Let $K$ be a field, and let $L$ be the splitting field of $x^4 -2$ over $K$. Find $Aut(L/K)$ when $K=\Bbb{Q}$, $K=\Bbb{F_3}$ and, $K=\Bbb{F_7}$.
I'm very new to the Galois theory. I know that the four roots of $x^4 - 2$ are $\pm\sqrt[4]{2},\pm i\sqrt[4]{2}$ and that $i=1/2\cdot i \sqrt[4]{2} \cdot (\sqrt [4] {2})^3$ and so the splitting field $L$ is $\Bbb{Q}(\sqrt [4] {2},i)$ but I'm having trouble showing each $Aut(L/K)$.
On
For the finite fields case: say $K = \mathbb{F}_3$. Then roots of the polynomial $x^4-2$ are precisely the $x$ such that $x^3 \equiv 2 \bmod 3$. Since $2$ is a cube root of unity in $\mathbb{F}_3$, the solutions of $x^4-2$ are precisely the 8th roots of unity. Now finite extensions of $\mathbb{F}_3$ are of the form $\mathbb{F}_{3^n}$ consisting of $3^n-1$ roots of unity. So $L = \mathbb{F}_{3^n-1}$, where $n$ is the smallest positive integer such that 8 divides $3^n-1$, and it's clear that $n$ should be 2. So $Aut(L/K) = \mathbb{F}_2$. The case $K = \mathbb{F}_7$ is similar.
First show that $F=\Bbb Q(\sqrt[4]2)$ has degree $4$ over $\Bbb Q$. It should be obvious that $|L:F|=2$ though, so $|L:\Bbb Q|=8$.
The group $G=\text{Aut}(L/K)$ has order $8$ and acts as permutations on the set $\{\sqrt[4]2,i\sqrt[4]2,-\sqrt[4]2,-i\sqrt[4]2\}$. But we cannot achieve all permutations: if $\sqrt[4]2$ goes to $\alpha$ then $-\sqrt[4]2$ must go to $-\alpha$. Which permutations can we achieve?