Finite, normal extension of odd degree.

Let $\mathbb{Q} \subseteq E$ be a finite normal extension. Prove that if $(E : \mathbb{Q})=n$ is odd then $E\subseteq \mathbb{R}$.

My attempt:

I am using the fact that a finite normal extension is a splitting field for some polynomial $f$ over $\mathbb{Q}$. Then $E = \mathbb{Q}(a_{1}, a_{2}, .....,a_{n})$, where $a_{i}$ is a root of $f$.

Moreover I know, that the degree $(\mathbb{Q}(a_{i}):\mathbb{Q})$ must divide $(E:\mathbb{Q})=n$ for all $i$, so it has to be odd too.

If any of $a_{i}$ was a complex number, then the degree $(\mathbb{Q}(a_{i}):\mathbb{Q})$ would be $2$ and we would get a contradiction. But is it enough to say that $E\subseteq \mathbb{R}$? Whether my way of thinking correct?

Solution 1:

Hints:

1. If $z\in E$ is arbitrary, then use normality of $E/\Bbb{Q}$ to show that the complex conjugate $\overline{z}\in E$, too.
2. So the restriction of complex conjugation to $E$ is an automorphism of $E$. What possibilities have we got for its order as an automorphism of $E$?
3. Consequently, what possibilities are there for the degree $[E:E\cap\Bbb{R}]$. Can you eliminate all but one of them? Observe that $E\cap\Bbb{R}$ is the fixed field of complex conjugation.