Degree extension $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}(\sqrt[3]{2})]$

Question

Find degree of extension $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}(\sqrt[3]{2})]$.

My approach was the following:

Consider the polynomial $x^2-2\in \mathbb{Q}[x]$ and $\sqrt{2}$ is its root;

this shows that $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}(\sqrt[3]{2})]\leq 2$.

I think that this equals $2$, but I cannot prove it rigorously.

I would be very grateful if anyone can show how to solve this problem.

BTW, please do not use any Galois theory, because I am not familiar with it yet.

Answer 1

$\Bbb Q(\sqrt[3]2,\sqrt2)=\Bbb Q(\sqrt[6]2)$, so has degree $6$ over $\Bbb Q$ (Eisenstein). But $\Bbb Q(\sqrt[3]2)$ has degree $3$ over $\Bbb Q$.

Answer 2

The only alternative would be that $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}(\sqrt[3]{2})]=1$, i.e., that $\sqrt{2}\in \Bbb Q(\sqrt[3]{2})$ and hence $\Bbb Q(\sqrt{2}\subseteq \Bbb Q(\sqrt[3]{2})$. However, this is impossible, because of $$ 3=[\Bbb Q(\sqrt[3]{2}):\Bbb Q]=[\Bbb Q(\sqrt[3]{2}):\Bbb Q(\sqrt{2})]\cdot [\Bbb Q(\sqrt{2}):\Bbb Q]. $$

Answer 3

Use the tower law. You know that $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}]=6$.

This is because $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$

and $[\mathbb{Q}(\sqrt[2]{2}):\mathbb{Q}]=2$, and 2 and 3 are coprime, which in turn implies $[\mathbb{Q}(\sqrt[3]{2},\sqrt{2}):\mathbb{Q}]=6$.

Now use the tower law to deduce the degree is 2.