Exegesis. degree of $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 })$ over $\mathbb{Q}$ [1505BB]

We have $$ [\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}] = [\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}(\sqrt[3]{ 2 })] [\mathbb{Q}(\sqrt[3]{ 2 }) : \mathbb{Q}] = 2 \times 3 = 6. $$ $[\mathbb{Q}(\sqrt[3]{ 2 }) : \mathbb{Q}] = 3$ is because $m_{\sqrt[3]{ 2 }, \mathbb{Q}}(x) = x^{3} - 2$, which has degree $3$, and then by Proposition 1407.

Similarly, $x^{2} + 3$ is monic and irreducible in $\mathbb{Q}[x]$, so $m_{\sqrt{ -3 }, \mathbb{Q}}(x) = x^{2} + 3$. Moreover, $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[3]{ 2 })$, hence we must have $m_{\sqrt{ -3 }, \mathbb{Q}(\sqrt[3]{ 2 })}(x) \mid m_{\sqrt{ -3 }, \mathbb{Q}}(x)$, which means either $\deg m_{\sqrt{ -3 }, \mathbb{Q}(\sqrt[3]{ 2 })}(x) = 1$ or $2$. The former case implies $x - \sqrt{ -3 } \in \mathbb{Q}\left( \sqrt[3]{ 2 } \right)[x]$, which means $\sqrt{ -3 } \in \mathbb{Q}\left( \sqrt[3]{ 2 } \right)$. But $\sqrt{ -3 }$ is a pure imaginary number, it cannot exist in $\mathbb{Q}\left( \sqrt[3]{ 2 } \right)$. Hence $\deg m_{\sqrt{ -3 }, \mathbb{Q}\left( \sqrt[3]{ 2 } \right)}(x) = 2$, which means $[\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) : \mathbb{Q}(\sqrt[3]{ 2 })] = 2$.

It is easy to see that $\mathbb{Q}(\sqrt{ -3 }), \mathbb{Q}(\theta_{1}), \mathbb{Q}(\theta_{2})$ and $\mathbb{Q}(\theta_{3})$ are all proper subfields of $\mathbb{Q}(\sqrt{ 2 }, \sqrt{ -3 })$ and $$ [\mathbb{Q}(\theta_{j}) : \mathbb{Q}] = 3 \quad (j = 1, 2, 3), \qquad [\mathbb{Q}(\sqrt{ -3 }) : \mathbb{Q}] = 2. $$