Find $[ \, \mathbb Q ( \sqrt[3]3 , \eta ) : \mathbb Q \, ] $

Question

Let $K=\mathbb Q ( \sqrt[3]3 , \eta )$ where $\eta = (e^{\frac{\pi}3i})^2$.

I want to find $[K: \mathbb Q ( \sqrt[3]3 )]$

This is apparently $[K: \mathbb Q ( \sqrt[3]3 )] \le 2$ but I don't know why!!

I am guessing the number $2$ comes from the fact that $[\mathbb Q(\eta ): \mathbb Q]=2$ but still don't know how.

Answer 1

Note that $\eta=i$ and consider the tower $$\mathbb{Q}\subseteq\mathbb{Q}(\sqrt[3]{3})\subseteq\mathbb{Q}(\sqrt[3]{3})(i)=K.$$

Since $i\notin\mathbb{Q}(\sqrt[3]{3})$ the degree of $i$ over $\mathbb{Q}(\sqrt[3]{3})$ is at least $2$. But $i$ satisfies the polynomial $x^2+1$ so the degree is at most $2$. Hence $[K:\mathbb{Q}(\sqrt[3]{3})]=2$ and clearly $[\mathbb{Q}(\sqrt[3]{3}):\mathbb{Q}]=3$ since $\sqrt[3]{3}$ has minimal polynomial $x^3-3$ over $\mathbb{Q}$. It follows that $[K:\mathbb{Q}]=3(2)=6$.

Edit: OP changes $\eta=e^{2\pi i/3}$

Similar reasoning applies. Note that the degree of $\eta$ over $\mathbb{Q}(\sqrt[3]{3})$ is still $2$ because $\eta\notin\mathbb{Q}(\sqrt[3]{3})$ and $\eta$ satsifies the polynomial $x^2+x+1$. Like before the degree of $\sqrt[3]{3}$ over $\mathbb{Q}$ is $3$. So $[K:\mathbb{Q}]=3(2)=6$.

Answer 2

You're basically looking for $K=\mathbb{Q}(i, \sqrt[3]{3})$. $[K:\mathbb{Q}]=[K:\mathbb{Q}(\sqrt[3]{3})][\mathbb{Q}(\sqrt{3}[3]):\mathbb{Q}] \leq 6$. Why is that? $[K:\mathbb{Q}(\sqrt[3]{3})]=[(\mathbb{Q}(\sqrt[3]{3}))(i):\mathbb{Q}(\sqrt[3]{3})]$, which is less or equal to $2$, since $i$ satsifies $x^{2}+1$ in $\mathbb{Q}(\sqrt[3]{3})[x]$. Now, the degre $[K:\mathbb{Q}]$ is divisible by $2$ and $3$ as it has subfields of degree $2$ and $3$ over $\mathbb{Q}$ and hence by $6$, since $gcd(2,3)=1$. Hence the extension has degree 6.