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Unfortunately I can't yet post a comment so hopefully it'll be OK if I'll add it as a question of my own:

I'm trying to solve this one: Finite, normal extension of odd degree.
Can the following be a proper solution?

Assume by contradiction that $E\not\subseteq \mathbb R$.
Then there's an $a\in E, a\notin \mathbb R$ such that $\epsilon(a)\neq a$, where $\epsilon$ is the complex conjugation restricted to $E$ $\Rightarrow$ $\epsilon\in Gal(E/\mathbb Q)$.
But $|Gal(E/\mathbb Q)|=|E/\mathbb Q|=n$ is odd and $|\epsilon|=2\nmid n \Rightarrow$ By Lagrange's theorem $\epsilon\notin Gal(E/\mathbb Q)$. Contrediction. Then $E\subseteq \mathbb R$.

Khal
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    "Assume by contradiction that $\mathbb R \subset E$" - don't you actually mean $E \not \subset \mathbb R$? – lisyarus Jul 01 '19 at 09:12
  • Thanks, Edited.. – Khal Jul 01 '19 at 09:14
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    There seems to be some steps missing. Why is $e(E) = E$? – Dirk Jul 01 '19 at 09:14
  • I'm not sure where you're pointing at – Khal Jul 01 '19 at 09:18
  • Shouldn't the starting assumption be $E \not\subseteq R$? – Saswat Padhi Jul 01 '19 at 09:34
  • @Khal There are fields $E\subseteq\Bbb{C}$ such that $\epsilon(E)\neq E$. For example $E=\Bbb{Q}(e^{2\pi i/3}\root3\of2)$. Such a field cannot be a Galois extension of $\Bbb{Q}$ though. Do you see why? Take a peek at my answer to the linker question :-) – Jyrki Lahtonen Jul 01 '19 at 10:30
  • Thanks @JyrkiLahtonen, but I'm not sure what the problem with my solution seems to be. I do understand that when the extension is galois, $\epsilon(a)\in Gal(E/\mathbb Q)$ and in the example you posted here, it wouldn't be the case. I feel like the problem is with the formalization of my answer, it's not coherent enough.. Can you be more specific? – Khal Jul 01 '19 at 10:41
  • My only complaint would be that you need to show that $\epsilon(a)\in E$ for all $a\in E$. Otherwise you don't have $\epsilon \in Gal(E/\Bbb{Q})$. – Jyrki Lahtonen Jul 01 '19 at 12:04
  • Or, rather, you claim that $\epsilon(a)\neq a$ implies that $\epsilon$ is in the Galois group. The points are: 1) you need another argument to show why $\epsilon$ is in the Galois group, and 2) $\epsilon(a)\neq a$ is needed to conclude that $\epsilon\neq id_E$. – Jyrki Lahtonen Jul 01 '19 at 12:07
  • I'm not sure how to show point #1. Regarding point #2, $E$ is normal, thus all $\mathbb Q$-conjugates of $a$ are in $E$. Further, $E/F$ is separable, thus all $\mathbb Q$-conjugates of $a$ are different. Assuming $\epsilon(a)\neq a$ makes $\epsilon\neq id_E$.. hopefully :-) – Khal Jul 01 '19 at 13:34
  • I give up.. @JyrkiLahtonen, can you please clarify why these 2 points you mentioned are valid? – Khal Jul 01 '19 at 17:50
  • Shouldn't this be posted as an answer rather than a new question? – David Jul 04 '19 at 09:46

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