that every algebraically closed field of characteristic $0$ is root-closed. Is the converse true? That is, is every root-closed field of characteristic $0$ algebraically closed?

Answer 1

Abel's impossibility theorem explicitly says "no". For instance, start with $K_0 = \Bbb Q\subseteq \Bbb C$. Then recursively define $K_i$ as the extension of $K_{i-1}$ by all roots of all polynomials of the form $x^n - k$, for $k\in K_{i-1}$. The union of all these $K_i$ will be a root-closed subfield of $\Bbb C$ (it is the smallest subfield of $\Bbb C$ where each non-zero element has all its $n$ $n$-th roots). It consists exactly of all complex numbers that can be reached from the rational numbers by some finite application of the four standard arithmetic operations as well as taking complex $n$-th roots.

Abel's theorem says that there are polynomials over the rationals whose roots are impossible to describe in this form. One such polynomial is $x^5-x-1$.