Algebraic number field without abelian extension 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!
 A: 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.
A: 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.
