Compute the splitting field and the Galois group of $x^4 - 5$ over $\mathbb{Q} (\sqrt{5})$. I believe the splitting field is easily found by the following. 
$x^4 - 5 = (x^2 - \sqrt{5})(x^2 + \sqrt{5})$, so the splitting field is $\mathbb{Q}(\sqrt[4]{5},i)$, or is this incorrect? 
Once I obtain this, how to do I find the Galois group of this polynomial over $\mathbb{Q}(\sqrt{5})$? 
Thanks! 
 A: It has been a while since I've done Galois theory, but here's a shot at it.
I agree with your identification of the splitting field. To find the Galois group over $\mathbb Q(\sqrt 5)$, it suffices to find the Galois group over $\mathbb Q$ and use the Galois correspondence. So, for fun, let's work out the entire Galois group over $\mathbb Q$.
The splitting field is over order $8$. We have the automorphisms
$$\alpha: \sqrt[4] 5 \rightarrow  i\sqrt[4] 5,$$
$$\beta: i\rightarrow -i,$$
and together these generate a a group of order 8, so they generate the entire Galois group. It remains to work out what group of order 8 this is. There are 5 groups of order 8. Let's check if $\alpha$ and $\beta$ commute, so we can determine whether the group is abelian. They don't! So let's check to see if they satisfy the generating for relations for our favorite non-abelian group of order $8$, the dihedral group. Indeed, it is not hard to see
$$\beta \alpha \beta = \alpha^3.$$
It suffices to check this on the two generating elements $(i, \sqrt[5]{5})$. Applying $\alpha^3$ changes $\sqrt[4]{5}$ to $-i\sqrt[4]{5}$, leaving us with $(i, - i\sqrt[4]{5})$. You can check that applying $\beta\alpha\beta$ (in that order!) does the same thing.
So the Galois group is $D_4$ (or $D_8$, depending on your notation). This can also be seen geometrically. The roots of the polynomial form a square in the complex plane (with vertices at $\sqrt[4]{5}, i\sqrt[4]{5}, -\sqrt[4]{5}, -i\sqrt[4]{5}$), and you can see that $\alpha$ rotates them and $\beta$ flips them across the real axis.
Now, using the Galois correspondence, we're looking for the subgroup that fixes $\mathbb Q(\sqrt 5)$. This is the subgroup generated by $\alpha^2$ and $\beta$. I get $\mathbb Z_2\times \mathbb Z_2$ for the Galois group over $\mathbb Q(\sqrt 5)$, since $\alpha^2$ and $\beta$ commute (check this!).
In retrospect, it seems simpler to just note that $\alpha^2$ and $\beta$ are commuting order 2 automorphisms in your desired (order 4) Galois group, so they generate the entire group.
