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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.

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I was under the impression that K((t)) is Kummerian for any algebraically closed field K of characteristic zero. Am I misinformed? –  Qiaochu Yuan Dec 28 '09 at 5:20
    
I feel like the term "kummerian" may not be ideal because kummer extensions "are a thing". –  Harry Gindi Dec 28 '09 at 5:22
    
@QY: You are absolutely correct. In my haste to get to the somewhat more interesting case in which K is real-closed, I guess I forgot to mention this. The Galois group is again Z-hat. –  Pete L. Clark Dec 28 '09 at 5:25
    
Let me just say that I find this question interesting, in connection with my more general interest in the maximal radical extension of any given field (the extension obtained by adjoining all roots of all elements of the basic field, not iteratively but all at once). –  Leonid Positselski Dec 28 '09 at 12:47

2 Answers 2

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...

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I had actually thought about pseudo-finite fields myself, and convinced myself that at least some of them are Kummerian. Here was my train of thought: (i) by Kummer theory, a field of characteristic 0 containing all roots of unity and with abelian Galois group is Kummerian. (ii) One can arrange for a pseudofinite field to contain all roots of unity, by making sure that for any given N, all but finitely many finite fields in an ultraproduct have N roots of unity. (Similarly, one can construct pseudofinite fields which do not contain all roots of unity.) ... –  Pete L. Clark Dec 28 '09 at 21:06
    
...Anyway, your approach is interesting: I agree that the statement that the unique field extension of degree n is radical is first order. This raises the following question: is the property of being Kummerian itself elementary? –  Pete L. Clark Dec 28 '09 at 21:09
    
I'm always confused with that elementary stuff :P There are infinitely many statements, one for each $n$, right? So are you aiming that somehow we can see it via merely finitely many statements. –  Lior Bary-Soroker Dec 28 '09 at 21:32
    
No, that's not what I mean (and yes, the terminology here is confusing). I just mean that the Kummerian fields form an elementary class, meaning that if K is Kummerian and K' is a field such that Th(K) = Th(K') -- i.e., a statement in the language of fields is true in K iff it is truein K' -- then K' is Kummerian. I think this is probably true, and I think it is probably not true that Kummerianity (yikes!) is finitely axiomatizable. –  Pete L. Clark Dec 28 '09 at 22:54
    
It is true if the absolute Galois group of a Kummerian field is f.g. as in all the examples. However I'm not sure it's true: what about $\mathbb{C}((t_1))((t_2))\cdots $? (the limit field of the field you suggested) I just throw it to, I'm not sure is Kummerian, but one the first thought it is –  Lior Bary-Soroker Dec 29 '09 at 6:49

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...

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