Exercise 7.2.1

Proof.

(a)

By Section 6.4.A (or Exercises 6.2.2 and 6.3.1), there exists $\sigma ,\tau \in \mathrm{Gal}(L∕\mathbb{Q})$ uniquely determined by $$\sigma (\omega )=\omega ,\sigma (\sqrt[3]{2})=\omega \sqrt[3]{2},$$

$$\tau (\omega )={\omega }^2,\tau (\sqrt[3]{2})=\sqrt[3]{2},$$

and $G=\mathrm{Gal}(L∕\mathbb{Q})=⟨\sigma ,\tau ⟩$.

Let $K=\mathbb{Q}(\sqrt[3]{2})$. We show that $\mathit{\sigma K}=\mathbb{Q}(\omega \sqrt[3]{2})$.

If $\beta \in \mathit{\sigma K},\beta =\sigma (\alpha ),\alpha \in K=\mathbb{Q}[\sqrt[3]{2}]$, thus $\alpha =p(\sqrt[3]{2}),p\in \mathbb{Q}[x]$, $\beta =\sigma (p(\sqrt[3]{2}))=p(\sigma (\sqrt[3]{2}))=p(\omega \sqrt[3]{2})\in \mathbb{Q}(\omega \sqrt[3]{2})$, consequently $\mathit{\sigma K}\subset \mathbb{Q}(\omega \sqrt[3]{2})$.

Conversely, if $\beta \in \mathbb{Q}(\omega \sqrt[3]{2})=\mathbb{Q}[\omega \sqrt[3]{2}]$, $\beta =p(\omega \sqrt[3]{2}),p\in \mathbb{Q}[x]$, then $\beta =\sigma (p(\sqrt[3]{2}))=\sigma (\alpha )$, where $\alpha =p(\sqrt[3]{2})\in \mathbb{Q}(\sqrt[3]{2})$, consequently $\mathbb{Q}(\omega \sqrt[3]{2})\subset \mathit{\sigma K}$.

$$\mathit{\sigma K}=\mathbb{Q}(\omega \sqrt[3]{2}).$$

As ${\sigma }^2(\sqrt[3]{2})={\omega }^2\sqrt[3]{2}$, we obtain similarly

$${\sigma }^{2}K=\mathbb{Q}({\omega }^2\sqrt[3]{2}),$$

and of course, $\mathit{eK}=K$. So $\mathbb{Q}(\sqrt[3]{2}),\mathbb{Q}(\omega \sqrt[3]{2}),\mathbb{Q}({\omega }^2\sqrt[3]{2})$ are conjugates fields of $K$ over $\mathbb{Q}$.

As $\mathit{\tau K}=K$, and $G=⟨\sigma ,\tau ⟩$, they are the only ones.

Conclusion:

the conjugate fields of $\mathbb{Q}(\sqrt[3]{2})$ in the extension $\mathbb{Q}\subset \mathbb{Q}(\sqrt[3]{2})$ are $\mathbb{Q}(\sqrt[3]{2}),\mathbb{Q}(\omega \sqrt[3]{2}),\mathbb{Q}({\omega }^2\sqrt[3]{2})$.

(b)

As $\sigma (\omega )=\omega$ and as $\sigma$ is the identity on $\mathbb{Q}$, $\mathit{\sigma \mathbb{Q}}(\omega )=\mathbb{Q}(\omega )$. Moreover $\mathit{\tau \mathbb{Q}}(\omega )=\mathbb{Q}({\omega }^2)$. Since ${\omega }^2=-1-\omega$, $\mathbb{Q}({\omega }^2)=\mathbb{Q}(\omega )$. As $\mathit{\sigma \mathbb{Q}}(\omega )=\mathbb{Q}(\omega ),\mathit{\tau \mathbb{Q}}(\omega )=\mathbb{Q}(\omega )$, and as $G=⟨\sigma ,\tau ⟩$, $\mathit{\lambda \mathbb{Q}}(\omega )=\mathbb{Q}(\omega )$ for all $\lambda \in \mathrm{Gal}(L∕F)$.

The only conjugate field of $\mathbb{Q}(\omega )$ is so $\mathbb{Q}(\omega )$.

Note: As $\mathbb{Q}\subset \mathbb{Q}(\omega )$ is a quadratic extension, thus a normal extension (Ex. 7.1.12), by Theorem 7.2.5, $K=\mathit{\sigma K}$ for all $\sigma \in \mathrm{Gal}(\mathbb{Q}(\omega ,\sqrt[3]{2})∕\mathbb{Q})$. We find again that the only conjugate field of $\mathbb{Q}(\omega )$ is $\mathbb{Q}(\omega )$.