Exercise 7.2.11

Proof.

(a)

We have proved in Exercise 6.1.2 that $|\mathrm{Gal}(L∕\mathbb{Q})|=4$, and $$\mathrm{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})∕\mathbb{Q})=\{1_L,\sigma ,\tau ,\mathit{\sigma \tau }\}.$$

where

$\begin{array}{cc}\sigma (\sqrt{2})=\sqrt{2},\hfill & \sigma (\sqrt{3})=-\sqrt{3},\hfill \\ \tau (\sqrt{2})=-\sqrt{2},\hfill & \tau (\sqrt{3})=\sqrt{3}.\hfill \end{array}$

and (Ex. 6.2.1) that $G=\mathrm{Gal}(L∕\mathbb{Q})\simeq \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2∕Z$.

(b)

The subgroups of $G=\mathrm{Gal}(L∕\mathbb{Q})$ are $\{e\},G,⟨\sigma ⟩=\{e,\sigma \},⟨\tau ⟩=\{e,\tau \},⟨\mathit{\sigma \tau }⟩=\{e,\mathit{\sigma \tau }\}.$

(c)

We obtain the right diagram from the left diagram by the map $H\mapsto L_H$. Explicitely:

$L_{\{e\}}=L$, and as $\mathbb{Q}\subset L$ is Galois, $L_G=\mathbb{Q}$.

As $(1,\sqrt{3})$ is a basis of $L$ over $\mathbb{Q}(\sqrt{2})$, a basis of the $\mathbb{Q}$-vector space $L$ is $(1,\sqrt{2},\sqrt{3},\sqrt{6})$. Let $\alpha =a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}(a,b,c,d\in \mathbb{Q})$ any element of $L$. Then

$$\begin{array}{llll}\hfill \sigma (\alpha )=\alpha & ⟺a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}& \hfill & \\ \hfill & ⟺c=d=0& \hfill & \\ \hfill & ⟺\alpha \in \mathbb{Q}(\sqrt{2})& \hfill & \end{array}$$

thus $L_{⟨\sigma ⟩}=\mathbb{Q}(\sqrt{2})$. We verify similarly $L_{⟨\tau ⟩}=\mathbb{Q}(\sqrt{3})$.

We compute $L_{⟨\mathit{\sigma \tau }⟩}$:

$$\begin{array}{llll}\hfill (\mathit{\sigma \tau })(\alpha )=\alpha & ⟺a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}& \hfill & \\ \hfill & ⟺b=c=0& \hfill & \\ \hfill & ⟺\alpha \in \mathbb{Q}(\sqrt{6})& \hfill & \end{array}$$

We obtain the left diagram from the right diagram by the map $K\mapsto \mathrm{Gal}(L∕K)$. For instance, the only elements of $G$ who fix $\mathbb{Q}(\sqrt{2})$ are $e$ and $\sigma$.

(d)

$G$ is Abelian, so all its subgroups are normal.

This implies (Theorem 7.2.5) that $\mathbb{Q}(\sqrt{2})$ equals all of its conjugates and so is a normal extension of $\mathbb{Q}$. Same conclusion for $\mathbb{Q}(\sqrt{3}),\mathbb{Q}(\sqrt{6})$.