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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?

essay
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user140436
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  • Is $Q$ supposed to be the field of rational numbers, $R$ the reals? – essay May 07 '14 at 18:01
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    What do you know about polynomials of odd degree over the reals? – Nicky Hekster May 07 '14 at 18:04
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    Also, if $a_i$ was a complex number, then $[\mathbb{Q}(a_i), : , \mathbb{Q}]$ would be even, not necessarily equal to $2$, as we don't know, a priori, that $a_i = r_i + i q_i$ where $r_i$, $q_i$ are in $\mathbb{Q}$. – Nicholas Stull May 07 '14 at 18:07
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    If $a_i$ should be one of the complex roots of $x^3-2$, then $[\Bbb{Q}(a_i):\Bbb{Q}]$ would be equal to three. But the field $\Bbb{Q}(a_i)$ is then not normal, because neither of the other roots is in there. Thus it is possible for a non-real field to be an odd degree extension of the rationals. Normality is the key here. – Jyrki Lahtonen May 07 '14 at 18:25
  • In step 3 of my hints below you need bits from Galois theory. Have you covered that yet? – Jyrki Lahtonen May 07 '14 at 18:51
  • No, I have not the Galois theory yet, so probably I should do this the other way. – user140436 May 07 '14 at 20:36
  • Ok. Then you could try to use Nicky Hekster's hint instead. Have you done the result that all finite extensions are simple? In other words: do you know that there exists a number $\alpha\in E$ such that $E=\Bbb{Q}(\alpha)$? What can you say about the field $\Bbb{Q}(\beta)$, where $\beta$ is a real root of the minimal polynomial of $\alpha$? First prove that such a number $\beta$ exists! – Jyrki Lahtonen May 08 '14 at 06:34

1 Answers1

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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.
Jyrki Lahtonen
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  • I don't quite get your answer;

    do you mean that $ [E: E\cap \mathbb{R}] $ has to be odd? That is true by the multiplication formula. And it would imply that the order of conjugation is $ 1 $ if the extension $ E \cap \mathbb{R} \subset E $ was Galois (because the Galois group would be of odd order).

    But do we know that it is?

    I'm trying to understand this problem better. I would appreciate some more clarification

     – Jytug May 17 '15 at 22:36
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    @Jytug: Consider the order of complex conjugation $\sigma$ as an automorphism of $E$. Because $\sigma^2(z)=z$ for all $z\in\Bbb{C}$ the order of $\sigma$ must be a factor of two. That order also has to be a factor of $[E:\Bbb{Q}]$, If the order is one, then all the elements of $E$ must be fixed points of $\sigma$. – Jyrki Lahtonen May 18 '15 at 05:07