Exercise 6.3.3

Proof. We have already proved (Ex. 5.1.13) that $f=x^2-3$ is irreducible over $\mathbb{Q}[\sqrt{2}]$. $f$ is so the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{2})$, thus $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=2$.

As $g=x^2-2$ is irreducible over $\mathbb{Q}$, $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=2$, therefore

Moreover, $h=x^2+1$ has no real root, a fortiori $h$ has no root in $\mathbb{Q}(\sqrt{2},\sqrt{3})$. As $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)=2$, $h$ is irreducible over $\mathbb{Q}(\sqrt{2},\sqrt{3})$, $h$ is the minimal polynomial of $i$ over $\mathbb{Q}(\sqrt{2},\sqrt{3})$, thus $[\mathbb{Q}(\sqrt{2},\sqrt{3},i):\mathbb{Q}(\sqrt{2},\sqrt{3})]=2$, and by the Tower Theorem, and the equality $\mathbb{Q}(\sqrt{2},\sqrt{3},i)=\mathbb{Q}(i,\sqrt{2},\sqrt{3})$,

$L=\mathbb{Q}(i,\sqrt{2},\sqrt{3})$ is the splitting field of $p=(x^2+1)(x^2-2)(x^2-3)$ over $\mathbb{Q}$, and $p=(x-i)(x+i)(x-\sqrt{2})(x+\sqrt{2})(x-\sqrt{3})(x+\sqrt{3})$ is separable. By theorem 6.2.1, we obtain

If $\sigma \in \mathrm{Gal}(L∕\mathbb{Q})$, $\sigma (i)=\pm i,\sigma (\sqrt{2})=\pm \sqrt{2},\sigma (\sqrt{3})=\pm \sqrt{3}$. As $|\mathrm{Gal}(L∕\mathbb{Q})|=8$, all of these possibilities occur: there exist 8 $\mathbb{Q}$-automorphisms of $L$ satisfying these equalities. As $L=\mathbb{Q}(i,\sqrt{2},\sqrt{3})$, $\sigma \in \mathrm{Gal}(L:\mathbb{Q})$ is uniquely determined by the images of these 3 elements.

In particular, there exist ${\sigma }_1,{\sigma }_2,{\sigma }_3\in \mathrm{Gal}(L:\mathbb{Q})$ defined by

and $1_L,{\sigma }_1,{\sigma }_2,{\sigma }_3,{\sigma }_1{\sigma }_2,{\sigma }_1{\sigma }_3,{\sigma }_2{\sigma }_3,{\sigma }_1{\sigma }_2{\sigma }_3$ give distinct images to $i,\sqrt{2},\sqrt{3}$, thus

Therefore

Note that ${\sigma }_1{\sigma }_2={\sigma }_2{\sigma }_1$ since they give the same images to $i,\sqrt{2},\sqrt{3}$. Similarly ${\sigma }_1{\sigma }_3={\sigma }_3{\sigma }_1$ and ${\sigma }_2{\sigma }_3={\sigma }_3{\sigma }_2$. Thus $G$ is abelian, generated by 3 elements of order 2, with ${\sigma }_2\notin ⟨{\sigma }_1⟩,{\sigma }_3\notin ⟨{\sigma }_1,{\sigma }_2⟩$. Therefore $G$ is the direct sum of the 3 subgroups $\{e,{\sigma }_i\},i=1,2,3$, of degree 2 :

Some instructions Sage and Gap to verify these results :

f=(x-i-sqrt(2)-sqrt(3))*(x-i-sqrt(2)+sqrt(3))*(x-i+sqrt(2)-sqrt(3))
*(x-i+sqrt(2)+sqrt(3))*(x+i-sqrt(2)-sqrt(3))*(x+i-sqrt(2)+sqrt(3))
*(x+i+sqrt(2)-sqrt(3))*(x+i+sqrt(2)+sqrt(3));f

$\left(x+\sqrt{3}+\sqrt{2}+i\right)\left(x+\sqrt{3}+\sqrt{2}-i\right)\left(x+\sqrt{3}-\sqrt{2}+i\right)\left(x+\sqrt{3}-\sqrt{2}-i\right)$
$\left(x-\sqrt{3}+\sqrt{2}+i\right)\left(x-\sqrt{3}+\sqrt{2}-i\right)\left(x-\sqrt{3}-\sqrt{2}+i\right)\left(x-\sqrt{3}-\sqrt{2}-i\right)$

g=f.expand();g

$x^8-16x^6+88x^4+192x^2+144$

g.factor()

$x^8-16x^6+88x^4+192x^2+144$

x=polygen(QQ,’x’)
K.<z> = NumberField(x^8-16*x^6+88*x^4+192*x^2+144)
G = K.galois_group();G

$⟨(1,2)(3,4)(5,6)(7,8),(1,3)(2,4)(5,7)(6,8),(1,5)(2,6)(3,7)(4,8)⟩$

With Gap :

G:=Group((1,2)(3,4)(5,6)(7,8),(1,3)(2,4)(5,7)(6,8),(1,5)(2,6)(3,7)(4,8));
StructureDescription(G);

$C_2\times C_2\times C_2$.

As $x^8-16x^6+88x^4+192x^2+144$ is irreductible over $\mathbb{Q}$, $[\mathbb{Q}(i+\sqrt{2}+\sqrt{3}):\mathbb{Q}]=8=[L:\mathbb{Q}]$, and since $\mathbb{Q}(i+\sqrt{2}+\sqrt{3})\subset L$, $L=\mathbb{Q}(i+\sqrt{2}+\sqrt{3})$.

These results imply that $L=\mathbb{Q}(i,\sqrt{2},\sqrt{3})$ is the splitting field of the irreducible polynomial $x^8-16x^6+88x^4+192x^2+144$, that $i+\sqrt{2}+\sqrt{3}$ is a primitive element of $\mathbb{Q}\subset L$, and that $\mathrm{Gal}(L∕\mathbb{Q})\simeq C_2\times C_2\times C_2.$ □