Compute the Galois Group of $\mathbb{Q}\left(2^{\frac{1}{4}}\right)$ over $\mathbb{Q}$. I need to see where exactly the basis elements are being mapped. I refered MORANDI and got to know that the any $\mathbb{Q}-$automorphism will permute the roots . So does that mean that each such permutation will give me one $\mathbb{Q}-$automorphism?
How do I get a basis for this extension? Since I am starting galois theory I would request to give an easy explaination without using the symmetric groups here. I refered other answers but wasn't able to understand.
First note that $\sqrt[4]2$ is a root of the polynomial $f(x)=x^4-2$. The roots of $f$ are $\sqrt[4]2,i\sqrt[4]2,-\sqrt[4]2,-i\sqrt[4]2$.
Let $\sigma\in\newcommand\Gal{\operatorname{Gal}}\Gal(\Bbb Q(\sqrt[4]2)/\Bbb Q)$. Then $\sigma(\sqrt[4]2)$ is a root of $f$. Since $\sigma(\sqrt[4]2)\in\Bbb Q(\sqrt[4]2)\subseteq\Bbb R$, we have $\sigma(\sqrt[4]2)\in\{\sqrt[4]2,-\sqrt[4]2\}$. Since $\sigma$ is dermined by $\sigma(\sqrt[4]2)$, it follows that $\Gal(\Bbb Q(\sqrt[4]2)/\Bbb Q)$ has order two, hence it's isomorphic to the cyclic group of oder $2$.