Degree of $\mathbb{Q}(\sqrt[3]{2},\sqrt{2})$ over $\mathbb{Q}$

Question

What is the degree of $\mathbb{Q}(\sqrt[3]{2},\sqrt{2})$ over $\mathbb{Q}$

Corrected:

My proposed basis is {$1,\sqrt{2}, \sqrt[3]{2}, 2^{\frac{2}{3}}, 2^{\frac{5}{6}}, 2^{\frac{1}{6}}$}

Thus my claim is that the degree is $6$.

Is my proposed basis correct ? If not do point out my mistake(s) (with explanation if possible).

If the proposed basis is correct, how do I show linear in-dependency ? (I tried the linear algebra argument but got stuck).

Any help or insight is deeply appreciated

Answer 1

You can use the degree of tower of extensions and that $\gcd(2,3)=1$.

Let $K = \mathbb{Q}(\sqrt[3]{2})$, $L = \mathbb{Q}(\sqrt[3]{2},\sqrt{2}),F= \mathbb{Q}(\sqrt{2})$. Show that $x^3-2$ is irreducible and $[K:\mathbb{Q}] = 3$.

Assume $\sqrt{2} \in K$ so that $K = L$ and $[L:\mathbb{Q}] = 3$. But $L=F(\sqrt[3]{2})$ and $[L:\mathbb{Q}]=[L:F][F:\mathbb{Q}] = 2 [L:F]$, a contradiction.

Hence $\sqrt{2} \not\in \mathbb{Q}(\sqrt[3]{2})$ so that $X^2-2$ is irreducible over $K$, $\ \ L =K(\sqrt{2}) \cong K[x]/(x^2-2)$, $\ \ [L:K] = 2$ and $[L:\mathbb{Q}]=[L:K][K:\mathbb{Q}]=6$