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Let E be a subfield of $mathbb{C}$, which is a Galois-extension of $mathbb{Q}$. Let $E_0=E cap mathbb{R}$.
Show that:

i) $[E:E_0]leq2$

ii) Is $[E_0:mathbb{Q}]$ always a Galois-extension?

iii) Let $E_0:mathbb{Q}$ be a Galois-extension and let p be a generating element of $operatorname{Gal}(E/E_0$). p can be also read as an element of $operatorname{Gal}(E/mathbb{Q}$). Show that $psigma=sigma p$ for all $sigma in operatorname{Gal}(E/mathbb{Q})$

For the first one, I can show that $[E:E_0]$ is a galois-extension and that $[E:E_0]$ divides $[E:mathbb{Q}]$. I know $[mathbb{C}:mathbb{R}]=2$, should I use the degree formula for extensions somehow?

For the second one, I would say no, but I can not think of a counter example.
For the third one I really need a hint because I am not sure what I have to show.

We will assume that $Eneq E_0$, which is equivalent to $Enotsubseteqmathbb{R}$.

i) Let $beta$ in $Esetminus E_0$ and let $f(X)$ in $E_0[X]$ its minimal polynomial over $E_0$. Because $E_0subseteq mathbb{R}$, we have that $f(X)$ is in $mathbb{R}[X]$ and so the conjugate $overline{beta}$ is a root of $f(X)$. Thus $overline{beta}$ is in $E$ because $E/E_0$ is normal. Thus conjugation define a morphism $p:xin Emapstooverline{x}in E$ and $E_0$ is fixed by $p$ so $p$ is in $text{Gal}(E/E_0)$.

By the Galois' correspondence $E_0 = E^{text{Gal}(E/E_0)}$ but we also have $E_0=E^{<p>}$. Then $text{Gal}(E/E_0)=<p>={text{id}_E,p}$ has order equal to $2$.

ii) Consider $E=mathbb{Q}[sqrt[3]{2},omega]$, where $omega$ is a primitive cube root of unity. Note that $E$ is the splitting field of the polynomial $X^3-2$ in $mathbb{Q}[X]$ so it is Galois. Check that $E_0=mathbb{Q}[sqrt[3]{2}]$, which is not Galois over $mathbb{Q}$.

iii) Note that the hypothesis $E_0/mathbb{Q}$ Galois is necessary. In the counter example of (ii), we have that $p(sqrt[3]{2})=sqrt[3]{2}$ and $p(omega)=omega^{-1}neq omega$. We know that there exists $sigma$ in $text{Gal}(E/mathbb{Q})$ such that $sigma(sqrt[3]{2})=omegasqrt[3]{2}$. So we have that
$$
sigma p(sqrt[3]{2})= sigma (sqrt[3]{2})= omegasqrt[3]{2} neq omega^{-1}sqrt[3]{2}=p(omega)p(sqrt[3]{2}) = p(omegasqrt[3]{2}) = p sigma (sqrt[3]{2})
$$

However, it is true in the case $E_0/mathbb{Q}$ Galois. In that case we have that $sigma(E_0)=E_0$ and so, for every $a$ in $E_0$, we have that $sigma (a)$ is in $E_0$. Then $p sigma(a)=sigma (a) = sigma p(a)$.

On the other hand, let $beta$ in $E$ such that $E_0[beta]=E$ and let $alpha = p(beta)$. So $sigma(alpha) = sigma p(beta)$.

Let $f(X)=X^2+bX+c$ in $E_0[X]$ the minimal polynomial over $E_0$ of $beta$ and $alpha$. Because $sigma(E_0)=E_0$ we have that $sigma f(X) := X^2+sigma(b)X+sigma(c)$ is in $E_0[X]$ and it is the minimal polynomial over $E_0$ of $sigma(beta)$ and $sigma(alpha)$. Note that $sigma(beta)$ is not in $E_0$ so $psigma(beta)=sigma(alpha)$.

We have proven that $p sigma(beta)=sigma (alpha) = sigma p(beta)$. Thus $psigma$ and $sigma p$ coincide on $E_0$ and $beta$. Now, because $E=E_0[beta]$ they are equal.