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?
The Galois group of the algebraic closure of $\mathbb{Q}$ over the maximal abelian extention of $\mathbb{Q}$
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$\begingroup$
abstract-algebra
galois-theory
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0The only thing that pops into my mind is something unnecessarily complicated. Let me think about it a bit more. – 2012-11-11
1 Answers
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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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0Sure you have to be aware that for $n\ge5$, the only nontrivial abelian quotient is cyclic of order two. – 2012-11-11