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 $p\sigma=\sigma p$ 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.

  • 1
    $\begingroup$ $E/E_0$ is Galois, $E_0$ is the fixed field of some subgroup of $Aut(E)$. Which one ? $\endgroup$ – reuns Dec 11 '18 at 17:09

We will assume that $E\neq E_0$, which is equivalent to $E\not\subseteq\mathbb{R}$.

i) Let $\beta$ in $E\setminus E_0$ and let $f(X)$ in $E_0[X]$ its minimal polynomial over $E_0$. Because $E_0\subseteq \mathbb{R}$, 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 $p:x\in E\mapsto\overline{x}\in E$ 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 $\sigma(\sqrt[3]{2})=\omega\sqrt[3]{2}$. So we have that $$ \sigma p(\sqrt[3]{2})= \sigma (\sqrt[3]{2})= \omega\sqrt[3]{2} \neq \omega^{-1}\sqrt[3]{2}=p(\omega)p(\sqrt[3]{2}) = p(\omega\sqrt[3]{2}) = p \sigma (\sqrt[3]{2}) $$

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 $p\sigma(\beta)=\sigma(\alpha)$.

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


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