This is sort of a random, spur of the moment question, but here goes:

We define [with apologies to Conan the Barbarian] a field K to be $\textbf{Kummerian}$ if there exists an index set I, and functions $x: I \rightarrow K, n: I \rightarrow \mathbb{Z}^+$ such that the algebraic closure of K is equal to $K[(x(i)^{\frac{1}{n(i)}})_{i \in I}]$. More plainly, the algebraic closure is obtained by adjoining roots of elements of the ground field, not iteratively, but all at once.

Questions:

QI) Is there a classification of Kummerian fields?

QII) What about a classification of "Kummerian (topological) groups", i.e., the absolute Galois groups of Kummerian fields?

Here are some easy observations:

1) An algebraically closed or real-closed field is Kummerian. In particular, the groups of order 1 and 2 are Galois groups of Kummerian fields. By Artin-Schreier, these are the only finite absolute Galois groups, Kummerian or otherwise.

2) A finite field is Kummerian: the algebraic closure is obtained by adjoining roots of unity. Thus $\hat{\mathbb{Z}}$ is a Kummerian group.

3) An algebraic extension of a Kummerian field is Kummerian. Thus the class of Kummerian groups is closed under passage to closed subgroup. Combining with 2), this shows that any torsionfree procyclic group is Kummerian. On the other hand, the class of Kummerian groups is certainly not closed under passage to the quotient, since $\mathbb{Z}/3\mathbb{Z}$ is not a Kummerian group.

4) A Kummerian group is metabelian: i.e., is an extension of one abelian group by another. This follows from Kummer theory, using the tower $\overline{K} \supset K^{\operatorname{cyc}} \supset K$, where $K^{\operatorname{cyc}}$ is the extension obtained by adjoining all roots of unity.

In particular no local or global field (except $\mathbb{R}$ and $\mathbb{C}$) is Kummerian.

5) The field $\mathbb{R}((t))$ is Kummerian. Its absolute Galois group is the profinite completion of the infinite dihedral group $\langle x,y \ | \ x^2 = 1, \ xyx^{-1} = y^{-1} \rangle$. In particular a Kummerian group need not be abelian.

Can anyone give a more interesting example?

ADDENDUM: In particular, it would be interesting to see a Kummerian group that does not have a finite index abelian subgroup or know that no such exists.