# Subfields of $\mathbb{C}$ Galois extension

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.

• $E/E_0$ is Galois, $E_0$ is the fixed field of some subgroup of $Aut(E)$. Which one ? – 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^{

}$$

. Then $$\text{Gal}(E/E_0)=

=\{\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.