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Let $K$ be the subfield of $\mathbb{C}$ generated by all the $n$-th roots of unity for all $n$ over $\mathbb{Q}$. Let $\bar{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$. Is $Gal(\bar{\mathbb{Q}}/K)$ an infinite non-abelian group?

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yes, very. As you know, $K$ is the max ab of $\mathbb Q$, and this is relatively small extension, as these things go. – Lubin Nov 11 '12 at 6:55
@Lubin I guessed so. I would like to know the proof. – Makoto Kato Nov 11 '12 at 7:08
The only thing that pops into my mind is something unnecessarily complicated. Let me think about it a bit more. – Lubin Nov 11 '12 at 7:35
up vote 3 down vote accepted

Here’s an argument that’s fairly crude, to show that $K$ has nonabelian extensions of arbitrarily large degree, which will answer your question positively, but I’m sure that many others can give better.

We know that “generically”, whatever that means, all $n$-th degree polynomials over $\mathbb Q$ have Galois group $\mathcal S_n$, the full symmetric group on $n$ letters. For each $n$, let $F^0_n$ be the splitting field of one of these, and let $F_n=KF^0_n$, an extension of $K$ that you see has Galois group $\mathcal A_n$, the $n$-th alternating group. And there you are.

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Could you explain why $Gal(F_n/K) = A_n$? – Makoto Kato Nov 11 '12 at 9:21
Sure, it’s because $K\cap F^0_n$ is quadratic over $\mathbb Q$. Then apply the theorem about the Galois group of $KF^0_n$ over $K$ versus the Galois group of $F^0_n$ over $\mathbb Q$, which I'm sure you know. I always think of it as the Theorem on Natural Irrationalities, but I may be misattributing here. – Lubin Nov 11 '12 at 20:54
Thanks. "$K \cap F_n^0$ is quadratic over $\mathbb{Q}$" I think this is reduced to the following question.… Here is a proof of the existence of a Galois extension $F_n^0/\mathbb{Q}$ whose Galois group is $S_n$… – Makoto Kato Nov 11 '12 at 22:53
Sure you have to be aware that for $n\ge5$, the only nontrivial abelian quotient is cyclic of order two. – Lubin Nov 11 '12 at 23:30

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