How to show that the degree of this field extension is 12? I need to show that $[\mathbb{Q}(2^{1/4},2^{1/6}):\mathbb{Q}]$ is a field extension of degree $12$. It is possible to show that the degree is at least $12$ because it is divisible by $6$ and $4$ by finding the minimal polynomial of the simple field extensions of $2^{1/4}$ and $2^{1/6}$, but I am not sure how to bound the inequality in the other direction.
Another way to approach this problem might just be to explicitly find the basis, but I think there should be a way to find a bound on the inequality.
 A: You've already done the hard work. Let $K = \mathbb{Q}(2^{1/4}, 2^{1/6})$, let $\alpha = 2^{1/12}$, and let $L = \mathbb{Q}(\alpha)$. Clearly, $K \subset L$, since $2^{1/4} = \alpha^{3}$ and $2^{1/6} = \alpha^{2}$. You've already shown that $[K:\mathbb{Q}] \geqslant 12$, and it's not hard to show (e.g., via Eisenstein) that $[L:\mathbb{Q}] = 12$ by computing the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Since $K$ is a $\mathbb{Q}$-subspace of $L$, we conclude that $[K:\mathbb{Q}] \leqslant 12$, which gives the desired equality.
A: (1) $[\mathbb{Q}(2^{1/4}):\mathbb{Q}]=4$ since $2^{1/4}$ satisfies irreducible
polynomial $x^4-2$ of degree $4$ over $\mathbb{Q}$.
(2) $[\mathbb{Q}(2^{1/6}):\mathbb{Q}]=6$ since $2^{1/6}$ satisfies irreducible
polynomial $x^6-2$ of degree $6$ over $\mathbb{Q}$.
(3) From degrees of extensions  $\mathbb{Q}(2^{1/4})$ and $\mathbb{Q}(2^{1/6})$ over $\mathbb{Q}$, none of them is contained in other.
(4) $\mathbb{Q}(2^{1/2})$ is contained in both fields, and again, $[\mathbb{Q}(2^{1/2}):\mathbb{Q}]=2$.
(5) From (3) and (4), $\mathbb{Q}(2^{1/4}) \cap \mathbb{Q}(2^{1/6})$ is precisely $\mathbb{Q}(2^{1/2})$.
(6) Draw lattice diagram of fields $\mathbb{Q}(2^{1/4}, 2^{1/6})$, $\mathbb{Q}(2^{1/6})$, $\mathbb{Q}(2^{1/4})$ and (intersection) $\mathbb{Q}(2^{1/2})$ [looks like  diamond].
Degrees of parallel sides are same; you can easily deduce your claim.
A: Degrees of field extensions are multiplicative. Can you find $[\mathbb{Q}(2^{1/4}):\mathbb{Q}]$ by explicitly producing the minimal polynomial of $2^{1/4}$ over $\mathbb{Q}$? Not too hard. You'll find the degree of this extension is $4$.
Then can you find $[\mathbb{Q}(2^{1/4},2^{1/6}):\mathbb{Q}(2^{1/4})]$ by finding the minimal polynomial of $2^{1/6}$ over $\mathbb{Q}(2^{1/4})$? Since you know the degree of the full extension should be $12$, the degree of this extension should be $3$. So perhaps a polynomial of degree $3$. To show that the polynomial you get is irreducible over $\mathbb{Q}(2^{1/4})$, simply find its roots in $\mathbb{C}$ and note that they do not lie in $\mathbb{Q}(2^{1/4})$.
