Help with computing Galois group of $x^4 - 3$. Let $f(x) = x^4 - 3$. I believe $Gal(f(x)) = Gal(\mathbb{Q}(\sqrt[4]{3}, i)/\mathbb{Q})$, and then we have 
$$\sigma_1 = 
          \begin{cases}
          \sqrt[4]{3} \rightarrow \zeta^n\sqrt[4]{3}, n = 0,1,2,3\\
          i \rightarrow i
          \end{cases},$$
and complex conjugation as another automorphism. Then our group is of order 8? How do I determine where $\sigma_1$ takes $\zeta$? And does that help me determine the group? I'm not sure how to handle it from here.
Thanks!
 A: Well note that first we need to extend into $\mathbb{Q}(\sqrt{3})$ so that we write
$$
x^4-3 = (x^2 - \sqrt{3})(x^2 + \sqrt{3})
$$
and then we need to extend into $\mathbb{Q}(\sqrt{3},\sqrt[4]{3}) = \mathbb{Q}(\sqrt[4]{3})$ since
$$
(x^2 - \sqrt{3})(x^2 + \sqrt{3}) = (x - \sqrt[4]{3})(x + \sqrt[4]{3})(x^2 + \sqrt{3})
$$
Now if for the sake of seeing where to extend to next, let's work in $\mathbb{C}$ and use the quadratic formula to find the linear factors of $x^2 + \sqrt{3}$:
$$
x_0 = \frac{\pm\sqrt{-4\sqrt{3}}}{2} = \pm i \sqrt[4]{3}
$$
so clearly we need to extend into $\mathbb{Q}(\sqrt[4]{3},i)$.
Now note that this is a separable polynomial (since in it's splitting field there are no repeated roots). This tells us that, denoting $E$ the extension field and $F$ the base field we have
$$
\left| \mathrm{Gal}(E/F) \right| = [E : F]
$$
Since for this polynomial $E = \mathbb{Q}(\sqrt[4]{3},i), F = \mathbb{Q}$ and we have
$$
[\mathbb{Q}(\sqrt[4]{3}) : \mathbb{Q}] = 4, [\mathbb{Q}(\sqrt[4]{3},i) : \mathbb{Q}(\sqrt[4]{3})] = 2 \implies [\mathbb{Q}(\sqrt[4]{3},i) : \mathbb{Q}] = 8
$$
Now an explanation for these above equalities would be needed for a valid proof, I'll leave that to you, but a hint would be to think of the minimal polynomials of the elements that are being adjoined.
So now we know that the size of $\mathrm{Gal}(\mathbb{Q}(\sqrt[4]{3},i) / \mathbb{Q})$ is $8$.
Now for the automorphisms, note that $i \to \pm i$ so you automatically have two automorphisms. Now we have that $\sqrt[4]{3}$ can be sent to any other fourth root of $3$. So since there are four choices for our roots, and two choices for $i$ we have our $8$ automorphisms. Note that defining the automorphisms for just $i$ and $\sqrt[4]{3}$ are sufficient for defining the entire automorphisms (why?). These automorphisms are as follow:
$$
\begin{pmatrix} 3^{1 / 4} \to 3^{1 / 4} \\ i \to i \end{pmatrix}
\begin{pmatrix} 3^{1 / 4} \to i3^{1 / 4} \\ i \to i \end{pmatrix}
\begin{pmatrix} 3^{1 / 4} \to -3^{1 / 4} \\ i \to i \end{pmatrix}
\begin{pmatrix} 3^{1 / 4} \to -i3^{1 / 4} \\ i \to i \end{pmatrix} \\
\begin{pmatrix} 3^{1 / 4} \to 3^{1 / 4} \\ i \to -i \end{pmatrix}
\begin{pmatrix} 3^{1 / 4} \to i3^{1 / 4} \\ i \to -i \end{pmatrix}
\begin{pmatrix} 3^{1 / 4} \to -3^{1 / 4} \\ i \to -i \end{pmatrix}
\begin{pmatrix} 3^{1 / 4} \to -i3^{1 / 4} \\ i \to -i \end{pmatrix} \\
$$
I hope this helps you understand how to determine the group. I usually find the size of the group to see how many elements I need to look for. Next I would think about where the adjoined elements can go. From there we can create any possible automorphism.
Also another comment: If you put $s = \begin{pmatrix} \sqrt[4]{3} \to i\sqrt[4]{3} \\ i \to -i \end{pmatrix}, r = \begin{pmatrix} \sqrt[4]{3} \to i\sqrt[4]{3} \\ i \to i \end{pmatrix}$ we see that this group is isomorphic to $D_4$!
