Exercise 6.3.2

Proof.

(a)

$\mathrm{Gal}(\mathbb{Q}(i,\sqrt{2})∕\mathbb{Q})=\{1_{\mathbb{Q}},\sigma ,\tau ,\mathit{\sigma \tau }\}$, where $$\begin{array}{cc}\sigma (i)=-i,\hfill & \sigma (\sqrt{2})=\sqrt{2},\hfill \\ \tau (i)=i,\hfill & \tau (\sqrt{2})=-\sqrt{2}.\hfill \end{array}$$

As every $g\in \mathrm{Gal}(\mathbb{Q}(i,\sqrt{2})∕\mathbb{Q})$ satisfies $g^2=1_{\mathbb{Q}}$,

$$\mathrm{Gal}(\mathbb{Q}(i,\sqrt{2})∕\mathbb{Q})\simeq \mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$$

(Klein’s ViertelGruppe: cf Exercise 6.2.1 and Example 6.2.2 for more details).

If we number the roots by ${\alpha }_1=i,{\alpha }_2=-i,{\alpha }_3=\sqrt{2},{\alpha }_4=-\sqrt{2}$, then $(1,2)$ corresponds to $\sigma$, and $(3,4)$ to $\tau$.

As subgroup of $S_4$, $\mathrm{Gal}(Q(i,\sqrt{2})∕\mathbb{Q})$ is represented by

$$\{(),(1,2),(3,4),(1,2)(3,4)\}=⟨(1,2),(3,4)⟩\simeq \mathrm{Gal}(\mathbb{Q}(i,\sqrt{2})∕\mathbb{Q}).$$

(b)

$$\begin{array}{llll}\hfill f& =x^4-2& \hfill & \\ \hfill & =(x^2-\sqrt{2})(x^2+\sqrt{2})& \hfill & \\ \hfill & =(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x+i\sqrt[4]{2})(x-i\sqrt[4]{2})& \hfill & \end{array}$$

The splitting field of $f$ over $\mathbb{Q}$ is thus $L=\mathbb{Q}(\sqrt[4]{2},i\sqrt[4]{2})=\mathbb{Q}(i,\sqrt[4]{2})$. $f$ is separable, since $f$ has simple roots in its splitting field. $L$ is so the splitting field over $\mathbb{Q}$ of a separable polynomial, therefore by Theorem 6.2.1,

$$|\mathrm{Gal}(L:\mathbb{Q})|=[L:\mathbb{Q}].$$

$f$ is irreducible over $\mathbb{Q}$ by the Sch $ö$nemann-Eisenstein Criterion with $p=2$. As $f$ is irreducible over $\mathbb{Q}$,

$$[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=4,$$

and $x^2+1$ is irreducible over $\mathbb{Q}(\sqrt[4]{2})$, since it is of degree 2, without root in $\mathbb{Q}(\sqrt[4]{2})\subset ℝ$, thus

$$[\mathbb{Q}(i,\sqrt[4]{2}):\mathbb{Q}(\sqrt[4]{2})]=2.$$

Consequently,

$$[\mathbb{Q}(i,\sqrt[4]{2}):\mathbb{Q}]=[\mathbb{Q}(i,\sqrt[4]{2}):\mathbb{Q}(\sqrt[4]{2})][\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=8,$$

and so

$$|\mathrm{Gal}(L:\mathbb{Q})|=8.$$

If $\sigma \in \mathrm{Gal}(L∕\mathbb{Q})$, as $i$ is a root of $x^2+1\in \mathbb{Q}[x]$, and $\sqrt[4]{2}$ a root of $x^4-2\in \mathbb{Q}[x]$, then $\sigma (i)=\pm i$, and $\sigma (\sqrt[4]{2})=i^k\sqrt[4]{2},k=0,1,2,3$.

As $\sigma$ is uniquely determined by the images of $i,\sqrt[4]{2}$, and as $|\mathrm{Gal}(L:\mathbb{Q})|=8$, these 8 possibilities occur, thus $G=\mathrm{Gal}(L∕\mathbb{Q})=\{{\sigma }_{j,k}|0\le j\le 1,0\le k\le 3\}$, where ${\sigma }_{j,k}$, which is identity on $\mathbb{Q}$, is determined by

$${\sigma }_{j,k}(i)={(-1)}^{j}i,{\sigma }_{j,k}(\sqrt[4]{2})=i^k\sqrt[4]{2}.$$

Write $\tau :L\to L,z\mapsto \bar{z}$ the complex conjugation restricted to $L$. $\tau$ is a ring homomorphism and an involution, thus $\tau$ is a field automorphism of $L$, which is identity on $\mathbb{Q}$, so $\tau \in G$. Moreover

$$\tau (i)=-i,\tau (\sqrt[4]{2})=\sqrt[4]{2}.$$

Let $\sigma \in \mathrm{Gal}(L∕\mathbb{Q})$ defined by

$$\sigma (i)=i,\sigma (\sqrt[4]{2})=i\sqrt[4]{2}.$$

Then $\tau ={\sigma }_{1,0},\sigma ={\sigma }_{0,1}$.

As ${\tau }^2=e$ and $\tau \ne e$ (where $e=1_L$), the order of $\tau$ is 2.

${\sigma }^4(i)=i$ and ${\sigma }^4(\sqrt[4]{2})=\sqrt[4]{2}$, thus ${\sigma }^4=e$. As ${\sigma }^2(\sqrt[4]{2})=i^2\sqrt[4]{2}=-\sqrt[4]{2},{\sigma }_2\ne e$, thus the order of $\sigma$ is 4.

$$|\tau |=2,|\sigma |=4.$$

As $\tau (i)=-i,\tau \notin ⟨\sigma ⟩$. Thus the subgroup $⟨\sigma ,\tau ⟩$ of $G$ contains at least 5 elements, so is equal to $G$ by Lagrange’s Theorem:

$$G=⟨\sigma ,\tau ⟩.$$

As the index of $H=⟨\sigma ⟩$ in $G$ is 2, and $\tau \notin H$, $G=H\cup \mathit{\tau H}$ :

$$G=\mathrm{Gal}(L∕\mathbb{Q})=\{1_L,\sigma ,{\sigma }^2,{\sigma }^3,\tau ,\mathit{\tau \sigma },\tau {\sigma }^2,\tau {\sigma }^3\}.$$

If we number the roots of $f$ by ${\alpha }_k=i^{k-1}\sqrt[4]{2}$, for $k=1,2,3,4$, then $\tau$ corresponds to the transposition $(2,4)$, and $\sigma$ to the cycle $(1,2,3,4)$ :

$$G\simeq ⟨(1,2,3,4),(2,4)⟩\subset S_4.$$

If we number the 4 summits of a square by 1,2,3,4 in the direct orientation, then $\sigma$ corresponds to a rotation of angle $\pi ∕2$, and $\tau$ to a symmetry with respect to the diagonal $[1,3]$. They generate the group of isometry of the square, which is the dihedral group $D_8$, defined also by generators and relations:

$$G=⟨\sigma ,\tau ⟩,{\sigma }^4={\tau }^2=e,\mathit{\tau \sigma }={\sigma }^{-1}\tau .$$

(Since ${\tilde{\sigma }}^{-1}\tilde{\tau }=(1,4,3,2)(2,4)=(1,4)(2,3)=(2,4)(1,2,3,4)=\tilde{\tau }\tilde{\sigma }$.)

As a verification, the following GAP instruction confirm the result $D_8$ :

     G:= Group((1,2,3,4),(2,4));
     StructureDescription(G);
       "D8"