Exercise 6.3.1

Proof. $L=\mathbb{Q}(\omega ,\sqrt[3]{2})$ is the splitting field over $\mathbb{Q}$ of $f=x^3-2$.

By Exercise 6.2.2, there exist $\sigma ,\tau \in \mathrm{Gal}(L∕\mathbb{Q})$ such that

Number the roots of $f$ by ${\alpha }_1=\sqrt[3]{2},{\alpha }_2=\omega \sqrt[3]{2},{\alpha }_3={\omega }^2\sqrt[3]{2}$.

Then $\sigma ({\alpha }_1)={\alpha }_2,\sigma ({\alpha }_2)={\alpha }_3,\sigma ({\alpha }_3)={\alpha }_1$. If we write $\tilde{\sigma }=(1,2,3)$, then for $i=1,2,3,\sigma ({\alpha }_i)=\sigma ({\alpha }_{\tilde{\sigma }(i)})$, so the 3-cycle $\tilde{\sigma }=(1,2,3)$ corresponds to $\sigma$.

$\tau ({\alpha }_1)={\alpha }_1,\tau ({\alpha }_2)={\alpha }_3,\tau ({\alpha }_3)={\alpha }_2$, so $\tilde{\tau }=(2,3)$ corresponds to $\tau$.

As $S_3$ is generated by $\tilde{\sigma },\tilde{\tau }$, $\mathrm{Gal}(L∕\mathbb{Q})\simeq S_3$ is generated by $\sigma ,\tau$. □