Algebraic number field without abelian extension

Question

What are examples of algebraic extensions $K / \Bbb Q$ such that $K$ has no non-trivial (finite or infinite) abelian extension? For instance, can we have $K \neq \overline{\Bbb Q}$?

I think that we must have $\sqrt d, e^{2 \pi i /n} \in K$ for any $d,n \in \Bbb Z^*$ (in particular $K$ cannot be finite over $\Bbb Q$ I think). But typically the maximal abelian extension of $\Bbb Q$ does have non-trivial abelian extensions!

I guess any field that is closed under taking roots should work ? — mercio, Jul 10 '18 at 20:44

score 2 · Accepted Answer · answered Jul 14 '18 at 10:27

The Uchida paper introduces without comment a Galois extension $\Omega/\mathbf Q$ which "has no abelian extension". This triggered in my mind the suspicion that the considerations about the Schur multiplier of a perfect group were perhaps irrelevant. And indeed they are.

An infinite extension as $\Omega$ must contain $K_1 :=\mathbf Q^{ab}$, and recursively, the chain of ascending extensions defined by $K_{i+1} = K_i^{ab}$. In terms of Galois groups, this means that one has to introduce the derived series of $G_{\mathbf Q}$ defined by $G_{\mathbf Q}^{i+1}=[G_{\mathbf Q}^i,G_{\mathbf Q}^i]$, and the problem of the existence of $\Omega \neq \bar {\mathbf Q}$ is equivalent to the question whether the derived series stabilizes non trivially. The Wikipedia article on derived series states : For a finite group, the derived series terminates in a perfect group, which may or may not be trivial. For an infinite group, the derived series need not terminate at a finite stage, and one can continue it to infinite ordinal numbers via transfinite recursion, thereby obtaining the transfinite derived series, which eventually terminates at the perfect core of the group. That's all we needed.

Many thanks for your help :-) $$ $$ 1) What is the Galois group of $K_2$ over $K_1$, in terms of the derived subgroups $G_{\bf Q}^i$ ?I'm not sure since $K_2$ is not Galois over $\bf Q$. $$ $$ 2) And so is the perfect core [= largest perfect subgroup] $H$ of $G_Q$ trivial or not (as closed subgroup of $G_Q$)? $$ $$ 3) I guess that you are saying that $H = Gal(\bar Q / \Omega)$ with $\Omega$ without any abelian extension, right? (I guess that there might be other $\Omega$ with this properties, also). — Alphonse, Jul 14 '18 at 12:56
Thank you ; I think that together with your nice comments, I can accept your answer :-) — Alphonse, Jul 14 '18 at 18:29

score 1 · Answer 2 · answered Jul 12 '18 at 16:46

1

For simplicity, let us stick to finite extensions of $\mathbf Q$. To put your question in context, recall the inverse Galois problem : is every finite group $G$ realizable as the Galois group of a normal extension of $\mathbf Q$ ? The most general positive result is the famous theorem of Shafarevitch (1954) : yes if $G$ is solvable. But this result is in the opposite direction w.r.t. your question, because a solvable extension is obtained by "piling up" abelian extensions, in more mathematical terms, in solving successive embedding problems with abelian kernels.

In the direction of your question (no abelian subextension), the most natural approach would be to try to realize non abelian simple groups (the complete list of these is known) as Galois groups over $\mathbf Q$. For example, among the classical families such as $PSL_n (q)$ or $PSU_n (q)$, the realization is possible when $q=p^f$ is small w.r.t. $n$. But for instance the problem remains open for $PSL_2 (5^f)$ if $f>2$, $PSL_2 (3^f)$ if $f>3$, $PSL_3 (3^f)$ if $f>2$, $PSL_4 (2^f)$ if $f>1$. On the other hand, one knows how to realize all the simple sporadic groups except the Mathieu group $M_{23}$.

One larger and deeper approach would be to study the absolute Galois group of $\mathbf Q$ using topological/geometric methods (coverings of the projective line $\mathbf P^1$, Hurwitz spaces, Grothendieck's "dessins d'enfants",...). An accessible account is Pierre Debes' report, "Autour du problème inverse de Galois", Bicentenaire de la naissance d'Evariste Galois, IHP-SMF, October 2011.

answered Jul 12 '18 at 16:46

nguyen quang do

14,578

Thank you! Just to be clear: I wanted $K/\Bbb Q$ such that if $L/K$ is abelian then $L=K$. So I don't understand why you are talking about "subextensions". – Alphonse Jul 12 '18 at 19:08
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Sorry, I misunderstood "subextension of K" instead of "extension of K", but this was perhaps because your question doesn't make sense : any number field K admits non trivial abelian extensions, which are described by class field theory. In particular cases, e.g . when K contains the group mu_n of n-th roots of unity (this is always the case for n=2), Kummer theory (which is purely algebraic) even tells you that the Galois group over K of the maximal abelian extension of exponent n of K is isomorphic to Hom (K/K^n, mu_n) – nguyen quang do Jul 12 '18 at 19:45
Yes, but what about infinite extension $K / \Bbb Q$ (as mentioned in my question)? – Alphonse Jul 12 '18 at 19:47
Is it possible to have $K \neq \overline{\Bbb Q}$ (infinite algebraic over $\Bbb Q$), without any abelian extension? – Alphonse Jul 12 '18 at 20:46
Sorry, do you have an answer to my above comment? Thank you ver much! – Alphonse Jul 13 '18 at 08:17
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Your original question is a difficult one. In general, finite and infinite algebraic extensions of Q seem to behave quite differently. One example: as I said, the main "conjecture" of inverse Galois theory is that every finite group should be realizable as a Galois group over Q. But on the other hand, it follows easily from the Kronecker-Weber theorem that an abelian profinite group of rank 2 without torsion is not realizable. Back to your question, all I can suggest is to consider an infinite algebraic extension K as the fixed field of a closed subgroup H of $G_Q$. – nguyen quang do Jul 13 '18 at 17:19
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Then "your" property is that H has no non trivial quotient, i.e. H = closure of its commutator subgroup, i.e. H is a perfect group. We can get a hold on such an H via its Schur mutiplier $M(H):=H_2(H, Z)$. If a group H is described by generators and relations, say H = F/R (with F free), then $M(H) = (R \cap [F,F])/[F,R]$. If moreover H is perfect, you have an exact sequence $1 \to M(H)= R/[F,R] \to [F,F]/[F,R] \to H \to 1$ . A starting point would be Tate's thm. that $M(G_Q)$ is trivial (this is true for the algebraic closure of any number field). – nguyen quang do Jul 13 '18 at 17:21
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The next natural step would be to try to catch M(H) from this, but in general, for a subgroup H of a group G, the image of the restriction map $M(G) \to M(H)$ is not well known. I'm stuck here for the moment. – nguyen quang do Jul 13 '18 at 17:21
Thank you so much for your comments. Would you mind to copy them into your answer? Moreover, I was confused because this paper is talking about a Galois extension $\Omega / \Bbb Q$ that "has no abelian extension", without making any further comment. Do they mean it should be easy to find such $\Omega$ (which is my question)? – Alphonse Jul 13 '18 at 19:39
Moreover, would you know an answer to https://math.stackexchange.com/questions/2689504/p-primary-part-of-an-abelian-group ? Thank you :) – Alphonse Jul 14 '18 at 09:47

Algebraic number field without abelian extension

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