Number of automorphisms 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.

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:


*

*$i\mapsto i$

*

*$\sqrt[4]{3}\mapsto\sqrt[4]{3}$

*$\sqrt[4]{3}\mapsto i\sqrt[4]{3}$

*$\sqrt[4]{3}\mapsto -\sqrt[4]{3}$

*$\sqrt[4]{3}\mapsto -i\sqrt[4]{3}$


*$i\mapsto -i$

*

*(same sublist as above)



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 $0\le a,b,c,d\le 3$.
