Order of a non-normal Galois Group

Question

I have $Q(\sqrt[4] 2) / Q$ which is a non-normal extension field. I have already proved that. Now I am supposed to find the order of the Galois group for this. I am unsure how to proceed (and frankly thinking that I may be making it harder than it really is). I keep looking back at my notes for the Fundamental Theorem of Galois Theory, specifically the part that states $[L:F] = \frac {|Gal(E/F|}{|Gal(E/L)|}$. Will $[L:F]$ = $|Gal(L/F)|$ when L is not a normal extension? Can $E = Q(\sqrt[4] 2, \zeta_4), L=Q(\sqrt[4] 2)$, and $F = Q? $Is there a different way that I need to look at this?

Answer 1

$\newcommand{\R}{\mathbb{R}}$$\newcommand{\Q}{\mathbb{Q}}$Note the following.

1. An element of the Galois group is completely determined by its action on $\alpha = \sqrt[4]{2}$.

2. An element of the Galois group takes $\alpha$ to another root of the minimal polynomial of $\alpha$ over $\Q$.

3. This minimal polynomial is $m = x^{4} - 2$.

4. The only roots of $m$ in $\Q(\alpha)$ are $\alpha, -\alpha$, as $\Q(\alpha) \subseteq \R$, and the other roots are non-real.

5. Hence the Galois group has order $2$.