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I want to show the existence of a field $E$ not containing $\sqrt{2}$, such that any finite extension of $E$ in $\overline{\mathbb{Q}}$ is cyclic.

I think the maximal field not containing $\sqrt{2}$ should work.

I think that any finite extension of $E$ must contain $2^{\frac{1}{2^n}}$, but I am not sure about it. How can I show that any finite extension of $E$ is cyclic?

Jyrki Lahtonen
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permutation_matrix
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  • Such that any subextension $F/\mathbb Q$ is cyclic? Or any extension $F/E$? – qualcuno Apr 25 '21 at 06:38
  • such that $\frac{F}{E}$ is cyclic. – permutation_matrix Apr 25 '21 at 06:42
  • Some thoughts that I can't make into a proof: assume $F/E$ normal, then by maximality we get $\sqrt{2} \in F$ and thus $2 \mid \mathbf{Gal}(F/E)$. Maximality gives that $E(\sqrt{2})$ is the minimal subextension of $F/E$. Maybe considering a $2$-Sylow we can see that the Galois group is actually a $2$-group. But then again, that as stated is not what you are looking for. Another idea: if $M_n$ is the maximal field not containing $2^{1/2^n}$ then $M_1 \subset M_2 \subset \cdots$. Put $M := \bigcup_n M_n$. A finite extension $F/M$ must - by maximality contain all roots $2^{1/2^n}$. – qualcuno Apr 25 '21 at 07:26
  • (cont.) then we get a filtration $M \subset M(\sqrt{2}) \subset M(\sqrt{2}^{1/2^n}) \subset \cdots F$. It'd be nice if in each step the subextensions are proper: this would imply a contradiction (we would have $[F: M] = \infty$) and thus there are no finite extensions of $M$, which maybe one can relate to having extensions of $E$ containing $2^n$-th roots of $2$. Again, this is more like thinking out loud than a solution, but I felt like it may be related to the problem. – qualcuno Apr 25 '21 at 07:30
  • It seems to me that this older thread is closely related. – Jyrki Lahtonen Apr 25 '21 at 07:42
  • @guidoar Something like the argument I used in the older thread? Great minds think alike etc :-) – Jyrki Lahtonen Apr 25 '21 at 07:47
  • @JyrkiLahtonen Haha, that does look like what I was hoping to be true :P I always mix-up the variance of the assignment $H \mapsto E^H$ of the fundamental theorem though :/ If I understand your answer correctly, this is precisely what OP is looking for, no? So maybe this should be marked as a duplicate? – qualcuno Apr 25 '21 at 07:54
  • @JyrkiLahtonen, I couldn't understand that solution – permutation_matrix Apr 25 '21 at 12:20
  • @JyrkiLahtonen, can we discuss? – permutation_matrix Apr 25 '21 at 12:20
  • At which point did I lose you? I suspect that at least the "group theoretic" part of my solution can be streamlined. I don't remember everything I had in mind back in the day, but it feels like I simply cooked up something to make it complete :-) – Jyrki Lahtonen Apr 26 '21 at 05:44
  • You wrote that a $p-group$ with unique maximal subgroup $H$ is abelian. I did it another way. If we consider $x \in G\setminus H$, then $ = G $, since $ \subset H $ is not possible(because of uniqueness and maximality of $H$).Hence it is cyclic. Hence it is abelian as well. Apart from that, I could not understand the notations used. I couldn't get the idea what you were exactly trying to show. Hence I want to discuss. – permutation_matrix Apr 26 '21 at 07:03
  • Does my answer help ? – rae306 Apr 28 '21 at 07:58
  • Yes, I got the idea. I am working on the details. – permutation_matrix Apr 28 '21 at 10:48

1 Answers1

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By taking the Galois closure if necessary, we may assume that $F/E$ is finite Galois. By construction of $E$, there is a unique minimal subextension of $F/E$, which is $E(\alpha)$. Under the Galois correspondence, it corresponds to a unique maximal subgroup $H$ of $\operatorname{Gal}(F/E)$. We have now reduced to the following group theoretic statement:

Let $G$ be a finite group with a unique maximal subgroup. Then $G$ is cyclic.

See here for a proof.

Even more is true: $G$ is cyclic of prime power order, $E(\alpha)/E$ is a $\mathbf{Z}/p\mathbf{Z}$-extension and $\operatorname{Gal}(\overline{\mathbf{Q}}/E)\cong \mathbf{Z}_p$.

rae306
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