This is related to "A radical extension with a non-radical subextension".

The problem is I can't understand why $L=\mathbb{Q}(\zeta_7+\zeta_7^{-1})$ is not a radical extension given in the solution above-mentioned.

Can anyone explain, or even better, give a simpler example of a radical extension with a non-radical subextension?

I am curious if there is something special about "7" here. I.e. would $\mathbb{Q}(\zeta_3)$ and $\mathbb{Q}(\zeta_3+\zeta_3^{-1})$ work as counter-examples.

Thanks a lot.

What I tried:

I can see that $\mathbb{Q}(\zeta_7)$ is a radical extension since $\zeta_7^7=1\in\mathbb{Q}$.

Using the same logic, so $L=\mathbb{Q}(\zeta_7+\zeta_7^{-1})$ is not a radical extension since $(\zeta_7+\zeta_7^{-1})^n\notin\mathbb{Q}$ for any power $n$? It seems possible to prove that by binomial theorem. Is that the correct logic?


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    $\begingroup$ One problem with your proposal with $\zeta_3$ is that $$ \zeta_3 + \zeta_3^{-1} = \frac{-1 + i\sqrt{3}}{2} + \frac{-1 - i\sqrt{3}}{2} = -1 \in \mathbb{Q} \, . $$ $\endgroup$ – André 3000 Dec 6 '16 at 7:41
  • $\begingroup$ @SpamIAm Oh yes, you are absolutely right. $\endgroup$ – yoyostein Dec 6 '16 at 7:50

That logic is not correct.

For example $\Bbb{Q}(1+\sqrt2)$ is a radical extension because it is equal to $\Bbb{Q}(\sqrt2)$. Yet, $(1+\sqrt2)^n\notin\Bbb{Q}$ for any $n>0$.

With $K=\Bbb{Q}(\zeta_7+\zeta_7^{-1})$ you can argue as follows. $[K:\Bbb{Q}]=3$ is a prime, so there are no non-trivial intermediate fields. Also, $K=\Bbb{Q}(z)$ for any $z\in K\setminus\Bbb{Q}$. Assume that here $z^n=q\in\Bbb{Q}$ for some integer $n$. We know that

  1. $K/\Bbb{Q}$ is Galois.
  2. The Galois conjugates, $z_2$ and $z_3$, of $z=z_1$ are of the form $z_j=z\zeta_n^{k_j}, j=2,3,$ where $\zeta_n=e^{2\pi i/n}$ and $k_2,k_3$ are some integers in the range $0<k_2<k_3<n$. This is because those conjugates must also be zeros of $x^n-q$.
  3. But the numbers $z_j/z_1=\zeta_n^{k_j}\in K, j=2,3.$ Because $K\subset \Bbb{R}$ we must have $\zeta_n^{k_j}=\pm1$ because there are no other real roots of unity.
  4. This implies that $z_2,z_3=\pm z_1$ contradicting the fact that the conjugates must all be distinct.
  • $\begingroup$ Thanks. How do we know $K/\mathbb{Q}$ is Galois? $\endgroup$ – yoyostein Dec 6 '16 at 15:44
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    $\begingroup$ @yoyostein: $\Bbb{Q}(\zeta_7)/\Bbb{Q}$ is Galois with an abelian Galois group $G\simeq C_6$. An intermediate field $K$ is Galois over the bottom field iff the subgroup $H=\operatorname{Gal}(\Bbb{Q}(\zeta_7)/K)$ is a normal subgroup of $G$. As $G$ is abelian ... $\endgroup$ – Jyrki Lahtonen Dec 6 '16 at 15:48
  • $\begingroup$ As $G$ is abelian, all subgroups are normal. Nice! I was thinking in terms of splitting field, got stuck in the process.. $\endgroup$ – yoyostein Dec 6 '16 at 15:51

Improving on Jyrki Lahtonen's answer, one might simply note that the only totally real number fields which are generated by radicals over $ \mathbf Q $ are generated by square roots of rationals. The reason for this is that all higher roots of rationals have conjugates which are not real, thus an embedding into $ \mathbf C $ may be defined by extending.

More concretely, let $ K $ be a totally real number field and let $ q^{1/n} \in K $ with $ q \in \mathbf Q $. $ q^{1/n} $ is a root of $ X^n - q $, thus its other conjugates are of the form $ \zeta_n^k q^{1/n} $. Define $ \mathbf Q(q^{1/n}) \to \mathbf C $ by sending $ q^{1/n} \to \zeta_n^k q^{1/n} $ for a conjugate, and use isomorphism extension to extend this to an embedding $ K \to \mathbf C $. Its image must be real, thus $ q^{1/n} $ and $ \zeta^k_n q^{1/n} $ are both real. From this, it follows that $ \zeta^k_n = \pm 1 $, and taking norms gives that $ q^{2/n} $ is rational.

Now, we may note that $ \mathbf Q(\zeta_7 + \zeta_7^{-1}) $ is a totally real field (just look at how $ \textrm{Gal}(\mathbf Q(\zeta_7)/\mathbf Q) $ acts on it), however its degree is not a power of $ 2 $. Thus, it cannot be generated by radicals over $ \mathbf Q $.

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    $\begingroup$ I liked this response. As a comment, the extension is totally real because any Galois extension of $\Bbb Q$ of odd degree is totally real. $\endgroup$ – Lubin Dec 7 '16 at 15:08
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    $\begingroup$ I did not mention this because the argument in the answer works for primes that are $ 1 $ mod $ 4 $ also, for instance, $ \mathbf Q(\zeta_{13} + \zeta_{13}^{-1}) $ cannot be generated by radicals over $ \mathbf Q $ either, by exactly the same argument. (Of course, this one actually does contain square roots: $ \sqrt{13} $, for instance.) It is an easy way to see this in this specific case, though. $\endgroup$ – Starfall Dec 7 '16 at 16:34

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