"Kummerian" fields? 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.
 A: Recall that a field $K$ is pseudo finite if (1) $K$ is perfect (2) the absolute Galois group of $K$ is $\hat{\mathbb{Z}}$, and (3) every absolutely irreducible non-void variety defined over $K$ has a $K$-rational point. 
Ax proved that large finite fields have the same elementary theory as a pseudo finite field.  
I guess that the statement: the unique extension of degree $n$ is generated by a root of an element of the field is elementary, hence from since finite fields are Kummerian, so are pseudo finite fields are Kummerian.
If I got it right so far, this gives an abundance of examples using Jarden's theorem (non of them explicit): Choose a Galois automorphism $\sigma$ in the absolute Galois group of $\mathbb{Q}$. Then with probability one, the fixed field of $\sigma$ in the algebraic closure is pseudo finite. (Recall that Galois groups are profinite, hence compact, hence equipped with Haar probability measure.)
Jarden's theorem holds in more generality, namely when we replace $\mathbb{Q}$ with any countable Hilbertian field (for some examples see here).
Anyway, I guess it doesn't answer any of your questions, in particular, it doesn't give any new absolute Galois group. But I thought some more examples might be interesting...
A: Here is one more easy construction: let $K_n = \mathbb{C}((t_1))\ldots((t_n))$, an $n$ times iterated Laurent series field over the complex numbers.  Then the absolute Galois group of 
$K_n$ is $\hat{\mathbb{Z}}^n$, obtained by adjoining all roots of the $t_i$'s.  Thus $\hat{\mathbb{Z}}^n$ is a 
Kummerian group.  Combined with 3) above, I believe this shows that any topologically finitely generated torsionfree abelian profinite group is Kummerian.  
ADDENDUM: I now believe that an infinite abelian profinite group is Kummerian iff it is torsionfree.  I'll sleep on this and see if someone else can say more...
