The degree extension $Q(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2})$

Question

So as in the title i am looking for the extension $Q(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2})$. First of all i know that $[Q(\sqrt[3]{2}):Q]=3$ as $x^3-3$ is irreducable and so is the minimal polynomial. I also know that $[Q(\sqrt{2},\sqrt[3]{2}):Q(\sqrt[3]{2})]=2$ By the fact that $(a+b\sqrt[3]{2}+c\sqrt[3]{2}^2)^2=2$ is never the case and so $x^2-2$ is the minimal polynomial. Now the problem is finding $[Q(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2}):Q(\sqrt{2},\sqrt[3]{2})]$. It is not obvious to me that $(a+b2^{5/6}+c2^{1/2}+d2^{1/3})^4=2$ is never the case. I dont know how to prove it is impossible, algebraically it is awful. I know that $x^2-\sqrt{2}$ would work but i dont know how to prove it is irreducable... I am doing it the wrong way?

Answer 1

Note that $\sqrt2\in \Bbb Q(\sqrt[4]{2})$, so you can actually ignore $\sqrt2$. Now, we have $$ [\Bbb Q(\sqrt[3]2,\sqrt[4]{2}):\Bbb Q]=[\Bbb Q(\sqrt[3]2,\sqrt[4]{2}):\Bbb Q(\sqrt[3]{2})]\cdot[\Bbb Q(\sqrt[3]2):\Bbb Q]\\ =3[\Bbb Q(\sqrt[3]2,\sqrt[4]{2}):\Bbb Q(\sqrt[3]{2})] $$ but also $$ [\Bbb Q(\sqrt[3]2,\sqrt[4]{2}):\Bbb Q]=[\Bbb Q(\sqrt[3]2,\sqrt[4]{2}):\Bbb Q(\sqrt[4]{2})]\cdot[\Bbb Q(\sqrt[4]2):\Bbb Q]\\ =4[\Bbb Q(\sqrt[3]2,\sqrt[4]{2}):\Bbb Q(\sqrt[4]{2})] $$ So $[\Bbb Q(\sqrt[3]2,\sqrt[4]{2}):\Bbb Q]$ is divisible by both $3$ and $4$ and must therefore be divisible by $12$. Finally, $[\Bbb Q(\sqrt[3]2,\sqrt[4]{2}):\Bbb Q(\sqrt[4]{2})]\leq3$ since $x^3-2$ is a polynomial over $\Bbb Q(\sqrt[4]{2})$, so answer can't be more than $12$ either.