Exercise 6.2.2

Proof.

(a)

As $L=\mathbb{Q}(\omega ,\sqrt[3]{2})$, Proposition 6.1.4(b) shows that $\sigma \in \mathrm{Gal}(L∕\mathbb{Q})$ is uniquely determined by $\sigma (\omega ),\sigma (\sqrt[3]{2})$.

Moreover, by theorem 6.1.4 (a), $\sigma (\omega )$ is a root of $f=x^2+x+1$, whose roots are $\omega ,{\omega }^2$, and $\sigma (\sqrt[3]{2})$ is a root of $g=x^3-2$ whose roots are $\sqrt[3]{2},\omega \sqrt[3]{2},{\omega }^2\sqrt[3]{2}$.

Then Exercise 6.1.1 shows that

$$|\mathrm{Gal}(L∕\mathbb{Q})|\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=6.$$

(b)

$L$ is the splitting field of the separable irreducible polynomial $g=x^3-2\in \mathbb{Q}[x]$. Indeed, $g$ is irreducible over $\mathbb{Q}$ since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=3$ and $g$ has no root in $\mathbb{Q}$. Moreover $g$ is separable since its roots in $ℂ$ are $\sqrt[3]{2},\omega \sqrt[3]{2},{\omega }^2\sqrt[3]{2}$ which are distinct.

By theorem 6.2.1, $|\mathrm{Gal}(L∕\mathbb{Q})|=[L:\mathbb{Q}]$, and by Exercise 5.1.8, $[L:\mathbb{Q}]=2\times 3=6$, therefore

$$|\mathrm{Gal}(L∕\mathbb{Q})|=[L:\mathbb{Q}]=6.$$

If all possible combinations for $\sigma (\omega )$ and $\sigma (\sqrt[3]{2})$ don’t actually occur, then $|\mathrm{Gal}(L∕\mathbb{Q})|<6$, which is false, so all possible combinations occur.