A faster way to find the quadratic extensions of a field extension? The usual method I have for finding Galois correspondence goes like this : 
Say we have the Splitting field of $x^4-3$, i.e. $\Bbb Q(i,\sqrt[4]{3})$.
Then it's generating automorphisms are those that permute the elements of the roots of the minimal polynomial while keeping the other fixed, in this case they are $\sigma:i\rightarrow i, \sqrt[4]{3}\rightarrow i\sqrt[4]{3} $ and 
$\tau :i\rightarrow - i, \sqrt[4]{3}\rightarrow \sqrt[4]{3} $ they generate the eight automorphisms of the Galois group:
$\phi_1:i\rightarrow i, \sqrt[4]{3}\rightarrow \sqrt[4]{3} $
$\phi_2:i\rightarrow i, \sqrt[4]{3}\rightarrow i\sqrt[4]{3} $
$\phi_3:i\rightarrow  i, \sqrt[4]{3}\rightarrow -\sqrt[4]{3} $
$\phi_4:i\rightarrow  i, \sqrt[4]{3}\rightarrow -i\sqrt[4]{3} $
$\phi_5:i\rightarrow - i, \sqrt[4]{3}\rightarrow \sqrt[4]{3} $
$\phi_6:i\rightarrow - i, \sqrt[4]{3}\rightarrow i\sqrt[4]{3} $
$\phi_7:i\rightarrow - i, \sqrt[4]{3}\rightarrow -\sqrt[4]{3} $
$\phi_8:i\rightarrow - i, \sqrt[4]{3}\rightarrow -i\sqrt[4]{3} $
Now say we wanted to find all quadratic extensions , then the way I know how to it is to look at these automorphism , see which are order 4 ( as they'll correspond to a fixed field of degree 2 ) and test the elements of $\Bbb Q(i,\sqrt[4]{3})$ with each automorphism one by one and see what remains fixed... But this is a horribly long and tedious way Can anyone show me how to it more efficiently ?
I also know that we can look at the Galois group directly see which subgroups are order 4 and test the automorphisms that way but it amounts to the same thing. I feel their may be some method that will involve the minimal polynomials and how the split in different fields but I can't quite figure it out.
 A: You have already found that the Galois group is generated by $\sigma$ and $\tau$, which have orders $4$ and $2$, respectively, and it is not hard to see that $\sigma\tau=\tau\sigma^{-1}$. This means the Galois group is isomorphic to the dihedral group of order $8$.
The quadratic subfields of the splitting field correspond precisely to the subgroups of index $2$ in the Galois group, i.e. the subgroups of order $4$. By basic group theory there are precisely three such subgroups.
As noted in the comments, it is easy to spot a few quadratic subfields of $\Bbb{Q}(i,\sqrt[4]{3})$. The field contains the elements $i$, $(\sqrt[4]{3})^2$ and $i(\sqrt[4]{3})^2$, and hence it contains the quadratic subfields $\Bbb{Q}(i)$, $\Bbb{Q}(\sqrt{3})$ and $\Bbb{Q}(\sqrt{-3})$. It is a simple exercise to show that these are distinct, and by the argument above these must be  all quadratic subfields.
It is then perhaps a nice exercise to formally verify that you indeed obtain these fields as the fixed fields of the three subgroups of index $2$ in the Galois group. But I hope this shows that this method is in general not the easiest way to find the desired subfields; it is often easier to look at the known elements of the subfield.
A: Given a galois extension N/K with group G, it would of course be desirable to have the maximal amount of information on G in order to determine the subextensions of N/K. But shortcuts are possible in case the structure of G is amenable enough. For example, if G is a p-group, i.e. a group of order equal to a power of a prime p, let L/K be the compositum of all the subextensions which are cyclic of degree p over K. Obviously $Gal(L/K)\cong G/[G,G]G^p$, where $[G,G]G^p$ is the so called Frattini subgroup, generated by the commutators and the $p$-th powers. If we consider $G/[G,G]G^p$ as a vector space of dimension $d$ over $\mathbf F_p$, the number of subgroups of order $p$ contained in it is straightforwardly equal to $(p^d-1)/(p-1)$. Note also that the center of a $p$-group is always non trivial.
In your case here, $p=2$ and $G$ is the dihedral group $D_8$, whose center has necessarily order 2 (why ?), so that the Frattini quotient has order 4. It follows that the number of quadratic subextensions is 3 . These are naturally $\mathbf Q(i), \mathbf Q(\sqrt3)$ and $\mathbf Q(i\sqrt 3)$. 
