Determining the degree of a root of unity over a cyclotomic expansion

Question

For $\xi_{n} = e^{2\pi i/n}$ , determine the degree of $\xi_{7}$ over the field $\Bbb{Q}(\xi_{3})$.

How would I approach this problem? I'm having trouble starting this problem and can't find any resources that could help me. Any help will be great, thanks.

Depending on how much you know about cyclotomic fields the observation $\Bbb{Q}(\xi_3,\xi_7)=\Bbb{Q}(\xi_{21})$ may or may not help. — Jyrki Lahtonen, May 27 '16 at 09:32
@JyrkiLahtonen $\Bbb{Q}(\xi_{21})$ has order 21 over $Q$, do I use that piece of info? — James Blake, May 27 '16 at 09:35
Err check a few facts first. Then you can use that piece of info. — Jyrki Lahtonen, May 27 '16 at 09:36
$\phi(21)=?????$ But I 'm redacting my suggestion to the extent that answering your question is more or less equivalent to proving the dimension formula for cyclotomic extensions of the rationals in this case. A lower brow approach is to be preferred. — Jyrki Lahtonen, May 27 '16 at 09:38

score 3 · Answer 1 · edited Apr 13 '17 at 12:19

Extended hints:

You probably know that $[\Bbb{Q}(\xi_7):\Bbb{Q}]=6$ and $[\Bbb{Q}(\xi_3):\Bbb{Q}]=2$, so you can deduce (how?) that the desired degree is either $3$ or $6$. The former case can happen only, if $\xi_3\in\Bbb{Q}(\zeta_7)$.
The Galois group $G=Gal(\Bbb{Q}(\xi_7):\Bbb{Q})\simeq\Bbb{Z}_7^*$ is cyclic of order six. Therefore, by Galois correspondence, it has a unique subfield of degree two.
By this answer (or many others) we see that $\sqrt{-7}\in \Bbb{Q}(\xi_7)$, so the unique quadratic subfield of $\Bbb{Q}(\xi_7)$ is $\Bbb{Q}(\sqrt{-7})$.
So if you can show that $\xi_3\notin \Bbb{Q}(\sqrt{-7})$, then it follows that $\xi_3\notin \Bbb{Q}(\xi_7)$, and you are done.

Determining the degree of a root of unity over a cyclotomic expansion

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