I'm having difficulties with understanding what automorphisms of field extensions are. I have the splitting field $L=\mathbb{Q}(\sqrt[4]3,i)$ of $X^4-3$ over the rationals.
Now I have to find $\#\mathrm{Aut}(L)$. Is this different from $\#\mathrm{Aut}_\mathbb{Q}(L)$? Also, how do I find the value of any of these? I know that the number of automorphisms is bounded above by $[\ L:\mathbb{Q}\ ]$ (which is $8$, if I'm correct), but other than that I'm stuck.
On
We have that the splitting field of the monic polynomial $f(x) = x^4 - 3$ is $L = \mathbb{Q}(\sqrt[4]{3},i)$. The four roots of $f(x)$ are precisely $\sqrt[4]{3},-\sqrt[4]{3}, i\sqrt[4]{3}$ and $-i\sqrt[4]{3}$. This extension is clearly Galois over $\mathbb{Q}$ as it is the splitting field of an irreducible polynomial. Hence, we have $$|\mathrm{Aut}(L / \mathbb{Q})| = [L : \mathbb{Q}] = 8.$$
Recall that the Galois group is uniquely determined by the action of the generators of $L$, namely $\sqrt[4]{3}$ and $i$. We can choose to map $\sqrt[4]{3}$ to one of the four roots above and fixing $i$, or we can map $i \mapsto -i$. This gives $4 \times 2 = 8$ choices, as claimed.
One can further investigate each possibilities and find the order of all such map $\sigma \in \mathrm{Aut}(L / \mathbb{Q})$ to realize $\mathrm{Gal}(L / \mathbb{Q})$ as the dihedral group of order 8, $D_8$.
An automorphism is determined by the permutation it induces on the sets $\{\sqrt[4]{3},i\sqrt[4]{3},-\sqrt[4]{3},-i\sqrt[4]{3}\}$ and $\{i,-i\}$. The permutation's restriction to the first set is determined by where it sends $\sqrt[4]{3}$ and how it acts on $i$ in the second set. This yields a list of eight possible candidates:
We want to know which of these eight permutations define an automorphism. Thanks to linearity it suffices to argue that $\phi(i^a\sqrt[4]{3}^bi^c\sqrt[4]{3}^d)=\phi(i^a\sqrt[4]{3}^b)\phi(i^c\sqrt[4]{3}^d)$ for appropriate $0\le a,b,c,d\le 3$.